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http://stackoverflow.com/questions/3678970/what-is-the-most-succinct-scala-way-to-reverse-a-map | 1,394,954,232,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678702045/warc/CC-MAIN-20140313024502-00055-ip-10-183-142-35.ec2.internal.warc.gz | 134,423,958 | 16,095 | What is the most succinct Scala way to reverse a Map?
What is the most succinct Scala way to reverse a Map? The Map may contain non-unique values.
EDIT:
The reversal of `Map[A, B]` should give `Map[B, Set[A]]` (or a MultiMap, that would be even better).
-
Define what is to happen if a given value is present under more than one key. – Randall Schulz Sep 9 '10 at 17:28
@Randall: See the edit. – missingfaktor Sep 9 '10 at 17:54
If you can lose duplicate keys:
``````scala> val map = Map(1->"one", 2->"two", -2->"two")
map: scala.collection.immutable.Map[Int,java.lang.String] = Map((1,one), (2,two), (-2,two))
scala> map.map(_ swap)
res0: scala.collection.immutable.Map[java.lang.String,Int] = Map((one,1), (two,-2))
``````
If you don't want access as a multimap, just a map to sets, then:
``````scala> map.groupBy(_._2).mapValues(_.keys.toSet)
res1: scala.collection.immutable.Map[
java.lang.String,scala.collection.immutable.Set[Int]
] = Map((one,Set(1)), (two,Set(2, -2)))
``````
If you insist on getting a `MultiMap`, then:
``````scala> import scala.collection.mutable.{HashMap, Set, MultiMap}
scala> ( (new HashMap[String,Set[Int]] with MultiMap[String,Int]) ++=
| map.groupBy(_._2).mapValues(Set[Int]() ++= _.keys) )
res2: scala.collection.mutable.HashMap[String,scala.collection.mutable.Set[Int]]
with scala.collection.mutable.MultiMap[String,Int] = Map((one,Set(1)), (two,Set(-2, 2)))
``````
-
+1 for succinctness, and for introducing me to mapValues :-) – Rodney Gitzel Sep 9 '10 at 18:25
a better starting would be clearer though, e.g. Map(1->"one", 2->"two",3->"two",4->"two") – Rodney Gitzel Sep 9 '10 at 18:26
@Rodney - Okay, okay, I'll show the overlap case! – Rex Kerr Sep 9 '10 at 18:32
Thanks a lot! :-) – missingfaktor Sep 9 '10 at 18:46
``````scala> val m1 = Map(1 -> "one", 2 -> "two", 3 -> "three", 4 -> "four")
m1: scala.collection.immutable.Map[Int,java.lang.String] = Map((1,one), (2,two), (3,three), (4,four))
scala> m1.map(pair => pair._2 -> pair._1)
res0: scala.collection.immutable.Map[java.lang.String,Int] = Map((one,1), (two,2), (three,3), (four,4))
``````
Edit for clarified question:
``````object RevMap {
def
main(args: Array[String]): Unit = {
val m1 = Map("one" -> 3, "two" -> 3, "three" -> 5, "four" -> 4, "five" -> 5, "six" -> 3)
val rm1 = (Map[Int, Set[String]]() /: m1) { (map: Map[Int, Set[String]], pair: (String, Int)) =>
map + ((pair._2, map.getOrElse(pair._2, Set[String]()) + pair._1)) }
printf("m1=%s%nrm1=%s%n", m1, rm1)
}
}
% scala RevMap
m1=Map(four -> 4, three -> 5, two -> 3, six -> 3, five -> 4, one -> 3)
rm1=Map(4 -> Set(four, five), 5 -> Set(three), 3 -> Set(two, six, one))
``````
I'm not sure this qualifies as succinct.
-
+1 for the working solution. :-) – missingfaktor Sep 9 '10 at 18:46 | 945 | 2,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | latest | en | 0.579373 |
https://questions.llc/questions/1647040 | 1,701,462,207,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100304.52/warc/CC-MAIN-20231201183432-20231201213432-00061.warc.gz | 541,929,654 | 6,681 | # What percentage of 50 is 12.5
A. 4%
B. 25%
C. 40%
D. 15%
Question ID
1647040
4
Created
Rating
1
Views
629
2289829
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## To find the percentage of one number in relation to another number, divide the first number by the second number and then multiply the result by 100.
In this case, to find what percentage of 50 is 12.5, divide 12.5 by 50 and then multiply the result by 100:
(12.5 ÷ 50) × 100 = 0.25 × 100 = 25
So, the answer is B. 25%.
3397711
Created
Rating
0
## To calculate what percentage of 50 is 12.5, you need to divide 12.5 by 50 and then multiply the result by 100.
Let's do the calculation:
12.5 ÷ 50 = 0.25
0.25 × 100 = 25
So, 12.5 is 25% of 50.
Therefore, the answer is B. 25%.
4147090
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## 50x = 12.5
x = ?
Multiply the decimal answer by 100 to find the percent. | 293 | 838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-50 | latest | en | 0.902719 |
https://openstax.org/books/college-physics-2e/pages/33-2-the-four-basic-forces | 1,723,733,517,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00261.warc.gz | 336,203,070 | 99,495 | College Physics 2e
# 33.2The Four Basic Forces
College Physics 2e33.2 The Four Basic Forces
## Learning Objectives
By the end of this section, you will be able to:
• State the four basic forces.
• Explain the Feynman diagram for the exchange of a virtual photon between two positive charges.
• Define QED.
• Describe the Feynman diagram for the exchange of a between a proton and a neutron.
As first discussed in Problem-Solving Strategies and mentioned at various points in the text since then, there are only four distinct basic forces in all of nature. This is a remarkably small number considering the myriad phenomena they explain. Particle physics is intimately tied to these four forces. Certain fundamental particles, called carrier particles, carry these forces, and all particles can be classified according to which of the four forces they feel. The table given below summarizes important characteristics of the four basic forces.
Force Approximate relative strength Range +/−1 Carrier particle
Gravity $10 − 38 10 − 38$ $∞ ∞$ + only Graviton (conjectured)
Electromagnetic $10 − 2 10 − 2$ $∞ ∞$ $+ / − + / −$ Photon (observed)
Weak force $10 − 13 10 − 13$ $+ / − + / −$ $W+,W−,Z0W+,W−,Z0$ (observed2)
Strong force $11$ $+ / − + / −$ Gluons (conjectured3)
Table 33.1 Properties of the Four Basic Forces
Figure 33.4 The first image shows the exchange of a virtual photon transmitting the electromagnetic force between charges, just as virtual pion exchange carries the strong nuclear force between nucleons. The second image shows that the photon cannot be directly observed in its passage, because this would disrupt it and alter the force. In this case it does not get to the other charge.
Figure 33.5 The Feynman diagram for the exchange of a virtual photon between two positive charges illustrates how the electromagnetic force is transmitted on a quantum mechanical scale. Time is graphed vertically while the distance is graphed horizontally. The two positive charges are seen to be repelled by the photon exchange.
Although these four forces are distinct and differ greatly from one another under all but the most extreme circumstances, we can see similarities among them. (In GUTs: the Unification of Forces, we will discuss how the four forces may be different manifestations of a single unified force.) Perhaps the most important characteristic among the forces is that they are all transmitted by the exchange of a carrier particle, exactly like what Yukawa had in mind for the strong nuclear force. Each carrier particle is a virtual particle—it cannot be directly observed while transmitting the force. Figure 33.4 shows the exchange of a virtual photon between two positive charges. The photon cannot be directly observed in its passage, because this would disrupt it and alter the force.
Figure 33.5 shows a way of graphing the exchange of a virtual photon between two positive charges. This graph of time versus position is called a Feynman diagram, after the brilliant American physicist Richard Feynman (1918–1988) who developed it.
Figure 33.6 is a Feynman diagram for the exchange of a virtual pion between a proton and a neutron representing the same interaction as in Figure 33.3. Feynman diagrams are not only a useful tool for visualizing interactions at the quantum mechanical level, they are also used to calculate details of interactions, such as their strengths and probability of occurring. Feynman was one of the theorists who developed the field of quantum electrodynamics (QED), which is the quantum mechanics of electromagnetism. QED has been spectacularly successful in describing electromagnetic interactions on the submicroscopic scale. Feynman was an inspiring teacher, had a colorful personality, and made a profound impact on generations of physicists. He shared the 1965 Nobel Prize with Julian Schwinger and S. I. Tomonaga for work in QED with its deep implications for particle physics.
Why is it that particles called gluons are listed as the carrier particles for the strong nuclear force when, in The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited, we saw that pions apparently carry that force? The answer is that pions are exchanged but they have a substructure and, as we explore it, we find that the strong force is actually related to the indirectly observed but more fundamental gluons. In fact, all the carrier particles are thought to be fundamental in the sense that they have no substructure. Another similarity among carrier particles is that they are all bosons (first mentioned in Patterns in Spectra Reveal More Quantization), having integral intrinsic spins.
Figure 33.6 The image shows a Feynman diagram for the exchange of a $π+π+$ between a proton and a neutron, carrying the strong nuclear force between them. This diagram represents the situation shown more pictorially in Figure 33.4.
There is a relationship between the mass of the carrier particle and the range of the force. The photon is massless and has energy. So, the existence of (virtual) photons is possible only by virtue of the Heisenberg uncertainty principle and can travel an unlimited distance. Thus, the range of the electromagnetic force is infinite. This is also true for gravity. It is infinite in range because its carrier particle, the graviton, has zero rest mass. (Gravity is the most difficult of the four forces to understand on a quantum scale because it affects the space and time in which the others act. But gravity is so weak that its effects are extremely difficult to observe quantum mechanically. We shall explore it further in General Relativity and Quantum Gravity). The $W+,W−W+,W−$, and $Z0Z0$ particles that carry the weak nuclear force have mass, accounting for the very short range of this force. In fact, the $W+,W−W+,W−$, and $Z0Z0$ are about 1000 times more massive than pions, consistent with the fact that the range of the weak nuclear force is about 1/1000 that of the strong nuclear force. Gluons are actually massless, but since they act inside massive carrier particles like pions, the strong nuclear force is also short ranged.
The relative strengths of the forces given in the Table 33.1 are those for the most common situations. When particles are brought very close together, the relative strengths change, and they may become identical at extremely close range. As we shall see in GUTs: the Unification of Forces, carrier particles may be altered by the energy required to bring particles very close together—in such a manner that they become identical.
### Footnotes
• 1 + attractive; ‑ repulsive; $+/−+/−$ both.
• 2Predicted by theory and first observed in 1983.
• 3Eight proposed—indirect evidence of existence. Underlie meson exchange. | 1,473 | 6,755 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-33 | latest | en | 0.865465 |
http://blogs.msdn.com/b/tilovell/archive/2014/08/11/more-tokenizing.aspx | 1,411,365,708,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657136545.14/warc/CC-MAIN-20140914011216-00058-ip-10-234-18-248.ec2.internal.warc.gz | 35,208,522 | 15,423 | So I flippantly said 'write a helper function that captures the right pattern for tokenizing' last post... But when you sit down to think about it, a helper function feels like the opposite of what you logically do when you are implementing a finite state machine... because there is no way to have helper functions be case statements in a switch block!
i.e. you can do this:
if (t = MatchStringLiteral()) return t;
if (t =TryMatchSymbol('+=')) return t;
if (t = TryMatchSymbol('+')) return t;
And that almost looks clean. (But not really clean - it seems like the whole if (t = blah return t pattern) cannot be abstracted out further, in C# anyway) but then if you ever programmed C your next thought is probably 'Would it perform well? I thought tokenizers are supposed to be implemented by lots of switch statements because those are fast?'
Well, you can try to mix it up:
switch(peeknextchar())
{
case '+':
if (t = TryMatchSymbol('+=')) return t;
if (t = TryMatchSymbol('+')) return t;
case 'a..z': if (t == TryMatchIdent()) return t; //actually I don't think you can do case 'a..z' in C# unfortunately
....
- but really this is just getting less and less elegant. And breaking the golden rule of optimization which is measure first. :p
Perhaps there should be a golden rule of abstraction too. Anyone know what it is? I don't think I've heard one yet. The layer of indirection saying comes to mind, but it's hardly prescriptive of an ideal.
If I were to invent one, perhaps it would be 'abstractions should be solutions to an expressiveness problem in your code'
So anyway...
How can we write expressive C# code for tokenizing things? Not worrying about whether it is optimizable...
Well. Regex are an obvious way of specifying patterns of things to accept without writing loops etc... but they work less well for symbols where you have to know all the escape characters...
You might even be able to optimize... see? I can't stop worrying about optimizing. :)
So anyway... these tokens I normally think of as outputs to the tokenizing problem. But maybe the token itself also describes how to solve the problem?
Imagine the increment by token '+='. You can have an instance of the token at some points in the document. They could even all be the same singleton instance if you don't need to know their position in the document. The singleton could also know how to match itself to the next part of the untokenized input stream and say whether it is the next token or not.
There's a whole bunch of conceptual overloading going on here, saying one object knows how to perform many tasks. I feel like this overloading is sort of a natural thing to do in javascript, because it's prototype based, and you can come up with ad-hoc solutions if you suddenly realize a need for separate token instances with different text per instance for string literal tokens, or you realize you need to record position on some tokens. You just start doing it, and de-singletonize your object graph as necessary to express the required variation.
You don't have to think up types and interfaces that plan in advance the separation of roles and responsibilities... that's kinda nice.
In C# OTOH I am like 'do I need to have both a Token type and a TokenGenerator type? And some subclass that implements both might be good, for the singleton-happy case?'
I.e.:
` interface IToken { string Text { get; set; } } interface ITokenGenerator { Func<TokenizingState, IToken> Match { get; set; } } class Stok: IToken, ITokenGenerator // 'SymbolToken' or 'SingletonToken' - covers all simple symbolic tokens: + - * { } ( ) . , ; : // also +=, -=, *=, /=, => { public Stok(string tokText) { this.Text = tokText; this.Match = (s) => s.StartsWith(this.Text) ? this : null; } public string Text { get; set; } public Func<TokenizingState, IToken> Match { get; set; } }`
`Now I can model my tokenizer as just a collection of token generators. Hopefully... [spot any obvious problems yet? :) ]` | 939 | 4,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | latest | en | 0.918345 |
https://stanford.library.sydney.edu.au/archives/win2006/entries/logic-intuitionistic/ | 1,638,449,842,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362219.5/warc/CC-MAIN-20211202114856-20211202144856-00424.warc.gz | 607,555,996 | 20,002 | This is a file in the archives of the Stanford Encyclopedia of Philosophy.
# Intuitionistic Logic
First published Wed Sep 1, 1999; substantive revision Tue Jan 13, 2004
Intuitionistic logic encompasses the principles of logical reasoning which were used by L. E. J. Brouwer in developing his intuitionistic mathematics, beginning in [1907]. Because these principles also underly Russian recursive analysis and the constructive analysis of E. Bishop and his followers, intuitionistic logic may be considered the logical basis of constructive mathematics.
Philosophically, intuitionism differs from logicism by treating logic as a part of mathematics rather than as the foundation of mathematics; from finitism by allowing (constructive) reasoning about infinite collections; and from platonism by viewing mathematical objects as mental constructs with no independent ideal existence. Hilbert's formalist program, to justify classical mathematics by reducing it to a formal system whose consistency should be established by finitistic (hence constructive) means, was the most powerful contemporary rival to Brouwer's developing intuitionism. In his 1912 essay Intuitionism and Formalism Brouwer correctly predicted that any attempt to prove the consistency of complete induction on the natural numbers would lead to a vicious circle.
Brouwer rejected formalism per se but admitted the potential usefulness of formulating general logical principles expressing intuitionistically correct constructions, such as modus ponens. Formal systems for intuitionistic propositional and predicate logic and arithmetic were developed by Heyting [1930], Gentzen [1935] and Kleene [1952]. Gödel [1933] proved the equiconsistency of intuitionistic and classical theories. Kripke [1965] provided a semantics with respect to which intuitionistic logic is correct and complete.
## 1. Rejection of Tertium Non Datur
Intuitionistic logic can be succinctly described as classical logic without the Aristotelian law of excluded middle (LEM): (A ¬A), but with the law of contradiction (¬A → (AB)). Brouwer [1908] observed that LEM was abstracted from finite situations, then extended without justification to statements about infinite collections. For example, if x, y range over the natural numbers 0, 1, 2, … and B(x) abbreviates the property (there is a y > x such that both y and y+2 are prime numbers), then we have no general method for deciding whether B(x) is true or false for arbitrary x, so ∀x(B(x) ¬B(x)) cannot be asserted in the present state of our knowledge. And if A abbreviates the statement ∀xB(x), then (A ¬A) cannot be asserted because neither A nor (¬A) has yet been proved.
One may object that these examples depend on the fact that the Twin Primes Conjecture has not yet been settled. A number of Brouwer's original "counterexamples" depended on problems (such as Fermat's Last Theorem) which have since been solved. But to Brouwer the general LEM was equivalent to the a priori assumption that every mathematical problem has a solution — an assumption he rejected, anticipating Gödel's incompleteness theorem by a quarter of a century.
The rejection of LEM has far-reaching consequences. On the one hand,
• Intuitionistically, Reductio ad absurdum only proves negative statements, since ¬¬AA does not hold in general. (If it did, LEM would follow by modus ponens from the intuitionistically provable ¬¬ (A ¬A).)
• Not every propositional formula has an intuitionistically equivalent disjunctive or conjunctive normal form.
• Not every predicate formula has an intuitionistically equivalent prenex form.
• Pure intuitionistic logic is axiomatically incomplete. Infinitely many intermediate axiomatic extensions of intuitionistic propositional and predicate logic are contained in classical logic.
On the other hand,
• Every intuitionistic proof of a closed statement of the form A B can be effectively transformed into an intuitionistic proof of A or an intuitionistic proof of B, and similarly for closed existential statements.
• Classical logic is finitistically interpretable in the negative fragment of intuitionistic logic.
• Intuitionistic arithmetic can consistently be extended by axioms (such as Church's Thesis) which contradict classical arithmetic, enabling the formal study of recursive mathematics.
• Brouwer's controversial intuitionistic analysis, which conflicts with LEM, can be formalized and shown consistent relative to a classically and intuitionistically correct subtheory.
## 2. Intuitionistic First-Order Predicate Logic
Formalized intuitionistic logic is naturally motivated by the informal Brouwer-Heyting-Kolmogorov explication of intuitionistic truth, outlined in Section 2 of the article on Constructive Mathematics in this Encyclopedia. The constructive independence of the logical operations &, , →, ¬, ∀, ∃ contrasts with the classical situation, where e.g., (A B) is equivalent to ¬(¬A & ¬B), and ∃xA(x) is equivalent to ¬∀x¬A(x). From the B-H-K viewpoint, a sentence of the form (A B) asserts that either a proof of A, or a proof of B, has been constructed; while ¬(¬A & ¬B) asserts that an algorithm has been constructed which would effectively convert any pair of constructions proving ¬A and ¬B respectively, into a proof of a known contradiction.
Following is a Hilbert-style formalism H, from Kleene [1952], for intuitionistic first-order predicate logic IQC. The language L has predicate letters P, Q(.),… of all arities and individual variables a,b,c,… (with or without subscripts 1,2,…), as well as symbols &, , →, ¬, ∀, ∃ for the logical connectives and quantifiers, and parentheses (, ). The prime formulas of L are expressions such as P, Q(a), R(a,b,a) where P, Q(.), R(…) are 0-ary, 1-ary and 3-ary predicate letters respectively; that is, the result of filling each blank in a predicate letter by an individual variable symbol is a prime formula. The (well-formed) formulas of L are defined inductively as follows.
• Each prime formula is a formula.
• If A and B are formulas, so are (A & B), (A B), (AB) and ¬A.
• If A is a formula and x is a variable, then ∀xA and ∃xA are formulas.
In general, we use A, B, C as metavariables for well-formed formulas and x,y,z as metavariables for individual variables. Anticipating applications (for example to intuitionistic arithmetic) we use s,t as metavariables for terms; in the case of pure predicate logic, terms are simply individual variables. An occurrence of a variable x in a formula A is bound if it is within the scope of a quantifier ∀x or ∃x, otherwise free. Intuitionistically as classically, "(AB)" abbreviates "(AB) & (BA)," and parentheses are omitted when this causes no confusion.
There are three rules of inference:
• Modus Ponens: From A and (AB), conclude B.
• ∀-Introduction: From (CA(x)), where x is a variable which does not occur free in C, conclude (C → ∀xA(x)).
• ∃-Elimination: From (A(x) → C), where x is a variable which does not occur free in C, conclude (∃xA(x) → C).
The axioms are all formulas of the following forms, where in the last two schemas the subformula A(t) is the result of substituting an occurrence of the term t for every free occurrence of x in A(x), and no variable free in t becomes bound in A(t) as a result of the substitution.
• A → (BA).
• (AB) → ((A → (BC)) → (AC)).
• A → (BA & B).
• A & BA.
• A & BB.
• AA B.
• BA B.
• (AC) → ((BC) → (A BC)).
• (AB) → ((A → ¬B) → ¬A).
• ¬A → (AB).
• xA(x) → A(t).
• A(t) → ∃xA(x).
A proof is any finite sequence of formulas, each of which is an axiom or an immediate consequence, by a rule of inference, of (one or two) preceding formulas of the sequence. Any proof is said to prove its last formula, which is called a theorem or provable formula of first-order intuitionistic predicate logic. A derivation of a formula E from a collection F of assumptions is any sequence of formulas, each of which belongs to F or is an axiom or an immediate consequence, by a rule of inference, of preceding formulas of the sequence, such that E is the last formula of the sequence. If such a derivation exists, we say E is derivable from F.
