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https://www.daniweb.com/programming/software-development/threads/335102/separate-each-of-double-variable | 1,548,174,450,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583857993.67/warc/CC-MAIN-20190122161345-20190122183345-00146.warc.gz | 763,735,482 | 12,338 | double i,j,k;
k=(i%j);
The program generates a error in second line and compiler show a message I can not use % in double variable. But I need to separate the each digit of a double variable how I can do this?
Try spacing everything out? And maybe try initializing all three variables to zero out of good practice.
Maybe use sprintf, thanks to it you can have digits in char table.
But I need to separate the each digit of a double variable how I can do this?
A floating-point representations is inexact; so very often you can compute a more accurate result of an operation on real numbers in your head than your computer can do with floating-point numbers. When it comes to 'each digit of a double variable' (where by 'digit' you mean 'decimal digit'), things get even murkier. The floating-point representation of a real number typically uses a radix of 2; though the external representation is decimal (to base 10). We expect 0.01 to be exactly representable; the digits are 0, 0, and 1 followed by zeroes. However 0.01 may not be exactly representable (in limited memory) using a binary radix.
With those caveats in place, write out the double value (to the precision that you require) to a std::ostringstream and pick off the digits from the characters in the std::string.
``````#include <iostream>
#include <iomanip>
#include <sstream>
#include <string>
#include <locale>
int main()
{
double a = 10.35 ;
std::cout << std::fixed << std::setprecision(16) << a << '\n' ;
std::ostringstream stm ;
stm << std::fixed << std::setprecision(16) << a ;
std::string str = stm.str() ;
std::cout << "decimal digits: " ;
for( std::string::size_type i = 0 ; i < str.size() ; ++i )
if( std::isdigit( str[i], stm.getloc() ) )
{
int digit = str[i] - '0' ;
std::cout << digit << ' ' ;
}
std::cout << '\n' ;
}`````` | 456 | 1,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-04 | latest | en | 0.745805 |
https://encyclopediaofmath.org/index.php?title=Inductive_limit&printable=yes | 1,624,434,748,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488536512.90/warc/CC-MAIN-20210623073050-20210623103050-00326.warc.gz | 209,444,914 | 7,307 | Inductive limit
A construction that first appeared in set theory, and then became widely used in algebra, topology and other areas of mathematics. An important special case of an inductive limit is the inductive limit of a directed family of mathematical structures of the same type. Let $C$ be a directed pre-ordered set, that is, a reflexive transitive relation $\prec$ is defined on $C$ and for any two elements $\alpha , \beta \in C$ there exists an element $\gamma \in C$ such that $\alpha \prec \gamma$ and $\beta \prec \gamma$. Suppose further that a structure $A _ \alpha$ is associated with each $\alpha \in C$( for definiteness, suppose that the $A _ \alpha$ are groups) and that for each $\alpha \prec \beta$ homomorphisms $\phi _ {\alpha \beta } : A _ \alpha \rightarrow A _ \beta$ are given satisfying the two conditions: $\phi _ {\alpha \alpha } = 1 _ {A _ \alpha }$ for any $\alpha \in C$ and $\phi _ {\alpha \beta } \phi _ {\beta \gamma } = \phi _ {\alpha \gamma }$ for any $\alpha \prec \beta \prec \gamma$ in $C$. An equivalence relation $\sim$ is introduced on the set $\overline{A}\; = \cup _ {\alpha \in C } A _ \alpha$: The element $x \in A _ \alpha$ is equivalent to $y \in A _ \beta$ if $x \phi _ {\alpha \gamma } = y \phi _ {\beta \gamma }$ for some $\gamma$. The quotient set $A = \overline{A}\; / \sim$ can then be endowed with a group structure: If $x \in A _ \alpha$, $y \in A _ \beta$ and $\alpha \prec \gamma$, $\beta \prec \gamma$, then the product of the equivalence classes represented by $x$ and $y$ is defined to be the equivalence class with representative $( x \phi _ {\alpha \gamma } ) ( y \phi _ {\beta \gamma } )$. The resulting group $A$ is called the inductive limit of the family of groups $A _ \alpha$. There exists for each $\alpha \in C$ a natural homomorphism $\phi _ \alpha : A _ \alpha \rightarrow A$ associating to an element $x \in A _ \alpha$ its equivalence class. The group $A$ together with its homomorphisms $\phi _ \alpha$ has the following property: For any system of homomorphisms $\psi _ \alpha : A _ \alpha \rightarrow B$, $\alpha \in C$, for which $\psi _ \alpha = \phi _ {\alpha \beta } \psi _ \beta$ for $\alpha \prec \beta$, there exists a unique homomorphism $\psi : A \rightarrow B$ such that $\psi _ \alpha = \phi _ \alpha \psi$ for any $\alpha \in C$.
A generalization of the above construction of an inductive limit is the notion of an inductive limit (direct limit or colimit) of a functor. An object $A$ of a category $\mathfrak K$ is called an inductive limit of the covariant functor $F : \mathfrak D \rightarrow \mathfrak K$ if:
1) there exist morphisms $\phi _ {D} : F ( D) \rightarrow A$, where $D \in \mathop{\rm Ob} \mathfrak D$, such that $F ( \alpha ) \phi _ {D _ {1} } = \phi _ {D}$ for any morphism $\alpha : D \rightarrow D _ {1}$ in $\mathfrak D$; and
2) for any family of morphisms $\psi _ {D} : F ( D) \rightarrow B$, where $D \in \mathop{\rm Ob} \mathfrak D$, such that $F ( \alpha ) \psi _ {D _ {1} } = \psi _ {D}$ for any $\alpha : D \rightarrow D _ {1}$ in $\mathfrak D$, there exists a unique morphism $\gamma : A \rightarrow B$ such that $\psi _ {D} = \phi _ {D} \gamma$, $D \in \mathop{\rm Ob} \mathfrak D$.
An inductive limit is denoted by $( A , \phi _ {D} ) = \lim\limits _ \rightarrow F$ or $A = \lim\limits _ \rightarrow F$ or $A = \lim\limits _ \rightarrow F ( D )$. An inductive limit of a contravariant functor $F : \mathfrak D \rightarrow \mathfrak K$ is defined as an inductive limit of the covariant functor $F ^ { * }$ from the dual category $\mathfrak D ^ {*}$ into the category $\mathfrak K$.
Every pre-ordered set $C$ can be regarded as a category whose objects are the elements of $C$ and whose morphisms are all pairs $( \alpha , \beta )$ for which $\alpha , \beta \in C$ and $\alpha \prec \beta$, with the obvious law of composition. In an arbitrary category $\mathfrak K$, a family of objects $A _ \alpha$, $\alpha \in C$, and morphisms $\phi _ {\alpha \beta } : A _ \alpha \rightarrow A _ \beta$, where $\alpha \prec \beta$, can be regarded as the image of a functor $F : C \rightarrow \mathfrak K$ if $\phi _ {\alpha \alpha } = 1$ and $\phi _ {\alpha \beta } \phi _ {\beta \gamma } = \phi _ {\alpha \gamma }$ for $\alpha \prec \beta \prec \gamma$. If $\mathfrak K$ is the category of sets (groups, topological spaces, etc.), then the inductive limit of the functor $F : C \rightarrow \mathfrak K$ coincides with the construction of the inductive limit defined above.
If $\mathfrak D$ is a small discrete category, then an inductive limit of any functor $F$ from $\mathfrak D$ into an arbitrary category $\mathfrak K$ is a coproduct of the objects $F ( D)$, $D \in \mathfrak D$. In particular, if $\mathfrak D$ is empty, then an inductive limit is a left null or an initial object of $\mathfrak K$. Cokernels of pairs of morphisms of any category $\mathfrak K$ are inductive limits of functors defined on the category with two objects $X$ and $Y$ and four morphisms $1 _ {X} , 1 _ {Y}$ and $\alpha , \beta : X \rightarrow Y$.
Every covariant functor $F$ from an arbitrary small category $\mathfrak D$ into a category $\mathfrak K$ has an inductive limit if and only if $\mathfrak K$ has coproducts and cokernels of pairs of morphisms.
References
[1] P.S. Aleksandrov, "Topological duality theorems" Trudy Mat. Inst. Steklov. , 48 (1955) pp. Chapt. 1 (In Russian) [2] I. Bucur, A. Deleanu, "Introduction to the theory of categories and functors" , Wiley (1968) [3] M.Sh. Tsalenko, E.G. Shul'geifer, "Fundamentals of category theory" , Moscow (1974) (In Russian)
In the main article above, $x \phi _ {\alpha \gamma }$ stands for $\phi _ {\alpha \gamma } ( x)$: the value of the homomorphism $\phi _ {\alpha \gamma }$ on the element $x$.
Similarly, composition of homomorphisms is written backwards: $\phi _ {\alpha \beta } \psi _ \beta$ means first apply $\phi _ {\alpha \beta }$ and then $\psi _ \beta$.
In English, the term "inductive limit" is usually restricted to limits over directed pre-ordered sets, the more general categorical concept being called a colimit. "Cokernels of pairs of morphisms" are commonly called coequalizers.
Dual to the notion of an inductive limit is that of a projective limit, also called an inverse limit.
References
[a1] S. MacLane, "Categories for the working mathematician" , Springer (1971) pp. Chapt. IV, Sect. 6; Chapt. VII, Sect. 7
How to Cite This Entry:
Inductive limit. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Inductive_limit&oldid=47336
This article was adapted from an original article by M.Sh. Tsalenko (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | 1,966 | 6,719 | {"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} | 3.171875 | 3 | CC-MAIN-2021-25 | latest | en | 0.817144 |
https://stats.stackexchange.com/questions/509475/showing-convergence-to-a-non-degenerate-distribution | 1,632,455,901,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057496.18/warc/CC-MAIN-20210924020020-20210924050020-00602.warc.gz | 580,455,149 | 32,289 | # Showing convergence to a (non-degenerate) distribution [closed]
The random variables $$X_1$$, ..., $$X_n$$ are i.i.d with density
$$$$f(x) = \begin{cases} \frac{24}{x^4}, & \text{if}\ x\geq 2 \\ 0, & \text{if x < 2} \end{cases}$$$$. Define $$M_n$$ to be $$\max(x_1, ..., x_n)$$. Find the constant c s.t $$M_n/n^c$$ converges to a non-degenerate distribution and obtain the CDF for this limiting distribution.
What I have done so far:
Find the CDF of f(x) which gives me $$F(x) = \begin{cases}1-8x^{-3}, x\geq 2, \\ 0 \text{ otherwise} \end{cases}$$.
Find the CDF of the max order statistic ie. $$M_n$$, which is $$$$P(M_n \leq x) = P(X_1 < x, ..., X_n < x) = \prod_{i=1}^{n} F_x(x) = \begin{cases} (1-\frac{8}{x^3})^{n}, x \geq 2 \\ 0 \text{ otherwise} \end{cases}$$$$
By definition of convergence in distribution, $$$$\lim_{n->\infty} F_{X_n}(t) = F(t)$$$$
Since our goal is to find a converging distribution in $$M_n/n^c$$, let $$Y_n = M_n/n^c$$. $$$$P(Y_n < y) = P(\frac{M_n}{n^c} < y) = P(M_n < n^cy) = (1-\frac{8y^{-3}}{n^{3c}})^{n}$$$$. If we take the limit of $$(1-\frac{8y^{-3}}{n^{3c}})^{n}$$ at c = 1/3, we will have $$e^{-8y^{-3}}$$ by the identity of $$\lim_{n-> \infty} (1-\lambda/n)^n= e^{-\lambda}$$. So it converges to a non degenerate distribution (with it being in the form of the survivor function of an exponential distribution?)
Any directions or comments will be greatly appreciated! (this was a past exam question and I am taking it out for self-study purposes).
• Questions where the unique best answer is "yes" or "no" are not considered suitable here. Could you specify what kind of guidance you are looking for?
– whuber
Feb 20 at 15:19 | 585 | 1,674 | {"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": 16, "wp-katex-eq": 0, "align": 0, "equation": 4, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-39 | latest | en | 0.739666 |
https://www.gradesaver.com/textbooks/math/other-math/CLONE-547b8018-14a8-4d02-afd6-6bc35a0864ed/chapter-7-measurement-test-page-528/28 | 1,679,852,729,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00608.warc.gz | 893,811,682 | 12,057 | ## Basic College Mathematics (10th Edition)
Use unit fractions to convert 125 lb to kilograms. $125lb\times\frac{0.45kg}{1lb}=56.25$ | 41 | 133 | {"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} | 3.234375 | 3 | CC-MAIN-2023-14 | latest | en | 0.631602 |
https://www.acmicpc.net/problem/5128 | 1,526,846,381,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863684.0/warc/CC-MAIN-20180520190018-20180520210018-00481.warc.gz | 664,301,950 | 15,770 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 128 MB 5 4 2 66.667%
## 문제
Recently, Tim has discovered a time period in the future in which Medieval artifacts, particularly those associated with the various armies of the time period, are the most valuable. So Tim plans to travel to the Middle Ages, find some chainmail, lances, etc., and then make a fortune off of them. The only problem is that he’s not likely to have an acceptable currency, so in order to acquire the things he needs, he must barter for them. In the interest of time, Tim wants to determine how many trades he will need to make before he actually executes those trades. That way, he can spend his time acquiring other items which require less trades.
Note: The time-machine can only hold at most 5 items. This means that Tim will not make a trade if he will even temporarily have to carry more than 5 items.
## 입력
The first line is the number K of input data sets, followed by the K data sets, each of the following form:
The first line contains four integers 1 ≤ M ≤ 20, the maximum number of trades Tim is willing to make, 1 ≤ H ≤ 5, the number of items Tim has to start with, 1 ≤ W ≤ 5, the number of items Tim wants, 1 ≤ T ≤ 20, the number of different trades.
This is followed by one line containing H words, the names of the items Tim has. Then one line containing W words, the names of the items Tim wants.
This is followed by T pairs of lines, each pair specifying a possible trade. The first line of a pair contains a number 1 ≤ gt ≤ 5, the number of items Tim will give away in the trade, followed by the names of the gt items. The second line contains a number 1 ≤ at ≤ 5, the number of items Tim will acquire in the trade, followed by the names of the at items.
## 출력
For each data set, output “Data Set x:” on a line by itself, where x is its number. If it is possible to acquire all of the wanted items by only making M trades or less, output the fewest number of trades that Tim will have to make. Otherwise, output “Impossible.”. Each data set should be followed by a blank line.
## 예제 입력 1
2
4 3 2 3
coin butterknife suit
sword chainmail
1 suit
1 chainmail
3 butterknife butterknife butterknife
1 sword
1 coin
2 butterknife butterknife
1 1 1 2
pvcpipe
lance
1 pvcpipe
1 shield
1 shield
1 lance
## 예제 출력 1
Data Set 1:
3
Data Set 2:
Impossible. | 607 | 2,334 | {"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} | 2.5625 | 3 | CC-MAIN-2018-22 | longest | en | 0.93903 |
http://www.weeklystandard.com/Content/Public/Articles/000/000/004/084xsudm.asp | 1,448,495,495,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398446218.95/warc/CC-MAIN-20151124205406-00318-ip-10-71-132-137.ec2.internal.warc.gz | 791,744,794 | 25,380 | # The Fall of Troy
## Brad Pitt, Eric Bana, and Wolfgang Petersen take on the IIiad and lead Hollywood's assault on the classics.
12:00 AM, May 14, 2004 • By JONATHAN V. LAST
Widget tooltip
IF YOU SPEND ENOUGH TIME watching Inside the Actor's Studio and listening to DVD commentary tracks, you learn that, since they have very little other work to do, actors are forever agonizing over "choices." Of course everyone involved in the production of cinema makes choices and these thousands of choices must align just-so in order to create a good movie. The master film editor Walter Murch likened the making of these choices to playing a game of Negative 20 Questions:
[The game] involves, say, four people: Michael, Anthony, Walter, and Aggie. . . . When Michael leaves the room, the three remaining players don't communicate with one another at all. Instead, each of them silently decides on an object [in the room]. Then they call Michael back in. . . .
Michael asks Walter: Is the object bigger than a breadbox? Walter--who has picked the alarm clock--says, No. Now Anthony has chosen the sofa, which is bigger than a breadbox. And since Michael is going to ask him the next question, Anthony must quickly look around the room and come up with something else--a coffee cup!--which is smaller than a breadbox. So when Michael asks Anthony, If I emptied out my pockets could I put their contents in this object? Anthony says, Yes.
Now Aggie's choice may have been the small pumpkin carved for Halloween, which could also contain Michael's keys and coins, so when Michael says, Is it edible? Aggie says, Yes. That's a problem for Walter and Anthony, who have chosen inedible objects: they now have to change their selection to something edible, hollow, and smaller than a breadbox.
So a complex vortex of decision making is set up, a logical but unpredictable chain of ifs and thens. To end successfully, the game must produce, in fewer than 20 questions, an object that satisfies all of the logical requirements: smaller than a breadbox, edible, hollow, etc.
In filmmaking, Negative 20 Questions begins with a story idea and each member of the cast and crew then contributes a series of if-then choices. If these choices line up in an interesting and non-contradictory way, then a good movie is produced. (In case you're wondering, the original creator of Negative 20 Questions was not Murch, but physicist John Wheeler.)
AS A GAME of Negative 20 Questions, the movie Troy is fascinating. Written by Homer, adapted by David Benioff, and directed by Wolfgang Petersen, Troy begins with the Iliad as its source and then spins outward as the various players make their choices: If the Iliad is going to be a movie, then it needs a big budget. If it's going to be a movie with a big budget, then it needs a movie star as Achilles. If it's going to be a big budget movie with a star as Achilles, then Achilles and his attendant Patroclus can't be lovers. And so on.
There are two choices, however, which stand out above the rest. The first is the decision to cast the Greeks as villains and the Trojans as the movie's heroes. In Troy, Paris and Helen are madly in love, and she leaves Sparta with him of her own volition. Menelaus--a wife-beater if ever there was one--enlists Agamemnon's help to bring Helen back to him, so he can slit her throat. Agamemnon, a brutal, power-mad tyrant, agrees to go to war and is thrilled that Helen has given him a pretext to attack Troy, since he has long coveted the city. And Achilles is a brooding, philosophical hunk. No longer a tragedy in which heroic equals are pitted against one another by Paris's vanity, Troy is the story of the conquest of a good and noble people by an army of crafty barbarians.
Then there's Petersen's approach toward the gods: Troy supposes that they don't actually exist. You'll recall that the gods were somewhat integral to the Iliad. Aphrodite, Hera, and Athena go to Paris asking him to mediate their dispute, and it is Aphrodite who gives Helen to the young prince as a bribe, setting the tale in motion. The gods directly intercede in many battles during the long siege. Achilles' invulnerability comes from the fact that his mother--the divine sea-nymph Thetis--dipped him in the River Styx.
Yet in Troy the gods never make their presence known. All of the events are brought about by earthly actors. Although most of the characters proclaim their devotion to the gods, the more cynical among them seem to believe that the gods are a myth for the weak-minded. We meet Achilles' mother, who appears quite mortal, and besides which, it isn't even clear that Achilles is invulnerable. Troy is the Iliad told by agnostics; the gods are AWOL and those who believe in them are dupes.
If you work in Hollywood, then religious believers are idiots. Even if they're only pagans.
IN TRUTH, once you get past these two big revisions, Troy has much to recommend it. | 1,130 | 4,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | latest | en | 0.949069 |
https://aakashsrv1.meritnation.com/ask-answer/expert_answers/class-10/gr10 | 1,656,541,693,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00266.warc.gz | 126,607,133 | 9,793 | Subject: English, asked 5 hours, 1 minute ago
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What are you looking for? | 487 | 1,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-27 | longest | en | 0.927197 |
http://cantorsattic.info/index.php?title=Huge&diff=3917&oldid=3465 | 1,632,771,543,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00685.warc.gz | 11,808,802 | 17,902 | # Difference between revisions of "Huge"
Huge cardinals (and their variants) were introduced by Kenneth Kunen in 1972 as a very large cardinal axiom. Kenneth Kunen first used them to prove that the consistency of the existence of a huge cardinal implies the consistency of $\text{ZFC}$+"there is a $\omega_2$-saturated $\sigma$-ideal on $\omega_1$". It is now known that only a Woodin cardinal is needed for this result. However, the consistency of the existence of an $\omega_2$-complete $\omega_3$-saturated $\sigma$-ideal on $\omega_2$, as far as the set theory world is concerned, still requires an almost huge cardinal.[1]
## Definitions
Their formulation is similar to that of the formulation of superstrong cardinals. A huge cardinal is to a supercompact cardinal as a superstrong cardinal is to a strong cardinal, more precisely. The definition is part of a generalized phenomenon known as the "double helix", in which for some large cardinal properties n-$P_0$ and n-$P_1$, n-$P_0$ has less consistency strength than n-$P_1$, which has less consistency strength than (n+1)-$P_0$, and so on. This phenomenon is seen only around the n-fold variants as of modern set theoretic concerns. [2]
Although they are very large, there is a first-order definition which is equivalent to n-hugeness, so the $\theta$-th n-huge cardinal is first-order definable whenever $\theta$ is first-order definable. This definition can be seen as a (very strong) strengthening of the first-order definition of measurability.
### Elementary embedding definition
The elementary embedding definitions are somewhat standard. Let $j:V\rightarrow M$ be a nontrivial elementary embedding of $V$ into a transitive class $M$ with critical point $\kappa$. Then:
• $\kappa$ is almost n-huge with target $\lambda$ iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length less than $\lambda$ (that is, $M^{<\lambda}\subseteq M$).
• $\kappa$ is n-huge with target $\lambda$ iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length $\lambda$ ($M^\lambda\subseteq M$).
• $\kappa$ is almost n-huge iff it is almost n-huge with target $\lambda$ for some $\lambda$.
• $\kappa$ is n-huge iff it is n-huge with target $\lambda$ for some $\lambda$.
• $\kappa$ is super almost n-huge iff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is almost n-huge with target $\lambda$ (that is, the target can be made arbitrarily large).
• $\kappa$ is super n-huge iff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is n-huge with target $\lambda$.
• $\kappa$ is almost huge, huge, super almost huge, and superhuge iff it is almost 1-huge, 1-huge, etc. respectively.
Note: $M^{<\lambda}$ and $M^{\lambda}$ can also be denoted $^{<\lambda}M$ and $^{\lambda}M$.
### Ultrahuge cardinals
A cardinal $\kappa$ is $\lambda$-ultrahuge for $\lambda>\kappa$ if there exists a nontrivial elementary embedding $j:V\to M$ for some transitive class $M$ such that $j(\kappa)>\lambda$, $M^{j(\kappa)}\subseteq M$ and $V_{j(\lambda)}\subseteq M$. A cardinal is ultrahuge if it is $\lambda$-ultrahuge for all $\lambda\geq\kappa$. [1] Notice how similar this definition is to the alternative characterization of extendible cardinals. Furthermore, this definition can be extended in the obvious way to define $\lambda$-ultra n-hugeness and ultra n-hugeness, as well as the "almost" variants.
### Hyperhuge cardinals
A cardinal $\kappa$ is $\lambda$-hyperhuge for $\lambda>\kappa$ if there exists a nontrivial elementary embedding $j:V\to M$ for some inner model $M$ such that $\mathrm{crit}(j) = \kappa$, $j(\kappa)>\lambda$ and $^{j(\lambda)}M\subseteq M$. A cardinal is hyperhuge if it is $\lambda$-hyperhuge for all $\lambda>\kappa$.[3, 4]
$\lambda$-hyperhuge and hyperhuge are other names for $2$-fold $\lambda$-supercompact and $2$-fold supercompact.
### Huge* cardinals
A cardinal $\kappa$ is $n$-huge* if for some $\alpha\gt\kappa$, $\kappa$ is the critical point of an elementary embedding $j: V_\alpha\prec V_\beta$ such that $j^n(\kappa)\lt\alpha$.[5]
Hugeness* variant is formulated in a way allowing for a virtual variant consistent with $V=L$: A cardinal $\kappa$ is virtually $n$-huge* if for some $\alpha\gt\kappa$, in a set-forcing extension, $\kappa$ is the critical point of an elementary embedding $j: V_\alpha\prec V_\beta$ such that $j^n(\kappa)\lt\alpha$.[5]
### Ultrafilter definition
The first-order definition of n-huge is somewhat similar to measurability. Specifically, $\kappa$ is measurable iff there is a nonprincipal $\kappa$-complete ultrafilter, $U$, over $\kappa$. A cardinal $\kappa$ is n-huge with target $\lambda$ iff there is a normal $\kappa$-complete ultrafilter, $U$, over $\mathcal{P}(\lambda)$, and cardinals $\kappa=\lambda_0<\lambda_1<\lambda_2...<\lambda_{n-1}<\lambda_n=\lambda$ such that:
$$\forall i<n(\{x\subseteq\lambda:\text{order-type}(x\cap\lambda_{i+1})=\lambda_i\}\in U)$$
Where $\text{order-type}(X)$ is the order-type of the poset $(X,\in)$. [1] $\kappa$ is then super n-huge if for all ordinals $\theta$ there is a $\lambda>\theta$ such that $\kappa$ is n-huge with target $\lambda$, i.e. $\lambda_n$ can be made arbitrarily large. If $j:V\to M$ is such that $M^{j^n(\kappa)}\subseteq M$ (i.e. $j$ witnesses n-hugeness) then there is a ultrafilter $U$ as above such that, for all $k\leq n$, $\lambda_k = j^k(\kappa)$, i.e. it is not only $\lambda=\lambda_n$ that is an iterate of $\kappa$ by $j$; all members of the $\lambda_k$ sequence are.
As an example, $\kappa$ is 1-huge with target $\lambda$ iff there is a normal $\kappa$-complete ultrafilter, $U$, over $\mathcal{P}(\lambda)$ such that $\{x\subseteq\lambda:\text{order-type}(x)=\kappa\}\in U$. The reason why this would be so surprising is that every set $x\subseteq\lambda$ with every set of order-type $\kappa$ would be in the ultrafilter; that is, every set containing $\{x\subseteq\lambda:\text{order-type}(x)=\kappa\}$ as a subset is considered a "large set."
As for hyperhugeness, the following are equivalent:[4]
• $\kappa$ is $\lambda$-hyperhuge;
• $\mu\lt\lambda$ and a normal, fine, $\kappa$-complete ultrafilter exists on $[\mu]^\lambda_{∗\kappa} := \{s\subseteq μ:|s|=\lambda, |s\cap\kappa|\lt\kappa, \mathrm{otp}(s\cap\lambda)\lt\kappa\}$;
• $\mathbb{L}_{\kappa,\kappa}$ is $[\mu]^\lambda_{∗\kappa}$-$\kappa$-compact for type omission.
### Coherent sequence characterization of almost hugeness
Almost huge cardinals can be characterized via coherent sequences of ultrafilters, so that the direct limit of the ultrapowers witnesses the almost hugeness of $\kappa$ and each individual ultrapower witnesses a $j_\gamma: V\prec M_\gamma$ with $\gamma\lt j(\kappa)$ and $M^\gamma\subseteq M$. $\kappa$ is almost-huge if there is some inaccessible $\lambda\gt\kappa$ and a coherent sequence of normal ultrafilters $\langle \mathcal{U}_\gamma|\kappa\le\gamma\lt\lambda\rangle$ over $\mathcal{P}_\kappa(\gamma)$ such that the corresponding embeddings $j_\gamma : V\to M_\gamma\cong Ult(V, \mathcal{U}_\gamma)$ and $k_{\gamma,\delta}: M_\gamma\to M_\delta$ satisfy: if $\kappa\le\gamma\lt\lambda$ and $\gamma\le\alpha\lt j_\gamma(\kappa)$, then there is $\delta$ such that $\gamma\le\delta\lt\lambda$ and $k_{\gamma,\delta}(\alpha)=\delta$.[1]
Because each $\mathcal{U}_\gamma$ is a normal fine ultrafilter, $M_\gamma^\gamma\subseteq M_\gamma$ and $j_\gamma(\kappa)\gt\gamma$. Also, if $\hat M$ is the direct limit and $\hat j: V\prec\hat M$, then for every $\beta\lt\hat j(\kappa)$, $M^\beta\subseteq\beta$ and $\text{crit}j=\kappa$. $j=k_\gamma\circ j_\gamma$ for $k_\gamma$ the factor embedding $k_\gamma: M_\gamma\prec\hat M$, and $\lambda=\hat j(\kappa)$.
### $C^{(n)}$-$m$-huge cardinals
(this section from [6], information about almost-huge and superhuge (and other, if noted) from 2019 extended arXiv version)
Elementary-embedding definitions:
• $κ$ is $C^{(n)}$-$m$-huge iff it is $m$-huge and $j(κ) ∈ C^{(n)}$ ($C^{(n)}$-huge if it is huge and $j(κ) ∈ C^{(n)}$ ($m=1$)).
• $κ$ is $C^{(n)}$-$m$-almost-huge iff it is almost-huge and $j(κ) ∈ C^{(n)}$.
• $κ$ is $C^{(n)}$-superhuge iff it is superhuge and there are witnessing embeddings with arbitrarily large $j(κ) ∈ C^{(n)}$.
Equivalent definitions in terms of normal measures:
• $κ$ is $C^{(n)}$-$m$-huge iff it is uncountable and there is a $κ$-complete normal ultrafilter $\mathcal{U}$ over some $\mathcal{P}(λ)$ and cardinals $κ = λ_0 < λ_1 < . . . < λ_m = λ$, with $λ_1 ∈ C^{(n)}$ and such that for each $i < m$, $\{x ∈ \mathcal{P}(λ) : ot(x ∩ λ_{i+1}) = λ_i \} ∈ \mathcal{U}$.
• $κ$ is $C^{(n)}$-almost-huge iff there is an inaccessible $λ ∈ C^{(n)}$ greater than $κ$ and a coherent sequence of normal ultrafilters $\langle \mathcal{U}_γ : κ ≤ γ < λ \rangle$ over $\mathcal{P}_κ(γ)$ such that the corresponding embeddings $j_γ : V → M_γ ≅ Ult(V, \mathcal{U}_γ )$ and $k_{γ,δ} : M_γ → M_δ$ satisfy: if $κ ≤ γ < λ$ and $γ ≤ α < j_γ (κ)$, then there is $δ$ such that $γ ≤ δ < λ$ and $k_{γ,δ}(α) = δ$. (See [1], 24.11 in 1994 edition for details.)
• $κ$ is $C^{(n)}$-superhuge iff for every $α$ there is a $κ$-complete fine and normal ultrafilter $\mathcal{U}$ over some $\mathcal{P}(λ)$, with $λ ∈ C^{(n)}$ and such that $\{x ∈ \mathcal{P}(λ) : ot(x) = κ \} ∈ \mathcal{U}$.
It follows that
• “$κ$ is $C^{(n)}$-$m$-huge” and “$κ$ is $C^{(n)}$-almost-huge” are $Σ_{n+1}$ expressible and
• “$κ$ is $C^{(n)}$-superhuge” is $Π_{n+2}$ expressible.
Every huge cardinal is $C^{(1)}$-huge. Every superhuge cardinal is $C^{(1)}$-superhuge.
Hierarchy:
• The first $C^{(n)}$-$m$-huge cardinal is not $C^{(n+1)}$-$m$-huge, for all $m$ and $n$ greater or equal than $1$. For suppose $κ$ is the least $C^{(n)}$-$m$-huge cardinal and $j : V → M$ witnesses that $κ$ is $C^{(n+1)}$-$m$-huge. Then since “x is $C^{(n)}$-$m$-huge” is $Σ_{n+1}$ expressible, we have $V_{j(κ)} \models$ “$κ$ is $C^{(n)}$-$m$-huge”. Hence, since $(V_{j(κ)})^M = V_{j(κ)}$, $M \models$ “$∃_{δ < j(κ)}(V_{j(κ)} \models$ “δ is huge”$)$”. By elementarity, there is a $C^{(n)}$-$m$-huge cardinal less than $κ$ in $V$ – contradiction.
• Similar argumentation shows that he first $C^{(n)}$-superhuge cardinal is not $C^{(n+1)}$-superhuge.
• If $κ$ is $C^{(n)}$-$2$-huge, then there is a $κ$-complete normal ultrafilter $\mathcal{U}$ over $κ$ such that $\{α < κ : V_κ \models$ “$α$ is $C^{(n)}$-superhuge” $\} ∈ \mathcal{U}$.
• If $κ$ is $C^{(n)}$-huge, then it is $C^{(n)}$-almost-huge and there is a $κ$-complete normal ultrafilter $\mathcal{U}$ over $κ$ such that $\{α < κ : α$ is $C^{(n)}$-almost-huge$\} ∈ \mathcal{U}$
• Every $C^{(n)}$-almost-huge cardinal is $C^{(n)}$-superstrong, so it belongs to $C^{(n)}$.
• Therefore, taking into account that $C^{(n)}$-huge is $Σ_{n+1}$ expressible, the first $C^{(n)}$-huge cardinal is smaller than the first $C^{(n+1)}$-almost-huge cardinal (provided both exist).
Relations with other large cardinals:
• If $κ$ is $C^{(n)}$-superhuge, then $κ$ is $C^{(n)}$-extendible (in particular $κ ∈ C^{(n+2)}$) and there is a $κ$-complete normal ultrafilter $\mathcal{U}$ over $κ$ such that $\{α < κ : α$ is $C^{(n)}$-extendible$\} ∈ \mathcal{U}$. $κ$ is also $C^{(n)}$-supercompact.
• Assuming $\mathrm{I3}(κ, δ)$, if $δ$ is a limit cardinal (instead of a successor of a limit cardinal – Kunen’s Theorem excludes other cases), it is equal to $sup\{j^m(κ) : m ∈ ω\}$ where $j$ is the elementary embedding. Then $κ$ and $j^m(κ)$ are $C^{(n)}$-$m$-huge and $C^{(n)}$-superhuge (inter alia) in $V_δ$, for all $n$ and $m$.
• If $κ$ is $C^{(n)}$-$\mathrm{I3}$, then it is $C^{(n)}$-$m$-huge, for all $m$, and there is a ($κ$-complete — from 2019 version) normal ultrafilter $\mathcal{U}$ over $κ$ such that $\{α < κ : α$ is $C^{(n)}$-$m$-huge for every $m\} ∈ \mathcal{U}$.
## Consistency strength and size
Hugeness exhibits a phenomenon associated with similarly defined large cardinals (the n-fold variants) known as the double helix. This phenomenon is when for one n-fold variant, letting a cardinal be called n-$P_0$ iff it has the property, and another variant, n-$P_1$, n-$P_0$ is weaker than n-$P_1$, which is weaker than (n+1)-$P_0$. [2] In the consistency strength hierarchy, here is where these lay (top being weakest):
• measurable = 0-superstrong = 0-huge
• n-superstrong
• n-fold supercompact
• (n+1)-fold strong, n-fold extendible
• (n+1)-fold Woodin, n-fold Vopěnka
• (n+1)-fold Shelah
• almost n-huge
• super almost n-huge
• n-huge
• super n-huge
• ultra n-huge
• (n+1)-superstrong
All huge variants lay at the top of the double helix restricted to some natural number n, although each are bested by I3 cardinals (the critical points of the I3 elementary embeddings). In fact, every I3 is preceeded by a stationary set of n-huge cardinals, for all n. [1]
Similarly, every huge cardinal $\kappa$ is almost huge, and there is a normal measure over $\kappa$ which contains every almost huge cardinal $\lambda<\kappa$. Every superhuge cardinal $\kappa$ is extendible and there is a normal measure over $\kappa$ which contains every extendible cardinal $\lambda<\kappa$. Every (n+1)-huge cardinal $\kappa$ has a normal measure which contains every cardinal $\lambda$ such that $V_\kappa\models$“$\lambda$ is super n-huge” [1], in fact it contains every cardinal $\lambda$ such that $V_\kappa\models$“$\lambda$ is ultra n-huge”. Furthermore, if $\kappa$ is 2-huge, this measure contains all $\lambda$ such that $V_\kappa\vDash$“$\lambda$ is hyperhuge”.
Every n-huge cardinal is m-huge for every m<n. Similarly with almost n-hugeness, super n-hugeness, and super almost n-hugeness. Every almost huge cardinal is Vopěnka (therefore the consistency of the existence of an almost-huge cardinal implies the consistency of Vopěnka's principle). [1] Every ultra n-huge is super n-huge and a stationary limit of super n-huge cardinals. Every super almost (n+1)-huge is ultra n-huge and a stationary limit of ultra n-huge cardinals.
In terms of size, however, the least n-huge cardinal is smaller than the least supercompact cardinal (assuming both exist). [1] This is because n-huge cardinals have upward reflection properties, while supercompacts have downward reflection properties. Thus for any $\kappa$ which is supercompact and has an n-huge cardinal above it, $\kappa$ "reflects downward" that n-huge cardinal: there are $\kappa$-many n-huge cardinals below $\kappa$. On the other hand, the least super n-huge cardinals have both upward and downward reflection properties, and are all much larger than the least supercompact cardinal. It is notable that, while almost 2-huge cardinals have higher consistency strength than superhuge cardinals, the least almost 2-huge is much smaller than the least super almost huge.
While not every $n$-huge cardinal is strong, if $\kappa$ is almost $n$-huge with targets $\lambda_1,\lambda_2...\lambda_n$, then $\kappa$ is $\lambda_n$-strong as witnessed by the generated $j:V\prec M$. This is because $j^n(\kappa)=\lambda_n$ is measurable and therefore $\beth_{\lambda_n}=\lambda_n$ and so $V_{\lambda_n}=H_{\lambda_n}$ and because $M^{<\lambda_n}\subset M$, $H_\theta\subset M$ for each $\theta<\lambda_n$ and so $\cup\{H_\theta:\theta<\lambda_n\} = \cup\{V_\theta:\theta<\lambda_n\} = V_{\lambda_n}\subset M$.
Every almost $n$-huge cardinal with targets $\lambda_1,\lambda_2...\lambda_n$ is also $\theta$-supercompact for each $\theta<\lambda_n$, and every $n$-huge cardinal with targets $\lambda_1,\lambda_2...\lambda_n$ is also $\lambda_n$-supercompact.
If $n\ge 1$ and $\mathrm{I}_4^{n+1}(\kappa)$, then $\kappa$ is $n$-huge and {$\alpha < \kappa | \alpha$ is $n$-huge} has measure 1.[7]
For $2$-huge $κ$, $V_κ$ is a model of $\mathrm{ZFC}$+“there are proper class many hyper-huge cardinals”.[3] Hyper-huge cardinals are extendible limits of extendible cardinals.[8]
Without the axiom of choice $\omega_1$ can be huge. [2]
An $n$-huge* cardinal is an $n$-huge limit of $n$-huge cardinals. Every $n + 1$-huge cardinal is $n$-huge*.[5]
As for virtually $n$-huge*:[5]
• If $κ$ is virtually huge*, then $V_κ$ is a model of proper class many virtually extendible cardinals.
• A virtually $n+1$-huge* cardinal is a limit of virtually $n$-huge* cardinals.
• A virtually $n$-huge* cardinal is an $n+1$-iterable limit of $n+1$-iterable cardinals. If $κ$ is $n+2$-iterable, then $V_κ$ is a model of proper class many virtually $n$-huge* cardinals.
• Every virtually rank-into-rank cardinal is a virtually $n$-huge* limit of virtually $n$-huge* cardinals for every $n < ω$.
### The $\omega$-huge cardinals
A cardinal $\kappa$ is almost $\omega$-huge iff there is some transitive model $M$ and an elementary embedding $j:V\prec M$ with critical point $\kappa$ such that $M^{<\lambda}\subset M$ where $\lambda$ is the smallest cardinal above $\kappa$ such that $j(\lambda)=\lambda$. Similarly, $\kappa$ is $\omega$-huge iff the model $M$ can be required to have $M^\lambda\subset M$.
Sadly, $\omega$-huge cardinals are inconsistent with ZFC by a version of Kunen's inconsistency theorem. Now, $\omega$-hugeness is used to describe critical points of I1 embeddings.
## Relative consistency results
### Hugeness of $\omega_1$
In [3] it is shown that if $\text{ZFC +}$ "there is a huge cardinal" is consistent then so is $\text{ZF +}$ "$\omega_1$ is a huge cardinal" (with the ultrafilter characterization of hugeness).
### Cardinal arithmetic in $\text{ZF}$
If there is an almost huge cardinal then there is a model of $\text{ZF+}\neg\text{AC}$ in which every successor cardinal is a Ramsey cardinal. It follows that (1) for all inner models $W$ of $\text{ZFC}$ and every singular cardinal $\kappa$, one has $\kappa^{+W} < \kappa^+$ and that (2) for all ordinal $\alpha$ there is no injection $\aleph_{\alpha+1}\to 2^{\aleph_\alpha}$. This in turn imply the failure of the square principle at every infinite cardinal (and consequently $\text{AD}^{L(\mathbb{R})}$, see determinacy). [4]
## In set theoretic geology
If $\kappa$ is hyperhuge, then $V$ has $<\kappa$ many grounds (so the mantle is a ground itself).[3] This result has been strenghtened to extendible cardinals[8]. On the other hand, it s consistent that there is a supercompact cardinal and class many grounds of $V$ (because of the indestructibility properties of supercompactness).[3]
## References
1. Kanamori, Akihiro. The higher infinite. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex
2. Kentaro, Sato. Double helix in large large cardinals and iteration of elementary embeddings. Annals of Pure and Applied Logic 146(2-3):199-236, May, 2007. www DOI bibtex
3. Usuba, Toshimichi. The downward directed grounds hypothesis and very large cardinals. Journal of Mathematical Logic 17(02):1750009, 2017. arχiv DOI bibtex
4. Boney, Will. Model Theoretic Characterizations of Large Cardinals. arχiv bibtex
5. Gitman, Victoria and Shindler, Ralf. Virtual large cardinals. www bibtex
6. Bagaria, Joan. $C^{(n)}$-cardinals. Archive for Mathematical Logic 51(3--4):213--240, 2012. www arχiv DOI bibtex
7. Corazza, Paul. The gap between $\mathrm{I}_3$ and the wholeness axiom. Fund Math 179(1):43--60, 2003. www DOI MR bibtex
8. Usuba, Toshimichi. Extendible cardinals and the mantle. Archive for Mathematical Logic 58(1-2):71-75, 2019. arχiv DOI bibtex
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# HKUST 2013- Calling All Applicants
Author Message
Intern
Status: Preparing for GMAT
Joined: 27 Jan 2012
Posts: 11
Location: Germany
GMAT 1: 690 Q39 V45
WE: Project Management (Investment Banking)
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Kudos [?]: 7 [1] , given: 2
Re: HKUST 2013- Calling All Applicants [#permalink]
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18 Jan 2013, 04:16
1
KUDOS
Congrats. I'll have my interviews next week, applied in R2. Which round did you apply in?
Intern
Joined: 17 Sep 2012
Posts: 13
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Kudos [?]: 2 [0], given: 0
Re: HKUST 2013- Calling All Applicants [#permalink]
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18 Jan 2013, 08:18
nmje2k1 wrote:
Congrats. I'll have my interviews next week, applied in R2. Which round did you apply in?
Thanks! Goodluck! I applied in Round 1.
Intern
Joined: 12 Sep 2010
Posts: 33
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Kudos [?]: 7 [0], given: 3
Re: HKUST 2013- Calling All Applicants [#permalink]
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20 Jan 2013, 00:51
Hi eybrj2!!
I applied in 2nd round around jan 01,2013.
Intern
Joined: 09 Oct 2012
Posts: 6
Location: Thailand
GPA: 3.29
WE: Project Management (Consumer Products)
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Kudos [?]: 5 [0], given: 3
Re: HKUST 2013- Calling All Applicants [#permalink]
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21 Jan 2013, 20:15
Congrats blueram. Could you share us your profile and how did interview go? It would be helpful for those who want to apply. Thanks
Intern
Status: Preparing for GMAT
Joined: 27 Jan 2012
Posts: 11
Location: Germany
GMAT 1: 690 Q39 V45
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Re: HKUST 2013- Calling All Applicants [#permalink]
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24 Jan 2013, 10:28
Had my first interview today. I think it went fairly well. The only "standard" question was "why is now the right time". Other than that they were fairly specific questions based on my application. Fairly laid back interview but by no means casual. Overall a good experience and made me want to attend HKUST even more.
Manager
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Concentration: General Management
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Re: HKUST 2013- Calling All Applicants [#permalink]
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25 Jan 2013, 10:27
nmje2k1 wrote:
Had my first interview today. I think it went fairly well. The only "standard" question was "why is now the right time". Other than that they were fairly specific questions based on my application. Fairly laid back interview but by no means casual. Overall a good experience and made me want to attend HKUST even more.
may i ask was that the "Part 1. Background and Motivation" or "Part 2. Potential " of the interview?
and who was the interviewer?
Intern
Status: Preparing for GMAT
Joined: 27 Jan 2012
Posts: 11
Location: Germany
GMAT 1: 690 Q39 V45
WE: Project Management (Investment Banking)
Followers: 0
Kudos [?]: 7 [0], given: 2
Re: HKUST 2013- Calling All Applicants [#permalink]
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25 Jan 2013, 11:53
I was a bit confused about that myself. I booked the "potential" interview for a date before the other one. But my feeling is that they then switched the topics so that they would be in the planned order. It felt like a background/motivation interview.
Intern
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Re: HKUST 2013- Calling All Applicants [#permalink]
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01 Feb 2013, 22:10
Hi Guys,
I am not sure if this is the right thread to answer this question. But I will still go ahead and post the question.
I want to know how good is Msc Information Systems Management from HKUST
I just took my GMAT and did not score very well.
OA Score - 640
Quant - 47
Verbal - 29
Overall Experience - 4 years in information technology
Degree - MCA
Can I apply for the ISM HKUST course, or should I consider taking GMAT again.
Director
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Re: HKUST 2013- Calling All Applicants [#permalink]
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07 Feb 2013, 01:22
Sikarwar786 wrote:
Hi Guys,
I am not sure if this is the right thread to answer this question. But I will still go ahead and post the question.
I want to know how good is Msc Information Systems Management from HKUST
I just took my GMAT and did not score very well.
OA Score - 640
Quant - 47
Verbal - 29
Overall Experience - 4 years in information technology
Degree - MCA
Can I apply for the ISM HKUST course, or should I consider taking GMAT again.
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Re: HKUST 2013- Calling All Applicants [#permalink]
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07 Feb 2013, 01:25
1
KUDOS
I'm in!!! Got the e-mail this morning.
It's conditional, they want me to send in some documents has anyone else been asked to do that?
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Re: HKUST 2013- Calling All Applicants [#permalink]
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07 Feb 2013, 11:44
nmje2k1 wrote:
I'm in!!! Got the e-mail this morning.
It's conditional, they want me to send in some documents has anyone else been asked to do that?
Congratulations! Yes, they asked me to send hard copies of certain documents as well as verified my references. As long as your documents are accurate and reflect what you stated in your application, it should not be an issue. Welcome to the class!
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Re: HKUST 2013- Calling All Applicants [#permalink]
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07 Feb 2013, 12:48
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Re: HKUST 2013- Calling All Applicants [#permalink]
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16 Feb 2013, 03:19
Hi Everyone,
Not much activity here. I applied in the phase -2 of application deadline bang on the deadline day, and no I/V notification yet. Makes me nervous.
Has anybody else heard about their i/v?
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Re: HKUST 2013- Calling All Applicants [#permalink]
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17 Feb 2013, 11:04
CountGarlic wrote:
Hi Everyone,
Not much activity here. I applied in the phase -2 of application deadline bang on the deadline day, and no I/V notification yet. Makes me nervous.
Has anybody else heard about their i/v?
It is over 50 days after submitting my application and have not heard anything.
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Re: HKUST 2013- Calling All Applicants [#permalink]
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17 Feb 2013, 23:32
How would you guys compare HKUST vs. CUHK MBA?
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Re: HKUST 2013- Calling All Applicants [#permalink]
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18 Feb 2013, 00:10
Marsp wrote:
How would you guys compare HKUST vs. CUHK MBA?
Got rejected by HKUST. Waiting interview invite by HKU.
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Re: HKUST 2013- Calling All Applicants [#permalink]
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18 Feb 2013, 00:29
Hi When did you get the notification from HKUST
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Re: HKUST 2013- Calling All Applicants [#permalink]
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18 Feb 2013, 00:35
gspk wrote:
Hi When did you get the notification from HKUST
If you are asking about Inrocks then he/she got his/her rejection mail on or before 12th of Jan.
Gyan
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Re: HKUST 2013- Calling All Applicants [#permalink]
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18 Feb 2013, 00:48
Jealous of those guys who got admits by HKUST
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Re: HKUST 2013- Calling All Applicants [#permalink]
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18 Feb 2013, 01:00
Yes I just saw the thread you posted. Good luck In rocks with your other applications
Re: HKUST 2013- Calling All Applicants [#permalink] 18 Feb 2013, 01:00
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# HKUST 2013- Calling All Applicants
Moderators: OasisGC, aerien, BrushMyQuant
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®. | 3,170 | 10,377 | {"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} | 2.703125 | 3 | CC-MAIN-2016-50 | latest | en | 0.803198 |
https://linuxhint.com/min-max-arduino-function-array/ | 1,726,063,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00717.warc.gz | 331,690,110 | 46,479 | Arduino
# Get Minimum and Maximum Value of an Arduino Array Using min() and max() Function
The min() and max() functions are two of the most useful functions in Arduino, which are used to find the minimum and maximum value in an array respectively. In this article, we will discuss the use of these functions and how you can use them to find the minimum and maximum value of an Arduino array.
## Introduction to min() Function – Arduino
The min() function is a built-in function in the Arduino programming language. It’s used to return the smallest value among two or more values. min() takes two or more arguments and returns the smallest value.
Syntax
The syntax of the min() function is quite simple. It takes two or more arguments as input and returns the smallest value. Here is the syntax of the function:
min(x, y)
## Parameter
The function can take two or more arguments, and it returns the smallest value among them.
x: This is the first number to be compared. It can be of any data type.
y: This is the second number to which the first will be compared. It can be of any data type.
Any data type can be used for arguments, like integer or float.
## Returns
The min() function compares the two arguments and returns the smallest one. Function arguments determine the data type of the returned value.
Note: Don’t use any other function inside the min() function brackets as this may lead to incorrect output results.
## Example Code
Following code explains use of min() function in Arduino code:
int a = 10;
int b = 5;
int c = min(a, b); // set c to the minimum of a and b
void setup() {
Serial.begin(9600); // initialize serial communication
Serial.print("Minimum of 10 and 5 is: ");
Serial.println(c); // print the value of c to the serial monitor
}
void loop() {
}
This code initializes two variables a and b with values of 10 and 5, respectively. The min() function is then used to set the value of c to the minimum of a and b, which is 5. The value of c is printed using the Serial.println().
Output
In output we can see a minimum of two numbers that is 10 and 5.
## Introduction to max() Function – Arduino
The Arduino max() function is a mathematical function that is used to determine the highest value in a set of given values. It returns the highest value among two or more input values.
Syntax
The syntax of the Arduino max() function is as follows:
max(x, y)
## Parameter
The max() function takes two or more parameters, which are the values to be compared to determine the highest value.
x: This is the first number to be compared. It can be of any data type.
y: This is the second number to which the first will be compared. It can be of any data type.
## Returns
The max() function returns the highest value among the set of given values. Return values will have the same datatype as input values.
Note: Don’t use any other function inside the min() function brackets as this may lead to incorrect output results.
## Example Code
Following code explains use of max() function in Arduino code:
int a = 10;
int b = 5;
int c = max(a, b); // set c to the maximum of a and b
void setup() {
Serial.begin(9600); // initialize serial communication
Serial.print("Maximum of 10 and 5 is: ");
Serial.println(c); // print the value of c to the serial monitor
}
void loop() {
}
Above code initializes two variables a and b with values of 10 and 5, respectively. The max() function is then used to set the value of c to the maximum of a and b, which is 10. The value of c is printed using the Serial.println().
Output
In output we can see a maximum of two numbers that is 10 and 5.
## Arduino Code to Get Minimum and Maximum Value of an Arduino Array Using min() and max() Function
Here is a simple Arduino code that uses the min() and max() functions to find the minimum and maximum values in an array:
void setup() {
Serial.begin(9600); // Initialize serial communication
int myArray[] = { 1, 2, 3, 4, 5 }; // Define the array with 5 elements
int minValue = min(myArray[0], myArray[1]); // Initialize the minimum value with the first two elements
int maxValue = max(myArray[0], myArray[1]); // Initialize the maximum value with the first two elements
// Loop through the rest of the elements to find the minimum and maximum values
for (int i = 2; i < 5; i++) {
minValue = min(minValue, myArray[i]);
maxValue = max(maxValue, myArray[i]);
}
// Print the minimum and maximum values
Serial.print("Minimum value: ");
Serial.println(minValue);
Serial.print("Maximum value: ");
Serial.println(maxValue);
}
void loop() {
}
In this code, we define an array myArray with 5 elements. We then initialize the minValue and maxValue variables with the first two elements of the array using the min() and max() functions.
We then use a for loop to iterate through the rest of the elements in the array and update the minValue and maxValue variables accordingly using the min() and max() functions.
Finally, we print the minValue and maxValue variables using the Serial.print().
Output
In output the minimum and maximum value from the array is printed.
## Conclusion
The min() and max() functions in Arduino can find the minimum and maximum values in an array. By initializing the minimum and maximum values and then using a for loop to iterate through the rest of the elements, we can easily find the maximum and minimum values from any array. For detailed description of min() and max() function read the article. | 1,243 | 5,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.738671 |
https://www.itwissen.info/en/Markov-process.html | 1,643,438,199,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300573.3/warc/CC-MAIN-20220129062503-20220129092503-00061.warc.gz | 849,695,724 | 6,467 | # Markov process
Markov processes, further notations: Markov or Markoff processes, represent a tool for modeling stochastic processes
. The definition of Markov processes goes back to the Russian mathematician Andrej Markov (1856 - 1922). Markov processes are important for the description of processes in physics, natural sciences, economics and biology. Markov processes characterize a stochastic process
. A distinction is made between continuous and discrete processes. If it is a discrete process, it is also called a Markov chain.States and state transitions form the basic concept of Markov processes. Since this can be better imagined for discrete, at most countably many states and it also fits better to the discrete approach of waiting systems
, the further representation is limited to Markov chains. In the modeling, suitable states and state transitions must be defined. The consideration of chains also has the advantage that they can be graphically represented as state graphs.
Conditions for the stochastic and the Markov process
A stochastic process is called a Markov process if these conditions are fulfilled (formula 1).The Markov property now states that a next state always depends only on the just current one and not on the previous state, especially not on the starting state. The current state thus contains everything that needs to be known about the past.
Structure of a Markov chain
The Markov chain is used to describe the stochastic occupancy behavior of the queue. The circles model the possible occupancy states of the queue and the directed marked edges of the graph
(arrows) model the state transitions, where the transition from state "i" to state "j" occurs with the transition probability p(ij) i .
States in the Markov chain
The Markov chain shown is characterized by the fact that each state has transitions only to its immediate neighbors, but not to more distant states. Since the number of states is finite, all states have exactly two transitions to their neighbors. Two states each have only one transition to their neighbors.
Transition probability
In the discrete-time case shown, each occupancy state also has one transition to itself with the transition probability (formula 2). In general, the future random behavior of a Markov chain at any time "t" depends only on the even active state, but not on the past random behavior that activated the state. This directly applies to the occupancy behavior of the memory of a queueing system as well.
Informationen zum Artikel
Englisch: Markov process Updated at: 14.01.2010 #Words: 481 Links: Translations: DE | 524 | 2,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-05 | latest | en | 0.944808 |
http://www.disboards.com/showthread.php?t=3109098&page=3 | 1,419,807,717,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447559834.31/warc/CC-MAIN-20141224185919-00061-ip-10-231-17-201.ec2.internal.warc.gz | 393,032,476 | 21,472 | The DIS Discussion Forums - DISboards.com Stripping . . . everyone's doin' it!
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05-10-2013, 08:20 PM #31 NoleFan DIS Veteran Join Date: Mar 2013 Posts: 721 . Last edited by NoleFan; 05-10-2013 at 08:29 PM.
05-10-2013, 08:51 PM #32
Deb & Bill
DVC-Trivia Contest, Apr-2006: Honorable Mention
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Quote:
Originally Posted by NoleFan I would not offer \$12 more pp. factor in about \$5-\$6 pp MF & \$5 pp differential between stripped vs. loaded. I would not pay the same price pp for a stripped vs. loaded contract. If I had to buy 100 points via transfer, it would cost around \$10-\$12 pp (\$5-\$6 mf + \$5 discount for stripped contract). That's the math on the approx. \$12 Same resort, same contract size ie. stripped \$70pp NO MF loaded \$75pp +MF
But you aren't paying \$5-12 per point more for every year you have the points to use. Your logic is still bad.
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05-10-2013, 09:02 PM #33
NoleFan
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Quote:
Originally Posted by Deb & Bill But you aren't paying \$5-12 per point more for every year you have the points to use. Your logic is still bad.
Certainly, everyone is entitled to their own opinion. So, please share yours & why if you feel my logic is bad. If you had same resort/same # point contract, what would you offer for stripped vs. loaded & why?? Use any \$, just curious on price differential & justification.
05-11-2013, 09:34 AM #34 sweetdana DIS Veteran Join Date: Sep 2009 Location: Michigan Posts: 2,790 A Friend of mine listed his AK With the broker I don't remember which one but it was one of the big 3. They told him to sell his current Points with a transfer for 10-12\$, then list it. They said he wouldn't get the 10 or \$12 more per point and DVC. Would more likely not ROFR. Based on his goals of what he wanted to make this worked out for him, and listed it 8\$ less pp after selling his 300 points at 10.50. He contemplated selling his next year points, bet others are doing this. __________________ There is only 1 thing I love more than holding my DD's hand.. It is holding her hand and watching her grow up. ... added a new DD for double the fun!
05-11-2013, 02:46 PM #35
Deb & Bill
DVC-Trivia Contest, Apr-2006: Honorable Mention
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Location: USA
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Quote:
Originally Posted by NoleFan Certainly, everyone is entitled to their own opinion. So, please share yours & why if you feel my logic is bad. If you had same resort/same # point contract, what would you offer for stripped vs. loaded & why?? Use any \$, just curious on price differential & justification.
Divide the total number of years you have left on the contract starting with the current UY into the price per point you are paying. Subtract one year's value from the asking price per point. So if you are paying \$60 per point for 40 more years that is \$1.50 per point. Subtract \$1.50 from the asking price per point. And ask the owner to pay the dues on the used points. You'll get that money in the contract, but you will have to pay it when they are due. You don't get a credit from DVC.
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OKW and VWL Homes
Stop the madness. No more DVC construction.
Bring back Vacation Magic. Dump Disney Files!
Stopped drinking the Kool-aid long ago.
05-11-2013, 04:28 PM #36
Galun
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Join Date: Apr 2013
Posts: 68
Quote:
Originally Posted by Deb & Bill Divide the total number of years you have left on the contract starting with the current UY into the price per point you are paying. Subtract one year's value from the asking price per point. So if you are paying \$60 per point for 40 more years that is \$1.50 per point. Subtract \$1.50 from the asking price per point. And ask the owner to pay the dues on the used points. You'll get that money in the contract, but you will have to pay it when they are due. You don't get a credit from DVC.
Why would you only subtract \$1.50 when you can immediately flip the current year points for a net gain of \$5-6 per point?
05-11-2013, 04:54 PM #37 NoleFan DIS Veteran Join Date: Mar 2013 Posts: 721 Sample 100 point contract: With current year points Contract Price 100 x \$75 = \$7,500 + MF 100 x \$5.50 = \$550 Net cost (prior to closing costs) \$8,050 WithOUT current year points Contract Price 100 \$70 = \$7,000 + MF 0 (no current points) Net Cost (prior to closing costs) \$7,000 Cost difference \$1,050 If the sellers sold the points prior to sale (on stripped contract) or if buyer had to go buy points to fill in current year missing points @ \$10-\$12/pp = \$1,000-\$1,200 If the seller sold the points, they would recover their current year MF & adjust for the contracted "discount" If the seller included the points, BUYER would pay MF & seller would get a higher price/pp for the contract. Seller might benefit from stripped contract since lower broker fee & depending on what they get for renting or transferring points. However, someone may pay a premium for a contract with current UY points. Dollar for dollar this seems to make sense & allow for fair pricing amongst all parties. Last edited by NoleFan; 05-11-2013 at 05:02 PM.
05-12-2013, 12:50 PM #38
cm8
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Quote:
Originally Posted by sweetdana A Friend of mine listed his AK With the broker I don't remember which one but it was one of the big 3. They told him to sell his current Points with a transfer for 10-12\$, then list it. They said he wouldn't get the 10 or \$12 more per point and DVC. Would more likely not ROFR. Based on his goals of what he wanted to make this worked out for him, and listed it 8\$ less pp after selling his 300 points at 10.50. He contemplated selling his next year points, bet others are doing this.
05-12-2013, 01:15 PM #39
Missyrose
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Join Date: Jan 2009
Location: Northern Va.
Posts: 7,329
Quote:
Originally Posted by NoleFan Sample 100 point contract: With current year points Contract Price 100 x \$75 = \$7,500 + MF 100 x \$5.50 = \$550 Net cost (prior to closing costs) \$8,050 WithOUT current year points Contract Price 100 \$70 = \$7,000 + MF 0 (no current points) Net Cost (prior to closing costs) \$7,000 Cost difference \$1,050 If the sellers sold the points prior to sale (on stripped contract) or if buyer had to go buy points to fill in current year missing points @ \$10-\$12/pp = \$1,000-\$1,200 If the seller sold the points, they would recover their current year MF & adjust for the contracted "discount" If the seller included the points, BUYER would pay MF & seller would get a higher price/pp for the contract. Seller might benefit from stripped contract since lower broker fee & depending on what they get for renting or transferring points. However, someone may pay a premium for a contract with current UY points. Dollar for dollar this seems to make sense & allow for fair pricing amongst all parties.
Bolding is mine. But what many have argued on this thread and others, is that there isn't enough of a premium put on a contract with current (or fully loaded) points that would benefit the sellers instead of transferring out the current points and selling the contract stripped. If you saw fully loaded contracts going for \$15-\$20 more per point or even contracts going for \$10-\$15 more per point than stripped deals, then you could say there were premiums put on those types of contracts. But there simply isn't in the current market.
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05-12-2013, 01:50 PM #40
NoleFan
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Quote:
Originally Posted by Missyrose Bolding is mine. But what many have argued on this thread and others, is that there isn't enough of a premium put on a contract with current (or fully loaded) points that would benefit the sellers instead of transferring out the current points and selling the contract stripped. If you saw fully loaded contracts going for \$15-\$20 more per point or even contracts going for \$10-\$15 more per point than stripped deals, then you could say there were premiums put on those types of contracts. But there simply isn't in the current market.
I totally agree that someone would pay a premium for a loaded contract (I recently did) & a loaded contract probably would sell faster.
05-12-2013, 01:58 PM #41 bobbiwoz I'm happy to dance with youI have 2 opinionsWe had tried usual things to keep them away, including something like coyote urine, it is sold in hardware stores by us Join Date: Aug 2003 Location: Washington Township OH and Cape May NJ Posts: 61,535 I recently paid \$96 per BLT point that had 96 (100 point contract) banked points. I have bought 4 resale contracts over a 9 year period, and I would not even look at stripped contracts. IF you just keep looking, what you want will show up. __________________
05-12-2013, 02:11 PM #42
NoleFan
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Quote:
Originally Posted by bobbiwoz I recently paid \$96 per BLT point that had 96 (100 point contract) banked points. I have bought 4 resale contracts over a 9 year period, and I would not even look at stripped contracts. IF you just keep looking, what you want will show up.
That's a great price for a smaller, loaded BLT contract- congrats!
Since you've been in the DVC market for so long, do you have any thoughts or advice on ROFR trends (if any)?
Last edited by NoleFan; 05-12-2013 at 02:17 PM.
05-12-2013, 06:51 PM #43
agie65
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Quote:
Originally Posted by Missyrose Bolding is mine. But what many have argued on this thread and others, is that there isn't enough of a premium put on a contract with current (or fully loaded) points that would benefit the sellers instead of transferring out the current points and selling the contract stripped. If you saw fully loaded contracts going for \$15-\$20 more per point or even contracts going for \$10-\$15 more per point than stripped deals, then you could say there were premiums put on those types of contracts. But there simply isn't in the current market.
Even a fully loaded contract can not demand \$15-20 premium, you will be close to direct purchase price, also direct comes with extra benefit. Some resale are asking as much as direct with and without point, don't know who will take a bite.
05-12-2013, 06:56 PM #44
NoleFan
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Quote:
Originally Posted by agie65 Even a fully loaded contract can not demand \$15-20 premium, you will be close to direct purchase price, also direct comes with extra benefit. Some resale are asking as much as direct with and without point, don't know who will take a bite.