Intuitionistic propositional logic IPC is the subtheory which results when the language is restricted to formulas built from proposition letters P, Q, R,… using the propositional connectives &, , → and ¬, and only the propositional postulates are used. Thus the last two rules of inference and the last two axiom schemas are absent from the propositional theory.
If, in the given list of axiom schemas for intuitionistic propositional or first-order predicate logic, the law of contradiction:
• ¬A → (AB).
is replaced by the law of double negation:
• ¬¬AA.
(or,equivalently, by LEM), a formal system for classical propositional or first-order predicate logic results. Since the law of contradiction is a classical theorem, intuitionistic logic is contained in classical logic. In a sense, classical logic is also contained in intuitionistic logic; see Section 4.1 below.
The Hilbert-style system H is useful for metamathematical investigations of intuitionistic logic, but its forced linearization of deductions and its preference for axioms over rules make it an awkward instrument for establishing derivability. A natural deduction system I for intuitionistic predicate logic results from the deductive system D, presented in Section 3 of the entry on Classical Logic in this Encyclopedia, by omitting the symbol and rules for identity, and replacing the classical rule (DNE) of double negation elimination by the intuitionistic negation elimination rule
• (INE) If F entails A and F entails ¬A, then F entails B.
While identity can of course be added to intuitionistic logic, for applications (e.g., to arithmetic) the equality symbol is generally treated as a distinguished predicate constant satisfying nonlogical axioms (e.g., the primitive recursive definitions of addition and multiplication) in addition to reflexivity, symmetry and transitivity. Identity is decidable, intuitionistically as well as classically, but intuitionistic extensional equality is not always decidable; see the discussion of Brouwer's continuum in the article on Constructive Mathematics in this Encyclopedia. The keys to proving that H is equivalent to I are modus ponens and its converse, the
Deduction Theorem: If B is derivable from A and possibly other formulas F, with all variables free in A held constant in the derivation (that is, without using the second or third rule of inference on any variable x occurring free in A, unless the assumption A does not occur in the derivation before the inference in question), then (AB) is derivable from F.
This fundamental result, roughly expressing the rule (→I) of I, can be proved for H by induction on the definition of a derivation. The other rules of I hold for H essentially by modus ponens, which corresponds to (→E) in I. To illustrate the usefulness of the Deduction Theorem, consider the (apparently trivial) theorem schema (AA) of IPC. A correct proof in H takes five lines:
• 1. A → (AA)
• 2. (A → (AA)) → ((A → ((AA) → A)) → (AA))
• 3. (A → ((AA) → A)) → (AA)
• 4. A → ((AA) → A)
• 5. AA
where 1, 2 and 4 are axioms and 3, 5 come from earlier lines by modus ponens. However, A is derivable from A (as assumption) in one obvious step, so the Deduction Theorem allows us to conclude that a proof of (AA) exists. (In fact, the formal proof of (AA) just presented is part of the constructive proof of the Deduction Theorem!)
It is important to note that, in the definition of a derivation from assumptions in H, the assumption formulas are treated as if all their free variables were universally quantified, so that ∀x A(x) is derivable from the hypothesis A(x). However, the variable x will be varied (not held constant) in that derivation, by use of the rule of ∀-introduction; and so the Deduction Theorem cannot be used to conclude (falsely) that A(x) → ∀x A(x) (and hence, by ∃-elimination, ∃x A(x) → ∀x A(x)) are provable in H. As an example of a correct use of the Deduction Theorem for predicate logic, consider the implication ∃x A(x) → ¬∀x¬A(x). To show this is provable in IQC, we first derive ¬∀x¬A(x) from A(x) with all free variables held constant:
• 1. ∀x¬A(x) → ¬A(x)
• 2. A(x) → (∀x¬A(x) → A(x))
• 3. A(x) (assumption)
• 4. ∀x¬A(x) → A(x)
• 5. (∀x¬A(x) → A(x)) → ((∀x¬A(x) → ¬A(x)) → ¬∀x¬A(x))
• 6. (∀x¬A(x) → ¬A(x)) → ¬∀x¬A(x)
• 7. ¬∀x¬A(x))
Here 1, 2 and 5 are axioms; 4 comes from 2 and 3 by modus ponens; and 6 and 7 come from earlier lines by modus ponens; so no variables have been varied. The Deduction Theorem tells us there is a proof P in IQC of (A(x) → ¬∀x¬A(x)), and one application of ∃-introduction converts P into a proof of ∃x A(x) → ¬∀x¬A(x). The converse is not provable in IQC, as shown in Section 5.1 below.
## 3. Intuitionistic Number Theory (Heyting Arithmetic)
Intuitionistic (Heyting) arithmetic HA and classical (Peano) arithmetic PA share the same first-order language and the same non-logical axioms; only the logic is different. In addition to the logical connectives, quantifiers and parentheses and the individual variables a, b, c, … (with metavariables x, y, z as usual), the language L(HA) of arithmetic has a binary predicate symbol =, individual constant 0, unary function constant S, and finitely or countably infinitely many additional constants for primitive recursive functions including addition and multiplication; the precise choice is a matter of taste and convenience. Terms are built from variables and 0 using the function constants; in particular, each natural number n is expressed in the language by the numeral n obtained by applying S n times to 0 (e.g., S(S(0)) is the numeral for 2). Prime formulas are of the form (s = t) where s,t are terms, and compound formulas are obtained from these as usual.
The logical axioms and rules of HA are those of first-order intuitionistic predicate logic IQC. The nonlogical axioms include the reflexive, symmetric and transitive properties of =, primitive recursive defining equations for each function constant, the axioms characterizing 0 as the least natural number and S as a one-to-one function:
• x¬(S(x) = 0),
• xy(S(x) = S(y) → x = y),
the extensional equality axiom for S:
• xy (x = yS(x) = S(y)),
and the (universal closure of the) schema of mathematical induction, for arbitrary formulas A(x):
• A(0) & ∀x(A(x) → A(S(x))) → ∀x A(x).
Extensional equality axioms for all the other function constants are derivable by mathematical induction from the equality axiom for S and the primitive recursive function axioms.
For theories T based on intuitionistic logic, if E is an arbitrary formula of L(T) then by definition
• E is decidable in T if and only if T proves (E ¬E).
• E is stable in T if and only if T proves (¬¬EE).
Decidability implies stability, but not conversely; in IPC, the negation of any prime formula is stable but not decidable. However, using mathematical induction in HA, one can prove ∀xy(x = y ¬(x = y)), and so
• Prime formulas (and hence quantifier-free formulas) are decidable and stable in HA.
Peano arithmetic PA comes from Heyting arithmetic HA by adding LEM or (¬¬AA) to the list of logical axioms, i.e., by using classical instead of intuitionistic logic. The two theories are proof-theoretically equivalent, as will be shown in the next section. Each is capable of (numeralwise) expressing its own proof predicate. By Gödel's famous Incompleteness Theorem, if HA is consistent then neither HA nor PA can prove its own consistency.
## 4. Basic Proof Theory
### 4.1 Translating classical into intuitionistic logic
A fundamental fact about intuitionistic logic is that it has the same consistency strength as classical logic. For propositional logic this was first proved by Glivenko [1929].
Glivenko's Theorem: An arbitrary propositional formula A is classically provable, if and only if ¬¬A is intuitionistically provable.
Glivenko's Theorem does not extend to predicate logic, although an arbitrary predicate formula A is classically provable if and only if ¬¬A is provable in intuitionistic predicate logic plus the "double negation shift" schema
(DNS)x¬¬B(x) → ¬¬∀x B(x).
The more sophisticated negative translation of classical into intuitionistic theories, due independently to Gödel and Gentzen, associates with each formula A of the language L another formula g(A) (with no or ∃), such that
• (I) Classical predicate logic proves Ag(A).
• (II) Intuitionistic predicate logic proves g(A) ↔ ¬¬g(A).
• (III) If classical predicate logic proves A, then intuitionistic predicate logic proves g(A).
The proofs are straightforward from the following inductive definition of g(A):
• g(P) is ¬¬P, if P is prime.
• g(A & B) is (g(A) & g(B)).
• g(A B) is ¬(¬g(A) & ¬g(B)).
• g(AB) is (g(A) → g(B)).
• gA) is ¬g(A).
• g(∀xA(x)) is ∀x g(A(x)).
• g(∃xA(x)) is ¬∀x¬g(A(x)).
For each formlula A, g(A) is provable intuitionistically if and only if A is provable classically. In particular, if (B & ¬B) were classically provable for some formula B, then (g(B) & ¬g(B)) (which is g(B & ¬B)) would in turn be provable intuitionistically. Hence
• (IV) Classical and intuitionistic predicate logic are equiconsistent.
The negative translation of classical into intuitionistic number theory is even simpler, since prime formulas of intuitionistic arithmetic are stable. Thus g(s=t) can be take to be (s=t), and the other clauses are unchanged. The negative translation of any instance of mathematical induction is another instance of mathematical induction, and the other nonlogical axioms of arithmetic are their own negative translations, so
• (I), (II), (III) and (IV) hold also for number theory.
Gödel [1933] interpreted these results as showing that intuitionistic logic and arithmetic are richer than classical logic and arithmetic, because the intuitionistic theory distinguishes formulas which are classically equivalent, and has the same consistency strength as the classical theory.
Direct attempts to extend the negative interpretation to analysis fail because the negative translation of the countable axiom of choice is not a theorem of intuitionistic analysis. However, it is consistent with intuitionistic analysis, including Brouwer's controversial continuity principle, by the functional version of Kleene's recursive realizability (Section 5.2 below).
### 4.2 Admissible rules of intuitionistic logic
Gödel [1932] observed that intuitionistic propositional logic has the disjunction property:
(DP) If (A B) is a theorem, then A is a theorem or B is a theorem.
Gentzen [1935] established the disjunction property for closed formulas of intuitionistic predicate logic. From this it follows that if intuitionistic logic is consistent, then (P ¬P) is not a theorem if P is prime. Kleene [1945, 1952] proved that intuitionistic first-order number theory also has the related (cf. Friedman [1975]) existence property:
(ED) If ∃xA(x) is a closed theorem, then for some closed term t, A(t) is a theorem.
The disjunction and existence properties are special cases of a general phenomenon peculiar to nonclassical theories. The admissible rules of a theory are the rules under which the theory is closed. In classical propositional logic, if (A is provable) implies (B is provable), then (AB) is provable; thus every classically admissible propositional rule is classically derivable. Intuitionistically the situation is more interesting.
Harrop [1960] observed that intuitionistic propositional logic IPC is closed under the rule
A → (B C)) / (¬AB) AC),
while the corresponding implication is not provable intuitionistically. Rybakov showed that the collection IPR of admissible rules of IPC is decidable, but does not have a finite basis (a subcollection from which every admissible rule is derivable). Building on work of Ghilardi, Iemhoff [2001] proved a conjecture of de Jongh and Visser which gives a recursively enumerable basis for IPR. Visser [2002] showed that these are also the admissible propositional rules of HA, and of HA extended by Markov's Principle MP (defined in Section 5.1 below).
While the admissible rules of intuitionistic predicate logic have not been completely characterized, they are known to include Markov's Rule for decidable predicates:
x(A(x) ¬A(x)) & ¬∀x¬A(x) / ∃x A(x)
and the following Independence-of-Premise Rule (where y is assumed not to occur free in A(x)):
x(A(x) ¬A(x)) & (∀x A(x) → ∃y B(y)) / ∃y (∀x A(x) → B(y)).
The corresponding implications (MP and IP respectively), which are not provable intuitionistically, are verified by Gödel's "Dialectica" interpretation of HA (cf. Section 6 below). So is the implication (CT) corresponding to one of the most interesting admissible rules of Heyting arithmetic, let us call it the Church-Kleene Rule:
If ∀xy A(x,y) is a closed theorem of HA then there is a number n such that, provably in HA, the partial recursive function with Gödel number n is total and maps each x to a y satisfying A(x,y).
Combining Markov's Rule with the negative translation gives the result that classical and intuitionistic arithmetic prove the same formulas of the form ∀xy A(x,y) where A(x,y) is quantifier-free. In general, if A(x,y) is provably decidable in HA and if ∀xy A(x,y) is a closed theorem of classical arithmetic, the conclusion of the Church-Kleene Rule holds even in intuitionistic arithmetic.
Here is a proof that the rule ∀x (A B(x)) / A x B(x) (where x is not free in A) is not admissible for HA, if HA is consistent. Gödel numbering provides a quantifier-free formula G(x) which (numeralwise) expresses the predicate "x is the code of a proof in HA of (0 = 1)." By intuitionistic logic with the decidability of quantifier-free arithmetical formulas, HA proves ∀x(∃yG(y) ¬G(x)). However, if HA proved ∃yG(y) x¬G(x) then by the disjunction property, HA must prove either ∃yG(y) or ∀x¬G(x). The first case is impossible, by the existence property with the consistency assumption and the fact that HA proves all true quantifier-free sentences. But the second case is also impossible, by Gödel's second incompleteness theorem, since ∀x¬G(x) expresses the consistency of HA.
## 5. Basic Semantics
### 5.1 Kripke semantics for intuitionistic logic
Intuitionistic systems have inspired a variety of interpretations, including Beth's tableaus, Rasiowa and Sikorski's topological models, formulas-as-types, Kleene's recursive realizabilities, and the Kleene and Aczel slashes. Kripke's [1965] possible-world semantics, with respect to which intuitionistic predicate logic is complete and consistent, most resembles classical model theory.
A Kripke structure K for L consists of a partially ordered set K of nodes and a domain function D assigning to each node k in K an inhabited set D(k), such that if kk′, then D(k) ⊆ D(k′). In addition K has a forcing relation determined as follows.
For each node k let L(k) be the language extending L by new constants for all the elements of D(k). To each node k and each 0-ary predicate letter (each proposition letter) P, either assign f(P,k) = true or leave f(P,k) undefined, consistent with the requirement that if kk′ and f(P,k) = true then f(P,k′) = true also. Say that
k forces P if and only if f(P,k) = true.
To each node k and each (n+1)-ary predicate letter Q(…), assign a (possibly empty) set T(Q,k) of (n+1)-tuples of elements of D(k) in such a way that if kk′ then T(Q,k) ⊆ T(Q,k′). Say that
k forces Q(d0,…,dn) if and only if (do,…,dn) ∈ T(Q,k).
Now define forcing for compound sentences of L(k) inductively as follows:
• k forces (A & B) if k forces A and k forces B.
• k forces (A B) if k forces A or k forces B.
• k forces (AB) if, for every k′ ≥ k, if kforces A then kforces B.
• k forces ¬A if for no k′ ≥ k does kforce A.
• k forcesxA(x) if for every k′ ≥ k and every dD(k′), kforces A(d).
• k forcesxA(x) if for some dD(k), k forces A(d).
Any such forcing relation is consistent and monotone:
• for no sentence A and no k does k force both A and ¬A.
• if kk′ and k forces A then k′ forces A.
Kripke's Soundness and Completeness Theorems establish that a sentence of L is provable in intuitionistic predicate logic if and only if it is forced by every node of every Kripke structure. Thus to show that (¬∀x¬P(x) → ∃xP(x)) is intuitionistically unprovable, it is enough to consider a Kripke structure with K = {k, k′}, k < k′, D(k) = D(k′) = {0}, T(P,k) empty but T(P,k′) = {0}. And to show the converse is intuitionistically provable (without actually exhibiting a proof), one only needs the consistency and monotonicity properties of arbitrary Kripke models, with the definition of forcing.
Kripke models for languages with equality may interpret = at each node by an arbitrary equivalence relation, subject to monotonicity. For applications to intuitionistic arithmetic, normal models (those in which equality is interpreted by identity at each node) suffice because equality of natural numbers is decidable.
Propositional Kripke semantics is particulary simple, since an arbitrary propositional formula is intuitionistically provable if and only if it is forced by the root of every Kripke model whose frame (the set K of nodes together with their partial ordering) is a finite tree with a least element (the root). Each terminal node or leaf of a Kripke model is a classical model, because a leaf forces every formula or its negation. Only those proposition letters which occur in a formula E, and only those nodes k' such that kk', are relevant to deciding whether or not k forces E. Such considerations allow us to associate effectively with each formula E of L(IPC) a finite class of finite Kripke structures which will include a countermodel to E if one exists. Since the class of all theorems of IPC is recursively enumerable, we conclude that
• IPC is effectively decidable. There is a recursive procedure which determines, for each propositional formula E, whether or not E is a theorem of IPC, concluding with either a proof of E or a Kripke countermodel.
Familiar non-intuitionistic logical schemata correspond to structural properties of Kripke models, for example
• DNS holds in every Kripke model with finite frame.
• (AB) (BA) holds in every Kripke model with linearly ordered frame. Conversely, every propositional formula which is not derivable in IPC + (AB) (BA) has a Kripke countermodel with linearly ordered frame.
• If x is not free in A then (∀x(AB(x)) → (A x B(x))) holds in every Kripke model K with constant domain (so that D(k) = D(k′) for all k, k′ in K). The same is true for MP.
Kripke models are a powerful tool for establishing properties of intuitionistic formal systems; cf. Troelstra and van Dalen [1988], Smorynski [1973], de Jongh and Smorynski [1976] and Iemhoff [2001].
### 5.2 Realizability semantics for Heyting arithmetic
One way to implement the B-H-K explication of intuitionistic truth for arithmetic is to associate with each sentence E of HA some collection of numerical codes for algorithms which could establish the constructive truth of E. Following Kleene [1945], a number e realizes a sentence E of the language of arithmetic by induction on the logical form of E:
• e realizes (r = t), if (r = t) is true.
• e realizes (A & B), if e codes a pair (f,g) such that f realizes A and g realizes B.
• e realizes AB, if e codes a pair (f,g) such that if f = 0 then g realizes A, and if f > 0 then g realizes B.
• e realizes AB, if, whenever f realizes A, then the eth partial recursive function is defined at f and its value realizes B.
• e realizes ¬A, if no f realizes A.
• e realizesx A(x), if, for every n, the eth partial recursive function is defined at n and its value realizes A(n).
• e realizesx A(x), if e codes a pair (n,g) and g realizes A(n).
An arbitrary formula is realizable if some number realizes its universal closure. Observe that not both A and ¬A are realizable, for any formula A. The fundamental result is
Nelson's Theorem [1947]. If A is derivable in HA from realizable formulas F, then A is realizable.
Some nonintuitionistic principles can be shown to be realizable. For example, Markov's Principle (for decidable formulas) can be expressed by the schema
(MP)x(A(x) ¬A(x)) & ¬∀x¬A(x) → ∃x A(x).
MP is realizable by an argument which uses Markov's Principle informally. But realizability is a fundamentally nonclassical interpretation. In HA it is possible to express an axiom of recursive choice CT (for "Church's Thesis"), which contradicts LEM but is (constructively) realizable. Hence by Nelson's Theorem, HA + MP + CT is consistent.
Kleene used a variant of number-realizability to prove HA satisfies the Church-Kleene Rule; the same argument works for HA with MP and/or CT. In Kleene and Vesley [1965] and Kleene [1969], functions replace numbers as realizing objects, establishing the consistency of formalized intuitionistic analysis and its closure under a second-order version of the Church-Kleene Rule.
De Jongh [1970] combined realizability with Kripke modeling to show that intuitionistic predicate logic is arithmetically complete for HA. If, to each n-place predicate letter P(…), a formula f(P) of L(HA) with n free variables is assigned, and if the formula f(A) of L(HA) comes from the formula A of L by replacing each prime formula P(x1,…,xn) by f(P)(x1,…,xn), then f(A) is called an arithmetical substitution instance of A. A uniform assignment of simple existential formulas to predicate letters suffices to prove
De Jongh's Theorem. If a sentence A of the language L is not provable in IQC, then some arithmetical substitution instance of A is not provable in HA.
For example, if P(x,y) expresses "x codes a proof in HA of the formula with code y," then ∀y (∃x P(x,y) ¬∃x P(x,y)) is unrealizable, hence unprovable in HA, and so is its double negation. (The proof of de Jongh's Theorem for IPC does not need realizability; cf. Smorynski [1973]. For example, Rosser's form of Gödel's Incompleteness Theorem provides a sentence C of L(HA) such that PA proves neither C nor ¬C, so by the disjunction property HA cannot prove (C ¬C).)
Without claiming that number-realizability coincides with intuitionistic arithmetical truth, Nelson observed that for each formula A of L(HA) the predicate "y realizes A" can be expressed in HA by another formula (abbreviated "y re A"), and the schema A ↔ ∃y (y re A) is consistent with HA. Troelstra [1973] showed that HA + (A ↔ ∃y (y re A)) is equivalent to HA + ECT, where ECT is a strengthened form of CT. In HA + MP + ECT, which Troelstra considers to be a formalization of Russian recursive mathematics (RUSS in the article on Constructive Mathematics in this Encyclopedia), every formula of the form (y re A) has an equivalent "classical" prenex form A′(y) consisting of a quantifier-free subformula preceded by alternating "classical" quantifiers of the forms ¬¬∃x and ∀z¬¬, and so ∃y A′(y) is a kind of prenex form of A.