The premium should be around \$5/pp + buyer pays MF (approx. \$5.50) so overall cost would be around \$10.50 more pp than stripped. Can add a few dollars if buyer wants but not much more.
05-12-2013, 07:07 PM #45
Deb & Bill
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Quote:
Originally Posted by Galun Why would you only subtract \$1.50 when you can immediately flip the current year points for a net gain of \$5-6 per point?
Don't buy a contract to sell them Use them.
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All times are GMT -5. The time now is 06:01 PM. | 4,032 | 16,527 | {"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} | 2.515625 | 3 | CC-MAIN-2014-52 | latest | en | 0.914635 |
http://pockettorch.net/book/ordinary-differential-equations/ | 1,555,998,682,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578593360.66/warc/CC-MAIN-20190423054942-20190423080942-00170.warc.gz | 137,791,373 | 10,586 | INTRODUCTION TO THEORY OF ORDINARY DIFFERENTIAL EQUATION
Author: V. DHARMAIAH
Publisher: PHI Learning Pvt. Ltd.
ISBN:
Category: Mathematics
Page: 420
View: 294
This systematically-organized text on the theory of differential equations deals with the basic concepts and the methods of solving ordinary differential equations. Various existence theorems, properties of uniqueness, oscillation and stability theories, have all been explained with suitable examples to enhance students’ understanding of the subject. The book also discusses in sufficient detail the qualitative, the quantitative, and the approximation techniques, linear equations with variable and constants coefficients, regular singular points, and homogeneous equations with analytic coefficients. Finally, it explains Riccati equation, boundary value problems, the Sturm–Liouville problem, Green’s function, the Picard’s theorem, and the Sturm–Picone theorem. The text is supported by a number of worked-out examples to make the concepts clear, and it also provides a number of exercises help students test their knowledge and improve their skills in solving differential equations. The book is intended to serve as a text for the postgraduate students of mathematics and applied mathematics. It will also be useful to the candidates preparing to sit for the competitive examinations such as NET and GATE.
Ordinary Differential Equations
Author: Vladimir I. Arnold
Publisher: Springer Science & Business Media
ISBN:
Category: Mathematics
Page: 338
View: 740
Few books on Ordinary Differential Equations (ODEs) have the elegant geometric insight of this one, which puts emphasis on the qualitative and geometric properties of ODEs and their solutions, rather than on routine presentation of algorithms. From the reviews: "Professor Arnold has expanded his classic book to include new material on exponential growth, predator-prey, the pendulum, impulse response, symmetry groups and group actions, perturbation and bifurcation." --SIAM REVIEW
Ordinary Differential Equations
Author: Edward L. Ince
Publisher: Courier Corporation
ISBN:
Category: Mathematics
Page: 576
View: 262
Among the topics covered in this classic treatment are linear differential equations; solution in an infinite form; solution by definite integrals; algebraic theory; Sturmian theory and its later developments; much more. "Highly recommended" — Electronics Industries.
Ordinary Differential Equations
Second Edition
Author: Philip Hartman
Publisher: SIAM
ISBN:
Category: Differential equations
Page: 612
View: 247
Ordinary Differential Equations covers the fundamentals of the theory of ordinary differential equations (ODEs), including an extensive discussion of the integration of differential inequalities, on which this theory relies heavily. In addition to these results, the text illustrates techniques involving simple topological arguments, fixed point theorems, and basic facts of functional analysis. Unlike many texts, which supply only the standard simplified theorems, this book presents the basic theory of ODEs in a general way. This SIAM reissue of the 1982 second edition covers invariant manifolds, perturbations, and dichotomies, making the text relevant to current studies of geometrical theory of differential equations and dynamical systems. In particular, Ordinary Differential Equations includes the proof of the Hartman-Grobman theorem on the equivalence of a nonlinear to a linear flow in the neighborhood of a hyperbolic stationary point, as well as theorems on smooth equivalences, the smoothness of invariant manifolds, and the reduction of problems on ODEs to those on "maps" (Poincaré). Audience: readers should have knowledge of matrix theory and the ability to deal with functions of real variables.
Ordinary Differential Equations
A First Course
Author: D. Somasundaram
Publisher: CRC Press
ISBN:
Category: Mathematics
Page: 295
View: 359
Though ordinary differential equations is taught as a core course to students in mathematics and applied mathematics, detailed coverage of the topics with sufficient examples is unique. Written by a mathematics professor and intended as a textbook for third- and fourth-year undergraduates, the five chapters of this publication give a precise account of higher order differential equations, power series solutions, special functions, existence and uniqueness of solutions, and systems of linear equations. Relevant motivation for different concepts in each chapter and discussion of theory and problems-without the omission of steps-sets Ordinary Differential Equations: A First Course apart from other texts on ODEs. Full of distinguishing examples and containing exercises at the end of each chapter, this lucid course book will promote self-study among students.
Advanced Ordinary Differential Equations
Author: Kurt Otto Friedrichs
Publisher: CRC Press
ISBN:
Category: Mathematics
Page: 205
View: 696
Ordinary Differential Equations
Methods and Applications
Author: W. T. Ang
Publisher: Universal-Publishers
ISBN:
Category: Mathematics
Page: 204
View: 884
This introductory course in ordinary differential equations, intended for junior undergraduate students in applied mathematics, science and engineering, focuses on methods of solution and applications rather than theoretical analyses. Applications drawn mainly from dynamics, population biology and electric circuit theory are used to show how ordinary differential equations appear in the formulation of problems in science and engineering. The calculus required to comprehend this course is rather elementary, involving differentiation, integration and power series representation of only real functions of one variable. A basic knowledge of complex numbers and their arithmetic is also assumed, so that elementary complex functions which can be used for working out easily the general solutions of certain ordinary differential equations can be introduced. The pre-requisites just mentioned aside, the course is mainly self-contained. To promote the use of this course for self-study, suggested solutions are not only given to all set exercises, but they are also by and large complete with details.
Asymptotic Expansions for Ordinary Differential Equations
Author: Wolfgang Wasow
Publisher: Courier Corporation
ISBN:
Category: Mathematics
Page: 374
View: 915
"A book of great value . . . it should have a profound influence upon future research."--Mathematical Reviews. Hardcover edition. The foundations of the study of asymptotic series in the theory of differential equations were laid by Poincaré in the late 19th century, but it was not until the middle of this century that it became apparent how essential asymptotic series are to understanding the solutions of ordinary differential equations. Moreover, they have come to be seen as crucial to such areas of applied mathematics as quantum mechanics, viscous flows, elasticity, electromagnetic theory, electronics, and astrophysics. In this outstanding text, the first book devoted exclusively to the subject, the author concentrates on the mathematical ideas underlying the various asymptotic methods; however, asymptotic methods for differential equations are included only if they lead to full, infinite expansions. Unabridged Dover republication of the edition published by Robert E. Krieger Publishing Company, Huntington, N.Y., 1976, a corrected, slightly enlarged reprint of the original edition published by Interscience Publishers, New York, 1965. 12 illustrations. Preface. 2 bibliographies. Appendix. Index.
A Course in Ordinary Differential Equations
Author: Bindhyachal Rai
Publisher: CRC Press
ISBN:
Category: Mathematics
Page: 463
View: 506
Designed as a text for both under and postgraduate students of mathematics and engineering, A Course in Ordinary Differential Equations deals with theory and methods of solutions as well as applications of ordinary differential equations. The treatment is lucid and gives a detailed account of Laplace transforms and their applications, Legendre and Bessel functions, and covers all the important numerical methods for differential equations.
An Introduction to Ordinary Differential Equations
Author: Earl A. Coddington
Publisher: Courier Corporation
ISBN:
Category: Mathematics
Page: 320
View: 507
A thorough, systematic first course in elementary differential equations for undergraduates in mathematics and science, requiring only basic calculus for a background. Includes many exercises and problems, with answers. Index. | 1,669 | 8,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-18 | latest | en | 0.897074 |
http://codeforces.com/topic/57355/en4 | 1,579,537,332,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598800.30/warc/CC-MAIN-20200120135447-20200120164447-00519.warc.gz | 36,390,956 | 17,424 | Sqrt-tree: answering queries in O(1) with O(loglogN) preprocessing.
Revision en4, by gepardo, 2018-01-11 21:58:21
Hello, Codeforces!
Some time ago I invented an interesting data structure and I'd like to share it with the community. Maybe, it was known before, and if you knew about it before my blog post, please share the link to the source.
# What can the data structure do?
Given an array a that contains n elements and the operation op that satisfies some properties:
1. Associative property: (x op y)op z = x op(y op z).
2. Commutative property (optional, not required): x op y = y op x.
So, such operations as gcd, min, max, sum, multiplication, and, or, xor, etc. satisfy these conditions.
Also we have some queries l, r. For each query, we need to find al op al + 1 op ... op ar. Let's call such queries q(l, r).
My data structure can process such queries in O(1) time with preprocessing time and memory.
# How it works?
## 1. Sqrt
Let's do a sqrt-decomposition. We divide our array in blocks, each block has size . For each block, we compute:
1. Answers to the queries that lie in the block and begin at the beginning of the block (prefix-op)
2. Answers to the queries that lie in the block and end at the end of the block (suffix-op)
And we'll calculate another one thing:
1. between[i, j] i ≤ j -- answer to the query that begins at the start of block i and ends at the end of block j. Note that we have blocks, so the size of this array will be .
Let's see the example.
Let op be + (we calculate sum on the segment) and we have the following array a:
1 2 3 4 5 6 7 8 9
It will be divided onto three blocks: 1 2 3, 4 5 6 and 7 8 9.
For first block prefix-op is 1 3 6 and suffix-op is 6 5 3.
For second block prefix-op is 4 9 15 and suffix-op is 15 11 6.
For third block prefix-op is 7 15 24 and suffix-op is 24 17 9.
between array is:
6 21 45
- 15 39
- - 24
It's obvious to see that these arrays can be easily calculated in O(n) time and memory.
We already can answer some queries using these arrays. If the query doesn't fit into one block, we can divide it onto three parts: suffix of a block, then some segment of contiguous blocks and then prefix of some block. We can answer a query by dividing it into three parts and taking op of some value from suffix-op, then some value from between, then some value from prefix-op.
But if we have queries that entirely fit into one block, we cannot process them using these three arrays. So, we need to do something.
## 2. Tree
We cannot answer only the queries that entirely fit in one block. But what if we build the same structure as described above for each block? Yes, we can do it. And we do it recursively, until we reach the block size of 1 or 2. Answers for such blocks can be calculated easily in O(1).
So, we get a tree. Each node of the tree represents some segment of the array. Node that represents array segment with size k has children -- for each block. Also each node contains the three arrays described above for the segment it contains. The root of the tree represents the entire array. Nodes with segment lengths 1 or 2 are leaves.
Also it's obvious that the radius of this tree is , because if some vertex of the tree represents an array with length k, then its children have length . , so decreases two times every layer of the tree and so its height is . The time for building and memory usage will be , because every element of the array appears exactly once on each layer of the tree.
Now we can answer the queries in . We can go down on the tree until we meet a segment with length 1 or 2 (answer for it can be calculated in O(1) time) or meet the first segment in which our query doesn't fit entirely into one block. See the first section on how to answer the query in this case.
OK, now we can do per query. Can it be done faster?
## 3. Optimizing the query complexity
One of the most obvious optimization is to binary search the tree node we need. Using binary search, we can reach the complexity per query. Can we do it even faster?
The answer is yes. Let's assume the following two things:
1. Each block size is a power of two.
2. All the blocks are equal on each layer.
To reach this, we can add some zero elements to our array so that its size becomes a power of two. | 1,095 | 4,299 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-05 | latest | en | 0.889785 |
https://discuss.pytorch.org/t/cross-entropy-loss-math-under-the-hood/79749 | 1,653,112,582,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662538646.33/warc/CC-MAIN-20220521045616-20220521075616-00219.warc.gz | 268,007,073 | 5,556 | # Cross Entropy Loss Math under the hood
@ptrblck could you help me?
Hi everyone!
Please, someone could explain the math under the hood of Cross Entropy Loss in PyTorch?
I was performing some tests here and result of the Cross Entropy Loss in PyTorch doesn’t match with the result using the expression below:
I took some examples calculated using the expression above and executed it using the Cross Entropy Loss in PyTorch and the results were not the same.
I am trying this example here using Cross Entropy Loss from PyTorch:
``````probs1 = torch.tensor([[
[
[ 0.1, 0.3],
[0.4, 0.5]
],
[
[0.2, 0.3],
[0.4, 0.5]
],
[
[0.7, 0.4],
[0.2, 0.0]
]
]])
target = torch.tensor([
[
[2, 2],
[0, 1]
]
])
print(torch.sum(probs1, dim=0))
print(probs1.shape)
criterion = nn.CrossEntropyLoss()
loss = criterion(probs1, target)
loss
result -> tensor(0.9488)
``````
Each pixel along the 3 channels corresponds to a probability distribution…there is a probability distribution for each position of the tensor…and the target has the classes for each distribution.
How can I know if this loss is beign computed correctly?
Best regards,
Matheus Santos.
Hello Matheus!
The issue is that pytorch’s `CrossEntropyLoss` doesn’t exactly match
the conventional definition of cross-entropy that you gave above.
Rather, it expects raw-score logits as it inputs, and, in effect, applies
`softmax()` to the logits internally to convert them to probabilities.
(`CrossEntropyLoss` might better have been named
`CrossEntropyWithLogitsLoss`.)
To check this, you could apply the logit function, `log (p / (1 - p))`
to convert your `probs1` tensor, and then run that through
`CrossEntropyLoss`.
Best.
K. Frank
1 Like
Note that you are not using `nn.CrossEntropyLoss` correctly, as this criterion expects logits and will apply `F.log_softmax` internally, while `probs` already contains probabilities, as @KFrank explained.
So, let’s change the criterion to `nn.NLLLoss` and apply the `torch.log` manually.
This approach is just to demonstrate the formula and shouldn’t be used, as `torch.log(torch.softmax())` is less numerically stable than `F.log_softmax`.
Also, the default reduction for the criteria in PyTorch will calculate the average over the observations, so lets use `reduction='sum'`.
Given that you’ll get:
``````criterion = nn.NLLLoss(reduction='sum')
loss = criterion(torch.log(probs1), target)
``````
The manual approach from your formula would correspond to:
``````# Manual approach using your formula
one_hot = F.one_hot(target, num_classes = 3)
one_hot = one_hot.permute(0, 3, 1, 2)
ce = (one_hot * torch.log(probs1 + 1e-7))[one_hot.bool()]
ce = -1 * ce.sum()
``````
While the manual approach from the PyTorch docs would give you:
``````# Using the formula from the docs
loss_manual = -1 * torch.log(probs1).gather(1, target.unsqueeze(1))
loss_manual = loss_manual.sum()
``````
We should get the same results:
``````print(loss, ce, loss_manual)
> tensor(2.8824) tensor(2.8824) tensor(2.8824)
``````
which looks correct.
1 Like
Hey, thank you so much for all explanations here @KFrank and @ptrblck !!!
I was performing some tests using tensors with lower dimensions to ensure that the loss result is correct and, due to this, expand to tensors with higher dimensions and do not worry about the possibility that the loss value is wrong.
I was actually using nn.CrossEntropyLoss () in the wrong way, I apologize for that.
Now I understood how to use it !! | 911 | 3,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.883454 |
https://www.plantengineering.com/single-article/transitional-root-cause-analysis/d8fdecff6b21251898531173a4eab567.html | 1,511,000,350,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804724.3/warc/CC-MAIN-20171118094746-20171118114746-00664.warc.gz | 832,652,185 | 18,072 | # Transitional Root Cause Analysis
## Discussing Root Cause Analysis (RCA) can be made easier using Shon Isenhour's tree diagram.
09/14/2017
When I discuss RCA I use a method called Transitional Root Cause Analysis or TRCA for short.
It is made up of 10 tools that can be explained and understood in a very short period of time.
In the next few minutes I will demonstrate both the simplicity and rules for use for 3 of the 10 and explain why we consider them transitional in nature.
In this blog we will use what I categorize as the tree methods. These are three tools that build out into a tree root like structure. So let's take a look: the first one is called the "5 why" method. It is very common in industry today and is a very basic root cause tool. It is created simply by asking the question why multiple times to create one causal chain. It creates a simple main tap root to build off of.
Now, if we take the 5 why diagram and branch it out by adding more elements at each level then we get a better representation of all of the causes that come together to create an effect. This transition of the 5 why is known as a fault tree. This method allows us to easily see all factors that led to a failure, but sometimes we need to show a bit more information to make the graphic more meaningful.
If for instance, an effect can only occur when all of its causes exist at the same place and in the same moment in time then we use the word "and" at that junction of the roots. If we eliminate either one of the cause then we can eliminate the effect. On the flip side, if either of the causes could precipitate the effect then the word "or" would be placed at the junction. This would be read as this or that could cause the effect above. This allows you to see that both possibilities must be addressed to prevent the cause.
These three tree methods transition from one to the next by adding one simple new feature as needed during the root cause process. First, we take a "5 why" and branch it to get the fault tree then we add in the "and and" or "or" to get logic tree. Three powerful tools that build on each other to get you to the lowest cost solution that mitigates the risk.The other Transitional tools work very much in the same way and allow us to use the right tool for the job instead of trying to use a screwdriver as a hammer.
Shon Isenhour, content marketing, Eruditio LLC. This article originally appeared on Eruditio's website. Eruditio is a CFE Media content partner.
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This course explains how maintaining power and communication systems through emergency power-generation systems is critical. | 1,264 | 6,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-47 | longest | en | 0.952152 |
http://betterlesson.com/lesson/resource/1754605/71452/warm-up-rubric | 1,487,529,840,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170249.75/warc/CC-MAIN-20170219104610-00181-ip-10-171-10-108.ec2.internal.warc.gz | 29,500,939 | 20,726 | Warm up Rubric
# Statistical Studies
Unit 10: Modeling Data with Statistics and Probability
Lesson 2 of 6
## Big Idea: Class generated scenarios provide the context for discussions about statistical population, sample, and type of study.
Print Lesson
4 teachers like this lesson
Standards:
Subject(s):
Math, Statistics, experiment, survey, sample, observational study, Algebra 2, master teacher project, population, sample biase
50 minutes
### Amelia Jamison
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http://www.jiskha.com/members/profile/posts.cgi?name=Kirk&page=2 | 1,369,522,237,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706470197/warc/CC-MAIN-20130516121430-00034-ip-10-60-113-184.ec2.internal.warc.gz | 536,143,101 | 3,528 | Saturday
May 25, 2013
# Posts by Kirk
Total # Posts: 65
physics
A 70g steel bullet moves at 200m/s just before imbedding itself into a thick anchored wood post. If 70% of the initial kinetic energy goes into heating the bullet (steel c = 460 J/kg-C)...how hot does the bullet become if it initially is at 26 degrees Celsius? What would be th...
physics
an ideal fluid in a pipe of diameter 14cm is moving at 6.0m/s. If the incompressible fluid density is 1.05grams/cc, what is the flow rate in kg/s and what is the speed of flow if the pipe narrows to 4.0cm radius?
physics
A wire of diameter 0.65mm and length 6.0m stretches 10mm when 5.5kg is hung from it. What is the Young's modulus for the wire and what is the effective spring constant for the stretching wire
MAT101
Thank you, I was stuck on this for several hours to how to write this out, thank you again!
MAT101
Suppose you have a cookie stand, and when you charge \$3 per cookie box you sell 200 boxes. But when you raise your price to \$4 you only sell 120 boxes. Write the equation for the number of boxes you sell as a function of the price you charge. Denote "b" for the numbe...
Geometry
The Great Pyramid of Cheops is a right square pyramid with height of 148 m and a square base with a perimeter of 940 m. The Transamerica Building in San Francisco has the basic shape of a right square pyramid that has a square base with a perimeter of 140 m and a height of 260...
Geometry
Find the cost of carpeting the following rectangular rooms: a. Dimensions: 6.5 m * 4.5 m; cost = \$13.85>m2 b. Dimensions: 15 ft * 11 ft; cost = \$30>yd2
Math
Explain how you would find the volume of an irregular shape.
Math
John claimed he had a garden twice as large as Al’s rectangular- shaped garden that measured 15 ft by 30 ft. When they visited John’s rectangular-shaped garden, they found it measured 18 ft by 50 ft. Al claimed that it could not be twice as large since neither the le...
Math
Which of the following properties do not change under a size transformation? Explain how you can be sure of your answers. a. Distance between points b. Angle measure c. Parallelism; that is, if two lines are parallel to each other, then their images are parallel to each other.
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https://codereview.stackexchange.com/questions/195360/performance-of-dynamic-fibonacci-generator-and-dead-rabbits/195367 | 1,718,900,364,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00075.warc.gz | 150,910,902 | 46,074 | # Performance of dynamic Fibonacci generator and dead rabbits
I would like to improve my code style and its efficiency.
I'm just moving beyond the standard recursive approach to a lot of simple algorithms.
This post has two code snippets with slightly different but related problems. They're both in Python.
The first snippet is just to produce Fibonacci numbers. I think I've done this with O(n) time complexity. Do you have suggestions on how to make my code more readable and more efficient?
# Using bottom-up dynamic programming approach, define all fibonacci numbers
def fib(num):
# First and second generations are trivial
if num == 1 or num == 2:
return 1
# If not trivial, keep going:
# Set up array
allGenerations = [None] * (num + 1)
allGenerations[1] = 1
allGenerations[2] = 1
# Fill in the array until you reach the given generation
for i in range(3, num+1):
allGenerations[i] = allGenerations[i - 1] + allGenerations[i - 2]
return allGenerations[num]
The second snippet tries to use what I learned from the first snippet to solve a related problem. The numbers represent living rabbits. If rabbits lived forever, their population sizes would follow a fibonacci sequence. In this alteration, rabbits live for a fixed amount of time (input m). Newborn rabbits keep living, adult rabbits produce themselves and one offspring, and rabbits that are about to die just produce one offspring (see http://rosalind.info/problems/fibd/).
I'm not sure how to analyze the time complexity of this problem, but it seems high? So questions are (1) What is the time complexity of my solution? and (2) How can I improve efficiency and readability?
def rabbit(n, m):
# First two generations are trivial
if n == 1 or n == 2:
return 1
# If not trivial, keep going:
# Set up array
allGenerations = [None] * (n + 1)
# Each index in the generations will be another array whose indicies represent the ages of the rabbits in that generation
allGenerations[1] = [1]
allGenerations[2] = [0, 1]
# Bottom-up filling of generations
for i in range(3, n + 1):
# Initalize answer as a list of number of rabbits at each age
answer = [None] * (i)
# Get the previous generation
previous = allGenerations[i - 1]
# Initalize the number of newborns
newborns = 0
# From age 1 (first reproductive age) to either age at death
# or the oldest in the previous generation (whichever comes first), produce newborn rabbits
for j in range(1, min(m, len(previous))):
newborns += previous[j]
# The 0-index of the answer represents newborn rabbits
answer[0] = newborns
# Move every element in the previous generation up one index into the new generation
for k in range(0, len(previous)):
answer[k + 1] = previous[k]
# Put this answer into the list of all answers
allGenerations[i] = answer
# Return all living rabbits
return sum(answer[:m])
# Some testcases
import unittest
class mainTest(unittest.TestCase):
def test(self):
self.assertEqual(rabbit(1, 3), 1)
self.assertEqual(rabbit(2, 3), 1)
self.assertEqual(rabbit(3, 3), 2)
self.assertEqual(rabbit(4, 3), 2)
self.assertEqual(rabbit(5, 3), 3)
self.assertEqual(rabbit(6, 3), 4)
self.assertEqual(rabbit(7, 3), 5)
self.assertEqual(rabbit(25, 31), 75025)
self.assertEqual(rabbit(30, 30), 832040)
self.assertEqual(rabbit(35, 29), 9227437)
self.assertEqual(rabbit(40, 28), 102333267)
self.assertEqual(rabbit(45, 27), 1134880302)
if __name__ == '__main__':
unittest.main(argv=[''], exit = False)
• You probably already realize this, but iterative Fibonacci is easy and only requires O(1) space and O(n) time: a += b; swap(a,b);. Or unroll it with a+=b; b+=a;. If you're not going to return a list / array of Fib(0..n), then don't make one in the first place. (Unless it's as an exercise) Commented May 29, 2018 at 7:17
# naming variable
naming the variables clearly and correctly goes a long way in documenting your code. n and m are not clear. I would name them generation and max_age or something
# generators
For the simple fibonacci series, instead of storing everything in lists, only store what you need, and yield the results instead of returning the complete list.
And instead of keeping track of how far you are in your list, use an endless generator to yield a stream of fibonacci numbers, and the nth itertools recipe
def fib(a=0, b=1):
while True:
yield a
a, b = b, a + b
can be found in almost all python handbooks
from itertools import islice
def nth(iterable, n, default=None):
"""
Returns the nth item or a default value
https://docs.python.org/3/library/itertools.html#itertools-recipes
"""
return next(islice(iterable, n, None), default)
def fib_n(n):
return nth(fib(), n)
# dying rabbits
The same approach can be used for the second example, but here you can use a double ended queue with a fixed length instead of a tuple.
def rabbits(max_age):
generations = deque(islice(fib(1,1), max_age), maxlen=max_age)
while True:
yield generations[0]
new_borns = generations[-1]
generations.append(sum(generations) - new_borns)
def rabbit(generation, max_age):
return nth(rabbits(max_age), generation - 1)
The generation - 1 is to account for python's 0-indexing
# dynamic solution:
another approach, using recursion and memoization is more in line with dynamic programming.
from functools import lru_cache
@lru_cache(None)
def rabbit_dynamic(generation, max_age):
# print(f'calling rabbit_dynamic({generation}, {max_age})')
generation -= 1 # 0-indexing
if generation < 1:
return 0
if generation < max_age:
return nth(fib(1,1), generation)
return sum(rabbit_dynamic(generation - i, max_age) for i in range(1, max_age))
# timings
since you wondered about performance:
%timeit rabbit_OP(45, 27)
183 µs ± 8.68 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit rabbit(45, 27)
40.6 µs ± 3.1 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
so a 4.5 time speedup by dropping the intermediary lists
rabbit_dynamic.cache_clear()
%timeit -n 1 -r 1 rabbit_dynamic(45,27)
150 µs ± 0 ns per loop (mean ± std. dev. of 1 run, 1 loop each)
%timeit rabbit_dynamic.cache_clear(); rabbit_dynamic(45,27)
155 µs ± 7.51 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
so about the same as your implementation
## cache
The caching has a huge influence:
rabbit_dynamic.cache_info()
CacheInfo(hits=399, misses=44, maxsize=None, currsize=44)
without caching:
@lru_cache(0)
def rabbit_dynamic_zero_cache(generation, max_age):
# print(f'calling rabbit_dynamic({generation}, {max_age})')
generation -= 1
if generation < 0:
return 0
if generation< max_age:
return nth(fib(1,1), generation)
return sum(rabbit_dynamic_zero_cache(generation - i, max_age) for i in range(1, max_age))
%timeit -n 1 -r 1 rabbit_dynamic_zero_cache(45, 27)
179 ms ± 0 ns per loop (mean ± std. dev. of 1 run, 1 loop each)
so about 1000 times slower (about 125ms without any lru_cache)
rabbit_dynamic_zero_cache.cache_info()
CacheInfo(hits=0, misses=67185, maxsize=0, currsize=0)
because of the repeated asking for the same answer (67k function calls instead of 443)
• For the cached version, maybe clearing the cache should be part of the loop that gets timed. Commented May 29, 2018 at 10:18
• You are correct. If I can find an easy way to do so that does not include the time the cache clearing itself takes Commented May 29, 2018 at 10:25
• Maybe just subtract the cache clearing time? Commented May 29, 2018 at 17:01
• Thank you for the response! While the other answers to this question have more upvotes (and are also very useful) this answer most comprehensively reviews my code as a whole, so I'm going to mark this one as the answer. Commented May 29, 2018 at 19:51
With regard to your time complexity:
With your calculation of the nth Fibonacci number, you could do it in $O(1)$ time by using the relation:
$$F_n = \left\lfloor \frac{\varphi^n}{\sqrt{5}} + \frac{1}{2} \right\rfloor$$
So the 484th Fibonacci number would be equal to:
$$\left\lfloor \frac{\varphi^{484}}{\sqrt{5}} + \frac{1}{2} \right\rfloor \approx 6.317 \times 10^{100}$$
If you do not want to use the floor function, you can achieve the exact same result by manipulating the equation and using the nearest-integer function: $\left[\frac{\varphi^n}{\sqrt5}\right]$
This will speed up your program if you intend on calculating very large Fibonacci numbers.
Note: when calculating $\varphi^n$, use math.pow(x,y) because it uses floating point exponentiation and is always $O(1)$, whereas pow(x,y) and x**y can vary. For calculating $\varphi$ itself, there are many methods, here is an example.
• You're going to have rounding errors if you implement this formula with floats. Also, math.pow(x,y) won't be faster than x**y if x (φ) is a float. Commented May 29, 2018 at 11:51
• What about these: Matrix Exponentiation or Fast Doubling (IME, the latter is terribly fast when n is not too low): nayuki.io/page/fast-fibonacci-algorithms Commented May 29, 2018 at 15:52
• @flornquake math.pow(x,y) always converts both it's arguments to float per the docs. Likewise, math.pow(x,y) calls the C implementation of pow, which will likely be a constant time operation (unless you're using a rare architecture that uses an inefficient algorithm). However, you are correct on the possibility of rounding errors with $\varphi$, but I'm sure Python has plenty of arbitrary precision libraries. @Rudy That is a very useful source, thank you (I like that it includes the code). Commented May 29, 2018 at 18:12
The loop
# Move every element in the previous generation up one index into the new generation
for k in range(0, len(previous)):
answer[k + 1] = previous[k]
makes the next allGenerations list one element longer than the previous one. After n iterations you have lists of lengths 1, 2, ... n, making them $O(n^2)$ total. Just populating them all has a quadratic time complexity.
The good news is you don't need to make separate lists for each iteration. Do not index rabbits with age, index them with the birthdate instead. Once the first m generations are populated, the rest is just
for date in range(m, n):
rabbits.append(sum(rabbits[(date - m):])
with $O(nm)$ time complexity.
The next step is to notice that sum is also redundant. The next bunch of newborns is produced by last bunch of newborns, plus all the breeders of the last round, minus those who'd passed away. The breeders of the last round would obviously produce as much as they did at the last round, which is exactly the number of last round newborns:
rabbits.append(2*rabbits[-1] - rabbits[date - m])
with $O(n)$ time complexity.
The problem statement asks for all numbers. Your functions however only return the final tally. The first snippet should return allGenerations rather than allGenerations[n]. | 2,837 | 10,764 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-26 | latest | en | 0.892487 |
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Solving More Difficult Proportions with an Unknown
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# Solving More Difficult Proportions with an Unknown
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Author: Graham Pardun
##### Description:
This lesson will present how to solve for unknown variables in more difficult proportion problems.
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## Solving Awesome Proportions With Variables In Them
7/4x=(2x+8)/3 might look scary, but it's not really that scary. In this lesson, I'll start with the very basic concept of a proportion (e.g., 3/4=6/8 -- something you learned in 5th grade, probably), and build up to the seemingly-scary problems with variables in them -- no worries! | 315 | 1,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-43 | longest | en | 0.907554 |
https://hsm.stackexchange.com/questions/5666/why-are-canonical-coordinates-canonical/6033 | 1,620,457,128,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988850.21/warc/CC-MAIN-20210508061546-20210508091546-00317.warc.gz | 326,241,130 | 41,369 | # Why are canonical coordinates canonical?
Canonical coordinates are
coordinates $q_i$ and $p_i$ in phase space that are used in the Hamiltonian formalism. The canonical coordinates satisfy the fundamental Poisson bracket relations: $$\{q_i,q_j\} = 0, \quad \{p_i,p_j\} = 0, \quad \{q_i,p_j\} = \delta_{ij},$$
-Wikipedia.
From this it is clear that we get canonical transformations and, with quantum mechanics, canonical commutation relations.
When did these coordinates start being called canonical and what was the reasoning?
Such coordinates were called canonical because they are those in which equations of motion (or, of the hamiltonian flow of a function $H$) take the “canonical form” $$\frac{dq_i}{dt}=\frac{\partial H}{\partial p_i}, \qquad \frac{dp_i}{dt}=-\frac{\partial H}{\partial q_i}$$ first written by Poisson (1809, pp. 272, 313), Lagrange (1810, p. 350), and Hamilton (1835, p. 98). Actually, none of them called that form canonical yet: according to the Encyklopädie (1935, p. 573) the first person to do so was Jacobi (1837, pp. 65-66):
On trouve au moyen de ce théorème, par le calcul même, des éléments dont les valeurs différentielles, dans le mouvement troublé, prennent la forme simple qu'elles ont dans le théorème, forme que je désigne dans mon mémoire sous le nom de canonique.
... and Thomson-Tait (1867, p. 254) commented:
This is the celebrated “canonical form” of the equations of motion of a system, though why it has been so called it would be hard to say.
As to the practice of calling the coordinates or “elements” $p_i, q_i$ themselves “canonical”, it seems to originate also with Jacobi in the aforementioned memoir, completed in 1838 but only published posthumously (1862, p. 128; German translation: 1906, p. 153):
Systema elementorum, quae in modum praecedentium per aequationes differentiales canonicas determinantur, et ipsum dicere convenit canonicum elementorum systema.
While this practice did not universally catch on at first (Thomson-Tait, Poincaré (1893), Whittaker (1917) or the Encyklopädie call the equations canonical, but not the variables), it was adopted by e.g. Donkin (1862, pp. v, 550), Tisserand (1868, p. 258; 1889, p. 164), Schering (1873, p. 23), Lie (1874, p. 258), Routh (1892, pp. 304-306), Dziobek (1892, pp. 102-103), Charlier (1902, pp. 56-58), Dirac (1925, p. 651), etc. In fact it must have leaked somehow between 1838 and 1862, for Cayley has it already in (1858, p. 9):
18. There is, however, one important point which requires to be adverted to. Lagrange, in the memoir of 1810, and the second edition of the ‘Mécanique Analytique,’ remarks, that for a particular system of arbitrary constants, viz., if $\alpha,...$ denote the initial values of the coordinates $\xi,..$ and $\lambda,..$ denote the initial values of $\smash[b]{\frac{dT}{d\xi'}},...$ then the equations for the variations of the elements take the very simple form $$\frac{d\alpha}{dt}=-\frac{d\Omega}{d\lambda}..,\quad \frac{d\lambda}{dt}=\frac{d\Omega}{d\alpha},...$$ This is, in fact, the original idea and simplest example of a system of canonical elements; viz. of a system composed of pairs of elements, $\alpha, \lambda$, the variations of which are given in the form just mentioned.
• +1 for the huge work of finding and linking all the references. – Antoine Feb 13 '17 at 18:05
I was told they are called "canonical" because of their prevalence and comparative simplicity. I believe the etymology is from the Greek "kanon", meaning "standard" or "model", or "usual".
I was told by the professor in a graduate physics course that the equations were called "canonical" because they were so perfect that they could be laws of the church, that is canon laws. At the time of their creation, the church was the most powerful entity extant. Thus, they were perfect enough to be church laws or "canonical" laws.
• That just sounds like an explanation of where the word "canonical," which is not a term in physics only, comes from. It does not sound like an authoritative explanation of its use in this situation rather than its general etymology. – KCd Jan 26 '18 at 15:59
A more appropriate translation of canon (kanon) is "law". Consider for example the dictionary definition of "canonical hours": Canonical hours: certain stated times of the day, fixed by ecclesiastical laws, and appropriated to the offices of prayer and devotion;
• Currently, this post does not answer the question as stated. Could you perhaps expand it to do so? – Danu May 20 '17 at 21:05 | 1,220 | 4,525 | {"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": 1, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-21 | latest | en | 0.721826 |
https://bepurebeauty.com/skin-diseases/frequent-question-what-is-the-mass-of-one-mole-of-na.html | 1,660,932,153,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573744.90/warc/CC-MAIN-20220819161440-20220819191440-00579.warc.gz | 152,995,041 | 17,363 | # Frequent question: What is the mass of one mole of Na?
Contents
## What is the average mass of one Na atom?
The Elements, sorted by Atomic Mass
Atomic Number Symbol Atomic Weight (amu, g/mol)
10 Ne 20.179
11 Na 22.98977
12 Mg 24.305
13 Al 26.98154
## How many atoms are in 3 moles of sodium Na?
3 mol Na × (6.022×10^23 atoms Na/1mol Na) = 1.807×10^24 atoms Na.
## What is the molar mass of 2NaCl?
The molar mass and molecular weight of 2NaCl is 116.8855.
## What is the mass by sodium Na in NaCl atomic mass of Na 23 g mol Cl 35 g mol?
This would mean the masses of Na and Cl are 22.99g and 35.45g respectively. Therefore, the percent composition of NaCl is 39.3% sodium and 60.7% chloride.
THIS IS IMPORTANT: When should I worry about a cancerous mole? | 241 | 767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-33 | latest | en | 0.829865 |
http://www.star.bristol.ac.uk/~mbt/stilts/sun256/layer-polygon.html | 1,653,317,834,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558030.43/warc/CC-MAIN-20220523132100-20220523162100-00141.warc.gz | 110,740,381 | 4,550 | Next Previous Up Contents
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#### 8.3.12 `polygon`
Draws a closed polygon given an array of coordinates that define its vertices. In fact this plot requires the position of the first vertex supplied as a positional value in the usual way (e.g. `X` and `Y` coordinates) and the second, third etc vertices supplied as an array using the `otherpoints` parameter.
Invocation might therefore look like "`xN=x1 yN=y1 otherpointsN=array(x2,y2, x3,y3, x4,y4)`".
Usage Overview:
``` layerN=polygon useposN=true|false polymodeN=outline|border|fill|cross|star
thickN=<int-value>
<pos-coord-paramsN> otherpointsN=<array-expr> inN=<table>
ifmtN=<in-format> istreamN=true|false icmdN=<cmds>
```
All the parameters listed here affect only the relevant layer, identified by the suffix `N`.
Positional Coordinate Parameters:
The positional coordinates `<pos-coord-paramsN>` give a position for each row of the input table. Their form depends on the plot geometry, i.e. which plotting command is used. For a plane plot (`plot2plane`) the parameters would be `xN` and `yN`. The coordinate parameter values are in all cases strings interpreted as numeric expressions based on column names. These can be column names, fixed values or algebraic expressions as described in Section 10.
Example:
``` stilts plot2sky in=big_tab_VIR_VIS_CSA_public.fits
icmd='select ALTITUDE>4e4&&ALTITUDE<4.3e4'
layer=polygon polymode=fill
lon=LON_CENTER lat=LAT_CENTER
otherpoints=array(lon_corner_1,lat_corner_1,lon_corner_2,lat_corner_2)
`icmdN = <cmds>` (ProcessingStep[])
Specifies processing to be performed on the layer N input table as specified by parameter `inN`. The value of this parameter is one or more of the filter commands described in Section 6.1. If more than one is given, they must be separated by semicolon characters (";"). This parameter can be repeated multiple times on the same command line to build up a list of processing steps. The sequence of commands given in this way defines the processing pipeline which is performed on the table.
Commands may alteratively be supplied in an external file, by using the indirection character '@'. Thus a value of "`@filename`" causes the file `filename` to be read for a list of filter commands to execute. The commands in the file may be separated by newline characters and/or semicolons, and lines which are blank or which start with a '`#`' character are ignored.
`ifmtN = <in-format>` (String)
Specifies the format of the input table as specified by parameter `inN`. The known formats are listed in Section 5.1.1. This flag can be used if you know what format your table is in. If it has the special value `(auto)` (the default), then an attempt will be made to detect the format of the table automatically. This cannot always be done correctly however, in which case the program will exit with an error explaining which formats were attempted. This parameter is ignored for scheme-specified tables.
[Default: `(auto)`]
`inN = <table>` (StarTable)
The location of the input table. This may take one of the following forms:
• A filename.
• A URL.
• The special value "`-`", meaning standard input. In this case the input format must be given explicitly using the `ifmtN` parameter. Note that not all formats can be streamed in this way.
• A scheme specification of the form `:<scheme-name>:<scheme-args>`.
• A system command line with either a "`<`" character at the start, or a "`|`" character at the end ("`<syscmd`" or "`syscmd|`"). This executes the given pipeline and reads from its standard output. This will probably only work on unix-like systems.
In any case, compressed data in one of the supported compression formats (gzip, Unix compress or bzip2) will be decompressed transparently.
`istreamN = true|false` (Boolean)
If set true, the input table specified by the `inN` parameter will be read as a stream. It is necessary to give the `ifmtN` parameter in this case. Depending on the required operations and processing mode, this may cause the read to fail (sometimes it is necessary to read the table more than once). It is not normally necessary to set this flag; in most cases the data will be streamed automatically if that is the best thing to do. However it can sometimes result in less resource usage when processing large files in certain formats (such as VOTable). This parameter is ignored for scheme-specified tables.
[Default: `false`]
`otherpointsN = <array-expr>` (String)
Array of coordinates giving the points of the vertices defining the polygon to be drawn. These coordinates are given as an interleaved array by this parameter, e.g. (x1,y1, x2,y2, y3,y3). The basic position for the row being plotted either is or is not included as the first vertex, according to the setting of the `usepos` parameter.
Some expression language functions that can be useful when specifying this parameter are `array()` and `parseDoubles()`.
The value is an array-valued algebraic expression based on column names as described in Section 10. Some of the functions in the Arrays class may be useful here.
`polymodeN = outline|border|fill|cross|star` (PolygonShape)
Polygon drawing mode. Different options are available, including drawing an outline round the edge and filling the interior with colour.
The available options are:
• `outline`: draws a line round the outside of the polygon
• `border`: draws a line butting up to the outside of the polygon; may look better for adjacent shapes, but more expensive to draw
• `fill`: fills the interior of the polygon
• `cross`: draws a line round the outside of the polygon and lines between all the vertices
• `star`: draws a line round the outside of the polygon and lines from the nominal center to each vertex
[Default: `outline`]
`shadingN = auto|flat|translucent|transparent|density|aux|weighted <shade-paramsN>` (ShapeMode)
Determines how plotted objects in layer N are coloured. This may be influenced by how many objects are plotted over each other as well as the values of other parameters. Available options (Section 8.4) are:
• `auto`
• `flat`
• `translucent`
• `transparent`
• `density`
• `aux`
• `weighted`
Each of these options comes with its own set of parameters to specify the details of how colouring is done.
[Default: `auto`]
`thickN = <int-value>` (Integer)
Controls the line thickness used when drawing polygons. Zero, the default value, means a 1-pixel-wide line is used. Larger values make drawn lines thicker, but note changing this value will not affect all shapes, for instance filled polygons contain no line drawings.
[Default: `0`]
`useposN = true|false` (Boolean)
Determines whether the basic positional coordinates are included as one of the polygon vertices or not. The polygon has N+1 vertices if true, or N vertices if false, where N is the number of vertices supplied by the array coordinate. If false, the basic position is ignored for the purposes of drawing the polygon.
[Default: `true`]
Next Previous Up Contents
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STILTS - Starlink Tables Infrastructure Library Tool Set | 1,699 | 7,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-21 | latest | en | 0.612959 |
https://discourse.mc-stan.org/t/multivariate-priors-for-dichotomous-variables/12098 | 1,653,653,963,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00307.warc.gz | 266,981,800 | 8,414 | # Multivariate priors for dichotomous variables
I am not really into bayesian statistics, but I face a problem of which I think you guys are probably into.
I have a set of dichotomous Variables A,B,C,… and I have assumptions about their probabilities P(A), P(B),… and also about the probabilities of each pair P(A&B), P(A&C), P(B&C), …
Or in other words, I have their variance-covariance matrix.
But I have no further assumptions about P(A&B&C) for example.
From this, I would like to construct the joint distribution of my variables with the maximum entropy.
How would you do this? Or how would you construct a reasonable joint distribution with this information? Is there an easy way to do this in stan?
I hope this question is not too obvious, but I did not find the answer in BDA 3, so if you would like to recommend literature, I would also be happy.
Hi @Maximilian_Ernst! Welcome to the Stan forums!
So, just to make sure I get this right: You have data A, B, C, which are binary and correlated. I think the lingo is usually, that when you assign a prior to parameters you speak of priors and when you assign a distribution to data you usually call it likelihood. It’s not hugely important, since these two are entangled anyways and the distinction between variable and data can get blurry, too.
If you have something like a variance-covariance matrix, you can fit a multivariate Probit regression (scroll down a bit). Specifying priors can be a bit tricky, but usually in this model you treat the correlation matrix (variances are set to 1 to identify the model) as a parameter and estimate it.
Is that what you are looking for? If not feel free to follow up with questions, or maybe some details on the problem or what you are trying to achieve.
Cheers! :)
Max
Hi @Max_Mantei !
Thanks! Yes, I would like to assign a distribution to data.
I will take a look at the link later (since it takes me a while to understand it), but I maybe can already try to give some details.
My problem is as follows: I have a bunch of diseases A,B,C,… and from the scientific literature, I know their probabilities (12% of people get disease A in their lifetime) and their pairwise relations (e.g. If you have disease A, it is more likely to also have disease B).
I have no real dataset at all! Only those published probabilities.
What I would like to estimate is how likely it is to have disease A and B and C and not D and not F … all together. Unfortunately, I can not directly compute them from the information I have - but the information I have restricts the possible values of those probabilities.
Therefore, I would like to estimate a joint distribution of all my diseases that is in accordance with the information I have. As I said, there are many distributions that satisfy those conditions, and with linear programming, I would be able to obtain the range of possible values.
However, I think the distribution that satisfies those conditions and has the maximum entropy would be the best one for further use in my model.
Ah ok!
What does the variance-covariance matrix that you have look like?
Or do you “only” have conditional probabilities, such as Pr(B | A) = 0.2
Yes, I have “only” conditional probabilities, but as P(A) determines the variance of A, and P(A|B) and P(B) determine their pairwise joint distribution, I think I also have their variance-covariance matrix.
I need the model to work for arbitrary diseases, but if it would help I could post a realistic example.
I see. I was just thinking about doing something like this (edit: I hope you know some R, sorry for assuming that):
# a little helper
freq_pos <- function(x) {
sum(x > 0) / length(x)
}
# assume some marginal probs
p_A <- 0.3
p_B <- 0.6
p <- c(p_A, p_B)
# convert to z-score
mu <- qnorm(p)
# number of simulations
n_sims <- 1e6
# diagonal correlation matrix, implying Pr(B) = Pr(B|A)
S_uncorr <- diag(c(1, 1))
# draw from multivariate Normal
Y_uncorr <- MASS::mvrnorm(n = n_sims, mu = mu, Sigma = S_uncorr)
# compute marginal probs
apply(Y_uncorr, 2, freq_pos)
# identifier for the case A = 1
A_true <- Y_uncorr[, 1] > 0
# computing Pr(B|A)
apply(Y_uncorr[A_true,], 2, freq_pos)
# specify positive correlation, implying Pr(B) < Pr(B|A)
S_corr <- matrix(c(1, 0.5,
0.5, 1), nrow = 2)
Y_corr <- MASS::mvrnorm(n = n_sims, mu = mu, Sigma = S_corr)
apply(Y_corr, 2, freq_pos)
A_true <- Y_corr[, 1] > 0
apply(Y_corr[A_true,], 2, freq_pos)
I’m just not sure how to cleverly map from specific conditional probabilities to correlations… But if this is figured out, then generating data is as easy as drawing from a (multivariate) Normal distribution.
That being said… there is probably a more straightforward way to do all this…
2 Likes
Let p(A) and p(B) denote the marginal probabilities of events A and B and let p(A, B) denote the joint probability of these events.
Then the correlation coefficient is
\rho_{AB} = \frac{p(A, B) - p(A)p(B)}{\sqrt{p(A) \left[1-p(A) \right] p(B) \left[1-p(B) \right]}}
Probably there is a more straightforward way, but I think this way would be also much more complicated… so maybe your approach would be a good idea.
However, I tried it for 3 variables - but it does not give the right conditional probabilities.
Has somebody an idea why?
# 3 Variables -------------------------------------------------------------
cor_bin <- function(p1, p2, p12){
(p12-p1*p2)/sqrt(p1*(1-p1)*p2*(1-p2))
}
var_bin <- function(p){
sqrt(p*(1-p))
}
p_A <- 0.2
p_B <- 0.2
p_C <- 0.3
mu <- qnorm(c(p_A, p_B, p_C))
p_AcondB <- 0.3 # positive correlation
p_AcondC <- 0.1 # negative correlation
p_BcondC <- 0.2 # no correlation
p_AandB <- p_B*p_AcondB
p_AandC <- p_C*p_AcondC
p_BandC <- p_C*p_BcondC
cor_AB <- cor_bin(p_A, p_B, p_AandB)
cor_AC <- cor_bin(p_A, p_C, p_AandC)
cor_BC <- cor_bin(p_B, p_C, p_BandC)
S <- matrix(c(1, cor_AB, cor_AC,
cor_AB, 1, cor_BC,
cor_AC, cor_BC, 1), nrow = 3)
n_sims <- 1e6
Y_corr <- MASS::mvrnorm(n = n_sims, mu = mu, Sigma = S)
apply(Y_corr, 2, freq_pos)
A_true <- Y_corr[, 1] > 0
B_true <- Y_corr[, 2] > 0
C_true <- Y_corr[, 3] > 0
apply(Y_corr[C_true,], 2, freq_pos) #P(A|C) should be 0.1, but is
#estimated as 0.15
Hey! Well, the correlation of the two binary variables A, B that Max pointed out is something different from the correlation of the latent variables in the multivariate Probit approach.
Right now I don’t really have time to think that much about it, but if there’s a way to express the correlation induced by the marginal and conditional probabilities in terms of Kendall’s \tau, then I can point you to a paper, where they discuss a conversion for the latent variable approach. (I’m on my phone right now, so, I will link that paper probably tomorrow.)
1 Like
I think I found the correlation we were looking for:
At first glance, this seems to give a usefull distribution. I will now start to test it, but I have a good feeling about it :)
@Max_Mantei thank you really, really much for the idea!
# polychoric correlation solution -----------------------------------------
library(tidyverse)
library(polycor)
p_A <- 0.2
p_B <- 0.2
p_C <- 0.3
mu <- qnorm(c(p_A, p_B, p_C))
p_AcondB <- 0.3 # positive correlation
p_AcondC <- 0.1 # negative correlation
p_BcondC <- 0.2 # no correlation
p_AandB <- p_B*p_AcondB
p_AandC <- p_C*p_AcondC
p_BandC <- p_C*p_BcondC
## obtain latent correlations
corl_AB <- polychor(matrix(c(
1 - p_A - p_B + p_AandB, p_B - p_AandB,
p_A - p_AandB, p_AandB
), nrow = 2))
corl_AC <- polychor(matrix(c(
1 - p_A - p_C + p_AandC, p_C - p_AandC,
p_A - p_AandC, p_AandC
), nrow = 2))
corl_BC <- polychor(matrix(c(
1 - p_B - p_C + p_BandC, p_C - p_BandC,
p_B - p_BandC, p_BandC
), nrow = 2))
S_lat <- matrix(c(1, corl_AB, corl_AC,
corl_AB, 1, corl_BC,
corl_AC, corl_BC, 1), nrow = 3)
n_sims <- 1e6
Y_corr <- MASS::mvrnorm(n = n_sims, mu = mu, Sigma = S_lat)
daty <- Y_corr %>% as.data.frame()
names(daty) <- c("A", "B", "C")
daty %<>% mutate_all(~ifelse(. > 0, 1, 0))
get_p <- function(df){
probs <-
data.frame(name = c("p(A)",
"p(B)",
"p(C)",
"p(A,B)",
"p(A,C)",
"p(B,C)",
"p(A,B,C)"),
value = c(
sum(df$A)/nrow(df), sum(df$B)/nrow(df),
sum(df$C)/nrow(df), sum(df$A&df$B)/nrow(df), sum(df$A&df$C)/nrow(df), sum(df$C&df$B)/nrow(df), sum(df$A&df$B&df$C)/nrow(df)
)
)
return(probs)
}
get_p(daty)
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# Lecture24 - ECO220Y Lecture 24 Introduction to Hypothesis...
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ECO220Y Lecture 24 Introduction to Hypothesis Testing Migiwa Tanaka Reading:11.1 1
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Outline Introduction to Hypothesis Testing Testing population mean when the population variance is known Rejection Region Method p-value Method One-tail vs. two-tail Test Calculating the probability of a type II error 2
Hypothesis Testing Step1: Identify hypotheses about a population parameter Step2: Collect a sample Step3: Conduct hypothesis testing: Conclude whether the hypotheses set up in step 1 is true or not, using evidence found in the sample. A. Calculate a test statistic B. Compare A. with the decision criteria 3
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Analogy with Criminal Trial (1) Hypothesis Testing C i i l T i l Criminal Trial Unknown: Population P Unknown: Defendant’s Guilt Parameter Construct hypotheses about population parameter Construct hypotheses about defendant’s guilt population parameter. Collect a sample from the population Collect evidence relevant to the crime. Calculate test statistics. Based on the evidence in a sample, test hypothesis. Based on the evidence, test hypothesis 4
Hypotheses All hypotheses are constructed in a pair. 1 N ll H th i (H ) I iti l ti b t th 1. Null Hypothesis (H 0 ): Initial presumption about the population, which is not based on the data. Status quo. 2. Alternative Hypothesis (H 1 ): The statement that can be proven by the data. New regime. It is also called research hypothesis hypothesis.
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Lecture24 - ECO220Y Lecture 24 Introduction to Hypothesis...
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Ask a homework question - tutors are online | 499 | 2,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.800196 |
https://www.physicsforums.com/threads/quotient-topology.392187/ | 1,653,232,057,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00254.warc.gz | 1,073,254,664 | 15,289 | # Quotient topology
## Homework Statement
verify that R, the reals, quotiented by the equivalence relation x~x+1 is S^1
## The Attempt at a Solution
All i can think of is to draw a unit square and identify sides like the torus, but this would be using IxI, a subset of R^2, and gives a cylinder at best... Obviously I am missing the point, so any help would be great.
What I have done above is actually on the right path, if the horizontals of the square are identified we get a cylinder, if the vertical height is shrunk to nothing, so we are working in just R, then we get a circle. Then the result follows. I think the logic I present is correct but is there a more mathematically concise way to present this?
Fredrik
Staff Emeritus
Gold Member
Sounds like you should try to write down a function from from R/~ into S1 (or from S1 into R/~) and then try to prove that it's a homeomorphism. You mentioned the "quotient topology" in the title. I take it that means that the open sets on R/~ are defined to be the preimages $\pi^{-1}(U)$ of open sets U in R, where $\pi:\mathbb R\rightarrow\mathbb R/\sim$ is the function that takes a number to its equivalence class: $\pi(x)=[x]$.
Ok so the quotient R/~ = [0,1) for the relation x~x+1?
Then defining the map f:[0,1)---S^1 via
f(x)=exp(2*Pi*x*i)
for x in [0,1).
Yes I am working on the definition that open sets in the preimage are open defines continuity and so give definition of homeomorphism.
I take by showing that S^1 is homeomorphic to unit interval this shows that it has the same topology as the real line?
Fredrik
Staff Emeritus
What I'm suggesting is that you define a topology on R/~ by saying that the function $\pi$ is continuous. Then you you show that your f is continuous with respect to that that topology on R/~. | 466 | 1,794 | {"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} | 3.71875 | 4 | CC-MAIN-2022-21 | latest | en | 0.943629 |
http://www.physicsforums.com/showthread.php?t=266793 | 1,408,531,911,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500804220.17/warc/CC-MAIN-20140820021324-00353-ip-10-180-136-8.ec2.internal.warc.gz | 543,443,894 | 8,781 | Emf and internal resistance of a battery
by StephenDoty
Tags: battery, internal, resistance
P: 267 When switch S in the figure is open, the voltmeter V of the battery reads 3.07 V. When the switch is closed, the voltmeter reading drops to 2.95 V, and the ammeter A reads 1.70 A. Assume that the two meters are ideal, so they do not affect the circuit. Find the value of the emf. (see picture posted below) the equation would be E-Ir-IR or E= I(r+R) E=Ir however I am totally confused on how to find the internal resistance value,r, when the switch is open and the value of R when the switch is close. Any help would be appreciated. Thanks. I apprecaite the help. Stephen Attached Thumbnails
Mentor
P: 41,429
Quote by StephenDoty Find the value of the emf.
I'm a bit puzzled by this problem, since the battery EMF is given. Are you being asked to find R and r?
the equation would be E-Ir-IR or E= I(r+R)
That's one equation. E and I are given.
Write another equation describing the given voltage reading across the battery terminals when the switch is closed.
P: 267 I have two equations there. Are they right? What would the emf be? the 3.07V or 2.95V? And how would you find the r and R? Thanks Stephen
Mentor
P: 41,429
Emf and internal resistance of a battery
Quote by StephenDoty I have two equations there. Are they right?
You must have snuck the second one in after I had posted. The one I saw is certainly correct. The other is just Ohm's law.
What would the emf be? the 3.07V or 2.95V?
The terminal voltage across the battery (which is what the voltmeter measures) is the EMF minus the voltage drop across the internal resistance (which is given by Ir). When the switch is open, what's the current through the circuit?
And how would you find the r and R?
By combining two equations. The one I quoted from your first post and another that you should write for the terminal voltage when current flows.
P: 267 So use 3.07 for E in E=Ir and 1.7 for I thus r=3.07/1.7=1.81? Then to find the emf you do the voltage 3.07 + (1.7)(1.81)?Then to find R would you use 2.95=I(r+R)=(1.7)(1.81+R)???? Are these right?
Mentor
P: 41,429
Quote by StephenDoty So use 3.07 for E in E=Ir and 1.7 for I thus r=3.07/1.7=1.81?
No. Reread what I said about EMF, terminal voltage, and internal resistance. Note that 3.07 is measured when the switch is open.
P: 267 The terminal voltage across the battery (which is what the voltmeter measures) is the EMF minus the voltage drop across the internal resistance (which is given by Ir). Well doesn't this mean that the emf = the voltage measured by voltmeter +the voltage drop, Ir? and to find r you use E=Ir or 3.07=(1.7)*r? If this is not right I need a little bit more info? The concept is not sinking in.
Mentor
P: 41,429
Quote by StephenDoty Well doesn't this mean that the emf = the voltage measured by voltmeter +the voltage drop, Ir? and to find r you use E=Ir or 3.07=(1.7)*r?
No. Note that the terminal voltage (in this setup) is the voltage measured by the voltmeter. So that means: Measured voltage = EMF - Ir.
And the current (I) depends on whether the switch is open or closed.
P: 267 From your equation EMF = Measured voltage +Ir right? And since the switch is not closed what would the current be when the voltmeter measures 3.07V? And then how would you find r from the equation E=Ir? Would you use the 3.07V for E and the current found above for I? If this is not right, would you please guide me to the right direction. Thanks.
Mentor
P: 41,429
Quote by StephenDoty From your equation EMF = Measured voltage +Ir right?
Yep.
And since the switch is not closed what would the current be when the voltmeter measures 3.07V?
You tell me. When the switch is open you have an open circuit.
P: 267 I have no idea my professor only went over closed circuits. Since the circuit is not closed, I do not see how a current exists. But since the voltmeter is giving a measurement there is obviously a current. Need a starting step. V=Ir or I = V/r, but since we do not know r, I do not see how we can solve it.
Mentor
P: 41,429
Quote by StephenDoty Since the circuit is not closed, I do not see how a current exists.
That's just the point: With an open circuit, there is no current. I = 0.
But since the voltmeter is giving a measurement there is obviously a current.
Not true. The voltmeter measures voltage, not current. You don't need a current to have a voltage.
Related Discussions Introductory Physics Homework 1 General Physics 2 Advanced Physics Homework 4 Introductory Physics Homework 14 Introductory Physics Homework 2 | 1,256 | 4,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2014-35 | latest | en | 0.948475 |
http://stackoverflow.com/questions/6444716/how-to-write-nested-loop-problem-using-parallel-strategies-in-haskell | 1,440,894,394,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064590.32/warc/CC-MAIN-20150827025424-00088-ip-10-171-96-226.ec2.internal.warc.gz | 219,363,701 | 20,457 | # How to write nested loop problem using parallel strategies in Haskell
I'm playing with parallel strategies and wondering if I'm doing the following the right way. Java code:
`````` double x = 0.0;
double[] arr = new double[2000];
for (int i = 0; i < arr.length; i++)
arr[i] = i;
for (int i = 0; i < arr.length; i++) {
x += arr[i] * 5;
for (int j = i + 1; j < arr.length; j++)
x -= arr[j] * 3;
}
``````
Haskell program which uses parallel strategies to compute the result:
`````` n = 2000
ns = [0..n-1]
segments = chunk 100 ns
chunk n [] = []
chunk n xs = ys : chunk n zs
where (ys,zs) = splitAt n xs
parCompute = foldl' (+) 0 (map (\ts -> compute ts) segments `using` parList rdeepseq)
compute ts = foldl' addfunc 0 ts
where
addfunc acc i = (acc + x) - (foldl' minusfunc 0 [(i+1)..(n-1)])
where
x = (ns!!i) * 5
minusfunc acc' j = (acc' + x')
where
x' = (ns!!j) * 3
main = print parCompute
``````
My questions are:
• is it right to use foldl' here? I thought since all computations need to be done to get the result, I should force evaluate.
• is there a better way to use segments? what common patterns are present in this problem that i can exploit?
• what other strategies could be applied to this problem? Also, any possibility of parallelising just using `par` and `seq` primitives.
-
style comment: nested wheres are ugly. – rampion Jun 22 '11 at 18:44
I don't think this bits of code are equivalent. The code "x -= arr[j] * 3" gets applied to all items after position "i" in the list but in the haskell code the equivalent only gets applied to the local chunk, not all values past position "i". Or maybe I read this wrong. – Tim Perry Jun 22 '11 at 18:58
@Tim: It does give the same answer. I've checked the Haskell's result against Java. – vis Jun 23 '11 at 7:59
In this case, I'm glad I was wrong. – Tim Perry Jun 23 '11 at 18:11
Ok, let's use REPA (REgular Parallel Arrays) this time and compare it with the parListChunk method (since the java example uses an array not a list) :
``````module Main where
import Control.Parallel.Strategies
import Data.List (tails)
import System.Environment (getArgs)
import qualified Data.Array.Repa as R
import qualified Data.Array.Repa.Shape as RS
chunksize = 100
parListCompute :: [Int] -> [Int]
parListCompute ts = (computes `using` parListChunk chunksize rseq)
where
computes = zipWith f ts (tail (tails ts))
f t tls = 5 * t - 3 * sum tls
parRepaCompute :: R.Array R.DIM1 Int -> R.Array R.DIM1 Int
parRepaCompute arr = R.force \$ computes
where
computes = R.map f arr
f x = 5*x - 3*(sumRest (x+1) 0)
sumRest x acc | x > (RS.size . R.extent \$ arr) = acc
| otherwise = sumRest (x+1) (acc+x)
main = do
(s:_) <- getArgs
case s of
"1" -> putStrLn . show .sum \$ parListCompute l
"2" -> putStrLn . show . R.sum \$ parRepaCompute r
where l = [1..70000]
r = R.fromList (R.Z R.:. (length l)) l
``````
And here is the result:
``````~/haskell\$ ghc --make nestloop.hs -O2 -rtsopts -threaded
[1 of 1] Compiling Main ( nestloop.hs, nestloop.o )
haskell\$ time ./nestloop 1 +RTS -N4
-342987749755000
real 0m5.115s
user 0m19.870s
sys 0m0.170s
~/haskell\$ time ./nestloop 2 +RTS -N4
[-342987749755000]
real 0m1.658s
user 0m3.670s
sys 0m0.070s
``````
I hope you'll like this comparison.
-
Thanks for the comparison. It is definitely useful. I will wait to see if we have other alternatives before choosing which answer is best in my case. – vis Jun 24 '11 at 9:49
Here is how I would translate your Java program into a parallel Haskell program:
``````parCompute ts = sum (computes `using` parListChunk 100 rseq)
where
computes = zipWith f ts (tail (tails ts))
f t tls = 5 * t - 3 * sum tls
``````
First off - yes, introducing strictness is a good idea here. On the other hand, GHC is smart enough to spot this as well! In fact, whether you use `foldl`, `foldl'` or simply `sum`, the generated code is exactly the same.
For evaluating the list in segments, you can simply use the chunking strategy as indicated above. The amount of work each chunk represents might vary wildly, however, so you could try to even it out by making bigger chunks for the end of the list. Apart from that, I don't think there is much room for improvement here.