While HA is a proper part of classical arithmetic, the intuitionistic attitude toward mathematical objects results in a theory of real numbers (INT in the article on Constructive Mathematics)) diverging from the classical. Kleene's function-realizability, developed to prove the consistency of INT, changes the interpretation of arithmetical formulas. For example, ¬¬∀x (A(x) ¬A(x)) is function-realizable for every arithmetical formula A(x). In the language of analysis, Markov's Principle and the negative translation of the countable axiom of choice are among the many non-intuitionistic principles which are function-realizable (by classical arguments) and hence consistent with INT; cf. Kleene [1965], Vesley [1972] and Moschovakis [2003]. Abstract and concrete realizability semantics for a wide variety of formal systems have been developed and studied by logicians and computer scientists; cf. Troelstra [1998] and van Oosten [2000]. Variations of the basic notions are useful for establishing relative independence of the nonlogical axioms in theories based on intuitionistic logic.
The article on L. E. J. Brouwer in this Encyclopedia discusses Brouwer's philosophy and mathematics, with a chronology of his life and a selected list of publications. The best way to learn more is to read the original papers. Benacerraf and Putnam's essential source book contains Brouwer [1912] (in English translation), Brouwer [1949] and Dummett [1975]. The third edition [1971] of Heyting's classic [1956] is an attractive introduction to intuitionistic philosophy, logic and mathematical practice. As part of the formidable project of editing and publishing Brouwer's Nachlass, van Dalen [1981] provides a comprehensive view of Brouwer's own intuitionistic philosophy. The English translation, in van Heijenoort [1969], of Brouwer's [1927] (with a fine introduction by Parsons) is still an indispensable reference for Brouwer's theory of the continuum. Veldman [1990] is an authentic modern example of traditional intuitionistic mathematical practice.
Kleene and Vesley's [1965] gives a careful axiomatic treatment of intuitionistic analysis, a proof of its consistency relative to a classically correct subtheory, and an extended application to Brouwer's theory of real number generators. Kleene's [1969] formalizes the theory of partial recursive functionals, enabling precise formalizations of the function-realizability interpretation used in [1965] and of a related q-realizability interpretation which gives the Church-Kleene Rule for intuitionistic analysis.
An alternative to realizability semantics for intuitionistic arithmetic is Gödel's [1958] "Dialectica" interpretation, which associates with each formula B of L(HA) a quantifier-free formula BD in the language of intuitionistic arithmetic of all finite types. The "Dialectica" interpretation of B, call it BD, is ∃Yx BD(Y,x). If B is a closed theorem of HA, then BD(F,x) is provable for some term F in Gödel's theory T of "primitive recursive" functionals of higher type. The translation from B to BD requires the axiom of choice (at all finite types), MP and IP, so is not strictly constructive; however, the number-theoretic functions expressible by terms F of T are precisely the provably recursive functions of HA (and of PA). The interpretation was extended to analysis by Spector [1962]; cf. Howard [1973]. Clear expositions, and additional references, are to be found in Troelstra's introduction to the English translation in Gödel [1990] of the original Dialectica article, and in Avigad and Feferman [1998].
Another line of research in intuitionistic logic concerns Brouwer's very controversial "creating subject counterexamples" to principles of classical analysis (such as Markov's Principle) which could not be refuted on the basis of the theory FIM of Kleene and Vesley [1965]. By weakening Kleene's form of Brouwer's principle of continuous choice, and adding an axiom he called Kripke's Schema (KP), Myhill managed to formalize the creating subject arguments. KP is inconsistent with FIM, but Vesley found an alternative principle (Vesley's Schema (VS)) which can consistently be added to FIM and implies all the counterexamples for which Brouwer required a creating subject. Krol and Scowcroft gave topological consistency proofs for intuitionistic analysis with Kripke's Schema and weak continuity.
Troelstra's [1973], Beeson's [1985] and Troelstra and van Dalen's [1988] stand out as the most comprehensive studies of intuitionistic and semi-intutionistic formal theories, using both constructive and classical methods, with useful bibliographies. Troelstra's [1998] presents formulas-as-types and (Kleene and Aczel) slash interpretations for propositional and predicate logic, as well as abstract and concrete realizabilities for a multitude of applications. Martin-Löf's constructive theory of types [1984] (cf. Section 3.4 of Constructive Mathematics) provides a general framework within which intuitionistic reasoning continues to develop.
## Bibliography
• Avigad, J. and Feferman, S., 1998, "Gödel's functional ("Dialectica") interpretation," Chapter V of S. Buss, ed., 1998: 337-405.
• Bar-Hillel, Y., ed., 1965, Logic, Methodology and Philosophy of Science, North Holland Publishing, Amsterdam.
• Beeson, M. J., 1985, Foundations of Constructive Mathematics, Springer, Berlin.
• Benecerraf, P. and Hilary Putnam, eds., 1983, Philosophy of Mathematics: Selected Readings, 2nd Edition, Cambridge University Press, Cambridge.
• Brouwer, L. E. J., 1907, "On the Foundations of Mathematics," Thesis, Amsterdam; English translation in A. Heyting, ed., 1975: 11-101.
• Brouwer, L. E. J., 1908, "The Unreliability of the Logical Principles," English translation in Heyting, ed., 1975: 107-111.
• Brouwer, L. E. J., 1912, "Intuitionism and Formalism," English translation by A. Dresden in Bull. Amer. Math. Soc. 20 (1913): 81-96, reprinted in Benacerraf and Putnam, eds., 1983: 77-89; also reprinted in Heyting, ed., 1975: 123-138.
• Brouwer, L. E. J., 1923, 1954, "On the significance of the principle of excluded middle in mathematics, especially in function theory," "Addenda and corrigenda," and "Further addenda and corrigenda," English translation in van Heijenoort, ed., 1967: 334-345.
• Brouwer, L. E. J., 1927, "Intuitionistic reflections on formalism," originally published in 1927, English translation in van Heijenoort, ed., 1967: 490-492.
• Brouwer, L. E. J., 1948, "Consciousness, philosophy and mathematics," originally published (1948), reprinted in Benacerraf and Putnam, eds., 1983: 90-96.
• Buss, S., Ed., 1998, Handbook of Proof Theory, Elsevier, Amsterdam and New York.
• Crossley, J., and M. A. E. Dummett, eds., 1965, Formal Systems and Recursive Functions. North-Holland Publishing, Amsterdam.
• van Dalen, D., ed., 1981, Brouwer's Cambridge Lectures on Intuitionism, Cambridge University Press, Cambridge.
• Dummett, M., 1975, "The philosophical basis of intuitionistic logic," originally published (1975), reprinted in Benacerraf and Putnam, eds., 1983: 97-129.
• Friedman, H., 1975, "The disjunction property implies the numerical existence property," Proc. Nat. Acad. Sci. 72: 2877-2878.
• Gentzen, G., 1934-5, "Untersuchungen Über das logische Schliessen," Math. Zeitschrift 39: 176-210, 405-431.
• Gödel, K., 1932, "Zum intuitionistischen Aussagenkalkül," Anzeiger der Akademie der Wissenschaftischen in Wien 69: 65-66.
• Gödel, K., 1933, "Zur intuitionistischen Arithmetik und Zahlentheorie," Ergebnisse eines mathematischen Kolloquiums 4: 34-38.
• Gödel, K., 1958, "Über eine bisher noch nicht benützte Erweiterung des finiten Standpunktes," Dialectica 12: 280-287. Reproduced with an English translation in Gödel, 1990: 241-251.
• Gödel, K., 1990, Collected Works, Vol. II, eds. S. Feferman et. al., Oxford University Press, Oxford.
• Glivenko, V., 1929, "Sur qulques points de la logique de M. Brouwer," Academie Royale de Belgique, Bulletins de la classe des sciences, ser. 5, vol. 15: 183-188.
• Harrop R., 1960, "Concerning formulas of the types AB C, A → (Ex)B(x) in intuitionistic formal systems," Jour. Symb. Logic 25: 27-32.
• van Heijenoort, J., ed., 1967, From Frege to Gödel: A Source Book In Mathematical Logic 1879-1931, Harvard University Press, Cambridge, MA.
• Heyting, A., 1930, "Die formalen Regeln der intuitionistischen Logik," in three parts, Sitzungsber. preuss. Akad. Wiss.: 42-71, 158-169.
• Heyting, A., 1956, Intuitionism: An Introduction, North-Holland Publishing, Amsterdam. Third Revised Edition (1971).
• Heyting, A., ed., 1975, L. E. J. Brouwer: Collected Works 1: Philosophy and Foundations of Mathematics, Elsevier, Amsterdam and New York.
• Howard, W. A., 1973, "Hereditarily majorizable functionals of finite type," in Troelstra, ed., 1973: 454-461.
• Iemhoff, R., 2001, "On the admissible rules of intuitionistic propositional logic," Jour. Symb. Logic 66: 281-294.
• de Jongh, D. H. J., 1970, "The maximality of the intuitionistic propositional calculus with respect to Heyting's Arithmetic," Jour. Symb. Logic 36: 606.
• de Jongh, D. H. J., and Smorynski, C., 1976, "Kripke models and the intuitionistic theory of species," Annals of Mathematical Logic 9: 157-186.
• Kleene, S. C., 1945, "On the interpretation of intuitionistic number theory," Jour. Symb. Logic 10: 109-124.
• Kleene, S. C., 1965, "Classical extensions of intuitionistic mathematics," in Y. Bar-Hillel, ed., 1965: 31-44.
• Kleene, S. C., 1952, Introduction to Metamathematics, Van Nostrand, Princeton.
• Kleene, S. C., 1969, Formalized Recursive Functionals and Formalized Realizability, Memoirs of the American Mathematical Society 89.
• Kleene, S. C. and Vesley, R. E., 1965, The Foundations of Intuitionistic Mathematics, Especially in Relation to Recursive Functions, North-Holland Publishing, Amsterdam.
• Kripke, S. A., "Semantical analysis of intuitionistic logic," in J. Crossley and M. A. E. Dummett, eds., 1965: 92-130.
• Martin-Löf, P., 1984, Intuitionistic Type Theory, Notes by Giovanni Sambin of a series of lectures given in Padua, June 1980, Bibliopolis, Napoli.
• Moschovakis, J. R., 2003, "Classical and constructive hierarchies in extended intuitionistic analysis," Jour. Symb. Logic 68: 1015-1043.
• Smorynski, C. A., 1973, "Applications of Kripke models," in Troelstra, ed., 1973: 324-391.
• Spector, C., 1962, "Provably recursive functionals of analysis: a consistency proof of an analysis by an extension of principles formulated in current intuitionistic mathematics," Recursive Function Theory: Proceedings of Symposia in Pure Mathematics, Vol. 5, J. C. E. Dekker, ed.,: 1-27.
• Troelstra, A. S., ed., 1973, Metamathematical Investigation of Intuitionistic Arithmetic and Analysis, Lecture Notes in Mathematics 344, Springer-Verlag, Berlin. Corrections and additions available from the editor.
• Troelstra, A. S., "Realizability," Chapter VI of S. R. Buss, ed., 1998: 407-473.
• Troelstra, A. S., Introductory note to 1958 and 1972, in Gödel, 1990: 217-241.
• Troelstra, A. S. and van Dalen, D., 1988, Constructivism in Mathematics: An Introduction, in two volumes, North-Holland Publishing, Amsterdam.
• Veldman, W., 1990, "A survey of intuitionistic descriptive set theory," in P. P. Petkov, ed., Mathematical Logic, Proceedings of the Heyting Conference, Plenum Press, New York and London: 155-174.
• Vesley, R. E., 1972, "Choice sequences and Markov's principle," Compositio Mathematica 24: 33-53.
• Visser, A., 2002, "Substitutions of Sigma01 sentences: explorations between intuitionistic propositional logic and intuitionistic arithmetic," Annals of Pure and Applied Logic 114: 227-271.
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# Integration of Algebraic and Trigonometric Functions
The following table contains integrated examples of basic algebraic and trigonometric formulas.
Ln means natural logarithm
∫dx x ∫xn.dx xn+1 / (n+1) ∫axn.dx a . xn+1 / (n+1) ∫(axn + b).dx= ∫axn.dx + ∫b.dx a.xn+1 / (n+1) + b.x ∫(ax + b)n.dx (ax+b)n+1 / a(n+1) ∫dx / (ax + b)= 1/a . ∫a.dx / (ax + b) 1/a . Ln(ax+b) ∫1/x . dx Ln(x) ∫1/(x + b)½ . dx 2(x+b)½ ∫1/(ax + b)½ . dx 2(ax+b)½ / a ∫1/(x² - a²) . dx -Acoth(x/a) / a or Ln[(x-a)/(x+a)] / 2a ∫1/(a² - x²) . dx Atanh(x/a) / a or Ln[(a+x)/(a-x)] / 2a ∫1/(a² + x²) . dx Atan(x/a) / a ∫(x² + a²)½ . dx ½x(x² + a²)½ + ½a² . Asinh(x/a) or ½x(x² + a²)½ + ½a² . Ln([x+(x² + a²)½] / a) ∫ƒ'(x)/ƒ(x) . dx = Ln(ƒ(x)) Note: If the numerator = the differential of the denominator then the inverse of the denominator is the logₑ of the denominator. So multiply the equation by the differential of the denominator and 'Logₑ' the result d(u.v) / dx u.v = ∫u.dv/dx.dx +∫v.du/dx.dx = ∫u.dv +∫v.du ∫u.dv = u.v - ∫v.du ∫ax . dx ax . loga(e) ∫ex . dx ex ∫Sin(x) . dx –Cos(x) ∫Cos(x) . dx Sin(x) ∫Tan(x) . dx –Ln(Cos(x)), orLn(Sec(x)) ∫Cot(x) . dx Ln(Sin(x)) ∫Sec(x) . dx Ln(Tan(¼π + ½x)) ∫Cosec(x) . dx Ln(Tan(½x)) ∫Sinh(x) . dx Cosh(x) ∫Cosh(x) . dx Sinh(x) ∫Tanh(x) . dx Ln(Cosh(x)) ∫Coth(x) . dx Ln(Sinh(x)) ∫Sin(ax) . dx –Cos(ax) / a ∫Sin(ax + b) . dx –Cos(ax + b) / a ∫Cos(ax) . dx Sin(ax) / a ∫Cos(ax + b) . dx Sin(ax + b) / a ∫Tan(ax) . dx Ln(Sec(ax)) / a ∫Sinh(ax) . dx Cosh(ax) / a ∫Cosh(ax) . dx Sinh(ax) / a ∫Sin(x).Cos(x) . dx -¼Cos(2x) ∫Sec(x).Tan(x) . dx Sec(x) ∫Csc(x).Cot(x) . dx –Csc(x) ∫1 / (a² – x²)½ . dx Asin(x/a), or–Acos(x/a) ∫1 / (a² + x²) . dx Asec(x/a) / a, or–Acsc(x/a) / a ∫1 / x(x² – a²)½ . dx Asec(x/a) / a, or–Acsc(x/a) / a ∫1 / (x² + a²)½ . dx Asinh(x/a), orLn(x+(x²+a²)½ / a) ∫1 / (x² – a²)½ . dx Acosh(x/a), orLn(x+(x²–a²)½ / a) ∫1 / (a² – x²) . dx Atanh(x/a) / a, orLn((a+x)/(a–x)) / 2a ∫1 / (x² – a²) . dx –Acoth(x/a) / a, orLn((a–x)/(a+x)) / 2a ∫1 / x(a² – x²)½ . dx –Asech(x/a) / a, or–Ln((a + (a²–x²)½) / x) / a ∫1 / x(a² + x²)½ . dx –Acsch(x/a) / a, or–Ln((a + (a²+x²)½) / x) / a ∫Sin²(x) . dx ½(x – ½.Sin(2x)) ∫Cos²(x) . dx ½(x + ½.Sin(2x)) ∫Tan²(x) . dx Tan(x) – x ∫Csc²(x) .dx –Cot(x) ∫Sec²(x) . dx Tan(x) ∫Cot²(x) . dx –(Cot(x) + x) ∫(x² – a²)½ .dx ½.x(x²–a²)½ – a².Acosh(x/a)/2, or½.x(x²–a²)½ – a²(logₑ((x+(x²–a²)½ / a) / 2 ∫(x² + a²)½ .dx ½.x(x²+a²)½ + a².Asinh(x/a)/2, or½x(x²+a²)½ + a²(logₑ((x+(x²+a²)½ / a) / 2 ∫(a² – x²)½ .dx ½.a².Asin(x/a) + ½.x(a² – x²)½ ∫Sin²(ax) ½x – ¼Sin(2ax)/a ∫x.Sin(ax).dx Sin(ax)/a² – x.Cos(ax)/a ∫x².Sin(ax) -x².Cos(ax)/a + 2.x.Sin(ax)/a² + 2Cos(ax)/a³ ∫x².Sin²(ax) x³/6 – ¼.x².Sin(2ax)/a – ¼x.Cos(2ax)/a² + ⅛Sin(2ax)/a³ ∫x³.Sin(ax) -x³.Cos(ax)/a + 3x².Sin(ax)/a² + 6.x.Cos(ax)/a³ – 6.Sin(ax)/a⁴ ∫Cos²(ax) ¼Sin(2ax)/a + ½x ∫x.Cos(ax).dx x.Sin(ax)/a + Cos(ax)/a² ∫x².Cos(ax) x².Sin(ax)/a + 2.x.Cos(ax)/a² – 2.Sin(ax)/a³ ∫x².Cos²(ax) ¼.x².Sin(2ax)/a + x³/6 + x.Cos(2ax) / 4a² – ⅛Sin(2ax)/a³ ∫x³.Cos(ax) x³.Sin(ax)/a + 3x².Cos(ax)/a² – 6.x.Sin(ax)/a³ – 6.Cos(ax)/a⁴ ∫Sin(x).Cos(x) -¼.Cos(2x)
## Worked Examples
The following table contains a number of examples worked through by CalQlata engineers from time to time.
The table may not yet be complete but will be eventually. We are adding new integral workings as we resolve them.
Note: there are a number of different ways to integrate these formulas, we have simply listed the methods we have used.
Typical Integration by Substitution: Problem: ∫(a + b.x²)⁰˙⁵ . dx set: m = √a; n = √b; x = m/n . Tan(θ) {i.e. θ = Atan[x.n/m]} note: Sec²(θ) = 1+Tan²(θ) ∫(m² + n².x²)⁰˙⁵ . dx = ∫(m² + n².m²/n² . Tan²[θ])⁰˙⁵ . dθ = ∫(m² + m² . Tan²[θ])⁰˙⁵ . dθ = ∫(m².(1 + Tan²[θ]))⁰˙⁵ . dθ = ∫(m².Sec²[θ])⁰˙⁵ . dθ = ∫m.Sec[θ] . dθ = m∫Sec[θ] . dθ = m . Ln(Tan[¼π + ½θ]) {see ∫Sec[x].dx above} substitute back: for x: m . Ln(Tan[¼π + ½{Atan[x.n/m]}]) for a & b: √a . Ln(Tan[¼π + ½{Atan[x.√b/√a]}]) ∫(a + b.x²)⁰˙⁵ . dx = √a . Ln(Tan[¼π + ½.Atan[x.√(b/a)]]) ∫Sin²(x).dx Sin²(x) = Sin(x).Sin(x) = ½(Cos(x–x) – Cos(x+x)) = ½(Cos(0) – Cos(2x)) = ½(1 – Cos(2x)) = ½ – ½Cos(2x) ∫Sin²(x) = ∫(½ – ½Cos(2x)).dx = ∫½.dx – ∫½Cos(2x).dx = ½∫dx – ½∫Cos(2x).dx = ½.x – ½.Sin(2x)/2 ∫Sin²(x) = ½x – ¼Sin(2x) ∫Sin²(ax).dx Sin²(ax) = Sin(ax).Sin(ax) = ½(Cos(ax–ax) – Cos(ax+ax)) = ½(Cos(0) – Cos(2ax)) = ½(1 – Cos(2ax)) = ½ – ½Cos(2ax) ∫Sin²(ax) = ∫(½ – ½Cos(ax)).dx = ∫½.dx – ∫½Cos(2ax).dx = ½∫dx – ½∫Cos(2ax).dx = ½x – ½Sin(2ax)/2a ∫Sin²(ax) = ½x – ¼Sin(2ax)/a ∫x.Sin(ax).dx (using integration by parts: ∫u.dv = uv - ∫v.du) u = x; dv = Sin(ax); du = dx; v = -Cos(ax)/a ∫x.Sin(ax).dx = x.-Cos(ax)/a – ∫-Cos(ax)/a.dx = -x.Cos(ax)/a + 1/a∫Cos(ax).dx = -x.Cos(ax)/a + 1/a.Sin(ax)/a = -x.Cos(ax)/a + Sin(ax)/a² ∫x.Sin(ax).dx = Sin(ax)/a² – x.Cos(ax)/a ∫x².Sin(ax) (using integration by parts: ∫u.dv = uv - ∫v.du) u = x²; dv = Sin(ax); du = 2x.dx; v = -Cos(ax)/a ∫x².Sin(ax) = x².-Cos(ax)/a - ∫-Cos(ax)/a . 2x.dx ∫x².Sin(ax) = -x².Cos(ax)/a + 2/a∫x.Cos(ax).dx ∫x.Cos(ax).dx u = x; dv = Cos(ax); du =dx; v = Sin(ax)/a ∫x.Cos(ax).dx = x . Sin(ax)/a – ∫Sin(ax)/a . dx = x . Sin(ax)/a – 1/a∫Sin(ax).dx = x . Sin(ax)/a – 1/a-Cos(ax)/a.dx = x.Sin(ax)/a + Cos(ax)/a/a ∫x.Cos(ax).dx = x.Sin(ax)/a + Cos(ax)/a² ∫x².Sin(ax) = -x².Cos(ax)/a + 2/a . (x.Sin(ax)/a + Cos(ax)/a²) = -x².Cos(ax)/a + (2/a . x.Sin(ax)/a + 2/a . Cos(ax)/a²) = -x².Cos(ax)/a + (2x.Sin(ax)/a² + 2Cos(ax)/a³) ∫x².Sin(ax) = 2Cos(ax)/a³ + 2x.Sin(ax)/a² – x².Cos(ax)/a ∫x².Sin²(ax) Sin²(ax) = Sin(ax).Sin(ax) = ½(Cos(ax–ax) – Cos(ax+ax)) = ½(Cos(0) – Cos(2ax)) = ½(1 – Cos(2ax)) Sin²(ax) = ½ – ½Cos(2ax) (using integration by parts: ∫u.dv = uv - ∫v.du) u = x²; dv = ½ – ½Cos(2ax); du = 2x.dx; v = ½x – ¼Sin(2ax)/a ∫x².Sin²(ax) = x².(½x – ¼.Sin(2ax)/a) – ∫(½x – ¼.Sin(2ax)/a) . 2x.dx = ½x³ – ¼.x².Sin(2ax)/a – ∫(x² – ½.x.Sin(2ax)/a).dx = ½x³ – ¼.x².Sin(2ax)/a – ∫x².dx + ∫½.x.Sin(2ax)/a.dx = ½x³ – ¼.x².Sin(2ax)/a – ∫x².dx + 1 / 2a∫x.Sin(2ax).dx = ½x³ – ¼.x².Sin(2ax)/a – ⅓x³ + 1 / 2a∫x.Sin(2ax).dx ∫x².Sin²(ax) = x³/6 – ¼.x².Sin(2ax)/a + 1 / 2a∫x.Sin(2ax).dx ∫x.Sin(2ax).dx u = x; dv = Sin(2ax); du = dx; v = -Cos(2ax)/2a ∫x.Sin(2ax).dx = -x.Cos(2ax) / 2a – ∫-Cos(2ax) / 2a . dx = -x.Cos(2ax) / 2a + 1 / 2a∫Cos(2ax) . dx = -x.Cos(2ax) / 2a + 1 / 2a.Sin(2ax) / 2a ∫x.Sin(2ax).dx = -x.Cos(2ax) / 2a + Sin(2ax) / 4a² ∫x².Sin²(ax) = x³/6 – ¼.x².Sin(2ax)/a + 1 / 2a . (-x.Cos(2ax) / 2a + Sin(2ax) / 4a²) = x³/6 – ¼.x².Sin(2ax)/a + (-x.Cos(2ax) / 4a² + ⅛Sin(2ax)/a³) ∫x².Sin²(ax) = x³/6 – ¼.x².Sin(2ax)/a – ¼x.Cos(2ax)/a² + ⅛Sin(2ax)/a³ ∫x³.Sin(ax) (using integration by parts: ∫u.dv = uv - ∫v.du) u = x³; dv = Sin(ax); du = 3.x².dx; v = -Cos(ax)/a ∫x³.Sin(ax) = x³.-Cos(ax)/a – ∫-Cos(ax)/a . 3x².dx ∫x³.Sin(ax) = -x³.Cos(ax)/a + 3/a∫x².Cos(ax).dx ∫x².Cos(ax).dx u = x²; dv = Cos(ax); du =2x.dx; v = Sin(ax)/a ∫x².Cos(ax).dx = x².Sin(ax)/a – ∫Sin(ax)/a . 2x.dx = x².Sin(ax)/a – 2/a∫Sin(ax) . x.dx ∫x².Cos(ax).dx = x².Sin(ax)/a – 2/a∫x.Sin(ax).dx ∫x.Sin(ax).dx u = x; dv = Sin(ax); du =dx; v = -Cos(ax)/a ∫x.Sin(ax).dx = x . -Cos(ax)/a – ∫-Cos(ax)/a . dx = -x.Cos(ax)/a + 1/a∫Cos(ax).dx = -x.Cos(ax)/a + Sin(ax)/a/a ∫x.Sin(ax).dx = -x.Cos(ax)/a + Sin(ax)/a² ∫x².Cos(ax).dx = x².Sin(ax)/a – 2/a . (-x.Cos(ax)/a + Sin(ax)/a²) = x².Sin(ax)/a – (2/a.-x.Cos(ax)/a + 2/aSin(ax)/a²) = x².Sin(ax)/a – (2.-x.Cos(ax)/a² + 2.Sin(ax)/a³) ∫x².Cos(ax).dx = x².Sin(ax)/a + 2.x.Cos(ax)/a² – 2.Sin(ax)/a³ ∫x³.Sin(ax) = -x³.Cos(ax)/a + 3/a . (x².Sin(ax)/a + 2.x.Cos(ax)/a² – 2.Sin(ax)/a³) = -x³.Cos(ax)/a + (3/a . x².Sin(ax)/a + 3/a . 2.x.Cos(ax)/a² – 3/a . 2.Sin(ax)/a³) = -x³.Cos(ax)/a + (3x².Sin(ax)/a² + 6.x.Cos(ax)/a³ – 6.Sin(ax)/a⁴) ∫x³.Sin(ax) = -x³.Cos(ax)/a + 3x².Sin(ax)/a² + 6.x.Cos(ax)/a³ – 6.Sin(ax)/a⁴ ∫Cos²(x).dx Cos²(x) = Cos(x).Cos(x) = ½(Cos(x+x) + Cos(x-x)) = ½(Cos(2x) + Cos(0)) = ½(Cos(2x) + 1) = ½Cos(2x) + ½ ∫Cos²(x) = ∫(½Cos(2x) + ½).dx = ∫½Cos(2x).dx + ∫½.dx = ½∫Cos(2x).dx + ½∫dx = ½.Sin(2x)/2 + ½.x ∫Cos²(x) = ¼Sin(2x) + ½x ∫Cos²(ax).dx Cos²(ax) = Cos(ax).Cos(ax) = ½(Cos(ax+ax) + Cos(ax-ax)) = ½(Cos(2ax) + Cos(0)) = ½(Cos(2ax) + 1) = ½Cos(2ax) + ½ ∫Cos²(ax) = ∫(½Cos(ax) + ½).dx = ∫½Cos(2ax).dx + ∫½.dx = ½∫Cos(2ax).dx + ½∫dx = ½Sin(2ax)/2a + ½x ∫Cos²(ax) = ¼Sin(2ax)/a + ½x ∫x.Cos(ax).dx (using integration by parts: ∫u.dv = uv - ∫v.du) u = x; dv = Cos(ax); du = dx; v = Sin(ax)/a ∫x.Cos(ax).dx = x.Sin(ax)/a – ∫Sin(ax)/a.dx = x.Sin(ax)/a – 1/a∫Sin(ax).dx = x.Sin(ax)/a – 1/a.-Cos(ax)/a = x.Sin(ax)/a + Cos(ax)/a² ∫x.Cos(ax).dx = x.Sin(ax)/a + Cos(ax)/a² ∫x².Cos(ax) (using integration by parts: ∫u.dv = uv - ∫v.du) u = x²; dv = Cos(ax); du =2x.dx; v = Sin(ax)/a ∫x².Cos(ax) = x².Sin(ax)/a – ∫Sin(ax)/a . 2x.dx ∫x².Cos(ax) = x².Sin(ax)/a – 2/a∫x.Sin(ax).dx ∫x.Sin(ax).dx u = x; dv = Sin(ax); du =dx; v = -Cos(ax)/a = x . -Cos(ax)/a – ∫-Cos(ax)/a . dx = -x.Cos(ax)/a + 1/a∫Cos(ax).dx = -x.Cos(ax)/a + 1/a.Sin(ax)/a ∫x.Sin(ax).dx = -x.Cos(ax)/a + Sin(ax)/a² ∫x².Cos(ax) = x².Sin(ax)/a – 2/a . (-x.Cos(ax)/a + Sin(ax)/a²) = x².Sin(ax)/a – (2/a.-x.Cos(ax)/a + 2/a.Sin(ax)/a²) = x².Sin(ax)/a – (2.-x.Cos(ax)/a² + 2.Sin(ax)/a³) ∫x².Cos(ax) = x².Sin(ax)/a + 2.x.Cos(ax)/a² – 2.Sin(ax)/a³ ∫x².Cos²(ax) Cos²(ax) = Cos(ax).Cos(ax) = ½(Cos(ax+ax) + Cos(ax-ax)) = ½(Cos(2ax) + Cos(0)) = ½(Cos(2ax) + 1) Cos²(ax) = ½Cos(2ax) + ½ (using integration by parts: ∫u.dv = uv - ∫v.du) u = x²; dv = ½Cos(2ax) + ½; du = 2x.dx; v = ¼Sin(2ax)/a + ½x ∫x².Cos²(ax) = x².(¼.Sin(2ax)/a + ½x) – ∫(¼Sin(2ax)/a + ½x) . 2x.dx = ¼.x².Sin(2ax)/a + ½x³ – ∫(½.x.Sin(2ax)/a + x²).dx = ¼.x².Sin(2ax)/a + ½x³ – ∫½.x.Sin(2ax)/a.dx – ∫x².dx = ¼.x².Sin(2ax)/a + ½x³ – ∫x².dx – 1 / 2a∫x.Sin(2ax).dx = ¼.x².Sin(2ax)/a + ½x³ – ⅓x³ – 1 / 2a∫x.Sin(2ax).dx ∫x².Cos²(ax) = ¼.x².Sin(2ax)/a + x³/6 – 1 / 2a∫x.Sin(2ax).dx ∫x.Sin(2ax).dx u = x; dv = Sin(2ax); du = dx; v = -Cos(2ax) / 2a ∫x.Sin(2ax).dx = -x.Cos(2ax) / 2a - ∫-Cos(2ax) / 2a . dx = -x.Cos(2ax) / 2a + 1 / 2a∫Cos(2ax) . dx = -x.Cos(2ax) / 2a + 1 / 2a.Sin(2ax) / 2a . dx ∫x.Sin(2ax).dx = -x.Cos(2ax) / 2a + Sin(2ax) / 4a² ∫x².Cos²(ax) = ¼.x².Sin(2ax)/a + x³/6 – 1 / 2a . (-x.Cos(2ax) / 2a + Sin(2ax) / 4a²) = ¼.x².Sin(2ax)/a + x³/6 – (-x.Cos(2ax) / 4a² + Sin(2ax) / 8a³) ∫x².Cos²(ax) = ¼.x².Sin(2ax)/a + x³/6 + x.Cos(2ax) / 4a² – ⅛Sin(2ax)/a³ ∫x³.Cos(ax) (using integration by parts: ∫u.dv = uv - ∫v.du) u = x³; dv = Cos(ax); du =3x².dx; v = Sin(ax)/a ∫x³.Cos(ax) = x³.Sin(ax)/a – ∫Sin(ax)/a . 3x².dx ∫x³.Cos(ax) = x³.Sin(ax)/a – 3/a∫x².Sin(ax).dx ∫x².Sin(ax).dx u = x²; dv = Sin(ax); du =2x.dx; v = -Cos(ax)/a ∫x².Sin(ax).dx = x².-Cos(ax)/a – ∫-Cos(ax)/a . 2x.dx = -x².Cos(ax)/a + 2/a∫Cos(ax) . x.dx ∫x².Sin(ax).dx = -x².Cos(ax)/a + 2/a∫x.Cos(ax).dx ∫x.Cos(ax).dx u = x; dv = Cos(ax); du =dx; v = Sin(ax)/a ∫x.Cos(ax).dx = x . Sin(ax)/a – ∫Sin(ax)/a . dx = x . Sin(ax)/a – 1/a∫Sin(ax).dx = x.Sin(ax)/a + Cos(ax)/a/a ∫x.Cos(ax).dx = x.Sin(ax)/a + Cos(ax)/a² ∫x².Sin(ax).dx = -x².Cos(ax)/a + 2/a . (x.Sin(ax)/a + Cos(ax)/a²) = -x².Cos(ax)/a + (2/a . x.Sin(ax)/a + 2/a . Cos(ax)/a²) = -x².Cos(ax)/a + (2.x.Sin(ax)/a² + 2.Cos(ax)/a³) ∫x².Sin(ax).dx = -x².Cos(ax)/a + 2.x.Sin(ax)/a² + 2.Cos(ax)/a³ ∫x³.Cos(ax) = x³.Sin(ax)/a – 3/a . (-x².Cos(ax)/a + 2.x.Sin(ax)/a² + 2.Cos(ax)/a³) = x³.Sin(ax)/a – (3/a . -x².Cos(ax)/a + 3/a . 2.x.Sin(ax)/a² + 3/a . 2.Cos(ax)/a³) = x³.Sin(ax)/a – (-3x².Cos(ax)/a² + 6.x.Sin(ax)/a³ + 6.Cos(ax)/a⁴) ∫x³.Cos(ax) = x³.Sin(ax)/a + 3x².Cos(ax)/a² – 6.x.Sin(ax)/a³ – 6.Cos(ax)/a⁴ ∫Sin(x).Cos(x).dx Sin(x).Cos(x) = ½(Sin(x+x) + Sin(x-x)) = ½(Sin(2x) + Sin(0)) = ½(Sin(2x) + 0) = ½Sin(2x) ∫Sin(x).Cos(x) = ∫½Sin(2x).dx = ½∫Sin(2x).dx = ½.-Cos(2x)/2 ∫Sin(x).Cos(x) = -¼.Cos(2x) ∫Tan²(x).dx Tan²(x) = Sec²(x) – 1 ∫Tan²(x) = ∫(Sec²(x) – 1).dx = ∫Sec²(x).dx – ∫dx ∫Tan²(x) = Tan(x) – x
Colour Coding is provided in the above table to assist with the flow/sequencing of some of the more complex calculations. | 6,530 | 12,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.289259 |
https://askworksheet.com/balancing-equations-worksheet-3rd-grade/ | 1,722,984,962,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640523737.31/warc/CC-MAIN-20240806224232-20240807014232-00840.warc.gz | 86,598,323 | 27,096 | # Balancing Equations Worksheet 3rd Grade
Balancing chemical equations worksheet in word. Balancing equations worksheet third grade 2.
Balancing Equations Balancing Equations Common Core State Standards Math Expressions
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Balancing equations worksheet 3rd grade. It has an answer key attached on the second page. The 2nd worksheet involves addition and subtraction only. Introduce them to beginner algebra and help them develop their deductive reasoning skills.
March 20 2020. This is a set of 43 worksheets that teach students how to balance addition equations. Addition and balancing equations worksheet for 3rd grade children.
Balancing chemical equations word equations worksheet. The 3rd and 4th worksheets are more challenging and include multiplication and larger numbers. Balancing chemical equations worksheet number 3.
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8th grade balancing equations worksheet. This balancing equations worksheet uses basic subtraction to train students to balance equations by finding the missing values. Train students in beginner algebra with this simple activity of finding the missing values to balance both sides of the equation.
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Home worksheets balancing equations worksheet third grade 3 this balancing equations worksheet uses subtraction and multiples of ten. The printables are also available as pdf files. Extremely beneficial for kindergarten grade one or grade two level.
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Balancing equations worksheet grade 8. The activities are varied and all worksheets are no prep printables more details listed below.
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Balancing Equations Worksheets Balancing Equations Equations Practices Worksheets | 819 | 4,387 | {"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} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.916251 |
https://icsehelp.com/ml-aggarwal-class-9-icse-understanding-apc-solutions/ | 1,716,149,226,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00369.warc.gz | 270,465,288 | 18,142 | # ML Aggarwal Class-9 ICSE Understanding APC Solutions
ML Aggarwal Class-9 ICSE Understanding APC Maths Solutions . Solutions of Exercise ,MCQ , Chapter Test of Class-9 ICSE Understanding Maths of ML Aggarwal Avichal Publications Company (APC)
## ML Aggarwal Class-9 ICSE Understanding APC Solutions
Although There are many edition in ICSE series But Understanding Maths is more popular than any other writer. Hence ML Aggarwal Solutions for ICSE Class-9 available for you. Visit official website CISCE for detail information about ICSE Board Class-9. We are providing valuable and important information for ICSE Maths Class 9th. Students with the help of word wide maths expert.
### How To Solve ICSE Class -9th Maths All Publication
• Plan a self made time table of Chapters which you can follow
• Keep yourself away from obstacle such as Phone Call, Game , Over Browsing on Net
• Read the Concept of Certain Chapter carefully
• Focus on Formulas used
• Attention on when and which formulas is suitable for certain questions
• Practice Example given of your textbook
• Also Practice example of other famous publications
• Now try to solve Exercise
• If feel any problems then view our Solutions given in Sequence of textbook
• Solving model paper of ML Aggarwal is also best
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### Chapter-Wise Solutions of ML Aggarwal for ICSE Class-9
Get Solutions of Latest Edition of ML Aggarwal ICSE Class 9 Solutions for Understanding Maths. Hence we provides step by step solutions for ML Aggarwal Maths for Class 9 ICSE .So You can view the Understanding ICSE Mathematics Class 9 ML Aggarwal Solved Questions with Exercise, MCQ Chapter Test Option.
### Get ML Aggarwal Solutions for ICSE Class 9 Maths
1. Expansions 3.1, Exe-3.2, MCQs, Ch-Test
2. Indices Exe-8, MCQs, Ch-Test
3. Logarithms Exe-9.1, Exe-9.2, MCQs, Ch-Test
4. Triangles, Exe 10.1, Exe-10.2, Exe 10.3, Exe-10.4, MCQs, Ch-Test
5. Mid Point Theorem, Exe-11, MCQs, Ch-Test
6. Pythagoras Theorem, Exe-12, MCQs, Ch-Test
7. Rectilinear Figures, Exe-13.1, Exe-13..2,
8. Theorems on Area, Exe-14, MCQs, Ch-Test
9. Circle, Exe-15.1, Exe-15.2, MCQs, Ch-Test
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### 7 thoughts on “ML Aggarwal Class-9 ICSE Understanding APC Solutions”
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2. Why is it saying content is protected when I am trying to open the book?Pandey Sir are the books available on your website only for the children who study in your coaching? | 716 | 2,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.824104 |
https://infogalactic.com/info/Decimal_sign | 1,540,335,695,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583517495.99/warc/CC-MAIN-20181023220444-20181024001944-00312.warc.gz | 706,378,332 | 24,031 | Decimal mark
(Redirected from Decimal sign)
For the proper choice of the decimal mark in English language Wikipedia articles, see Wikipedia:Manual of Style (dates and numbers)#Decimal points.
"Decimal period" redirects here. For the length of the repeating sequence in the decimal expansion of a rational number, see Repeating decimal.
A decimal mark is a symbol used to separate the integer part from the fractional part of a number written in decimal form.
Different countries officially designate different symbols for the decimal mark. The choice of symbol for the decimal mark also affects the choice of symbol for the thousands separator used in digit grouping, so the latter is also treated in this article.
In mathematics the decimal mark is a type of radix point, a term that also applies to number systems with bases other than ten. Conversationally, the decimal point is commonly referred to as a "decimal dot".
History
In the Middle Ages, before printing, a bar ( ¯ ) over the units digit was used to separate the integral part of a number from its fractional part, e.g. 9995 (meaning, 99.95 in decimal point format). This practice derived from the decimal system used in Indian mathematics[1] and was popularized by the Persian mathematician Al-Khwarizmi,[2] when Latin translation of his work on the Indian numerals introduced the decimal positional number system to the Western world. His Compendious Book on Calculation by Completion and Balancing presented the first systematic solution of linear and quadratic equations in Arabic. A similar notation remains in common use as an underbar to superscript digits, especially for monetary values without a decimal mark, e.g. 9995. Later, a separator (a short, roughly vertical, ink stroke) between the units and tenths position became the norm among Arab mathematicians, e.g. 99ˌ95. When this character was typeset, it was convenient to use the existing comma (99,95) or full stop (99.95) instead.
Gerbert of Aurillac marked triples of columns with an arc (called a "Pythagorean arc") when using his Hindu–Arabic numeral-based abacus in the 10th century. Fibonacci followed this convention when writing numbers such as in his influential work Liber Abaci in the 13th century.[3]
In France, the full stop was already in use in printing to make Roman numerals more readable, so the comma was chosen.[4] Many other countries, such as Italy, also chose to use the comma to mark the decimal units position.[5] It has been made standard by the ISO for international blueprints.[citation needed] However, English-speaking countries took the comma to separate sequences of three digits. In some countries, a raised dot or dash (upper comma) may be used for grouping or decimal mark (see examples below); this is particularly common in handwriting.
In the United States, the full stop or period (.) was used as the standard decimal mark.