-
Thanks for this. Your solution looks good and is useful but I'm not quite sure about the generated code being the same for foldl, foldl'. foldl' forces the accumulated result to be evaluated completely after each application of function. In some cases, it gives better speedup than using foldl but unsure in what way the generated code could be the same. – vis Jun 24 '11 at 9:46 | 1,397 | 4,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-35 | longest | en | 0.862095 |
https://www.proofwiki.org/wiki/Set_of_Integer_Multiples_of_GCD | 1,685,901,348,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650201.19/warc/CC-MAIN-20230604161111-20230604191111-00053.warc.gz | 1,022,371,955 | 10,818 | # Set of Integer Multiples of GCD
Jump to navigation Jump to search
## Theorem
Let $m, n \in \Z$.
Let $m \Z$ denote the set of integer multiples of $m$
Then:
$m \Z \cup n \Z \subseteq \gcd \set {m, n} \Z$
where $\gcd$ denotes greatest common divisor.
## Proof
Let $x \in m \Z \cup n \Z$.
Then either:
$m \divides x$
or:
$n \divides x$
In both cases:
$\gcd \set {m, n} \divides x$
and so:
$x \in \gcd \set {m, n} \Z$
Hence by definition of subset:
$m \Z \cup n \Z \subseteq \gcd \set {m, n} \Z$
$\blacksquare$ | 187 | 528 | {"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} | 3.3125 | 3 | CC-MAIN-2023-23 | latest | en | 0.469489 |
https://coinswitch.co/calculators/lumpsum-calculator | 1,726,187,934,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651506.7/warc/CC-MAIN-20240913002450-20240913032450-00840.warc.gz | 155,365,058 | 14,266 | # Lumpsum Calculator
Take control of your investment journey with our Lump Sum Calculator, offering clarity on potential returns for one-time contributions.
Total Investment
500
1Cr
Expected Return (P.A)
%
1%
20%
Period
Yr
1Y
30Y
Invested
₹0
Estimated Returns
₹0
Growth
%
5 Years Value
₹0
0
Invested
Returns
Lumpsum investing is a popular way of investing in mutual funds. Here, the amount involved is typically large, different from the systematic, regular investing offered by the SIP mode. Many mutual fund schemes allow investors to use this one-time investment option. However, many first-time investors prefer to know the estimated returns before they dip their toes into the risky, yet lucrative world of mutual funds. Not to worry, we have your back as you can simply use our lumpsum calculator to estimate the returns from your mutual fund investment.
## Understanding the lumpsum calculator
A lumpsum mutual fund calculator is designed to make complex return calculations easy to help you estimate the expected return from your lumpsum mutual fund investment. CoinSwitch’s lumpsum calculator is intuitive and easy to use. Just plug in the numbers to get an appropriate estimate of returns at the click of a button.
## Using the lumpsum calculator
The CoinSwitch lumpsum calculator is simple and easy to use. The plug-in fields are self-explanatory. Follow the order mentioned below to key in your inputs to estimate the returns on your lumpsum investments:
• The amount of one-time or lumpsum investment
• The expected rate of return
• The desired holding period of investment
Double-check your inputs to ensure their accuracy. If the inputs you enter are accurate, you will get the estimated redemption value of your investment in just a few seconds.
## Calculating MF returns the traditional way
The estimation of the redemption value of your lumpsum investment follows the financial concept of the future value of money. This means that the formula will estimate the value of your lumpsum investment at the end of your investment period.
The future value of money is a core finance concept that requires a good understanding of finance to calculate it using Excel or a normal calculator. That is why we have designed the lumpsum calculator for you. Still, we have also included the lumpsum investment estimation formula below for the mathematically inclined.
A = P (1 + r) ^ n
Where:
A = Value at the end of the investment period
P = Lumpsum investment amount
R = Rate of interest
N = Investment period in years
Let’s put the formula to use for better understanding. If you plan to invest ₹20,00,000 in a mutual fund scheme for 10 years at an interest rate of 14% expected rate of return per annum, your redemption value at the end of 10 years will be:
A = 20,00,000(1+14)^10
That is, the redemption value would be ₹74,14,443. Of the redemption value, ₹20,00,000 will be your investment and ₹54,14,443 will be the returns on your investment.
Money and investment are important topics that need serious consideration and accuracy even if you want only an estimate. Investing a large amount in mutual funds requires comparing multiple mutual fund schemes. However, it is easier said than done as calculating the estimated redemption value of various products can get complex, tedious, and time-consuming.
This is where a lumpsum calculator can be of help to you. Some of the benefits of using a lumpsum calculator are as follows:
• Minimizes errors
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Lumpsum investment schemes have different underlying assets and different rates of return. A calculator will help you estimate the returns. You can select the scheme that helps you achieve your financial goals based on the estimated returns of each scheme.
• Easily available online
The CoinSwitch Lumpsum Calculator is available online. This means you can access the calculator from any corner of the world with internet connectivity.
• Saves time
Calculating the estimated returns for your lumpsum investment can be time-consuming. The calculator above will save you time while giving you accurate results.
## Conclusion
The whole point of investing is to grow your savings. Selecting a mutual fund scheme to invest a lumpsum amount is no child’s play. The lumpsum calculator can come in handy here as it provides you with the estimated redemption value of your investment across various mutual fund schemes. Use the calculator above before making an investment decision.
## FAQs
1.How different are lumpsum investments and SIP investments?
A lumpsum investment is a relatively large, one-time investment in a mutual fund scheme. An SIP, on the other hand, is one where you invest a fixed amount of money in a mutual fund scheme at regular intervals.
2.Which is more beneficial in the long run - lumpsum or SIP?
SIP will suit you if you are an individual who receives a fixed monthly income, like a salary. On the other hand, if you get a windfall like a large inheritance or huge profits from a business, you could consider a lumpsum investment. Yet, both have their advantages and disadvantages. Generally speaking, SIPs need time to build a large corpus as the investment amount is relatively small. However, a lumpsum investment will help you grow your corpus faster.
3.How can I make mutual fund investments?
There are plenty of online investment platforms available in India such as CoinSwitch. You can also invest in mutual fund schemes directly through the fund websites. Alternatively, you can also make investments through mutual fund distributors.
4.Can I trust mutual fund calculators completely?
While mutual fund calculators can provide you with a ballpark estimate of the returns you can expect for a specific scheme, the numbers will not be as accurate as a fixed deposit calculator.
5.Can you suggest a financial product for making a lumpsum investment
You can put your funds in a mutual fund scheme of your choice. Alternatively, you can also open a bank FD and park your lumpsum funds there. | 1,237 | 6,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.897296 |
https://www.physicsforums.com/threads/basis-for-subspace.322458/ | 1,532,017,020,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591140.45/warc/CC-MAIN-20180719144851-20180719164851-00070.warc.gz | 955,048,197 | 15,505 | # Homework Help: Basis for subspace
1. Jun 29, 2009
### JG89
1. The problem statement, all variables and given/known data
Suppose T is a linear operator on R^4 such that T(a,b,c,d) = (a + b, b - c, a + c, a + d). Find a basis for the T-cyclic subspace of R^4 generated by z = (1, 0, 0, 0)
2. Relevant equations
3. The attempt at a solution
I found a basis, but I don't think the method I used was the most efficient way. To save you the tedious calculations, I'll post an outline of what I did.
First denote the T-cyclic subspace in question by W. Then since W is a subspace of R^4,
1<= dim(W) <= 4. Now dim(W) cannot be equal to 1, for then any single vector in W would be a basis for W (since a set consisting of a single vector is linearly independent) and (1, 0, 0, 0) does NOT span W (I've computed enough vectors in W to know this). So we have
2<= dim(W) <= 4. The set { (1, 0, 0, 0), (1, 0, 1, 1) } is linearly independent, both those vectors are in W, and so if dim(W) = 2, then that set should span W, but it doesn't.
So we have 3 <= dim(W) <=4. What I did now was take 4 vectors that span W and showed that they aren't linearly independent, and so this must mean that dim(W) = 3, and so I took those 4 vectors that I found that span W, removed one, and showed that the other 3 are linearly independent, which is my basis.
Last edited: Jun 29, 2009
2. Jun 29, 2009
### Staff: Mentor
I don't know the answer to your question, but some of the things you have written are incorrect.
1. The set {0} is a subspace of R4 (and lots of other vector spaces), but the dimension of this subspace is 0.
2. The set {0} is linearly dependent, not linearly independent. The equation a0 = 0 has an infinite number of solutions.
3. Jun 29, 2009
### JG89
1. The dimension of W cannot be 0, because then W = {0} which I know isn't true.
2. I should have said that a set consisting of a single NON-ZERO vector is linearly independent. Either way, W still cannot be one dimensional since it isn't spanned by (1, 0, 0, 0)
Surely there has to be a way to find a basis without going through all the stuff I did?
4. Jun 29, 2009
### Dick
Your description of what you are doing is a little unclear, but if you mean that {z,T(z)} is linearly independent that's ok. Then you looked at {z,T(z),T(T(z))}, right? Just keep adding T^n(z) for increasing values of n until you get a linearly dependent set. Then you know you've added one too many.
5. Jun 29, 2009
### JG89
Beautiful - that's much simpler than what I was trying to do. Thanks.
6. Jun 29, 2009
### JG89
Just one question...
You said to keep increasing the value of n for T^n(z) until I get a linearly dependent set. Say the set {z, T(z), T^2(z)} is linearly independent and also suppose that when I choose n = 3, then the set {z, T(z), T^2(z), T^3(z)} is linearly dependent. That doesn't necessarily mean that any set consisting of 4 vectors from the T-cyclic subspace is going to be linearly dependent, does it? So then is it possible that there is another value of n, say n = 56, so that the set {z, T(z), T^2(z), T^56(z)} is linearly independent?
7. Jun 29, 2009
### Dick
Good question. If {z,T(z),T^2(z),T^3(z)} is linearly dependent then you can write T^3(z)=a*z+b*T(z)+c*T^2(z). What does that tell you about T^4(z) etc?
8. Jun 29, 2009
### JG89
If T^3(z) = a*z+b*T(z)+c*T^2(z) then T^4(z) = a*T(z) + b*T^2(z) + c*T^3(z)
= a*T(z) + b*T^2(z) + c*(a*z+b*T(z)+c*T^2(z))
= (c*a)*z + a*T(z) + b*T^2(z) + c*a*T(z) + c^2*T^2(z)
with at least one of a, b, or c being non-zero (For if a = b = c =0 then T^4(z) would be the zero vector and the set {z, T^(z), T^2(z), T^4(z)} would be linearly dependent).
If a is non-zero, then T^4(z) = a*T(z) + ... the rest with a being non-zero
If b is non-zero, then T^4(z) = b*T^2(z) + ... the rest with b being non-zero
If c is non-zero, then T^4(z) = c^2*T^2(z) + ... the rest with c^2 being non-zero
And so T^4(z) can be written as a linear combination of the vectors z, T(z), T^2(z) and so the set {z, T(z), T^2(z), T^4(z)} is linearly dependent, and by induction the set {z, T(z), T^2(z), T^n(z)} is linearly dependent for all positive integers n >= 3
Last edited: Jun 29, 2009
9. Jun 29, 2009
### Dick
That's a bit overcomplicated. But sure, if T^3(z) can be expressed in terms of T^2(z), T(z) and z, then so can T^4(z) and so on. Just apply T and replace powers greater than 2 with linear combinations of T^2(z), T(z) and z. | 1,447 | 4,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-30 | latest | en | 0.969036 |
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Question
# The plates S and T of an uncharged parallel plate capacitor are connected across a battery. The battery is then disconnected and the charged plates are now connected in a system as shown in the figure. The system shown is in equilibrium. All the strings are insulating and massless. The magnitude of charge on one of the capacitor plates is [Area of plates = A]:√2mgA∈0√mgA∈0√2mgA∈0k√4mgA∈0k
A
2mgA0
B
4mgA0k
C
mgA0
D
2mgA0k
Solution
Verified by Toppr
#### Force on metal plate S due to electrostatic force by plate T is Q22AϵoForce exerted on plate S by spring is mgIn equilibrium Q22Aϵo=mgorQ=√2mgAϵo
4
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Q1
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is: [Area of plates = A]
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Q2
The plates S and T of an uncharged parallel plate capacitor are connected across a battery. The battery is then disconnected and the charged plates are now connected in a system as shown in the figure. The system shown is in equilibrium. All the strings are insulating and massless. The magnitude of charge on one of the capacitor plates is [Area of plates = A]:
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Q3
The plates S and T of an uncharged parallel plates capacitor are connected across a battery. The battery is then disconnected and the charged plates are now connected in a system as shown in the figure. The system shown is in equilibrium. All the strings are insulating and massless. The magnitude of charge on one of the capacitor plates is pC.
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Q4
The plates S and T of an uncharged parallel plate capacitor are connected across a battery. The battery is then disconnected and the charged plates are now coonnected in a system shown is in equilibrum. All the strings are insulating and massless. The magnitude of charge on one of the capacitor plates is
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Q5
The plates S and T of an uncharged parallel plates capacitor are connected across a battery. The battery is then disconnected and the charged plates are connected as shown in the figure. The sytem shown is in equilibrium. All the strings are massless. The magnitude of charge on one of the capacitor plates is [Area of plates =A]
View Solution | 607 | 2,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.942676 |
https://metanumbers.com/113 | 1,701,423,821,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00752.warc.gz | 443,090,420 | 7,338 | # 113 (number)
113 is an odd three-digits prime number following 112 and preceding 114. In scientific notation, it is written as 1.13 × 102. The sum of its digits is 5. It has a total of one prime factor and 2 positive divisors. There are 112 positive integers (up to 113) that are relatively prime to 113.
## Basic properties
• Is Prime? yes
• Number parity odd
• Number length 3
• Sum of Digits 5
• Digital Root 5
## Name
Name one hundred thirteen
## Notation
Scientific notation 1.13 × 102 113 × 100
## Prime Factorization of 113
Prime Factorization 113
Prime number
Distinct Factors Total Factors Radical ω 1 Total number of distinct prime factors Ω 1 Total number of prime factors rad 113 Product of the distinct prime numbers λ -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ 4.72739 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 113 is 113. Since it has only one prime factor, 113 is a prime number.
## Divisors of 113
1, 113
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ 2 Total number of the positive divisors of n σ 114 Sum of all the positive divisors of n s 1 Sum of the proper positive divisors of n A 57 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 10.6301 Returns the nth root of the product of n divisors H 1.98246 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 113 can be divided by 2 positive divisors (out of which none is even, and 2 are odd). The sum of these divisors (counting 113) is 114, the average is 57.
## Other Arithmetic Functions (n = 113)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 112 Total number of positive integers not greater than n that are coprime to n λ 112 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 30 Total number of primes less than or equal to n r2 8 The number of ways n can be represented as the sum of 2 squares
There are 112 positive integers (less than 113) that are coprime with 113. And there are approximately 30 prime numbers less than or equal to 113.
## Divisibility of 113
m n mod m
2 1
3 2
4 1
5 3
6 5
7 1
8 1
9 5
113 is not divisible by any number less than or equal to 9.
## Classification of 113
• Arithmetic
• Prime
• Deficient
• Polite
### By Shape (2D, centered)
• Centered Square
• Prime Power
• Square Free
## Base conversion 113
Base System Value
2 Binary 1110001
3 Ternary 11012
4 Quaternary 1301
5 Quinary 423
6 Senary 305
8 Octal 161
10 Decimal 113
12 Duodecimal 95
20 Vigesimal 5d
36 Base36 35
## Basic calculations (n = 113)
### Multiplication
n×y
n×2 226 339 452 565
### Division
n÷y
n÷2 56.5 37.666 28.25 22.6
### Exponentiation
ny
n2 12769 1442897 163047361 18424351793
### Nth Root
y√n
2√n 10.6301 4.83459 3.26039 2.57404
## 113 as geometric shapes
### Circle
Diameter 226 710 40115
### Sphere
Volume 6.04399e+06 160460 710
### Square
Length = n
Perimeter 452 12769 159.806
### Cube
Length = n
Surface area 76614 1.4429e+06 195.722
### Equilateral Triangle
Length = n
Perimeter 339 5529.14 97.8609
### Triangular Pyramid
Length = n
Surface area 22116.6 170047 92.2641
## Cryptographic Hash Functions
md5 73278a4a86960eeb576a8fd4c9ec6997 e993215bfdaa515f6ea00fafc1918f549119f993 6c658ee83fb7e812482494f3e416a876f63f418a0b8a1f5e76d47ee4177035cb 2b7e9caccc6e0ce596d8d56c524c88d4f451ebd89bc07dfb8880ae7b78e22f137cb48833871b5f0e3a65c75a1188873af63eee6a0ce4bbe8af387b1dd1019ba1 aace8937b61eb63bba5f8653cac3a539c8f9ef91 | 1,279 | 3,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-50 | latest | en | 0.843183 |
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1. Registered User
Join Date
Jan 2013
Posts
53
## Unanswered: Combine number of months and days to determine new date
Hello All!
Am trying to consolidate all these formulas into one. Please see below:
First formula determines an effective date based on a 1/1/2015 hire date with a 6 month waiting period
Code:
`DATE(YEAR(A1),MONTH(A1)+B8,DAY(A1))`
=12/1/2014
From here I need to determine the last day in the month from the formula above
Code:
`EOMONTH(C8,0)`
=12/31/2014
Based on the date generated above, need to add an addtional 30 days waiting period
Code:
`EOMONTH(A11+29,0)+1`
=2/1/2015
Any suggestions greatly appreciated it!
2. Registered User
Join Date
Jan 2013
Posts
53
If anyone can suggest a vba function that would be helpful as well.
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Posted by4 years ago
Archived
## If 2 space ships accelerated away from each other at 1/2 the speed of light, what would they see behind them?
Two identical space ships are travelling in space at 0 m/s relative to each other. They both face away from each other and then instantly accelerate to 1/2 the speed of light away from each other. I know that in a Newtonian universe they would be travelling at the speed of light away from each other, but special relativity says otherwise. If each ship had a window out facing the rear, what would they see as the ship went from stationary (relative to the other ship), to 1/2 speed of light?
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level 1
Aerospace | Quantum Field Theory242 points · 4 years ago
Let's answer this question two ways. First, let's set up the question differently. Imagine a space station, with two space ships. The space ships then leave the space station, moving 0.5c with respect to the space station. The question is: what does someone on the space station see, and what does someone on the space ship see?
Well, someone on the space station has it easy. He will simply see the ships separating at 1c, as each are leaving away from him at 0.5c. This is not a problem, since he is not seeing any single massive object leaving at a speed greater than c. I understand that this part might be obvious, but it gives us a reference point.
Now, what does someone on a spaceship see? Well, they will see the other space ship moving away at 0.8c. This can be calculated using the relativistic velocity-addition formula. Where does this come from? Well, as Einstein showed, the speed of light is the same in all reference frames. There are several consequences of this- the ones being important here are time dilation and length contraction. What this means is, a person on the space ship and a person on the space station will not measure time or distances the same way. The person on the space ship will have his clock running slower, and will measure distances shorter, than the person on the space station. So, he is measuring that the second ship is closer to him, and it took a different amount of time to get there, and thus will measure his speed to be less than the speed of light.
level 2
[deleted]
5 points · 4 years ago
Why would a SHIP person and a STATION person see things differently? What is their reference point but to each other?
What I mean is, discard one ship. Now you have just one ship and one station, parked together in space. What is their reference point? Just each other. So when the ship take off at 0.5c or 0.8c or whatever, how is that any different from the STATION taking off at that speed?
I fear I'm confusing the question, or just sounding dumb. So lemme 'splain further...
The SHIP and STATION are together in the black void. They are lightyears away from any solar system, any star. So there's little to reference by. Furthermore, as I understand it, the universe is constantly expanding, and increasingly so - speeding up its rate of expansion. No? So there's really no "standing still" in space. Standing still compared to what?
So when ONE of those objects suddenly "moves" away from the other, how would a person know which of them is moving? How would physics know, and thereby cause a person to perceive time differently?
Thank you science, if you can help me understand (probably a simple, obvious answer that will make me feel dumb, I'm sure).
level 3
Aerospace | Quantum Field Theory7 points · 4 years ago
It is a far thing from a dumb question- it is a very, very good question. First off, you are right that there isn't a "right answer" and a "wrong answer." You are also right that if the station measures the ship moving away from it at 0.5c, then the ship will measure the station moving away from it at 0.5c. So, they will agree in that regard. It is only with the addition of a third object that a discrepancy arises. With a second space ship, thing are no longer symmetric, since the station sees a symmetric picture and each space ship sees an unsymmetric one. The thing is, no person will ever measure a massive object moving faster than the speed of light, but there is no problem measuring two objects separating at a speed greater than 'c.'
Now, let's say you just have one ship and one station. You are correct, if the ship simply flies away from the station, the station will say "the ship's clock is moving slower" and the ship will say "the station's clock is moving slower." Seemingly a contradiction. However, if the ship and the station never meet back up, no contradiction has taken place- because there is never a way to synchronize their clocks. But if instead, the ship turns around and comes back to the station, suddenly the situation is no longer symmetric- the ship had to accelerate in order to turn around and come back. This is known as the twin paradox and explains how this paradox can be resolved.
level 4
1 point · 4 years ago · edited 4 years ago
Follow-up question on the Twin Paradox: Station and ship situation.
The ship accelerates away, reaches 0.9 c, flies for a while, slows down to relative 0 again. The impossible objective observer would now say: "The station experienced 1 year, the ship only experienced 1 day." True?
Now the ship accelerates towards the station to 0.9 c, flies for a while, slows down just in time to 0 again. They compare clocks and say / the objective observer says: "The station experienced 2 years, the ship only experienced 2 days." True?
Could you please point out what my thinking problem is?
EDIT:
Or does the clock on the ship, when it flies away and decelerates to 0, suddenly tick a lot faster and hence catches up?
EDIT 2:
I think I get it. The acceleration process itself is only experienced by the ship, not by the station. This means that they are not really equals.
level 5
2 points · 4 years ago
Could you please point out what my thinking problem is?
There is no problem, the crew on the ship would be younger than the people on the station when they return (your numbers are off, of course, but the general idea is ok).
That is the key plot device of Planet of the Apes (the 1968 film version at least).
level 5
0 points · 4 years ago
The impossible objective observer would now say: "The station experienced 1 year, the ship only experienced 1 day." True?
Impossibility of the objective observer means precisely that we can't say "objective observer would say X".
level 6
1 point · 4 years ago
Not really. A second space station, located exactly in the middle of station 1 and the space ship, all of which are now stationary in relation to each other and are hence in the same frame of reference, could send a radio message to station 1 and the ship at a point in time that was calculated in advance (the moment when the ship would reach speed 0). "Hey. Give me your current time." Once the message had arrived at the two points, and they would have sent their answer, the station could say what the impossible objective observer said.
level 7
0 points · 4 years ago
No, they really can't. They can say what the quite possible observer who is at rest with respect to the station and the ship could say, but so can the observers on the station and the ship! If the ship and the station (after they are at rest wrt each other) send the messages to each other asking for the other's current time, and take into account that the messages took equal time to travel back and forth, they'll agree that
"The station experienced 1 year, the ship only experienced 1 day."
but that doesn't make it objective.
level 8
-1 points · 4 years ago
Oh, you were making a philosophical (or whatever) statement earlier! I thought this was about science. My bad.
level 9
1 point · 4 years ago
No, I wasn't. Other observers, who are equally "objective" to that space station in the middle, would say "the station experienced 30 days, and the ship experienced 1 hour", or "the station experienced 1 minute and the ship experienced 1 second".
level 3
3 points · 4 years ago
You're right that when the ship is at constant velocity there's no preference between the ship's reference frame and the station's reference frame: they both see each other in a symmetrical manner.
Note that accelerations make a difference, because they don't apply symmetrically. If the ship accelerates, that is measurable in the ship's own reference frame. People on the ship would feel the acceleration, while people on the station woudln't feel anything. This is the reason for the twin paradox for example: it's not so much that the ship is going fast, but mostly that at some point it's turning around (i.e. accelerating).
level 2
Solar Astrophysics | Computer Vision44 points · 4 years ago
Well, if you want to be very careful about it, each space ship would see the exhaust from the other one's rockets, highly blue-shifted...
level 3
35 points · 4 years ago
Why would it be blue-shifted if the ships are moving away from each other? I was under the impression that it was red-shifted when things moved away from each other. For example, galaxies moving away from each other.
level 4
4 points · 4 years ago · edited 4 years ago
The exhaust would be travelling out of the back of the space ship, back towards the viewer
(Potentially the exhaust would also get in the way of the view of the red shifted rocket...)
Edit; ssjsonic1 makes a fair point below, I was trying to explain why someone would say the exhaust was blue shifted, while totally forgetting about the overall system. D'oh!
level 5
30 points · 4 years ago
The exhaust from ship 2 would be moving away from ship 2 towards ship one, but this is all within the reference frame of the ship-exhaust system which is moving at 0.8c from ship 1. The exhaust should be red-shifted.
In other words, the exhaust doesn't need to be moving towards ship 1 for ship 2 to accelerate, it just needs to be moving away from ship 1 less quickly than ship 2 is moving away from ship 1.
The point is moot however, as a ship moving at a constant velocity has no exhaust. :P
level 6
4 points · 4 years ago
The point is moot however, as a ship moving at a constant velocity has no exhaust. :P
Well, it has to overcome the friction of space doesn't it?
level 7
8 points · 4 years ago
Sure.
level 7
2 points · 4 years ago
What friction of space? You mean inertia?
level 8
7 points · 4 years ago
Space is mostly empty. It is not totally empty.
level 8
4 points · 4 years ago
There is not much 'friction' or 'drag' in space, but there is still a tiny bit due to the few atoms per cubic meter. The amount is practically unnoticeable to modern equipment, but at relativistic speeds you might need to accelerate to counteract it.
level 9
1 point · 4 years ago
Right but wouldn't drag (as you used) be a better word since it would be particles you are running into not ones you are sliding against? Or am I splitting hairs because I have a rudimentary knowledge of physics after just 1 year of college physics? My understanding is that drag is an inertial effect caused by you running into particles in front of you. You impart some of your energy into them and accelerate them in a direction roughly parallel to that in which you are moving (ignoring angle of surfaces and glancing impacts since they are essentially the same). Friction on the other hand is particles near you that you don't collide with but slow you through the electromagnetic interaction. It seems to me as though this would be a crucial difference but maybe in space when its so diffuse it doesn't matter.
Sorry if this is a chaotic mess of thoughts. I appreciate any clarity you can give.
level 10
2 points · 4 years ago
No, you're right. Drag is the more applicable term. I'm just very open-minded with terminology.
Friction, drag, and even object collisions are all the result of electromagnetic forces between atoms so, you know, pot-ay-to / tom-ah-to; it's all vegetables... and fruits. =P
level 8
[deleted]
3 points · 4 years ago
Well, space isn't empty, for one.
Not sure of the validity of this link, granted, but for a quick search it seemed to make the point fairly well.
I'm unsure how relevant the spattering of atoms and other such "stuff" in space would be compared to any life-sustaining craft in size traveling at sufficient speed (for example), but there should, in hypothesis, be some sense of friction in space. I'd hope someone far more knowledgeable on space would offer insights into just how littered with random stuff space is, even in "empty" space, since this is beyond my layman knowledge on the subject.
level 6
[deleted]
-1 points · 4 years ago
Red or Blue shift occurs when space/time is actually expanding or contracting the wavelength of light. Correct? As Red light has a longer wave length and Blue is shorter.
EDIT: Blue shift is actually purple I guess. (http://en.wikipedia.org/wiki/File:EM_Spectrum_Properties_edit.svg)
level 8
[deleted]
3 points · 4 years ago
level 5
[deleted]
1 point · 4 years ago(0 children)
level 2
13 points · 4 years ago
Ok, Follow up question. If the person on the station sees the ships separating at c, and each spaceship fired a missile, say at .5c relative to the spaceship, would the station guy see the missles seperating at 2c?
level 3
181 points · 4 years ago · edited 4 years ago
No. They would see them separate at 1.6c because 0.5c + 0.5c = 0.8c using the formula. So each missile moves away from the station at 0.8c. But each ship still sees its own missile moving away from itself at 0.5c.
EDIT: I'm being downvoted, but I assure you the differential velocity is 1.6c.
Space station looks left. There is a ship moving away at 0.5c and it fired a missile that appears to move at 0.8c.
Space station looks right and sees the same thing. From the perspective of the space station, the missiles travel apart at 1.6c. This does not violate special relativity. Nothing is moving faster than light. From the reference frame of one of the missiles, the other missile is moving away at almost c, but not more.
EDIT 2: For those confused, special relativity says that no object can move faster than c in any reference frame. Nothing in any of these reference frames is moving faster than c. There is no such restriction on derived quantities, so there's no law being broken that the two objects appear to move apart at 1.6c from the space station's reference frame. Another example is the motion of far-off stars as the earth rotates. Over the course of one night, a star overhead will appear to move across the sky. But if you use trigonometry to calculate how far it's moved, it turns out that from your perspective it has traveled many, many light years in a single night! Of course, there's no contradiction here because the star isn't actually moving that fast, but the apparent FTL motion of the star does not contradict of relativity.
level 4
Nuclear Magnetic Resonance50 points · 4 years ago
We're seeing hive-voting behaviour here - which is inevitable when a subreddit goes default...
This comment is entirely correct.
level 5
[deleted]
30 points · 4 years ago(1 child)
level 6
Nuclear Magnetic Resonance6 points · 4 years ago
Or possibly just misreading the question, which asks for the apparently recession speed to an observer in the station, rather than the relative speed to an observer on one of the missiles.
Not to mention that what you've described is basically the same as the top comment by Weed_O_Whirler, which mentions an apparent recession speed of 1c in the original scenario.
level 4
1 point · 4 years ago
Could the same be said of length contraction? For example, the observer sees the length of an object contract at relativistic speeds, but in reality it is the same length? (ladder paradox)
level 5
Particle Phenomenology | QCD | Computational Physics4 points · 4 years ago
No, length contraction is a real effect. What the observer sees in your example is just as much "in reality" as what you thought "in reality" was.
level 4
2 points · 4 years ago
Wouldn't that break the "top speed = c in all reference frames" rule? I can't say I know the answer for sure (so I don't really know if you're right or wrong), but the combined effect may not be additive in that way. o_O
level 5
Nuclear Magnetic Resonance8 points · 4 years ago
Wouldn't that break the "top speed = c in all reference frames" rule?
It does not. It is exactly the same as the original scenario. See the top comment of this thread:
Well, someone on the space station has it easy. He will simply see the ships separating at 1c, as each are leaving away from him at 0.5c. This is not a problem, since he is not seeing any single massive object leaving at a speed greater than c.
level 6
6 points · 4 years ago
Thanks for the clarification!
So it sounds like you're saying that no observer (on the station or either ship) sees anyone traveling faster than c, even if the arithmetic difference in speed between the missiles deceptively indicates a value above c.
One missile looking at the other would still see a sub-c speed (if even just barely).
This stuff always makes my brain tingle, but I appreciate you taking the time to explain it! :-)
level 7
Nuclear Magnetic Resonance2 points · 4 years ago
Correct.
level 8
0 points · 4 years ago
I just studied this concept last night for a extra-physics course, and now it's on reddit.
level 6
[deleted]
1 point · 4 years ago
Is this the same as saying, if you could somehow track individual photons, that photons moving in the exact opposite direction of each other move at a relative velocity of 2c from each other?
level 7
Nuclear Magnetic Resonance2 points · 4 years ago
Basically, yes. Shoot two photons at opposite directions, and you'll measure them to be 2 light-seconds apart after 1 second.
level 5
12 points · 4 years ago
Nope. Differential velocity doesn't count. No object is moving faster than light in the example.
level 6
9 points · 4 years ago
It may be useful here to refer to the scissor paradox, or the rotating beam of a lighthouse (or it's cosmic-sized analog, a pulsar). The point where the blades cross, or the point being illuminated can appear to be moving much faster than the speed of light, because nothing real is actually moving along that path.
level 4
1 point · 4 years ago
It may help to point out, that the fastest the space station can 'see' things moving apart, is 2c. For example, in the case of one photon going to the right, and 1 photon going to the left.
level 4
[deleted]
-7 points · 4 years ago(More than 10 children)
level 5
Nuclear Magnetic Resonance15 points · 4 years ago
The original comment is correct. You can - and do - apply the velocity addition formula in this case.
The observer at the station will see the two missiles separate at 0.8c +0.8c.
Note that what you've calculated is the relative speed of the two missiles as seen by an observer on one of the missiles, while the comment you replied to (and the question to which that comment is intended for) is about the apparent recession speed to an observer on the station.
level 6
[deleted]
0 points · 4 years ago
So hold up, you're saying that from the station, the missiles appear to be moving away from each other at over the speed of light? I thought that wasn't possible?
level 7
10 points · 4 years ago
From the perspective of the space station, one object is moving at +0.8c and the other is moving at -0.8c (assuming you define your axis appropriately). Neither object is moving faster than the speed of light, but they appear to be moving apart from each other at 1.6c. There's no rule being broken here. Nothing can actually move faster than c but derived quantities like differential velocity can exceed the speed of light.
level 8
[deleted]
4 points · 4 years ago
Huh. That's craziness. Frames of reference always trip me up.
level 8
Brazing Technical Engineering | Quality Engineering-1 points · 4 years ago · edited 4 years ago
what would something appearing to move at 1.6c look like to the stationary observer?
I realise now that they would be only moving away from each other, nothing is going >c
level 9
3 points · 4 years ago
I suggest you carefully read the comments in the tree again, nothing is moving at 1.6c.
level 9
-1 points · 4 years ago
It's not possible. Nothing can ever appear to move faster than c.
level 10
0 points · 4 years ago
Good thing then that it doesn't.
Imagine this: For some unexplained reason, someone uses their pocket calculator to add up the velocities of all objects they can observe. The sum: Greater than the speed of light. How is this a problem?
And that's almost exactly what's happening here.
level 5
1 point · 4 years ago
That is only from the reference frame of one of the missiles (assuming you set up the formula correctly). You don't use the equation for differential velocity.
level 3
Quantum Field Theory | Mathematical Physics12 points · 4 years ago
No. A missile fired at .5c relative to the spaceship in the forward direction from a ship traveling at .5c from the space station would be traveling at speed .8c relative to the space station.
The relativist formula for adding velocities v and w is (v+w)/{1+vw/c2}.
level 3
Aerospace | Quantum Field Theory2 points · 4 years ago
No, they would not.
Again, let's look at it from both person's perspectives.
From a person on the space ship, he is at rest, and he shoots a missile at 0.5c. He sees that missile moving at 0.5c in relation to him. A person on the space station however, sees the space ship moving at 0.5c, and via relativistic effects, sees the missile leaving the space ship at 0.3c, for a grand total of 0.8c.
Well, what if he shot it even faster, so that according to the space station it was moving in front of him at 0.5c? Well- he couldn't. Because in his own frame that would mean he was launching a missile at c, which is impossible.
level 2
2 points · 4 years ago
I have a question along similar lines that's been bothering me for a while. It has to do with time dilation.
We know that a person traveling in the spaceship leaving earth will experience "slower time" relative to Earth.
So let's say our spaceman wants to go from Earth to a star 1 LY away. He will travel in his space ship at .1C. That means it should take him 10 years to get to the star. He takes a picture and comes back, reversing course and coming home. His entire trip takes 20 years.
According to the twin paradox, the people on earth should have experienced a time span much greater than 20 years. I don't know the formula, but let's say it takes 100 years for him to come back..
So, if we were on Earth, and we were watching him traveling 1 year and .1C, what will we actually see? If he was actually traveling .1C relative to earth, shouldn't we expect him back in 20 years? From what I understand about the twin paradox, we on Earth will have aged significantly. Doesn't that mean, assuming it takes him 100 years, that he was actually traveling at .02C (1LY/100 years)?
*This is a serious question of mine. Maybe I'm understanding the theory wrong?
level 3
Aerospace | Quantum Field Theory3 points · 4 years ago
First, at only 0.1c (which is still incredibly fast by our standards) the effect of time dilation would be almost unnoticeable. It has a gamma factor of 0.995, so if as measured by the traveller he travelled for 1 year, a person on Earth would measure he travelled for 1.005 years, or a little less than 2 extra days.
OK, but to the heart of your question. Whenever you give a velocity, you must also give a reference frame of who is measuring the velocity. And if you give a distance, you have to say who is measuring the distance. So, if someone on Earth measures the distance to the star as being 1 ly, then someone on the spaceship would measure that distance as a little shorter (this is due to length contraction). So, according to the guy on the spaceship, he didn't travel a round trip of 2 ly's, he travelled less than that- which is why someone on Earth measures it takes more time for him to travel than someone on the space ship.
level 4
1 point · 4 years ago
Yes, I'm assuming that we did all the measurements on Earth first, and launched the ship at a speed relative to earth.
Basically, for the astronomer it would be 20 years, but for the spaceman it would feel like 20 years -2 days, and the distance would be 1ly - x?
level 2
[deleted]
1 point · 4 years ago · edited 4 years ago
Is it correct to say that a stationary object is able to observe greater then c? Seeing that entire galaxies are/will be effectively moving away from us greater than c. (http://www.universetoday.com/13808/how-can-galaxies-recede-faster-than-the-speed-of-light/)
In the same thought: If an stationary observer can observe (or kind of observe) an object moving at C (say perpendicular) that means while the object is moving at C respect to the stationary but moving less than C according to itself due to time dilation?
level 3
Particle Phenomenology | QCD | Computational Physics2 points · 4 years ago
Things are different in general relativity, because there you have the expansion of space itself to deal with, and that isn't restricted to be less than any particular rate. Space can expand in such a way that the distance between two objects increases at a rate greater than c.
One way I like to think of the "speed limit" rule in GR is that no two objects can pass each other at a speed of c or greater. That's not a perfect "translation" of the rule, but it works in a lot of cases.
level 4
[deleted]
1 point · 4 years ago
I always liked the Relativistic Rocket approach to describe the difference between apparent and effective velocity.
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
While apparent velocity never exceeds c the effective can.
level 2
1 point · 4 years ago
I have been reading a couple books this summer on General Relativity. I am starting to grasp and "fully" understand the concepts. One lingering question though: You say the person on the spaceship will have his clock run slower.. Ive never quite understood the context of this. Will they look down and see their watch run slower or is it just measuring time slower relative to the person on the space station?
level 3
Aerospace | Quantum Field Theory3 points · 4 years ago
No one will notice that their clock is running slower. And it isn't just that clocks run slower, but time itself passes slower. For instance, if the guy flew away from the space station really fast, turned around and then flew back, he will have aged less than someone who stayed on the station. But he doesn't notice this happening, to everyone their clock passes at "1 second per second" if that makes sense.
level 4
1 point · 4 years ago
So if there were clocks on the space station and on the spaceships the clock on the spaceships would run slower (because time passes slower) than the clock on the space station and when the spaceship comes back less time has passed on that clock, right?
If so what happens with the time on the clock of the spaceship that went in the different direction? From the perspective of the one spaceship time on the other spaceship must have gone slower and vice versa, right? But then what happens when comparing the times when both spaceships returned?
I hope it makes any sense what I am trying to ask.
level 5
Aerospace | Quantum Field Theory3 points · 4 years ago
Not that I don't want to help you out, but you are stumbling upon a very complicated topic, which confused quite a few bright minds. So, I present to you the Wikipedia article on the twin paradox which pretty much exactly deals with what you're asking. I feel it gives a better description that I could- but it will explain why both space ships will measure other people's time moving slower than theirs, but at the end of the day, the two space ships will have equal time pass (assuming they have symmetric paths), and it will be less than the space station (the TLDR is because the space ships have to accelerate, while the space station did not).
level 5
1 point · 4 years ago
It's actually quite simple. Ship 1 flies away, returns, compares clocks: The station experienced a year, the ship experienced a day. (These numbers don't really work, but the overall statement is correct.) Ship 2 does the same - and has the same result. Once both ships have returned to the station and compare clocks with each other, they'll have the same time.
The reason that station and ship(s) don't experience the same time effects is that the ship(s) go through the process of acceleration while the station does not.
level 4
1 point · 4 years ago
That was well put thanks. If I could ask one last question since I feel a peer redditor may be able to explain quicker and more to the point then surfing google. But what causes the time dilation? Ive seen examples of dropping the ball on the train and the distances that the ball travels are different from different frames of reference. And for one to travel the distance that it does from the given frame of reference.. the only explanation is time passed slower.. is that somewhat correct?
level 5
Aerospace | Quantum Field Theory4 points · 4 years ago
What "causes" time dilation is (best I can tell) impossible to say. Einstein theorized that the speed of light is the same in all reference frames. This has been experimentally verified. For the speed of light to remain a constant in all frames, there are certain consequences which can be calculated. Time dilation is one of them (that is, if two people are moving relative to each other, and they measure the speed of light to be the same, always, the only way this can happen is if their clocks are running at different speeds). Some others are length contraction and relativistic momentum.
Pretty much, it was observed that light is the same in all reference frames (see the Michselson-Moreley Experiment), and a direct consequence of that is time dilation.
level 6
1 point · 4 years ago
thank you sir. well explained and good links provided
level 2
1 point · 4 years ago
Man, you are smart. And the way you explained it, I actually understood. Thanks.
level 1
Biophysics82 points · 4 years ago
When properly adding relativistic velocities, two ships going half the speed of light away from each other in their centre of mass frame would be going at 80% the speed of light relative to one another.
level 2
Original Poster32 points · 4 years ago
Can you explain why this is? Additionally, if say you increased that speed to 99% of c for each ship will that make a difference? Or is it that no matter how much you add relativistic velocities you only end up with a percentage of c? The idea of achieving c relative between two objects isn't really the part I'm focussing on however. I really want to know what it would look like when they accelerate apart.
level 3
Biophysics55 points · 4 years ago
Every "weird" relativistic effect is a result of the fact that the speed of light is the same in all reference frames. The velocity addition formula is the only way to add velocities and keep the speed of light constant between different reference frames. As to what one would see out the back window, the other ship would get smaller and redder. If it had a giant clock on the back it would appear to be ticking slowly.
level 4
-4 points · 4 years ago
Searched for "red" and was not disappointed. The Doppler Red Shift is an important piece of information, since even galaxies that are moving away from us at near light speeds are visible, but red-shifted..
level 5
[deleted]
5 points · 4 years ago
Until they recede away from us faster than the speed of light and we can no longer see them.
level 6
-6 points · 4 years ago(4 children)
level 7
8 points · 4 years ago
You're partly correct, Galaxies may not be moving faster than the speed of light, but due to the expansion of the universe they may recede faster than that like Naejard stated. source
level 8
-3 points · 4 years ago
Thanks for the clarification, I forgot to take in acount the expansion of the universe.
level 7
[deleted]
1 point · 4 years ago
The expansion of space permits objects receding away from one another at faster than the speed of light because none is actually 'moving' faster than the speed of light: it's the space itself that is growing in-between.
This limit, past which objects recede away from us faster than the speed of light define the visible universe and the observable universe.
To quote Wikipedia (Metric expansion of space article):
It is thus possible for two objects to be stationary or moving at speeds below that of light, and yet to become separated in space by more than the distance light could have travelled, which can suggest the objects travelled faster than light. For example there are stars which may be expanding away from us (or each other) faster than the speed of light, and this is true for any object that is more than approximately 4.5 gigaparsecs away from us.
level 8
-1 points · 4 years ago
Yeah, i am well aware of that, i forgot to take the expansion of the universe in account. Sorry for the misinformation.
level 2
4 points · 4 years ago
Could you give me the formula for adding velocities or whatever you just did here?
level 3
Biophysics18 points · 4 years ago
For velocities v and u adding up to V, the formula is V=(v+u)/(1+vu/c2 ).
level 4
11 points · 4 years ago
Hopefully this helps clarify. It written the way you did made me a little unsure of the denominator
level 1
22 points · 4 years ago
This question has been answered correctly already, but I wanted to point out a cool game to help you experience what relativity is like.
A Slower Speed of Light
This game makes the speed of light comparable to everyday velocities, and it's really striking just how bizarre that would be. It's pretty cool.
level 1
[deleted]
9 points · 4 years ago
a core concept at speeds above 1% the speed of light is time dilation, which has been proven many different times, and is more generally known as einstein's theory of relativity. Basically when something starts moving at notable fractions of the speed of light with respect to another object (whether moving or "still", and it will change with any other perspective) the time that elapses between the two will vary.
For example, to test this, the united states synchronized two clocks, kept one "still" in a lab, while the other took off in an airplane going as fast as possible for hours on end, when the plane landed and the clocks were next to each other again, they were out of sync. This was because the moving clock's time was dilated.
As a more extreme example, certain atomic particles hurtling near the speed of light down towards the earth with an atomic decay rate of a fraction of a second were noted to reach further down towards the earth's crust as expected (instead of decaying in the atmosphere), but when this time dilation was taken into account, it made perfect sense and they were actually decaying at the correct rate for their relative speed.
Length contraction
The same thing happens for the length of an object. Lets say you have a 20ft long car, and a 15ft long garage, Because the time is dilated or "slowed down" at high speeds, the car's length is also "shortened" by the same fraction (1/2 the time rate means 1/2 the length), so at a high enough speed, the car (from the point of view of someone standing next to the garage) will fit perfectly inside. Now here's a tricky, yet vital part, from the perspective of the driver of the car, he feels like he is standing still and the garage appears to be moving towards him at a really high speed, so the garage's length is shortened. So the driver, being in a 20ft car couldn't possibly fit inside a garage that now appears to be less than 10feet.
Because of this amazing feature of the universe, you could have a set of twins heading towards a new planet in separate rocket ships, one at .5c and the other at .9 c, and because the faster ship gets there first (depending on how long they traveled) even though he would have to wait possibly years for his twin to arrive, when the second twin arrives he would notice he would be much younger than his "twin".
So on to your question. Your original "perspective" is from a still point, where two objects are moving away from you at 1/2 the speed of light. From this middle perspective you see they are moving from each other at the speed of light. But keep in mind that their time will be dilated (thus they will be in "Slo-mo" compared to you) so when you try to take their perspective and go into this slightly slower motion, when you look behind at the opposing ship it would only appear to be moving away from you at less than the speed of light.
Now for the most extreme example, because time is dilated more the faster you move, lets say you want to live forever, like literally live forever until the end of the universe, so you invent a rocket ship with no max speed. You take off, and accelerate as fast as possible without causing bodily harm, time starts to slow down for you and you age slower compared to people from your home planet, you accelerate up to .9c, and keep going until you're at .99c, then .999c, and so on. depending on how fast you accelerate, you could literally outlive all life, and stars in the entire universe, but to you, it would only FEEL like ~100years.
ELI actually 5: when things go really fast, they go into slow motion, something can't go the actual speed of light because as soon as you get close your time slows down, making it harder the closer you get. When you go really fast everything around you looks shorter (in the direction you're moving) like the lines on a road look shorter when you're on the highway. so when two space ships moving away from each other at a certain speed, when they look behind them it looks like the other person is moving slower than their two speeds put together. Why does this happen, magic. This is a complicated subject and difficult for a ELI5, spent about 2 weeks on it in a sophomore level physics major class. But i'd be glad to clarify further if anyone requests any extreme examples or the math behind everything.
level 1
Nuclear Magnetic Resonance3 points · 4 years ago
level 1
4 points · 4 years ago
How do you accelerate at a set speed?
level 1
2 points · 4 years ago · edited 4 years ago
As a physics student, I'd like to point out that none of these answers consider the relativistic Doppler effect, and instead are talking about the actual velocity of ship 2 in ship 1's reference frame.
Like the other answers say, ship 2 is moving at 0.8c from the first ship's reference frame, but since light is taking longer and longer to reach the first ship from the second ship as it speeds away, if you looked out the window of ship one you would actually see the second ship moving much slower than 0.8c, although 0.8c is the velocity it is 'actually' moving at (for you).
The formula for the relativistic Doppler shift is thus:
vobserved = vsource * squareroot[(1 + v/c) / (1 - v/c)]
where a positive v is towards the observer.
Plunging in our 0.8c answer for the calculated velocity of the other ship, we get that the observed velocity of ship two, the speed you would see it going if you looked out the window, is a mere 4/15 C, four fifteenths the speed of light.
edit: corrected below by diazona, thanks.
level 2
Particle Phenomenology | QCD | Computational Physics4 points · 4 years ago
Actually, the relativistic Doppler effect isn't quite what's going on here. The RDE relates the frequency of a cyclic process happening in the moving reference frame to the frequency at which the cycles are seen in the stationary reference frame (or vice-versa, as you know it doesn't matter which frame is labeled which). It takes into account two effects: time dilation, and also the travel time of light. But when you're looking at a spaceship trying to figure out how fast it's moving away from you, you don't care about time dilation. You're not looking at something happening on the ship to evaluate its frequency, you're only looking at how the ship moves.
If you e.g. draw a spacetime diagram with a ship moving at 0.8c that emits a light beam back to the origin, you should find that the correction factor accounting for only the finite speed of light is 1/(1 + v/c). So you'd actually see the ship moving at 1/1.8 = 4/9 of the speed of light.
Incidentally, in some cases an effect much like this can actually give an observed velocity faster than light: http://en.wikipedia.org/wiki/Superluminal_motion (for interested readers)
level 3
1 point · 4 years ago
Dang it, I knew I was off with something there. I've only finished first year physics at the University of Toronto, and we rushed through relativity, whereas the lecture course for non-science students I took on 'time', taught by a biology professor, dealt with it a lot but not in any depth.
Thanks for pointing out my error, and for the link, I'd seen some similar stories before and it's very cool.
level 2
[deleted]
1 point · 4 years ago
Mine technically did go into that with time dilation, but i did not bring up the actual equation.
level 1
1 point · 4 years ago
What if the ships turned on headlights? How fast would that light than travel?
level 2
7 points · 4 years ago
Light (in a vacuum) always travels at 1C regardless of reference frame.
level 3
1 point · 4 years ago
So if I was flying along in my spaceship at 1c, turned on my headlights, what would happen?
Is the light traveling 2c now? Or is it pooling at the front of my lightbulb and when I stop some insane deathray of stored-up bulb light is going to come blasting out?
level 4
9 points · 4 years ago
First, your spaceship has mass, so it's not travelling at the speed of light. But let's say it's going very close to it, 0.9 c relative to a space station. (Remember, speed is always relative, so specifying a speed is meaningless unless you say what that speed is relative to).
You turn on your headlights, and see the light moving away from you at c. People on the space station see your ship moving at 0.9 c and the light from your headlights moving at c. Everyone in the universe, no matter where they are or how fast they are moving, will measure the speed of that light as, well, the speed of light: c.
level 4
2 points · 4 years ago
I think what helped me figure it out is say you leave earth and you are going 1c (well really close to it, you cant actually go the speed of light) and then when you flip your lights on you see it leave at 1c. You think that must mean its going at 2c. Thats not true. If you were to get rid of everything else in the universe and now its just you in your space ship, how do you prove you are even going 1c. In that world you are still and the light goes out at 1c. Thats why its the theory of relativity because in essence you cant define speed in the universe. Everything is relative to something else. 1c is the max speed anyone can see anything move. The universe changes time and space to keep this true. So no the light moves away from you at 1c and outside observers see you moving at .9999 c and the light moving at 1c. Its hard to wrap your head around but its true.
level 5
1 point · 4 years ago
To make sure my common mind is working right and understanding you.
Considering the ship is moving at 1c, an its produces the light. Light is only moving away from the ship at 1c because the ship is the mass thats producing the light reguardless of the speed. Meaning if it was traveling say 5x the speed light then light would still be coming off the ship at 1c because it is the object in which is producing the light. Pretty sure I understood what you said with my simple little mind just trying to make sure I actually am. Unless I did a horrible job explaining how I think I understood it
level 6
1 point · 4 years ago
Im not completely sure if it matters whether or not the ship is producing the light. Im no expert on this stuff. It still throws me for a loop a lot. Seems like you've got it to me. The speed of light is essentially the cosmic speed limit. You'll never observe something going faster than light no matter what you reference. Their are weird loopholes around it. Like there may be particles that only go faster than light but never slower, There are weird things about bending space to kind of cheat the system and move faster then the speed of light without actually breaking the speed of light. Its all crazy.
level 7
1 point · 4 years ago
Well at least I understood correctly. Not good at math at all and oddly this kinda math stuff interest me but when the formulas an numbers come up I'm lost im the wind.
One of the things that I am personally not convinced of is if we could travel to another stare in a reasonable life time travel. That is considered traveling in the past. Yet what has me going bonkers is if it is reasonable enough and could travel back to earth who sees the effects of this so called time travel? Now I understand its depends on tye distance an perhaps time left an returned. To me it would seem time passed the same for both. Them again thats where all the lovely formulas kick in an sweep my ass under the carpet.
At least I learned one good thing that was beyond my knowledge.. ty
level 7
1 point · 4 years ago
Well at least I understood correctly. Not good at math at all and oddly this kinda math stuff interest me but when the formulas an numbers come up I'm lost im the wind.
One of the things that I am personally not convinced of is if we could travel to another stare in a reasonable life time travel. That is considered traveling in the past. Yet what has me going bonkers is if it is reasonable enough and could travel back to earth who sees the effects of this so called time travel? Now I understand its depends on the distance an perhaps time left an returned. To me it would seem time passed the same for both. Then again thats where all the lovely formulas kick in an sweep my ass under the carpet.
At least I learned one good thing that was beyond my knowledge.. ty
Edit: think I dbl posted. Sorry bout that if I did. Using my phone
level 4
[deleted]
1 point · 4 years ago
To add to jswhitten, when you get close to c, your time is dilated, you see everything around you as if it's moving at a faster rate of time, not kidding this actually happens. So if you're going .9c with headlights on, even though from an outside observer the light is only appears to be moving away from you at .1c, from your point of view you (because you have been slowed down) it looks like its moving away from you at 1c.
To help wrap you head around this crazy concept, forget about the outside observer, lets say there are no stars around you that you can see, because you're flying through space, you may have no way of telling if you're actually moving at all! you could technically be moving backwards compared to something. but in the universe there is no "still" point, anything could be considered still, or maybe everything is moving at close to the speed of light in one direction all together. Which is why at any "speed" light will always move away from you at c.
level 2
[deleted]
1 point · 4 years ago
you would see it moving away from you at the speed of light. an outside observer would also see it moving away at the speed of light. It's the theory of relativity and it has yet to be disproved
level 1
1 point · 4 years ago
Spaceships separating from each other at let's say 0.6c each. Spaceship A shines a laser pointer back at Spaceship B. How is it shown that that beam of light will reach Spaceship B?
level 2
2 points · 4 years ago
As explained in other answers, velocities don't just add up, they follow the formula (v1 + v2) / (1 + v1v2/c²).
So even though both ships are moving at 0.6c relatively to someone staying in the middle, to each other they are not going at 1.2c, but rather at 1.2c / (1 + 0.36c²/c²) = 0.88c, which is always slower than the speed of light, so light has no problem going from one ship to the other.
level 1
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level 1
[deleted]
-5 points · 4 years ago(0 children)
level 1
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-21 points · 4 years ago(0 children)
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-6 points · 4 years ago(0 children)
level 2
2 points · 4 years ago
first of all, in these toy problems typically the acceleration phase is ignored. we can pretend the spaceships were already going at .5c beforehand and cross each other at the space station. this is because acceleration is significantly trickier and introduces its own affects, but there is interesting physics even at constant velocities.
second of all, this isn't true. let's say the other spaceship is shining a light towards you. what you will see looking behind you is a spaceship traveling at .8c with respect to yourself, and the light emitted from the flashlight is blueshifted compared to how the light appears to the other spaceship.
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September 4, 2018 | 11,754 | 49,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-34 | latest | en | 0.967854 |
https://www.kth.se/student/kurser/kurs/SK1115 | 1,716,593,180,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058751.45/warc/CC-MAIN-20240524214158-20240525004158-00164.warc.gz | 749,178,414 | 15,295 | • Svenska
# SK1115 Electromagnetism and Waves 7.5 credits
### Choose semester and course offering
Choose semester and course offering to see current information and more about the course, such as course syllabus, study period, and application information.
## Application
### For course offering
Autumn 2024 Start 26 Aug 2024 programme students
### Application code
51148
Headings with content from the Course syllabus SK1115 (Autumn 2020–) are denoted with an asterisk ( )
## Content and learning outcomes
### Course contents
Electrical statistics: Electric power, field strength and potential, Gauss's theorem, electric field and potential in metals and dielectrics, capacitor, electrostatic energy.
Magnetic field: Origin, power effect, magnetic material, magnetic energy. Electromagnetic induction. Introduction to the relationship between electric and magnetic field and Maxwell's equations.
Mechanical waves: Basic wave concepts. Acoustics.
Electromagnetic waves: Generation, polarization, interference and diffraction, technical applications.
### Intended learning outcomes
After completing the course the student should be able to:
• solve technical problems in basic physics related to electric and magnetic fields, mechanical and electromagnetic waves, and assess the reasonableness of the solution
• use physical measurement methods and instruments, evaluate measurement data and report results and evaluate limitations.
## Literature and preparations
### Specific prerequisites
One-variable analysis (SF1625) and Algebra and geometry (SF1624) are recommended, but are read parallel to this course, hence no formal prerequisites.
### Recommended prerequisites
No information inserted
### Equipment
No information inserted
### Literature
No information inserted
## Examination and completion
If the course is discontinued, students may request to be examined during the following two academic years.
A, B, C, D, E, FX, F
### Examination
• INL1 - Hand in tasks, 1.5 credits, grading scale: P, F
• LAB1 - Laboratory experiments, 1.5 credits, grading scale: P, F
• TEN1 - Written exam, 4.5 credits, grading scale: A, B, C, D, E, FX, F
Based on recommendation from KTH’s coordinator for disabilities, the examiner will decide how to adapt an examination for students with documented disability.
The examiner may apply another examination format when re-examining individual students.
### Other requirements for final grade
The course is examined through a written exam (TEN1; 4.5 credits, grade scale A, B, C, D, E, Fx, F), assignments (INL1; 1.5 credits, grade scale P/F) and approved laboratory exercises (LAB1; 1 , 5 credits, grading scale P/F).
### Opportunity to complete the requirements via supplementary examination
No information inserted
Yes
### Ethical approach
• All members of a group are responsible for the group's work.
• In any assessment, every student shall honestly disclose any help received and sources used.
• In an oral assessment, every student shall be able to present and answer questions about the entire assignment and solution.
## Further information
### Course room in Canvas
Registered students find further information about the implementation of the course in the course room in Canvas. A link to the course room can be found under the tab Studies in the Personal menu at the start of the course.
Technology
### Education cycle
First cycle
No information inserted
### Contact
Jonas Sellberg (jonassel@kth.se) | 738 | 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} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.84186 |
https://www.modulargrid.net/e/doepfer-a-1-thru-zero-quadrature-vco-vcf-qvco | 1,611,815,071,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00267.warc.gz | 870,351,094 | 6,542 | # A-1?? Thru-Zero Quadrature VCO/VCF (QVCO)
## Doepfer
A-1?? QVCO prototype
The picture shows the preliminary front panel of the planned Thru-Zero Quadrature VCO (QVCO). The term "quadrature" means in this connection that the oscillator outputs sine and cosine waveforms simultaneously. The term "Thru-Zero" means that even "negative" frequencies are generated. But this a bit a misleading term as negative frequencies do not really exist. "Negative" means in this connection simply that the polarity of the waveform is reversed (thru-zero pictures will follow as soon as our pc-based oscilloscope is repaired).
The module will have two different control modes: exponential and linear. The exponential control mode is known from most of the VCO and VCF modules. The most common exponential scale is 1V/octave. In this mode thru-zero is not possible as the exponential function cannot become negative. In the exponential mode the lowest frequency is about 0.3Hz (i.e. period time ~ 3 seconds), the maximum frequency is about 5kHz for the prototype. In the linear control mode the relation between control voltage and frequency is linear: e.g. 0V = 0Hz (i.e. oscillation stops), 1V = 1kHz, 2V = 2kHz, 3V = 3kHz and so on. In this mode thru-zero is possible. The frequency range for the prototype is about -5kHz ... +5kHz.
After all the QVCO is a 12dB filter circuit in self-oscillation, optimized for constant output level over the entire frequency range (for many filter circuits the amplitude varies with the frequency in self-oscillation). This is why the module can be switched between VCO and VCF. In the VCF mode the audio input is used. The sine output becomes the low pass output and the cosine output becomes the band pass output. In the VCO mode the audio input is not used and the resonance control defines the output level and has also an effect on the waveforms. From about 2 to 7 the module generates low distortion sine/cosine waves with adjustable level. Beyond 5 the waveforms change more and more into triangles, i.e. odd harmonics are added. In the VCF mode the resonance control works as usual and the audio input .
http://www.doepfer.de/A100_QVCO.htm
• ? mA +12V
• ? mA -12V
• ? mA 5V
• Ø 3.00 (1 Votes) Average Rating
This Module is a prototype or in a concept phase.
8 HP Oscillator
submitted Feb 28th 2014, 14:29 by balpirol | last Change Dec 27th 2018, 09:03 by cds | 603 | 2,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-04 | latest | en | 0.908133 |
https://www.nag.com/numeric/nl/nagdoc_27.1/examples/baseresults/g01aece.r.html | 1,624,219,421,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488253106.51/warc/CC-MAIN-20210620175043-20210620205043-00011.warc.gz | 832,932,219 | 1,420 | ```
nag_stat_frequency_table (g01aec) Example Program Results
Problem 1
Number of cases 70
Number of classes, including extreme classes 7
Routine-supplied class boundaries
Successful call of nag_stat_frequency_table (g01aec)
*** Frequency distribution ***
Class Frequency
Up to 20.70 0
20.70 to 21.28 6
21.28 to 21.86 16
21.86 to 22.44 21
22.44 to 23.02 14
23.02 to 23.60 13
23.60 and over 0
Total frequency = 70
Minimum = 20.70
Maximum = 23.60
``` | 174 | 564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-25 | latest | en | 0.753498 |
icantbreatheblacklivesmatters.com | 1,623,976,441,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634576.73/warc/CC-MAIN-20210617222646-20210618012646-00132.warc.gz | 25,651,348 | 7,755 | Nctm Pan Balance 2021 // icantbreatheblacklivesmatters.com
# Pan BalanceEvaluated - Online Math.
Access NCTM Illuminations – Pan Balance – Shapes. Click on “Set 3”. Start by dragging one yellow to one side of the balance scale and one red to the other side until the scale is balanced. Ask students for observations and write their responses on the whiteboard. Be sure to ask students to explain their thinking, drawing out and. Demonstrate the use of the pan balance at the example Web site. Hand out the worksheet, Algebraic Expressions. Go through the instructions demonstrating how to use the balance. Have students work in pairs to answer the worksheet. Teacher Tips. Two other tools, Pan Balance - Numbers and Pan Balance - Shapes can help the students understand the concept. In Pan Balance-Shapes, students can explore equality which is an important component of algebra. Students put shapes on a balance to try to find equivalent relationships. The applet records what shapes are on the pan when an equivalent relationship is found. Once students have found many different combinations of equivalent relationships they. fractals tool, shape sorter, shape tool, shape pan balance visualizing algebra, fraction pie, and many more tools and activities spanning all levels NCTM's Electronic Examples for. We look at the data,has 0 rank in the world wide web. The website server is using IP address 199.83.128.95 and is hosted in. The Google page rank of this website is 7/10. Website ping to the server is timed at 1699 ms. This domain creation date on 0001-01-01. By thedomain, you can see that different countries, middleware.
Como usar a balança. Coloca pesos em cada um dos pratos da balança, arrastando as formas, de maneira que a balança fique equilibrada. Se quiseres ver o número de peças, pressiona Count Items. Scales Problems 1. Solve how much each geometric shape "weighs". You can use either pounds or kilograms. This is a pan balance or scales. Things go into the two "pans", and the heavier pan will go down, like in a seesaw. If the two things weigh the same, the balance stays balanced. a. The square weighs _____ b. The square weighs _____ c. Solve simple linear equations using a balance beam representation.
students use a modified pan balance to investigate the equivalence of two numeric expressions. In the third “A mathematics curriculum should be well- part, middle school students use a further enhanced pan articulated across the grades.”NCTM, 2000, p. 16 balance, with accompanying graphing tool, to consider An important feature of the Principles and Stan- equivalence of two symbolic. NCTM publishes five journals. All are available in print and online versions. Teaching Children Mathematics, an official journal of the National Council of Teachers of Mathematics NCTM, supports improvement of pre-K–6 mathematics education by serving as a resource for teachers so as to provide more and better mathematics for all students. It is a forum for the exchange of mathematics idea,. 11/5/2014 1 Illuminate Your Classroom and Teach Conceptually Using Free Virtual Manipulatives 2014 NCTM Regional Conference – Indianapolis. One example of such lesson plans created by the NCTM is pan balance-shapes virtual manipulative online tools which reinforce the equality concept. In this applet, students virtually interact with the algebraic concept of equality by placing shapes in the right and left side of the balance pan.