Decimal separator in a British print from 1839[6]
In the nations of the British Empire, although the full stop could be used in typewritten material and its use was not banned, the interpunct (also known as the decimal point, point, or mid-dot) was preferred for the decimal mark in printing technologies that could accommodate it, e.g. 99·95.[7] However, as the mid-dot was already in common use in the mathematics world to indicate multiplication, the SI rejected its use as the decimal mark. During the beginning of British metrication in the late 1960s, and with impending currency decimalisation, there was widespread debate in the United Kingdom as to whether the decimal comma or decimal point should be preferred; the British Standards Institution and some sectors of industry advocated the comma, and the Decimal Currency Board advocated for the point. In the event, the point was decided on by the Ministry of Technology in 1968.[8]
When South Africa adopted the metric system, it adopted the comma as its decimal mark,[9] although a number of house styles, including some English-language newspapers, such as The Sunday Times continue to use the full stop.[citation needed]
The three most spoken international auxiliary languages, Ido, Esperanto, and Interlingua all use the comma as the decimal mark. Interlingua has used the comma as its decimal mark since the publication of the Interlingua Grammar in 1951.[10] Esperanto also uses the comma as its official decimal mark, while thousands are separated by non-breaking spaces: 12 345 678,9. Ido's Kompleta Gramatiko Detaloza di la Linguo Internaciona Ido (Complete Detailed Grammar of the International Language Ido) officially states that commas are used for the decimal mark while full stops are used to separate thousands, millions, etc. So the number 12,345,678.90123 (in American notation) for instance, would be written 12.345.678,90123 in Ido. The 1931 grammar of Volapük by Arie de Jong uses the comma as its decimal mark, and (somewhat unusually) uses the middle dot as the thousands separator (12·345·678,90123).[11]
In 1958, disputes between European and American delegates over the correct representation of the decimal mark nearly stalled the development of the ALGOL computer programming language.[12] ALGOL ended up allowing different decimal marks, but most computer languages and standard data formats (e.g. C, Java, Fortran, Cascading Style Sheets (CSS)) specify a dot.
The 22nd General Conference on Weights and Measures declared in 2003 that "the symbol for the decimal marker shall be either the point on the line or the comma on the line". It further reaffirmed that "numbers may be divided in groups of three in order to facilitate reading; neither dots nor commas are ever inserted in the spaces between groups"[13] e.g. 1 000 000 000. This usage has therefore been recommended by technical organizations, such as the United States' National Institute of Standards and Technology.
Digit grouping
For ease of reading, numbers with many digits may be divided into groups using a delimiter.[14] In some countries, these "digit group separators" are only employed to the left of the decimal mark; in others, they are also used to separate numbers with a long fractional part. An important reason for grouping is that it allows rapid judgement of the number of digits, via subitizing (telling at a glance) rather than counting – contrast 1,000,000,000 with 1000000000 for a short billion.
The groups created by the delimiters tend to follow the use of the local language, which varies. In European languages, large numbers are read in groups of thousands and the delimiter (which occurs every three digits when it is used) may be called a "thousands separator". In East Asian cultures, particularly China, Japan, and Korea, large numbers are read in groups of myriads (10,000s) but the delimiter commonly separates every three digits.[citation needed] The Indian numbering system is somewhat more complex: it groups the rightmost three digits in a similar manner to European languages but then groups every two digits thereafter: 1.5 million would accordingly be written 15,00,000 and read as "15 lakh".[citation needed]
The convention for digit group separators varies but usually seeks to distinguish the delimiter from the decimal mark. Typically, English-speaking countries employ commas as the delimiter—10,000—and other European countries employ periods or spaces: 10.000 or 10 000. Because of the confusion that can result in international documents, the superseded SI/ISO 31-0 standard advocates the use of spaces[15] and the International Bureau of Weights and Measures and International Union of Pure and Applied Chemistry advocate the use of a "thin space" in "groups of three".[16][17] Within the United States, the American Medical Association's widely followed AMA Manual of Style also calls for a thin space.[14] In some online encoding environments (for example, ASCII-only) a thin space is not practical or available, in which case a regular word space or no delimiter are the alternatives. The Metrication Board proposed this system for the United Kingdom and, while it is not universally adopted, it is standard within the country's construction industry.[citation needed]
Digit group separators can occur either as part of the data or as a mask through which the data is displayed. This is an example of the separation of presentation and content. In many computing contexts, it is preferred to omit digit group separators from the data and instead overlay them as a mask (an input mask or an output mask). Common examples include spreadsheets and databases in which currency values are entered without such marks but are displayed with them inserted. Similarly, phone numbers can have hyphens or spaces as a mask rather than as data.
In some programming languages, it is possible to group the digits in the program's source code to make it easier to read; see Integer literal: Digit separators. Ada, D, Java, Perl and Ruby use the underscore (_) character for this purpose. All these languages allow writing seven hundred million as 700_000_000. C++14 allows the use of the apostrophe (') for digit grouping and therefore seven hundred million can be represented as 700'000'000.
Exceptions to digit grouping
The International Bureau of Weights and Measures states that "when there are only four digits before or after the decimal marker, it is customary not to use a space to isolate a single digit".[16] Likewise, some manuals of style state that thousands separators should not be used in normal text for numbers from 1000 to 9999 inclusive where no decimal fractional part is shown (in other words, for four-digit whole numbers), whereas others use thousands separators, and others use both. For example, APA style stipulates a thousands separator for "most figures of 1,000 or more" except for page numbers, binary digits, temperatures, etc.
There are always common-sense exceptions to digit grouping, such as years, postal codes and ID numbers of predefined nongrouped format, which style guides usually point out.
In non-base-10 numbering systems
In binary (base-2), a full space can be used between groups of four digits, corresponding to a nibble, or equivalently to a hexadecimal digit. Alternatively, binary digits may be grouped by threes, corresponding to an octal digit. Similarly, in hexadecimal (base-16), full spaces are usually used to group digits into twos, making each group correspond to a byte.
Influence of calculators and computers
In countries with a decimal comma, the decimal point is also common as the "international" notation because of the influence of devices, such as electronic calculators, which use the decimal point. Most computer operating systems allow selection of the decimal mark and programs that have been carefully internationalized will follow this, but some programs ignore it and a few are even broken by it.[which?]
Hindu–Arabic numeral system
Decimal marks:[citation needed]
Point "."
Comma ","
Data unavailable
Countries using Arabic numerals with decimal point
Countries where a dot "." is used as decimal mark include:
• Australia
• Botswana
• British West Indies
• Brunei
• China, People's Republic of
• Dominican Republic
• Egypt
• Ghana
• Guatemala
• Honduras
• Hong Kong
• India
• Ireland
• Israel
• Japan
• Jordan
• Kenya
• Korea, North
• Korea, South
• Lebanon
• Luxembourg (uses both marks officially)
• Macau (in Chinese and English text)
• Malaysia
• Malta
• Mexico
• Mongolia
• Nepal
• New Zealand
• Nicaragua
• Nigeria
• Pakistan
• Palestine
• Panama
• Philippines
• Singapore
• Sri Lanka
• Switzerland (for Swiss currency)
• Taiwan
• Tanzania
• Thailand
• Uganda
• United Kingdom
• United States (including insular areas)
• Zimbabwe
Countries using Arabic numerals with decimal comma
Countries where a comma "," is used as decimal mark include:
• Albania
• Algeria
• Andorra
• Angola
• Argentina
• Armenia
• Austria
• Azerbaijan
• Belarus
• Belgium
• Bolivia
• Bosnia and Herzegovina
• Brazil
• Bulgaria
• Cameroon
• Chile
• Colombia
• Costa Rica
• Croatia (comma used officially, but both forms are in use)[citation needed]
• Cuba
• Cyprus
• Czech Republic
• Denmark
• East Timor
• Estonia
• Faroes
• Finland
• France
• Germany
• Georgia
• Greece
• Greenland
• Hungary
• Iceland
• Indonesia
• Italy
• Kazakhstan
• Kosovo
• Kyrgyzstan
• Latvia
• Lebanon
• Lithuania
• Luxembourg (uses both marks officially)
• Macau (in Portuguese text)
• Macedonia
• Moldova
• Mongolia
• Morocco
• Mozambique
• Namibia
• The Netherlands
• Norway
• Paraguay
• Peru
• Poland
• Portugal
• Romania
• Russia
• Serbia
• Slovakia
• Slovenia
• South Africa
• Spain
• Switzerland (other than Swiss currency)
• Sweden
• Tunisia
• Turkey
• Ukraine
• Uruguay
• Uzbekistan
• Venezuela
• Vietnam
Other numeral systems
Unicode defines a decimal separator key symbol (⎖ in hex U+2396, decimal 9110) which looks similar to the apostrophe. This symbol is from ISO/IEC 9995 and is intended for use on a keyboard to indicate a key that performs decimal separation.
In the Arab world, where Eastern Arabic numerals are used for writing numbers, a different character is used to separate the integer and fractional parts of numbers. It is referred to as an Arabic decimal separator (٫) in Unicode. An Arabic thousands separator (٬) also exists.
In Persian, the decimal mark is called momayyez, which is written like a forward slash—there is a small difference between the "comma" character used in sentences and the momayyez (٫) used to separate sequences of three digits. To separate sequences of three digits, a comma or blank space may be used; however this is not a standard.[18][19][20]
In English Braille, the decimal point, , is distinct from both the comma, , and the period / full stop .
Examples of use
The following examples show the decimal mark and the thousands separator in various countries that use the Hindu–Arabic numeral system.
Style Countries
1,234,567.89 Australia, Canada (English-speaking, unofficial), China, Hong Kong, Ireland, Israel, Japan, Korea, Malaysia, Mexico, New Zealand, Pakistan, Philippines, Singapore, Taiwan, Thailand, United Kingdom, United States
1234567.89 SI style (English version), Canada (English-speaking), China, Sri Lanka
1234567,89 SI style (French version), Albania, Austria, Belgium, Bosnia, Brazil, Bulgaria, Canada (French-speaking), Czech Republic, Denmark, Estonia, Finland, France, Germany, Greece, Hungary, Italy, Latin Europe, Lithuania, Netherlands (non-currency numbers, see below), Norway, Poland, Portugal, Romania, Russia, Serbia, Slovakia, Slovenia, South Africa, Spain, Sweden, Ukraine
1,234,567·89 Ireland, Korea, Malaysia, New Zealand, Philippines, Singapore, Taiwan, Thailand, United Kingdom, United States (older, typically hand written)
1.234.567,89 Turkey, Austria, Brazil, Denmark, Germany, Greece, Indonesia, Italy, Netherlands (currency), Portugal, Romania, Russia, Slovenia, Spain (older),[21] Sweden (not recommended)
1˙234˙567,89 Italy (handwriting)
12,34,567.89 India (see Indian Numbering System)
1'234'567.89 Switzerland (printed, computing, currency, international requisite, everyday use)
1'234'567,89 Switzerland (handwriting)
1.234.567'89 Spain (handwriting)
123,4567.89 China (based on powers of 10,000, see Chinese numerals)
• In Albania, Belgium, Bosnia, Estonia, France, Hungary, Poland, Slovakia and much of Latin Europe as well as French Canada: 1234567,89 (In Spain, in handwriting it is also common to use an upper comma: 1.234.567'89)[citation needed]
• In Brazil, Denmark, Germany, Greece, Indonesia, Italy, Netherlands, Portugal, Romania, Russia, Slovenia, Sweden and much of Europe: 1234567,89 or 1.234.567,89. In handwriting, 1˙234˙567,89 is also seen, but never in Brazil, Denmark, Germany, the Netherlands, Portugal, Romania, Russia, Slovenia or Sweden. In Italy a straight apostrophe is also used in handwriting: 1'234'567,89. In the Netherlands, the points thousands separator is preferred in currency amounts, and the space for other large numbers[citation needed].
• In Germany and Austria, thousands separators were occasionally denoted by alternating uses of comma and point, e.g. 1.234,567.890,12[22][23] for "one billion 234 million ..."
• In Switzerland: There are two cases. 1'234'567.89 is used for currency values. An apostrophe as thousands separator along with a "." as decimal mark. For other values the SI style 1234567,89 is used with a "," as decimal mark. The apostrophe is also the most common variety for non-currency values: 1'234'567,89, though this usage is officially discouraged.
• In Ireland, Israel, Japan, Korea (both), Malaysia, New Zealand, the Philippines, Singapore, Taiwan, Thailand, the United Kingdom, and the United States: 1,234,567.89 or 1,234,567·89; the latter is generally found only in older, and especially handwritten, documents. Australia used this style up until the 1970s; now it uses the SI style.[24]
• In English Canada: There are two cases. The preferred method for currency values is \$4,000.00 while for numeric values it is 1234567.89; however, commas are also sometimes used although no longer taught in school or used in official publications.[25]
• SI style: 1 234 567.89 or 1 234 567,89 (in their own publications the dot is used in the English version and the comma in the French version).
• In China, comma and space are used to mark digit groups because dot is used as decimal mark. There is no universal convention on digit grouping, so both thousands grouping and no digit grouping can be found. Although names for large numbers in Chinese are based on powers of 10,000 (Trad / Simp: 萬/万) (e.g. the next new word (T/S: 億/亿) is for 108), grouping is never done for every 4 digits. Japan is similar, although when grouping by myriads numeral kanji are frequently used as separators: 1億2345萬6789 / 1亿2345万6789.
• In India, due to a numeral system using lakhs (lacs) (1,23,456 equal to 123 456) and crores (1,23,45,678 equal to 12 345 678), comma is used at levels of thousand, lakh and crore, for example, 10 million (1 crore) would be written as 1,00,00,000. In Pakistan, there is a greater tendency to use the standard western system, while using the Indian numbering system when conducting business in Urdu.
South Asian Value Value
One 1
Ten 10
Hundred 100
Thousand 1,000
Lakh 100,000
Crore 10,000,000
Arab 1,000,000,000
Kharab 100,000,000,000
Unicode characters
Used with Western Arabic numerals (0123456789):
• U+0020 SPACE (HTML ` `)
• U+0027 ' APOSTROPHE (HTML `'`<dot-separator> `'`)
• U+002C , COMMA (HTML `,`)
• U+002E . FULL STOP (HTML `.`)
• U+00B7 · MIDDLE DOT (HTML `·`<dot-separator> `·`)
• U+2009 THIN SPACE (HTML ` `<dot-separator> ` `)
• U+202F NARROW NO-BREAK SPACE (HTML ` `)
• U+02D9 ˙ DOT ABOVE (HTML `˙`)
Used with Eastern Arabic numerals (٠١٢٣٤٥٦٧٨٩):
• U+066B ٫ ARABIC DECIMAL SEPARATOR (HTML `٫`)
• U+066C ٬ ARABIC THOUSANDS SEPARATOR (HTML `٬`)
Used with keyboards:
• U+2396 DECIMAL SEPARATOR KEY SYMBOL (HTML `⎖`) (resembles an apostrophe)
References
1. Reimer, L., and Reimer, W. Mathematicians Are People, Too: Stories from the Lives of Great Mathematicians, Vol. 2. 1995. pp. 22-22. Parsippany, NJ: Pearson Education, Inc. as Dale Seymor Publications. ISBN 0-86651-823-1.
2. Khwarizmi, Abu Jafar Muhammad ibn Musa al-, Oxford Islamic Studies Online
3. Devlin, Keith (2011). The Man of Numbers: Fibonacci's Arithmetic Revolution. New York: Walker & Company. pp. 44–45. ISBN 9780802779083.
4. Enciclopedia Universal Santillana, 1996 by SANTILLANA S.A., Barcelona, Spain. ISBN 84-294-5129-3. Comma, def.2: "coma: MAT. Signo utilizado en los números no enteros para separar la parte entera de la parte decimal o fraccionaria; p.ej., 2,123."
5. Enciclopedia Universal Santillana, 1996 by SANTILLANA S.A., Barcelona, Spain. ISBN 84-294-5129-3. Comma, def.2: "coma: MAT. Signo utilizado en los números no enteros para separar la parte entera de la parte decimal o fraccionaria; p.ej., 2,123."
6. Thomas Henderson (1839-01-03). "On the Parallax of α Centauri" (PDF). Memoirs of the Royal Astronomical Society, Vol. 11, p. 64. Retrieved 2015-03-26. Scan published by Harvard-Smithsonian Center for Astrophysics
7. Reimer, L., and Reimer, W. Mathematicians Are People, Too: Stories from the Lives of Great Mathematicians, Vol. 1. 1990 p. 41. Parsippany, NJ: Pearson Education, Inc. as Dale Seymor Publications. ISBN 0-86651-509-7.
8. "Victory on Points". Nature. 218 (5137): 111. 1968. doi:10.1038/218111c0.
9. Government Notice R. 1144, Government Gazette 4326, 5 July 1974
10. Grammar of Interlingua: Parts of Speech – Numerals
11. Gramat Volapüka. Cathair na Mart: Evertype. ISBN 978-1-904808-94-7
12. Perlis, Alan, The American Side of the Development of ALGOL, ACM SIGPLAN Notices, August 1978.
13. [1], Resolution 10.
14. Iverson, Cheryl, et al. (eds) (2007). AMA Manual of Style (10th ed.). Oxford, Oxfordshire: Oxford University Press. p. 793. ISBN 978-0-19-517633-9.
15. "Decimals Score a Point on International Standards". 2006-11-22. Retrieved 2008-11-27.
16. International Bureau of Weights and Measures. "Rules and style conventions for expressing values of quantities": "[2]".
17. "Guidelines for drafting IUPAC technical reports and recommendations". 2007. Retrieved 2008-11-27.
18. Pournader, Roozbeh (2000-10-15). "Persian decimal separator". Unicode Mail List Archive. Unicode Consortium. Retrieved 2008-06-21.
19. "The Decimal Numeral". Academic Grammar of New Persian. Archived from the original on 2006-06-20. Retrieved 2006-06-19.
20. Descriptive Grammar of New Persian (archived)
21. Diccionario panhispánico de dudas, when writing numbers more than four figures, these three will be grouped into three, starting from the right, and separating the groups by whitespace. (Exceptions: Never written with periods, commas or white separation numbers that refer to years, pages, verses, urban roads, postal codes, legal articles, decrees or laws.)
22. Röll. "Enzyklopädie des Eisenbahnwesens". Retrieved 26 August 2014., entry "Union Pacific-Eisenbahn", largest numbers in table
23. Röll. "Enzyklopädie des Eisenbahnwesens". Retrieved 26 August 2014., entry "Bilanz", sums in last table
24. Steinle, Vicki. "Teaching and Learning about Decimals". Retrieved 20 April 2012.
25. Government of Canada (2011-01-13). "Public Service Commission Style Guide". Retrieved 2013-09-25. | 5,523 | 22,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-43 | latest | en | 0.896916 |
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PHYS809_11F_second_exam_solutions
# PHYS809_11F_second_exam_solutions - November 9th 11th 2011...
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PHYS809 Fall 2011 Second Take-home Exam November 9 th – 11 th , 2011 Please write your name on each answer sheet. Since this is a take home exam, there are a few rules that you must follow. You cannot discuss or communicate about the exam with any one by any means. You cannot look for or use solutions to the problems on the internet. You cannot use printed solutions to Jackson problems. You can use your textbook, notes, and the online course notes. You can use Mathematica or similar software, but you must say when you have used it. The problems do not have equal weight. 1. A ring of radius a has a uniformly distributed total charge Q . The ring is in a concentric spherical cavity of radius b (> a ) . Outside the cavity is dielectric of relative permittivity . r ε (a) Find the potential and electric field everywhere as expansions in Legendre polynomials. (b) Find the surface polarization charge density induced on the surface of the cavity. What is the total surface polarization charge? (c) The ring is replaced by a uniformly charged disk of radius a and total charge Q . Use your solution to part (a) to find the potential everywhere. Find the surface polarization charge density induced on the surface of the cavity. What is the total surface polarization charge? Compare with the result in part (b). (a) We can find the potential of the ring in the absence of dielectric by first finding the potential on the axis of symmetry. This is ( ) 2 2 0 1 . 4 Q z z a πε Φ = + Series expansions are ( ) 2 2 1 0 0 1 , 2 4 n n n z Q z z n πε < + = > - Φ = where ( ) ( ) min , , max , . z z a z z a < > = = Inserting the Legendre polynomials, we find the potential everywhere is ( ) ( ) 2 2 2 1 0 0 1 , cos , 2 4 n n n n r Q r P r n θ θ πε < + = > - Φ = where ( ) ( ) min , , max , . r r a r r a < > = = To take into account the dielectric, we add a potential of form ( ) ( ) 2 1 2 2 1 0 0 , cos , 4 n n n n n r Q r P r θ α θ πε < + = > Φ =
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where ( ) ( ) min , , max , . r r b r r b < > = = Since the normal component of D is continuous at , r b = we have ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 0 0 2 2 2 2 2 2 0 0 1 1 2 1 cos 2 cos 2 1 1 2 1 cos 2 1 cos . 2 n n n n n n n n r n n n n n n a n P n P b b n a n P n P b b n θ α θ ε θ α θ + = = + = = - - + + = - - + - + This leads to ( )( ) ( ) 2 2 1 2 1 1 .
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{[ snackBarMessage ]} | 828 | 2,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-17 | latest | en | 0.888327 |
https://www.crazy-numbers.com/en/4349 | 1,642,524,655,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300934.87/warc/CC-MAIN-20220118152809-20220118182809-00181.warc.gz | 760,237,024 | 4,918 | Discover a lot of information on the number 4349: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 4349
Is 4349 a prime number? Yes
Is 4349 a perfect number? No
Number of divisors 2
List of dividers 1, 4349
Sum of divisors 4350
Prime factorization 4349
Prime factors 4349
## How to write / spell 4349 in letters?
In letters, the number 4349 is written as: Four thousand three hundred and forty-nine. And in other languages? how does it spell?