Mit „Pan Balance – Numbers“ ist das Bauen von gleichwertigen Rechnungen als Vorübung zum Verständnis von Gleichungen möglich. “Ein aktiv spielerischer Zugang zu Gleichungen ist natürlich auch mit einer Balkenwaage mit verschiedenen Objekten und einem Säckchen, in dem solche Objekte versteckt sind, sinnvoll. Balance word problems from Math Kangaroo Algebraic Reasoning Game - a weighing scales game that practices algebraic reasoning Interactive Pan Balance with Shapes Balance Beam Activity A virtual balance that provides balance puzzles where student is to find the weights of various figures, practicing algebraic thinking. Includes three levels. Gleichungen mit dem Waagemodell erforschen. Gleichung in die beiden Waagschalen eintippen und durch Bewegen des Schiebereglers x die Waagschalen ins Gleichgewicht bringen. NCTM - National Council of Teachers of Mathematics December 3 at 8:30 AM · December 3 is Giving Tuesday and The Mathematics Education Trust MET hopes you will consider making a. Teaching Resources: Solving Equations using the Plate Strategy For "Plate" read "Cover-Up" Many are tougher than required at this stage Java WisWeb Pan Balance - Expressions Put.
Practice Activity-NCTM Pan Balance and Open Middle Properties of Numbers Challenge Section 4 CUNY HSE Framework Math – Unit 7 The Math Practitioner Prove It and Creating True Equations Civics It Up-An Online Resource for Teachers of IEL-CE. Civics it Up! Handout 2-Webquest 1 Civics it Up! Handout 4- Case Study Civics It Up. To make it harder, students can make equations balance each other such as 32=41. Also, as students learn multiplication and division they can use the operations to make balanced equations as well. This is a great activity that would be best used after a lesson using a real pan balance to demonstrate equality. The fact that this game is online. will take to balance the scale. Materials: • Pattern blocks all the same thickness • Pan balance with bucket/lid • Number Path Directions: 1. Have one child designated as the “hider” and the rest as “seekers.” The Hider should choose several of the same shape to hide on one side of the balance for example, 4 trapezoids. 2. The. 6th Grade Math Resources. Unit. 6: Intro to Algebra. Name of Resource. Link or Location. Description. That Quiz.. On-line quiz website.
• I always introduce the concept of equality using NCTM’s pan balance and balancing shapes activities. These applets are an easy and mobile-friendly digital exploration for students to figure out the weights of various shapes, and an excellent lead-in to equation solving or systems of equations. Scenarios for Use. Whole group instruction.
• 02.03.2015 · Learn how to use the software, Pan Balance Shapes, with Ms. Fraylick and Ms. McCoy! illuminations./activit. 6.EE.B5 6TH GRADE EXPRESSIONS AND.
• The Illuminations Pan Balance applet can be used as a mathematical model to represent and understand quantitative relationships, which supports one of the NCTM Algebra standards. The interactive point-plotting feature allows students to explore the graphic implications of changing the input or x-value of two different equations. The applet also.
• Name of Tech Tool: Illuminations - Pan Balance - Expressions Brief Description of Tech Tool: Students explore numeric expressions and balance using a scale and graphical representation. Younger students can compare whole and decimal values, while older students can explore graphical representation and changing x-values.
• Interactive, Manipulative, Publisher, NCTM Illuminations Build up to algebraic thinking by exploring this balance tool using shapes of unknown weight. Challenge yourself to find the weight of each shape in one of six built-in sets or a random set.
## Pan Balance - Shapes - MathsLinks.
- The "reset balance" and "count items" tools can tell your how many of each shape is on the pan. - There are also a "random" button to make the game more settings. Powered by Create your own unique website with customizable templates. Balance Scales Unit 6: Algebra Grade Level 4 Overview In this activity, students find an unknown quantity, explore the relationship between quantities, and enhance their understanding of the equal sign as equivalence and a balance point. Key Standards M4A1. Students will represent and interpret mathematical relationships in quantitative expressions. b. Represent unknowns using symbols, such. Pan Balance. This is an interactive website where students explore equivalence. Students fill in numbers or expressions on either side of the scale to see if they balance the scale and also can see graphically how the expressions compare. This could be used to introduce the idea of equivalent expressions as an individual exploration activity or a class activity. Students could also use this.
### Online Manipulatives and Applets - Math Cats.
Thank you to all of you who attended my session at the NCTM 2012 Annual Meeting and Exposition. I have listed the apps and links below. Please contact me. | 1,821 | 8,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-25 | latest | en | 0.870201 |
http://www.developersite.org/question-25083 | 1,556,086,075,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578633464.75/warc/CC-MAIN-20190424054536-20190424080536-00417.warc.gz | 230,728,171 | 5,946 | plot - Ellipse around the data in MATLAB - DeveloperSite- developersite.org
# plot - Ellipse around the data in MATLAB
Keywords：matlab
Question:
I would like to reproduce the following figure in MATLAB:
There are two classes of points with X and Y coordinates. I'd like to surround each class with an ellipse with one parameter of standard deviation, which determine how far the ellipse will go along the axis.
The figure was created with another software and I don't exactly understand how it calculates the ellipse.
Here is the data I'm using for this figure. The 1st column is class, 2nd - X, 3rd - Y. I can use `gscatter` to draw the points itself.
``````A = [
0 0.89287 1.54987
0 0.69933 1.81970
0 0.84022 1.28598
0 0.79523 1.16012
0 0.61266 1.12835
0 0.39950 0.37942
0 0.54807 1.66173
0 0.50882 1.43175
0 0.68840 1.58589
0 0.59572 1.29311
1 1.00787 1.09905
1 1.23724 0.98834
1 1.02175 0.67245
1 0.88458 0.36003
1 0.66582 1.22097
1 1.24408 0.59735
1 1.03421 0.88595
1 1.66279 0.84183
];
gscatter(A(:,2),A(:,3),A(:,1))
``````
FYI, here is the SO question on how to draw ellipse. So, we just need to know all the parameters to draw it.
Update:
I agree that the center can be calculated as the means of X and Y coordinates. Probably I have to use principal component analysis (`PRINCOMP`) for each class to determine the angle and shape. Still thinking...
Consider the code:
``````%# generate data
num = 50;
X = [ mvnrnd([0.5 1.5], [0.025 0.03 ; 0.03 0.16], num) ; ...
mvnrnd([1 1], [0.09 -0.01 ; -0.01 0.08], num) ];
G = [1*ones(num,1) ; 2*ones(num,1)];
gscatter(X(:,1), X(:,2), G)
axis equal, hold on
for k=1:2
%# indices of points in this group
idx = ( G == k );
%# substract mean
Mu = mean( X(idx,:) );
X0 = bsxfun(@minus, X(idx,:), Mu);
%# eigen decomposition [sorted by eigen values]
[V D] = eig( X0'*X0 ./ (sum(idx)-1) ); %#' cov(X0)
[D order] = sort(diag(D), 'descend');
D = diag(D);
V = V(:, order);
t = linspace(0,2*pi,100);
e = [cos(t) ; sin(t)]; %# unit circle
VV = V*sqrt(D); %# scale eigenvectors
e = bsxfun(@plus, VV*e, Mu'); %#' project circle back to orig space
%# plot cov and major/minor axes
plot(e(1,:), e(2,:), 'Color','k');
%#quiver(Mu(1),Mu(2), VV(1,1),VV(2,1), 'Color','k')
%#quiver(Mu(1),Mu(2), VV(1,2),VV(2,2), 'Color','k')
end
``````
# EDIT
If you want the ellipse to represent a specific level of standard deviation, the correct way of doing is by scaling the covariance matrix:
``````STD = 2; %# 2 standard deviations
conf = 2*normcdf(STD)-1; %# covers around 95% of population
scale = chi2inv(conf,2); %# inverse chi-squared with dof=#dimensions
Cov = cov(X0) * scale;
[V D] = eig(Cov);
``````
I'd try the following approach:
1. Calculate the x-y centroid for the center of the ellipse (x,y in the linked question)
2. Calculate the linear regression fit line to get the orientation of the ellipse's major axis (angle)
3. Calculate the standard deviation in the x and y axes
4. Translate the x-y standard deviations so they're orthogonal to the fit line (a,b)
I'll assume there is only one set of points given in a single matrix, e.g.
``````B = A(1:10,2:3);
``````
you can reproduce this procedure for each data set.
1. Compute the center of the ellipsoid, which is the mean of the points. Matlab function: `mean`
2. Center your data. Matlab function `bsxfun`
3. Compute the principal axis of the ellipsoid and their respective magnitude. Matlab function: `eig`
The successive steps are illustrated below:
``````Center = mean(B,1);
Centered_data = bsxfun(@minus,B,Center);
[AX,MAG] = eig(Centered_data' * Centered_data);
``````
The columns of AX contain the vectors describing the principal axis of the ellipsoid while the diagonal of MAG contains information on their magnitude. To plot the ellipsoid, scale each principal axis with the square root of its magnitude.
Hope this helps.
A. | 1,267 | 3,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-18 | longest | en | 0.799651 |
https://gamedev.stackexchange.com/questions/113315/how-to-find-a-material-through-a-raycast-in-unity-5/113321 | 1,713,560,355,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817455.17/warc/CC-MAIN-20240419203449-20240419233449-00247.warc.gz | 235,067,785 | 35,536 | # How to find a material through a raycast in Unity 5?
I have an empty object with a script on it and I want to be able to move the empty over other objects and then have it print out the name of the material for the triangle that it's hovering above. (Its for something else, but I only need help getting to that stage, I can do the rest from that point). I found this on the topic but the answer is very vauge. I finished steps 1 and 2 but I can't figure out number 3. Here is my code so far...
void Update (){
Ray ray = new Ray(transform.position, new Vector3(0, -100, 0));
RaycastHit hit;
if(Physics.Raycast(ray, out hit)){
print ("Triangle Index: " + hit.triangleIndex);
Mesh mesh = hit.collider.gameObject.GetComponent<MeshFilter>().mesh;
print ("Submesh Count: " + mesh.subMeshCount);
}
}
Could someone show me how to finish it off so that when I move the empty over different objects, it would print out the name of the material per triangle?
Thanks, Matthew
I'd recommend testing empirically to confirm, but the second answer at this link says that triangle data for submeshes is indexed sequentially in order of submesh index. That means all the triangles for submesh 0 have lower indices than those in submesh 1, etc...
So, you can iterate over the submeshes until you find the one your triangle belongs within:
MeshCollider collider = hit.collider as MeshCollider;
// Remember to handle case where collider is null because you hit a non-mesh primitive...
Mesh mesh = collider.sharedMesh;
// There are 3 indices stored per triangle
int limit = hit.triangleIndex * 3;
int submesh;
for(submesh = 0; submesh < mesh.subMeshCount; submesh++)
{
int numIndices = mesh.GetTriangles(submesh).Length;
if(numIndices > limit)
break;
limit -= numIndices;
}
Material material = collider.GetComponent<MeshRenderer>().sharedMaterials[submesh];
This is pretty ugly, but still neater than the accepted answer at that link, which visits every triangle to test if it's the one the ray hit. ;) | 462 | 2,000 | {"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} | 2.625 | 3 | CC-MAIN-2024-18 | latest | en | 0.833264 |
https://www.coursehero.com/file/6879767/Ch3-Homework-ANSWERS/ | 1,604,090,718,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911229.96/warc/CC-MAIN-20201030182757-20201030212757-00202.warc.gz | 664,116,421 | 59,652 | Ch.3 Homework ANSWERS - 1 Short-term solvency ratios Current ratio = Current assets Current liabilities Current ratio 2008 = \$56,260 \$38,963 = 1.44
# Ch.3 Homework ANSWERS - 1 Short-term solvency ratios...
This preview shows page 1 - 3 out of 11 pages.
1. Short-term solvency ratios: Current ratio = Current assets / Current liabilities Current ratio 2008 = \$56,260 / \$38,963 = 1.44 times Current ratio 2009 = \$60,550 / \$43,235 = 1.40 times Quick ratio = (Current assets – Inventory) / Current liabilities Quick ratio 2008 = (\$56,260 – 23,084) / \$38,963 = 0.85 times Quick ratio 2009 = (\$60,550 – 24,650) / \$43,235 = 0.83 times Cash ratio = Cash / Current liabilities Cash ratio 2008 = \$21,860 / \$38,963 = 0.56 times Cash ratio 2009 = \$22,050 / \$43,235 = 0.51 times Asset utilization ratios: Total asset turnover = Sales / Total assets Total asset turnover = \$305,830 / \$321,075 = 0.95 times Inventory turnover = Cost of goods sold / Inventory Inventory turnover = \$210,935 / \$24,650 = 8.56 times
Receivables turnover = Sales / Accounts receivable Receivables turnover = \$305,830 / \$13,850 = 22.08 times Long-term solvency ratios: Total debt ratio = (Total assets – Total equity) / Total assets Total debt ratio 2008 = (\$290,328 – 176,365) / \$290,328 = 0.39 Total debt ratio 2009 = (\$321,075 – 192,840) / \$321,075 = 0.40 Debt-equity ratio = Total debt / Total equity Debt-equity ratio 2008 = (\$38,963 + 75,000) / \$176,365 = 0.65 Debt-equity ratio 2009 = (\$43,235 + 85,000) / \$192,840 = 0.66 Equity multiplier = 1 + D/E Equity multiplier 2008 = 1 + 0.65 = 1.65 Equity multiplier 2009 = 1 + 0.66 = 1.66 Times interest earned= EBIT / Interest Times interest earned= \$68,045 / \$11,930 = 5.70 times Cash coverage ratio = (EBIT + Depreciation) / Interest
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Want to read all 11 pages? | 591 | 1,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-45 | latest | en | 0.781429 |
https://th.tradingview.com/script/gThgCf69-Discrete-Fourier-Transform-Overlay-wbburgin/ | 1,696,136,278,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510781.66/warc/CC-MAIN-20231001041719-20231001071719-00731.warc.gz | 601,895,467 | 100,819 | # Discrete Fourier Transform Overlay [wbburgin]
The discrete Fourier transform (DFT) overlay uses a discrete Fourier transform algorithm to identify trend direction. This is a simpler interpretation that only uses the magnitude of the first frequency component obtained from the DFT algorithm, but can be useful for visualization purposes. I haven't seen many Fourier scripts on TradingView that actually have the magnitude plotted on the chart (some have lines, for instance, but that makes it difficult to look into the past or to see previous lines).
The DFT is a mathematical transformation that decomposes a time-domain signal into its constituent frequency components. By applying the DFT to OHLC data, we can interpret the periodicities and trends present in the market. I've designed the overlay so that you can choose your source for the Fourier transform, as well as the length.
Settings and Configuration
The "Fourier Period" is the transform length of the DFT algorithm. This input indicates the number of data points considered for the DFT calculation. For example, if this input is set to 20, the DFT will be performed on the most recent 20 data points of the input series. The transform length affects the resolution and accuracy of the frequency analysis. A shorter transform length may provide a broader frequency range but with less detail, while a longer transform length can provide finer frequency resolution but may be computationally more intensive (I recommend using under 100 - anything above that might take too much time to load on the platform).
The "Fourier X Series" is the source you want the Fourier transform to be applied to. I have it set in default to the close.
"Kernel Smoothing" is the bar-start of the rational quadratic kernel used to smooth the frequency component. Think of it just like a normal moving average if you are unfamiliar with the concept, it functions similarly to the "length" value of a moving average.
Premium strategies, indicators, and algorithms: www.patreon.com/wbburgin/membership
Use my strategies on CryptoRobotics: cryptorobotics.co/?trade=f23b09
Backtesting of certain algorithms (for fun): tinyurl.com/yc7jwj8h
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01-08-2013, 03:35 AM #93
Technatic
Semi-Pro
Join Date: Oct 2009
Posts: 452
Quote:
To give an example why this doesn't work lets consider three racquets, all 70 cm long and with a mass of 250 gram evenly distributed. Then: to A: Add 50 g evenly distributed to B: Add 50 g to the midpoint to C: Add 25 g each to the top and bottom You end up with three racquets that have that same mass (300 g) and the same balance point. So if you try to calculate the swing weight from these two values you will end up with the same swing weight, independent of which formula you use. With your method all three will be identical.
I misunderstood the matter at first:
If you add weights in one point you havd to add the Moment of inertia of that weight separately to get the right total SW.
meaning:
In case of adding 25 gram at both ends: a
The added SW with pivot point at 10 cm= 25*(10^2 + (Length - 10)^2 gr >>> 25 * Length^2
In case of the 50 gram in the center
The added SW = 50 * (bal-10)^2
If the balance point is at 35 cm and the length is 70 cm the added SW's are 90 and 31250 kgcm^2.
So this is the same difference as you mention between 350 and 290 kg cm
This is what the last section of the SW advisor calculates when you add weight.
We are testing this system and comparing it with actual SW tests on for quite some time now and we think that it is quite accurate.
Quote:
Hi Technatic - What is that 8 sided racquet? Is from late 80's? Any info would be appreciated! Thanks - Jack
It was a sample from a manufacturer, more than 20 years old, I don't think that it was ever produced.
Last edited by Technatic; 01-08-2013 at 05:59 AM. | 447 | 1,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2015-22 | latest | en | 0.938808 |
https://platformrnnb.web.app/kapps39165wi/what-does-pulse-rate-equal-gox.html | 1,638,110,565,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00419.warc.gz | 538,004,951 | 6,229 | ## What does pulse rate equal
Heart rate is the speed of the heartbeat measured by the number of contractions ( beats) of the heart per minute (bpm). The heart rate can vary according to the body's physical needs, including the need to absorb oxygen and excrete carbon dioxide. It is usually equal or close to the pulse measured at any peripheral point. The heart rate is the number of times the heart beats in the The pulse rate is exactly equal to the heartbeat, 18 Nov 2018 Pulse rates vary from person to person. Your pulse is lower when you are at rest and increases when you exercise (more oxygen-rich blood is
Heart rate is a measurement of how many times your heart beats in one minute. Here’s how to check heart rate so you can optimize workouts or identify concerning symptoms. We’ll give you five A heart rate is the number of times the heart beats per minute.The heart rate changes throughout a person’s life, according to their age, their fitness, and even whether they are frightened. What do heart rate and oxygen levels mean? At first, it might feel a little overwhelming to see your baby’s heart rate and oxygen levels but this data is meant to empower you, not confuse you. We hope this info will shed a little light on what heart rate and oxygen levels mean for you and your baby. What is a heart rate? 4. Myth: If my heart rate is normal, my blood pressure is fine. Sometimes your heart rate and your blood pressure go hand in hand. For example, when you exercise, or get angry or scared, they both When it comes to your heart rate, it's a bit like the speed of your car. What you want is not too fast, not too slow, and not too erratic. In fact, most of the time, heart rhythm and pace are not things you need to think about. And unless something u Individuals with certain heart conditions where the heart does not efficiently pump blood with each contraction, they may have a pulse that is lower than the measured heart rate. Heart rate can
## Individuals with certain heart conditions where the heart does not efficiently pump blood with each contraction, they may have a pulse that is lower than the measured heart rate. Heart rate can
28 Feb 2019 Roberts explains that a lower heart rate will deliver the same blood If you experience symptoms such as irregular heart beat, dizziness, slope, but which would have different origins at 70 C. Type E is the same as type C in that the pulse rates lie on a straight line, but has been separated from the What is a normal heart rate? This will allow you to work out your average daytime blood pressure and this may be a better way of assessing your risk of future A normal heart rate for a healthy adult is between 60 and 100 beats per minute. as children grow, and usually by the teen years the heart rate is in the same range as an adult's. Atrial fibrillation is the most common type of fast heartbeat.
### In all of these circumstances, the heart rate increase is a normal response. Likewise, the sinus node signals the heart to slow down during rest or relaxation. We see patients who are concerned because their heart rate stays elevated in the range of 100 to 130 beats per minute.
23 Oct 2018 A low resting heart rate and high maximum heart rate are associated with Your pulse, both at rest and during exercise, can reveal your risk for 7 Feb 2019 Also, do not take the pulses on both sides of the neck at the same time. Doing so can slow the flow of blood to the head and lead to fainting. Once 18 Dec 2019 The left ventricle does the bulk of the work, pumping your blood through The easiest way to measure heart rate is to find your pulse and count the prediction equation is: Maximal heart rate is equal to 220 minus your age.
### Your resting heart rate (RHR) is the number of times your heart beats per minute ( bpm) The body needs fewer heartbeats to pump the same amount of blood. Your pulse rate will rise if you do any activity, get up, or eat, drink, or smoke.
28 Feb 2019 Roberts explains that a lower heart rate will deliver the same blood If you experience symptoms such as irregular heart beat, dizziness, slope, but which would have different origins at 70 C. Type E is the same as type C in that the pulse rates lie on a straight line, but has been separated from the What is a normal heart rate? This will allow you to work out your average daytime blood pressure and this may be a better way of assessing your risk of future A normal heart rate for a healthy adult is between 60 and 100 beats per minute. as children grow, and usually by the teen years the heart rate is in the same range as an adult's. Atrial fibrillation is the most common type of fast heartbeat.
## Determining your heart rate is easy; just take your pulse and count the beats for a "What's even more important than a single heart rate is the trend," she says. ( Important note: Heart rate and blood pressure aren't the same thing, and they
23 Oct 2018 A low resting heart rate and high maximum heart rate are associated with Your pulse, both at rest and during exercise, can reveal your risk for 7 Feb 2019 Also, do not take the pulses on both sides of the neck at the same time. Doing so can slow the flow of blood to the head and lead to fainting. Once 18 Dec 2019 The left ventricle does the bulk of the work, pumping your blood through The easiest way to measure heart rate is to find your pulse and count the prediction equation is: Maximal heart rate is equal to 220 minus your age. Basically, discrepancy in the rate cannot happen as the origin of the pulse in both sides is the same, the heart. But actually we may see in some conditions Pulse equality refers to whether the pulse force is comparable on both sides of the body. For example, palpate the radial pulse on the right and left wrist at the
Determining your heart rate is easy; just take your pulse and count the beats for a full minute. But that info is most useful if you track it over time and tell your doctor about any substantial 4. Myth: If my heart rate is normal, my blood pressure is fine. Sometimes your heart rate and your blood pressure go hand in hand. For example, when you exercise, or get angry or scared, they both Heart rate is the speed of the heartbeat measured by the number of contractions (beats) of the heart per minute (bpm). The heart rate can vary according to the body's physical needs, including the need to absorb oxygen and excrete carbon dioxide. It is usually equal or close to the pulse measured at any peripheral point. In all of these circumstances, the heart rate increase is a normal response. Likewise, the sinus node signals the heart to slow down during rest or relaxation. We see patients who are concerned because their heart rate stays elevated in the range of 100 to 130 beats per minute. | 1,484 | 6,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.92263 |
https://sthalles.github.io/logistic-regression/ | 1,716,164,496,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00505.warc.gz | 485,434,416 | 8,230 | ## Introduction
In the last post, we tackled the problem of Machine Learning classification through the lens of dimensionality reduction. We saw how Fisher’s Linear Discriminant can project data points from higher to smaller dimensions. The projection follows two principles.
• It maximizes the between-class variance.
• It minimizes the within-class variance.
Even though Fisher’s method is not (in essence) a discriminant, we built one by modeling a class conditional distribution using a Gaussian. First, we found the prior class probabilities p(Ck). Then, we used Bayes theorem to find the posterior class probabilities p(Ck|x). Here, x is the input vector and Ck is the label for class k.
In short, we can categorize ML models based on the way they classify data. There are two types: generative and discriminative methods.
Generative methods learn the posterior class probabilities explicitly. As opposed to it, discriminative algorithms learn the posterior class probabilities directly.
Intuitively, it has a nice geometrical interpretation. For each class, generative models are concerned to find a probabilistic representation of the data. On the contrary, discriminative algorithms focus on separating the data by decision boundaries.
In other words, generative models try to explain the data by building a statistical model for each class.
On the other hand, the goal of a discriminative algorithm is to find an optimal decision boundary that separates the classes. Thus, as long as there is a decision surface, such models do not care about the distributions of the data.
Take a binary classification problem as an example. Given an input vector x, we need to decide to which class Ck, x is most likely to belong to. To make this decision, both types of ML algorithms need a way to compute the posterior probability p(Ck|x) from the training data.
For Fisher’s, we explicitly learned posterior class probabilities using a Gaussian. Once we found it, we used decision theory to determine class membership for x.
For a discriminative model, the posterior p(Ck|x) will be directly derived. In this case, once we have the posterior, we can use decision theory and assign x to the most probable class.
## Logistic Regression
Before we begin, make sure you follow along with these Colab notebooks.
Logistic Regression is probably the best known discriminative model. As such, it derives the posterior class probability p(Ck|x) implicitly.
For binary classification, the posterior probabilities are given by the sigmoid function σ applied over a linear combination of the inputs ϕ.
Similarly, for multiclass problems, we can estimate the posterior using the softmax function. Like the sigmoid, softmax normalizes a given vector to probabilities — values between 0 and 1.
Let’s begin with the case of binary classification.
## Binary Logistic Regression
For an M-dimensional input feature-vector, Logistic Regression has to learn M parameters.
Take the SVHN dataset as an example. Each RGB image has a shape of 32x32x3. Thus, logistic regression needs to learn 32x32x3=3072 parameters.
To find the optimal weight values, we usually optimize the cross-entropy error function.
The cross-entropy, or the negative logarithm of the likelihood, measures how far the model’s predictions are from the labels. It increases when the predicted values deviate from the targets and decrease otherwise.
Assuming target values t to be either 0 or 1, cross-entropy is defined as:
Here, N denotes the total number of instances in the dataset and yᵢ are the model’s probabilities.
Cross-entropy compares 2 probability distributions. Because of that, it is important to note that the output of logistic regression is interpreted as probabilities — even during learning.
Taking the derivative of the cross-entropy with respect to the weight vector w, we get the gradient.
Note that to compute this derivative, we need the derivative of the sigmoid function w.r.t weights w.
Luckily, one nice property of the sigmoid is that we can express its derivative in terms of itself.
The gradient is a vector-valued function. In fact, the gradient is a linear transformation that maps input vectors to other vectors of the same shape.
The gradient captures the derivative of a whole multi-variable function. Each one of its values denotes the direction in which we can change one specific weight so that we can reach the maximum point of a function. Thus, the gradient represents the direction of steepest ascent.
## Softmax Regression
For multiclass classification, only a few things change. Now, we can model the posterior probabilities using a softmax function.
Since logistic regression treats its predictions as probabilities, we need to change the way we represent our labels.
Up to this point, the target/label vector is represented as a vector of integers. In this vector, each value represent a different class. If we want them to be equally valued probabilities, they need to be between 0 and 1. To do this, we can change their representation to one-hot-encodings.
This time, for inputs with M features and K different classes, logistic regression learns MxK parameters. We can view it as the following matrix.
Now, we can proceed similarly to the case of binary classification. First, we take the derivative of the softmax with respect to the activations. Then, the negative logarithm of the likelihood gives us the cross-entropy function for multi-class classification.
In practice, cross-entropy measures the distance between two vectors of probabilities. One, that comes out of the softmax. And a second containing the one-hot-encoding representations of the true target values.
Note the difference between the error functions used for binary and multiclass classification. In reality, they are really the same thing.
The binary cross-entropy treats the targets as scalars. They take either 0 or 1. For multiclass problems, targets are represented as one-hot-encoded vectors.
Finally, we take the gradient of the error function w.r.t the weights w and obtain the following gradient vectors.
## Iterative Reweighted Least Squares
Different from linear regression, logistic regression does not have a closed form solution. In other words, for linear regression, we can solve for a point of gradient equal 0 with the following equation:
For logistic regression, such a closed-form equation does not exist. Thanks to the nonlinearity we apply on the linear combination of the inputs.
However, the relationship between the loss function and the parameters w still gives us a concave error function. Thus, we can rest assured that there is only a unique minimum on the error surface.
As a result, we can solve for it using an iterative technique such as gradient descent or Newton-Raphson.
If we choose gradient descent, we have everything set.
In plain English, the direction of steepest descent. As such, we can iteratively update the weightsw as:
Note the minus sign there. It represents the fact that we are going downhill as opposed to uphill.*
However, we can do a little better.
Gradient descent offers the direction of steepest descent to the next critical point. And the learning rate controls the length (magnitude) of the step we take in that direction. Take a look at the following image though.
Especially in (c), depending on the magnitude of the step we take, we might get our loss function to rather increasing. To avoid this problem, we take advantage of the information given by the second derivative.
In practice, instead of only taking the derivative of the cross-entropy, we also take the derivative of its derivative. The second derivative, described by f’’(x), gives information about the curvature of a function.
Intuitively, if:
• f’’(x)=0, then there is no curvature.
• f’’(x)<0, it means that the function curves downward.
• f’’(x)>0, there is an upward curvature in the function.
With that information, the update step, for minimizing the cross-entropy takes the form of:
This update equation is called Newton-Raphson.
Note that it multiplies the inverse of the matrix H⁻¹ by the gradient. H, or the Hessian, stores the second derivatives of the cross-entropy w.r.t the weights w.
Let’s now dive into the code.
Taking this toy dataset as an example.
There are 210 points in this 2D coordinate system with 3 classes: blue, red and green circles. Since the number of classes is greater than 2, we can use Softmax Logistic Regression.
First, to introduce the bias variables to our model, we can perform a simple transformation called: fixed basis function. This is done by simply adding a column full of 1s to the input. For binary classification, it makes the corresponding weight value w₀ to play the role of the bias. For multiclass, the first column of the weight matrix act as the biases.
Then, we can create a Logistic Regression object.
Following sklearn’s based API, we can fit and evaluate the model
Note that we can choose between Newton’s and Gradient descent update rules. While Newton’s method tends to converge faster, it needs to compute and store a full Hessian in each iteration. Besides, the Hessian needs to be invertible — for parameter update.
For a matrix to be invertible, there are some constraints that must be true.
First, H has to be a square matrix. Second, the columns of H need to be linearly independent. It means that for any column i of H, i cannot be represented as a linear combination of any other column j.
Since for really large matrices these constraints are likely not to hold, we can use pseudo-inverse techniques here. That is what the function pinv(H) does on the code below.
That is the same to say that the columns of H span the coordinate system. Or that the determinant of H is non-zero.
Even though for smaller datasets, this might not be a problem, the Hessian tends to grow as the number of feature and classes increase. To have an idea, using inputs with M features and a dataset with K classes. The full Hessian has shape [MK, MK]; which for this example mean: [9x9] — remember, we added a new feature column to the inputs.
For the CIFAR-10 dataset, each RGB image has a shape of 32x32x3. That means storing and inverting a square matrix of shape [30720x30720]. Using float 32-bit precision, the Hessian requires 3.775 GB (gigabytes).
To close up, have a look at the fitted model using the toy dataset and Newton’s method. The crosses are the test data.
Enjoy.
In summary, generative models are a class of ML algorithms that learn the class probabilities explicitly.
• They usually perform well with fewer training examples.
• They can deal with missing data.
• Can be used for supervised and unsupervised learning.
Discriminative models learn the class probabilities implicitly.
• In general, they require more labeled data.
• They often have fewer assumptions and fewer parameters.
• But, can only be used for supervised training.
For binary classification, Logistic Regression uses the sigmoid function to represent the posterior probabilities. For multiclass classification, it uses softmax.
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https://mathoverflow.net/questions/312256/sidon-sets-and-diophantine-equation | 1,544,686,421,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824525.29/warc/CC-MAIN-20181213054204-20181213075704-00328.warc.gz | 657,410,968 | 30,311 | # Sidon Sets and Diophantine Equation
Suppose $$X$$ is a subset of $$\{1, \cdots, n\}$$ such that the equation $$ax_i+bx_j=cx_k+dx_{\ell}$$ where $$a+b=c+d,$$ $$a,b,c,d \in \mathbb{N}$$ and $$x_i, x_j, x_k, x_{\ell} \in X,$$ has only trivial solution. A solution is trivial if $$x_i=x_j=x_{k}=x_{\ell}.$$
What can we say about the size of $$X?$$ Is this possible that $$|X|\geq n^{1-o(1)}$$?
I think the answer is related to Sidon Sets, but I could not find any references. Any help is greatly appreciated.
• are $a,b,c,d$ fixed? Else there are solutions like $a=c,b=d$, $x_i=x_k$, $x_j=x_l$. – Fedor Petrov Oct 7 at 19:53
• also maybe you need the reverse inequality for $|X|$? – Fedor Petrov Oct 7 at 20:01
• @FedorPetrov: No, I need a lower bound. – Kim Oct 7 at 20:02
• Then "is it true" must be read as "is it possible"? And what is $\epsilon$? – Fedor Petrov Oct 7 at 20:07
• what does it mean "$\epsilon$ is a positive constant depends on $n$?" – Fedor Petrov Oct 7 at 20:53
Such $$X$$ do indeed exist, and are explicitly constructed in I.Z. Ruzsa, Solving a linear equation in a set of integers, Acta Arith., LXV.3 (1993), pp. 259-282, Theorem 7.5. The whole paper is devoted to upper and lower bounds on sizes of solution-free sets to equations such as the one you are asking about. A mathscinet forward search from that paper should yield further results. | 452 | 1,370 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-51 | longest | en | 0.867788 |
https://mathoverflow.net/questions/389653/groups-with-three-conjugacy-classes-that-define-an-ordering/389669 | 1,627,224,647,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151699.95/warc/CC-MAIN-20210725143345-20210725173345-00391.warc.gz | 399,057,313 | 30,680 | # Groups with three conjugacy classes that define an ordering
Consider the following property for a group $$(\mathcal{G},\cdot,1)$$:
There are exactly three conjugacy classes $$\{1\}$$, $$\mathcal{C}_1$$, $$\mathcal{C}_2$$ in $$\mathcal{G}$$, and we have $$\mathcal{C}_1 \mathcal{C}_1 \subseteq \mathcal{C}_1$$ and $$\mathcal{C}_2=\mathcal{C}^{-1}_1$$.
Note that the only finite groups with exactly three conjugacy classes are the cyclic group of order $$3$$ and the symmetric group of order $$6$$. Those do not satisfy the property above. So any such group must be infnite, hence also non-abelian.
In fact, given such a group $$\mathcal{G}$$, define $$x if and only if $$y \cdot x^{-1} \in \mathcal{C}_1$$. This defines a linear order on $$\mathcal{G}$$ for which it is bi-ordered, i.e. with $$\forall x,y,z \in \mathcal{G},\ y>1 \Longleftrightarrow x \cdot y \cdot z> x \cdot z$$. So it is perhaps best to think of those groups as linearly bi-ordered groups where any two strictly positive elements are conjugated.
I suspect not so many such groups are known. So my question is: are there known examples of such groups? Or better yet: have they been studied to some extent?
One could also expect such groups to be related to the first-order theory $$T$$ of linearly bi-ordered non-trivial groups with trivial center. The question would be: does any algebraically closed model of $$T$$ satisfy the property above? Indeed I believe that László Fuchs proved that in the case of partially bi-ordered, lattice ordered groups, the algebraically closed models have the property that any two strictly positive elements are conjugated. In that case however there may still be more conjugacy classes than three because not every element must be positive or negative.
• As far as I'm aware, this is an open problem: see Problem 3.31 of arxiv.org/abs/0906.2621. – shane.orourke Apr 8 at 14:43
• In fact, you can even say that such a group cannot be finitely generated, since f.g. bi-orderable groups are indicable (i.e., admit an epimorphism onto $\mathbb Z$), hence they must contain infinitely many conjugacy classes. – Ashot Minasyan Apr 8 at 15:10
• What do you mean by an "algebraically closed model of $T$"? – Alex Kruckman Apr 8 at 15:15
• @shane.orourke : Thanks for this article, a lot of related and more general problems there! Strangely the reference for Problem 3.31 is the "Black Swamp Problem Book" which I cannot find, so I'll try to find another reference and post it here. – nombre Apr 8 at 15:28
• @AlexKruckman I mean a model $\mathcal{M}$ of $T$ such that if $\mathcal{N}\supseteq \mathcal{M}$ is a model of $T$ and $\varphi[x]$ is a conjunction of equations which has a solution in $\mathcal{N}$, then it has a solution in $\mathcal{M}$. So it's a little weaker than being existentially closed. – nombre Apr 8 at 15:31 | 793 | 2,837 | {"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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-31 | longest | en | 0.903929 |
https://cboard.cprogramming.com/tech-board/166585-very-fast-hashing-single-integer-post1229446.html?s=74f9446413e4c1620663172c2737c879 | 1,623,898,308,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626465.55/warc/CC-MAIN-20210617011001-20210617041001-00215.warc.gz | 160,203,638 | 14,909 | # Thread: Very fast hashing of single integer
1. ## Very fast hashing of single integer
I am looking for a very fast (at most a few CPU cycles) algorithm to hash an 64-bit integer into another 64-bit integer.
The main objective is that similar input values should produce very different output values.
It does not need to be cryptographically secure.
One obvious approach is to pre-compute a random number for each possible input value, and use that as the hash, but the table would be way too big for 64-bit.
It can be done for each part (for example, 16-bit parts) instead, with the results xor-ed together, but that's quite a few instructions -
Code:
```rand0[4][65536]; // all randomly generated
uint16_t parts[4];
parts[0] = x & 0xffff;
parts[1] = (x >> 16) & 0xffff;
parts[2] = (x >> 32) & 0xffff;
parts[3] = (x >> 48) & 0xffff;
hash = rand[0][parts[0]] ^ rand[1][parts[1] ^ rand[2][parts[2] ^ rand[3][part[3]]```
It also has the problem that the big rand table takes up 2MB, and will generate many cache misses.
Does anyone know of a faster way?
Any algorithm with L1 misses will probably be too slow, since an L1 miss is about 10 cycles, or about 40 simple arithmetic instructions (4 instructions per cycle on most modern Intel CPUs).
2. Also, the input domain is numbers with up to 8 arbitrary bits set.
3. O_o
You aren't mixing any bits of the source that way.
*shrug*
I guess one would try using smaller primes to mutate bytes or words within the string if one required a hash in the fewest instructions possible.
Code:
```unsigned long long Hash
(
unsigned long long fValue
)
{
const unsigned long long kPrime = (~0 - 83 + 1);
const unsigned long long kMultiplier = 0x00F100F100F100F1;
fValue ^= fValue * kMultiplier;
fValue ^= fValue * kPrime;
return(fValue);
}```
I don't really like such an idea without serious testing, but I like not using the actual value in the mix even less.
Soma
4. Is something like MurmurHash too slow?
5. Originally Posted by phantomotap
O_o
You aren't mixing any bits of the source that way.
*shrug*
I guess one would try using smaller primes to mutate bytes or words within the string if one required a hash in the fewest instructions possible.
Code:
```unsigned long long Hash
(
unsigned long long fValue
)
{
const unsigned long long kPrime = (~0 - 83 + 1);
const unsigned long long kMultiplier = 0x00F100F100F100F1;
fValue ^= fValue * kMultiplier;
fValue ^= fValue * kPrime;
return(fValue);
}```
I don't really like such an idea without serious testing, but I like not using the actual value in the mix even less.
Soma
Thanks, will definitely give that a try.
Is there anything fundamentally wrong with not using bits in the input, if it satisfies all the properties of a good hash (or is there a property that doesn't satisfy)?
6. Originally Posted by MutantJohn
Is something like MurmurHash too slow?
I have never heard of it but just looked it up. Pretty cool!
It does seem to be a little too slow, though.
7. Is there anything fundamentally wrong with not using bits in the input, if it satisfies all the properties of a good hash (or is there a property that doesn't satisfy)?
O_o
I realize that the circumstances would be unlikely, but the generated values could conceivably cancel out in any of several ways.
Let's pretend that `rand[1][\$1] ^ rand[2][\$2] ^ rand[3][\$3]' evaluates to `\$4' regardless of the corresponding values.
If any other `rand[1][\$4] ^ rand[2][\$5] ^ rand[3][\$6]' evaluates to `\$4', the lowest bits are solely responsible for changing the hashed value.
Yes. I again do realize the circumstances would be unlikely. I just don't like the idea than any bit could ever be excluded from changing the computed hash.
Soma
8. Originally Posted by phantomotap
O_o
I realize that the circumstances would be unlikely, but the generated values could conceivably cancel out in any of several ways.
Let's pretend that `rand[1][\$1] ^ rand[2][\$2] ^ rand[3][\$3]' evaluates to `\$4' regardless of the corresponding values.
If any other `rand[1][\$4] ^ rand[2][\$5] ^ rand[3][\$6]' evaluates to `\$4', the lowest bits are solely responsible for changing the hashed value.
Yes. I again do realize the circumstances would be unlikely. I just don't like the idea than any bit could ever be excluded from changing the computed hash.
Soma
Is there a theoretical basis for this being a bad thing?
In this case, I wouldn't say the higher bits are excluded from changing the computed hash. If they were different, the computed hash would be different, so they had just as much of a role in determining the hash as the lower bits.
9. Is there a theoretical basis for this being a bad thing?
O_o
Yes. Of course, the research is focused on cryptography.
Yes. I know that you don't care about the strength of the function.
I only said I don't like the idea.
In this case, I wouldn't say the higher bits are excluded from changing the computed hash. If they were different, the computed hash would be different, so they had just as much of a role in determining the hash as the lower bits.
No. The higher order bits would not play any role in determining the hash in situation I described.
You should really just read the explanation again if you don't understand the situation I described, but I will offer an example.
Code:
```// ...
Table[1][0] = 641671;
// ...
Table[2][0] = 751345;
// ...
Table[3][0] = 179318;
// ...```
Code:
```Value = Value & 0xffff;
// ...
Word[0] = Value & 0xffff; // Word[0] = Value;
Word[1] = (Value >> 16) & 0xffff; // Word[1] = 0;
Word[2] = (Value >> 32) & 0xffff; // Word[2] = 0;
Word[3] = (Value >> 48) & 0xffff; // Word[3] = 0;
// ...
Fragment = Table[1][Word[1]] ^ Table[2][Word[2]] ^ Table[3][Word[3]]; // Fragment = 0;
// ...
Hash = Fragment ^ Table[Word[0]]; // Hash = Table[Word[0]];```
Code:
```// ...
Table[3][(1 << 15)] = 179318;
// ...```
Code:
```Value = Value & 0xffff;
// ...
Value |= 1 << 63;
// ...
Word[0] = Value & 0xffff; // Word[0] = Value;
Word[1] = (Value >> 16) & 0xffff; // Word[1] = 0;
Word[2] = (Value >> 32) & 0xffff; // Word[2] = 0;
Word[3] = (Value >> 48) & 0xffff; // Word[3] = 1 << 15;
// ...
Fragment = Table[1][Word[1]] ^ Table[2][Word[2]] ^ Table[3][Word[3]]; // Fragment = 0;
// ...
Hash = Fragment ^ Table[Word[0]]; // Hash = Table[Word[0]];```
Code:
```Value1 = Value & 0xffff;
Value2 = Value1 | (1 << 63);
Hash(Value1) == Hash(Value2);```
As I said, the circumstances would be unlikely.
I still don't like the idea that any bit could ever be excluded from changing the computed hash.
Soma
10. I have read your example multiple times, and still don't think I get it.
Are you essentially trying to say Table[3][1 << 15] can happen to be equal to Table[3][0]? Because that's the only case the 2 Fragments can be equal, assuming the other entries of the table are the same.
Isn't that essentially the same as finding a collision in the 3-words hash function (that generates Fragment)?
Does this property make collisions more likely? Since all the table entries are randomly generated, I would think that makes this function more uniformly distributed, not less?
11. Out of curiosity, is this too many instructions for you?
Code:
```int hash( int k )
{
k *= 357913941;
k ^= k << 24;
k += ~357913941;
k ^= k >> 31;
k ^= k << 31;
return k;
}```
12. Originally Posted by MutantJohn
Out of curiosity, is this too many instructions for you?
Code:
```int hash( int k )
{
k *= 357913941;
k ^= k << 24;
k += ~357913941;
k ^= k >> 31;
k ^= k << 31;
return k;
}```
That seems fast enough (extended to 64-bit). What's the story behind that?
13. Oh, it's just some code I found from the authors of gFlip3D so I don't know much about it but it seems quite condensed. I also trust the authors' wisdom. Their other code is pretty good so I'm trusting that this as well.
This is the code in pure source form :
Code:
```// originally CUDA code
__forceinline__ __device__ float hash( int k )
{
k *= 357913941;
k ^= k << 24;
k += ~357913941;
k ^= k >> 31;
k ^= k << 31;
return int_as_float( k );
}```
Source : 3D Delaunay Triangulation on the GPU
The code is in KerPredicates.cu of gDel3D.
Though I'm not sure about it being originally intended for floating point numbers. I don't know if that'd make a good integer hash. I guess the only way would be to test it.
14. Originally Posted by MutantJohn
Oh, it's just some code I found from the authors of gFlip3D so I don't know much about it but it seems quite condensed. I also trust the authors' wisdom. Their other code is pretty good so I'm trusting that this as well.
This is the code in pure source form :
Code:
```// originally CUDA code
__forceinline__ __device__ float hash( int k )
{
k *= 357913941;
k ^= k << 24;
k += ~357913941;
k ^= k >> 31;
k ^= k << 31;
return int_as_float( k );
}```
Source : 3D Delaunay Triangulation on the GPU
The code is in KerPredicates.cu of gDel3D.
Though I'm not sure about it being originally intended for floating point numbers. I don't know if that'd make a good integer hash. I guess the only way would be to test it.
Ah! Good old CUDA.
The constants are 0x155555..., so it basically shifts the number by all even amounts, then xor them together. And to make sure the lower bits aren't too static, the upper bits are mixed into lower bits.
I suppose that could work.
Will just have to verify experimentally that this doesn't cause too many collisions.
15. Does this property make collisions more likely?
O_o
Let me try a different approach to explaining the reason behind my comment: despite having a 64bit result, you don't really have a 64bit hash. The 64bit result of each component is just an arbitrarily lengthened 16bit hash. You have four 16bit hashes which you've combined with the `XOR' operation. I'm basically guaranteed to find a collision in 262144 attempts.
Soma | 2,669 | 9,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-25 | latest | en | 0.862058 |
http://math.stackexchange.com/questions/207583/is-this-question-erroneous-stationary-points/207591 | 1,469,448,400,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824226.79/warc/CC-MAIN-20160723071024-00135-ip-10-185-27-174.ec2.internal.warc.gz | 151,786,677 | 17,158 | # Is this question erroneous? (Stationary points)
Using the second partial derivative test, I have found (-1,1) to be a saddle point but this option is not available in the MCQ. Have I made a mistake?
The person who set the question insists that (-1,1) is a minimum, not saddle, because:
"We are talking about the local situation at the 2 points. Since the y coordinate of the 2 points are the same at y=1, that means we pass a vertical plane through the 3-D surface, so it becomes a 2-D curve in terms of x. So using 2nd derivative test (as stated explicitly in the question), it leads to local minimum. "
-
The given function can be written as $$f(x,y)=(1-x^2)^2 -(y-1)^2 -7\ .$$ It follows by inspection that $(0,1)$ is a local maximum. Now when $x=-1$ then $1-x^2=0$, and for all $x$ in a punctured neighborhood of $-1$ one has $(1-x^2)^2>0$. It follows that $f(x,1)>f(-1,1)$ for these $x$; on the other hand we have $f(-1,y)<f(-1,1)$ for all $y\ne 1$. So $(-1,1)$ is definitely a saddle point.
Since $(0,1)$ is a maximum, only option 2. and 5. are left. Since $(-1,1)$ is not a minimum (5.), and further you found it's a saddle point, which is neither a local minimum nor local maximum ($\to$ inconclusive), I would go for option 2. | 370 | 1,242 | {"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} | 3.4375 | 3 | CC-MAIN-2016-30 | latest | en | 0.933227 |
https://physics.stackexchange.com/questions/680911/physical-meaning-of-the-terms-in-reynolds-transport-theorem?noredirect=1 | 1,718,865,165,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00669.warc.gz | 402,445,133 | 42,204 | # Physical meaning of the terms in Reynolds transport theorem
Reynolds transport theorem gives the time variation of the amount of a magnitude $$\mathbf f = \mathbf f(\mathbf x,t)$$ contained within a time-dependent control volume $$\Omega(t)$$ by
$$\frac{d}{d t} \int_{\Omega(t)} \mathbf{f} d V =\underbrace{\int_{\Omega(t)} \frac{\partial \mathbf{f}}{\partial t} d V}_{1}+\underbrace{\int_{\partial \Omega(t)}\left(\mathbf{v}^{b} \cdot \mathbf{n}\right) \mathbf{f} d A}_{2}$$
where $$\mathbf n(\mathbf x,t)$$ is the outward-pointing unit normal vector, $$\mathbf x$$ is a point in the region and is the variable of integration, $$dV$$ and $$dA$$ are volume and surface elements at $$\mathbf x$$, and $$\mathbf v^b(\mathbf x,t)$$ is the velocity of the area element.
I see that the term $$2$$ is the amount of $$\mathbf f$$ that flows across the surface of the control volume $$\partial \Omega(t)$$. However, what would the meaning of the term 1 be, and how does it differ from the LHS term?
I have seen this related question, but it does not fully answer my doubt.
• The LHS is a rate of change which includes the changing boundary. Term 1 is a rate of change which does not account for the changing boundary. These are my observations as a novice, hopefully they helps you down the right path. Commented Dec 6, 2021 at 11:07
It's actually pretty simple.
LHS: Rate of change in a quantity $$\bf f$$ in a volume $$\Omega$$ is equal to
(1): the rate of change in quantity $$\bf f$$ in the volume itself, plus
(2): the net influx of the quantity across the boundaries of $$\Omega$$
The easiest example is mass. If you have a fluid flowing, and an imaginary box (the control volume), the total change in mass inside the box equals the change in density of the fluid inside (usually zero for liquids or low-speed flows of gases) plus the total of the inflow and outflow of fluid across the boundaries.
In the case of mass which cannot have a source or sink in a 3 dimensional fluid, this means that e.g. if you have density decreasing in a control volume, like due to an expanding gas, any decrease in the amount of mass inside the volume must correspond to a net outflow across the boundaries. The mass has to go somewhere. In an incompressiblec flow of a liquid, say, the density doesn't change, so your term (1) is zero, and the equation will require that any inflow must be balanced by an equal outflow. "What goes in must come out elsewhere."
Other things that can be tracked on this way are: momentum, which gives us the velocity equations; energy, which gives us the energy/heat transfer equation; entropy, which gives us yet another equation, usually used more in compressible or turbulent flows.
Unlike mass, momentum and energy can have sources and sinks in the flow, which is why we need to add terms accounting for forces (viscosity, pressure, gravity) that add or remove momentum, and heat sources (viscous dissipation, external heating, internal heat generation due to chemical reaction) that add or remove energy.
But the full set of Navier Stokes equations are taking 3 conversation laws:
Mass is conserved
Momentum is conserved
Energy is conserved
And then saying "each one of these three in any control volume is the sum of the change in that quantity in the CV, plus the net influx/outflux to the CV."
Perhaps the simplest way to understand the difference between the two terms is to place oneself in a situation for which the integral noted 1 is zero, ie in steady state.
Imagine for example a nozzle with a gas which enters on the left with a low speed and this same gas which leaves on the right with a high speed. If we are in steady state, the quantity of gas in the nozzle does not change, nor does its momentum. If we apply Reynolds' theorem to the momentum of the nozzle, the integral (1) represents the derivative with respect to time of the momentum in the nozzle. But there is always the same momentum inside this nozzle: it does not change and therefore the integral (1) is zero.
Now isolate in thought the matter that is in the nozzle at any given time and follow it for a short while. It is a closed system. He moved a little left and right. The small part on the right has a high speed while on the left, we "lost" a part which had a low speed. In total, the momentum of the closed system that we follow over time has changed and its derivative is non-zero. It is the LHS of the Reynolds equation.
It should be understood that even in steady state, while the fields do not depend on time, a closed system moves within these fields and the quantities associated with it can vary. We find the same idea for the Eulerian acceleration which can be non-zero even if the velocity field is permanent. A fluid particle moves within this permanent field and its speed changes over time.
Hope it can help and sorry for my poor english.
I have the answer to your question. First, I'll note that while the wiki proof is no doubt correct, there are much better, i.e. comprehensible, illustrated proofs available at https://www.youtube.com/watch?v=hpCYVluvC_o and https://www.youtube.com/watch?v=hpCYVluvC_o The conceptual proof is easy but the math symbolism obscures it big time in the absence of illustrative diagrams, IMO.
The lhs of the RTT in the wiki proof represents the time derivative of integral of f over the domain while first integral on the rhs integrates the partial derivative of f over the domain. When the theorem is proved the integral on the right also has the derivative outside the integral and the integrals look the same, but they are not.
To apply the theorem to determine the lhs of the Navier-Stokes equations, for example, the derivative is moved inside the integral on the right, justified by the Leibniz integration rule, and evaluated for in infinitesimal control volume.
If you could move the derivative under the integral of the lhs of the RTT then the integrals for the control volume would be equal. However, you cannot do that because boundaries of the lhs integral change with time and the simple form of the Leibniz integration rule used on the rhs integral does not apply to the lhs integral. | 1,432 | 6,177 | {"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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-26 | latest | en | 0.898381 |
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+ (Xa) ^_^'s Discussions (543)
Discussions Replied To (12) Replies Latest Activity
"ye assignment just practice k lye hai ... iskay marks nai hain "
+ (Xa) ^_^ replied Jan 28, 2012 to MTH301 assignment 04 NON-GRADED, only for practice
4 Jan 29, 2012
"you have to solve it further .... "t" ki value nikal k x, y and z mein dalni hai ...…"
+ (Xa) ^_^ replied Jan 5, 2012 to MTH301 Assignment 03..... Due date: January 04, 2012
5 Jan 5, 2012
"Q 2 ??? why ? you only have to submit Q5"
+ (Xa) ^_^ replied Nov 22, 2011 to MTH301 Calculas II Assignment no 2
20 Nov 22, 2011
"in which part ur facing difficulty???"
+ (Xa) ^_^ replied Nov 20, 2011 to MTH301 Calculas II Assignment no 2
20 Nov 22, 2011
"ok thnx for confirming the ans :)"
+ (Xa) ^_^ replied Nov 20, 2011 to MTH301 Calculas II Assignment no 2
20 Nov 22, 2011
"that's good :)"
+ (Xa) ^_^ replied Nov 20, 2011 to MTH301 Calculas II Assignment no 2
20 Nov 22, 2011
"@ASIF .... Solve the question, if u face any difficulty u can ask ... my be i can he…"
+ (Xa) ^_^ replied Nov 19, 2011 to MTH301 Calculas II Assignment no 2
20 Nov 22, 2011
"mera answer minimum aya hai.... let me know when you solve Q5 ... i just need to con…"
+ (Xa) ^_^ replied Nov 18, 2011 to MTH301 Calculas II Assignment no 2
20 Nov 22, 2011
"who solved Q#5 ? i need to confirm my answer. is it maximum, minimum or saddle?"
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https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH104/December_2010/Question_03 | 1,716,617,055,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058773.28/warc/CC-MAIN-20240525035213-20240525065213-00803.warc.gz | 514,305,063 | 11,686 | # Science:Math Exam Resources/Courses/MATH104/December 2010/Question 03
MATH104 December 2010
Other MATH104 Exams
### Question 03
A city's supply of an herb comes from many small home businesses. The supply q of the herb is a function of the street price p. The Elasticity of Supply is by definition equal to
${\displaystyle -{\frac {p}{q}}{\frac {{\textrm {d}}q}{{\textrm {d}}p}}.}$
Suppose that a city's weekly supply q of the herb, in millions of grams, is related to the street price, p, in dollars per gram, by the equation
${\displaystyle \displaystyle {}p^{3}+p=2q^{3}+q^{2}+10.}$
Calculate the elasticity of supply when p=3. Please simplify. Note that when p=3, we have q=2.
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.
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# Calculations Concerning Rotry Cement Kiln
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### Rotary Kiln Design Calculations
basic calculations of rotary kiln design Prominer . heat calculation in rotary kiln provesprojektde cement rotary kiln calculation goedkoopcdpersen nl Cement kiln heat balance calculation cement kilns design features of rotary kilns design features of rotary cement kilns the shell of the kiln is made of mild steel plate mild steel is the only viable material for the purpose but presents the ...
### Modelling And Optimization Of A Rotary Kiln Direct
rotary kiln. This is determined by (i) the maximum flow rate of the proposed burden through a kiln, and (ii) the residence time at temperatures that would allow sufficient reduction to take place. A calculation method was developed that allows for the prediction of the bed profile and residence time in a rotary kiln.
### Heat Balance Analysis In Cement Rotary Kiln Science
Apr 02, 2019 The purpose of this study is optimizing the air and fuel quantities at kiln considering design parameters of the cement plant by keeping adequate safety factors at each level of calculations to assure that neither production rate nor quality of the clinker vary. Analysis of Heat balance was used to determine the sources of heat loss from the kiln system.
### Rotary Kilns For Cement Plants Flsmidth
modern cement plant. The 2-base rotary kiln is a statically determined beam system, where the support load is always known. It is a shorter and wider kiln, with a reduced slope to maintain the material retention time, similar to a 3-base kiln. The larger diameter reduces
### Rotary Cement Kiln Simulator Rocks Integrated Modeling
May 01, 2007 Rotary kiln. The partially calcined raw meal is passed slowly to the rotary kiln where the clinkerization reactions occur. In the initial part of the kiln the remaining calcination occurs. Other solid–solid and solid–liquid clinkerization reactions take place as the solid bed moves towards the burner. Part of the solids melts in the kiln.
### The Effects Of Rotary Kiln Operating Conditions And
Rotary kilns have been important for decades as calciners for various processes, especially cement production. More recently, rotary kilns have been adopted as an effective means for the incineration of hazardous wastes. This is par-ticularly true of solid hazardous wastes such as sludges, sorbents, and contaminated soils. Rotary kilns are able to
### Impact Of Coating Layers In Rotary Cement Kilns Numerical
Mar 01, 2020 The rotary kiln selected for the simulations is a generic model which has been created on the basis of actual industrial kilns and burners. It is based on the furnace modeled and described in . The generic rotary kiln is regarded as part of a modern cement plant with preheater and calciner. The kiln has a total length of 40 m.
### Waste Incineration And Emission Control By
kilns like preheated kilns and precalciner kilns the gas temperature in the burning zone is about 2.000 C, at mid-kiln it is about 1.700 C, and at the kiln exit it is about 1.100 C. The gas retention time is about 5 seconds. The large size of kilns and the quantity …
### A Study On The Failure Of Steel Chains In Rotary Cement Kilns
Jan 30, 2018 Kufa cement plant depends on a wet process production with 4881 chains inside the rotary cement kiln, and each chain includes 45 rings. Heavy oil has been used as a fuel during cement kiln continuous operation. Two types of steel grades are in service, DIN 1.4742 grade with ring dimension of (80 ϕ 20 thick) usually located at the first four meters of the kiln chains curtain.
### Cement Final Report 70514 Final4
Cement Kiln Report (FINAL – 7/14/2006) 1-1 1.0 INTRODUCTION This study was conducted to explore potential NOx emissions reduction strategies for cement kilns in Ellis County. This project included assessing existing NOx control technologies as well as new technologies that have not been previously considered by TCEQ.
### Numerical Modelling Of The Calcination Process In A
additives and ground to the final product cement. A precalciner kiln system is the basis for this study. It normally consists of a preheater, a precalciner (also known as a calciner), a rotary kiln and a cooler. The kiln feed (i.e. the raw meal) is heated in the preheater and …
### Parametric Studies Of Cement Production Processes
The cement industry is one of the most intensive energy consumers in the industrial sectors. The energy consumption represents 40% to 60% of production cost. Additionally, the cement industry contributes around 5% to 8% of all man-made CO 2 emissions. Physiochemical and thermochemical reactions involved in cement kilns are still not well understood because of their complexity.
### Rotary Kiln Design Calculations
basic calculations of rotary kiln design Prominer . heat calculation in rotary kiln provesprojektde cement rotary kiln calculation goedkoopcdpersen nl Cement kiln heat balance calculation cement kilns design features of rotary kilns design features of rotary cement kilns the shell of the kiln is made of mild steel plate mild steel is the only viable material for the purpose but presents the ...
### Heat Balance Analysis In Cement Rotary Kiln Science
Apr 02, 2019 The purpose of this study is optimizing the air and fuel quantities at kiln considering design parameters of the cement plant by keeping adequate safety factors at each level of calculations to assure that neither production rate nor quality of the clinker vary. Analysis of Heat balance was used to determine the sources of heat loss from the kiln system.
### Linear Identification Of Rotary White Cement
figure 1, a schematic of the kiln with its cyclone pre-heater is shown. Figure 1: Rotary Cement kiln Process. In this paper we will use a black box identification procedure for modelling the Saveh white cement kiln. It is a 65 m long, 4.7 m diameter kiln with 4 stage double string pre-heater and water immersion cooler.
### Identification Of Refractory Material Failures
The heading “Identification of failure of refractory materials in cement materials” appearing on top of each page (page 1 to page 6) should be in same font throughout the text such as the one appearing on page 7 and for the rest of the next.
### The Effects Of Rotary Kiln Operating Conditions And
Rotary kilns have been important for decades as calciners for various processes, especially cement production. More recently, rotary kilns have been adopted as an effective means for the incineration of hazardous wastes. This is par-ticularly true of solid hazardous wastes such as sludges, sorbents, and contaminated soils. Rotary kilns are able to
### Rotary Kiln Design Calculations Rotary Kiln Design
The rotary kiln consists of cylindrical shell, supporting device, supporting device of thrust roller, driving unit, moveable kiln head, sealing device on the tail of kiln, coal dust pipe device, etc. 2.Why you should choose our rotary kiln 9 These new technologies provide our rotary kiln with strong visual sense, convenient operation and reliable performance. and complete sets of machinery for ...
### Simulation Of A Cement Plant Using Thermochemical And
Keywords: Thermochemical process modelling, Cement manufacturing process, Rotary kiln 1. Introduction Most of today’s world wide cement production is carried out in rotary kiln plants (see Taylor, 1990). A powdery raw mixture of mainly CaCO 3, SiO 2, Fe 2O 3 and Al 2O 3 is passed through a multi-stage cyclone preheater and then fed into the ...
### Rotary Kiln Calculations Rotary Kiln Calculations
The rotary kiln consists of cylindrical shell, supporting device, supporting device of thrust roller, driving unit, moveable kiln head, sealing device on the tail of kiln, coal dust pipe device, etc. 2.Why you should choose our rotary kiln 9 These new technologies provide our rotary kiln with strong visual sense, convenient operation and reliable performance. and complete sets of machinery for ...
### Cement European Commission
Cement 110 of kiln. It remains for ca. 10-20 min at this temperature and is cooled as quickly as possible after leaving the kiln. In Europe, cement clinker is predominantly burnt in rotary kilns; shaft kilns are seldom used. Rotary kilns are refractory-lined tubes with a diameter up to about 6 m. They are
### Duoflex Burner Flsmidth
rotary cement kiln must fulfil the follow-ing requirements: Key benefits - Robust design - Low primary air consumption - Adjustable swirl - Adjustable air nozzle area - Central fuel injection. 3 • The burner must be able to fire coal, coke, fuel oil and natural gas or any
### Kiln Drive Application Considerations Cimentec
In the cement manufacturing process the kiln is at the heart of the operation and its design and application requirements are unique to the cement industry. Its main use is in the chemical transformation of raw mix into clinker. Speed variation is required to control ... converts this torque into a rotary …
### Temperature Scanning Of Rotary Kiln Shell Fluke Process
Rotary kilns are traditionally large steel pipes, several meters in diameter and up to 100 m in length (in larger facilities, however, kiln lengths of over 100 m are not uncommon). To protect the steel from intense heat, however, each kiln is lined with bricks composed of refractory material, which is typically a blend of various ceramic compounds.
### Cement Rotary Kiln International Cement Review
CEMENT ROTARY KILN Questions & Answers Question-1: What is the maximum continuous shell temperature a kiln stands without permanent damage to the shell? Answer-1: The maximum recommended kiln shell temperature varies by plant, by country and by kiln manufacturer, despite the fact that most kiln shells are made of low alloy carbon steel.
### Modelling And Optimization Of A Rotary Kiln Direct
rotary kiln. This is determined by (i) the maximum flow rate of the proposed burden through a kiln, and (ii) the residence time at temperatures that would allow sufficient reduction to take place. A calculation method was developed that allows for the prediction of the bed profile and residence time in a rotary kiln.
### Study Of A Full Scale Oxyfuel Cement Rotary Kiln
Apr 01, 2019 The flow and combustion in the cement rotary kiln have been simulated using the ANSYS Fluent 17.2 RANS code. The rotary kiln model dimensions have been setup according to a full scale 3000 t/d of cement with a total length of 60 m and inner diameter of 3.76 m, except from the clinker sintering zone (the first 20 m from the burner end of the rotary kiln) where the coating thickness …
### Rotary Kilns For Cement Plants Flsmidth
modern cement plant. The 2-base rotary kiln is a statically determined beam system, where the support load is always known. It is a shorter and wider kiln, with a reduced slope to maintain the material retention time, similar to a 3-base kiln. The larger diameter reduces
### Cement Rotary Kiln Calculation Goedkoopcdpersennl
cement kiln heat balance calculation. Cement Kilns: Design features of rotary kilns Design features of rotary cement kilns , The shell of the kiln is made of mild steel plate Mild steel is the only viable material for the purpose, but presents the problem that the maximum temperature of the feed inside the kiln is over 1400 C, while the gas temperatures reach 1900 C [Chat Online]...