4349 in other languages
Write 4349 in english Four thousand three hundred and forty-nine
Write 4349 in french Quatre mille trois cent quarante-neuf
Write 4349 in spanish Cuatro mil trescientos cuarenta y nueve
Write 4349 in portuguese Quatro mil trezentos quarenta e nove
## Decomposition of the number 4349
The number 4349 is composed of:
2 iterations of the number 4 : The number 4 (four) is the symbol of the square. It represents structuring, organization, work and construction.... Find out more about the number 4
1 iteration of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
1 iteration of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9
Other ways to write 4349
In letter Four thousand three hundred and forty-nine
In roman numeral MMMMCCCXLIX
In binary 1000011111101
In octal 10375
In US dollars USD 4,349.00 (\$)
In euros 4 349,00 EUR (€)
Some related numbers
Previous number 4348
Next number 4350
Next prime number 4357
## Mathematical operations
Operations and solutions
4349*2 = 8698 The double of 4349 is 8698
4349*3 = 13047 The triple of 4349 is 13047
4349/2 = 2174.5 The half of 4349 is 2174.500000
4349/3 = 1449.6666666667 The third of 4349 is 1449.666667
43492 = 18913801 The square of 4349 is 18913801.000000
43493 = 82256120549 The cube of 4349 is 82256120549.000000
√4349 = 65.94694837519 The square root of 4349 is 65.946948
log(4349) = 8.3777012125976 The natural (Neperian) logarithm of 4349 is 8.377701
log10(4349) = 3.6383894076653 The decimal logarithm (base 10) of 4349 is 3.638389
sin(4349) = 0.86025389702375 The sine of 4349 is 0.860254
cos(4349) = 0.50986589673703 The cosine of 4349 is 0.509866
tan(4349) = 1.6872159964592 The tangent of 4349 is 1.687216 | 770 | 2,454 | {"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} | 3.78125 | 4 | CC-MAIN-2022-05 | latest | en | 0.761022 |
https://programmer.ink/think/61fc33fbe7d83.html | 1,718,335,887,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00202.warc.gz | 441,009,326 | 6,967 | # (teaching-08-01) 20203 digital opening and closing (maximum number of questions with fixed answers)
Posted by l053r on Thu, 03 Feb 2022 20:57:59 +0100
## Design idea:
<(teaching-06) maximum number of addition and subtraction methods within 20 (optimized version 20220122 vs Python addition and subtraction method within 20) >The code can randomly generate up to addition questions and addition questions within x. A Xia tried to modify the code to make the list questions and quantity of "5 on / off" and "10 on / off".
Main solutions:
1. There are several addition questions with sum equal to 5: 0 + 5 = 5 ， 1 + 4 = 5... (opening and closing of 5, opening and closing of 6, opening and closing of 10)
2. There are several subtraction questions with a difference equal to 5 (set the subtracted number to 10): 10-5=5 9-4=5 ……()
## usage method:
### 2. py code
```# -*- coding: utf-8 -*-
"""
@author: Ma Qingxin
@file: Asha modification exercise
How many questions are there in addition questions with sum equal to 5.py
How many subtraction questions with a difference equal to 5(Set subtracted).py
@time: 2022/2/3 20:15
# Special note
1.There are two number addition questions with an answer equal to 5. There are several questions in total (within 5))？
2.How many questions are there for the two digit subtraction questions with an answer equal to 5?(Set subtracted)
"""
import random
from re import X
from tkinter import Y
#Set questions:
choice=int(input('-----------------------------------\n'
'You need X Integer addition or integer subtraction within?\n'
'------------------------------------\n'))
regNum = int(input('Please enter the number of formulas to be generated:\n'))# Enter the required quantity
# Generate a list from 0 to the arithmetic maximum sum. Because the range function package does not include the right, if you want to include the arithmetic maximum sum, you must + 1
# The range function starts from 0, so that the list elements are consistent with the list index, reducing the difficulty of the algorithm
numList = [x for x in range(0,plusAnswer+1)]# Establish an integer list. Assume that summax = 20, summax + 1 equals 21, but the actual range is 1-20. Ensure that the list index is consistent with the list elements
# The resultList list is used to save the compliance arithmetic expression of the last demand quantity
resultList = []# List of stored results
# Cycle from the second element of the list to the last element of the list
for x in numList[0:]:# 0 equals the number 1
# Cycle from the x-th element of the list to the last element of the list
for y in numList[x:]:
# In the first addition, xy is equal and X + y < 20. At the beginning of the list element, there are two equal elements. Because the addition exchange law will produce repetition, it is distinguished separately
tempStr = str(x) + ' + ' + str(y) + ' = ' #Addition type 1 x+y = (x=y)
resultList.append((tempStr, (x+y)))# For questions with answers, such as 2 + 2 = 4, lines 70 and 72, select 0 or 1 to print the results with or without answers
tempStr = str(y) + ' + ' + str(x) + ' = ' #Addition type 1 x+y = (x=y)
resultList.append((tempStr, (y+x)))# Questions with answers, such as 2 + 2 = 4, line 70 and
selectList = []# Selection list
if regNum > len(resultList):#If (quantity of input formula) is greater than (quantity of result list)#If the number of input questions is greater than the actual demand, shuffle the cards,
print(f'Your demand is greater than the maximum formula generation quantity! The maximum number of generated formulas is{len(resultList)}') # Add the statistical value of the maximum non repeating formula. If the number of questions entered is greater than the number of lists in which the results are stored, add LEN to the list
i = len(resultList)# The number of i is equal to the number of lists in which the results are stored
for _ in resultList:# The value is in the list of cyclic storage results
# selectList = random.sample(resultList,i)
selectList.append(_) #To select the list, you need to add the contents of the circular storage result list
selectList.sort() #In positive order from small to large, 0 + 5 = 5, 1 + 4 = 5
# selectList.sort(reverse=True) #Flashback from large to small 5 + 0 = 5 4 + 1 = 5
# for _ in selectList:
# # Optional printing with and without answers 8
# else:#If the number of input questions is less than the actual demand, the code will be automatically selected at random and will not be sorted,
# i = regNum
# selectList = random.sample(resultList,i)
# for _ in selectList:
# # You can print with or without answers
# # print(f'{_[0]}{_[1]}')# With answer 2
print(f'The number of questions that generate non repeated addition formulas{len(selectList)}')# Select the number of lists
print(f'The limited number of maximum unrepeatable addition formulas{len(resultList)}')# Number of result lists
# Save as TXT file (in the default open folder, manually copy to Word 1)
str_title = 'Sum equals%d The addition questions are common%d topic.txt' % (plusAnswer, len(resultList)) # Save file name 60
with open(str_title,'w') as f:# Open TXT file
for a in selectList:# #Loop through the contents of the answer 878
# f.write(str(a[0])+'\n')# Display 0 + 5 = 1 + 4=
f.write(str(a[0])+str(a[1])+'\n')# Display 0 + 5 = 5 1 + 4 = 5
f.close()# #Turn off TXT
elif choice==2: # subtraction
subAnswer = int(input('Please enter the answer you want to subtract: '))# Enter the number of answers you want for the subtraction question, such as 5. For example, how many questions are there for the question with the answer of 5
sumMax=int(input('Please enter the number of subtracted (maximum): '))# For example, 10-5 = 5, 9-4 = 5, input the first number X10, 9 of the subtraction question
regNum = int(input('Please enter the number of formulas to be generated: '))# Enter the required quantity
# Generate a list from 0 to the arithmetic maximum sum. Because the range function package does not include the right, if you want to include the arithmetic maximum sum, you must + 1
# The range function starts from 0, so that the list elements are consistent with the list index, reducing the difficulty of the algorithm
numList = [x for x in range(0,sumMax+1)]# Establish an integer list with a value range of 5-10
# The resultList list is used to save the compliance arithmetic expression of the last demand quantity
resultList = []# List of stored results
# Cycle from the second element of the list to the last element of the list
for x in numList[0:]:# 0 equals the number 1
# Cycle from the x-th element of the list to the last element of the list
for y in numList[x:]:
# The first subtraction, xy is not equal, x > y
if x - y == subAnswer and y!=x:# The first condition of y-x is x+y < 20, and X is not equal to y (equal to 0)
tempStr = str(x) + ' - ' + str(y) + ' = '#Subtraction type 1 y-x=
resultList.append((tempStr, (x-y)))# Questions with answers, such as 3-2 = 1,
elif y - x ==subAnswer: #Subtraction conditions include y+x less than 20 and Y - x > 0
tempStr = str(y) + ' - ' + str(x) + ' = '#Subtraction type 2 y-x = (second condition of y-x Y-X > 0)
resultList.append((tempStr, (y-x)))# Questions with answers, such as 2 + 2 = 4,
selectList = []# Selection list
if regNum > len(resultList):#If (quantity of input formula) is greater than (quantity of result list)#If the number of input questions is greater than the actual demand, shuffle the cards,
print(f'Your demand is greater than the maximum formula generation quantity! The maximum number of generated formulas is{len(resultList)}') # Add the statistical value of the maximum non repeating formula. If the number of questions entered is greater than the number of lists in which the results are stored, add LEN to the list
i = len(resultList)# The number of i is equal to the number of lists in which the results are stored
for _ in resultList:# The value is in the list of cyclic storage results
# selectList = random.sample(resultList,i)
selectList.append(_) #To select the list, you need to add the contents of the circular storage result list
# At this time, it is arranged in positive order from small to large, 5 - 0 = 5, 6 - 1 = 5
sorted(selectList)#Positive order from small to large 5 - 0 = 5 6 - 1 = 5
# reversed(selectList)#Positive order from small to large 5 - 0 = 5 6 - 1 = 5
for _ in selectList:
# Optional printing with and without answers 8
else:#If the number of input questions is less than the actual demand, the code will be automatically selected at random and will not be sorted,
i = regNum
selectList = random.sample(resultList,i)
for _ in selectList:
# You can print with or without answers
# Verify the number of generated formulas 70
print(f'The number of questions that generate non repeated subtraction formulas{len(selectList)}')# Select the number of lists
print(f'The maximum number of non repeating subtractions{len(resultList)}')# Number of result lists
# Save as TXT file (in the default open folder, manually copy to Word 1)
str_title = 'Difference equals%d Subtraction (subtraction)%d)share%d topic.txt' % (subAnswer, sumMax,len(resultList))
with open(str_title,'w') as f:# Open TXT file
for a in selectList:# #Cycle through the contents of the answer 8
# f.write(str(a[0])+'\n')# Display 10-5 = 9-5=
f.write(str(a[0])+str(a[1])+'\n')# Display 10-5 = 5 9-5 = 5
f.close()# #Turn off TXT
else:
print('The function you entered does not exist.')
```
## Long term objectives:
This code design is mainly to prepare for the problem type of division and combination of mathematics in large class within 10. Through fine-tuning the code content, we can further interpret and understand the content of Mr. Ma's source code, so as to provide the basis for subsequent design. Thank you, Mr. Ma. | 2,572 | 9,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.847564 |
http://mathhelpforum.com/advanced-algebra/32775-echelon-form-print.html | 1,529,559,411,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00292.warc.gz | 207,562,134 | 3,007 | # Echelon Form
• Mar 31st 2008, 09:13 PM
ah-bee
Echelon Form
What is the precise definition of a non-square matrix in echelon form? Say we have something like:
1, 2, 3, 4, 5, 6 | 7
2, 3, 4, 5, 6 ,7 | 8
3, 4 ,5, 6, 7, 8 | 9
How would the echelon form of that look like? Don't worry about explaining that set of equations having so and so amount of parameters and infinite solutions. I'm just wondering how the echelon form of a non-square matrix is defined.
• Mar 31st 2008, 09:18 PM
mr fantastic
Quote:
Originally Posted by ah-bee
What is the precise definition of a non-square matrix in echelon form? Say we have something like:
1, 2, 3, 4, 5, 6 | 7
2, 3, 4, 5, 6 ,7 | 8
3, 4 ,5, 6, 7, 8 | 9
How would the echelon form of that look like? Don't worry about explaining that set of equations having so and so amount of parameters and infinite solutions. I'm just wondering how the echelon form of a non-square matrix is defined.
It might look something like
* * * * * * | *
0 0 * * * * | *
0 0 0 0 0 * | *
for example where * denotes non-zero entries.
• Mar 31st 2008, 09:29 PM
ah-bee
does that rule hold for every non-square matrix though? having the last row in the matrix being zeros except for the last term and is there any restrictions to how many zeros u can have in the preceding row? in your case u have used 0 0 **** but not 0 0 0 ***, why is this?
• Apr 1st 2008, 02:01 AM
mr fantastic
Quote:
Originally Posted by ah-bee
does that rule hold for every non-square matrix though? having the last row in the matrix being zeros except for the last term and is there any restrictions to how many zeros u can have in the preceding row? in your case u have used 0 0 **** but not 0 0 0 ***, why is this?
It was only an example. I assumed you understood the technical definition of row echelon form - read this. | 558 | 1,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-26 | latest | en | 0.928263 |
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Theorem itgabs 23821
Description: The triangle inequality for integrals. (Contributed by Mario Carneiro, 25-Aug-2014.)
Hypotheses
Ref Expression
itgabs.1 ((𝜑𝑥𝐴) → 𝐵𝑉)
itgabs.2 (𝜑 → (𝑥𝐴𝐵) ∈ 𝐿1)
Assertion
Ref Expression
itgabs (𝜑 → (abs‘∫𝐴𝐵 d𝑥) ≤ ∫𝐴(abs‘𝐵) d𝑥)
Distinct variable groups: 𝑥,𝐴 𝜑,𝑥 𝑥,𝑉
Allowed substitution hint: 𝐵(𝑥)
Proof of Theorem itgabs
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 itgabs.1 . . . . . . . . . . . 12 ((𝜑𝑥𝐴) → 𝐵𝑉)
2 itgabs.2 . . . . . . . . . . . 12 (𝜑 → (𝑥𝐴𝐵) ∈ 𝐿1)
31, 2itgcl 23770 . . . . . . . . . . 11 (𝜑 → ∫𝐴𝐵 d𝑥 ∈ ℂ)
43cjcld 14156 . . . . . . . . . 10 (𝜑 → (∗‘∫𝐴𝐵 d𝑥) ∈ ℂ)
5 iblmbf 23754 . . . . . . . . . . . . . . 15 ((𝑥𝐴𝐵) ∈ 𝐿1 → (𝑥𝐴𝐵) ∈ MblFn)
62, 5syl 17 . . . . . . . . . . . . . 14 (𝜑 → (𝑥𝐴𝐵) ∈ MblFn)
76, 1mbfmptcl 23624 . . . . . . . . . . . . 13 ((𝜑𝑥𝐴) → 𝐵 ∈ ℂ)
87ralrimiva 3105 . . . . . . . . . . . 12 (𝜑 → ∀𝑥𝐴 𝐵 ∈ ℂ)
9 nfv 1993 . . . . . . . . . . . . 13 𝑦 𝐵 ∈ ℂ
10 nfcsb1v 3691 . . . . . . . . . . . . . 14 𝑥𝑦 / 𝑥𝐵
1110nfel1 2918 . . . . . . . . . . . . 13 𝑥𝑦 / 𝑥𝐵 ∈ ℂ
12 csbeq1a 3684 . . . . . . . . . . . . . 14 (𝑥 = 𝑦𝐵 = 𝑦 / 𝑥𝐵)
1312eleq1d 2825 . . . . . . . . . . . . 13 (𝑥 = 𝑦 → (𝐵 ∈ ℂ ↔ 𝑦 / 𝑥𝐵 ∈ ℂ))
149, 11, 13cbvral 3307 . . . . . . . . . . . 12 (∀𝑥𝐴 𝐵 ∈ ℂ ↔ ∀𝑦𝐴 𝑦 / 𝑥𝐵 ∈ ℂ)
158, 14sylib 208 . . . . . . . . . . 11 (𝜑 → ∀𝑦𝐴 𝑦 / 𝑥𝐵 ∈ ℂ)
1615r19.21bi 3071 . . . . . . . . . 10 ((𝜑𝑦𝐴) → 𝑦 / 𝑥𝐵 ∈ ℂ)
17 nfcv 2903 . . . . . . . . . . . 12 𝑦𝐵
1817, 10, 12cbvmpt 4902 . . . . . . . . . . 11 (𝑥𝐴𝐵) = (𝑦𝐴𝑦 / 𝑥𝐵)
1918, 2syl5eqelr 2845 . . . . . . . . . 10 (𝜑 → (𝑦𝐴𝑦 / 𝑥𝐵) ∈ 𝐿1)
204, 16, 19iblmulc2 23817 . . . . . . . . 9 (𝜑 → (𝑦𝐴 ↦ ((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) ∈ 𝐿1)
214adantr 472 . . . . . . . . . . 11 ((𝜑𝑦𝐴) → (∗‘∫𝐴𝐵 d𝑥) ∈ ℂ)