### Impact Of Coating Layers In Rotary Cement Kilns Numerical
Mar 01, 2020 The rotary kiln selected for the simulations is a generic model which has been created on the basis of actual industrial kilns and burners. It is based on the furnace modeled and described in . The generic rotary kiln is regarded as part of a modern cement plant with preheater and calciner. The kiln has a total length of 40 m.
### Cement Rotary Kiln International Cement Review
CEMENT ROTARY KILN Questions & Answers Question-1: What is the maximum continuous shell temperature a kiln stands without permanent damage to the shell? Answer-1: The maximum recommended kiln shell temperature varies by plant, by country and by kiln manufacturer, despite the fact that most kiln shells are made of low alloy carbon steel.
### Modelling And Optimization Of A Rotary Kiln Direct
rotary kiln. This is determined by (i) the maximum flow rate of the proposed burden through a kiln, and (ii) the residence time at temperatures that would allow sufficient reduction to take place. A calculation method was developed that allows for the prediction of the bed profile and residence time in a rotary kiln.
### Study Of A Full Scale Oxyfuel Cement Rotary Kiln
Apr 01, 2019 The flow and combustion in the cement rotary kiln have been simulated using the ANSYS Fluent 17.2 RANS code. The rotary kiln model dimensions have been setup according to a full scale 3000 t/d of cement with a total length of 60 m and inner diameter of 3.76 m, except from the clinker sintering zone (the first 20 m from the burner end of the rotary kiln) where the coating thickness …
### Rotary Kilns For Cement Plants Flsmidth
modern cement plant. The 2-base rotary kiln is a statically determined beam system, where the support load is always known. It is a shorter and wider kiln, with a reduced slope to maintain the material retention time, similar to a 3-base kiln. The larger diameter reduces
### Cement Rotary Kiln Calculation Goedkoopcdpersennl
cement kiln heat balance calculation. Cement Kilns: Design features of rotary kilns Design features of rotary cement kilns , The shell of the kiln is made of mild steel plate Mild steel is the only viable material for the purpose, but presents the problem that the maximum temperature of the feed inside the kiln is over 1400 C, while the gas temperatures reach 1900 C [Chat Online]...
### Impact Of Coating Layers In Rotary Cement Kilns Numerical
Mar 01, 2020 The rotary kiln selected for the simulations is a generic model which has been created on the basis of actual industrial kilns and burners. It is based on the furnace modeled and described in . The generic rotary kiln is regarded as part of a modern cement plant with preheater and calciner. The kiln has a total length of 40 m. | 3,231 | 14,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-31 | latest | en | 0.902185 |
https://rd.springer.com/article/10.1007/s10958-007-0164-8 | 1,521,548,902,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647406.46/warc/CC-MAIN-20180320111412-20180320131412-00314.warc.gz | 719,683,035 | 18,931 | Journal of Mathematical Sciences
, Volume 143, Issue 1, pp 2773–2790
# Factorization of the R-matrix. I
• S. E. Derkachov
Article
## Abstract
We study the general rational solution of the Yang-Baxter equation with the symmetry algebra sℓ(3). The R-operator which acts in the tensor product of two arbitrary representations of the symmetry algebra can be represented as a product of simpler “building blocks,” R-operators. The R-operators are constructed explicitly and have a simple structure. In such a way, we construct the general rational solution of the Yang-Baxter equation with the symmetry algebra s sℓ(3). To illustrate the factorization in the simplest situation, we treat also the sℓ(2) case. Bibliography: 23 titles.
## Keywords
Tensor Product Invariant Subspace Symmetry Algebra Casimir Operator Invariant Solution
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.
## References
1. 1.
P. P. Kulish and E. K. Sklyanin, “On solutions of the Yang-Baxter equation,” Zap. Nauchn. Semin. LOMI, 95, 129 (1980).
2. 2.
M. Jimbo, “Introduction to the Yang-Baxter equation,” Int. J. Mod. Phys, A4, 3759 (1983); Yang-Baxter Equation in Integrable Systems, Adv. Ser. Math. Phys., 10, World Scientific, Singapore (1990).
3. 3.
V. G. Drinfeld, “Hopf algebras and Yang-Baxter equation,” Soviet Math. Dokl., 32, 254 (1985); “Quantum Groups,” in: Proc. Int. Congress Math., Berkeley (1986), AMS, Providence (1987), pp. 798.Google Scholar
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V. G. Drinfeld, “Quasi-Hopf algebras,” Leningrad. Math. J., 1, 1419 (1990).
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V. Terras, “Drinfeld twists and functional Bethe ansatz,” Lett. Math. Phys., 48, 263 (1999); H. Pfeiffer, “Factorizing twists and the universal R-matrix of the Yangian Y (sℓ 2),” J. Phys. A: Math. Gen., 33, 8929 (2000).
6. 6.
P. P. Kulish and E. K. Sklyanin, “Quantum spectral transform method. Recent developments,” Lect. Notes Phys., 151, 61 (1982); L. D. Faddeev, “How algebraic Bethe ansatz works for integrable model,” Les-Houches Lect. (1995), hep-th/9605187; E. K. Sklyanin, “Quantum inverse scattering method. Selected topics,” in: Mo-Lin Ge (ed.), Quantum Group and Quantum Integrable Systems (Nankai Lectures in Mathematical Physics), World Scientific, Singapore (1992), pp. 63–97; hep-th/9211111.
7. 7.
P. P. Kulish, N. Yu. Reshetikhin, and E. K. Sklyanin, “Yang-Baxter equation and representation theory,” Lett. Math. Phys., 5, 3930–403 (1981).
8. 8.
N. MacKay, “Rational R-matrices in irreducible representations,” J. Phys., A 24, 4017 (1991); R.-B. Zhang, M. Gould, and A. Bracken, “From the representation of the braid group to solutions of the Yang-Baxter equation,” Nucl. Phys., B 354, 625 (1991); M. Gould and Y-Z. Zhang, “ R-matrices and the tensor product graph method,” hep-th/0205071.
9. 9.
E. K. Sklyanin, “Private communication.”Google Scholar
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D. P. Zhelobenko, Compact Lie Groups and Their Representations, AMS, Providence, Rhode Island (1973).
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J. J. de Swart, “The Octet model and its Clebsch-Gordan coeficients,” Rev. Mod. Phys., 916 (1963).Google Scholar
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M. Shifman, ITEP Lectures on Particle Physics and Field Theory, 2, 775–875 (1999); World. Sci. Lect. Notes Phys., 62.
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P. P. Kulish and N. Yu. Reshetikhin, “On GL(3)-invariant solutions to the Yang-Baxter equation and the associated quantum systems,” Zap. Nauchn. Semin. LOMI, 120, 92 (1982).
14. 14.
A. Molev, “Yangians and their applications,” (2002), math.QA/0211288.Google Scholar
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A. Molev, M. Nazarov, and G. Olshanski, “Yangian and classical Lie algebras,” Russian Math. Surveys, 51:2, 205–282 (1996).
16. 16.
P. P. Kulish, “Yang-Baxter equation and reflection equations in integrable models,” hep-th/9507070.Google Scholar
17. 17.
S. I. Alishauskas and P. P. Kulish, “Spectral resolution of the su(3)-invariant solutions to the Yang-Baxter equation,” Zap. Nauchn. Semin. LOMI, 145, 3 (1985).
18. 18.
E. K. Sklyanin, “Classical limits of the Yang-Baxter equation,” J. Soviet. Math., 40, 93 (1988).
19. 19.
S. Derkachov, D. Karakhanyan, and R. Kirschner, “Universal R-matrix as an integral operator,” Nucl. Phys., B 618, 589 (2001).
20. 20.
L. N. Lipatov, “High-energy asymptotics of multicolor QCD and exactly solvable lattice models,” JETP Lett., 59, 596 (1994); L. N. Lipatov, “Duality symmetry of reggeon interactions in multicolor QCD,” Nucl. Phys., B 548, 328 (1999).Google Scholar
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L. D. Faddeev and G. P. Korchemsky, “High-energy QCD as a completely integrable model,” Phys. Lett., B342, 311 (1995).Google Scholar
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D. Karakhanian and R. Kirschner, “Conserved currents of the three-reggeon interaction,” hep-th/9902147; “High-energy scattering in gauge theories and integrable spin chains,” hep-th/9902031; Fortschr. Phys., 48, 139 (2000).Google Scholar
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C-N. Yang, “Some exact results for the many-body problem in one dimension with repulsive delta-function interaction,” Phys. Rev. Lett., 19, 1312 (1967). | 1,604 | 5,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-13 | longest | en | 0.752015 |
https://testbook.com/objective-questions/ml/mcq-on-cost-of-capital--5f916a15595206050362e888 | 1,722,747,865,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640389685.8/warc/CC-MAIN-20240804041019-20240804071019-00181.warc.gz | 468,207,023 | 52,389 | # Cost of Capital MCQ Quiz in मल्याळम - Objective Question with Answer for Cost of Capital - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on May 4, 2024
നേടുക Cost of Capital ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Cost of Capital MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്സി, റെയിൽവേ, യുപിഎസ്സി, സ്റ്റേറ്റ് പിഎസ്സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക
## Top Cost of Capital MCQ Objective Questions
#### Cost of Capital Question 1:
ABC Ventures is private equity investor considering investing Rs. 1,000 million in the equity of XYZ ltd. ABC Ventures requires a return of 30% on investment with planned holding period of 5 years. The likely debt of XYZ ltd. will be Rs. 1000 million and cash balance of Rs. 300 million with a projected EBITDA of Rs. 1500 million for the year 5. The desired ownership share of ABC Ventures in XYZ ltd., given FVF0.30,5 = 3.713, will be:
1. 37.13%
2. 30.00%
3. 37.90%
4. 7.13%
#### Answer (Detailed Solution Below)
Option 3 : 37.90%
#### Cost of Capital Question 1 Detailed Solution
The correct answer is 37.90%.
Key Points Assuming an interest rate of 10% for the debt of XYZ Ltd., we can proceed with the calculations.
Given:
• Investment amount: Rs. 1,000 million
• Required return: 30% per year
• Holding period: 5 years
• Debt of XYZ Ltd.: Rs. 1,000 million
• Cash balance of XYZ Ltd.: Rs. 300 million
• Projected EBITDA of XYZ Ltd. for year 5: Rs. 1,500 million
• FVF0.30,5 (Future Value Factor at 30% for 5 years): 3.713
• Interest rate for XYZ Ltd.'s debt: 10%
• First, let's calculate the net cash flows for year 5:
• Interest Expense = Debt * Interest Rate
• = Rs. 1,000 million * 10% = Rs. 100 million
Net Cash Flow (Year 5) = EBITDA - Interest Expense + Cash Balance
= Rs. 1,500 million - Rs. 100 million + Rs. 300 million
= Rs. 1,700 million
To calculate the present value of the net cash flows for year 5:
PV (Year 5) = Net Cash Flow (Year 5) / (1 + Required Return)^5
= Rs. 1,700 million / (1 + 0.30)^5
= Rs. 1,700 million / 2.48832
= Rs. 683.27 million
Next, let's calculate the present value of the investment amount:
PV (Investment Amount) = Investment Amount / (1 + Required Return)^5
= Rs. 1,000 million / (1 + 0.30)^5
= Rs. 1,000 million / 2.48832
= Rs. 402.65 million
Finally, the desired ownership share can be calculated using the present value of the investment amount and the present value of the net cash flows for year 5:
Ownership Share = PV (Investment Amount) / (PV (Investment Amount) + PV (Year 5))
= Rs. 402.65 million / (Rs. 402.65 million + Rs. 683.27 million)
≈ 0.3790or 37.90%
Therefore, the desired ownership share of ABC Ventures in XYZ Ltd. would be approximately 37.90%.
#### Cost of Capital Question 2:
What is the tax shield if a firm has Rs. 100 depreciation and Rs. 50 amortization, and the tax bracket is 30 percent ?
1. 45
2. 15
3. 30
4. 50
Option 1 : 45
#### Cost of Capital Question 2 Detailed Solution
The correct answer is 45.
Key Points Tax Shield:
• In finance, the tax shield is the reduction in taxable income resulting from tax-deductible expenses, such as depreciation and amortization. The tax shield represents the tax savings that a company receives from deducting these expenses from its taxable income.
• To calculate the tax shield in this scenario, we can use the following formula:
• Tax Shield = Tax Rate x Tax-Deductible Expenses
• where the tax-deductible expenses include items such as depreciation and amortization.
• In this scenario, the tax-deductible expenses are Rs. 100 (depreciation) + Rs. 50 (amortization) = Rs. 150. The tax rate is given as 30 percent.
• Thus, the tax shield would be:
• Tax Shield = 0.30 x Rs. 150 = Rs. 45
• This means that the company can reduce its taxable income by Rs. 150 due to depreciation and amortization expenses, resulting in a tax savings of Rs. 45 (i.e., 30% of Rs. 150) due to the tax shield.
#### Cost of Capital Question 3:
Net Income approach to valuation is based on which of the following assumption?
1. There are taxes and cost of debt is less than cost of equity
2. There are taxes and use of debt does not change the risk perception of investors
3. There are no taxes and use of debt changes the risk perception of investors
4. There are no taxes and cost of debt is less than the cost of equity and the use of debt does not change the risk perception of investors.
#### Answer (Detailed Solution Below)
Option 4 : There are no taxes and cost of debt is less than the cost of equity and the use of debt does not change the risk perception of investors.
#### Cost of Capital Question 3 Detailed Solution
Key Points
Net Income approach of firm's valuation:
According to the theory, a company's value can be raised by lowering its total cost of capital, which is calculated using the weighted average cost of capital. This can be accomplished by using more debt, which is a less expensive form of funding than equity.
Important Points Assumptions
• Diminishing debt will not change the confidence/perception of the investors
• There are no expenses like the cost of floatation, transaction cost, or corporate dividend taxes.
• Cost of raising debt is cheaper as compared to cost of equity due to tax benefit
• All investors have access to information about all important companies since the capital market is functioning flawlessly
• All investors are rational and seek to minimize risk while maximizing profit.
• Uniform dividend payout ratio i.e. 1 must be used by all businesses
• In theory, debt and equity are the only sources of funding that are taken into account.
Weighted Average Cost of Capital
WACC, or weighted average cost of capital, is the cost of equity and debt divided by the total amount of capital raised from all sources.
$$WACC = \frac{\text{(Required rate of Return}\times {\text {Amount of Equity})+}\text{Cost of Debt} \times{\text{Amount of Debt}}}{Debt + Equity}$$
#### Cost of Capital Question 4:
According to the traditional approach, what is the effect of increase in degree of leverage on the valuation of a firm?
1. Increases
2. Decreases
3. Remains unaffected
4. Increases first and then decreases
#### Answer (Detailed Solution Below)
Option 4 : Increases first and then decreases
#### Cost of Capital Question 4 Detailed Solution
The correct answer is Increases first and then decreases Key Points
• Traditional theory of capital structure:
• The traditional theory of capital structure says that for any company or investment there is an optimal mix of debt and equity financing that minimizes the WACC and maximizes value.
• Under this theory, the optimal capital structure occurs where the marginal cost of debt is equal to the marginal cost of equity.
• This theory depends on assumptions that imply that the cost of either debt or equity financing vary with respect to the degree of leverage.
• The traditional theory of capital structure says that a firm's value increases to a certain level of debt capital, after which it tends to remain constant and eventually begins to decrease if there is too much borrowing
• In essence, the firm faces a trade-off between the value of increased leverage against the increasing costs of debt as borrowing costs rise to offset the increased value. Beyond this point, any additional debt will cause the market value and to increase the cost of capital
#### Cost of Capital Question 5:
Which of the sources of finance has an implicit cost of capital?
1. Equity share capital
2. Preference share capital
3. Debentures
4. Retained earnings
#### Answer (Detailed Solution Below)
Option 4 : Retained earnings
#### Cost of Capital Question 5 Detailed Solution
Implicit cost
1. They are a specific type of opportunity cost: the cost of resources already owned by the firm that could have been put to some other use.
2. For example, an entrepreneur who owns a business could use her labor to earn income at a job.
3. Implicit cost is that cost that has already been occurred but is not shown or reported as a separate expense.
4. Its an opportunity cost that arises when a company uses internal sources towards a project without any compensation for the utilization of the resource.
5. Retained earnings is an internal source.
Explicit Cost:
• Explicit costs are out-of-pocket costs for a firm—for example, payments for wages and salaries, rent, or materials.
• The implicit cost of retained earnings is the return that could have been earned by the investors, had the profit been distributed to them.
• Except for the retained earning, all other sources of funds have explicit costs of capital.
#### Cost of Capital Question 6:
Which of the following source of finance has an implicit cost of capital?
1. Equity share capital
2. Debentures
3. Retained earnings
4. Preference share capital
#### Answer (Detailed Solution Below)
Option 3 : Retained earnings
#### Cost of Capital Question 6 Detailed Solution
The correct answer is Retained Earnings.
Key Points Retained earnings has an implicit cost of capital. This is because retained earnings are essentially the profits that the company has chosen not to distribute to its shareholders. This means that the shareholders have forgone the opportunity to earn a return on this capital by investing it elsewhere.
The implicit cost of capital of retained earnings is the opportunity cost of the shareholders, i.e., the return that they could have earned if they had invested the money elsewhere. This return is typically estimated using the weighted average cost of capital (WACC) of the company.
The other sources of finance listed do not have an implicit cost of capital:
• Debt financing: The company pays a fixed interest rate to the lenders for the use of the debt. This interest expense is an explicit cost, not an implicit cost.
• Equity financing: The company issues shares to investors in exchange for cash. The investors expect to earn a return on their investment, either through dividends or capital gains. This return is an explicit cost, not an implicit cost.
Therefore, the correct answer is retained earnings.
#### Cost of Capital Question 7:
The cost in the process of raising fund through equity is known as ______ .
1. Financial risk
2. Cost of debt
3. Floating cost
4. Cost of capital
#### Answer (Detailed Solution Below)
Option 3 : Floating cost
#### Cost of Capital Question 7 Detailed Solution
The correct answer is Floating costs.
Key Points
• Floating costs:
• These costs refer to the expenses incurred during the process of issuing new shares or securities in the financial market.
• They include underwriting fees, legal fees, registration fees, and other costs associated with the marketing of new stock.
• Floating costs are critical for financial enterprises as they impact the net proceeds from the issuance of equity.
• Understanding and managing these costs are crucial for setting the pricing of the issued securities to ensure the fundraising is cost-effective.
• Financial risk:
• Refers to the uncertainty associated with financial loss that a company might face due to its financial operations.
• Financial risk does not directly relate to the costs of raising funds but rather to the impact of financial decisions and market conditions.
• Cost of debt:
• Represents the effective rate that a company pays on its borrowed funds from financial institutions or through bond issuance.
• While important, it specifically refers to borrowing costs and not to equity financing.
• Cost of capital:
• Encompasses the overall return that a company must earn on its investment projects to maintain the value of its stock and satisfy its creditors.
• This cost includes both debt and equity costs, providing a comprehensive measure of the expense of financing the company’s operations.
#### Cost of Capital Question 8:
Which kind of risk affects the opportunity cost of capital?
1. Non-diversifiable risk
2. The risk inherent in "riskless" portfolios such as broad market holdings
3. Diversifiable risk
4. Both non-diversifiable and diversifiable risk
#### Answer (Detailed Solution Below)
Option 1 : Non-diversifiable risk
#### Cost of Capital Question 8 Detailed Solution
The correct answer is Non-diversifiable risk.
Key PointsNon-diversifiable Risk:
• Opportunity cost of capital is the return that could have been earned by investing in an alternative investment with equivalent risk. Non-diversifiable risk, also known as systematic risk, is the risk that is inherent in the entire market or a whole asset class, and cannot be diversified away.
• Examples of non-diversifiable risk include macroeconomic factors such as inflation, interest rates, and geopolitical events, as well as industry-specific factors such as regulatory changes and technological disruptions.
• Since non-diversifiable risk is inherent in the entire market or asset class, it affects all investors and all investments within that market or asset class. As a result, it affects the opportunity cost of capital for all investors, as the risk-adjusted return required to invest in a particular asset or market will be higher if the non-diversifiable risk is perceived to be higher.
• In other words, non-diversifiable risk increases the required return on investment, which in turn affects the opportunity cost of capital. Investors will require a higher return to compensate them for the increased risk, which will lead to a higher opportunity cost of capital.
hence, the correct answer is Non-diversifiable risk.
#### Cost of Capital Question 9:
Company's current price of share is Rs. 60 and dividend per share is Rs. 4 . If its capitalisation rate is 12 percents. What is dividend growth rate?
1. 10%
2. 20%
3. 3.2%
4. 5%
Option 4 : 5%
#### Cost of Capital Question 9 Detailed Solution
The correct answer is 5%.
Key Points
• Dividend growth calculates the annualized average rate of increase in the dividends paid by a company.
• Calculating the dividend growth rate is necessary for using a dividend discount model for valuing stocks.
• A history of strong dividend growth could mean future dividend growth is likely, which can signal long-term profitability.
• Formula of Gordon Growth Model :
Now it is given that P = 60, D1 = 4, r = 0.12, g = ?
60 = 4 / (0.12 - g)
0.12 - g = 4 / 60
0.12 - g = 0.6667
0.12 - 0.667 = g
So g = 0.547 or 5 % (approximately)
Hence, the correct answer is 5%.
#### Cost of Capital Question 10:
Match the following:
LISTI LIST II A. Realized Yield Method 1. Cost of equity share capital B. Taxation 2. Cost of equity capital C. Cost of total capital employed 3. Cost of debt capital D. Dividend Growth is a consideration 4. Weighted cost of capital
1. A-4, B-2, C-3, D-1
2. A-2, B-3, C-4, D-1
3. A-1, B-2, C-3, D-4
4. A-4, B-3, C-2, D-1
#### Answer (Detailed Solution Below)
Option 2 : A-2, B-3, C-4, D-1
#### Cost of Capital Question 10 Detailed Solution
The correct answer is A-2, B-3, C-4, D-1.
Key Points
LISTI LIST II A. Realized Yield Method 2. Cost of equity capital B. Taxation 3. Cost of debt capital C. Cost of total capital employed 4. Weighted cost of capital D. Dividend Growth is a consideration 1. Cost of Equity share capital
Important Points A - Realized Yield Method: This method is used to calculate the cost of equity capital. It takes into account the actual returns earned by investors who have invested in the company's shares, rather than using estimates or assumptions about future returns.
B - Taxation: Taxation is an important consideration in calculating the cost of debt capital. The after-tax cost of debt is used to determine the weighted average cost of capital (WACC) of a company, which takes into account the cost of both equity and debt capital.
C - Cost of total capital employed: The cost of total capital employed is the overall cost of financing a company's operations, which includes both debt and equity capital. This cost is used to determine the company's WACC.
D - Dividend Growth is a consideration: The growth rate of dividends is an important consideration in calculating the cost of equity capital. The cost of equity is the return required by investors to invest in the company's shares, and it is based on the expected future dividends and the growth rate of those dividends.
In summary, the matching is as follows:
A matches with 2 because the Realized Yield Method is used to calculate the cost of equity capital.
B matches with 3 because taxation is an important consideration in calculating the cost of debt capital.
C matches with 4 because the cost of total capital employed is used to determine the company's WACC.
D matches with 1 because the growth rate of dividends is an important consideration in calculating the cost of equity share capital. | 4,297 | 16,835 | {"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} | 3.359375 | 3 | CC-MAIN-2024-33 | latest | en | 0.39305 |
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Excel assignment | Statistics homework help.
• Type all your answers in a single MS word document. Copy and paste relevant Excel output reports, graphs, and/or tables in the same document. You need not retype the questions, but label your answers with the question numbers, e.g., 1)______
• Please keep the Excel file containing all of your analyses and results.
• You will be graded based on the completeness and correctness of your answers. Simply submitting the Excel output, table, and/or plots without any explanation will not be sufficient.
BASEBALL FORECAST
Paul Raymond, a math savvy baseball manager, would like to apply his knowledge in statistics to develop a multiple regression model to forecast pitching performances for starting pitchers for this baseball season. He intended to use the baseball statistics from the last year season to build the model. He understood that the initial variable selection was the most important aspect of developing a regression model. He knew that, if he didn’t have good predictor variables, he wouldn’t end up with useful predicting equations.
Paul had spent considerable amount of time to download the baseball statistics for starting pitchers from the last year season. He also decided to include only starting pitchers who had pitched at least 100 innings during the last season. The data for the 138 pitchers selected is presented in the Excel filename “DS312Fall21 Excel assignment#2_data.xlsx”
Paul decided that Earned Run Average (ERA) is the best indicator of performance and so wanted to develop a regression model to predict this variable. He chose the six potential predictor variables as follow,
WHIP: Number of walks plus hits given up per inning pitched
CMD: Command of pitches, the ratio strikeouts/walks
K/9: How many batters a pitcher strikes out per game (nine innings pitched)
HR/9: Opposition homeruns per gram (nine innings pitched)
OBA: Opposition batting average
THROWS: Right-handed pitcher (1) or left-handed pitcher (0)
Examine the correlation matrix, which predictor variables are highly correlated with ERA. Are there potential correlations among predictor variables? Explain in detail.
Based on the results from part 1, which predictor variable(s) should be excluded from the model? Use the remaining predictor variables to develop a multiple regression model to predict Earned Run Average (ERA). Present the Excel summary report and do the following tasks:
2.1. conduct the t-test (using α = 0.01) to determine which predictor variables, if any, are significant. Provide the details of the test and comment on the results of the test.
2.2. conduct the F-test (using α = 0.01) to test the overall significance of the model. Provide the details of the test and comment on the results of the test.
2.3. comment on the values of r2 and adjust r2
3. Based on the results from part 2, if we want to have a simplified regression model with fewer predictor variables, which variables should be dropped? Explain clearly why.
4. Based on your conclusion from part 3, develop the simplified regression model with the predictor variables you decide to keep. Evaluate the model by conducting the t-test and F-test. Comment on the values of r2 and adjust r2. Comment on how good this model is compared to the model in part 2.
Excel assignment | Statistics homework help
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posted by .
How do i enter my answer?
so the question is Simplify:
√3:(63x^5/(72y^6))
and for my answer i got
x√3:(7x^2/(2y^2))
but my work page says: Enter example: For 4√3:(2), enter 4(2)^(1/3) so how do i enter my answer
• algebra2 -
If ã3:(63x^5/(72y^6)) mean
ã 3 /( 63 x ^ 5 / ( 72 y ^ 6 ) )
then :
ã 3 / ( 63 x ^ 5 / ( 72 y ^ 6 ) ) =
ã 3 * 72 y ^ 6 / ( 63 x ^ 5 ) =
ã 3 * 9 * 8 * y ^ 6 / ( 9 * 7 x ^ 5 ) =
8 ã 3 * y ^ 6 / 7 x ^ 5
• algebra2 -
ã =square root
• algebra2 -
simplify
fourth root sqrt 32x^11y^15/fourth root sqrt 2x^3y^-2
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I just want to see if my answers are correct and if no, can you please show your work and explain how to do it?
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posted by .
Ann sells apples and Beth sells oranges. Today, half the number of apples than Ann sold was fifteen less than thrice as much the number of oranges that Beth sold.
I just want it to know; how do I convert it into an algebraic expression or equation? Is confusing and I don't get it. Can someone please help?
Hope this helps :)
1/2A=3o-15
Thanks! :-)
## Similar Questions
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## Matrix Visual Metadata Levels need to ignore blanks
Hi,
I have a problem in matrix visual and i hope maybe someone can help me on this.
In this example I have 2 markets called (Zara and Zara Home) and they have different locations. I have created 10 leves of location and a market can have 1, 2, etc levels depends or the details (picure below and only the 01/01/2019 but the example has value from Dec 2018 to Jan 2019)
So, you can see that the 2 markets doesn't have all the levels because they don't need. So, I need to have the sum of VALOR and do the matrix visual (picture below) and then i will have the values by all the metadata and dates.
The problem here is that we can see all the levels (including blanks) and I only want to see the levels with names. For that, i have tried (but these don't solve all the problems).
• First i have created 2 measures in order to know the number of the level and the maximum levels that the market has.
```Asset_Max_lentgh = value(max(Table1[PATHLENGH]))
Asset_path = ISFILTERED(Table1[LEVEL 1])+ISFILTERED(Table1[LEVEL 2])+ISFILTERED(Table1[LEVEL 3])+ISFILTERED(Table1[LEVEL 4])+ISFILTERED(Table1[LEVEL 5])+ISFILTERED(Table1[LEVEL 6])+ISFILTERED(Table1[LEVEL 7])+ISFILTERED(Table1[LEVEL 8])+ISFILTERED(Table1[LEVEL 9])+ISFILTERED(Table1[LEVEL 10])+ISFILTERED(Table1[TIENDA])```
And the result of this:
Knowing the position of the levels I can 2 different measures
1. VALUE TYPE 1
`VALUE TYPE 1 = IF(Table1[Asset_path]>Table1[Asset_Max_lentgh],BLANK(),sum(Table1[VALOR]))`
In this case, i'm calculating the value only in the leves that the asset path is less or equal to the max lenthg. Result:
So, as you see the system automaticaly ignore the leves that doesn't have value. I lost the name, but it can solve it puting in the next level of the last level that the market has a text.
But the big problem is that here i can see the date hierarchy becasue the visual remove the next levels.
2. VALUE TYPE 2
`VALUE TYPE 2 = IF(AND(Table1[Asset_path]>Table1[Asset_Max_lentgh],Table1[Asset_path]<11),BLANK(),SUM(Table1[VALOR]))`
Here I'm doing a mixure. I need to have the value in the levels that they have name and apart of that need the values in the other metatada apart of levels (tienda, dates, etc) and the result is
But the visual doesn't remove the levels in blank as the Visual for Value Type 1. It seems that the visual remove the blank rows only if there aren't exist any row with values in the bottom.
Can someone help me on this?
*Apart of that i can't upload a file pbi, i don't know why
9 REPLIES 9
Senior Member
## Re: Matrix Visual Metadata Levels need to ignore blanks
How large is your original dataset?
Member
## Re: Matrix Visual Metadata Levels need to ignore blanks
hi @tex628
In rows... millions. and if you refer for levels of metadata... only 10 (is the limit in the database)
Senior Member
## Re: Matrix Visual Metadata Levels need to ignore blanks
Ì dont think it's possible to do what your aiming for, but im not entirely certain.
This might be close to a solution but i'm not sure it's possible with the size of your dataset.
If you in the query loop through your columns to identify which are empty and which hold a value you can place "ZARA" in the highest empty column in the table. In this case that would mean that you place the value "ZARA" in the Level 6 column, since it's lacking a value.
This might not work with how u want it, but its the closest that i can think of
Member
## Re: Matrix Visual Metadata Levels need to ignore blanks
Yes, i did something similar, but the problem is that the levels in blank can have the name of the market or blank. but the problem in the visual is the repeteated blanks or same name that i need to remove.
In reaility in the measure VALUE TYPE 1 i have solved this problem, but the system will not show the other rows like (year, month)
Member
## Re: Matrix Visual Metadata Levels need to ignore blanks
Yes @tex628, i did something similar, but the problem is that the levels in blank can have the name of the market or blank. but the problem in the visual is the repeteated blanks or same name that i need to remove.
In reaility in the measure VALUE TYPE 1 i have solved this problem, but the system will not show the other rows like (year, month)
Senior Member
## Re: Matrix Visual Metadata Levels need to ignore blanks
Do this:
Create calculated columns in powerquery and replace the Levels that you currently have:
`Level 1_ = if [Level 1] <> null [Level 1] else [TIENDA]`
```Level 2_ = if [Level 1_] = [TIENDA] then null
else if [Level 2] <> null then [Level 2] else [TIENDA]```
```Level 3_ = if [Level 2_] = [TIENDA] or [Level 2_] = null then null
else if [Level 3] <> null then [Level 3] else [TIENDA]```
Do this all the way upp to Level 11!
I hope u understand how im thinking
I dont think you can ignore steps in a heirarchy when you have values further down, thats why i dosnt think you can get your VALUE TYPE measure to work! :-/
Member
## Re: Matrix Visual Metadata Levels need to ignore blanks
@tex628 yes, this is that what i did in reality. the problem is not the levels. when i put the visual for tienda it is ok, the problem is when i need to put the same levels and down the dates or another attribute that i will hav ein the future. As the visual will not ignore the blanks becasue there will be values after levels..
It is a big problem, thanks for your time
Senior Member
## Re: Matrix Visual Metadata Levels need to ignore blanks
I have a kind of a hard time understanding exactly what you're saying.
But if you're doing what im proposing you will have no blanks before "ZARA"
This is how i want the result to look:
I'm sorry, i'm not sure that I'm understanding you correctly :/
Member
## Re: Matrix Visual Metadata Levels need to ignore blanks
Yes, you are right, but the problem is not the Levels. the problem is when I need to put something more in the rows visual like Dates, or any metadata that i will have in the future.
I have tried a lot of things and i think that the only solutions is to create more levels taking the info from metadata... but the problem that it is not dinamic and need to create columns levels. | 1,594 | 6,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-22 | longest | en | 0.904139 |
https://dafyomi.co.il/pesachim/points/ps-ps-058.htm | 1,611,776,616,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704832583.88/warc/CC-MAIN-20210127183317-20210127213317-00432.warc.gz | 284,847,482 | 5,317 | 12TH CYCLE DEDICATIONS:
PESACHIM 58 (16 Adar) - dedicated by Avi Berger of in memory of his father, Reb Pinchas ben Reb Avraham Yitzchak, on the day of his Yahrzeit.
1)
WHEN WE OFFER THE TAMID
(a)
(Mishnah): The [afternoon] Tamid is slaughtered at eight and a half hours of the day and it is offered (i.e. the Hakravah is finished) at nine and a half hours;
(b)
On Erev Pesach it is slaughtered at seven and a half hours and it is offered at eight and a half hours [to allow more time for Pesach, which is offered after the Tamid], whether Erev Pesach is on a weekday or Shabbos;
(c)
[The only exception is] when Erev Pesach is on Friday (the Pesach must be offered earlier, to allow time to cook it before Shabbos -] the Tamid is slaughtered at six and a half hours and it is offered at seven and a half hours, and afterwards Korban Pesach.
(d)
(Gemara) Question: What is the source of this (to offer the Tamid between eight and a half and nine and a half hours)?
(e)
Answer #1 (R. Yehoshua ben Levi): "Es ha'Keves Echad Ta'aseh ba'Boker v'Es ha'Keves ha'Sheni Ta'aseh Bein ha'Arbayim" - the plural suggests splitting the afternoon [before and after offering it] into two equal parts:
1.
We allow two and a half hours beforehand, two and a half afterwards, and one hour to offer it.
(f)
Objection (Rava - Mishnah): On Erev Pesach it is slaughtered at seven and a half and it is offered at eight and a half hours, whether Erev Pesach is on a weekday or Shabbos;
1.
If the Torah requires the Tamid to be at eight and a half hours, we could not make it earlier!
(g)
Answer #2 (Rava): Mid'Oraisa, the Tamid may be offered from when shadows point east (after six hours; mid'Rabanan we wait another half hour, until they clearly point east - see note 36 in Appendix);
1.
We learn from Bein ha'Arbayim - from when the sun is in Ma'arav (the west);
2.
On other days we delay the Tamid two hours until eight and a half hours, to allow more time to offer Nedarim and Nedavos (voluntary Olos or Shelamim), for they cannot be offered after the Tamid (Tosfos - we do not delay more, lest we finish [Hadlakas ha'Menorah and Ketores] close to sundown);
i.
"V'Hiktir Aleha Chelvei ha'Shelamim" - Aleha (by offering the afternoon Tamid) Hashlem (finish) all the Korbanos [of the day].
3.
On Erev Pesach, when Pesach is offered after the Tamid, it is slaughtered one hour earlier than usual, at seven and a half hours;
4.
When Erev Pesach is on Friday, the Pesach must be [offered and also] roasted before Shabbos - roasting does not override Shabbos - therefore, we offer the Tamid as early as possible, at six and a half hours.
2)
THE ARGUMENT OF R. YISHMAEL AND R. AKIVA
(a)
(Beraisa - R. Yishmael): We offer the Tamid on Shabbos the same [time] as during the week;
(b)
R. Akiva says, it is offered like it is on Erev Pesach.
(c)
Question: What does this mean?
(d)
Answer #1 (Abaye): R. Yishmael teaches that we offer the Tamid on Erev Pesach on Shabbos the same as when Erev Pesach is on a weekday (seven and a half hours);
1.
R. Akiva says, it is offered like when Erev Pesach is on Erev Shabbos (six and a half hours - there is no reason to delay it, since Nedarim and Nedavos may not be brought on Shabbos).
2.
Our Mishnah equates Erev Pesach on a weekday and on Shabbos (we offer at seven and a half hours) - it is like R. Yishmael.
3.
Question: What do they argue about?
4.
Answer: They argue about whether or not Musafim are offered before or after Bazichei Levonah (spoons of frankincense on Lechem ha'Panim that are offered on Shabbos. We do not offer Musafim before six hours, for the Torah says "Uv'Yom ha'Shabbos," it does not mention morning):
i.
R. Yishmael holds that Musafim are offered before the Levonah - the Musafim are offered at six hours, the Levonah is offered at seven, and the Tamid at seven and a half;
ii.
R. Akiva holds that Musafim are offered after the Levonah - the Levonah is offered at five hours, the Musafim at six, and the Tamid at six and a half.
(e)
Objection (Rava): R. Akiva did not say that it is offered like when Erev Pesach is on Erev Shabbos - he only said 'like on Erev Pesach'!
(f)
Answer #2 (Rava): R. Yishmael teaches that we offer the Tamid on Erev Pesach on Shabbos the same as on a regular weekday (eight and a half hours);
1.
R. Akiva says, it is offered like on a regular Erev Pesach (seven and a half hours, to allow time to offer all the Pesachim afterwards).
2.
Our Mishnah equates Erev Pesach on a weekday and on Shabbos - it is like R. Akiva.
3.
Question: What do they argue about?
4.
Answer: R. Yishmael is concerned lest the meat rot if left too long in the heat (we may not roast it until Motza'ei Shabbos; even though one of the miracles in the Mikdash was that Kodesh meat never rotted (Avos 5:5), we do not rely on miracles);
i.
R. Akiva is not concerned.
58b----------------------------------------58b
(g)
Question: If we are not concerned, we should offer it at six and a half hours (there are no Nedarim and Nedavos to wait for)!
(h)
Answer: R. Akiva holds that Musafim are offered before the Levonah - the Musafim are offered at six hours, the Levonah at seven, and the Tamid at seven and a half.
(i)
Objection (Rabah bar Ula): R. Yishmael did not say 'it is offered on Erev Pesach on Shabbos like on a weekday' - he only said 'it is offered on Shabbos...'!
(j)
Answer #3 (Rabah bar Ula): R. Yishmael teaches that we offer the Tamid on a regular Shabbos like on a regular weekday (eight and a half hours);
1.
R. Akiva says, it is offered like on a regular Erev Pesach (seven and a half hours).
2.
Our Mishnah equates Erev Pesach on a weekday and on Shabbos - it is like both of them.
3.
Question: What do they argue about?
4.
Answer: R. Yishmael decrees not to offer it earlier lest it be offered earlier on a weekday, and people will not be able to offer Nedarim and Nedavos afterwards (R. Chananel; Rashi - perhaps they will offer them afterwards and they will be Pesulim); R. Akiva does not decree.
(k)
Question: If we do not decree, we should offer it at six and a half hours!
(l)
Answer: R. Akiva holds that Musafim are offered before the Levonah - the Musafim are offered at six hours, the Levonah at seven, and the Tamid at seven and a half.
(m)
Question (Beraisa - R. Yishmael): All year round the Tamid is slaughtered at eight and a half hours and it is offered at nine and a half; on Erev Pesach it is slaughtered at seven and a half and offered at eight and a half; when it falls on Shabbos, it is like when it is on Monday;
1.
R. Akiva says, it is like on Erev Pesach.
2.
We understand according to Abaye (R. Yishmael teaches that when Erev Pesach falls on Shabbos, it is like when it is on Monday) - but this is difficult for Rava (he says that R. Yishmael delays it, lest the Pesach rot)!
(n)
Answer: It does not mean [when it falls on Shabbos,] it is like when it falls on Monday, rather, it is like a regular Monday.
(o)
Question (Beraisa - R. Yishmael): When it falls on Shabbos, it is like a regular day of the year;
1.
R. Akiva says, it is like a regular Erev Pesach.
2.
We understand according to Rava - but this is difficult for Abaye!
(p)
Answer: R. Yishmael does not mean it is like a regular day of the year, rather, it is like a regular year [when Pesach is on a weekday]; R. Akiva says, it is like an Erev Pesach on Erev Shabbos.
3)
THE FIRST AND LAST KORBANOS OF THE DAY
(a)
(Beraisa) Question: What is the source that no Korban may precede the morning Tamid?
(b)
Answer: "V'Orach Aleha ha'Olah" - the morning Tamid goes [directly] on the wood, no Korban precedes it.
(c)
Question: How does the verse show this [that the Olah is the morning Tamid]?
(d)
Answer (Rava): "Ha'Olah" refers to the first Olah [that the Torah discusses].
(e)
(Beraisa) Question: What is the source that no Korban may follow the afternoon Tamid?
(f)
(g)
Question: How do we learn this from the verse?
(h)
Answer #1 (Abaye): [Chalavim of] Shelamim may be offered on (after) it (the afternoon Tamid), not after another Korban [offered afterwards. We must say that Zerikah of the Shelamim preceded the Tamid; the Chelev may be offered later, even at night.]
(i)
Objection: According to this, Olos [or other Korbanos] may be offered after the Tamid, as long as we do not offer Shelamim!
(j)
Answer #2 (Rava): We read "ha'Shelamim" like Hashlem - Aleha (by offering the afternoon Tamid) Hashlem (finish) all the Korbanos. | 2,420 | 8,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-04 | latest | en | 0.94262 |
http://electrical-engineering-portal.com/when-transformer-has-a-stomach-pain-and-wants-to | 1,529,698,704,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864795.68/warc/CC-MAIN-20180622201448-20180622221448-00552.warc.gz | 102,759,570 | 12,821 | # When a transformer has stomach pain and wants to…
Home / Technical Articles / When a transformer has stomach pain and wants to…
During normal operation, transformer internal structures and windings are subjected to mechanical forces due to the magnetic forces. These forces are illustrated in Figure 1.
By designing the internal structure very strong to withstand these forces over a long period of time, service life can be extended. However, in a large transformer during a “through fault” (fault current passing through a transformer), forces can reach millions of pounds, pulling the coils up and down and pulling them apart 60/50 times per second.
Notice in Figure 1 that the internal low-voltage coil is being pulled downward, while the high-voltage winding is pulled up, in the opposite direction.
At the same time, the right-hand part of the figure shows that the high- and low-voltage coils are being forced apart.
Keep in mind that these forces are reversing 50/60 times each second. It is obvious why internal structures of transformers must be built incredibly strong.
Many times, if fault currents are high, these forces can rip a transformer apart and cause electrical faults inside the transformer itself. This normally results in arcing inside the transformer that can result in explosive failure of the tank, throwing flaming oil over a wide area.
There are protective relaying systems to protect against this possibility, although explosive failures do occur occasionally.
## How to prevent pain
### Through Fault – Short Circuit withstand considerations
The windings are subject to both radial and axial forces related to the current and flux interactions. Radial forces in the inner winding (normally the LV winding) are in compression while the outer winding (normally the HV winding) forces are in tension.
Design of the windings and bracing must consider the magnitude of these forces and provide adequate strength to withstand them without significant mechanical deformation which could result in a dielectric failure.
The picture below is an example ofa free bucking mechanical failure of an inner winding resulting from radial forces in compression on the winding. Note, even though there is mechanical failure, there wasn’t a dielectric failure of this winding.
Flux fields are dependent of the balance of the ampere turn distribution of the HV and LV windings. When the ampere turns of the HV and LV windings are equal and balanced, the only forces are radial.
DETC taps (De Energized Tap Changer) in the HV windings and LTC (Load Tap Changer) operation result in changes in the ampere turn distribution resulting in axial forces. If the HV and LV windings are not aligned axially or one winding is physically shorter than the other, ampere turn balance is significantly affected and axial forces are magnified.
Autotransformers, low impedance, motor starting duty, transformers with multiple voltages by reconnecting the transformer windings in series and parallel configurations, three winding transformers with dual secondary windings for start up or unit auxiliary service at power plants – all can result in increased axial and radial forces during a short circuit and require special consideration.
### Primary and assembly process (VIDEO)
###### References:
• Presentation on Short-circuit Forces by H. Jin Sim, CTO of SPX-Waukesha
• Transformers Basics, Maintenance and Diagnostics – U.S. Dpt. of the Interior Bureau of Reclamation
• The Design and Performance of Circular Disc, Helical and Layer Windings for Power Transformer Applications – David L. Harris, P.E.
### SEARCH: Articles, software & guides
#### Edvard Csanyi
Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry fascilities. Professional in AutoCAD programming. Present on
1. SZHJ
Feb 27, 2017
Detail description of short circuit test along with calculations of fault current
2. r k mohapatra
Jun 27, 2016
LV coil carrying high current requires thick conductor and it has high radial force withstand capability . hv coil is subjected to elongating force and the that is normally met by tensile strength of the conductor .
so , radial force is normally not a problem for transformer under through fault currnt
pl give opinion—-
thanx—mohapatra
3. r k mohapatra
Jun 27, 2016
buckling is a combination of radial and axial force . if axial force is controlled by support mechanism – then buckling can be avoided .
axial force can be controlled by clamping the inner and outer coils together with then help of top plate , bottom plate and clamping rods .
radial force can be controlled by selecting the no. of radial supports | 984 | 4,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-26 | latest | en | 0.937172 |
http://www.prepjunkie.com/web/test.php?name=mock6_pmt | 1,585,828,353,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506959.34/warc/CC-MAIN-20200402111815-20200402141815-00471.warc.gz | 281,598,200 | 369,871 | Question 1
A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-z plane. If the current in the loop is i. The resultantmagnetic field due to the two semicircular parts at their common centre is
A : µ0i / 2√2 R
B : µ0i / 2R
C : µ0i / 4R
D : µ0i / √2R
•
.
Solution :
B = √(Bxy2 + Bxz2)
= √((µ0i / 4R)2 + (µ0i / 4R)2)
0i√2 / 4R)2
0i / 2√2 R
• Question 2
Two following figure shows a logic gate circuit with two inputs A and B and the output Y. the voltage wave forms of A, B and Y are as given.
The logic gate is:?
A : NOR
B : OR
C : AND
D : NAND
•
.
Solution :
It is clear from given logic circuit, that output Y is low when both the inputs are high, otherwise it is high. Thus logic circuit is NAND gate.
• Question 3
Two parallel metal plates having charges +Q and -?Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will
A : become zero
B : increase
C : decrease
D : remains same
•
.
Solution :
Electric field between two parallel plates placed in vacuum is given by
E = σ/ ε0
In a medium of dielectric constant K,
E'= σ / ε0K
For kerosene oil K > 1
so E' = E
• Question 4
The electric field at a distance 3R/2 from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R/2 from the centre of the sphere is
A : 0
B : E
C : E/2
D : E/3
•
.
Solution :
Electric field inside charged conductor is always zero.
• Question 5
Astudent measures the distance traversed in free fall of abody, initially at rest, in a given time. He uses this data to estimate g, the acceleration due to gravity. If themaximum percentage errors in measurement of the distance and the time are e1 and e2 respectively, the percentage error in the estimation of g is
A : e2 - e1
B : e1 + 2e2
C : e1 + e2
D : e1 - 2e2
•
.
Solution :
using h= ut + 1/2gt2
we get Δg/g =Δh/h -2Δt/t
For max permissible error
(Δg/g *100) =(Δh/h *100 ) +2*(Δt/t *100)
Δh/h *100 = e1
Δt/t *100 = e2
therefore (Δg/g *100) = e1 + 2e2
• Question 6
Whenmonochromatic radiation of intensity ? falls on ametal surface, the number of photoelectrons and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2?, the number of emitted electrons and their maximum kinetic energy are respectively
A : N and 2T
B : 2N and T
C : 2N and 2T
D : N and T
•
.
Solution :
The number of photoelectrons ejected is directly proportional to the intensity of incident light. Maximum kinetic energy is independent of intensity of incident light but depends upon the frequency of light. Hence option (2) is correct.
• Question 7
The electric field of an electromagnetic wave in free space is given by Ê = 10cos(107t + kx)ĵ V/m, where t and x are in seconds and metres respectively. It can be inferred that
(i) the wavelength λ is 188.4 m.
(ii) the wave number k is 0.33 rad/m
(iii) the wave amplitude is 10 V/m(iv) the wave is propagating along +x direction
Which one of the following pairs of statements is correct ?
A : iii and iv
B : i and ii
C : ii and iii
D : i and iii
•
.
Solution :
Comparing it with standard equation of e.m. wave,
E= E0cos(wt + kx)
c = vλ = wλ/2π
λ= 2πc /w ;
λ=188.4m
also c=w/k
k=w/c = 0.033
• Question 8
The speed of light in media M1 and M2 are 1.5 × 108 m/s and 2.0 × 108 m/s respectively.A ray of light enters from medium M1 to M2 at an incidence angle i. If the ray suffers total internal reflection, the value of i is:
A : Equal to sin-(2/3)
B : Equal to or less than sin-(3/5)
C : Equal to or greater than sin-(3/4)
D : Less than sin-(2/3)
•
.
Solution :
µ1 = c/v1 =2
µ2 = c/v2 =3/2
For TIR sin i ≥ sin C
sin C= µ2 / µ1 sin i ≥ µ21 ≥ (3/2)/2
Equal to or greater than sin-(3/4)
• Question 9
A ray of light is incident on a 60° prism at the minimum deviation position. The angle of refraction at the first face (i.e., incident face) of the prism is
A : 0
B : 30°
C : 45°
D : 60°
•
.
Solution :
Angle of prism, A = r1 + r2 For minimum deviation r1 = r2 = r
A = 2r
Given, A= 60°
so r = A/2 =30°
• Question 10
For transistor action
(1) Base, emitter and collector regions should have similar size and doping concentrations.
(2) The base region must be very thin and lightly doped.
(3) The emitter-base junction is forward biased and base-collector junction is reverse biased
(4) Both the emitter-base junction as well as the base-collector junction are forward biased
A : 4 and 1
B : 1 and 2
C : 2 and 3
D : 3 and 4
•
.
• Question 11
The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius R1 to another of radius R2 (R2 > R1) is
A : GmM(1/R12 - 1/R22 )
B : GmM(1/R1 - 1/R2 )
C : 2GmM(1/R1 - 1/R2 )
D : 1/2 *GmM(1/R1 - 1/R2 )
•
.
Solution :
KE = (-GmM/2R2) - (-GmM/2R1)
• Question 12
The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is(in deg)
A : 60
B : 15
C : 45
D : 60
•
.
Solution :
v= ucosΘ
given v = u/2
u/2 = ucosΘ
• Question 13
Froma circular disc of radius Randmass 9M, a small disc ofmassMand radius R/3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is
A : 40/9 * MR2
B : MR2
C : 4MR2
D : 4/9 * MR2
•
.
Solution :
Themoment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane is
I1 = 9/2 * MR2
I2 = 1/18 * MR2
I = I1 - I2
• Question 14
A particle moves in x-y plane according to rule x = asin wt and y = acos wt. The particle follows
A : an elliptical path
B : a circular path
C : a parabolic path
D : a straight line path inclined equally to x and y-axes
•
.
Solution :
x/a = sin wt
\ y/a = cos wt
square and add the 2 equations and get equation of circle
• Question 15
A closely wound solenoid of 2000 turns and area of cross-section 1.5 * 10-?4 m2 carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 * 10?-?2 tesla making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be
A : 3 * 10 -3 Nm
B : 1.5 * 10 -3 Nm
C : 1.5 * 10 -2 Nm
D : 3 * 10 -2 Nm
•
.
Solution :
Magnetic moment of the loop.
M = NIA = 2000 * 2 * 1.5 * 10-?4 = 0.6 J/T
torque t = MBsin30°
• Question 16
The decay constant of a radio isotope is λ. If A1 and A2 are its activities at times t1 and t2 respectively, the number of nuclei which have decayed during the time (t1 ? t2)
A : A1t1 - A2t2
B : A1 - A2
C : (A1 - A2) /λ
D : λ/(A1 - A2)
•
.
Solution :
(A1 = λN1 at t1
(A2 = λN2 at t2
Therefore, number of nuclei decayed during time interval (t1-t2) is
N1 - N2 =[A1-A2]/λ
• Question 17
A particle having a mass of 10-?2 kg carries a charge of 5 * 10?-8 C. The particle is given an initial horizontal velocity of 105 ms?-?1 in the presence of electric field E and magnetic field B. To keep the particle moving in a horizontal direction, it is necessary that
(i) B should be perpendicular to the direction of velocity and E should be along the direction of velocity (ii) Both B and E should be along the direction of velocity
(iii) Both B and E are mutually perpendicular and perpendicular to the direction of velocity.
(iv) B should be along the direction of velocity and E should be perpendicular to the direction of velocity
Which one of the following pairs of statements is possible
A : i and iii
B : iii and iv
C : ii and iii
D : ii and iv
•
.
• Question 18
The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is
A : 23.6MeV
B : 2.2MeV
C : 28.0MeV
D : 30.2MeV
•
.
Solution :
The binding energy per nucleon of a deuteron = 1.1 MeV
so Total binding energy = 2 * 1.1 = 2.2 MeV
so The binding energy per nucleon of a helium nuclei = 7 MeV
so Total binding energy = 4 * 7 = 28 MeV
so Hence, energy released
ΔE = (28 ? 2 * 2.2) = 23.6 MeV
• Question 19
The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be (the energy of the electron in nth state En = -13.6/n2 eV
A : 5.1 V
B : 12.1V
C : 17.2V
D : 7V
•
.
Solution :
Energy released when electron in the atom jumps from excited state (n = 3) to ground state (n = 1) is
E = hv
E3 - E1 = 12.1eV
therefore stopping potential
eV0 = hv - φ
=12.1-5.1
V0 = 7V
• Question 20
If cp and cv denote the specific heats (per unit mass) of an ideal gas of molecular weight M where R is the molecular weight constant, then
A : cp-cv = R/M2
B : cp - cv = R
C : cp - cv =R/M
D : cp - cv = MR
•
.
Solution :
Let CV and CP be molar specific heats of the ideal gas at constant volume and constant pressure, respectively, then
CP = Mcp and CV = Mcv
CP-CV = R
Mcp - Mcv = R
cp - cv =R/M
• Question 21
A condenser of capacity C is charged to a potential difference of V1. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V2 is
A : ( (C(V1 - V2)2)/L )1/2
B : C(V12 - V22)/L
C : C(V12 - V22)/L
D : (C(V12 - V22)/L )1/2
•
.
• Question 22
The dependence of acceleration due to gravity g on the distance r from the centre of the earth, assumed to be a sphere of radius R of uniform density is as shown in figures below. The correct figure is.
A : B :
C : D :
•
.
Solution :
The acceleration due to gravity at a depth d below surface of earth is
g'=GM/R2 (1 - d/R) = g(1 - d/R)
g' = 0 at d = R acceleration due to gravity is zero at the centre of earth . Thus, the variation in value g with r is for, r > R,
g' ∝ 1/r2
Here, R + h =r
rg'=gr/R
here R-d = r
g' ∝ r
Therefore, the variation of g with distance from centre of the earth will be as shown in the figure.
• Question 23
Asolid cylinder and a hollow cylinder, both of the samemass and same external diameter are relaeased from the same height at the same time on an inclined plane. Both roll down without slipping.Which one will reach the bottom first ?
A : Both together only when angle of inclination of plane is 45°
B : Both together
C : Hollow cylinder
D : Solid cylinder
•
.
Solution :
time taken to reach the bottom of inclined plane
t = √(2l(1 + K2/R2)/gsinΘ)
here l is the length of the incline plane
For solid cylinder K2 =R2/2
For Hollow cylinder = K2 =R2
Hence, solid cylinder will reach the bottom first.
• Question 24
The thermo e.m.f. E in volts of a certain thermocouple is found to vary with temperature difference Θ in °C between the two junctions according to the relation
E = 30Θ - Θ2/15
The neutral temperature for the thermo-couple will be
A : 450°
B : 400°
C : 225°
D : 30°
•
.
Solution :
E = 30Θ - Θ2/15
dE/dΘ = 0;
dE/dΘ = 30 -2Θ/15
Θ=15*30/2 =225°C
• Question 25
(i) Centre of gravity (C.G.) of a body is the point at which the weight of the body acts
(ii) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius
(iii) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be considered to be concentrated at its C.G.
(iv) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the C.G. of the body to the axis.
Which one of the following pairs of statements is correct ?
A : iv and i
B : i and ii
C : ii and iii
D : iii and iv
•
.
• Question 26
The magnetic moment of a diamagnetic atom is
A : much greater than one
B : one
C : between zero and one
D : equal to zero
•
.
Solution :
The magnetic momentum of a diamagnetic atom is equal to zero.
• Question 27
Two identical bar magnets are fixed with their centres at a distance d apart. a stationary charge Q is placed at P in between the gap of the two magnets at a distance D from the centre O as shown in the figure
The force on the charge Q is?
A : Zero
B : directed along OP
C : directed along PO
D : directed perpendicular to the plane of paper
•
.
Solution :
Magnetic field due to bar magnets exerts force on moving charges only. Since the charge is at rest, zero force acts on it.
• Question 28
Aparticle ofmass M, starting from rest, undergoes uniform acceleration. If the speed acquired in time T is V, the power delivered to the particle is
A : MV2/T
B : 1/2 * MV2/T2
C : MV2/T2
D : 1/2 * MV2/T
•
.
Solution :
Power delivered in time T is
P = F.V. = MaV
P = MVdV/dT
PT=MV2/2
• Question 29
A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocity ?. Two objects each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by
A : (M + 2m)w/2m
B : 2Mw/(M+2m)
C : (M+2m)w/M
D : Mw/(M+2m)
•
.
Solution :
As no external toruqe is acting about the axis, angular momentum of system remains conserved.
I1w1 = I2w2
w2 = Mw/(M+2m)
• Question 30
A monoatomic gas at pressure P1 and volume V1 is compressed adiabatically to 1/8th of its original volume. What is the final pressure of the gas
A : 64P1
B : P1
C : 16P1
D : 32P1
•
.
Solution :
Ideal gas equation, for an adiabatic process is
PVγ = constant
P1V1γ =P2V2γ
For monoatomic gas γ =5/3;
P1V15/3 = P2(V1/8 5/3)
• Question 31
Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is :
A : Mg < Ca < Cl < P
B : Cl < P < Mg < Ca
C : P < Cl < Ca < Mg
D : Ca < Mg < P < Cl
•
.
Solution :
The atomic radii decreases on moving from left to right in a period, thus order of sizes for Cl, P and Mg is Cl < P < Mg. Down the group size increases. Thus overall order is : Cl < P < Mg < Ca.
• Question 32
The reaction,is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression :
A : [(0.75)3 (0.25)] / [(1.00)2 (1.00)]
B : [(0.75)3 (0.25)] /[(0.50)2 (0.75)]
C : [(0.75)3 (0.25)] / [(0.50)2 (0.25)]
D : [(0.75)3 (0.25)] / [(0.75)2 (0.25)]
•
.
• Question 33
Which of the following expressions correctly represents the equivalent conductance at infinite dilution of Al2(SO4)3. Given that ÅAl3+ and ÅSO42- are the equivalent conductances at infinite dilution of the respective ions?
A : Al3+ + 3ÅSO42-
B : ÅAl3+ + ÅSO42-
C : Al3+ + 3ÅSO42-) * 6
D : 1/3 ÅAl3 + 1/2 + ÅSO42-
•
.
Solution :
At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards molar conductance of the electrolyte irrespective of the nature of the other ion with which it is associated
• Question 34
The pressure exerted by 6.0 g of methane gas in a 0.03m3 vessel at 129° C is (Atomic masses) : C = 12.01, H = 1.01 and R = 8.314 J K-?1 mol?-1) :
A : 215216Pa
B : 13409Pa
C : 41648Pa
D : 31684Pa
•
.
Solution :
T = 129°C = 129 + 273 = 402 K
V = 0.03m3
Molecular mass of CH4, M = 12.01 + 4 * 1.01 = 16.05
PV = nRT = RTw/M
P=41647.7 Pa
• Question 35
Match List I (Equations) with List II (Types of processes) and select the correct option.
List 1 Equations List II Types of process a)Kp > Q i)Non-spontaneous b)ΔG° < RT In Q ii)Equilibrium c)Kp = Q iii)Spontaneous and endothermic d)T = ΔH/ΔS vi)Spontaneous
A : a - (i), b - (ii), c - (iii), d - (iv)
B : a - (iii), b - (iv), c - (ii), d - (i)
C : a - (iv), (b -(i), c - (ii), d - (iii)
D : a - (ii), b - (i), c - (iv), d - (iii)
•
.
Solution :
When Kp > Q, rate of forward reaction > rate of backward reaction.
Reaction is spontaneous. When ΔG° < RT ln Q, ΔG° is positive, reverse reaction is feasible, thus reaction is non-spontaneous.
When Kp = Q, rate of forward reaction = rate of backward reaction
Reaction is in equilibrium. When TΔS > ΔH, ΔG will be negative only when ΔH = + ve
• Question 36
Among the following four compounds :
(i) Phenol
(ii) Methyl phenol
(iii)Meta-nitrophenol
(iv) Para-nitrophenol
The acidity order is :
A : (iv) > (iii) > (i) > (ii)
B : (iii) > (iv) > (i) > (ii)
C : (i) > (iv) > (iii) > (ii)
D : (ii) > (i) > (iii) > (iv)
•
.
Solution :
In phenols, the presence of electron releasing groups decreases the acidity, whereas presence of electron withdrawing groups increase the acidity, compared to phenol. Among the meta and para-nitrophenols, the latter is more acidic as the present of -NO2 group at para position stabilises the phenoxide ion to a greater extent than when it is present at meta position.
• Question 37
Among the following which one has the highest cation to anion size ratio?
A : CsI
B : CsF
C : LiF
D : NaF
•
.
Solution :
The order of size of given cations is Li+ < Na+ < Cs+ and the order of size of given anions is I?- > F-? Thus, when the cation is large and anion is smallest, the cation to anion size ratio is maximum. Hence, cation to anion size ratio is maximum for CsF.
• Question 38
Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be :
A : Infinite
B : 3 Joules
C : 9 Joules
D : Zero
•
.
Solution :
Since the ideal gas expands spontaneously into vacuum, Pext = 0, hence work done is also zero.
• Question 39
Which of the following species is not electrophilic in nature?
A : B :
C : D :
•
.
Solution :
Electrophiles are electron deficient species.Among the given, 3) has lone pair of electrons for donation, thus it is not electron deficient and hence, does not behave like an electrophile.
• Question 40
A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be : (h = 6.6 * 10-?34 Js)
A : 6.6 * 10-?32
B : 6.6 * 10-?34
C : 1.0 * 10-?35
D : 1.0 * 10-?32
•
.
Solution :
use de-Broglie equation, λ = h/mv
• Question 41
Which of the following is not a fat soluble vitamin ?
•
.
Solution :
Vitamin B complex is fat insoluble
• Question 42
Which of the statements about "Denaturation" given below are correct ?
Statements:
(a) Denaturation of proteins causes loss of secondary and tertiary structures of the protein.
(b) Denturation leads to the conversion of double strand of DNA into single strand
(c) Denaturation affects primary strucrture which gets distorted
•
.
Solution :
During denaturation secondary and tertiary structures of protein destroyed but primary structures remains intact.
• Question 43
Which has the maximum number of molecules among the following ?
•
.
Solution :
Moles of 1: 44/44 =1 →so number of molecules = Na
Moles of 2: 48/48 =1 →so number of molecules = Na
Moles of 3: 8/2=4 →so number of molecules = 4Na Moles of 4: 64/64 =1 →so number of molecules = Na
• Question 44
The half life of a substance in a certain eznzyme-catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L-1 to 0.04 mg L-1 ,is:
•
.
Solution :
Enzyme catalysed reactions are initially follow first order kinetics when concentration decreases 1.28 mg L-1to 0.04 mg L-1 .Then five half life completed
No. of half lives = 5
So, times required = 5 × 138 = 690 s
• Question 45
Which of the following compounds undergoes nucleophilic substitution reaction most easily ?