2221, 16mulcld 10273 . . . . . . . . . 10 ((𝜑𝑦𝐴) → ((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵) ∈ ℂ)
2322iblcn 23785 . . . . . . . . 9 (𝜑 → ((𝑦𝐴 ↦ ((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) ∈ 𝐿1 ↔ ((𝑦𝐴 ↦ (ℜ‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵))) ∈ 𝐿1 ∧ (𝑦𝐴 ↦ (ℑ‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵))) ∈ 𝐿1)))
2420, 23mpbid 222 . . . . . . . 8 (𝜑 → ((𝑦𝐴 ↦ (ℜ‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵))) ∈ 𝐿1 ∧ (𝑦𝐴 ↦ (ℑ‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵))) ∈ 𝐿1))
2524simpld 477 . . . . . . 7 (𝜑 → (𝑦𝐴 ↦ (ℜ‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵))) ∈ 𝐿1)
26 ovexd 6845 . . . . . . . 8 ((𝜑𝑦𝐴) → ((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵) ∈ V)
2726, 20iblabs 23815 . . . . . . 7 (𝜑 → (𝑦𝐴 ↦ (abs‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵))) ∈ 𝐿1)
2822recld 14154 . . . . . . 7 ((𝜑𝑦𝐴) → (ℜ‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) ∈ ℝ)
2922abscld 14395 . . . . . . 7 ((𝜑𝑦𝐴) → (abs‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) ∈ ℝ)
3022releabsd 14410 . . . . . . 7 ((𝜑𝑦𝐴) → (ℜ‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) ≤ (abs‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)))
3125, 27, 28, 29, 30itgle 23796 . . . . . 6 (𝜑 → ∫𝐴(ℜ‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) d𝑦 ≤ ∫𝐴(abs‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) d𝑦)
323abscld 14395 . . . . . . . . 9 (𝜑 → (abs‘∫𝐴𝐵 d𝑥) ∈ ℝ)
3332recnd 10281 . . . . . . . 8 (𝜑 → (abs‘∫𝐴𝐵 d𝑥) ∈ ℂ)
3433sqvald 13220 . . . . . . 7 (𝜑 → ((abs‘∫𝐴𝐵 d𝑥)↑2) = ((abs‘∫𝐴𝐵 d𝑥) · (abs‘∫𝐴𝐵 d𝑥)))
353absvalsqd 14401 . . . . . . . . . 10 (𝜑 → ((abs‘∫𝐴𝐵 d𝑥)↑2) = (∫𝐴𝐵 d𝑥 · (∗‘∫𝐴𝐵 d𝑥)))
363, 4mulcomd 10274 . . . . . . . . . 10 (𝜑 → (∫𝐴𝐵 d𝑥 · (∗‘∫𝐴𝐵 d𝑥)) = ((∗‘∫𝐴𝐵 d𝑥) · ∫𝐴𝐵 d𝑥))
3712, 17, 10cbvitg 23762 . . . . . . . . . . . 12 𝐴𝐵 d𝑥 = ∫𝐴𝑦 / 𝑥𝐵 d𝑦
3837oveq2i 6826 . . . . . . . . . . 11 ((∗‘∫𝐴𝐵 d𝑥) · ∫𝐴𝐵 d𝑥) = ((∗‘∫𝐴𝐵 d𝑥) · ∫𝐴𝑦 / 𝑥𝐵 d𝑦)
394, 16, 19itgmulc2 23820 . . . . . . . . . . 11 (𝜑 → ((∗‘∫𝐴𝐵 d𝑥) · ∫𝐴𝑦 / 𝑥𝐵 d𝑦) = ∫𝐴((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵) d𝑦)
4038, 39syl5eq 2807 . . . . . . . . . 10 (𝜑 → ((∗‘∫𝐴𝐵 d𝑥) · ∫𝐴𝐵 d𝑥) = ∫𝐴((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵) d𝑦)
4135, 36, 403eqtrd 2799 . . . . . . . . 9 (𝜑 → ((abs‘∫𝐴𝐵 d𝑥)↑2) = ∫𝐴((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵) d𝑦)
4241fveq2d 6358 . . . . . . . 8 (𝜑 → (ℜ‘((abs‘∫𝐴𝐵 d𝑥)↑2)) = (ℜ‘∫𝐴((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵) d𝑦))
4332resqcld 13250 . . . . . . . . 9 (𝜑 → ((abs‘∫𝐴𝐵 d𝑥)↑2) ∈ ℝ)
4443rered 14184 . . . . . . . 8 (𝜑 → (ℜ‘((abs‘∫𝐴𝐵 d𝑥)↑2)) = ((abs‘∫𝐴𝐵 d𝑥)↑2))
4526, 20itgre 23787 . . . . . . . 8 (𝜑 → (ℜ‘∫𝐴((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵) d𝑦) = ∫𝐴(ℜ‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) d𝑦)
4642, 44, 453eqtr3d 2803 . . . . . . 7 (𝜑 → ((abs‘∫𝐴𝐵 d𝑥)↑2) = ∫𝐴(ℜ‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) d𝑦)
4734, 46eqtr3d 2797 . . . . . 6 (𝜑 → ((abs‘∫𝐴𝐵 d𝑥) · (abs‘∫𝐴𝐵 d𝑥)) = ∫𝐴(ℜ‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) d𝑦)
4812fveq2d 6358 . . . . . . . . 9 (𝑥 = 𝑦 → (abs‘𝐵) = (abs‘𝑦 / 𝑥𝐵))
49 nfcv 2903 . . . . . . . . 9 𝑦(abs‘𝐵)
50 nfcv 2903 . . . . . . . . . 10 𝑥abs
5150, 10nffv 6361 . . . . . . . . 9 𝑥(abs‘𝑦 / 𝑥𝐵)
5248, 49, 51cbvitg 23762 . . . . . . . 8 𝐴(abs‘𝐵) d𝑥 = ∫𝐴(abs‘𝑦 / 𝑥𝐵) d𝑦
5352oveq2i 6826 . . . . . . 7 ((abs‘∫𝐴𝐵 d𝑥) · ∫𝐴(abs‘𝐵) d𝑥) = ((abs‘∫𝐴𝐵 d𝑥) · ∫𝐴(abs‘𝑦 / 𝑥𝐵) d𝑦)
5416abscld 14395 . . . . . . . . 9 ((𝜑𝑦𝐴) → (abs‘𝑦 / 𝑥𝐵) ∈ ℝ)
5516, 19iblabs 23815 . . . . . . . . 9 (𝜑 → (𝑦𝐴 ↦ (abs‘𝑦 / 𝑥𝐵)) ∈ 𝐿1)
5633, 54, 55itgmulc2 23820 . . . . . . . 8 (𝜑 → ((abs‘∫𝐴𝐵 d𝑥) · ∫𝐴(abs‘𝑦 / 𝑥𝐵) d𝑦) = ∫𝐴((abs‘∫𝐴𝐵 d𝑥) · (abs‘𝑦 / 𝑥𝐵)) d𝑦)
5721, 16absmuld 14413 . . . . . . . . . 10 ((𝜑𝑦𝐴) → (abs‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) = ((abs‘(∗‘∫𝐴𝐵 d𝑥)) · (abs‘𝑦 / 𝑥𝐵)))
583adantr 472 . . . . . . . . . . . 12 ((𝜑𝑦𝐴) → ∫𝐴𝐵 d𝑥 ∈ ℂ)
5958abscjd 14409 . . . . . . . . . . 11 ((𝜑𝑦𝐴) → (abs‘(∗‘∫𝐴𝐵 d𝑥)) = (abs‘∫𝐴𝐵 d𝑥))
6059oveq1d 6830 . . . . . . . . . 10 ((𝜑𝑦𝐴) → ((abs‘(∗‘∫𝐴𝐵 d𝑥)) · (abs‘𝑦 / 𝑥𝐵)) = ((abs‘∫𝐴𝐵 d𝑥) · (abs‘𝑦 / 𝑥𝐵)))
6157, 60eqtrd 2795 . . . . . . . . 9 ((𝜑𝑦𝐴) → (abs‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) = ((abs‘∫𝐴𝐵 d𝑥) · (abs‘𝑦 / 𝑥𝐵)))
6261itgeq2dv 23768 . . . . . . . 8 (𝜑 → ∫𝐴(abs‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) d𝑦 = ∫𝐴((abs‘∫𝐴𝐵 d𝑥) · (abs‘𝑦 / 𝑥𝐵)) d𝑦)
6356, 62eqtr4d 2798 . . . . . . 7 (𝜑 → ((abs‘∫𝐴𝐵 d𝑥) · ∫𝐴(abs‘𝑦 / 𝑥𝐵) d𝑦) = ∫𝐴(abs‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) d𝑦)
6453, 63syl5eq 2807 . . . . . 6 (𝜑 → ((abs‘∫𝐴𝐵 d𝑥) · ∫𝐴(abs‘𝐵) d𝑥) = ∫𝐴(abs‘((∗‘∫𝐴𝐵 d𝑥) · 𝑦 / 𝑥𝐵)) d𝑦)
6531, 47, 643brtr4d 4837 . . . . 5 (𝜑 → ((abs‘∫𝐴𝐵 d𝑥) · (abs‘∫𝐴𝐵 d𝑥)) ≤ ((abs‘∫𝐴𝐵 d𝑥) · ∫𝐴(abs‘𝐵) d𝑥))
6665adantr 472 . . . 4 ((𝜑 ∧ 0 < (abs‘∫𝐴𝐵 d𝑥)) → ((abs‘∫𝐴𝐵 d𝑥) · (abs‘∫𝐴𝐵 d𝑥)) ≤ ((abs‘∫𝐴𝐵 d𝑥) · ∫𝐴(abs‘𝐵) d𝑥))
6732adantr 472 . . . . 5 ((𝜑 ∧ 0 < (abs‘∫𝐴𝐵 d𝑥)) → (abs‘∫𝐴𝐵 d𝑥) ∈ ℝ)
687abscld 14395 . . . . . . 7 ((𝜑𝑥𝐴) → (abs‘𝐵) ∈ ℝ)
691, 2iblabs 23815 . . . . . . 7 (𝜑 → (𝑥𝐴 ↦ (abs‘𝐵)) ∈ 𝐿1)
7068, 69itgrecl 23784 . . . . . 6 (𝜑 → ∫𝐴(abs‘𝐵) d𝑥 ∈ ℝ)
7170adantr 472 . . . . 5 ((𝜑 ∧ 0 < (abs‘∫𝐴𝐵 d𝑥)) → ∫𝐴(abs‘𝐵) d𝑥 ∈ ℝ)
72 simpr 479 . . . . 5 ((𝜑 ∧ 0 < (abs‘∫𝐴𝐵 d𝑥)) → 0 < (abs‘∫𝐴𝐵 d𝑥))
73 lemul2 11089 . . . . 5 (((abs‘∫𝐴𝐵 d𝑥) ∈ ℝ ∧ ∫𝐴(abs‘𝐵) d𝑥 ∈ ℝ ∧ ((abs‘∫𝐴𝐵 d𝑥) ∈ ℝ ∧ 0 < (abs‘∫𝐴𝐵 d𝑥))) → ((abs‘∫𝐴𝐵 d𝑥) ≤ ∫𝐴(abs‘𝐵) d𝑥 ↔ ((abs‘∫𝐴𝐵 d𝑥) · (abs‘∫𝐴𝐵 d𝑥)) ≤ ((abs‘∫𝐴𝐵 d𝑥) · ∫𝐴(abs‘𝐵) d𝑥)))
7467, 71, 67, 72, 73syl112anc 1481 . . . 4 ((𝜑 ∧ 0 < (abs‘∫𝐴𝐵 d𝑥)) → ((abs‘∫𝐴𝐵 d𝑥) ≤ ∫𝐴(abs‘𝐵) d𝑥 ↔ ((abs‘∫𝐴𝐵 d𝑥) · (abs‘∫𝐴𝐵 d𝑥)) ≤ ((abs‘∫𝐴𝐵 d𝑥) · ∫𝐴(abs‘𝐵) d𝑥)))
7566, 74mpbird 247 . . 3 ((𝜑 ∧ 0 < (abs‘∫𝐴𝐵 d𝑥)) → (abs‘∫𝐴𝐵 d𝑥) ≤ ∫𝐴(abs‘𝐵) d𝑥)
7675ex 449 . 2 (𝜑 → (0 < (abs‘∫𝐴𝐵 d𝑥) → (abs‘∫𝐴𝐵 d𝑥) ≤ ∫𝐴(abs‘𝐵) d𝑥))
777absge0d 14403 . . . 4 ((𝜑𝑥𝐴) → 0 ≤ (abs‘𝐵))
7869, 68, 77itgge0 23797 . . 3 (𝜑 → 0 ≤ ∫𝐴(abs‘𝐵) d𝑥)
79 breq1 4808 . . 3 (0 = (abs‘∫𝐴𝐵 d𝑥) → (0 ≤ ∫𝐴(abs‘𝐵) d𝑥 ↔ (abs‘∫𝐴𝐵 d𝑥) ≤ ∫𝐴(abs‘𝐵) d𝑥))
8078, 79syl5ibcom 235 . 2 (𝜑 → (0 = (abs‘∫𝐴𝐵 d𝑥) → (abs‘∫𝐴𝐵 d𝑥) ≤ ∫𝐴(abs‘𝐵) d𝑥))
813absge0d 14403 . . 3 (𝜑 → 0 ≤ (abs‘∫𝐴𝐵 d𝑥))
82 0re 10253 . . . 4 0 ∈ ℝ
83 leloe 10337 . . . 4 ((0 ∈ ℝ ∧ (abs‘∫𝐴𝐵 d𝑥) ∈ ℝ) → (0 ≤ (abs‘∫𝐴𝐵 d𝑥) ↔ (0 < (abs‘∫𝐴𝐵 d𝑥) ∨ 0 = (abs‘∫𝐴𝐵 d𝑥))))
8482, 32, 83sylancr 698 . . 3 (𝜑 → (0 ≤ (abs‘∫𝐴𝐵 d𝑥) ↔ (0 < (abs‘∫𝐴𝐵 d𝑥) ∨ 0 = (abs‘∫𝐴𝐵 d𝑥))))
8581, 84mpbid 222 . 2 (𝜑 → (0 < (abs‘∫𝐴𝐵 d𝑥) ∨ 0 = (abs‘∫𝐴𝐵 d𝑥)))
8676, 80, 85mpjaod 395 1 (𝜑 → (abs‘∫𝐴𝐵 d𝑥) ≤ ∫𝐴(abs‘𝐵) d𝑥)
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∨ wo 382 ∧ wa 383 = wceq 1632 ∈ wcel 2140 ∀wral 3051 Vcvv 3341 ⦋csb 3675 class class class wbr 4805 ↦ cmpt 4882 ‘cfv 6050 (class class class)co 6815 ℂcc 10147 ℝcr 10148 0cc0 10149 · cmul 10154 < clt 10287 ≤ cle 10288 2c2 11283 ↑cexp 13075 ∗ccj 14056 ℜcre 14057 ℑcim 14058 abscabs 14194 MblFncmbf 23603 𝐿1cibl 23606 ∫citg 23607 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1871 ax-4 1886 ax-5 1989 ax-6 2055 ax-7 2091 ax-8 2142 ax-9 2149 ax-10 2169 ax-11 2184 ax-12 2197 ax-13 2392 ax-ext 2741 ax-rep 4924 ax-sep 4934 ax-nul 4942 ax-pow 4993 ax-pr 5056 ax-un 7116 ax-inf2 8714 ax-cc 9470 ax-cnex 10205 ax-resscn 10206 ax-1cn 10207 ax-icn 10208 ax-addcl 10209 ax-addrcl 10210 ax-mulcl 10211 ax-mulrcl 10212 ax-mulcom 10213 ax-addass 10214 ax-mulass 10215 ax-distr 10216 ax-i2m1 10217 ax-1ne0 10218 ax-1rid 10219 ax-rnegex 10220 ax-rrecex 10221 ax-cnre 10222 ax-pre-lttri 10223 ax-pre-lttrn 10224 ax-pre-ltadd 10225 ax-pre-mulgt0 10226 ax-pre-sup 10227 ax-addf 10228 ax-mulf 10229 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3or 1073 df-3an 1074 df-tru 1635 df-fal 1638 df-ex 1854 df-nf 1859 df-sb 2048 df-eu 2612 df-mo 2613 df-clab 2748 df-cleq 2754 df-clel 2757 df-nfc 2892 df-ne 2934 df-nel 3037 df-ral 3056 df-rex 3057 df-reu 3058 df-rmo 3059 df-rab 3060 df-v 3343 df-sbc 3578 df-csb 3676 df-dif 3719 df-un 3721 df-in 3723 df-ss 3730 df-pss 3732 df-nul 4060 df-if 4232 df-pw 4305 df-sn 4323 df-pr 4325 df-tp 4327 df-op 4329 df-uni 4590 df-int 4629 df-iun 4675 df-iin 4676 df-disj 4774 df-br 4806 df-opab 4866 df-mpt 4883 df-tr 4906 df-id 5175 df-eprel 5180 df-po 5188 df-so 5189 df-fr 5226 df-se 5227 df-we 5228 df-xp 5273 df-rel 5274 df-cnv 5275 df-co 5276 df-dm 5277 df-rn 5278 df-res 5279 df-ima 5280 df-pred 5842 df-ord 5888 df-on 5889 df-lim 5890 df-suc 5891 df-iota 6013 df-fun 6052 df-fn 6053 df-f 6054 df-f1 6055 df-fo 6056 df-f1o 6057 df-fv 6058 df-isom 6059 df-riota 6776 df-ov 6818 df-oprab 6819 df-mpt2 6820 df-of 7064 df-ofr 7065 df-om 7233 df-1st 7335 df-2nd 7336 df-supp 7466 df-wrecs 7578 df-recs 7639 df-rdg 7677 df-1o 7731 df-2o 7732 df-oadd 7735 df-omul 7736 df-er 7914 df-map 8028 df-pm 8029 df-ixp 8078 df-en 8125 df-dom 8126 df-sdom 8127 df-fin 8128 df-fsupp 8444 df-fi 8485 df-sup 8516 df-inf 8517 df-oi 8583 df-card 8976 df-acn 8979 df-cda 9203 df-pnf 10289 df-mnf 10290 df-xr 10291 df-ltxr 10292 df-le 10293 df-sub 10481 df-neg 10482 df-div 10898 df-nn 11234 df-2 11292 df-3 11293 df-4 11294 df-5 11295 df-6 11296 df-7 11297 df-8 11298 df-9 11299 df-n0 11506 df-z 11591 df-dec 11707 df-uz 11901 df-q 12003 df-rp 12047 df-xneg 12160 df-xadd 12161 df-xmul 12162 df-ioo 12393 df-ioc 12394 df-ico 12395 df-icc 12396 df-fz 12541 df-fzo 12681 df-fl 12808 df-mod 12884 df-seq 13017 df-exp 13076 df-hash 13333 df-cj 14059 df-re 14060 df-im 14061 df-sqrt 14195 df-abs 14196 df-clim 14439 df-rlim 14440 df-sum 14637 df-struct 16082 df-ndx 16083 df-slot 16084 df-base 16086 df-sets 16087 df-ress 16088 df-plusg 16177 df-mulr 16178 df-starv 16179 df-sca 16180 df-vsca 16181 df-ip 16182 df-tset 16183 df-ple 16184 df-ds 16187 df-unif 16188 df-hom 16189 df-cco 16190 df-rest 16306 df-topn 16307 df-0g 16325 df-gsum 16326 df-topgen 16327 df-pt 16328 df-prds 16331 df-xrs 16385 df-qtop 16390 df-imas 16391 df-xps 16393 df-mre 16469 df-mrc 16470 df-acs 16472 df-mgm 17464 df-sgrp 17506 df-mnd 17517 df-submnd 17558 df-mulg 17763 df-cntz 17971 df-cmn 18416 df-psmet 19961 df-xmet 19962 df-met 19963 df-bl 19964 df-mopn 19965 df-cnfld 19970 df-top 20922 df-topon 20939 df-topsp 20960 df-bases 20973 df-cn 21254 df-cnp 21255 df-cmp 21413 df-tx 21588 df-hmeo 21781 df-xms 22347 df-ms 22348 df-tms 22349 df-cncf 22903 df-ovol 23454 df-vol 23455 df-mbf 23608 df-itg1 23609 df-itg2 23610 df-ibl 23611 df-itg 23612 df-0p 23657 This theorem is referenced by: ftc1a 24020 ftc1lem4 24022 itgulm 24382 fourierdlem47 40892 fourierdlem87 40932 etransclem23 40996
Copyright terms: Public domain W3C validator | 8,614 | 11,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-27 | latest | en | 0.111011 |
https://www.geeksforgeeks.org/python-subgroups-of-ith-index-size-in-list/ | 1,642,758,028,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302740.94/warc/CC-MAIN-20220121071203-20220121101203-00419.warc.gz | 683,774,173 | 23,177 | # Python | Subgroups of i’th index size in list
• Last Updated : 18 Mar, 2019
Sometimes we need to group elements and grouping techniques and requirements vary accordingly. One such way to group the elements is by the i’th size in list which stores the dictionary of index keys with values as list elements of subsequent size i.
Input : [4, 7, 8, 10, 12, 15, 13, 17, 14, 5]
Output: {1: [4], 2: [7, 8], 3: [10, 12, 15], 4: [13, 17, 14, 5]}
Let’s discuss certain ways in which this can be done.
Method #1 : Using `islice()` + dictionary comprehension
The slice method can be used to group the chunks of list required to be made as values of the dictionaries which are then assigned to their destined index key using the dictionary comprehension.
`# Python3 code to demonstrate``# Subgrouping of i'th index size in list``# using islice() + dictionary comprehension``from` `itertools ``import` `islice`` ` `# initializing list``test_list ``=` `[``4``, ``7``, ``8``, ``10``, ``12``, ``15``, ``13``, ``17``, ``14``, ``5``]`` ` `# printing original list ``print``(``"The original list : "` `+` `str``(test_list))`` ` `# using islice() + dictionary comprehension``# Subgrouping of i'th index size in list``temp ``=` `iter``(test_list)``res ``=` `{key: val ``for` `key, val ``in` `((i, ``list``(islice(temp, i)))`` ``for` `i ``in` `range``(``1``, ``len``(test_list))) ``if` `val}`` ` `# printing result``print``(``"The grouped dictionary is : "` `+` `str``(res))`
Output :
The original list : [4, 7, 8, 10, 12, 15, 13, 17, 14, 5]
The grouped dictionary is : {1: [4], 2: [7, 8], 3: [10, 12, 15], 4: [13, 17, 14, 5]}
Method #2 : Using `itemgetter() + takewhile() + islice()`
In order to increase the computation speed, we introduce new functions to perform this particular task, takewhile and itemgetter functions which performs the task of grouping the sliced values.
`# Python3 code to demonstrate``# Subgrouping of i'th index size in list``# using itemgetter() + takewhile() + islice()``from` `itertools ``import` `islice, takewhile``from` `operator ``import` `itemgetter`` ` `# initializing list``test_list ``=` `[``4``, ``7``, ``8``, ``10``, ``12``, ``15``, ``13``, ``17``, ``14``, ``5``]`` ` `# printing original list ``print``(``"The original list : "` `+` `str``(test_list))`` ` `# using itemgetter() + takewhile() + islice()``# Subgrouping of i'th index size in list``temp ``=` `iter``(test_list)``res ``=` `{key: val ``for` `key, val ``in` ` ``takewhile(itemgetter(``1``), ((i, ``list``(islice(temp, i)))`` ``for` `i ``in` `range``(``1``, ``len``(test_list))))}`` ` `# printing result``print``(``"The grouped dictionary is : "` `+` `str``(res))`
Output :
The original list : [4, 7, 8, 10, 12, 15, 13, 17, 14, 5]
The grouped dictionary is : {1: [4], 2: [7, 8], 3: [10, 12, 15], 4: [13, 17, 14, 5]}
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http://dmtcs.episciences.org/volume/view/id/97 | 1,511,402,027,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806715.73/warc/CC-MAIN-20171123012207-20171123032207-00109.warc.gz | 85,055,341 | 9,847 | # Vol. 14 no. 2
### 1. The competition number of a generalized line graph is at most two
In 1982, Opsut showed that the competition number of a line graph is at most two and gave a necessary and sufficient condition for the competition number of a line graph being one. In this paper, we generalize this result to the competition numbers of generalized line graphs, that is, we show that the competition number of a generalized line graph is at most two, and give necessary conditions and sufficient conditions for the competition number of a generalized line graph being one.
Section: Graph Theory
### 2. On bipartite powers of bigraphs
The notion of graph powers is a well-studied topic in graph theory and its applications. In this paper, we investigate a bipartite analogue of graph powers, which we call bipartite powers of bigraphs. We show that the classes of bipartite permutation graphs and interval bigraphs are closed under taking bipartite power. We also show that the problem of recognizing bipartite powers is NP-complete in general.
Section: Graph Theory
### 3. On neighbour-distinguishing colourings from lists
An edge colouring of a graph is said to be neighbour-distinguishing if any two adjacent vertices have distinct sets of colours of their incident edges. In this paper the list version of the problem of determining the minimum number of colours in a neighbour-distinguishing colouring of a given graph is considered.
Section: Graph and Algorithms
### 4. Random Horn formulas and propagation connectivity for directed hypergraphs
We consider the property that in a random definite Horn formula of size-3 clauses over n variables, where every such clause is included with probability p, there is a pair of variables for which forward chaining produces all other variables. We show that with high probability the property does not hold for p <= 1/(11n ln n), and does hold for p >= (5 1n ln n)/(n ln n).