•
.
Solution :
The correct order of nucleophilic substitution reactions
A>D>B>C
• Question 46
Which of the following statements is incorrect ?
•
.
• Question 47
A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be :
•
.
Solution :
ΔTf = iKfm
HA → H+ + AH-
1-α → α α
1-0.3 → 0.3 0.3
i = 1 - 0.3 + 0.3 + 0.3
i = 1.3
ΔTf = 1.3 *1.86 *0.1 = 0.2418
Tf = 0 - 0.2418 = -0.2418°C
• Question 48
The rate of the reaction 2N2O5 → 4NO2 + O2 can be written in three ways :
-d[N2O5]/dt = k[N2O5]
d[NO2]/dt = k'[N2O5]
d[O2 ]/dt = k"[N2O5]
The relationship between k and k' and between k and k" are :
•
.
Solution :
-1/2 d[N2O5]/dt = 1/4 d[NO2]/dt = d[O2 ]/dt
1/2 k[N2O5] = 1/4 k'[N2O5] = k"[N2O5]
• Question 49
Which of the following carbonyls will have the strongest C-O bond ?
•
.
Solution :
As + ve charge on the central metal atom increases, the less readily the metal can donate electron density into the π* orbitals of CO ligand to weaken the C-O bond. Hence the C-O bond would be strongest in Mn(CO)6+
• Question 50
The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds :
•
.
Solution :
4 due to steric crowding
• Question 51
The IUPAC name of the following compound
•
.
• Question 52
According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon ?
•
.
Solution :
Energy of photon obtained from the transition n = 6 to n = 5 will have least energy.
ΔE = 13.6Z2(1/n12 - 1/n22)
• Question 53
A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y-) will be :
•
.
Solution :
Radius ratio of NaCl like crystal = r+ / r- = 0.414
r- = 100 / 0.414 =241.5
• Question 54
Consider the following processes and their ΔH((kJ/mol) :
1/2 A → B (+150)
3B → 2C + D -(125)
E + A → 2D (+350)
For B + D → E + 2C, ΔH will be
•
.
Solution :
by doing 2*eq(1) + eq(2) - eq(3) we will get the asked equation
ΔH = 300 -125 -350 =-175
• Question 55
Match the compounds given in List-I with List-II and select the suitable option using the code given below :
List1 List2 a)Benzaldehyde i)Phenolphthalein b)Phthalic anhydride ii)Benzoin condensation c)Phenyl benzoate iii)Oil of wintergreen d)Methyl salicylate (iv) Fries rearrangement
a,b,c and d are
•
.
• Question 56
Which of the following compounds is most basic ?
•
.
Solution :
In benzylamine, the lone pair on nitrogen is not delocalized with the benzene ring.
• Question 57
Which of the following structures is the most preferred and hence of lowest energy for SO3 ?
•
.
Solution :
Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species. Generally the lowest energy structure is the one with the smallest formal charges on the atoms.
• Question 58
A solution contains Fe2+, Fe3+ and I- ions. This solution was treated with iodine at 35°C. E° for Fe3+/Fe2+ is + 0.77 V and E° for I2/2I- = 0.536 V. The favourable redox reaction is
•
.
Solution :
Since the reduction potential of Fe3+/Fe2+ is greater than that of I2/2I- , Fe3+ will be reduced and I- will be oxidized
• Question 59
What is the value of electron gain enthalpy of Na+ if IE1 of Na = 5.1 eV ?
•
.
Solution :
Electron gain enthalpy is the ΔH for the process
Na+ + e- → Na
ΔH = -5.1eV
• Question 60
The unit of rate constant for a zero order reaction is :
•
.
• Question 61
In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride slats. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl- concentration is 0.10 M. What will be concentrations of Ag+ and Pb2+ be at equilibrium?
(Ksp for AgCl = 1.8 * 10-10, Ksp for PbCl2 = 1.7 * 10–5)
•
.
Solution :
[Ag+] = 1.8 * 10-10/0.1
[Pb2+] = 1.7 * 10-5/ (0.1)2
• Question 62
A bubble of air is underwater at temperature 15°C and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar, what will happen to the volume of the bubble?
•
.
Solution :
V ∝ T/P
V2 ∝ 298
V1 ∝ 288/1.5
V2 /V1 = 1.55
• Question 63
Match List-I with List-II for the compositions of substances and select the correct answer using the code given below the lists :
CaSO4.1/2H2O
Substances Composition a)Palster of paris i)CaSO4.2H2O b)Epsomite ii) c)kieserite iii)MaSO4.7H2O d)Gypsum (iv)MaSO4.H2O (v)CaSO4
•
.
• Question 64
The pairs of species of oxygen and their magnetic behaviours are noted below. Which of the following presents the correct description ?
•
.
Solution :
they contain one and two unpaired electrons each.
• Question 65
Consider the reactions :
The mechanisms of reactions (i) and (ii) are respectively :
•
.
Solution :
Rearrangements do not occur in the given substitution reactions. So carbocation intermediates are not involved. Hence the mechanism is SN2 in both cases.
• Question 66
Which of the following complex compounds will exhibit highest paramagnetic behaviour ?
(At. No. Ti 22, Cr = 24, Co = 27, Zn = 30)
•
.
Solution :
[Cr(NH3)6]3+ contain three unpaired electrons.
• Question 67
200mL of an aqueous solution of a protein contains its 1.26 g. The Osmotic pressure of this solution at 300K is found to be 2.57 * 10-?3 bar. The molar mass of protein will be (R = 0.083 L bar mol?-?1 K-?1) :
•
.
Solution :
πv=nST
M = W*S*T/π*v
• Question 68
Which of the following oxide is amphoteric ?
•
.
Solution :
because it reacts with acids as well as bases to form corresponding salts.
• Question 69
The following reactions take place in the blast furnace in the preparation of impure iron. Identify the reaction pertaining to the formation of the slag.
•
.
Solution :
Slag can be defined as a fusiblemass, which is obtained when a flux reacts with an infusible acidic or basic impurity present in the oxide ore.
CaO(s) (basic flux) + SiO2 (s) (acidic flux) → CaSiO3 (s) (slag)
• Question 70
An organic compound ?'A'? on treatment with NH3 gives ?'B'? which on heating gives ?'C?', ?'C?' when treated with Br2 in the presence of KOH produces ethylamine. Compound ?'A'? is :
•
.
• Question 71
The technique called gamete intrafallopian transfer (GIFT) is recommended for those females:
•
.
• Question 72
Which one of the following is a possibility for most of us in regard to breathing, by making a conscious effort?
•
.
Solution :
Eustachian tube connect middle ear cavity (Tympanic cavity) with pharynx
• Question 73
Bacillus thuringiensis forms protein crystals which contain insecticidal protein
•
.
• Question 74
Which one of the following pairs is wrongly matched while the remaining three are correct?
•
.
Solution :
Water hyacinth is offset.
• Question 75
Which one of the following diagrams represents the placentation in Dianthus?
•
.
Solution :
Free central placentation occurs in Dianthus
• Question 76
Which one of the following statements is totally wrong about the occurrence of notochord, while the other three are correct?
•
.
Solution :
Because get changed or replaced by vertibral column
• Question 77
Which one of the following animals may occupy more than one trophic levels in the same ecosystem at the same time?
•
.
Solution :
It feeds upon grains hence called primary consumer and also insects hence called secondary consumer
• Question 78
Both, hydrarch and xerarch successions lead to:
•
.
• Question 79
What happens during fertilisation in humans after many sperms reach close to the ovum?
•
.
• Question 80
About which day in a normal human menstrual cycle does rapid secretion of LH (Popularly called LH-surge) normally occurs?
•
.
• Question 81
The damping force on a oscillator is directly proportional to the velocity. The Unit of the constant of proportionality are
A : kgs-1
B : kgs
C : kgms-1
D : kgms-2
•
.
Solution :
F= Kv
Unit of K = kgms-2 /ms-1
• Question 82
The motion of a particle along a straight line is described by equation
x = 8 + 12t - t3
where x is in meter and t in second. The retardation of the particle when its velocity becomes zero is
A : 6 ms-2
B : 12 ms-2
C : 24 ms-2
D : 0
•
.
Solution :
x = 8 + 12t - t3
v = dx/dt = 12 - 3t2
a = dv/dt = -6t
putting v = 0
3t2 = 12
t = 2 sec
thus a = -6*2 = -12
retardation of 12 ms-2
• Question 83
The horizontal range and the maximum height of a projectile are equal . The angle of projection of the projectile is:
A : tan-1(2)
B : 45°
C : tan-1(1/4)
D : tan-1(4)
•
.
Solution :
tan Θ = 4H/R
given H = R
• Question 84
A particle has initial velocity (2ĩ + 3j̃) and acceleration (0.3ĩ + 0.2j̃).The magnitude of velocity after 10 seconds will be
A : 5 units
B : 9 units
C : 9√2
D : 5√2
•
.
Solution :
v = u + at
v = (2ĩ + 3j̃)+(0.3ĩ + 0.2j̃)*10
ṽ = 5ĩ + 5j̃
v = 5√2 units
• Question 85
A car of mass 1000 kg negotiates a banked curve of radius 90m on a frictionless road. If the banking angle is 45°, the speed of the car is
A : 5 ms-1
B : 10 ms-1
C : 20 ms-1
D : 30 ms-1
•
.
Solution :
v = √(rg tanΘ)
v = √(90*10*tan45°)
=30
• Question 86
A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 ms-1. It collides with a horizontal spring of force constant 200 Nm-1 . The maximum compression produced in the spring will be
A : 0.7m
B : 0.2m
C : 0.5m
D : 0.6m
•
.
Solution :
1/2 mv2 [1 + K2/R2]
1/2kx2
putting m = 3kg, v = 4 m/s,
K2/R2 = 1/2 (For solid cylinder), k = 200 Nm-1 we get
xmax = 0.6m
• Question 87
The potential energy of a particle in a force field is :
U = A/r2 - B/r , where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is
A : A/B
B : B/A
C : B/2A
D : 2A/B
•
.
Solution :
For Stable equilibrium, F = -dU/dr = 0
we get r = 2A/B
• Question 88
Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction.
A : Θ = tan-1(1/2) to the x-axis
B : Θ = tan-1(-1/2) to the x-axis
C : same as that of B
D : opposite to the of B
•
.
Solution :
Applying the law of conservation of momentum we have
in y-direction, m1 u sin Θ = m2 v/2 .....(i)
In x-direction, m1 u cos Θ = m2 v .....(ii)
• Question 89
Two persons of masses 55 kg and 65kg respectively, are at the opposite ends of a boat. The length of the boat is 30. m and weighs 100 kg. The 55 kg man walks up to the 65 kg man and sits with him. If the boat is in still water the center of mass of the system shift by:
A : zero
B : 0.75m
C : 3.0m
D : 2.3m
•
.
Solution :
Since net external force on the system is zero hence C.O.M. remains unchanged
• Question 90
ABC is an equilateral triangle with O as its centre F1,F2 and F3 represent three forces acting along the sides AB, BC and AC respectively. If the torque about O is zero then the magnitude of F3 is
A : (F1+F2)/2
B : 2(F1+F2)
C : F1+F2
D : F1-F2
•
.
Solution :
Taking anticlockwise torque
F1d + F2d - F3d = 0 (where d = perpendicular distance of the centre O from each side)
F3 = F1 + F2
• Question 91
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along
B : the tangent to the orbit
C : a line perpendicular to the plane of rotation
D : the line making an angle of 45° to the plane of rotation.
•
.
Solution :
L=r x mv
• Question 92
A spherical planet has a mass MP and diameter DP. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to
A : GMP/ DP2
B : 4GMPm/ DP2
C : 4GMP/ DP2
D : GMPm/ DP2
•
.
Solution :
g = GMP/ RP2
where RP = DP/2
• Question 93
A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth. The time period of another satellite of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is
A : 6√2
B : 6/√2
C : 5
D : 10
•
.
Solution :
According to Kepler's third law, T2 ∝ r3
T2/T1 = (r2/r1)3/2
Given T1 = 24 h r1 = R + h1 = R + 5R = 6R
r2 = R + h2 = R + 2R = 3R
On solving we get 6√2
• Question 94
The height at which the weight of a body becomes 1/16th , its weight on the surface of earth (radius R), is
A : 3R
B : 4R
C : 5R
D : 15R
•
.
Solution :
Wh/W = mg/mg = (R/(R+h))2 = 1/16
h=3R
• Question 95
Two sources of sound placed close to each other, are emitting progressive waves given by y1 = 4 sin 600 πt and y2 = 5 sin 608 πt: An observer located near these two sources of sound will hear
A : 8 beats per second with intensity ratio 81 : 1 between waxing and waning
B : 4 beats per second with intensity ratio 81 : 1 between waxing and waning
C : 4 beats per second with intensity ratio 25 : 16 between waxing and waning
D : 8 beats per second with intensity ratio 25 : 16 between waxing and waning
•
.
Solution :
Comapare both equations with y = a sin 2πft we get
f1=300Hz
f2=304Hz
Number of beats = f2-f1 = 4s-1
Imax/Imin = ((a1 + a2)/(a1 - a2))
((4+5)/(4-5))2 = 81/1
• Question 96
When a string is divided into three segments of length l1, l2 and l3, the fundamental frequencies of these three segments are v1, v2 and v3 respectively. The original functamental frequency (v) of the string is :
A : 1/v = 1/v1 + 1/v2 + 1/v3
B : 1/√v = 1/√v1 +1/√v2 +1/√v3
C : √v = √v1 + √v2 + √v3
D : v = v1 +v2 +v3
•
.
Solution :
Length of the string l = l1 + l2 + l3 v ∝ 1/l
• Question 97
One mole of an ideal gas goes from an initial state A to final state B via two processes : It first undergoes isothermal expansion from volume V to 3V and then its volume is reduced from 3V to V at constant pressure. The correct P-V diagram representing the two processes is:
A : B :
C : D :
•
.
Solution :
For isothermal expansion P-V curve is rectangular hyperbola (clockwise curve)
• Question 98
A thermodynamics system is taken trough the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is:
A : 1/2 PV
B : PV
C : 2PV
D : 4PV
•
.
Solution :
For given cyclic process, ΔU = 0
Q = W (From first law of thermodynamic Q = ΔU + W) Also W = - area enclosed by the curve = - P * 2V
Heat rejected = 2PV
• Question 99
Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which one of the following graphs represents the variation of temperature with time ?
A : B :
C : D :
•
.
Solution :
At first temperature will increase then there will be state change from liquid to gas.
• Question 100
If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q ?(&sigma is Stefan's constant)
A : (4πR2Q/σ)1/4
B : (Q/4πR2Qσ)1/4
C : Q/4πR2σ
D : (Q/4πR2Qσ)-1/2
•
.
Solution :
Using Stefan's law, the rate of energy production is Q = E * A
• Question 101
A coil of resistance 400Ω is placed in a magnetic filed. if the magnetic flux φ (wb) linked with the coil varies with times t(sec) as
φ = 50t2 + 4
The current is the coil at t = 2 sec is
A : 2A
B : 1A
C : 0.5A
D : 0.1A
•
.
Solution :
i = |(-dφ/dt)/R| = 0.5A
• Question 102
The current (I) in the inductance is varying with time according to the plot shown in figure.
Which one of the following is the correct variation of voltage with time in the coil ?
A : B :
C : D :
•
.
Solution :
|V|=|-Ldi/dt| therefore answer will be best represented by graph -2
• Question 103
In an electrical circuit R, L, C and an a.c. voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the ciruit is π/3 . If instead, C is removed from the circuit, the phase difference is again π/3 . The power factor of the circuit is
A : 1
B : √3/2
C : 1/2
D : 1/√2
•
.
Solution :
It is the condition of resonance therefore phase difference between voltage and current is zero and power factor cos &phi = 1
• Question 104
A ring is made of a wire having a resistance R0 = 12Ω. Find the points A and B, as shown in the figure, at which a current carrying conductor should be connected so that the resistance R of the sub circuit between these points is equal to 8/3Ω
A : l1/l2 = 3/8
B : l1/l2 = 1/2
C : l1/l2 = 5/8
D : l1/l2 = 1/3
•
.
Solution :
Since R∝l
According to problem R1 + R2 = 12Ω and (R1 * R2)/(R1 + R2) = 8/3Ω
On solving this we get R1 = 4Ω and R2 = 8Ω therefore the ratio l1/l2 =1/2
• Question 105
If voltage across a bulb rated 220 Volts. 100 Watt drops by 2.5% of its rated value the percentage of the rated value by which the power would decrease is
A : 5%
B : 10%
C : 20%
D : 2.5%
•
.
Solution :
Power P = V2/R
For small variation,ΔP/P * 100% = 2*ΔV/V * 100%
therefore power would decrease by 5%
• Question 106
In the circuit shown the cells A and B have negligible resistance. For VA = 12V, R1 = 500Ω and R = 100Ω the galvanometer (G) shows no deflection. The value of VB is
A : 12V
B : 6V
C : 4V
D : 2V
•
.
Solution :
By using kirchoff voltage law the potential difference across R is 2 volt
• Question 107
The electric field associated with an e. m. wave in vacuum is given by E = î 40 cos (kz – 6 × 108t) , where E, z and t are in volt/m, meter and seconds respectively. The value of wave vector k is :
A : 6 m-1
B : 3 m-1
C : 2 m-1
D : 0.5 m-1
•
.
Solution :
Equation of Electromagnetic wave is E = E0 cos (kz - ω t) and speed of EM wave V = ω/k
By comparing k = ω/v = 6 * 10 8/3* 10 8
• Question 108
What is the flux through a cube of side 'a' if a point charge of q is at one of its corner
A : q/ε0
B : (q/2ε0) * 6a2
C : 2q/ε0
D : q/8ε0
•
.
• Question 109
An electric dipole of moment 'p' is placed in an electric field of intensity 'E'. The dipole acquires a position such that the axis of the dipole makes an angle θ with the direction of the field. Assuming that the potential energy of the dipole to be zero when θ = 90°, the torque and the potential energy of the dipole will respectively be
A : pEsinθ , 2pEcosθ
B : pEcosθ,-pEsinθ
C : pEsinθ,-pEcosθ
D : pEsinθ,-2pEcosθ
•
.
Solution :
Toque = PEsinθ and Potential energy U = -PEcosθ
• Question 110
Four point charge -Q, -q, 2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero is
A : Q = q
B : Q = 1/q
C : Q = -q
D : Q = -1/q
•
.
Solution :
Potential V at centre of the square =
-kQ/r - kq/r + k2Q/r + k2q/r = 0
So Q = -q
• Question 111
A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It.
A : Will stay in north-south direction only
B : Will stay in east-west direction only
C : Will becomes rigid showing no movement
D : Will stay in any position
•
.
Solution :
Will remain in any position at geomagnetic north and south pole
• Question 112
A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampere rage. The value (in ohm) of necessary shunt will be
A : 1
B : 0.05
C : 0.001
D : 0.01
•
.
Solution :
If the resistance of the meter is G then the full scale deflection current ig = 25mV/G ampere
The Value of shunt required for converting it into ammeter of range 25 ampere is S = igG/i-ig
so 25mV/25
• Question 113
Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2I, respectively. The resultant magnetic field induction at the centre will be
A : µ0I/2R
B : µ0I/R
C : √5µ0I/2R
D : 0I/2R
•
.
Solution :
Magnetic field at the centre of circular current carrying coil of radius R and current I is =
µ0I/2R = B
Magnetic field at the centre of circular current carrying coil of radius R and current 2I is =
µ02I/2R = 2B
The resulting magnetic field will be =
√(B2+4B2)=√5B
• Question 114
An alternating electric field, of frequency v, is applied across the dees (radius = R) of a cyclotron that is being used to accelerated protons (mass = m). The operating magnetic field (B) used in the cyclotron and the kinetic energy (K) of the proton beam, produced by it, are given by
A : B = 2πmv/e and K = 2mπ2v2R2
B : B = mv/e and K = m2πvR2
C : B = mv/e and K = 2mπ2v2R
D : B = 2πmv/e and K = m2πvR2
•
.
Solution :
frequency v = eB/2πm so B = 2πmv/e
Kinetic energy K = 1/2 mv2
• Question 115
The magnifying power of a teloscope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are
A : 18 cm, 2 cm
B : 11 cm, 9 cm
C : 10 cm, 10 cm
D : 15 cm, 5 cm
•
.
Solution :
L = fo + fe = 20cm
M = fo/fe = 9
• Question 116
A ray of light is incident at an angle of incidence, i, on one face of a prism of angle A (assumed to be small) and emerges normally from the opposite face. It the refractive index of the prism is µ, the angle of incidence i, is nearly equal to
A : A/µ
B : A/2µ
C : µA
D : µA/2
•
.
• Question 117
A concave mirror of focal length 'f1' is placed at a distance of 'd' from a convex lens of focal length 'f2' A beam of light coming from infinity and falling on this convex lens- concave mirror combination returns to infinity. The dsitance 'd' must equal
A : 2f1 + f2
B : -2f1 + f2
C : f1 + f2
D : -f1 + f2
•
.
• Question 118
When a biconves lens of glass having refractive index 1.47 is dipped in a liqud, it acts as a plane sheet of glass. This implies that the liquid must have refractive index
A : greater than that of glass
B : less than that of glass
C : equal to that of glass
D : less than one
•
.
Solution :
1/f = (µ21 - 1)(1/R1 - 1/R2)
For power = 1/f = 0,
µ2 = µ1
• Question 119
An α-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m2. The de-Broglie wavelength associated with the particle will be
A : 10Å
B : 0.01Å
C :
D : 0.1Å
•
.
Solution :
radius R = mv/qB and de- Broglie wavelength λ = h/mv on solving we get λ= 0.01Å
• Question 120
Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is
A : 1.6 * 1015 Hz
B : 2.5 * 1015 Hz
C : 4 * 1015 Hz
D : 5 * 1015 Hz
•
. | 13,501 | 42,243 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-16 | latest | en | 0.829036 |
https://soohba.com/7998741/foreach-ul-li-as-an-infinity-condition | 1,603,749,434,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107892062.70/warc/CC-MAIN-20201026204531-20201026234531-00316.warc.gz | 530,610,531 | 6,002 | Tags
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# Foreach ul li as an infinity condition
By : William Farley
Date : October 18 2020, 03:08 PM
To fix this issue How's your model called? Lets suppose it's named Location. Add your relationship name to the \$with property in your Location model:
code :
protected \$with = ['get_sub'];
\$locations = Location::get();
protected \$with = ['parent'];
public function parent()
{
return \$this->belongsTo('App\Location', 'location_id', 'id');
}
Share :
## What do these three special floating-point values mean: positive infinity, negative infinity, NaN?
By : Mark Oberg
Date : March 29 2020, 07:55 AM
this one helps. This may be a good reference if you want to learn more about floating point numbers in Java.
Positive Infinity is a positive number so large that it can't be represented normally. Negative Infinity is a negative number so large that it cannot be represented normally. NaN means "Not a Number" and results from a mathematical operation that doesn't yield a number- like dividing 0 by 0.
code :
double a = Math.pow(10, 600) - Math.pow(10, 600); //==NaN
## Why cexp(+infinity+I*infinity)=+/-infinity+I*NaN in the C langage?
By : Littlemouse
Date : March 29 2020, 07:55 AM
like below fixes the issue The motivation is indeed given in the document linked by njuffa, http://www.open-std.org/jtc1/sc22/wg14/www/C99RationaleV5.10.pdf :
## C++ Infinity While Loop with condition
By : yongfeng kang
Date : March 29 2020, 07:55 AM
may help you . I am kinda new to C++, and I have encountered a problem in building a simple c++ addition calculator. Here is the code: , One simple solution would be to add
code :
std::string junk;
cin >> junk;
## Javascript - find max and min values using loop and array, -Infinity/Infinity
By : Jackson Metaform
Date : March 29 2020, 07:55 AM
like below fixes the issue Could someone explain to me how does this work. In find highest number method every number is larger than -Infinity. Why does it take the highest number? Same with lowest number finder, how does it selects lowest number, all of the numbers are smaller than Infinity. , The key is here.
code :
if (Numbers[i] > max) {
max = Numbers[i];
}
## how i do infinity parent - child foreach loop in angular?
By : user3012894
Date : March 29 2020, 07:55 AM
To fix this issue I want to do foreach for following json:- , You need use a recursive function
code :
showData(element,level,parent)
{
element.forEach(x=>
{
x.level=level
x.parent=parent
console.log(x.name,x.level,x.parent);
if (x.children)
this.showData(x.children,x.level+1,x.name)
})
}
this.showData(TREE_DATA,0,null) | 741 | 2,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-45 | latest | en | 0.794365 |
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###### Question:
what is a central idea of the poem talking in their sleep
### In the story Rikki-Tikki Tavi, how are Nagaina and Rikki-Tikki alike? Give at least 2 examples...will be marked brainliest.
In the story Rikki-Tikki Tavi, how are Nagaina and Rikki-Tikki alike? Give at least 2 examples...will be marked brainliest....
### You have a large tank filled with 175 gallons of water the tank springs a leak and water is pouring out at rate of 5 gallons per minute write and graph a linear function that shows how the amount of water in the tank depends on the number of minutes?
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### A 9.5 V battery supplies a 2.5 mA current to a circuit for 7.0 h . Part A How much charge has been transferred from the negative to the positive terminal? Express your answer with the appropriate units. Δq Δ q = nothing nothing SubmitRequest Answer Part B How much electric potential energy has been gained by the charges that passed through the battery? Express your answer with the appropriate units. ΔU Δ U = nothing nothing SubmitRequest Answer Provide Feedback Next
A 9.5 V battery supplies a 2.5 mA current to a circuit for 7.0 h . Part A How much charge has been transferred from the negative to the positive terminal? Express your answer with the appropriate units. Δq Δ q = nothing nothing SubmitRequest Answer Part B How much electric potential energy has bee...
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### Explaining the contributions of President Andrew Jackson and the Indian Removal Act of 1830 which led to the Trail of Tears. for another essay please help!!!!!!!
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### Why was Jefferson an Anti-Federalist? 1. He wanted the country to become a center of manufacturing. 2. He thought the central government was becoming too powerful. 3. He thought Washington wanted to become king. 4. He believed the president should hold more power.
Why was Jefferson an Anti-Federalist? 1. He wanted the country to become a center of manufacturing. 2. He thought the central government was becoming too powerful. 3. He thought Washington wanted to become king. 4. He believed the president should hold more power....
### Do sample problem 13.8 in the 8th ed Silberberg book. You add 1.4 kg of ethylene glycol (C2H6O2) antifreeze to 4,192 g of water in your car's radiator. What is the boiling point of the solution? The Kb for water is 0.512 °C/m. Enter to 2 decimal places.
Do sample problem 13.8 in the 8th ed Silberberg book. You add 1.4 kg of ethylene glycol (C2H6O2) antifreeze to 4,192 g of water in your car's radiator. What is the boiling point of the solution? The Kb for water is 0.512 °C/m. Enter to 2 decimal places....
### NEED HELP ON (1) (2) (3) PLEASE ASAPPPPPPPPPPPPPP
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### Gideon noted that three inches of rain had fallen on his patio overnight. The following day, scientists measured and found that the water table had only gained one inch of water. What explains the difference in water amounts? A. The water table rests on bedrock and the water is absorbed by the rock immediately after contact. B. The water in the water table evaporated very quickly. C. High winds caused an error in measurement. D. Most of the water is absorbed by soil and plants and never reac
Gideon noted that three inches of rain had fallen on his patio overnight. The following day, scientists measured and found that the water table had only gained one inch of water. What explains the difference in water amounts? A. The water table rests on bedrock and the water is absorbed by the rock...
### What is an antibiotic
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### Solve the following system of equations for x: y = 4 -x^2 X - y = 2
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### What type of government is Lincoln describing in his conclusion?
What type of government is Lincoln describing in his conclusion?... | 1,224 | 5,283 | {"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} | 2.921875 | 3 | CC-MAIN-2023-06 | latest | en | 0.930894 |
http://typedrawers.com/discussion/1665/fraction-slash | 1,542,804,174,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039748315.98/warc/CC-MAIN-20181121112832-20181121134832-00187.warc.gz | 363,342,879 | 13,564 | #### Howdy, Stranger!
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# Fraction Slash
Posts: 5
Hello,
Working on my second typeface and I'm doing some fractions right now. I've made numerators and denominators, but I was wondering if anyone could shed some light on the fraction slash for a beginner.
The way I had imagined them looking was like this: https://cl.ly/gdtr
But I recognize that in good typefaces, the fraction slash is diagonal, I suppose so that you can type out any long fraction you want.
My question then— is having a horizontal fraction slash something that ever happens—or does it seem to all of you more like a tacky idea to be abandoned?
Tagged:
• Posts: 442
Fractions with a horizontal bar instead of a diagonal one are called "en" or "nut" fractions in typography. The more common diagonal bar fraction is called an "em" or "vulgar" fraction. I wouldn't use the horizontal style unless I was designing a font for mathematical or other special use. You could of course include both, although EN fractions are not encoded in Unicode.
• Posts: 1,143
edited July 2016
The separate fraction slash slot should almost certainly have a diagonal bar.
For the precomposed fractions, you are welcome to structure them as you like. Vertically stacked “nut” fractions require smaller numerals and hence are a bit less legible at small text sizes, but they are otherwise a perfectly reasonable design option.
It is even possible to support arbitrary fractions as nut fractions, although the code is rather more complex.
• Posts: 800
edited July 2016
Matthew Carter is fond of nut fractions in his fonts. There was a version of ITC Galliard that Carter & Cone distributed back in the day that had an expert set for setting multi-digit nut fractions.
Matthew specified that the precomposed fractions in the updated Big Caslon be nut fractions. I proposed developing a {frac} feature that would also provide arbitrary nut fractions. I even produced a proof of concept.
(This was, coincidentally, around the same time John unveiled his Nutso, unknown to me when I first tackled this. My solution was slightly different, not relying upon {mark} for the arbitrary positioning among other things.)
But in the end FB decided not to pursue, mostly because it wouldn’t work in a default InDesign environment since it relies upon the World-Ready Composer (which is sadly not the default).
• Posts: 800
Correction: The ITC Galliard expert set had only precomposed fractions (half, thirds, quarters, fifths, sixths, and eighths).
It was Carter & Cone’s cut of Monticello that had a “Tabular” font with offset numerators for arbitrary nut fractions. Not sure you can get those forms in the commercial OpenType versions. If you can, I seriously doubt there’s a feature to implement them.
• Posts: 77
edited July 2016
As a graphic designer, a certain amount of quirkiness in a typeface might be just the thing I'm looking for, or it just might be the reason I choose to use another typeface. A horizontal fraction symbol is interesting, but more than likely in most situations, I'd regard it as an attention-diverting annoyance to be worked around.
If it's an all-around quirky typeface that you're designing, a horizontal fraction just might fit nicely as a natural continuation of the face's personality. On the other hand, if the horizontal fraction really does stand out as an oddity that leaves designers, like me, and readers scratching their collective heads over why it's there, well, that's another matter. If so, maybe including these out-of-the-ordinary fractions as alternates might be your best bet.
• Posts: 5
Thank you, all! This was all very helpful!
• Posts: 442
@Jens Kutilek -- That would be correct for math notation.
• Posts: 1,143
Yes, math notation everywhere uses nut fractions.
• Posts: 800
Nut fractions are far less common in non-scientific typographic settings these days. But I suspect they were a bit more common a century ago, in the metal era.
• Posts: 753
edited July 2016
In part that may be because a century ago fonts didn't routinely come in families with weights ranging to extra bold or black (the squeezed Pabst example above notwithstanding).
• Posts: 1,314
edited July 2016
I experimented with making arbitrary nut fractions, but there were issues in support from layout applications. It’s easier just to make pre-composed fractions that substitute in the Fractions feature (see below).
Lindsey, bear in mind that often the Fraction feature will not be used, so it’s a good idea to make the fraction glyph thin, so that it is matched to your basic lining figures, reduced to 66% size by layout apps (or whatever the default percentage is). And with appropriate sidebearings.
Nut fractions are appropriate for certain kinds of typeface, notably didones, because in this day and age, consistency is expected throughout a type family, and nut fractions are problematic in heavier weights of most other serifed genres, and sans.
Here is a fairly simple code for Fractions:
feature frac {
# Fractions
# Prebuilt stacking fractions, otherwise slashed fractions
sub one slash two by onehalf;
sub one slash three by onethird;
sub one slash four by onequarter;
sub one slash eight by oneeighth;
sub two slash three by twothirds;
sub three slash four by threequarters;
sub three slash eight by threeeighths;
sub five slash eight by fiveeighths;
sub seven slash eight by seveneighths;
sub @figs' slash @figs by @figs_numr;
sub @figs' @figs slash by @figs_numr;
sub @figs' @figs @figs slash by @figs_numr;
sub @figs_numr slash @figs' by @figs_dnom;
sub slash by fraction;
sub @figs space' [@figs_numr onehalf onethird onequarter oneeighth twothirds threequarters threeeighths fiveeighths seveneighths] by space.frac;
} frac;
Tal Leming has some excellent material on the subject:
http://opentypecookbook.com/common-techniques.html
• Posts: 1,314
Some fonts even omit the fraction slash.
• Posts: 800
I unearthed this as a walk down memory lane for those of us who worked with Postscript Type 1 expert sets back in the day. ;-)
• Posts: 1,096
For anyone interested in generating nut fractions easily, below is what I use to generate them. "fracbar" is a horizintal bar component glyph I make that is the midpoint.
fracbar+onesuperior+~fourinferior=onequarter.nut fracbar+onesuperior+~twoinferior=onehalf.nut fracbar+threesuperior+~fourinferior=threequarters.nut fracbar+onesuperior+~threeinferior=onethird.nut fracbar+twosuperior+~threeinferior=twothirds.nut fracbar+onesuperior+~eightinferior=oneeighth.nut fracbar+threesuperior+~eightinferior=threeeighths.nut fracbar+fivesuperior+~eightinferior=fiveeighths.nut fracbar+~sevensuperior+~eightinferior=seveneighths.nut fracbar+onesuperior+~fiveinferior=onefifth.nut fracbar+twosuperior+~fiveinferior=twofifths.nut fracbar+threesuperior+~fiveinferior=threefifths.nut fracbar+foursuperior+~fiveinferior=fourfifths.nut fracbar+zerosuperior+~zeroinferior=zeronut.nut | 1,714 | 7,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | longest | en | 0.955003 |
http://slideplayer.com/slide/736438/ | 1,696,171,072,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510903.85/warc/CC-MAIN-20231001141548-20231001171548-00763.warc.gz | 43,475,360 | 17,163 | # Solving One-Step Equations. Created by S. Koch - 2003 3x – 6y + 18 = 0 What are your coefficients? What is your constant? 3, -6 18.
## Presentation on theme: "Solving One-Step Equations. Created by S. Koch - 2003 3x – 6y + 18 = 0 What are your coefficients? What is your constant? 3, -6 18."— Presentation transcript:
Solving One-Step Equations
Created by S. Koch - 2003 3x – 6y + 18 = 0 What are your coefficients? What is your constant? 3, -6 18
What you do to one side of the equation must also be done to the other side to keep it balanced. An equation is like a balance scale because it shows that two quantities are equal.
Created by S. Koch - 2003 Solving one-step equations: A one-step equation means you only have to perform 1 mathematical operation to solve it. You can add, subtract, multiply or divide to solve a one-step equation.
What is the variable? To solve one step equations, you need to ask three questions about the equation: What operation is performed on the variable? What is the inverse operation? (The one that will undo what is being done to the variable)
Created by S. Koch - 2003 Solving an addition equation : t + 7 = 21 Get rid of the 7 by adding its opposite to both sides of the equation. t + 7 = 21 -7 t = 14 Example 1
Created by S. Koch - 2003 Solving a subtraction equation : g – 6 = 40 Get rid of the 6 by adding its opposite to both sides of the equation. g – 6 = 40 +6 g = 46 Example 2
Created by S. Koch - 2003 Solving a multiplication equation : 8n = 32 Get rid of the 8 by dividing both sides of the equation by 8. Here we undo multiplication by doing the opposite – division. 8n = 32 88 n = 4 Example 3
Created by S. Koch - 2003 Solving a division equation : k / 9 = 11 Get rid of the 9 by multiplying both sides of the equation by 9. Here we undo division by doing the opposite – multiplication. k / 9 = 11 x 9 k = 99 Example 4
Created by S. Koch - 2003 Homework Use your notes and show ALL steps.
Download ppt "Solving One-Step Equations. Created by S. Koch - 2003 3x – 6y + 18 = 0 What are your coefficients? What is your constant? 3, -6 18."
Similar presentations | 608 | 2,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2023-40 | latest | en | 0.94915 |
http://mamicode.com/info-detail-3079664.html | 1,601,483,792,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402127075.68/warc/CC-MAIN-20200930141310-20200930171310-00527.warc.gz | 75,804,351 | 6,124 | # 题目描述
``````输入：n = 3
``````
``````输入：n = 11
``````
``````0 <= n < 2^31
``````
# 代码实现
``````class Solution {
public:
int findNthDigit(int n) {
int n_digits = 1;
long digit_count = 0;
while(true) {
digit_count += 9 * pow(10, n_digits - 1) * n_digits;
if(digit_count >= n)
break;
n_digits++;
}
int pre_digit_count = digit_count - 9 * pow(10, n_digits - 1) * n_digits;
int nth_number = (n - pre_digit_count - 1) / n_digits;
int nth_digits = (n - pre_digit_count - 1) % n_digits;
int tar_number = pow(10, n_digits - 1) + nth_number;
int result = tar_number / pow(10, n_digits - nth_digits - 1);
result = result % 10;
return result;
}
};
``````
(0)
(0) | 244 | 655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.188047 |
http://isabelle.in.tum.de/repos/isabelle/rev/f132d13fcf75 | 1,556,217,093,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578732961.78/warc/CC-MAIN-20190425173951-20190425195951-00264.warc.gz | 88,546,980 | 6,993 | author blanchet Thu May 12 15:29:19 2011 +0200 (2011-05-12) changeset 42747 f132d13fcf75 parent 42746 7e227e9de779 child 42748 a37095d03074
use the same code for extensionalization in Metis and Sledgehammer and generalize that code so that it gracefully handles negations (e.g. negated conjecture), formulas of the form (%x. t) = u, etc.
``` 1.1 --- a/src/HOL/Metis_Examples/Clausify.thy Thu May 12 15:29:19 2011 +0200
1.2 +++ b/src/HOL/Metis_Examples/Clausify.thy Thu May 12 15:29:19 2011 +0200
1.3 @@ -12,6 +12,37 @@
1.4 declare [[unify_search_bound = 100]]
1.5 declare [[unify_trace_bound = 100]]
1.6
1.7 +sledgehammer_params [prover = e, blocking, timeout = 10]
1.8 +
1.9 +
1.10 +text {* Extensionality *}
1.11 +
1.12 +lemma plus_1_not_0: "n + (1\<Colon>nat) \<noteq> 0"
1.13 +by simp
1.14 +
1.15 +definition inc :: "nat \<Rightarrow> nat" where
1.16 +"inc x = x + 1"
1.17 +
1.18 +lemma "inc \<noteq> (\<lambda>y. 0)"
1.19 +sledgehammer [expect = some] (inc_def plus_1_not_0)
1.20 +by (metis inc_def plus_1_not_0)
1.21 +
1.22 +lemma "inc = (\<lambda>y. y + 1)"
1.23 +sledgehammer [expect = some] (inc_def)
1.24 +by (metis inc_def)
1.25 +
1.26 +lemma "(\<lambda>y. y + 1) = inc"
1.27 +sledgehammer [expect = some] (inc_def)
1.28 +by (metis inc_def)
1.29 +
1.30 +definition add_swap :: "nat \<Rightarrow> nat \<Rightarrow> nat" where
1.31 +"add_swap = (\<lambda>x y. y + x)"
1.32 +
1.33 +lemma "add_swap m n = n + m"
1.34 +sledgehammer [expect = some] (add_swap_def)
1.36 +
1.37 +
1.38 text {* Definitional CNF for facts *}
1.39
1.40 declare [[meson_max_clauses = 10]]
1.41 @@ -83,6 +114,7 @@
1.42 (r 3 0 \<and> r 3 1 \<and> r 3 2 \<and> r 3 3)"
1.43 by (metisFT rax)
1.44
1.45 +
1.46 text {* Definitional CNF for goal *}
1.47
1.48 axiomatization p :: "nat \<Rightarrow> nat \<Rightarrow> bool" where
1.49 @@ -108,6 +140,7 @@
1.50 (p 0 1 \<or> \<not> x) \<and> (p 1 0 \<or> \<not> x) \<and> (p a b \<or> \<not> x)"
1.51 by (metisFT pax)
1.52
1.53 +
1.54 text {* New Skolemizer *}
1.55
1.56 declare [[metis_new_skolemizer]]
```
``` 2.1 --- a/src/HOL/Tools/Meson/meson.ML Thu May 12 15:29:19 2011 +0200
2.2 +++ b/src/HOL/Tools/Meson/meson.ML Thu May 12 15:29:19 2011 +0200
2.3 @@ -22,6 +22,8 @@
2.4 val choice_theorems : theory -> thm list
2.5 val skolemize_with_choice_theorems : Proof.context -> thm list -> thm -> thm
2.6 val skolemize : Proof.context -> thm -> thm
2.7 + val extensionalize_conv : Proof.context -> conv
2.8 + val extensionalize_theorem : Proof.context -> thm -> thm
2.9 val is_fol_term: theory -> term -> bool
2.10 val make_clauses_unsorted: thm list -> thm list
2.11 val make_clauses: thm list -> thm list
2.12 @@ -610,12 +612,37 @@
2.13 skolemize_with_choice_theorems ctxt (choice_theorems thy)
2.14 end
2.15
2.16 +fun is_Abs (Abs _) = true
2.17 + | is_Abs _ = false
2.18 +
2.19 +(* Removes the lambdas from an equation of the form "t = (%x. u)". *)
2.20 +fun extensionalize_conv ctxt ct =
2.21 + case term_of ct of
2.22 + Const (@{const_name HOL.eq}, _) \$ t1 \$ t2 =>
2.23 + ct |> (if is_Abs t1 orelse is_Abs t2 then
2.24 + Conv.rewr_conv @{thm fun_eq_iff [THEN eq_reflection]}
2.25 + then_conv extensionalize_conv ctxt
2.26 + else
2.27 + Conv.comb_conv (extensionalize_conv ctxt))
2.28 + | _ \$ _ => Conv.comb_conv (extensionalize_conv ctxt) ct
2.29 + | Abs _ => Conv.abs_conv (extensionalize_conv o snd) ctxt ct
2.30 + | _ => Conv.all_conv ct
2.31 +
2.32 +val extensionalize_theorem = Conv.fconv_rule o extensionalize_conv
2.33 +
2.34 (* "RS" can fail if "unify_search_bound" is too small. *)
2.35 -fun try_skolemize ctxt th =
2.36 - try (skolemize ctxt) th
2.37 - |> tap (fn NONE => trace_msg ctxt (fn () => "Failed to skolemize " ^
2.38 - Display.string_of_thm ctxt th)
2.39 - | _ => ())
2.40 +fun try_skolemize_etc ctxt =
2.41 + Raw_Simplifier.rewrite_rule (unfold_set_const_simps ctxt)
2.42 + (* Extensionalize "th", because that makes sense and that's what Sledgehammer
2.43 + does, but also keep an unextensionalized version of "th" for backward
2.44 + compatibility. *)
2.45 + #> (fn th => insert Thm.eq_thm_prop (extensionalize_theorem ctxt th) [th])
2.46 + #> map_filter (fn th => try (skolemize ctxt) th
2.47 + |> tap (fn NONE =>
2.48 + trace_msg ctxt (fn () =>
2.49 + "Failed to skolemize " ^
2.50 + Display.string_of_thm ctxt th)
2.51 + | _ => ()))
2.52
2.53 fun add_clauses th cls =
2.54 let val ctxt0 = Variable.global_thm_context th
2.55 @@ -645,7 +672,7 @@
2.56 fun gocls cls = name_thms "Goal#" (map make_goal (neg_clauses cls));
2.57
2.58 fun skolemize_prems_tac ctxt prems =
2.59 - cut_facts_tac (map_filter (try_skolemize ctxt) prems) THEN' REPEAT o etac exE
2.60 + cut_facts_tac (maps (try_skolemize_etc ctxt) prems) THEN' REPEAT o etac exE
2.61
2.62 (*Basis of all meson-tactics. Supplies cltac with clauses: HOL disjunctions.
2.63 Function mkcl converts theorems to clauses.*)
```
``` 3.1 --- a/src/HOL/Tools/Meson/meson_clausify.ML Thu May 12 15:29:19 2011 +0200
3.2 +++ b/src/HOL/Tools/Meson/meson_clausify.ML Thu May 12 15:29:19 2011 +0200
3.3 @@ -10,7 +10,6 @@
3.4 val new_skolem_var_prefix : string
3.5 val new_nonskolem_var_prefix : string
3.6 val is_zapped_var_name : string -> bool
3.7 - val extensionalize_theorem : thm -> thm
3.8 val introduce_combinators_in_cterm : cterm -> thm
3.9 val introduce_combinators_in_theorem : thm -> thm
3.10 val cluster_of_zapped_var_name : string -> (int * (int * int)) * bool
3.11 @@ -87,16 +86,6 @@
3.12
3.13 (**** REPLACING ABSTRACTIONS BY COMBINATORS ****)
3.14
3.15 -val fun_cong_all = @{thm fun_eq_iff [THEN iffD1]}
3.16 -
3.17 -(* Removes the lambdas from an equation of the form "t = (%x. u)".
3.18 - (Cf. "extensionalize_term" in "Sledgehammer_Translate".) *)
3.19 -fun extensionalize_theorem th =
3.20 - case prop_of th of
3.21 - _ \$ (Const (@{const_name HOL.eq}, Type (_, [Type (@{type_name fun}, _), _]))
3.22 - \$ _ \$ Abs _) => extensionalize_theorem (th RS fun_cong_all)
3.23 - | _ => th
3.24 -
3.25 fun is_quasi_lambda_free (Const (@{const_name Meson.skolem}, _) \$ _) = true
3.26 | is_quasi_lambda_free (t1 \$ t2) =
3.27 is_quasi_lambda_free t1 andalso is_quasi_lambda_free t2
3.28 @@ -321,7 +310,7 @@
3.29 val (th, ctxt) = Variable.import true [th] ctxt |>> snd |>> the_single
3.30 val th = th |> Conv.fconv_rule Object_Logic.atomize
3.31 |> Raw_Simplifier.rewrite_rule (unfold_set_const_simps ctxt)
3.32 - |> extensionalize_theorem
3.33 + |> extensionalize_theorem ctxt
3.34 |> make_nnf ctxt
3.35 in
3.36 if new_skolemizer then
```
``` 4.1 --- a/src/HOL/Tools/Metis/metis_tactics.ML Thu May 12 15:29:19 2011 +0200
4.2 +++ b/src/HOL/Tools/Metis/metis_tactics.ML Thu May 12 15:29:19 2011 +0200
4.3 @@ -132,20 +132,9 @@
4.4 (if Config.get ctxt verbose then "\n" ^ loc ^ ": " ^ msg
4.5 else ""))
4.6
4.7 -(* Extensionalize "th", because that makes sense and that's what Sledgehammer
4.8 - does, but also keep an unextensionalized version of "th" for backward
4.9 - compatibility. *)
4.10 -fun also_extensionalize_theorem th =
4.11 - let val th' = Meson_Clausify.extensionalize_theorem th in
4.12 - if Thm.eq_thm (th, th') then [th]
4.13 - else th :: Meson.make_clauses_unsorted [th']
4.14 - end
4.15 -
4.16 fun neg_clausify ctxt =
4.17 single
4.18 #> Meson.make_clauses_unsorted
4.19 - #> maps (Raw_Simplifier.rewrite_rule (Meson.unfold_set_const_simps ctxt)
4.20 - #> also_extensionalize_theorem)
4.21 #> map Meson_Clausify.introduce_combinators_in_theorem
4.22 #> Meson.finish_cnf
4.23
```
``` 5.1 --- a/src/HOL/Tools/Sledgehammer/sledgehammer_atp_translate.ML Thu May 12 15:29:19 2011 +0200
5.2 +++ b/src/HOL/Tools/Sledgehammer/sledgehammer_atp_translate.ML Thu May 12 15:29:19 2011 +0200
5.3 @@ -355,24 +355,11 @@
5.4 subst_atomic (map (fn (j, T) => (Free (concealed_bound_name j, T), Bound j))
5.5 (0 upto length Ts - 1 ~~ Ts))
5.6
5.7 -(* Removes the lambdas from an equation of the form "t = (%x. u)".
5.8 - (Cf. "extensionalize_theorem" in "Meson_Clausify".) *)
5.9 -fun extensionalize_term t =
5.10 - let
5.11 - fun aux j (@{const Trueprop} \$ t') = @{const Trueprop} \$ aux j t'
5.12 - | aux j (t as Const (s, Type (_, [Type (_, [_, T']),
5.13 - Type (_, [_, res_T])]))
5.14 - \$ t2 \$ Abs (var_s, var_T, t')) =
5.15 - if s = @{const_name HOL.eq} orelse s = @{const_name "=="} then
5.16 - let val var_t = Var ((var_s, j), var_T) in
5.17 - Const (s, T' --> T' --> res_T)
5.18 - \$ betapply (t2, var_t) \$ subst_bound (var_t, t')
5.19 - |> aux (j + 1)
5.20 - end
5.21 - else
5.22 - t
5.23 - | aux _ t = t
5.24 - in aux (maxidx_of_term t + 1) t end
5.25 +fun extensionalize_term ctxt t =
5.26 + let val thy = Proof_Context.theory_of ctxt in
5.27 + t |> cterm_of thy |> Meson.extensionalize_conv ctxt
5.28 + |> prop_of |> Logic.dest_equals |> snd
5.29 + end
5.30
5.31 fun introduce_combinators_in_term ctxt kind t =
5.32 let val thy = Proof_Context.theory_of ctxt in
5.33 @@ -436,7 +423,7 @@
5.34 val t = t |> need_trueprop ? HOLogic.mk_Trueprop
5.35 |> Raw_Simplifier.rewrite_term thy
5.36 (Meson.unfold_set_const_simps ctxt) []
5.37 - |> extensionalize_term
5.38 + |> extensionalize_term ctxt
5.39 |> presimp ? presimplify_term thy
5.40 |> perhaps (try (HOLogic.dest_Trueprop))
5.41 |> introduce_combinators_in_term ctxt kind
``` | 3,553 | 10,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | longest | en | 0.338678 |
https://yutsumura.com/matrices-satisfying-hf-fh-2f/ | 1,611,802,503,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835583.91/warc/CC-MAIN-20210128005448-20210128035448-00681.warc.gz | 1,074,970,117 | 28,758 | Matrices Satisfying $HF-FH=-2F$
Problem 69
Let $F$ and $H$ be an $n\times n$ matrices satisfying the relation
$HF-FH=-2F.$
(a) Find the trace of the matrix $F$.
(b) Let $\lambda$ be an eigenvalue of $H$ and let $\mathbf{v}$ be an eigenvector corresponding to $\lambda$. Show that there exists an positive integer $N$ such that $F^N\mathbf{v}=\mathbf{0}$.
Hint.
1. For (a), take the trace of the both sides of the given relation.
2. For (b), show that if $F^k\mathbf{v}\neq \mathbf{0}$ then there are infinitely many eigenvalues, hence a contradiction.
Proof.
(a) The trace of the matrix $F$
Using the given relation we compute the trace of $F$ as follows.
By taking the trace of both sides we have
$\tr(-2F)=\tr(HF-FH).$
The right hand side is $-2\tr(F)$ and the left hand side is
\begin{align*}
\tr(HF-FH)&=\tr(HF)-\tr(FH)\\
&=\tr(HF)-\tr(HF)=0.
\end{align*}
Therefore we have $\tr(F)=0$.
(b) There exists an positive integer $N$ such that $F^N\mathbf{v}=\mathbf{0}$.
Since $\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $\lambda$ of $H$, we have $H\mathbf{v}=\lambda \mathbf{v}$ or equivalently $(H-\lambda I)\mathbf{v}=\mathbf{0}$.
Now we compute
\begin{align*}
& F(H-\lambda I)=FH-\lambda F\\
&=(HF+2F)-\lambda F=(H-(\lambda-2)I)F.
\end{align*}
Therefore we have
$F(H-\lambda I)=(H-(\lambda-2)I)F.$
Evaluating at $\mathbf{v}$, we obtain
$\mathbf{0}=F(H-\lambda I)\mathbf{v}=(H-(\lambda-2)I)F\mathbf{v}.$
If $F\mathbf{v} \neq \mathbf{0}$, then this equality implies that $F\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $\lambda-2$ of $H$. In this case we further calculate
\begin{align*}
\mathbf{0}&=F(H-(\lambda-2)I)F\mathbf{v} \\
&=(FH-(\lambda-2)F)F=(HF+2F-(\lambda-2)F)F\mathbf{v}\\
&=(H-(\lambda-4))F^2\mathbf{v}.
\end{align*}
If the vector $F^2\mathbf{v}\neq \mathbf{0}$, then this equality implies that $F^2\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $\lambda-4$ of $H$.
Repeating this procedure, we see that
$\mathbf{0}=(H-(\lambda-2k))F^k\mathbf{v}$ for all $k$.
Therefore, if $F^k\mathbf{v}$ is nonzero vector for all $k$, then there are infinitely many eigenvalues $\lambda-2k$ but this is impossible since $H$ is an $n \times n$ matrix and hence $H$ has at most $n$ eigenvalues. Therefore there exists $N$ such that $F^N\mathbf{v}=\mathbf{0}$.
Related Question.
See the problem “Matrices satisfying the relation HE-EH=2E” for similar questions.
As noted there, the relation $HF-FH=-2F$ comes from the Lie algebra $\mathfrak{sl}(2)$.
More from my site
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• Stochastic Matrix (Markov Matrix) and its Eigenvalues and Eigenvectors (a) Let $A=\begin{bmatrix} a_{11} & a_{12}\\ a_{21}& a_{22} \end{bmatrix}$ be a matrix such that $a_{11}+a_{12}=1$ and $a_{21}+a_{22}=1$. Namely, the sum of the entries in each row is $1$. (Such a matrix is called (right) stochastic matrix (also termed […]
• Trace of the Inverse Matrix of a Finite Order Matrix Let $A$ be an $n\times n$ matrix such that $A^k=I_n$, where $k\in \N$ and $I_n$ is the $n \times n$ identity matrix. Show that the trace of $(A^{-1})^{\trans}$ is the conjugate of the trace of $A$. That is, show that […]
• Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even Let $A$ be a real skew-symmetric matrix, that is, $A^{\trans}=-A$. Then prove the following statements. (a) Each eigenvalue of the real skew-symmetric matrix $A$ is either $0$ or a purely imaginary number. (b) The rank of $A$ is even. Proof. (a) Each […]
• Matrix $XY-YX$ Never Be the Identity Matrix Let $I$ be the $n\times n$ identity matrix, where $n$ is a positive integer. Prove that there are no $n\times n$ matrices $X$ and $Y$ such that $XY-YX=I.$ Hint. Suppose that such matrices exist and consider the trace of the matrix $XY-YX$. Recall that the trace of […]
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3 Responses
1. D says:
a simpler approach for B
$HF − FH= −2F$
If F is singular, then there must be at least one eigenvalue = 0, and we are done. (i.e. we are satisfied with natural number N = 1.)
Now suppose F is non-singular, and right multiply the equation by $F^{-1}$. This gives us
$H − FHF^{-1}= −2I$
now take the trace
$trace(H) − trace(FHF^{-1})= −2trace(I)$
$trace(H) − trace(F^{-1}FH)= −2n$
$trace(H) − trace(H)= −2n$
$0= −2n$
a contradiction. Hence F is singular and we’re done.
• Yu says:
Dear D,
Your argument is correct if $\lambda=0$. But $\lambda$ is any eigenvalue of $A$, and even if $A$ is singular, there could be a nonzero eigenvalue of $A$.
1. 07/29/2017
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Matrices Satisfying the Relation $HE-EH=2E$
Let $H$ and $E$ be $n \times n$ matrices satisfying the relation $HE-EH=2E.$ Let $\lambda$ be an eigenvalue of the...
Close | 2,025 | 6,091 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2021-04 | latest | en | 0.509399 |
https://auto.howstuffworks.com/fuel-efficiency/alternative-fuels/question707.htm/printable | 1,521,491,939,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647146.41/warc/CC-MAIN-20180319194922-20180319214922-00757.warc.gz | 526,491,796 | 42,293 | # How much corn would I need to fuel a cross-country trip with ethanol?
Alternative Fuel Vehicle Image Gallery Packaged seed corn sits in the barn of a seed dealer. How much of this would you need to drive across the country? Check out pictures of alternative fuel vehicles.
Scott Olson/Getty Images
With so much volatility in today's world oil market, many are seeking out alternative fuels to power cars. Some, including corn producers, have touted ethanol is a possible alternative fuel. Ethanol, or ethyl alcohol, is made by fermenting and distilling simple sugars from corn. Ethanol is sometimes blended with gasoline to produce gasohol. Ethanol-blended fuels account for 12 percent of all automotive fuels sold in the United States, according to the Renewable Fuels Association. In very pure forms, ethanol can be used as an alternative to gasoline in vehicles modified for its use.
In order to calculate how much corn you would have to grow to produce enough ethanol to fuel a trip across the country, there are a couple of basic factors we have to consider:
• Let's assume that you drive a Toyota Camry, the best-selling car in America in 2000. We know that the Toyota Camry with automatic transmission gets 30 miles per gallon of gas on the highway.
• Gasoline is more efficient than ethanol. One gallon of gasoline is equal to 1.5 gallons of ethanol. This means that same Camry would only get about 20 miles to the gallon if it were running on ethanol.
• We also need to know how far you are traveling: Let's say from Los Angeles to New York, which is 2,774 miles (4,464.2 km), according to MapQuest.com.
• Through research performed at Cornell University, we know that 1 acre of land can yield about 7,110 pounds (3,225 kg) of corn, which can be processed into 328 gallons (1240.61 liters) of ethanol. That is about 26.1 pounds (11.84 kg) of corn per gallon.
First, we need to figure out how much fuel we will need:
(METRIC: 4,464.2 km / 8.5 km per liter = 525.2 liters)
We know that it takes 26.1 pounds of corn to make 1 gallon of ethanol, so we can now calculate how many pounds of corn we need to fuel the Camry on its trip:
(METRIC: 525.2 liters * 3.13 kg = 1,642 kg)
You will need to plant a little more than a half an acre of corn to produce enough ethanol to fuel your trip.
According to the research from Cornell, you need about 140 gallons (530 liters) of fossil fuel to plant, grow and harvest an acre of corn. So, even before the corn is converted to ethanol, you're spending about \$1.05 per gallon.
"The energy economics get worse at the processing plants, where the grain is crushed and fermented," reads the Cornell report. The corn has to be processed with various enzymes; yeast is added to the mixture to ferment it and make alcohol; the alcohol is then distilled to fuel-grade ethanol that is 85- to 95-percent pure. To produce ethanol that can be used as fuel, it also has to be denatured with a small amount of gasoline.
The final cost of the fuel-grade ethanol is about \$1.74 per gallon. (Of course, a lot of variables go into that number.) | 722 | 3,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-13 | latest | en | 0.946731 |
http://spmath84109.blogspot.com/2009/12/riemers-scribepost-for-december-3-09.html | 1,490,764,601,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190183.80/warc/CC-MAIN-20170322212950-00256-ip-10-233-31-227.ec2.internal.warc.gz | 326,691,969 | 22,557 | ## Thursday, December 3, 2009
### Riemer's scribepost for December 3, 09
Question 21
Kate is going to put a patio in her backyard. The patio stones she is using each have an area of 1 m2. She has created the rectangular design shown.
a)What is the area of the patio?
The area of the patio is 56m.
14 x 4 = 56
b) What are the dimensions of another rectangular patio she could build with the same area?
Another rectangular patio she could build with the same area is 8 x 7.
8 x 7 = 56
c) Kate decides to make a patio with the same area but she wants it to be a square with whole number side lengths. Is this possible? Explain your reasoning.
I think that this isn't possible because it has a decimal and 56 is not a perfect square.
Question 25
A square digital photo on the computer has an area of 144 cm2.
a) What is the side length of the photo?
b) The photo is enlarged so that the side length is now 36 cm. What is the area of the enlarged photo?
36 x 36 = 1296 squared
c) How many times as large as the original area is the enlarged area? that it is not a perfect square.
RIEMERMUNOZ !
elijah841 said...
Hey Riemer ,
Good job ! But, for question 21, a) you should put why did you multiply the " 14 x 4 " together . Also, for c) you should explain why nmber 56 is not a perfect square .. like the sqaure root of a 56 is a decimal ...
sierra841 said...
Good job. I like how you took pictures, and had the time to add them into your scribe. Also, on question 25, you could put "144 cm squared" instead of "144cm2"
vanessa 8-41 said...
Good job Riemer ! I liked your use of colour and the way you organized your post. For question 21a), I agree with Elijah, you should explain how you got to that answer and show it somehow. Just a reminder, there is a squared 2 on character map. Your green that you chose, isn't such a great green to use mostly because its too bright, but I understand that you want it to stand out. Overall good job.
angaldrine said...
Good job Riemer i liked how you put photos in and colors. same with me i agree with elijah but you did an awesome job the pictures are clear and put a lot of effort. :D good job awesome work
emmanuel841 said...
Good Job Riemer,
You could have explained how, and why you got those numbers you did to find the areas of the patios, like elijah and vanessa said. Also for nextime, you should be aware of the colours you use. But overall, this was a great scribepost. | 657 | 2,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-13 | longest | en | 0.946449 |
http://www.exampleproblems.com/wiki/index.php/Function | 1,611,375,551,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703533863.67/warc/CC-MAIN-20210123032629-20210123062629-00132.warc.gz | 136,103,802 | 20,115 | Function
In mathematics, a function is a relation, such that each element of a set (the domain) is associated with a unique element of another (possibly the same) set (the codomain, not to be confused with the range). The concept of a function is fundamental to virtually every branch of mathematics and every quantitative science.
The terms function, mapping, map and transformation are usually used synonymously. The term operation is frequently used for binary functions; functions whose domain is a set of functions, or a vector space, are often called operators (see also operator (programming)).
Intuitive introduction
Essentially, a function is a "rule" or procedure that assigns an "output" value to each given "input" value. The following are examples of functions:
• In a group of people, each person has a favorite colour—from the set of red, orange, yellow, green, cyan, blue, indigo, or violet. Here, the input is the person, and the output is one of the 8 colours. The favorite colour is a function of the person. For example, John has favorite colour red, while Kim has favorite colour violet. Note that more than one person may be associated with a given colour (e.g., John and Kim may both like red), but one person cannot have more or less than one favorite color.
• A stone is dropped from different stories of a tall building. The dropped stone may take 2 seconds to fall from the second story, and 4 seconds to fall from the 8th story. Here, the input is the story, and the output is the number of seconds. The relevant function describes the relationship between the time it takes the stone to reach the ground and the story. (See acceleration)
The "rule" defining a function can be specified by a formula, a relationship, or simply a table listing the outputs against inputs. The most important feature of a function is that it is consistent, or deterministic, always producing the same output from a given input. In this way, a function may be thought of as a mechanism or "machine" (a "black box") consistently converting a given valid input into its unique associated output. In certain technical contexts, the input is often called the argument of the function, and the output the value of the function.
A very common type of function occurs when the argument (input) and the value (output) are both numbers, the functional relationship is expressed by a formula, and the value (output) of the function is obtained by direct substitution of the argument into the formula. Consider for example
${\displaystyle f(x)=x^{2}}$
which for any number x, assigns to x the associated value the square of x.
A straightforward generalization is to allow functions depending on several arguments. For instance,
${\displaystyle g(x,y)=xy}$
is a function which takes the input, two expressions x and y, and assigns to it its product (output), xy. It might seem that this is not really a function as we described above, because this "rule" depends on two inputs. However, if we think of the two inputs together as a single pair (x, y), then we can interpret g as a function -- the argument (unified single input) is the ordered pair (x, y), and the function value (output) is xy. Such functions whose input consists of ordered pairs are called "binary" or "2-ary".
In the sciences, we often encounter functions that are not given by (known) formulas. Consider for instance the temperature distribution on earth over time: this is a function which takes location and time as arguments and gives as output value the temperature at the indicated location at the indicated moment in time.
We have seen that the intuitive notion of function is not limited to computations using single numbers and not even limited to computations; the mathematical notion of function is still more general and is not limited to situations involving numbers. Rather, a function links a "domain" (set of inputs) to a "codomain" (set of possible outputs) in such a way that every element of the domain is associated to precisely one element of the codomain. Functions are abstractly defined as certain relations, as will be seen below. Because of this generality, the function concept is fundamental to virtually every branch of mathematics and the quantitative sciences.
History
As a mathematical term, "function" was coined by Leibniz in 1694, to describe a quantity related to a curve, such as a curve's slope or a specific point of a curve. The functions Leibniz considered are today called differentiable functions, and they are the type of function most frequently encountered by nonmathematicians. For this type of function, one can talk about limits and derivatives; both are measurements of the change of output values associated to a change of input values, and these measurements are the basis of calculus.
The word function was later used by Euler during the mid-18th century to describe an expression or formula involving various arguments, e.g. f(x) = sin(x) + x3.
During the 19th century, mathematicians started to formalize all the different branches of mathematics. Weierstrass advocated building calculus on arithmetic rather than on geometry, which favoured Euler's definition over Leibniz's (see arithmetization of analysis).
By broadening the definition of functions, mathematicians were then able to study "strange" mathematical objects such as continuous functions that are nowhere differentiable. These functions were first thought to be only theoretical curiosities, and they were collectively called "monsters" as late as the turn of the 20th century. However, powerful techniques from functional analysis have shown that these functions are in some sense "more common" than differentiable functions. Such functions have since been applied to the modelling of physical phenomena such as Brownian motion.
Towards the end of the 19th century, mathematicians started trying to formalize all of mathematics using set theory, and they sought to define every mathematical object as a set. Dirichlet and Lobachevsky independently and almost simultaneously gave the modern "formal" definition of function (see formal definition below).
In this definition, a function is a special case of a relation. In most cases of practical interest, however, the differences between the modern definition and Euler's definition are negligible.
The notion of function as a rule for computing, rather than a special kind of relation, has been formalized in mathematical logic and theoretical computer science by means of several systems, including the lambda calculus, the theory of recursive functions and the Turing machine.
Formal definition
Formally, a function f from a set X of input values to a set Y of possible output values (written as f : X → Y) is a relation between X and Y which satisfies:
1. f is total, or entire: for all x in X, there exists a y in Y such that x f y (x is f-related to y), i.e. for each input value, there is at least one output value in Y.
2. f is many-to-one, or functional: if x f y and x f z, then y = z. i.e., many input values can be related to one output value, but one input value cannot be related to many output values.
For each input value x in the domain, the corresponding unique output value y in the codomain is denoted by f(x).
A more concise expression of the above definition is the following: a function from X to Y is a subset f of the cartesian product X × Y, such that for each x in X, there is a unique y in Y such that the ordered pair (x, y) is in f.
The set of all functions f : X → Y is denoted by YX. Note that |YX| = |Y||X| (refer to Cardinal numbers).
A relation between X and Y that satisfies condition (1) is a multivalued function. Every function is a multivalued function, but not every multivalued function is a function. A relation between X and Y that satisfies condition (2) is a partial function. Every function is a partial function, but not every partial function is a function. In this encyclopedia, the term "function" will mean a relation satisfying both conditions (1) and (2), unless otherwise stated.
Consider the following three examples:
File:NotMap1.png This relation is total but not many-to-one; the element 3 in X is related to two elements b and c in Y. Therefore, this is a multivalued function, but not a function. File:NotMap2.png This relation is many-to-one but not total; the element 1 in X is not related to any element of Y. Therefore, this is a partial function, but not a function. 200px This relation is both total and many-to-one, and so it is a function from X to Y. Note that the emphasis is on "-to-one" as "many" may actually mean "one". The function can be given explicitly as f = {(1, D), (2, B), (3, A), (4, A)} or as ${\displaystyle f(x)=\left\{{\begin{matrix}D,&{\mbox{if }}x=1\\B,&{\mbox{if }}x=2\\A,&{\mbox{if }}x=3\\A,&{\mbox{if }}x=4.\end{matrix}}\right.}$
Domains, codomains, and ranges
X, the set of input values, is called the domain of f, and Y, the set of possible output values, is called the codomain. The range of f is the set of all actual outputs {f(x) : x in the domain}. Beware that sometimes the codomain is incorrectly called the range because of a failure to distinguish between possible and actual values.
Functions are named after their ranges, for example real functions and complex functions.
An endofunction is a function whose domain and range are identical.
In computer science, the datatypes of the arguments and return values specify the domain and codomain (respectively) of a subprogram. So the domain and codomain are constraints imposed initially on a function; on the other hand the range has to do with how things turn out in practice.
Injective, surjective and bijective functions
Main article: bijection, injection and surjection
Several properties of functions that are very useful have special names:
• Injective (one-to-one) functions send different arguments to different values; in other words, if x1 and x2 are members of the domain of f and f(x1) = f(x2), then x1 = x2.
• Surjective (onto) functions have their range equal to their codomain; in other words, if y is any member of the codomain of f, then there exists at least one x such that f(x) = y.
• Bijective functions are both injective and surjective; they are often used to show that the sets X and Y are the "same size" in some sense.
Images and preimages
The image of an element xX under f is the output f(x).
The image of a subset AX under f is the subset of Y formally defined by
f[A] = {f(x) | x in A}
Usually (when subsets of the domain are not at the same time elements of the domain) one writes f(A) instead of f[A].
(An old-fashioned notation writes f'x instead of f(x), and f''A instead of f[A].)
Notice that the range of f is the image f(X) of its domain. In our function above, the image of {2,3} under f is f({2, 3}) = {c, d} and the range of f is {c, d}.
The preimage (or inverse image) of a set BY under f is the subset of X defined by
f −1(B) = {x in X | f(x) ∈B}
For our function above, the preimage of {a, b} is f −1({a, b}) = {1}.
Graph of a function
The graph of a function f is the set of all ordered pairs(x, f(x)), for all x in the domain X. There are theorems formulated or proved most easily in terms of the graph, such as the closed graph theorem. If X and Y are real lines, then this definition coincides with the familiar sense of graph.
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This cubic polynomial is surjective but not injective.
Note that a binary relation on the two sets X and Y could be identified with an ordered triple (X,Y,G) where G is the graph of the relation or identified with its graph. In the latter case a function f is the same as its graph.
Examples of functions
(More can be found at list of functions.)
• The relation wght between persons in the United States and their weights at a particular time.
• The relation between nations and their capitals, if we exclude those nations that maintain multiple capitals [1].
• The relation sqr between natural numbers n and their squares n2.
• The relation ln between positive real numbers x and their natural logarithms ln(x). Note that the relation between real numbers and their natural logarithms is not a function because not every real number has a natural logarithm; that is, this relation is not total.
• The relation dist between points in the plane R2 and their distances from the origin (0,0).
• The relation grav between a point in the punctured plane R2 \ {(0,0)} and the vector describing the gravitational force that a certain mass at that point would experience from a certain other mass at the origin (0,0).
The most commonly used types of mathematical functions involve addition, division, exponents, logarithms, multiplication, polynomials, radicals, rationals, subtraction, and trigonometric expressions. These functions, and functions obtained by composing them, are sometimes collectively referred as elementary functions -- but the meaning of this term varies among different branches of mathematics. Examples of non-elementary functions (or special functions) are Bessel functions and the Gamma function.
Properties of functions
Functions can be
Ambiguous functions
An ambiguous function is a mathematical equation that can have more than one correct answer. For example, the square root of 4 can be either -2 or 2 as both answers squared would give 4.
Strictly speaking, an ambiguous function is not truly a function because a mathematical function is defined as having "a unique output to each given input". In fact, such "functions" are more properly termed relations.
n-ary function: function of several variables
Functions in applications are often functions of several variables, or multivariate functions: the values they take depend on a number of different factors. From a mathematical point of view all the variables must be made explicit in order to have a functional relationship - no 'hidden' factors are allowed. Then again, from the mathematical point of view, there is no qualitative difference between functions of one and of several variables. A function of three real variables is just a function that applies to triples of real numbers. The following paragraph says this in more formal language.
If the domain of a function is a subset of the Cartesian product of n sets then the function is called an 'n-ary function. For example, the relation dist has the domain R × R and is therefore a binary function. In that case dist((x,y)) is simply written as dist(x,y).
Another name applied to some types of functions of several variables is operation. In abstract algebra, operators such as "*" are defined as binary functions; when we write a formula such as x*y in this context, we are implicitly invoking the function *(x,y), but writing it in a convenient infix notation.
An important theoretical paradigm, functional programming, takes the function concept as central. In that setting, the handling of functions of several variables becomes an operational matter, for which the lambda calculus provides the basic syntax. The composition of functions (see under composing functions immediately below) becomes a question of explicit forms of substitution, as used in the substitution rule of calculus. In particular, a formalism called currying can be used to reduce n-ary functions to functions of a single variable.
Composing functions
The functions fX → Y and gY → Z can be composed by first applying f to an argument x and then applying g to the result. Thus one obtains a composite function g o f: X → Z defined by (g o f)(x) = g(f(x)) for all x in X. As an example, suppose that an airplane's height at time t is given by the function h(t) and that the oxygen concentration at height x is given by the function c(x). Then (c o h)(t) describes the oxygen concentration around the plane at time t.
Inverse function
If a function f:XY is bijective then the preimage of any element y in the codomain Y is a singleton. A function taking yY to its preimage f−1(y) is a well-defined function called the inverse of f and is denoted by f−1.
An example of an inverse function, for f(x) = 2x, is f(x)−1 = x/2. The inverse function is the function that "undoes" its original. See also inverse image.
Inverses are sometimes difficult or impossible to find. Consider f(x) = x2. The function f(x) = √x is not an inverse when the domain of f is R. (As -22 is 4, but √4 is either 2 or -2).
Restrictions and extensions
Suppose that ${\displaystyle X}$ is a subset of ${\displaystyle Y}$ and that
${\displaystyle f:Y\rightarrow Z}$
is a function. Let
${\displaystyle i:X\hookrightarrow Y}$
be the inclusion function
${\displaystyle i(x)=x\,}$
for all ${\displaystyle x}$ in ${\displaystyle X}$.
The restriction of ${\displaystyle f}$ to ${\displaystyle X,}$ denoted by ${\displaystyle f|_{X},}$ is the function
${\displaystyle f|_{X}=f\circ i.}$
Intuitively, this is the same function as ${\displaystyle f}$ except that we restrict the domain of ${\displaystyle f}$ to ${\displaystyle X}$.
An extension of a function
${\displaystyle g:X\to Z,}$
is a function
${\displaystyle f:Y\to Z}$
defined on a superset ${\displaystyle Y}$ of ${\displaystyle X}$ such that ${\displaystyle f|_{X}=g}$. Provided the domain of ${\displaystyle g}$ is not the universal set, ${\displaystyle g}$ always has lots of extensions.
Pointwise operations
If fX → R and gX → R are functions with common domain X and codomain is a ring R, then one can define the sum function f + g: X → R and the product function f × g: X → R as follows:
(f + g)(x) = f(x) + g(x)
(f × g)(x) = f(x) × g(x)
for all x in X.
This turns the set of all such functions into a ring. The binary operations in that ring have as domain ordered pairs of functions, and as codomain functions. This is an example of climbing up in abstraction, to functions of more complex types.
By taking some other algebraic structure A in the place of R, we can turn the set of all functions from X to A into an algebraic structure of the same type in an analogous way.
Computable and non-computable functions
The number of computable functions from integers to integers is countable, because the number of possible algorithms is. The number of all functions from integers to integers is higher: the same as the cardinality of the real numbers. This argument shows that there are functions from integers to integers that are not computable. For examples of noncomputable functions, see the articles on the halting problem and Rice's theorem.
Functions from the categorical viewpoint
In the formal definition, a function represents a relationship between its domain and its codomain, rather than just a rule for taking an input to an output. A generalisation of the notion of function is morphism in the context of category theory. A category is a collection of objects and morphisms, each morphism is an ordered triple (X, Y, f), where f is a rule connecting domain X and codomain Y, and X and Y are objects in the collection.
Ordinary functions are sometimes referred to as morphisms in a concrete category. | 4,393 | 19,185 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 24, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-04 | longest | en | 0.934945 |
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103 學年度第 2 學期統計學 期末會考
1. A Type II error is committed if we make:
A. a correct decision when the null hypothesis is false. B. a correct decision when the null hypothesis is true. C. an incorrect decision when the null hypothesis is false. D. an incorrect decision when the null hypothesis is true. 2. If a test of hypothesis has a Type I error probability of .05, this means that: A. if the null hypothesis is true, we don't reject if 5% of the time. B. if the null hypothesis is true, we reject it 5% of the time. C. if the null hypothesis is false, we don't reject it 5% of the time. D. if the null hypothesis is false, we reject it 5% of the time. 3. Which of the following is an appropriate null hypothesis? A. The mean of a population is equal to 60. B. The mean of a sample is equal to 60. C. The mean of a population is not equal to 60. D. All of these choices are true.
4. Which of the following conclusions is not an appropriate conclusion from a hypothesis test?
A. Reject H 0 . Sufficient evidence to support H 1 . B. Accept H 0 . Sufficient evidence to support H 0 . C. Fail to reject H 0 . Insufficient evidence to support H 1 . D. All of these choices are true. 5. The power of a test is measured by its capability of: A. rejecting a null hypothesis that is true. B. not rejecting a null hypothesis that is true. C. rejecting a null hypothesis that is false. D. not rejecting a null hypothesis that is false.
6. A major electronics store chain is interested in estimating the average amount its credit card customers spent on their first visit to the chain's new store in the mall. Fifteen credit card accounts were randomly sampled and analyzed with the following results: = \$50.50 and s 2 = 400. A 95% confidence interval for the average amount the credit card customers spent on their first visit to the chain's new store in the mall is:
A. \$50.50 ± \$9.09. B. \$50.50 ± \$10.12. C. \$50.50 ± \$11.08. D. None of these choices.
1
7. Based on sample data, the 90% confidence interval limits for the population mean are LCL = 170.86 and UCL = 195.42. If the 10% level of significance were used in testing the hypotheses H 0 : µ = 201 vs. H 1 : µ 201, the null hypothesis:
A. would be rejected. B. would be accepted. C. would fail to be rejected. D. would become H 0 : µ ≠ 201
8. A random sample of 25 observations is selected from a normally distributed population. The sample variance is 10. In the 95% confidence interval for the population variance, the upper limit is:
A. 19.355 B. 17.391 C. 17.11 D. 6.091
9. In a hypothesis test for the population variance, the hypotheses are H 0 : σ 2 = 100 vs. H 1 : σ 2 100. If the sample size is 15 and the test is being carried out at the 10% level of significance, the rejection region is:
A. χ 2 < 7.790 or χ 2 > 21.064. B. χ 2 < 6.571 or χ 2 > 23.685. C. χ 2 < 8.547 or χ 2 > 22.307. D. χ 2 < 7.261 or χ 2 < 24.996.
10. The use of the standard normal distribution for constructing confidence interval estimate for the population proportion p requires:
A. n and n(1 − ) both greater than 5. B. np and n(1 − p) both greater than 5. C. n(1 + ) and n(1 − ) both greater than 5. D. sample size greater than 5.
11. A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors' results in 83 who indicate that they recommend aspirin. The value of the test statistic (z) in this problem is approximately equal to:
A. −1.67 B. −2.33 C. −1.86 D. −.14
12. When determining the sample size needed for a proportion for a given level of confidence and sampling error, the closer to 0.50 that p is estimated to be:
A. the smaller the sample size required. B. the larger the sample size required. C. the sample size is not affected. D. the effect cannot be determined from the information given.
13. As a manufacturer of guitars, a major corporation wants to estimate the proportion of guitar players who are right-handed. How many golfers must be surveyed if they want to be within 0.02, with a 95% confidence? (Assume that there is no prior information that could be used as an estimate of .)
A. 600 B. 1101 C. 1200 D. 2401
2
14. The quantity
2
s p
is called the pooled variance estimate of the common variance of two unknown but equal population variances. It is the weighted average
of the two sample variances, where the weights represent the:
A. Sample variances. B. Sample standard deviations. C. Degrees of freedom for each sample. D. None of these choices.
15. Two independent samples of sizes 20 and 30 are randomly selected from two normally distributed populations. Assume that the population variances are
unknown but equal. In order to test the difference between the population means,
is:
A. Normal.
C. Student-t with 48 degrees of freedom.
B.
D.
, the sampling distribution of the sample mean difference,
Student-t with 50 degrees of freedom. None of these choices.
,
16. In testing the difference between the means of two normal populations using two independent samples when the population variances are unequal, the sampling distribution of the resulting statistic is:
A. Normal. B. Student-t. C. Approximately normal. D. Approximately Student-t.
17. If some natural relationship exists between each pair of observations that provides a logical reason to compare the first observation of sample 1 with the first observation of sample 2, the second observation of sample 1 with the second observation of sample 2, and so on, the samples are referred to as:
A. Matched pairs. B. Independent samples. C. Nonrandom samples. D. None of these choices.
18. When testing for the difference between two population variances with sample sizes of n 1 = 8 and n 2 = 10, the degrees of freedom are:
A. 8 and 10 B. 7 and 9 C. 2 and 18 D. 18 and 2
19. Which of the following statements is correct regarding the percentile points of the F-distribution?
A. F 0.05,10,20 = 1/F 0.95,10,20 B. F 0.05,10,20 = 1/F 0.05,20,10 C. F 0.95,10,20 = 1/F 0.05,20,10 D. F 0.95,10,20 = 1/F 0.95,20,10
3
20. For testing the difference between two population proportions, the pooled proportion estimate should be used to compute the value of the test statistic when the:
A. Populations are normally distributed. B. Sample sizes are small. C. Null hypothesis states that the two population proportions are equal. D. Samples are independently drawn from the populations.
21. Which of the following is a required condition for using the normal approximation to the binomial in testing the difference between two population proportions?
A.
C.
,
,
, and
B.
D. Choice b is the true requirement, but choice c is the one you actually check.
,
,
, and
22. Given the following information about a hypothesis test of the difference between two means based on independent random samples, which one of the
following is the correct rejection region at a significance level of 0.05? H 0 :
µ µ
A
B
, H 1 :
µ >µ
A
B
.
X
1
= 12 ,
X
2 = 9 , s 1 = 4, s 2 = 3.9, n 1 = 13, n 2 = 10.
A. Reject H 0 if Z > 1.96 B. Reject H 0 if Z > 1.645 C. Reject H 0 if t > 1.721 D. Reject H 0 if t > 2.08
23. Given the following information about a hypothesis test of the difference between two means based on independent random samples, what is the standard deviation of the difference between the two means? Assume that the samples are obtained from normally distributed populations having equal variances. H 0 :
µ µ
A
B
, H 1 :
µ >µ
A
B
.
X
1
= 12 ,
X
2 = 9 , s 1 = 5, s 2 = 3, n 1 = 13, n 2 = 9.
A. 2.823 B. 3.21 C. 4.313 D. 5.281
24. Given the following information about a hypothesis test of the difference between two means based on independent random samples, what is the calculated
.
value of the test statistic? Assume that the samples are obtained from normally distributed populations having equal variances. H 0 :
µ µ
A
B
, H 1 :
µ >µ
A
B
X
1
= 12 ,
X
2 = 9 , s 1 = 5, s 2 = 3, n 1 = 13, n 2 = 9.
A. t = 1.064 B. t = 1.604 C. t = 1.894 D. t = 2.824
25. Given the following information about a hypothesis test of the difference between two variances based on independent random samples, what is the calculated
value of the test statistic? Assume that the samples are obtained from normally distributed populations. H 0 :
5, s 2 = 3, n 1 = 13, n 2 = 9.
σ =σ
A
2
2
B
, H 1 :
σ σ
A
2
2
B
.
X
1
= 12 ,
X
2
= 9 , s 1 =
A. 1.333 B. 1.667 C. 2 D. 2.778
4 | 2,404 | 8,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-24 | latest | en | 0.914119 |
http://www.entisoft.com/ESTools/MathFinancial_ModifiedInternalRateOfReturn.HTML | 1,516,550,604,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890795.64/warc/CC-MAIN-20180121155718-20180121175718-00000.warc.gz | 427,121,782 | 2,146 | ## Modified Internal Rate Of Return FunctionMath Financial Class
```Public Function ModifiedInternalRateOfReturn( _
ByVal vValues As Variant _
, ByVal vFinanceRate As Variant _
, ByVal vReinvestRate As Variant _
) As Variant```
### Calculate the Internal Rate of Return for a series of periodic cash flows, when the payments and receipts are financed at different rates.
Example: What is the internal rate of return on an investment where \$10,000 is invested today, \$4,000 is withdrawn one year from today, and \$10,000 is withdrawn two years from today, assuming that 7% is paid to finance the investment and 5% is received on gains from reinvestment? Approximately 19.1638%.
` ModifiedInternalRateOfReturn(Array(-10000, 4000, 10000), .07, .05) = 0.191637528781298`
Example: What is the internal rate of return on an loan where \$10,000 is borrowed today, \$4,000 is repaid one year from today, and \$10,000 is repaid two years from today, assuming that the 7% is paid to finance the investment and 5% is received on gains from reinvestment? Approximately -5.9824%.
` ModifiedInternalRateOfReturn(Array(10000, -4000, -10000), .07, .05) = -0.0598244062629451`
See the ModifiedInternalRateOfReturnVerify Subroutine for more examples of this function.
``` InternalRateOfReturn Function
UnevenInternalRateOfReturn Function
XIRR Function
NetPresentValue Function
UnevenNetPresentValue Function
MIRR Function (Visual Basic)
MIRR Function (Microsoft Excel)```
Summary: The modified internal rate of return is the internal rate of return when payments are financed at one rate and receipts are financed at a different rate. This function uses the order of the values within the array vValues as the order of payments and receipts.
vValues: One dimensional array of values representing a table of periodic cash flows. The array must contain at least one payment (a negative value) and at least one receipt (a positive value). Every element of this array is examined and considered to be one cash flow, unlike some other Entisoft Tools functions which examine arrays beginning with element #1.
vFinanceRate: The interest rate paid to finance the investment. For example, 9% would be represented as 0.09.
vReinvestRate: The interest rate received on gains from reinvestment. For example, 11% would be represented as 0.11.
Return value: The return value will be a percentage between -100% (as represented by -1) and 1E12 (1,000,000,000,000). A rate of return of 17%, for example, would be returned as 0.17.
Algorithm: Function uses successive approximation to calculate the Internal Rate of Return. It will return Null if it cannot calculate the result to an accuracy of 2.22044604925031E-16 (D1MACH(4)) within 256 tries. In practice, this seldom happens.
v2.0 Addition: This function is new to this version of Entisoft Tools. | 679 | 2,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-05 | latest | en | 0.863644 |
https://diyodemag.com/education/what_the_tech_little_black_boxes | 1,723,159,358,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00664.warc.gz | 148,247,823 | 15,091 | What The Tech
# Integrated Circuits Explained
Bob Harper
Issue 22, May 2019
Most of our electronics now days include integrated circuits, but what are these mysterious black boxes and how do they work?
# INTEGRATED
We have often used ICs in Diyode Projects, and in fact, most of our projects use ICs rather than discrete components. So, what is meant by an IC? Integrated Circuits have been around for much longer than the ICs we currently think of.
In valve times, a Septode was a valve with seven electrodes, which actually was an integrated circuit with both a Triode (three electrodes) and a Pentode (five electrodes) in the one glass ‘Tube’ (aka Bottle or Valve). Now hopefully at least one reader has done the math and recognised that 3 + 5 = 8, not 7. (...recognising that the word ‘Septode’ suggests seven electrodes).
That’s where the word ‘Integrated’ comes in. The Anode of the Triode is also the Cathode of the Pentode! With one shared electrode serving between two components, we end up with seven electrodes.
One example, the 6BE6 valve, was a “Mixer” valve, including both an oscillator and an amplifier acting together to make the mixer. The bottom line is that the 6BE6 acts as “more than one component already connected together to perform a defined function”, which is a pretty good and simple definition for the term “Integrated Circuit”.
# WHAT IS AN IC?
Modern ICs are more complex, in general, than the aforementioned Electronic Valve. However there are simple ICs that we would not even consider today as integrated circuits, such as the LM394 (and it’s brothers and sisters of similar numbers) being a “Supermatched Pair” of two transistors in a single package. The original package being a 6-pin T05 package which looks like a metal capped transistor.
The LM394 was used in Instrument circuits where it was essential for a differential amplifier to use very well matched transistors in the input. In fact, the LM394 might be considered a predecessor to the now ubiquitous ‘Operational Amplifier’.
# LINEAR CIRCUITS
The LM741 Operational Amplifier is an example of a group of ICs known as ‘Linear’ ICs which simply means they have predominantly ‘Analogue’ functions. Analogue or Linear databooks listed all manner of ICs including Op-Amps (e.g. LM741 etc., Linear Power Supplies (e.g. LM723 etc.), Timers (e.g. LM555 etc.) and yes, the ‘Transistor Array’ ICs such as the LM394, ULN2003, and many others.
# SPECIAL FUNCTION
Another group of analogue or linear devices sprung up including Power Amplifiers and Radio Function Blocks that are often gathered together into a Data Book labelled as ‘Special Function’.
While the possible ‘functions’ would indeed require a data book to explain, they include Phase Locked Loop ICs, Radio Mixers, Radio Power Output Modules (RF-Blocks) commonly called Bricks, and so many other Radio operations they in turn got their very own Data Book of “RF Devices”.
Perhaps I should point out that at one time I had over thirty Motorola Data Books, and those of many other manufacturers as well including the almost essential National Semiconductors three volume set of Linear Devices.
Data Books are a great way to keep up with what is available, but unfortunately not commonly available today. Of course the internet has taken over, but another issue is that many of the electronics manufacturers have been taken over by less well known brand names, and in some cases, less reputable manufacturers.
Most data books have been replaced with an online store of data sheets, and application notes, with at times restricted access, or only as paid downloads.
# DIGITAL LOGIC
The request for this article came from our discussions on Digital Logic, and the fact that we mentioned so many ‘families’ of logic devices. As the logic families came out at different times, a Data Book would be produced for each family, e.g. TTL as a major digital logic family had sub-families of differing technology that while still TTL, was perhaps a faster type, lower power type, advanced type etc. and all had their own Data Book. We will go into the families of Logic Devices in another article however.
# Going ‘DIP’py
Although many early ICs did come in TO5 metal capped form, even 8-pin as the metal cap LM741 shown below attests, it became obvious that a new and simpler standard for integrated circuit packages was needed.
Many audio and RF ICs were made in a Single Inline Package (SIP) form. Although they took up less horizontal space, they had weight and vibration issues in sometimes violent applications such as industrial and military, where oscillations (vibrations) and shock loads could cause failure of the soldered joints to the pcb.
The SIPs did however start a standard that is still around today having lasted for decades, the 0.1” pin spacing, meaning that the legs of the ICs were spaced 0.1 inches apart, or an odd sounding 2.54mm spacing; for those of you that missed the era of inches and cubits!
Another option became more popular as being far more stable and also allowing for greater number of connections to the electronic ‘chip’; the actual Silicon semiconductor.
The DIP, (Dual Inline Package) was very flexible, yet still a good standard with two rows of pins 0.3” or 7.62mm between two rows of 0.1” spaced pins.
Note: when larger ICs such as ROMs, Microprocessors, and other ‘Large Scale Integration’ (LSI) ICs required more pins and a larger semiconductor chip, a Wide DIP; the WDIP of 0.6” width was introduced. The term is not usually necessary as the part number is almost always one or the other package.
PCB software may refer to WDIP, or perhaps define packages as DIP3 and DIP6.
Of course, our components have become more complex and grown many more legs so after the M68000 microcomputer was introduced with 64 legs in a WDIP package, other packages began to pop up and although they are standard packages, there are now many more ‘standards’ than necessary.
The DIP package can have as few as 6 legs usually representing an opto-coupler, an IC with an LED on one side and a phototransistor on the other side to provide electrical isolation between a high voltage and a low voltage circuit. Our new friend the LM394 can also be bought in a 6-pin package.
DIP devices with 8-pins include most commonly available Op-Amps, the LM555, and a lot of devices even including the ATtiny and some PIC micro-controllers.
Although not all combinations are used, DIP ICs up to at least 28 pins, on 0.3” spacing, are common enough with WDIP packages beginning at 24-pins and going up to the common 40-pin devices, and extending to the aforementioned M68000 (Motorola 68k) of 64-pins.
# ORIENTATION
With two rows of pins, or even on SIP devices with one row, the problem arises of how to make sure you know which pin is which, so you can place the IC the correct way around, or as we say in it’s correct orientation.
The original SIP devices were a slip of ceramic with the silicon chip grafted to the middle, legs attached to the ceramic and then wired to the chip itself by tiny strands of silver or gold wires.
Then to protect the device the whole thing, sparing the attachment parts of legs of course, was dipped in a ceramic slip and fired to set the protective coating. Then the part number (etc.) was printed on the coating, including a small round dot to indicate pin-1; which is typically the pin on the left when you are reading the printed information.
When epoxy or plastic variations were made, the dot became a dimple or round dent in the moulded plastic case, just above pin-1. Moulded SIP devices may also have been identified by a chamfer on the top left corner of the case, or along the corner between the front face and the left end of the device.
While some DIP devices have a similar dimple or painted dot in the corner near pin-1, and sometimes ICs are found with a similar chamfer on the corner beside pin-1, the most common marking is a ‘Notch’ being a half round notch missing from the end of the IC, or more commonly today, a half round dimple at the end of the IC between pin-1 and the highest numbered pin.
Every IC identifies pin-1 in some way, but the other pins are never numbered, and I sometimes feel like that! (Joking!)
# PIN NUMBERING
When I first started working with ICs I asked why (looking down on the IC with the text readable!) pin-1 was in the bottom left corner and the pins were numbered anti-clockwise around the IC. I was told by those more learned than I, “That’s just the way it is!”
It wasn’t until I started making PCB patterns, originally from stick on pads and black crape tape, that I recognised that those people that numbered the ICs were looking from below making pin-1 at figuratively the one o’clock position, (i.e. with two vertical columns of pins) and the pins were indeed numbered clockwise around the IC. This is even more obvious when looking at an 8-pin TO5 Metal Cap op-amp from below.
# DIP-SMDS
Surface Mount Devices are those intended to be soldered directly to the pcb without having any legs that require passing through the pcb. That’s great, and in itself solves several issues in making a pcb at home. No need to drill holes for example, but the technology also ignored hobbyists by ignoring 0.1” pin alignment, and instead started a competition into how small the gap between legs could become, and even for those with good eyesight, the next issue is the chip itself is almost too small to have a label.
Still, we sometimes need to identify a pin before we can predict what signal should appear on that pin. While some tiny ICs use the Notch/Dimple ID methods, some have a slight chamfer on the corner next to pin 1 and others have a painted dot. Get out the Jeweller’s Eye Glass!
There are too many square IC types with pins on each side to get into great detail, but it is worth pointing out that they are not always numbered from a corner as might be expected. Some micro-controller ICs are numbered from the centre of one side of pins, if odd numbered, or the pin to the left of centre if even numbered.
# PIN GRID ARRAY
Other ICs have no legs but a pattern of gold coloured raised dots on the bottom surface of the IC. They are intended to be surface mounted to a PCB, and unless you are designing with them you would have no reason to know the pinout.
Some ICs, including many CPUs, use BGA (Ball Grid Array) packages, which are dropped into an IC socket. This method makes them easy to get in and out if you need to replace or upgrade them.
# HEAT DISSIPATION
As we don’t yet have superconductor ICs, every IC has losses in operation, resulting in heat generation. Heat is energy. EH = I2RT, meaning that the heat generated and stored increases with time. Energy stored in a space results in an increase in some form of pressure or stress, whether pneumatic or hydraulic pressure, voltage, or in the case of Heat Energy, an increase in Temperature.
Thankfully heat flows from a position of high temperature to a position of low temperature, just as electrical energy flows from a high voltage to a low voltage. The thermal flow is indeed just like the electrical current, and perhaps you might have guessed that therefore electrical laws also work for Heat!
Very good if you did, Ohm’s law for example works for heat (now we’re using Fourier’s Law) as long as you remember which parameters are alike. Temperature(T) equates to Voltage(V), Heat(Q) equates to Charge(Q), Thermal Flow (ask your physics teacher) equates to Current(I) and Thermal Resistance (ask your physics teacher) equates to Resistance(R).
Essentially, the thermal flow depends on the Temperature difference between the hot spot and the ambient temperature of the air, or wherever the heat has to go, and the quantity of heat energy, and finally the thermal resistance of the path the heat must travel through, which like electrical resistance, depends upon the area, the length, and the material of the conductor.
# IC HEAT DISSIPATION
In an IC, the heat generated is therefore limited by the opportunity the package has to dissipate the heat, and therefore to limit the temperature rise. It is the temperature of the silicon chip that is most important, and the temperature is greatest within the silicon, and of more concern, within the semiconductor junctions themselves.
That is where the semiconductor layers are most prone to “meltdown”. NOT that the silicon will totally melt, but the junctions may get hot enough that under current flow, the junctions themselves may dissolve together into either a conductor or insulator, rather than a semiconductor.
If just one junction is effected, the IC may not perform as desired and may simply result in lower performance, or higher noise levels. In some cases the IC may simply cease all operations even though externally it looks fine. Of course, in heavy abuse, a real meltdown can occur resulting in visible damage to the package itself which at least tells the repair guy where to look. See where the smoke came from!
To assist in cooling the device, the IC designer uses all of the factors at hand, from reducing heat generated to increasing the opportunities for heat to dissipate through the shortest path, over the greatest area of the path, and with the best conductive properties of the path.
# IC HEAT DESIGN
The first option is to use a thermally conductive material for the case, one that is of course not electrically conductive. The material should be a minimum thickness for the voltage protection requirement, and present as large an area as possible to both the IC chip and the external environment.
Of course, there are other ways to convey the heat from the IC. You may have noticed IC pins designated as ‘NC’, meaning ‘Not Connected’. Mostly this is simply because those pins are not used for that particular version of an IC, pins 1,4 and 8 of some op-amps, for example.
However extra pins are often a way to transfer more heat out of the IC and ‘Audio Power’ ICs (without obvious heat sinks) may have four or more pins closest to the middle of the IC connected to ground, as a thermal sink, allowing heat to transfer to the copper ground plane which is also designed as a heat sink by using as much area of the copper layer as possible.
## SPEAKING OF HEAT SINKS...
SIP packages are often used for power devices, allowing all of the electrical connections on one side of the IC and a large Aluminium “flag” on the other side that can be mounted to a much larger Aluminium heat sink at the top.
The whole heat sink circuit can be engineered and calculations made using formulae very familiar to what people working with electrical circuits already use, except you will not need a return path.
For DIP devices, instead of pins, some power ICs in DIP format have two large ‘legs’ that are used as heat sink connections replacing any number of narrow pins with the one wide ‘flag’ either side of the IC. They are sometimes referred to as ‘Butterfly’ packages.
Small glue-on heat sinks can also be attached to the top to increase heat dissipation.
# STATIC ELECTRICITY
Another danger that ICs face is Electro-Static Discharge (ESD), or Static Electricity. In fact static electricity is fine by me, as long as it doesn’t get out. When static electricity is let loose it may have a significant current flow (i.e. no longer static!) and a significant voltage to push the current.
First, I must explain how Static Electricity arises in a circuit or component. If two insulators are rubbed together, the electrons on the surface of one insulator are transferred to the other insulator, having no option but to settle in and relax!
It must be said that the two insulators must be different materials, and as far apart on the ‘Tribo-electric’ scale as possible. Tribo-electric is a scale that determines a materials propensity to attract or reject electrons.
One insulator has more electrons than it should have while the other has less. As the number of electrons increase, the insulator builds an electronic charge, increasing the negative voltage on that insulator while increasing the positive voltage on the other insulator, and twice that voltage between the two insulators.
An employee in an electronics shop can generate tens of thousands of volts simply by shuffling along in plastic shoes on a wool carpet, for example. The employee literally has lethal fingers!
The moment the person reaches to pick up an electronic component, the electrostatic charge may discharge to the component resulting sometimes in an audible ‘crack’ or a visible spark, but often no indication at all.
Electric charge (Q) is determined from current and time (Q=IT) or from Capacity and Voltage (Q=CV), giving us a double formula of Q=CV=IT. You might have noticed that to carry a charge a person must become a capacitor for a period, and insulated shoes are enough to achieve this. Dry air as found in many sealed electronics labs, and many parts of Australia, helps greatly.
Now for any person to charge up to 10,000 volts is not difficult, and may not be noticeable though not desirable around electronics (or flammable vapours!)
Engineers have specified a “Simulated Human” for testing purposes as a 100pF capacitor in series with a 1500Ω Resistor. So let us charge our simulated human to 10,000 volts (DC of course), resulting is a charge of Q = CV = 100e-12 x 10e3 = 1µC, 1 micro Coulomb.
Now you might agree that 10,000 VDC would be sufficient to cook an IC, but to be thorough, let’s imagine that 10,000 VDC would actually cause the IC to conduct, but without current flow, there is no damage!
Of course, once the IC conducts, current flows, limited of course by the 1500Ω resistor.
Now we should do exponential calculations to be precise and we could do that, but even as a ‘Guesstimate’ using DC values, 10,000 V limited by a 1500Ω resistor, using no more maths than simple Ohm’s law, results in a maximum current flow of 6.667 Amps.
This only happens for a very short time, but through such sensitive electronics, is enough to do damage to very small circuit elements even if the current only lasts for T = Q/I = CV/I = 100e-12 X 10,000 /6.67 = 150nS! (Remember over simplified but useful!).
# PROTECTING ICS AGAINST ESD
Electro-Static Discharge (ESD) is therefore a big issue for small devices. Internal protection is possible but mostly to improve the defences rather than completely protect the device. The technology of the IC further complicates matters with CMOS being more sensitive than more conventional technology, but therefore tends to be designed to be better protected by internal diodes and such.
When testing devices in circuit, a good habit is to connect the ground of the PCB and of the testing device together, and connect both to a common bench ground if available.
Operators may wear a wrist strap attached to that ground, or regularly touch that ground to discharge themselves, especially before touching circuitry.
Tools such as the soldering iron may be double insulated with no ground, which in effect means a floating ground. Some workshops have used a technique of sliding an appropriate size of light spring over the soldering iron barrel with an attached flexible wire returning to the bench ground. Of course a ESD safe tool would be better.
Touching the soldering iron tip to a grounded point, bench ground, makes sure that an un-earthed soldering iron is at least temporarily discharged.
Note: In some workshops, it is common (but not recommended) practice to solder live circuits, mostly low voltage and low current circuits, and therefore an earthed soldering iron would actually present a hazard!
Other unexpected shorts can result from direct earthing/grounding, so one way to prevent unexpected shorts is to place a 1MΩ resistor in series with grounding leads. The resistor allows discharge while limiting any unintended shorts. Any good wrist strap and lead will also feature a series resistance just in case the lead comes into contact with dangerous voltages.
# IN SUMMARY
We have looked at ICs from a general perspective, from a distant perspective in fact without going into great detail on any one IC. However we looked at different groups of ICs, and where they are used or how they may be different, and how they are standardised.
We looked at issues like Thermal and ESD protection, and how manufacturers attempt to protect their devices. As you can see however, the number of IC devices with many varying requirements and technologies makes standardisation very difficult. | 4,490 | 20,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.961915 |
http://www.aspmessageboard.com/showthread.php?107649-Looping-several-columns | 1,477,138,025,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718957.31/warc/CC-MAIN-20161020183838-00027-ip-10-171-6-4.ec2.internal.warc.gz | 313,598,071 | 13,303 | Looping several columns
1. Junior Member
Join Date
Dec 1969
Posts
12
## Looping several columns
Problem!<BR><BR>I would like to know how to loop a 1 dimension array into 3 columns looping 1 column at a time (html-table).<BR><BR>I already know how to loop columns row by row but the sortoder is wrong then.<BR><BR>This is how I would like it to show:<BR>A E I<BR>B F J<BR>C G K<BR>D H L<BR><BR>And NOT like this (this I know how to do):<BR>A B C<BR>D E F<BR>G H I<BR>J K L<BR><BR>Please help me!
2. Senior Member
Join Date
Dec 1969
Posts
497
## RE: Looping several columns
Use UBound on the array to get the maxArray entry.<BR><BR>dim arry<BR><BR>columns = 3<BR>maxEntry = UBound(arry)<BR>'ceiling function<BR>if Int(maxEntry/columns)<(maxEntry/columns) then<BR>rows = Int(maxEntry/columns) + 1<BR>else<BR>rows = Int(maxEntry/columns)<BR>end if<BR><BR>for i = 0 to rows<BR>for j = 0 to columns-1<BR>response.write arry(i + (j*rows))<BR>next<BR>response.write "< b r >"<BR>next<BR><BR>I have not tested this... let me know if it bugs out<BR><BR>sponge
3. Junior Member
Join Date
Dec 1969
Posts
12
## RE: Looping several columns
Works great! Can't believe I couldĀ“nt work that out by myself!<BR><BR>Thanks!!
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• | 455 | 1,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-44 | latest | en | 0.772072 |
https://www.sawaal.com/coding-decoding-questions-and-answers/in-a-certain-code-language-student-is-written-as-tutdneshow-will-sources-be-writte-in-that-code-lang_7313 | 1,708,631,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.45/warc/CC-MAIN-20240222193722-20240222223722-00896.warc.gz | 1,008,433,063 | 14,282 | 139
Q:
# In a certain code language STUDENT is written as TUTDNES.How will SOURCES be writte in that code language?
A) SOURCES B) SUORECS C) SRUOCES D) SOURSEC
Explanation:
- There are 7 letters in the word
- The middle letter has been left intact
- The first and the last letters of the word have interchanged their position. Similarly, the second and the third letters and the fifth and sixth letters have interchanged their position.
Q:
If RESEARCH is $#!#%$&@ then SCARE is
A) !&%$# B) !@%$# C) !$%#& D) !@%#$
Explanation:
5 290
Q:
If 12@18 = -3, 18@6 = 6 and 16@8 = 4, then find the value of 6@10 = ?
A) -10 B) 8 C) 4 D)
Explanation:
5 2385
Q:
The following equation is incorrect. Which two signs should be interchanged to correct the equation?
10 + 14 x 20 - 7 ÷ 10 = 40
A) + and x B) + and ÷ C) ÷ and D) - and +
Explanation:
1 2394
Q:
In a certain code language, 'x' represents '+', '÷' represents 'x', '-' represents '÷' and '+' represents '-'. Find out the answer to the following question.
18 x 12 ÷ 5 - 10 + 10 = ?
A) 14 B) 6 C) 39 D) 2
Explanation:
3 240
Q:
In a certain code language, “NOTES” is written as “141520519” and “MAKAR” is written as “13111118”. How is “LOFTS” written in that code language?
A) 206191316 B) 121562019 C) 12150219 D) 1515121119
Explanation:
0 179
Q:
If Q = 10 and FAX = 50, then XEROX = ?
A) 45 B) 46 C) 49 D) 50
Explanation:
1 420
Q:
If 60#40 = 10, 40#20 = 6 and 60#90 = 15, then find the value of 40#30 = ?
A) 15 B) 12 C) 18 D) 7
Explanation:
1 1795
Q:
The following equation is incorrect. Which two signs should be interchanged to correct the equation?
8 ÷ 2 - 50 x 6 + 16 = 50
A) + and ÷ B) - and + C) + and x D) ÷ and x | 616 | 1,720 | {"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} | 3.625 | 4 | CC-MAIN-2024-10 | latest | en | 0.815025 |
http://www.numbersaplenty.com/9993317 | 1,558,696,670,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257605.76/warc/CC-MAIN-20190524104501-20190524130501-00316.warc.gz | 304,361,739 | 3,368 | Search a number
9993317 is a prime number
BaseRepresentation
bin100110000111…
…110001100101
3200210201020212
4212013301211
510024241232
6554105205
7150641015
oct46076145
920721225
109993317
115706144
12341b205
1320bb809
141481c45
15d25eb2
hex987c65
9993317 has 2 divisors, whose sum is σ = 9993318. Its totient is φ = 9993316.
The previous prime is 9993289. The next prime is 9993337. The reversal of 9993317 is 7133999.
9993317 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 9928801 + 64516 = 3151^2 + 254^2 .
It is a cyclic number.
It is not a de Polignac number, because 9993317 - 214 = 9976933 is a prime.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (9993337) by changing a digit.
It is a good prime.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 4996658 + 4996659.
It is an arithmetic number, because the mean of its divisors is an integer number (4996659).
Almost surely, 29993317 is an apocalyptic number.
It is an amenable number.
9993317 is a deficient number, since it is larger than the sum of its proper divisors (1).
9993317 is an equidigital number, since it uses as much as digits as its factorization.
9993317 is an evil number, because the sum of its binary digits is even.
The product of its digits is 45927, while the sum is 41.
The square root of 9993317 is about 3161.2208084852. The cubic root of 9993317 is about 215.3954646844.
The spelling of 9993317 in words is "nine million, nine hundred ninety-three thousand, three hundred seventeen". | 523 | 1,763 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-22 | latest | en | 0.889091 |
http://oeis.org/A125247 | 1,618,317,561,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072366.31/warc/CC-MAIN-20210413122252-20210413152252-00346.warc.gz | 67,432,030 | 4,859 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A125247 Numbers n whose abundance sigma(n) - 2n = -8. Numbers n whose deficiency is 8. 16
22, 130, 184, 1012, 2272, 18904, 33664, 70564, 85936, 100804, 391612, 527872, 1090912, 17619844, 2147713024, 6800695312, 34360655872, 549759483904, 1661355408388, 28502765343364, 82994670582016, 99249696661504, 120646991405056, 431202442356004, 952413274955776 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(19) > 10^12. - Donovan Johnson, Dec 08 2011 a(20) > 10^13. - Giovanni Resta, Mar 29 2013 a(30) > 10^18. - Hiroaki Yamanouchi, Aug 21 2018 a(20) <= 36028797958488064 ~ 3.6*10^16. Indeed, if k is in A057195 then 2^(k-1)*A168415(k) is in this sequence, and k=28 yields this upper bound for a(20) which is in any case a term of this sequence. - M. F. Hasler, Apr 27 2015 If n is in this sequence and p a prime not dividing n, then np is abundant if and only if p < sigma(n)/8 = n/4-1. For all n=a(k) except {22, 70564, 100804, 17619844}, there is such a p near this limit, such that n*p is a primitive weird number (A002975; in A258882 for the terms mentioned in the preceding comment). - M. F. Hasler, Jul 20 2016 Any term x of this sequence can be combined with any term y of A088833 to satisfy the property (sigma(x)+sigma(y))/(x+y) = 2, which is a necessary (but not sufficient) condition for two numbers to be amicable. - Timothy L. Tiffin, Sep 13 2016 Is there any odd number in this sequence? Is it possible to prove the contrary? - M. F. Hasler, Feb 22 2017 LINKS Hiroaki Yamanouchi, Table of n, a(n) for n = 1..29 EXAMPLE The abundance of 22 = (1+2+11+22)-44 = -8 MATHEMATICA Select[Range[10^6], DivisorSigma[1, #] - 2 # == -8 &] (* Michael De Vlieger, Jul 21 2016 *) PROG (PARI) for(n=1, 1000000, if(((sigma(n)-2*n)==-8), print1(n, ", "))) (MAGMA) [n: n in [1..2*10^7] | (DivisorSigma(1, n)-2*n) eq - 8]; // Vincenzo Librandi, Jul 22 2016 CROSSREFS Cf. A033880, A088833 (abundance 8). Sequence in context: A206418 A309923 A215626 * A249302 A095694 A233060 Adjacent sequences: A125244 A125245 A125246 * A125248 A125249 A125250 KEYWORD easy,nonn AUTHOR Jason G. Wurtzel, Nov 25 2006 EXTENSIONS a(13)-a(15) from Klaus Brockhaus, Nov 29 2006 a(16)-a(17) from Donovan Johnson, Dec 23 2008 a(18) from Donovan Johnson, Dec 08 2011 a(19) from Giovanni Resta, Mar 29 2013 a(20)-a(25) from Hiroaki Yamanouchi, Aug 21 2018 STATUS approved
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Last modified April 13 07:51 EDT 2021. Contains 342935 sequences. (Running on oeis4.) | 1,012 | 2,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-17 | latest | en | 0.704379 |
https://number.academy/236198 | 1,660,939,001,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573760.75/warc/CC-MAIN-20220819191655-20220819221655-00083.warc.gz | 391,770,254 | 12,405 | # Number 236198
Number 236,198 spell 🔊, write in words: two hundred and thirty-six thousand, one hundred and ninety-eight . Ordinal number 236198th is said 🔊 and write: two hundred and thirty-six thousand, one hundred and ninety-eighth. Color #236198. The meaning of number 236198 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 236198. What is 236198 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 236198.
## What is 236,198 in other units
The decimal (Arabic) number 236198 converted to a Roman number is (C)(C)(X)(X)(X)(V)MCXCVIII. Roman and decimal number conversions.
#### Weight conversion
236198 kilograms (kg) = 520722.1 pounds (lbs)
236198 pounds (lbs) = 107138.7 kilograms (kg)
#### Length conversion
236198 kilometers (km) equals to 146767 miles (mi).
236198 miles (mi) equals to 380124 kilometers (km).
236198 meters (m) equals to 774919 feet (ft).
236198 feet (ft) equals 71995 meters (m).
236198 centimeters (cm) equals to 92991.3 inches (in).
236198 inches (in) equals to 599942.9 centimeters (cm).
#### Temperature conversion
236198° Fahrenheit (°F) equals to 131203.3° Celsius (°C)
236198° Celsius (°C) equals to 425188.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
236198 seconds equals to 2 days, 17 hours, 36 minutes, 38 seconds
236198 minutes equals to 5 months, 3 weeks, 3 days, 38 minutes
### Codes and images of the number 236198
Number 236198 morse code: ..--- ...-- -.... .---- ----. ---..
Sign language for number 236198:
Number 236198 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 236198
### Multiplications
#### Multiplication table of 236198
236198 multiplied by two equals 472396 (236198 x 2 = 472396).
236198 multiplied by three equals 708594 (236198 x 3 = 708594).
236198 multiplied by four equals 944792 (236198 x 4 = 944792).
236198 multiplied by five equals 1180990 (236198 x 5 = 1180990).
236198 multiplied by six equals 1417188 (236198 x 6 = 1417188).
236198 multiplied by seven equals 1653386 (236198 x 7 = 1653386).
236198 multiplied by eight equals 1889584 (236198 x 8 = 1889584).
236198 multiplied by nine equals 2125782 (236198 x 9 = 2125782).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 236198
Half of 236198 is 118099 (236198 / 2 = 118099).
One third of 236198 is 78732,6667 (236198 / 3 = 78732,6667 = 78732 2/3).
One quarter of 236198 is 59049,5 (236198 / 4 = 59049,5 = 59049 1/2).
One fifth of 236198 is 47239,6 (236198 / 5 = 47239,6 = 47239 3/5).
One sixth of 236198 is 39366,3333 (236198 / 6 = 39366,3333 = 39366 1/3).
One seventh of 236198 is 33742,5714 (236198 / 7 = 33742,5714 = 33742 4/7).
One eighth of 236198 is 29524,75 (236198 / 8 = 29524,75 = 29524 3/4).
One ninth of 236198 is 26244,2222 (236198 / 9 = 26244,2222 = 26244 2/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
236198
#### Is Prime?
The number 236198 is not a prime number. The closest prime numbers are 236167, 236207.
#### Factorization and factors (dividers)
The prime factors of 236198 are 2 * 17 * 6947
The factors of 236198 are 1 , 2 , 17 , 34 , 6947 , 13894 , 118099 , 236198
Total factors 8.
Sum of factors 375192 (138994).
#### Powers
The second power of 2361982 is 55.789.495.204.
The third power of 2361983 is 13.177.367.188.194.392.
#### Roots
The square root √236198 is 486,002058.
The cube root of 3236198 is 61,814744.
#### Logarithms
The natural logarithm of No. ln 236198 = loge 236198 = 12,372426.
The logarithm to base 10 of No. log10 236198 = 5,373276.
The Napierian logarithm of No. log1/e 236198 = -12,372426.
### Trigonometric functions
The cosine of 236198 is 0,878572.
The sine of 236198 is 0,47761.
The tangent of 236198 is 0,543621.
### Properties of the number 236198
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 236198 in Computer Science
Code typeCode value
PIN 236198 It's recommendable to use 236198 as a password or PIN.
236198 Number of bytes230.7KB
CSS Color
#236198 hexadecimal to red, green and blue (RGB) (35, 97, 152)
Unix timeUnix time 236198 is equal to Saturday Jan. 3, 1970, 5:36:38 p.m. GMT
IPv4, IPv6Number 236198 internet address in dotted format v4 0.3.154.166, v6 ::3:9aa6
236198 Decimal = 111001101010100110 Binary
236198 Decimal = 110000000002 Ternary
236198 Decimal = 715246 Octal
236198 Decimal = 39AA6 Hexadecimal (0x39aa6 hex)
236198 BASE64MjM2MTk4
236198 MD56c7c319df5b78b996ba6d78918eb6bd5
236198 SHA10c5aed37c013d82413d4e480ea32f5966296356b
236198 SHA2242c289fed5ca4e560c657bd31ef268a466f156d7e8f91acb30533a268
236198 SHA256a12ba4d295e169467cbecd4afdef241039750c4190e1f5a65b0e466e3e679292
236198 SHA3841c6e76f552bbdfa399a1f5a959eaae91c009a16a23697fc31b2796ede571407fe272f30feddec4cec74fa34006263857
More SHA codes related to the number 236198 ...
If you know something interesting about the 236198 number that you did not find on this page, do not hesitate to write us here.
## Numerology 236198
### Character frequency in number 236198
Character (importance) frequency for numerology.
Character: Frequency: 2 1 3 1 6 1 1 1 9 1 8 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 236198, the numbers 2+3+6+1+9+8 = 2+9 = 1+1 = 2 are added and the meaning of the number 2 is sought.
## Interesting facts about the number 236198
### Asteroids
• (236198) 2005 WY78 is asteroid number 236198. It was discovered by Spacewatch from Obs. US National at Kitt Peak on 11/25/2005.
## № 236,198 in other languages
How to say or write the number two hundred and thirty-six thousand, one hundred and ninety-eight in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 236.198) doscientos treinta y seis mil ciento noventa y ocho German: 🔊 (Anzahl 236.198) zweihundertsechsunddreißigtausendeinhundertachtundneunzig French: 🔊 (nombre 236 198) deux cent trente-six mille cent quatre-vingt-dix-huit Portuguese: 🔊 (número 236 198) duzentos e trinta e seis mil, cento e noventa e oito Chinese: 🔊 (数 236 198) 二十三万六千一百九十八 Arabian: 🔊 (عدد 236,198) مئتان و ستة و ثلاثون ألفاً و مائةثمانية و تسعون Czech: 🔊 (číslo 236 198) dvěstě třicet šest tisíc sto devadesát osm Korean: 🔊 (번호 236,198) 이십삼만 육천백구십팔 Danish: 🔊 (nummer 236 198) tohundrede og seksogtredivetusindethundrede og otteoghalvfems Dutch: 🔊 (nummer 236 198) tweehonderdzesendertigduizendhonderdachtennegentig Japanese: 🔊 (数 236,198) 二十三万六千百九十八 Indonesian: 🔊 (jumlah 236.198) dua ratus tiga puluh enam ribu seratus sembilan puluh delapan Italian: 🔊 (numero 236 198) duecentotrentaseimilacentonovantotto Norwegian: 🔊 (nummer 236 198) to hundre og tretti-seks tusen, en hundre og nitti-åtte Polish: 🔊 (liczba 236 198) dwieście trzydzieści sześć tysięcy sto dziewięćdzisiąt osiem Russian: 🔊 (номер 236 198) двести тридцать шесть тысяч сто девяносто восемь Turkish: 🔊 (numara 236,198) ikiyüzotuzaltıbinyüzdoksansekiz Thai: 🔊 (จำนวน 236 198) สองแสนสามหมื่นหกพันหนึ่งร้อยเก้าสิบแปด Ukrainian: 🔊 (номер 236 198) двiстi тридцять шiсть тисяч сто дев'яносто вiсiм Vietnamese: 🔊 (con số 236.198) hai trăm ba mươi sáu nghìn một trăm chín mươi tám Other languages ...
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For this topic there are no comments. | 2,738 | 8,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-33 | latest | en | 0.665369 |
https://www.mytutor.co.uk/answers/54127/A-Level/Physics/Where-does-the-simple-harmonic-motion-equation-come-from-and-what-does-it-mean/ | 1,726,381,014,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00845.warc.gz | 835,791,146 | 15,292 | # Where does the simple harmonic motion equation come from and what does it mean?
We know that the displacement, x, is described by the equation x = Acos(ωt + Φ), where A is the amplitude of oscillation, ω is the angular frequency and Φ is the phase shift. The velocity, v, is the time derivative of displacement (v = dx/dt), so differentiating both sides with respect to time t gives v = -ωAsin(ωt + Φ). The acceleration, a, is the time derivative of velocity (a = dv/dt), so differentiating both sides with respect to time again gives a = -ω2Acos(ωt + Φ) = -ω2x. This is the defining equation of simple harmonic motion: it states that the acceleration is proportional (since ω2 is a constant) and in the opposite direction (due to the negative sign) to the displacement.
This can be more easily visualised by sketching the curves for displacement, velocity and acceleration. Assuming the phase shift Φ = 0, the displacement x = Acos(ωt) and is described by a cosine curve. The velocity is described by an upside-down sine curve, and the acceleration is described by an upside-down cosine curve. So the acceleration curve is the same as the displacement curve, but reflected in the x-axis. | 279 | 1,191 | {"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} | 3.6875 | 4 | CC-MAIN-2024-38 | latest | en | 0.922054 |
https://www.advancedlearners.com/bethesda/algebra/tutor/find.aspx | 1,722,807,566,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00455.warc.gz | 497,533,739 | 12,023 | ## Algebra Tutors
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## Ashley A
### Teaching Style
First and foremost, I believe it is important to establish a trusting relationship with those you teach. Throughout my experiences in the field of education, I have found the most fruitful of those relationships to be those in which I was able to work with a student or group of students regularly over an extended period of time to develop a routine and a strong relationship in which great amounts of learning and understanding could be accomplished. Although many people experience obstacles in learning Mathematics, I believe that by presenting multiple ways to approach a problem, every student can find a method that works for them. Every student should be allowed the resources and opportunity to realize that they CAN achieve their goals. I am here to help students who have had difficulty with Math in the past to succeed, and feel confident in both their abilities in Math, and in life.
### Experience Summary
As a student, I was always committed to learning and to achieving my goals. As a teacher and tutor, I strive to help others share the same love for learning and for understanding as I do. Now, I help others set goals, and work toward achieving them. I have worked with all ages, all ability levels, and various sized groups and have enjoyed each and every experience. In the past, I have primarily tutored in the subject of Mathematics, but am also trained by the Literacy Council to tutor reading and writing, and have enjoyed volunteering with that organization as well. I believe that I can help anyone enjoy and understand math, and help them feel better about themselves for it.
### Credentials
Type Subject Issued-By Level Year
Degree Mathematics UNC Chapel Hill BA 2005
## Davorin D
### Teaching Style
From my tutoring experience, I have noticed that students have trouble understanding the meaning of numbers and symbols on paper simply because no one has taught them how to visualize and interpret them in a real world situation. They also don't realize that they have plethora of resources and tools available to them to help them, yet they rarely utilize them. I try to give hints and clues to my students and let them obtain the right answers on their own instead of simply solving the problem for them. I believe this gives them a much better understand of the material.
### Experience Summary
Having recently graduated with a Bachelor's degree, I am continuing my education to obtain my Master's degree in mathematics. Number Theory will be my focus, as I intend to get involved in the encryption field. During my senior year as undergraduate, I have worked on campus as a teacher's assistant and at home as an online tutor for UNCG's iSchool program. I have privately tutored undergraduates needing help in pre-calculus and calculus. I enjoyed helping the students and showing them some of my own tricks and ways when it comes to solving the problems.
### Credentials
Type Subject Issued-By Level Year
Degree Mathematics with Concentration in Computer Science University of North Carolina at Greensboro BS 2007
## Lori L
### Teaching Style
I believe that students are wary of mathematics and statistics because they appear clinic and distant, with little to do with the “real” world. To educate I try to create dynamic and above all relevant “uses” for the lessons. My experiences have exposed me to different teaching styles and class formats and allowed me to develop a teaching philosophy that encompasses the best of all these methods. My philosophy is best described with reference to six primary concepts: 1. Knowledge conveyed in a relevant context 2. Interaction with each student and the material 3. Passion for teaching and the subject 4. Adaptability 5. Creativity in teaching methods and 6. Respect between the student and the teacher All students seek knowledge; it’s the teacher’s role to facilitate learning and guide them along the path. A successful lesson is one in which the student comes out seeing the world a little differently.
### Experience Summary
During my education and career, the teaching of others has featured prominently in my personal goals and life objectives. While attending high school, I tutored fellow students professionally in subjects ranging from basic algebra to complex calculus. My undergraduate degree was in the field of Mathematics and Statistics with an additional biological statistics certificate and at present I am completing my Masters of Science in statistics. My understandings of these fields lead to my recruit by a number of professors and researchers to provide assistance and advisement on statistical analysis of their projects. In addition to assisting my professors, I was also selected to be a Graduate Teaching Assistant, and was also selected to conduct "stand alone" courses. I have a strong passion for these subjects and I believe my years of teaching experience have given me the insight, patience and ability to convey the complex world of mathematics and statistics to my students.
### Credentials
Type Subject Issued-By Level Year
Degree Statistics University of North Florida Master's of Science (ABD) 2008
Certification Biostatistics University of West Florida Accredited 2006
Degree Mathematics University of West Florida Bachelor's of Science 2006
## Paul P
### Teaching Style
I am passionate about teaching, and love to work with school age and college level students to help them achieve their academic goals. I am available to help students learn and grow academically. My teaching style is facilitative, patient, easygoing and helpful. I strive to create a productive, safe, and caring environment for my students, where their learning experience is nurtured and productive. I have had the awesome privilege of being educated by many phenomenal teachers, and I believe in providing the same quality of teaching to all my students. It is my goal to make a profound difference in the lives of each of my students - one which help them to become successful adults in a complex business world.
### Experience Summary
As a secondary school and college level teacher, I am passionate about helping my students succeed academically. Having earned two bachelors degrees (A Bachelor of Education in Secondary Education and a Bachelor of Science in Computer Science/Mathematics), and a Master of Business Administration degree (GPA of 3.81/4.00), I have the academic and professional competencies which enable me to produce positive outcomes in a teaching environment. During the last 2 years, I have tutored students in Statistics, Mathematics, Biostatistics, and English. Also, I have taught for more than 8 years at the college level, teaching such subjects as Math, Information Technology, and Management Information Systems.
### Credentials
Type Subject Issued-By Level Year
Certification Secondary School Teacher Newfoundland Department of Education - Canada Secondary Education 1985
Degree Secondary Education Memorial University B. Ed 1984
Degree Computer Science/Mathematics Memorial University B. Sc. 1984
## Dara D
### Teaching Style
I believe in my students and their abilities to learn and synthesize their experiences. My students learn their subjects because I provide a variety of techniques to command their attention. I believe teaching is not just about giving students information but about reaching students who might "get lost" in the system without a guide and friend to help them along.
### Experience Summary
I have taught for over ten years in the public school system and learned how to "connect" with students. I have taught physics, chemistry, and mathematics at the high school level, and I have a wide range of teaching experiences in those fields. I have taught AP, honors, and standard classes. Using interesting movies followed by a lab to reinforce the concept is one of the ways I have used to reach students and make a difference in their lives.
### Credentials
Type Subject Issued-By Level Year
Certification Physics 6 - 12 State of Florida 9 credit hours current
Degree Mathematics University of Central Florida M.S. 1993
Degree Physics University of Central Florida B.S. 1990
## Redha R
### Teaching Style
I teach by example and I am methodical. I show a student how to solve a problem by going very slowly and by following a sequence of steps. I make sure that the student follows understands each step before moving on to the next one. I then ask the student to solve almost the exact same problem (with a slight change in numbers for example)so that the student learns the method and how to solve the problem by herself/himself. I am very patient but expect the student to be willing to learn. I tell students that learning a subject matter is more important than getting an A in a class - this is because 1. if you learn, chances are you won't forget (at least for a long while) and 2. you will get a good grade as a result.
### Experience Summary
I have been employed by the IBM corporation for the last 27 years. I held numerous positions in hardware development, software development, telecommunication network development, project management, solution architecture, performance analysis, and others. As much as I enjoy my job, I have passion in Mathematics. I develop new ideas related to my work and expand them into U.S. patents and external technical papers. I informally tutor family and friends attending high school or college. I was a Mathematics teaching assistant at the Univ. of Pittsburgh for 2 years and at the Univ. of Michigan at Ann Arbor for 3 years. I love teaching and sharing my knowledge with others. I have published numerous technical papers and hold numerous U.S. patents as well.
### Credentials
Type Subject Issued-By Level Year
Other Arabic Native speaker Fluent current
Other French Native speaker Fluent current
Degree Electrical Engineering Univeristy of Michigan P.h.D 1990
Degree Electrical Engineering and Mathematics University of Pittsburgh M.S. 1982
Degree Computer Science and Mathematics Univerisity of Pittsburgh B.S. 1980
## Gary G
### Teaching Style
My teaching style has been, for the most part, dictated by student response. I am comfortable teaching in a traditional lecture format, in a format that uses a cooperative learning approach exclusively, or in a hybrid format. The goal is for effective learning to take place, and I believe my strongest quality is to be able to adapt in such a way that best helps students reach their academic goals.
### Experience Summary
For eight years, I taught freshman and sophomore-level mathematics courses at Arizona State University. These courses included College Algebra, Pre-Calculus, Calculus, Finite Mathematics, and Elementary Mathematics Theory. Additionally, I have tutored students in these courses both in the Mathematics Department Tutor Center, and on my own personal time.
### Credentials
Type Subject Issued-By Level Year
Degree Mathematics Southern Illinois University MS 1999
Degree Mathematics Allegheny College BS 1996
## Lucille L
### Teaching Style
My students know I love math. I am enthusiastic, patient, and caring. I believe everyone can learn math given the right circumstances. I take interest in my students, I email them, I encourage them to do their homework. Through this personal interest, my students work to please the teacher. I also use different teaching styles: discovery learning, Look-Do-Learn, one-to-one instruction, critical thinking. Once a student, always a student for me. I go the extra mile.
### Experience Summary
From an early age I loved math and so when I graduated with a BA in Math and Latin, I went straight into teaching math at the high school level. During graduate school years, I taught math at the University of Toronto. While working with computer programming, I taught math in the evening division of Westbury College in Montreal. I have taught math in different countries: Jamaica, Canada, U.S. Virgin Islands, Nigeria-West African Educational Council, The Bahamas, and Florida.
### Credentials
Type Subject Issued-By Level Year
Degree Math University of Toronto M.Sc. 1966
## Jeff, B.
Weston, FL
Sharon is doing great... Best so far. Thanks.
## Kim, S.
Knoxville, TN
Things are going very well. Micah is great with both children and they really seem to be catching on.
## Deb, M.
Celebration, FL
Elle met with Randy and it went extremely well. Elle learned a great deal from Randy and looks forward to continuing to work with him. | 3,294 | 16,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.925735 |
http://rgtusyllabus.blogspot.com/2011/12/me-701b-computer-aided-egineering-and.html | 1,527,203,747,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866894.26/warc/CC-MAIN-20180524224941-20180525004941-00373.warc.gz | 249,607,212 | 17,601 | ### ME 701(B) Computer Aided Egineering and FEM Syllabus ME 7th sem RGTU/RGPV Mechanical Engineering(ME) Seventh-7th Sem syllabus
ME 701(B) Computer Aided Egineering and FEM Syllabus
RGTU/RGPV Computer Aided Egineering and FEM Syllabus
Mechanical Engineering(ME) Seventh-7th Semester Syllabus
ME 701(B) Computer Aided Egineering and FEM Course Contents:
Unit-I Introduction : Structural analysis, objectives, static, Dynamic and kinematics analyses, Skeletal and continuum structures, Modeling of infinite d.o.f. system into finite d.o.f. system, Basic steps in finite element problem formulation, General applicability of the method.
Unit-II Element Types and Characteristics : Discretization of the domain, Basic element shapes, Aspect ratio, Shape functions, Generalized co-ordinates and nodal shape functions; ID bar and beam elements, 2D rectangular and triangular elements; axis-symmetric elements.
Unit-III Assembly of Elements and Matrices : Concept of element assembly, Global and local coordinate systems, Band width and its effects, Banded and skyline assembly, Boundary conditions, Solution of simultaneous equations, Gaussian elimination and Choleksy decomposition methods, Numerical integration, One and 2D applications.
Unit-IV Higher Order and iso-parametric Elements: One dimensional quadratic and cubic elements, Use of natural co-ordinate system, Area co-ordinate system continuity and convergence requirements, 2D rectangular and triangular requirement.
Unit-V Static Analysis: Analysis of trusses and frames, Analysis of machine subassemblies, Use commercial software packages, Advantages and limitations
Unit-VI Dynamic Analysis: Hamilton's principle, Derivation of equilibrium, Consistent and lumped mass matrices, Derivation of mass matrices for ID elements, Determination of natural frequencies and mode shapes, Use of commercial software packages.
References:
1. Gokhle Nitin; et al; Practical Finite Element Analysis; Finite to Infinite, 686 Budhwar Peth, Pune.
2. Logan DL ; A First Course in Finite element Method; Cegage
3. Krishnamoorthy; Finite Element Analysis, theory and programming; TMH
4. Buchanan; Finite Element Analysis; Schaum series; TMH
5. Seshu P; Textbook of Finite Element Analysis; PHI.
6. Chennakesava RA; Finite Element Methods-Basic Concepts and App; PHI Learning
7. Reddy JN; An introduction to finite element method; TMH
8. Desai Chandrakant S et al; Introduction to finite element Method; CBS Pub
9. Hutton D; Fundamentals of Finite Element Analysis; TMH
10. Zienkiewicz; The finite element Method; TMH
11. Martin and Grahm; Introduction to finite element Analysis (Theory and App.)
12. Rao, S.S., The Finite Element Method in Engineering; Peragamon Press, Oxford.
13. Robert DC., David DM et al, Concepts and Application of Finite Element Analysis; John Wiley.
14. Chandrupatla, T.R. an Belegundu, A.D., Introduction to Finite Elements in Engineering,PHI
Mixx it! | More | 684 | 2,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-22 | latest | en | 0.748315 |
https://discourse.julialang.org/t/custom-likelihoods-in-turing-jl/16818/5 | 1,675,792,993,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500628.77/warc/CC-MAIN-20230207170138-20230207200138-00635.warc.gz | 215,514,277 | 13,238 | # Custom likelihoods in Turing.jl
Hi all-
I have two related questions. First, how do I implement a custom likelihood in turing.jl? Unfortunately, I was unable to implement a simple example based on the documentation, which is probably impeding my ability to identify problems in the second question: How to I implement a custom likelihood that has two potential complications? In my particular use case, my model has vectors of related parameters. This is convenient because the vector may have different sizes in different applications and it allows vectorization of certain operations. In some versions of the model, I would like to fix some elements in the vector, while allowing others to update with the data (see model definition below). The other complicating factor in my use case is that the likelihood requires computationally intensive operations, which cause it to scale linearly with the number of data. Thus, its dramatically faster to perform those operations once and then loop over the data to evaluate the loglikelihood. Otherwise, the model will run 2-3 orders of magnitude slower in my cases.
Below, I posted a toy example with essential features that are used in a more realistic application. Any help would be greatly appreciated.
using Turing,Distributions
import Distributions: pdf, logpdf
#In most applications, i need the parameters to be vectors
mutable struct mydistribution{T} <: ContinuousUnivariateDistribution
ν::T
α::T
end
function pdf(d::mydistribution,T::AbstractArray)
#do some expensive computations
likelihoods = fill(0.0,length(T))
for (i,t) in enumerate(T)
likelihoods[i] = sum(@. .5*pdf(Normal.(d.ν,d.α),t))
end
#return array of likelihoods
return likelihoods
end
#Returns sum of loglikelihood for all data
logpdf(d::mydistribution,T::AbstractArray) = sum(log.(pdf(d,T)))
@model myModel(data) = begin
α = fill(.85,3) #fixed parameter
ν1 ~ Normal(15,5)
ν2 ~ Normal(15,5)
N = length(data)
#fix the third element to zero
data ~ mydistribution([ν1,ν2,0.0],α)
end
# Settings of the Hamiltonian Monte Carlo (HMC) sampler.
iterations = 1000
ϵ = 0.05
τ = 10
data = rand(Normal(0,1),50)
chain = sample(myModel(data), HMC(iterations, ϵ, τ))
MethodError: no method matching mydistribution(::Array{Flux.Tracker.TrackedReal{Float64},1}, ::Array{Float64,1})
Closest candidates are:
mydistribution(::T, !Matched::T) where T at /home/dfish/Projects/ACT-R-Likelihood/FFT Models/LogNormalACTR/temp.jl:6
macro expansion at compiler.jl:56 [inlined]
#myModel_model#18(::Array{Float64,1}, ::Function, ::Turing.VarReplay.VarInfo, ::Turing.Sampler{HMCDA{Any}}) at utils.jl:297
myModel_model(::Turing.VarReplay.VarInfo, ::Turing.Sampler{HMCDA{Any}}) at utils.jl:297
#invokelatest#1 at essentials.jl:697 [inlined]
invokelatest at essentials.jl:696 [inlined]
runmodel!(::Function, ::Turing.VarReplay.VarInfo, ::Turing.Sampler{HMCDA{Any}}) at hmc_core.jl:65
#8 at back.jl:139 [inlined]
forward at back.jl:126 [inlined]
forward(::Function, ::Array{Real,1}) at back.jl:139
#67 at hmc_core.jl:10 [inlined]
#_leapfrog#75(::getfield(Turing, Symbol("##71#72")){Turing.VarReplay.VarInfo,Turing.Sampler{HMCDA{Any}}}, ::getfield(Turing, Symbol("##73#74")){Turing.Sampler{HMCDA{Any}}}, ::Function, ::Array{Real,1}, ::Array{Float64,1}, ::Int64, ::Float64, ::getfield(Turing, Symbol("##67#68")){Turing.VarReplay.VarInfo,Turing.Sampler{HMCDA{Any}},typeof(myModel_model)}) at hmc_core.jl:93
(::getfield(Turing, Symbol("#kw##_leapfrog")))(::NamedTuple{(:rev_func, :log_func),Tuple{getfield(Turing, Symbol("##71#72")){Turing.VarReplay.VarInfo,Turing.Sampler{HMCDA{Any}}},getfield(Turing, Symbol("##73#74")){Turing.Sampler{HMCDA{Any}}}}}, ::typeof(Turing._leapfrog), ::Array{Real,1}, ::Array{Float64,1}, ::Int64, ::Float64, ::Function) at none:0
macro expansion at logging.jl:312 [inlined]
#_hmc_step#83(::Function, ::Function, ::Function, ::Array{Real,1}, ::Float64, ::getfield(Turing, Symbol("##69#70")){Turing.VarReplay.VarInfo,Turing.Sampler{HMCDA{Any}},typeof(myModel_model)}, ::Function, ::Float64, ::Float64, ::Array{Float64,1}) at hmcda.jl:150
(::getfield(Turing, Symbol("#kw##_hmc_step")))(::NamedTuple{(:rev_func, :log_func),Tuple{getfield(Turing, Symbol("##71#72")){Turing.VarReplay.VarInfo,Turing.Sampler{HMCDA{Any}}},getfield(Turing, Symbol("##73#74")){Turing.Sampler{HMCDA{Any}}}}}, ::typeof(Turing._hmc_step), ::Array{Real,1}, ::Float64, ::Function, ::Function, ::Float64, ::Float64, ::Array{Float64,1}) at none:0
macro expansion at logging.jl:319 [inlined]
step(::Function, ::Turing.Sampler{HMCDA{Any}}, ::Turing.VarReplay.VarInfo, ::Bool) at hmcda.jl:82
macro expansion at logging.jl:309 [inlined]
#sample#94(::Int64, ::Bool, ::Nothing, ::Int64, ::Nothing, ::Function, ::Function, ::HMC{Any}) at hmc.jl:148
sample(::Function, ::HMC{Any}) at hmc.jl:104
top-level scope at none:0
include_string(::Module, ::String, ::String, ::Int64) at eval.jl:30
(::getfield(Atom, Symbol("##117#123")){String,Int64,String})() at eval.jl:94
withpath(::getfield(Atom, Symbol("##117#123")){String,Int64,String}, ::String) at utils.jl:30
withpath at eval.jl:46 [inlined]
#116 at eval.jl:93 [inlined]
with_logstate(::getfield(Atom, Symbol("##116#122")){String,Int64,String}, ::Base.CoreLogging.LogState) at logging.jl:397
with_logger at logging.jl:493 [inlined]
#115 at eval.jl:92 [inlined]
hideprompt(::getfield(Atom, Symbol("##115#121")){String,Int64,String}) at repl.jl:85
macro expansion at eval.jl:91 [inlined]
Wrapping the data in an array, but that caused other problems.
1 Like
This does not directly answer your question, but here is a version using DynamicHMC.jl & friends:
using DynamicHMC, Distributions, LogDensityProblems, TransformVariables
struct MyModel{Tα, Tp, Td}
α::Tα
νprior::Tp
data::Td
end
function (model::MyModel)(νs) # log posterior
loglikelihood = sum(logpdf(Normal(ν, α), x) for (α, ν) in zip(model.α, νs), x in model.data)
logprior = sum(logpdf(model.νprior, ν) for ν in νs)
loglikelihood + logprior
end
p = MyModel(fill(0.85, 3), Normal(15, 5), rand(Normal(0,1), 50)) # prior and data
p([1.0, 2.0, 3.0]) # test for random νs
P = TransformedLogDensity(as(Array, 3), p) # identity transformation
∇P = ForwardDiffLogDensity(P) # AD using ForwardDiff
chain, nuts = NUTS_init_tune_mcmc(∇P, 1000)
νs_posterior = get_position.(chain)
if you want to code your likelihood directly.
3 Likes
And you get very good mixing:
julia> using MCMCDiagnostics
julia> mapslices(effective_sample_size, get_position_matrix(chain); dims = 1)
1×3 Array{Float64,2}:
909.708 933.698 858.296
Thank you for taking the time to put together that code. It’s good to know that alternative approaches are available. I was wondering if you would be able to answer a few questions? Is pmap the preferred method for running multiple chains in parallel? For example,
rngs = [Random.seed!(i) for i in 1:Nchains]
output = pmap(x->NUTS_init_tune_mcmc(x,∇P, 1000),rngs)
When I pass an rng, it prints out a lot of data. So I’m guessing that I am doing it incorrectly.
Secondly, is it possible to output the chains into an MCMCChain.jl object for the purpose of plotting and diagnostics? This would be handy. Thanks!
NUTS_init_tune_mcmc returns the chain and the tuned sampler (so you could continue sampling). Very few people use that so I may remove it as the default, or introduce an alternative API. Currently you can just keep the first value, eg (continuing the above example)
using Random, Distributed
Nchains = 4
rngs = [Random.seed!(i) for i in 1:Nchains];
output = pmap(x -> first(NUTS_init_tune_mcmc(x, ∇P, 1000)), rngs)
works fine (diagnostic output is garbled though, as it is not ready for parallel chains yet).
Conversion to MCMCChain.jl is not supported by my packages (and I don’t know about others). I kind of prefer the bare-bones, modular approach. In practice, IMO the two most useful diagnostics are \hat{R} and ESS, both of which are supported by MCMCDiagnostics.
If you get errors, please make a self-contained MWE and I will look at it.
1 Like
Thanks for your help. The printing was a false alarm. I added a print statement to see what was happening in the loglikelihood function. I forgot to resubmit the code after I removed the print statement. So it looked like it was printing for another reason. Sorry for the noise.
For anyone who is following the code for parallel chains with DynamicHMC, Random.seed!() is not the correct method. It gives the same result for each chain. rngs = [MersenneTwister() for i in 1:Nchains] seems to be working properly.
Good point. Opened a docs issue to address this.
https://github.com/tpapp/DynamicHMCExamples.jl/issues/10
1 Like
I adapted your example to my code and encountered a type problem that I am not sure how to resolve. I did not realize that the sampler uses a forward diff object. The problem I am having is that (1) I reparameterize my model so that resulting parameters are a function of the parameters I input, and (2), I pass these parameters to other likelihood functions because my model is a mixture of convolutions. Aside from the type issue, I’m not sure if this would evaluate the forward difference at each level of the model, rather than just at the creation of mydistribution in the toy example below.
In the example below, the commented function i initially created crashes. I thought maybe if I wrap the log posterior calculations in a different function and extract the values, it might evaluate properly (presumably at two points), but evidently that does not work. Do you have any ideas?
using Distributions
import Distributions: pdf, logpdf
using DynamicHMC, Distributions, LogDensityProblems, TransformVariables,MCMCChain,Random
mutable struct mydistribution{T} <: ContinuousUnivariateDistribution
ν::T
α::T
μ::T
function mydistribution(ν,α)
n = new{typeof(ν)}()
n.ν = ν
n.α = α
n.μ = α./ν #some kind of transformation
return n
end
end
function pdf(d::mydistribution,T::AbstractArray)
#various intensive computations
likelihoods = fill(0.0,length(T))
for (i,t) in enumerate(T)
likelihoods[i] = sum(@. .5*pdf(Normal.(d.μ,d.α),t))
end
return likelihoods
end
logpdf(d::mydistribution,T::AbstractArray) = sum(log.(pdf(d,T)))
struct MyModel{Tα, Tp, Td}
α::Tα
νprior::Tp
data::Td
end
# function (model::MyModel)(νs) # log posterior
# #This crashes
# #dist = mydistribution([νs;0.0],model.α)
# loglikelihood = logpdf(dist,model.data)
# logprior = sum(logpdf(model.νprior, ν) for ν in νs)
# loglikelihood + logprior
# end
function (model::MyModel)(νs) # log posterior
posterior(model,νs)
end
function posterior(model,νs)
#This runs but does not update
#println(fieldnames(typeof(νs[1])))
dist = mydistribution([νs[1].value;νs[2].value;0.0],model.α)
loglikelihood = logpdf(dist,model.data)
logprior = sum(logpdf(model.νprior, ν.value) for ν in νs)
return loglikelihood + logprior
end
p = MyModel(fill(0.85, 3), Normal(15, 5), rand(Normal(0,.85), 50)) # prior and data
P = TransformedLogDensity(as(Array, 2), p) # identity transformation
∇P = ForwardDiffLogDensity(P) # AD using ForwardDiff
chain, nuts = NUTS_init_tune_mcmc(∇P, 1000)
νs_posterior = get_position.(chain)
mean(hcat(νs_posterior...),dims=2)
I don’t think it does; you are just getting an error message. I think it is informative.
AD should not be a concern, as long as you get your model (log) density working. You can test that by calling p with parameters.
Sorry if I am being dense here. Maybe you are referring to the extra # in the commented function? In the process of debugging, I accidently left an an extra # on one line of commented the function, but that is not the source of the problem. Sorry about that if that is the point of confusion. When I run the first function, I get:
function (model::MyModel)(νs) # log posterior
#This crashes
dist = mydistribution([νs;0.0],model.α)
loglikelihood = logpdf(dist,model.data)
logprior = sum(logpdf(model.νprior, ν) for ν in νs)
loglikelihood + logprior
end
p([1,1])
-104.30759593477619
So the test passes as that point. However, when I try to sample, I get the following error:
chain, nuts = NUTS_init_tune_mcmc(∇P, 1000)
MethodError: no method matching Float64(::ForwardDiff.Dual{ForwardDiff.Tag{getfield(LogDensityProblems, Symbol("##1#2")){TransformedLogDensity{TransformVariables.ArrayTransform{TransformVariables.Identity,1},MyModel{Array{Float64,1},Normal{Float64},Array{Float64,1}}}},Float64},Float64,2})
Closest candidates are:
Float64(::Real, !Matched::RoundingMode) where T<:AbstractFloat at rounding.jl:185
Float64(::T<:Number) where T<:Number at boot.jl:725
Float64(!Matched::Int8) at float.jl:60
...
convert(::Type{Float64}, ::ForwardDiff.Dual{ForwardDiff.Tag{getfield(LogDensityProblems, Symbol("##1#2")){TransformedLogDensity{TransformVariables.ArrayTransform{TransformVariables.Identity,1},MyModel{Array{Float64,1},Normal{Float64},Array{Float64,1}}}},Float64},Float64,2}) at number.jl:7
setindex!(::Array{Float64,1}, ::ForwardDiff.Dual{ForwardDiff.Tag{getfield(LogDensityProblems, Symbol("##1#2")){TransformedLogDensity{TransformVariables.ArrayTransform{TransformVariables.Identity,1},MyModel{Array{Float64,1},Normal{Float64},Array{Float64,1}}}},Float64},Float64,2}, ::Int64) at array.jl:769
transform_logdensity(::TransformVariables.ArrayTransform{TransformVariables.Identity,1}, ::MyModel{Array{Float64,1},Normal{Float64},Array{Float64,1}}, ::Array{ForwardDiff.Dual{ForwardDiff.Tag{getfield(LogDensityProblems, Symbol("##1#2")){TransformedLogDensity{TransformVariables.ArrayTransform{TransformVariables.Identity,1},MyModel{Array{Float64,1},Normal{Float64},Array{Float64,1}}}},Float64},Float64,2},1}) at TransformVariables.jl:144
macro expansion at Parameters.jl:732 [inlined]
logdensity at LogDensityProblems.jl:138 [inlined]
vector_mode_dual_eval(::getfield(LogDensityProblems, Symbol("##1#2")){TransformedLogDensity{TransformVariables.ArrayTransform{TransformVariables.Identity,1},MyModel{Array{Float64,1},Normal{Float64},Array{Float64,1}}}}, ::Array{Float64,1}, ::ForwardDiff.GradientConfig{ForwardDiff.Tag{getfield(LogDensityProblems, Symbol("##1#2")){TransformedLogDensity{TransformVariables.ArrayTransform{TransformVariables.Identity,1},MyModel{Array{Float64,1},Normal{Float64},Array{Float64,1}}}},Float64},Float64,2,Array{ForwardDiff.Dual{ForwardDiff.Tag{getfield(LogDensityProblems, Symbol("##1#2")){TransformedLogDensity{TransformVariables.ArrayTransform{TransformVariables.Identity,1},MyModel{Array{Float64,1},Normal{Float64},Array{Float64,1}}}},Float64},Float64,2},1}}) at apiutils.jl:35
macro expansion at Parameters.jl:733 [inlined]
When I change ν::T to ν::AbstractArray or ν::Any in mydistribution, I still get the problem.
This happens because you are allocating
which is of Float64s, and trying to save values in there, which the dual types of AD don’t convert to (because that would lose the derivative information). Instead, you should use generators, broadcasting, or map, like the first example I wrote.
Also, incidentally, don’t use pdf, always use logpdf, to avoid under/overflow.
Thanks for your patience. I usually do not delve into the internal workings of packages. I did not realize that logpdf was accepting and returning a Dual object.
This is good advice and I usually try to follow it. Unfortunately, there is no analytic solution to the loglikelihood function. I have to approximate the pdf numerically with fourier transforms and other operations, then take the log of the pdf. So in certain cases there might be numeric problems.
One problem in my particular use case is that broadcasting data over logpdf is prohibitively slow because of the computations I need to perform before finally obtaining the loglikelihood. On each call to pdf, I use fourier transforms along with other calculations to obtain densities for several values given a fixed set of parameters. From that point, I can use interpolation to get a very accurate approximation of a the density at any point in the domain. So if I do this on each call to logpdf via map or some other broadcasting method, performance suffers greatly.
Is the solution to redefine logpdf method so that it adds the log densities to the Dual object?
No worries, I am happy to help, I also do Bayesian inference, and this part of Julia is tricky.
Julia is powerful because it allows very generic programming, eg in the above case (ostensibly) seamlessly replacing Float64 <: Real with another <: Real type (eg ForwardDiff.Dual). Ideally, you would not need to care.
This requires a different approach if you are used to filling preallocated containers for output. Your best solution is to rearrange your algorithm so that it implicitly figures out the result types. I would recommend thinking hard about this; it is fine if you need an intermediate result for calculations, just generate that and use it.
In case that fails, there are various solutions:
1. Use the type of the first element, and then widen if necessary. This is very robust, but tricky to program and is frankly not the best introduction to the language. Base.collect and friends do this.
2. Use promotion to figure out. See the use of promote_ functions in the SparseArrays and LinearAlgebra standard libraries.
3. Use something quick and dirty. Eg
function use_preallocated(A::AbstractArray, B::AbstractArray)
T = typeof(one(eltype(A))/one(eltype(B)))
preallocated = Vector{T}(undef, len)
...
end
`
Thanks for the advice. I’ll give this some thought over the weekend and hopefully get closer to a solution next week. By chance, do you know of any examples or package code that might also be illuminating? I looked through the repositories for DynamicHMC and LogDensityProblems, but didn’t spot anything.
I started
but it is very basic at the moment. If you open an issue and describe the problem in detail, I am very happy to help more.
1 Like | 4,925 | 17,976 | {"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 | 3 | CC-MAIN-2023-06 | latest | en | 0.852128 |
https://mcqslearn.com/electronics/electric-circuit-analysis/quiz/quiz.php?page=2 | 1,718,900,532,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00633.warc.gz | 336,356,483 | 16,744 | Engineering Programs Courses
Electric Circuit Analysis Certification Exam Tests
Electric Circuit Analysis Practice Test 2
# Singularity Functions MCQ (Multiple Choice Questions) PDF - 2
Books:
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## Singularity Functions Questions and Answers : Quiz 2
MCQ 6:
The function that is 0 everywhere except at t=0, where it is undefined is known as
1. Unit step
2. Unit ramp
3. Unit impulse
4. Singularity
MCQ 7:
Y-parameters between two impedances are combined as a
1. series connection
2. hybrid connection
3. parallel connection
4. time invariant connection
MCQ 8:
A jump in the capacitor voltage requires an
1. impulse current
2. variable current
3. AC current
4. continuous current
MCQ 9:
Voltage or current source which have capability to supply energy to the whole network or circuit is
1. Passive Element
2. Active Element
3. Magnetic Element
4. Homogeneous Element
MCQ 10:
Magnetic amplifiers are devices where one winding is used to control the saturation of a magnetic core and hence alter the impedance of the other winding is similar to
1. transistor
2. transformer
3. motor
4. generator | 456 | 2,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.827778 |
http://mrlangemath.com/blog/?m=20141208 | 1,606,836,427,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674594.59/warc/CC-MAIN-20201201135627-20201201165627-00127.warc.gz | 62,317,923 | 7,341 | # Statistics Dec 8 Ch 7 Wrap Up
Compare and Discuss your Coins and Dice Answers—- Help the people that screwed it up finish up.
Work on Book assignments and Turn in HW Journals—-
DUE Assignments
Ch 7 Reading Guide Complete— Check it in
Open Bookwork
6.1 Simulation 1-4, 7-8
6.2 Probability Models 22, 26, 29, 33,36, 37, 39, 43, 45, 47, 49
6.3 General Probability Rules 65-6.70 71-73,76-77 79 , 81-83
7.1 Discrete and Continuous Random Variables 2,3,7, 9
7.2 Means and Variances of Random Variables 23, 24, 32, 34, 37, 38, 41
# Calculus AB Dec 8 Antiderivatives with U-Substitution
Get Mr. Lange’s UCLA Presentation Checklist and FRQs to Practice
Self Rate Yourself on All of the topics on the checklist 1-5…..
Where 1 means you have no clue what it means and 5 means you are an expert and can teach it to others
Finish grading Thanksgiving TH Exam
Homework- KhanAcademy– Master all Limits and Derivatives, Start working on Antiderivatives/Indefinite Integrals
Work on the HW From Friday’s Handout— posted today p 3 and 4
Work on your 4 Essays Due Friday | 333 | 1,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-50 | latest | en | 0.831662 |
https://m.everything2.com/title/Bi-Phase+encoding | 1,718,738,257,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00456.warc.gz | 342,306,933 | 9,501 | One form of encoding asynchronous binary data is BiPhase encoding (or Bi-Phase encoding). Given that a message consists of '0's and '1's, and that two different types of signal can be transmitted 'L' and 'H', bi-phase encoding consists of transmitting either 'LH' or 'HL' for every bit. The name comes from the fact that every bit period looks like one cycle of a digitized sine wave, either with phase 0 (HL) or phase π (LH).
The advantage of bi-phase encoding is that every pulse is either half a bit long or one bit long, so it is self-clocking; i.e. so long as the clock used to transmit and the clock used to receive are both reasonably constant, the receiver will never lose track of how many bit-periods (or, to be more specific, half-bit-periods) have elapsed since the beginning of the message. NRZ (Non-Return to Zero) encoding is not self-clocking; therefore, without bit stuffing or framing or some other mechanism, the resolution and accuracy of the receiver's clock may accumulate an error of greater than 1 bit period if the signal is unchanging for too long (imagine trying to guess whether a long beep is 100 seconds long or 101 seconds long). The disadvantage of a self-clocking signal is that it requires more bandwidth than NRZ.
The most obvious form of BiPhase encoding is fully-coherent bi-phase encoding, a.k.a. Manchester encoding. Fully-coherent (or fully coherent) means that the value of the underlying message corresponds directly with the (absolute) value of the signal; a bit '0' is represented by 'LH', and a bit 1 is represented by 'HL' (unless you ask someone from the other half, who will tell you that '0' is 'HL' and '1' is 'LH'). For example:
``` <--0--> <--0--> <--0--> <--1--> <--1--> <--1--> <--0--> <--1--> <--0--> <--1-->
H --- --- ------- --- --- ------- -------
L / H \ L / H \ L / H H \ L / H \ L / H \ L L / H H \ L L / H H \ L
L --- --- --- --- --- ------- ------- ---
```
A less obvious, but sometimes more useful, form of BiPhase is differentially-coherent bi-phase (sometimes called differential bi-phase, sometimes even just called bi-phase by people who don't realise Manchester encoding is also a form of bi-phase). Differentially coherent means that the value of the message corresponds to changes in the value of the signal. In this scheme, a '0' is represented by a cycle with the same phase as the previous cycle, and a '1' is represented by a cycle with the phase opposite from the previous cycle. (Again, if you talk to some others, they will tell you that a '0' corresponds to a phase change and a '1' corresponds to no phase change. Change on '1' has a mathematical nicety in that converting between fully-coherent and differentially-coherent is identical to changing between binary and gray code.) Differential bi-phase is often described in different terms; one may say that there is an edge (or a transition) in the middle of every bit, and an additional edge at the beginning of a '0' bit; or, there's an edge in the middle of every bit and an edge at the end of a '0' bit; or, there's an edge between every pair of bits and an edge in the middle of a '0' bit. All of these descriptions are equivalent, with the exception of the details of the very first and/or very last bits being transmitted; in RFID, commonly a message is transmitted again and again back-to-back; so even this detail is not significant. The same data in the above example looks rather different (this diagram assumes the signal is at H before the bit sequence shown):
``` <--0--> <--0--> <--0--> <--1--> <--1--> <--1--> <--0--> <--1--> <--0--> <--1-->
H - --- --- ------- ------- --- --- -------
\ L / H \ L / H \ L / H H \ L L / H H \ L / H \ L L / H \ L / H H \ L
L --- --- --- ------- --- ------- --- ---
```
A comparison of the two signals is worthwhile. For the fully-coherent signal, a regular square wave with a period equal to the bit period appears whenever a long string of '0's or a long string of '1's appears. A regular square wave with a period equal to twice the bit period appears whenever the bits are alternating. By contrast, in the differentially-coherent signal, a regular square wave with a period equal to the bit period only appears where there is a string of '0's, and a regular square wave with a period of twice the bit period appears when there is a string of '1's. Note that a message transmitted as fully-coherent bi-phase is physically indistinguishable from a message transmitted as differentially-coherent bi-phase. The '-coherent' part of the encoding is all about the interpretation of signals, not about the shape of the signals.
Other forms of encoding include NRZ and Miller encoding.
Log in or register to write something here or to contact authors. | 1,154 | 4,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-26 | latest | en | 0.899695 |
http://programmers.stackexchange.com/questions/140511/what-is-the-name-of-this-tree?answertab=active | 1,448,543,281,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447266.73/warc/CC-MAIN-20151124205407-00350-ip-10-71-132-137.ec2.internal.warc.gz | 198,519,751 | 17,962 | # What is the name of this tree?
It has a single root and each node has 0..N ordered sub-nodes . The keys represent a distinct set of paths. Two trees can only be merged if they share a common root. It needs to support, at minimum: insert, merge, enumerate paths.
For this tree:
```The
+--------+----------------+
| | |
cat cow dog
+ +--------+ +
| | | |
drinks jumps moos barks
+
|
milk
```
the paths would be:
• The cat drinks milk
• The cow jumps
• The cow moos
• The dog barks
It's a bit like a trie. What is it?
-
It's not bit like a trie, it is a trie. – vartec Mar 19 '12 at 17:03
@vartec: I think it would be more accurate to say that a trie could be one possible implementation. – Robert Harvey Mar 19 '12 at 17:06
@Daniel: Wikipedia says yes. – Robert Harvey Mar 19 '12 at 17:08
@RobertHarvey: It says the root is the empty string. Not so in my case. I see some other minor differences in the Wikipedia article, but it's probably similar enough to be--as you stated--a form of trie. – Daniel Mar 19 '12 at 17:10
The root.. well, you can add fake empty root, which wouldn't change much from logical point of view. Main difference is that in your example you deal with words instead of characters, but that also doesn't change much from logical point of view. – vartec Mar 19 '12 at 17:19
This is called a "rope" when applied specifically to strings, I believe.
http://en.wikipedia.org/wiki/Rope_(computer_science)
They offer vastly improved algorithmic complexity for many mutating operations over strings-as-arrays.
-
It's not a rope. A rope is just a "fat string;" it doesn't contain any logical relationships, per se. – Robert Harvey Mar 19 '12 at 19:36
This is very close to a Radix tree. The primary differences are that a normal radix tree wouldn't split on words, so the 'c' in both "cat" and "cow" would be the same node, and it only splits when necessary:
``````The
+-------------------------+
| |
c dog barks
+---------------+
| |
at drinks milk ow
+--------+
| |
jumps moos
``````
I might describe what you have as a modified Radix tree, that is forced to use spaces as a delimiter. Regardless, it is some sort of "tree", so that should be sufficient to describe it if you have some extra explanation as to its structure.
-
By Wikipedia, it looks like your tree is specified by the two properties arborescence and ordered tree (scroll down to find the definition "ordered tree or plane tree.")
-
So you would just call it a "tree"? – Daniel Mar 19 '12 at 17:52
Well, the second paragraph on the WP Tree page makes it sound like that, but no, "tree" is a more general term. A tree may or may not have these properties. The arborescence property is about having a single root and directed edges away from the root. The parent-child relationship forms these directed edges. The "ordered tree" property is about having an ordering among the children of a given node. – Sonia Mar 19 '12 at 18:02
One more clarification, note the two links under my answer are under (graph theory) and that the Tree page we're looking at now is under (data structure.) It probably is true that nearly all tree data structures have these two graph theory properties. If you're looking for the name of a data structure, this answer may not be the one you want. – Sonia Mar 19 '12 at 18:09
There are tree data structures where the arrows point towards the (multiple) roots. This happens in union-find, for example. I should have thought of that earlier. – Steve314 Mar 19 '12 at 18:13 | 911 | 3,649 | {"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} | 2.65625 | 3 | CC-MAIN-2015-48 | longest | en | 0.922406 |
https://examgold.com/2015/09/time-and-distance-important-facts-and-formulas.html | 1,721,292,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00773.warc.gz | 216,918,330 | 12,235 | Time and Distance – Important Facts and Formulas
By | September 25, 2015
TIME AND DISTANCE -> IMPORTANT FACTS AND FORMULAE
1. Speed = [Distance/Time],
Time=[Distance/Speed],
Distance = (Speed*Time)
2. x km/hr = [x*5/18] m/sec.
3. If the ratio of the speeds of A and B is a:b, then the ratio of the times taken by them to cover the same distance is 1/a : 1/b or b:a.
4. x m/sec = [x*18/5] km/hr.
5. Suppose a man covers a certain distance at x km/hr and an equal distance at y km/hr. then, the average speed during the whole journey is [2xy/x+y] km/hr. | 171 | 557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.888075 |
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# Is a*b*c divisible by 24? (1) a,b, and c are consecutive
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Is a*b*c divisible by 24? (1) a,b, and c are consecutive [#permalink]
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28 May 2012, 04:06
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Is a*b*c divisible by 24?
(1) a, b, and c are consecutive even integers
(2) a*b is divisible by 12
This problem is from GMATClub Quant tests.
I am confused in this problem. Since 0 is also an EVEN number so statement 1 cannot be true. Am i right?
[Reveal] Spoiler: OA
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Last edited by Bunuel on 28 May 2012, 23:39, edited 1 time in total.
Edited the question and added the OA
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Re: GMATClub number problem confusion [#permalink]
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28 May 2012, 06:45
Is the answer A?
Even if one of the even integers a, b, c is 0, the product a*b*c = 0 is divisible by 24.
0/24 = 0 => integer. So a * b * c is divisible by 24.
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Re: GMATClub number problem confusion [#permalink]
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28 May 2012, 07:42
Oops! here is the catch which you pointed out. 0 is divisible by 24. Now I got it.
Yes the answer is A. Thanks a lot for the explanation.
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Re: Is a*b*c divisible by 24? (1) a,b, and c are consecutive [#permalink]
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28 May 2012, 23:40
dexerash wrote:
Is a*b*c divisible by 24?
(1) a, b, and c are consecutive even integers
(2) a*b is divisible by 12
This problem is from GMATClub Quant tests.
I am confused in this problem. Since 0 is also an EVEN number so statement 1 cannot be true. Am i right?
Note that an integer $$a$$ is a multiple of an integer $$b$$ (integer $$a$$ is a divisible by an integer $$b$$) means that $$\frac{a}{b}=integer$$: so, as 0 divided by any integer (except zero itself) yields an integer then zero is a multiple of every integer (except zero itself).
Also on GMAT when we are told that $$a$$ is divisible by $$b$$ (or which is the same: "$$a$$ is multiple of $$b$$", or "$$b$$ is a factor of $$a$$"), we can say that:
1. $$a$$ is an integer;
2. $$b$$ is an integer;
3. $$\frac{a}{b}=integer$$.
BACK TO THE ORIGINAL QUESTION:
Is a*b*c divisible by 24?
(1) a, b, and c are consecutive even integers --> $$a=2k-2$$, $$b=2k$$ and $$c=2k+2$$ for some integer $$k$$ --> $$abc=(2k-2)2k(2k+2)=8(k-1)k(k+1)$$, now $$(k-1)$$, $$k$$, $$(k+1)$$ are 3 consecutive integers, which means that one of them must be a multiple of 3, thus $$abc$$ is divisible by both 8 and 3, so by 24. Sufficient.
Or even without the formulas: the product of 3 consecutive even integers will have 2*2*2=8 as a factor, plus out of 3 consecutive even integers one must be a multiple of 3, thus abc is divisible by both 8 and 3, so by 24.
(2) a*b is divisible by 12, clearly insufficient as no info about c (if ab=12 and c=1 answer will be NO but if ab=24 and c=any integer then the answer will be YES).
Hope it's clear.
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Re: Is a*b*c divisible by 24? (1) a,b, and c are consecutive [#permalink] 28 May 2012, 23:40
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# Is a*b*c divisible by 24? (1) a,b, and c are consecutive
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,634 | 5,340 | {"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} | 3.703125 | 4 | CC-MAIN-2017-04 | latest | en | 0.861778 |
http://petergabel.info/blog/ThereWeSATDown/ | 1,716,130,794,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00099.warc.gz | 25,300,686 | 8,827 | # There we SAT down
Keywords: Reasoning, SAT, Description Logics
SAT solving is a non-numeric constraint problem involving Boolean logic. SAT (short for "satisfiability") determines the ability for a propositional formula to be satisfied by some assignment of truth values to its variables. Its significance is wide:
• Many reasoning problems can be reduced to SAT or one of its variants or extensions. Configuration and logic verification problems are frequently addressed with SAT solvers.
• SAT is the quintessential problem in the study of complexity of computation. It has been shown to be NP-complete. Many problems have been shown to be as hard to solve as SAT and because SAT is well understood, it is strong evidence of the complexity theory conjecture that the class of problems designated as P is not equivalent to the class NP.
• Much progress has been made toward the creation of efficient SAT solvers for problems as they tend to arise in the real world. In other words, just because a problem is theoretically intractable across all possible problems does not imply that the problems that are actually encountered will be intractable given some approach.
• Many of the techniques employed in state-of-the-art SAT solvers can be applied to the implementation of other reasoning mechanisms such as Description Logic ("DL"), Temporal Logic and other theorem provers.
• Some of the sub-problems of inference within a DL, Temporal Logic or other theorem prover can be exported to a SAT solver.
This post addresses in brief a problem that is at the heart of DL reasoning and some techniques used to solve SAT constraint problems.
## Defining the problem formally
The Boolean Satisfiability (SAT) problem consists of determining a satisfying variable assignment, V, for a Boolean function, f, or determining that no such V exists. Most solvers operate on problems for which f is specified in conjunctive normal form (CNF). This form consists of the logical AND of one or more clauses, which consist of the logical OR of one or more literals. The literal comprises the fundamental logical unit in the problem, being merely an instance of a variable or its complement. All Boolean functions can be described in the CNF format. The advantage of CNF is that in this form, for f to be satisfied, each individual clause must be satisfied.
An aside
Logicians have two senses of the word "subsumes" which are, in fact, absolutely opposite in meaning! To a worker in Description Logic, "subsumes" means "is more general than or equal to." To most other logicians, "subsumes" means "implies" which would in fact be "less general than or equal to." Usually the sense is clear from the context, but care should be taken when reading papers written by authors whose inclinations are unknown to you. I always use the word subsumes in the DL sense and never use the word outside of a DL context.
The usual approaches to SAT are modifications of a method proposed by Davis, Putnam, Logemann, and Loveland for backtracking through a search space that is explored using Boolean constraint propagation (BCP). BCP is based on the observation that if a clause whose literals are all 0 save one is to be satisfied, then that single literal must be assigned the value 1. That variable assignment is propagated to all clauses forcing each to be satisfied (and eliminated), false (forcing backtracking over the assignment and to a previous choice point), unit (again all 0 save one and a candidate for BCP), or unknown (more than one literal unknown).
BCP is applied until there is a contradiction, all clauses have been satisfied and removed, or all remaining clauses have at least 2 unknown literals. A choice must be made for assignment, a choice point is recorded, and the variable assignment is propagated. Again, some clauses may be satisfied, false, unit, or have more than one unknown (with the remaining literals 0). BCP and choices ("splitting") continues until satisfiability has been determined.
Modern SAT solvers employ many refinements. Some of these are:
• Conflict analysis and non-chronological backtracking
• Recording conflicts so portions of a previously searched subtree are not searched again
• Heuristics to select which variable is to be assigned a value when BCP does not apply
• Splitting heuristics which choose which of the two phase assignments is made first for a variable
• Restarts where the solver abandons the current search tree
All the state-of-the-art versions of SAT are based on "lazy" data structures for clauses which "watch" literals in some form to efficiently recognize opportunities for BCP.
Various extensions and variant problems to SAT have been developed. The two most interesting directions are QSAT which adds universal quantification to selected variables and integration of SAT with numeric solvers. The latter extensions address problems involving Boolean functions over literals and mathematical propositions.
update needed
Discuss PPSZ and its variants
## Quantified Number Restrictions in DLs
Description Logics often support quantified number restrictions (QNRs). A complete description of how DL languages such as SHIQ(D) or SHOIQ(D) are implemented is beyond the scope of this post but some insight may be gained in exploring some aspects of the problem in isolation.
We can understand QNR processing as a "black box" that may either return that the constraints cannot be satisfied or a (minimal) set of assertions that must be true for satisfaction. Looking more closely at a simplified version of the problem we have the following:
Inputs:
• A set of concepts that are taxonomically related in a directed acyclic graph and a collection of collectively unsatisfiable sets of concepts.
• A set of roles that are organized in a role hierarchy as a directed acyclic graph and inverse role assertions.
• A set of distinct individuals. Each individual is labelled with those concepts that are currently known to (directly) subsume it and with those concepts that are currently known to be the most general that cannot subsume it (out of contradiction).
• A set of QNRs of the form =nR.C or =nR.C
Unsatisifiable Output:
• A minimal set of individuals and QNRs that led to the contradiction.
Satisfiable Output (non-deterministic):
• A set of assertions in the form of i:C (existing individual i is asserted as a concept C)
• A set of assertions in the forms =nR.C and =nR.C
The implementation of such a black box (and its equivalent as implemented as interleaved processing of the tableau of an ABOX consistency test) has been the subject of intensive research within the Description Logic community. A spectrum of solutions ranging from direct expansion of individuals in the tableau, to signature representations which merge processing for sets of unnamed, distinct but otherwise identical individuals, to direct solving using integer programming, have been proposed. As with reasoning problems generally, no single approach dominates.
One of the complications of the problem as I have posed it above is that we seek an "explanation" for unsatisfiability and a non-deterministic set of assertions for satisfiability. The careful reader will note that there may be more than one reason for unsatisfiability and more than one satisfiable set of assertions. The explanation of unsatisfiability is input to the non-chronological backtracking control; it is hard to determine what may be a "best" explanation, any is better than none. The satisfiable output actually must create choice points whenever alternative sets of assertions may exist. Backtracking will enumerate alternatives until unsatisfiability results.
## References
Babic, D.; Bingham, J.; Hu, A. J. (2006). "B-Cubing: New Possibilities for Efficient SAT-Solving" (PDF). IEEE Transactions on Computers. 55 (11): 1315. doi:10.1109/TC.2006.175.
Clarke, E.; Biere, A.; Raimi, R.; Zhu, Y. (2001). "Bounded Model Checking Using Satisfiability Solving". Formal Methods in System Design. 19: 7. doi:10.1023/A:1011276507260.
Davis, M., and Putnam, H. 1960. A computing procedure for quantication theory. Journal of the Association for Computing Machinery 7:201-215.
Davis, M.; Logemann, G.; and Loveland, D. 1962. A machine program for theorem-proving. Communications of the Association for Computing Machinery 5:394-397.
Michael R. Garey & David S. Johnson (1979). Computers and Intractability: A Guide to the Theory of NP-Completeness. W.H. Freeman. ISBN 0-7167-1045-5. A9.1: LO1 - LO7, pp. 259 - 260.
Giunchiglia, E., Narizzano, M., and Tacchella, A. Backjumping for Quantified Boolean Logic Satisfiability.
Giunchiglia, E.; Tacchella, A. (2004). Giunchiglia, Enrico; Tacchella, Armando (eds.). "Theory and Applications of Satisfiability Testing". Lecture Notes in Computer Science. 2919. doi:10.1007/b95238. ISBN 978-3-540-20851-8.
Goldberg, E., and Novikov, Y. 2002. BerkMin: a Fast and Robust SAT-Solver.
Carla P. Gomes; Henry Kautz; Ashish Sabharwal; Bart Selman (2008). "Satisfiability Solvers". In Frank Van Harmelen; Vladimir Lifschitz; Bruce Porter (eds.). Handbook of knowledge representation. Foundations of Artificial Intelligence. 3. Elsevier. pp. 89-134. doi:10.1016/S1574-6526(07)03002-7. ISBN 978-0-444-52211-5.
Lynce, I. and Marques-Silva, J. P. 2001. Efficient Data Structures for Fast SAT Solvers.
Marques-Silva, J.; Glass, T. (1999). "Combinational equivalence checking using satisfiability and recursive learning" (PDF). Design, Automation and Test in Europe Conference and Exhibition, 1999. Proceedings (Cat. No. PR00078). p. 145. doi:10.1109/DATE.1999.761110. ISBN 0-7695-0078-1.
Marques-Silva, J. P., and Sakallah, K. A. 1999. GRASP-A search algorithm for propositional satisability. IEEE Transactions on Computers 48(5):506-521.
Moskewicz, M.; Madigan, C.; Zhao, Y.; Zhang, L.; and Malik, S. 2001. Chaff: Engineering an efficient SAT solver. In Proceedings of the Design Automation Conference.
Rodriguez, C.; Villagra, M.; Baran, B. (2007). "Asynchronous team algorithms for Boolean Satisfiability" (PDF). 2007 2nd Bio-Inspired Models of Network, Information and Computing Systems. p. 66. doi:10.1109/BIMNICS.2007.4610083.
Sebastiani, R. Integrating SAT Solvers with Math Reasoners: Foundations and Basic Algorithms.
Van Gelder, A. 2001. Generalizations of Watched Literals for Backtracking Search.
Vizel, Y.; Weissenbacher, G.; Malik, S. (2015). "Boolean Satisfiability Solvers and Their Applications in Model Checking". Proceedings of the IEEE. 103 (11). doi:10.1109/JPROC.2015.2455034.
Zhang, H. 1997. SATO: An effcient propositional prover. In Proceedings of the International Conference on Automated Deduction, 272-275.
Zhang, H. and Stickel, M. E. 1999. Implementing the Davis-Putnam Method. | 2,522 | 10,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.947613 |
https://zbmath.org/?q=an:0572.35040 | 1,620,981,882,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991648.40/warc/CC-MAIN-20210514060536-20210514090536-00279.warc.gz | 1,142,511,660 | 9,402 | # zbMATH — the first resource for mathematics
On the number of positive solutions of weakly nonlinear elliptic equations when a parameter is large. (English) Zbl 0572.35040
We consider positive solutions of $$-\Delta u=\lambda f(u)$$ in D, $$u=0$$ on $$\partial D$$. We assume that $$a>0$$ (possibly $$+\infty)$$, f: [0,a]$$\to R$$ is $$C^ 1$$ such that $$f(t)>0$$ on (0,a) $$f(a)=0$$ if $$a<\infty$$ and $$f(t)\to C>0$$ as $$t\to \infty$$ if $$a=\infty$$. We only consider solutions such that u(x)$$\leq a$$ in D. We also assume that D is bounded in $$R^ n$$ with smooth boundary. If $$f(0)>0$$ or $$f'(0)>0$$ and some technical conditions are satisfied, we prove that there is a unique positive solution for large $$\lambda$$.
Now assume that $$f(0)=f'(0)=0$$ and that there is a $$p>1$$ such that $$y^{1-p}f'(y)\to bp$$ as $$y\to 0$$ where $$b>0$$. If $$p\leq n(n-2)^{-1}$$, we prove that there is a unique positive solution whose sup norm is not small and prove that the small positive solutions are largely detemied by the positive solutions of the equation $$-\Delta u=u^ p$$ in D, $$u=0$$ on $$\partial D.$$
We prove a similar result if $$p\in (n(n-2)^{-1}, (n+2)(n-2)^{-1})$$ under a global condition on f and give examples showing that this condition can not be entirely removed. We also give an example where $$a=\infty$$ and the set of positive solutions is not connected.
##### MSC:
35J65 Nonlinear boundary value problems for linear elliptic equations 47J05 Equations involving nonlinear operators (general) 35B30 Dependence of solutions to PDEs on initial and/or boundary data and/or on parameters of PDEs 35B40 Asymptotic behavior of solutions to PDEs
##### Keywords:
uniqueness; blowing up methods; positive solutions
Full Text: | 524 | 1,747 | {"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} | 3.0625 | 3 | CC-MAIN-2021-21 | latest | en | 0.826724 |
https://www.jiskha.com/questions/51018/use-6-of-the-digits-from-1-to-9-to-make-each-number-use-each-digit-only-once-a-the | 1,642,941,096,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304261.85/warc/CC-MAIN-20220123111431-20220123141431-00077.warc.gz | 871,139,617 | 4,942 | # math
use 6 of the digits from 1 to 9 to make each number. use each digit only once.
a) the greatest possible even number
1. 👍
2. 👎
3. 👁
1. would 987654 do it?
1. 👍
2. 👎
2. 987,654
1. 👍
2. 👎
3. (3w)^9+2=81
1. 👍
2. 👎
4. 987654
1. 👍
2. 👎
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https://simplywall.st/stocks/hk/banks/hkg-2388/boc-hong-kong-holdings-shares/news/boc-hong-kong-holdings-limited-hkg2388-is-trading-at-a-4-64-discount/ | 1,571,602,561,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986718918.77/warc/CC-MAIN-20191020183709-20191020211209-00092.warc.gz | 696,938,870 | 15,984 | latest
# BOC Hong Kong (Holdings) Limited (HKG:2388) Is Trading At A 4.64% Discount
Bank stocks such as 2388 are hard to value. This is because the rules banks face are different to other companies, which can impact the way we forecast their cash flows. For example, banks are required to hold more capital to reduce the risk to depositors. Emphasizing data points such as book values, on top of the return and cost of equity, may be useful for evaluating 2388’s valuation. Below I’ll determine how to value 2388 in a relatively accurate and uncomplicated method.
### Why Excess Return Model?
Let’s keep in mind two things – regulation and type of assets. Financial firms operating in Hong Kong face strict financial regulation. Furthermore, banks usually do not have substantial portions of physical assets on their balance sheet. This means the Excess Returns model is best suited for calculating the intrinsic value of 2388 rather than the traditional discounted cash flow model, which has more emphasis on things like capital expenditure and depreciation.
### The Calculation
The key belief for this model is that equity value is how much the firm can earn, over and above its cost of equity, given the level of equity it has in the company at the moment. The returns above the cost of equity is known as excess returns:
Excess Return Per Share = (Stable Return On Equity – Cost Of Equity) (Book Value Of Equity Per Share)
= (0.14% – 11%) x HK\$26.68 = HK\$0.72
Excess Return Per Share is used to calculate the terminal value of 2388, which is how much the business is expected to continue to generate over the upcoming years, in perpetuity. This is a common component of discounted cash flow models:
Terminal Value Per Share = Excess Return Per Share / (Cost of Equity – Expected Growth Rate)
= HK\$0.72 / (11% – 2.2%) = HK\$8.19
These factors are combined to calculate the true value of 2388’s stock:
Value Per Share = Book Value of Equity Per Share + Terminal Value Per Share
= HK\$26.68 + HK\$8.19 = HK\$34.87
This results in an intrinsic value of HK\$34.87. Relative to the present share price of HK\$33.25, 2388 is , at this time, trading in-line with its true value. This means there’s no real upside in buying 2388 at its current price. Valuation is only one part of your investment analysis for whether to buy or sell 2388. Analyzing fundamental factors are equally important when it comes to determining if 2388 has a place in your holdings.
### Next Steps:
For banks, there are three key aspects you should look at:
1. Financial health: Does it have a healthy balance sheet? Take a look at our free bank analysis with six simple checks on things like bad loans and customer deposits.
2. Future earnings: What does the market think of 2388 going forward? Our analyst growth expectation chart helps visualize 2388’s growth potential over the upcoming years.
3. Dividends: Most people buy financial stocks for their healthy and stable dividends. Check out whether 2388 is a dividend Rockstar with our historical and future dividend analysis.
For more details and sources, take a look at our full calculation on 2388 here.
To help readers see past the short term volatility of the financial market, we aim to bring you a long-term focused research analysis purely driven by fundamental data. Note that our analysis does not factor in the latest price-sensitive company announcements.
The author is an independent contributor and at the time of publication had no position in the stocks mentioned. For errors that warrant correction please contact the editor at editorial-team@simplywallst.com. | 802 | 3,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-43 | latest | en | 0.922096 |
http://www.physicsforums.com/showthread.php?t=699309 | 1,386,432,352,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163054973/warc/CC-MAIN-20131204131734-00078-ip-10-33-133-15.ec2.internal.warc.gz | 490,665,284 | 8,115 | # Calculating e=mc^2 using foot-pounds of force
by markteller
Tags: foot pounds, force, speed of light
P: 2 I have seen the on-line calculators, but would like to see how the formula works in full detail. I have seen the kilograms / meters / second example already. The second part of the question is, what is the actual formula for converting foot pounds of force to Newton meters? Again, the actual details. Thanks...
P: 335
Quote by markteller I have seen the on-line calculators, but would like to see how the formula works in full detail. I have seen the kilograms / meters / second example already. The second part of the question is, what is the actual formula for converting foot pounds of force to Newton meters? Again, the actual details. Thanks...
What sort of details are you looking for? For the first part, if you just want to calculate the rest energy, then you plug in ##m## and ##c## and get a number. Deriving the formula requires a bit more effort and can be found easily online. If you choose to meters, kilograms, seconds, then you get the energy in Joules. If you want it in ft*lbs then you use the conversion factor. According to the all powerful google, it is ##1\ J=0.7375ft\cdot lbs##.
If you wanted to, you could multiply the individual conversion factors to get there. That is, the conversion meters to feet and Newtons to lbs (of force) and get the same number.
P: 2 Thanks! I am looking to convert 1 kilogram of mass into foot-pounds using e=mc^2 and without any unnecessary conversions. I want to see how the final number is arrived at. I then want to see how foot-pounds are converted back to newton meters, which are more typical for e=mc^2. Again, I know there are online calculators, but they don't educate the mind :)
Mentor
P: 10,680
## Calculating e=mc^2 using foot-pounds of force
In the spirit of teaching a man how to fish versus simply giving him a fish, it looks like you might benefit from a tutorial on "how to convert units". Here's the first one I found with a Google search on that phrase: | 494 | 2,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2013-48 | longest | en | 0.917004 |
https://chemistry.stackexchange.com/questions/124659/the-relationship-between-the-temperatures-at-different-times-and-pressures-at-di/124664 | 1,606,660,202,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141198409.43/warc/CC-MAIN-20201129123729-20201129153729-00204.warc.gz | 236,371,204 | 29,079 | # The relationship between the temperatures at different times and pressures at different times is? [closed]
At the instant the boiling point is reached, the vapour pressure of the solution should be equal to atm. pressure. And since the boiling is continued the pressure will keep on increasing. Thus the relationship between the pressures will be be $$P_5 < P_{10} < P_{15}$$.
And since at boiling the temperature of the liquid remains constant the temperature should be equal at all interval. That is Option C should be correct.
But the given answer is option A which doesn't make any sense to me. Can you help me with the question?
• Boiling of a solution differs from boiling of a pure liquid a great deal. To begin with, there is no longer such thing as boiling point. – Ivan Neretin Dec 4 '19 at 16:15
• When making candy, the boiling temperature increases as water boils off and the sugar concentration in the liquid increases, see e.g. craftybaking.com/howto/candy-sugar-syrup-temperature-chart – Karsten Theis Dec 4 '19 at 16:56
As more of the solvent is boiled off, the molality of the solute increases, which increases the boiling point (as $$\Delta T_B = mK_b$$). Therefore, $$T_5 (Karsten Theis mentioned a good link in the comments giving a practical example)
Since the solution is boiled in an open vessel, the vapour pressure of the solvent will at all times be equal to the atmospheric pressure, as it is an isobaric process. Therefore, $$P_5=P_{10}=P_{15}=P_{\text{atm}}$$ | 360 | 1,496 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-50 | latest | en | 0.903743 |
http://www.shapeways.com/product/ZEFQMF3NL/bubble-pent-read-instructions | 1,480,733,052,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540804.14/warc/CC-MAIN-20161202170900-00191-ip-10-31-129-80.ec2.internal.warc.gz | 711,282,802 | 64,834 | <div class="sw-email-modal sw--display-block"> <div id="emailModalContentContainer"> <span class="noty_close sw--position-absolute sw--position-right sw--padding-top-3 sw--padding-right-3 icon-cancel sw--opacity-8"></span> <div class="sw-row"> <div class="sw-email-modal__copy sw--position-relative sw--display-block sw--padding-vert-4"> <p class="last sw--font-size-16">Sign up to get email alerts on discount promotions. There might be one very soon...</p> <form action="/register/email-signup" class="sw-email-modal__signup sw--position-relative" data-confirmation="emailConfirmationModal" data-sw-email-modal-form> <input type="text" class="sw-email-modal__signup-input sw--input-height__medium" placeholder="Email address" name="email" /> <input type="hidden" class="sw-email-modal__signup-input" name="location" value="/product/ZEFQMF3NL/bubble-pent-read-instructions" /> <input type="hidden" class="sw-email-modal__signup-input" name="confirmation" value="emailConfirmationModal" /> <input type="submit" class="btn-primary" value="Sign Up" /> <div id="emailModalFormError" class="text-error" style="display:none"></div> </form> </div> </div> </div>
# Bubble Pent - READ INSTRUCTIONS
##### Not For Sale
Bubble Pent is a pentagonal variation to Mosaic Block and Bubble Block. It is a boublesized corner-turning pillowed pentagon prism. As a result of the pillowing, the puzzle never exposes any ridges mid-turn. This makes it a child-safe puzzle :-).
Read more at the Twisty Puzzles Forum.
NOTE: Bubble Pent consist of two sets of pieces.
-1x Set 1 -- THIS
-1x Set 2
You need to order BOTH sets for a complete puzzle.
Please order a 3D-printed do-it-yourself puzzle kit from Shapeways at this page (check with Oskar about screws and stickers), or contact Oskar directly if you are interested in obtaining a fully colored, stickered and assembled sample of this puzzle.
##### What's in the Box
Bubble Pent - READ INSTRUCTIONS in White Strong & Flexible
Width Height Depth
13.828 cm 8.596 cm 4.148 cm
This model is 3D Printed in White Strong & Flexible: White nylon plastic with a matte finish and slight grainy feel.
Last updated on 01/10/2015 | 551 | 2,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-50 | latest | en | 0.551006 |
http://www.ck12.org/geometry/Rotations/lesson/Rotations/ | 1,462,413,980,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860125750.3/warc/CC-MAIN-20160428161525-00146-ip-10-239-7-51.ec2.internal.warc.gz | 428,936,714 | 34,335 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Rotations
## Transformations by which a figure is turned around a fixed point to create an image.
Estimated6 minsto complete
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Progress
Practice Rotations
Progress
Estimated6 minsto complete
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Rotations
What if you were given the coordinates of a quadrilateral and you were asked to rotate that quadrilateral about the origin? What would its new coordinates be? After completing this Concept, you'll be able to rotate a figure like this one in the coordinate plane.
### Watch This
Transformation: Rotation CK-12
### Guidance
A transformation is an operation that moves, flips, or otherwise changes a figure to create a new figure. A rigid transformation (also known as an isometry or congruence transformation) is a transformation that does not change the size or shape of a figure.
The rigid transformations are translations (discussed elsewhere), reflections (discussed elsewhere), and rotations (discussed here). The new figure created by a transformation is called the image. The original figure is called the preimage. If the preimage is , then the image would be , said “a prime.” If there is an image of , that would be labeled , said “a double prime.”
A rotation is a transformation where a figure is turned around a fixed point to create an image. The lines drawn from the preimage to the center of rotation and from the center of rotation to the image form the angle of rotation. In this concept, we will only do counterclockwise rotations.
While we can rotate any image any amount of degrees, and rotations are common and have rules worth memorizing.
Rotation of :
Rotation of :
Rotation of :
#### Example A
A rotation of clockwise is the same as what counterclockwise rotation?
There are around a point. So, an rotation clockwise is the same as a rotation counterclockwise.
#### Example B
A rotation of counterclockwise is the same as what clockwise rotation?
clockwise rotation.
#### Example C
Rotate , with vertices , and , about the origin. Find the coordinates of .
Use the rule above to find .
Transformation: Rotation CK-12
-->
### Guided Practice
1. Rotate .
2. Find the coordinates of after a rotation.
3. The rotation of a quadrilateral is shown below. What is the measure of and ?
1.
2. Using the rule, we have:
3. Because a rotation produces congruent figures, we can set up two equations to solve for and .
### Explore More
In the questions below, every rotation is counterclockwise, unless otherwise stated.
1. If you rotated the letter counterclockwise, what letter would you have?
2. If you rotated the letter clockwise, what letter would you have?
3. A clockwise rotation is the same as what counterclockwise rotation?
4. A clockwise rotation is the same as what counterclockwise rotation?
5. A counterclockwise rotation is the same as what clockwise rotation?
6. A counterclockwise rotation is the same as what clockwise rotation?
7. A counterclockwise rotation is the same as what clockwise rotation?
8. Rotating a figure is the same as what other rotation?
9. Does it matter if you rotate a figure clockwise or counterclockwise? Why or why not?
10. When drawing a rotated figure and using your protractor, would it be easier to rotate the figure counterclockwise or clockwise? Explain your reasoning.
Rotate each figure in the coordinate plane the given angle measure. The center of rotation is the origin.
Find the measure of in the rotations below. The blue figure is the preimage.
Find the angle of rotation for the graphs below. The center of rotation is the origin and the blue figure is the preimage. Your answer will be , or .
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 12.4.
### Vocabulary Language: English
Center of Rotation
Center of Rotation
In a rotation, the center of rotation is the point that does not move. The rest of the plane rotates around this fixed point.
Image
Image
The image is the final appearance of a figure after a transformation operation.
Origin
Origin
The origin is the point of intersection of the $x$ and $y$ axes on the Cartesian plane. The coordinates of the origin are (0, 0).
Preimage
Preimage
The pre-image is the original appearance of a figure in a transformation operation.
Rotation
Rotation
A rotation is a transformation that turns a figure on the coordinate plane a certain number of degrees about a given point without changing the shape or size of the figure.
Rigid Transformation
Rigid Transformation
A rigid transformation is a transformation that preserves distance and angles, it does not change the size or shape of the figure. | 1,040 | 4,769 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 2, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2016-18 | latest | en | 0.893767 |
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