Section: Combinatorics
### 5. Some results on stable sets for k-colorable P₆-free graphs and generalizations
This article deals with the Maximum Weight Stable Set (MWS) problem (and some other related NP-hard problems) and the class of P-6-free graphs. The complexity status of MWS is open for P-6-free graphs and is open even for P-5-free graphs (as a long standing open problem). Several results are known for MWS on subclasses of P-5-free: in particular, MWS can be solved for k-colorable P-5-free graphs in polynomial time for every k (depending on k) and more generally for (P-5, K-p)-free graphs […]
Section: Graph and Algorithms
### 6. On paths, trails and closed trails in edge-colored graphs
In this paper we deal from an algorithmic perspective with different questions regarding properly edge-colored (or PEC) paths, trails and closed trails. Given a c-edge-colored graph G(c), we show how to polynomially determine, if any, a PEC closed trail subgraph whose number of visits at each vertex is specified before hand. As a consequence, we solve a number of interesting related problems. For instance, given subset S of vertices in G(c), we show how to maximize in polynomial time the number […]
Section: Graph Theory
### 7. Graphs with many vertex-disjoint cycles
We study graphs G in which the maximum number of vertex-disjoint cycles nu(G) is close to the cyclomatic number mu(G), which is a natural upper bound for nu(G). Our main result is the existence of a finite set P(k) of graphs for all k is an element of N-0 such that every 2-connected graph G with mu(G)-nu(G) = k arises by applying a simple extension rule to a graph in P(k). As an algorithmic consequence we describe algorithms calculating minmu(G)-nu(G), k + 1 in linear time for fixed k.
Section: Graph Theory
### 8. Random Cayley digraphs of diameter 2 and given degree
We consider random Cayley digraphs of order n with uniformly distributed generating sets of size k. Specifically, we are interested in the asymptotics of the probability that such a Cayley digraph has diameter two as n -> infinity and k = f(n), focusing on the functions f(n) = left perpendicularn(delta)right perpendicular and f(n) = left perpendicularcnright perpendicular. In both instances we show that this probability converges to 1 as n -> infinity for arbitrary fixed delta is an […]
Section: Graph Theory
### 9. The asymmetric leader election algorithm with Swedish stopping: a probabilistic analysis
We study a leader election protocol that we call the Swedish leader election protocol. This name comes from a protocol presented by L. Bondesson, T. Nilsson, and G. Wikstrand (2007). The goal is to select one among n > 0 players, by proceeding through a number of rounds. If there is only one player remaining, the protocol stops and the player is declared the leader. Otherwise, all remaining players flip a biased coin; with probability q the player survives to the next round, with probability […]
Section: Analysis of Algorithms
### 10. The analysis of find and versions of it
In the running time analysis of the algorithm Find and versions of it appear as limiting distributions solutions of stochastic fixed points equation of the form X D = Sigma(i) AiXi o Bi + C on the space D of cadlag functions. The distribution of the D-valued process X is invariant by some random linear affine transformation of space and random time change. We show the existence of solutions in some generality via the Weighted Branching Process. Finite exponential moments are connected to […]
Section: Analysis of Algorithms
### 11. Immersion containment and connectivity in color-critical graphs
The relationship between graph coloring and the immersion order is considered. Vertex connectivity, edge connectivity and related issues are explored. It is shown that a t-chromatic graph G contains either an immersed Kt or an immersed t-chromatic subgraph that is both 4-vertex-connected and t-edge-connected. This gives supporting evidence of our conjecture that if G requires at least t colors, then Kt is immersed in G.
Section: Graph Theory
### 12. Acyclic chromatic index of fully subdivided graphs and Halin graphs
An acyclic edge coloring of a graph is a proper edge coloring such that there are no bichromatic cycles. The acyclic chromatic index of a graph is the minimum number k such that there is an acyclic edge coloring using k colors and is denoted by a'(G). A graph G is called fully subdivided if it is obtained from another graph H by replacing every edge by a path of length at least two. Fully subdivided graphs are known to be acyclically edge colorable using Δ+1 colors since they are properly […]
Section: Graph Theory
### 13. On 4-valent Frobenius circulant graphs
A 4-valent first-kind Frobenius circulant graph is a connected Cayley graph DLn(1, h) = Cay(Zn, H) on the additive group of integers modulo n, where each prime factor of n is congruent to 1 modulo 4 and H = {[1], [h], −[1], −[h]} with h a solution to the congruence equation x 2 + 1 ≡ 0 (mod n). In [A. Thomson and S. Zhou, Frobenius circulant graphs of valency four, J. Austral. Math. Soc. 85 (2008), 269-282] it was proved that such graphs admit 'perfect ' routing and gossiping schemes in some […]
Section: Graph Theory
### 14. Digraph complexity measures and applications in formal language theory
We investigate structural complexity measures on digraphs, in particular the cycle rank. This concept is intimately related to a classical topic in formal language theory, namely the star height of regular languages. We explore this connection, and obtain several new algorithmic insights regarding both cycle rank and star height. Among other results, we show that computing the cycle rank is NP-complete, even for sparse digraphs of maximum outdegree 2. Notwithstanding, we provide both a […]
Section: Automata, Logic and Semantics
### 15. A note on planar Ramsey numbers for a triangle versus wheels
For two given graphs G and H , the planar Ramsey number P R ( G; H ) is the smallest integer n such that every planar graph F on n vertices either contains a copy of G , or its complement contains a copy of H . In this paper, we determine all planar Ramsey numbers for a triangle versus wheels.
Section: Graph Theory
### 16. On quadratic threshold CSPs
A predicate P: {-1, 1}k →{0, 1} can be associated with a constraint satisfaction problem Max CSP(P). P is called ''approximation resistant'' if Max CSP(P) cannot be approximated better than the approximation obtained by choosing a random assignment, and ''approximable'' otherwise. This classification of predicates has proved to be an important and challenging open problem. Motivated by a recent result of Austrin and Mossel (Computational Complexity, 2009), we consider a natural subclass of […]
Section: Discrete Algorithms
### 17. Random graphs with bounded maximum degree: asymptotic structure and a logical limit law
For any fixed integer R≥2 we characterise the typical structure of undirected graphs with vertices 1,...,n and maximum degree R, as n tends to infinity. The information is used to prove that such graphs satisfy a labelled limit law for first-order logic. If R≥5 then also an unlabelled limit law holds.
### 18. On the algebraic numbers computable by some generalized Ehrenfest urns
This article deals with some stochastic population protocols, motivated by theoretical aspects of distributed computing. We modelize the problem by a large urn of black and white balls from which at every time unit a fixed number of balls are drawn and their colors are changed according to the number of black balls among them. When the time and the number of balls both tend to infinity the proportion of black balls converges to an algebraic number. We prove that, surprisingly enough, not every […]
### 19. Secure frameproof codes through biclique covers
For a binary code Γ of length v, a v-word w produces by a set of codewords {w1,...,wr}⊆Γ if for all i=1,...,v, we have wi∈{w1i,...,wri} . We call a code r-secure frameproof of size t if |Γ|=t and for any v-word that is produced by two sets C1 and C2 of size at most r then the intersection of these sets is nonempty. A d-biclique cover of size v of a graph G is a collection of v-complete bipartite subgraphs of G such that each edge of G belongs to at least d of these complete bipartite subgraphs. […]
Section: Graph Theory
### 20. Upper k-tuple domination in graphs
For a positive integer k, a k-tuple dominating set of a graph G is a subset S of V (G) such that |N [v] ∩ S| ≥ k for every vertex v, where N [v] = {v} ∪ {u ∈ V (G) : uv ∈ E(G)}. The upper k-tuple domination number of G, denoted by Γ×k (G), is the maximum cardinality of a minimal k-tuple dominating set of G. In this paper we present an upper bound on Γ×k (G) for r-regular graphs G with r ≥ k, and characterize extremal graphs achieving the upper bound. We also establish an upper bound on Γ×2 (G) […]
Section: Graph Theory | 2,520 | 10,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-47 | latest | en | 0.928365 |
https://seanet2017.com/calculator-390 | 1,674,945,576,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00841.warc.gz | 534,365,347 | 3,639 | Math homework help answers is a software program that supports students solve math problems. Our website can solving math problem.
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Scatter plot solver Fundamental trigonometric identities solver Solve the compound inequality Slope intercept form to standard form solver Purple math equation solver | 729 | 3,601 | {"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} | 4.6875 | 5 | CC-MAIN-2023-06 | latest | en | 0.963175 |
https://www.physicsforums.com/threads/decimal-precision-multiplying-rounding-and-adding.987224/ | 1,718,486,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00347.warc.gz | 825,719,171 | 19,104 | # Decimal Precision - Multiplying, Rounding, and Adding
• I
• pob
In summary, the conversation discusses the possibility of creating a value of $150.01 using two or more factors that are multiplied by$150 and rounded to two decimal places. The participants explore different scenarios and conclude that it is possible to create a value of $150.01 with certain combinations of factors and rounding. However, the conversation also mentions that in most cases, the sum of the rounded products will equal$150.
pob
Hi,
I have $150. This value is multiplied by 2 or more factors. Each factor is less than or equal to 1 and greater than or equal to 0. Each factor has a maximum of 9 decimal places. The sum of the factors equals 1. For example, the following set of factors meets these 3 conditions: 0.500033333 0.499966667 So does this set of factors: 0.099703630 0.095107000 0.035644140 0.264757680 0.050352750 0.144806740 0.145405230 0.110790870 0.053431960 Suppose I perform the following operations: 1. Multiply the$150 by each of the factors and store each product, storing up to 13 decimal places
2. Round each product from #1 to 2 decimal places
3. Sum the rounded products
Is it always true, that the sum of the rounded products will equal $150? Or is it possible to create a value such as$150.01?
So far, I've worked the problem backwards by supposing I had 2 factors that resulted in the following unrounded products:
75.005
74.995
And the following rounded products sum to $150.01, 75.01 75.00 However, I find it impossible to produce the products 75.005 and 74.995 using factors limited to 9 decimal places. The closest I get are the following factors: 0.500033333 0.499966667 Which, when multiplied by$150, yield the following products. These rounded products also sum to $150. 75.00499995 74.99500005 Given this example, I believe I've shown that, given 2 factors, the sum of the rounded products is always equal to$150. However, I'm trying to generalize this to any number of factors.
Thank you,
pob
You'll need at least three numbers I think.
0.33337 -> 50.0055 -> rounded to 50.01
0.33337 -> 50.0055 -> rounded to 50.01
0.33326 -> 49.9890 -> rounded to 49.99
Sum: 150.01.
All numbers are exact, I didn't add trailing zeroes.
0.9999 x 150 = 149.985, rounds to 149.99
0.0001 x 150 = 0.015, rounds to 0.02
pob and mfb
That's a good option I missed.
scottdave said:
0.0001 x 150 = 0.015, rounds to 0.02
It's noteworthy that we can use this 10,000 times and produce a sum of $200. We can also use 0.000 033 334, to give us a single cent, 29999 times. The remainder is not enough to give another cent, but the sum is$299.99. This is the maximum we can get. The minimum is $0, of course, just split it up enough. scottdave and pob And you could go more than 2 digits and find examples of inaccuracies. The fact is that any amount of rounding can lead to inaccurate results, given the proper conditions. Anyone who's worked with floating point arithmetic in computer programming should be familiar with this. Thank you for the answers pob said: Hi, I have$150. This value is multiplied by 2 or more factors. Each factor is less than or equal to 1 and greater than or equal to 0. Each factor has a maximum of 9 decimal places. The sum of the factors equals 1. For example, the following set of factors meets these 3 conditions:
0.500033333
0.499966667
So does this set of factors:
0.099703630
0.095107000
0.035644140
0.264757680
0.050352750
0.144806740
0.145405230
0.110790870
0.053431960
Suppose I perform the following operations:
1. Multiply the \$150 by each of the factors and store each product, storing up to 13 decimal places
2. Round each product from #1 to 2 decimal places
3. Sum the rounded products
Once upon a time when I was just out of school, still wet behind the ears...
I was coding for an accounting application. We had a list of values and a percentage charge back rate. We wanted to allocate the charge back to the individual list entries in such a way that things still added up -- the sum of the individual charge backs had to match the mandated percentage of the total.
Danged accountants.
The approach that I took was to first total up the line items and apply the charge back percentage to obtain a desired total charge back. Then I went through the list a second time computing each item's rounded fair share of the remaining charge back total and then deducting that line item from the remaining total and the remaining chargeback before proceeding to the next item.
By the time one comes to the last item in the list, the fair share is 100% and the totals are guaranteed to match to the penny.
Nobody ever talked to me about the possibility that the percentages on the individual line items could be out of whack in pathological cases. The books balanced and random data tends not to be pathological so nobody noticed. Remediation for that defect would be an interesting task.
[I was using double precision floating point for the totals. But I scaled everything up so that the floats encoded integer numbers of pennies. Floating point adds and subtracts are free from round-off error in that environment. It was 32 bit hardware so the vanilla integer data type would have risked overflow]
Last edited:
jim mcnamara
jbriggs444 said:
Once upon a time when I was just out of school, still wet behind the ears...
I was coding for an accounting application. We had a list of values and a percentage charge back rate. We wanted to allocate the charge back to the individual list entries in such a way that things still added up -- the sum of the individual charge backs had to match the mandated percentage of the total.
Danged accountants.
The approach that I took was to first total up the line items and apply the charge back percentage to obtain a desired total charge back. Then I went through the list a second time computing each item's rounded fair share of the remaining charge back total and then deducting that line item from the remaining total and the remaining chargeback before proceeding to the next item.
By the time one comes to the last item in the list, the fair share is 100% and the totals are guaranteed to match to the penny.
Nobody ever talked to me about the possibility that the percentages on the individual line items could be out of whack in pathological cases. The books balanced and random data tends not to be pathological so nobody noticed. Remediation for that defect would be an interesting task.
[I was using double precision floating point for the totals. But I scaled everything up so that the floats encoded integer numbers of pennies. Floating point adds and subtracts are free from round-off error in that environment. It was 32 bit hardware so the vanilla integer data type would have risked overflow]
I like that approach. We decided to sum the rounded products and subtract that sum from the true total, then add the difference to the largest rounded product. It's the same as yours except that our largest value makes up the difference whereas your last entry makes up the difference.
pob said:
I like that approach. We decided to sum the rounded products and subtract that sum from the true total, then add the difference to the largest rounded product. It's the same as yours except that our largest value makes up the difference whereas your last entry makes up the difference.
The method I used is slightly more subtle than "last entry makes up for all". If the total chargeback so far is randomly too low, the effect is to increase the chargeback on all of the remaining items, not just the last one. Similarly if the running total is randomly too high.
In retrospect, sorting the list from low to high would have been a nice touch as well.
## 1. How do I multiply numbers with different decimal precision?
In order to multiply numbers with different decimal precision, you first need to align the decimal points by adding zeros to the number with fewer decimal places. Then, you can multiply the numbers as usual.
## 2. What is rounding and why is it important in decimal precision?
Rounding is the process of approximating a number to a certain number of decimal places. It is important in decimal precision because it helps to simplify and make numbers easier to work with, especially when dealing with large or small numbers.
## 3. Can I add numbers with different decimal precision?
Yes, you can add numbers with different decimal precision. However, it is important to align the decimal points before adding to ensure accuracy.
## 4. How do I know how many decimal places to round to?
The number of decimal places to round to depends on the desired level of accuracy. For example, if you are dealing with money, it is common to round to two decimal places. However, for scientific calculations, you may need to round to more decimal places for greater precision.
## 5. What are some common mistakes to avoid when working with decimal precision?
Some common mistakes to avoid when working with decimal precision include forgetting to align the decimal points, rounding too early in the calculation process, and forgetting to account for significant figures. It is important to double check your work and be mindful of these potential errors.
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2K | 2,283 | 9,665 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-26 | latest | en | 0.924062 |
voting.readthedocs.io | 1,600,993,340,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400221382.33/warc/CC-MAIN-20200924230319-20200925020319-00739.warc.gz | 660,487,154 | 4,586 | # (Dis)proportionality Measures¶
The proportion module contains functions for measuring disproportionality in electoral systems.
voting.proportion.adjusted_loosemore_hanby(votes, seats, parties='votes')[source]
Calculate the adjusted Loosemore-Hanby index of disproportionality.
$N \sum_{i=1}^n |v_i - s_i|$
where N is $$\sum_{i=1}^n v_i$$ if parties == 'votes' or $$\sum_{i=1}^n s_i$$ if parties == 'seats'.
Parameters
• votes (list) – a list of vote counts
• seats (list) – a list of seat counts
• parties (str) – votes or seats to use to calculate the effective number of parties
voting.proportion.dhondt(votes, seats)[source]
Calculate the D’Hondt index of disproportionality.
$\max \frac{s_i}{v_i}$
Parameters
• votes (list) – a list of vote counts
• seats (list) – a list of seat counts
voting.proportion.gallagher(votes, seats)[source]
Calculate the Gallagher index of disproportionality.
$\sqrt{\frac{1}{2} \sum_{i=1}^n (v_i - s_i)^2}$
Parameters
• votes (list) – a list of vote counts
• seats (list) – a list of seat counts
voting.proportion.grofman(votes, seats, parties='votes')[source]
Calculate the Grofman index of disproportionality.
$N \sum_{i=1}^n |v_i - s_i|$
where N is $$\sum_{i=1}^n v_i^2$$ if parties == 'votes' or $$\sum_{i=1}^n s_i^2$$ if parties == 'seats'.
Parameters
• votes (list) – a list of vote counts
• seats (list) – a list of seat counts
• parties (str) – votes or seats to use to calculate the effective number of parties
voting.proportion.least_square(votes, seats)[source]
Calculate the least squares index of disproportionality.
$\sqrt{\sum_{i=1}^n (v_i - s_i)^2}$
Parameters
• votes (list) – a list of vote counts
• seats (list) – a list of seat counts
voting.proportion.lijphart(votes, seats)[source]
Calculate the Lijphart index of disproportionality.
$\max | v_i - s_i |$
Parameters
• votes (list) – a list of vote counts
• seats (list) – a list of seat counts
voting.proportion.loosemore_hanby(votes, seats)[source]
Calculate Loosemore-Hanby index of disproportionality.
$\frac{1}{2} \sum_{i=1}^n |v_i - s_i|$
Parameters
• votes (list) – a list of vote counts
• seats (list) – a list of seat counts
voting.proportion.rae(votes, seats)[source]
Calculate Rae’s index of disproportionality.
$\frac{1}{n} \sum_{i=1}^n |v_i - s_i|$
Parameters
• votes (list) – a list of vote counts
• seats (list) – a list of seat counts
voting.proportion.regression(votes, seats)[source]
Calculate the regression index of disproportionality.
$\frac{\sum_{i=1}^n v_i s_i}{\sum_{i=1}^n v_i^2}$
Parameters
• votes (list) – a list of vote counts
• seats (list) – a list of seat counts
voting.proportion.rose(votes, seats)[source]
Calculate the Rose index of proportionality.
$100 - \frac{1}{2} \sum_{i=1}^n |v_i - s_i|$
Parameters
• votes (list) – a list of vote counts
• seats (list) – a list of seat counts
voting.proportion.sainte_lague(votes, seats)[source]
Calculate the Sainte-Lague index of disproportionality.
$\sum_{i=1}^n \frac{(v_i - s_i)^2}{v_i}$
Parameters
• votes (list) – a list of vote counts
• seats (list) – a list of seat counts | 955 | 3,116 | {"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} | 3.15625 | 3 | CC-MAIN-2020-40 | latest | en | 0.784434 |
https://math.stackexchange.com/questions/3183436/interpret-model-estimates-after-log-transformation?noredirect=1 | 1,566,792,361,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330962.67/warc/CC-MAIN-20190826022215-20190826044215-00397.warc.gz | 550,538,286 | 30,207 | # Interpret model estimates after log transformation
I sat up a mixed-effects linear model with the dependent variable log-transformed (in oder to get it normal distributed and as ist is common with this kind of data in other publications). All fixed affects are significant and a consequent multiple comparissons of means works also great (significant differences of log10 values of different treatments). Now I wonder how to interpret the model estimates quantitatively. It seems not correct to just re-transform the estimates. Although the results are in the regular range and the means are in the same order as with untransformed values. [Only related topic I found is this][1] but I'm not sure if it is applicable to my case. The formula I use is:
log10(y) ~ = 0 + var1:var2 + var1:var2:cov1 + var1:var2:cov2
I'm working with R btw. Thanks for any help.
• This is a major problem as you noticed in my answer in the linked page. The key problem is that you measure $y$ and not $\log_{10}(y)$; so, if the errors are not very samll, you can face serious dangers. Can't you afford a nonlinear regression since you already have good (or at least reasonable) estimates of the parameters ? It would be much safer. – Claude Leibovici Apr 11 at 8:42
• @Claude: Thanks for your answer. I specified my question here: stats.stackexchange.com/questions/402415/…. If you would be so kind and have a look :) – Clem Snide Apr 12 at 7:23
• @Claude: How to set up a nonlinear regression on basis of my pre-work? – Clem Snide Apr 12 at 7:23
• No idea ! I am not a statistician ... even if I have some experience with data fitting. This expalins my comment. Just a stupid question : can you write $y=10^{xxx}$ ? – Claude Leibovici Apr 12 at 7:30 | 441 | 1,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-35 | latest | en | 0.900445 |
https://numbersdb.org/numbers/747567 | 1,604,190,017,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922746.99/warc/CC-MAIN-20201101001251-20201101031251-00194.warc.gz | 439,968,496 | 3,390 | # The Number 747567 : Square Root, Cube Root, Factors, Prime Checker
You can find Square Root, Cube Root, Factors, Prime Check, Binary, Octal, Hexadecimal and more of Number 747567. 747567 is written as Seven Hundred And Fourty Seven Thousand, Five Hundred And Sixty Seven. You can find Binary, Octal, Hexadecimal Representation and sin, cos, tan values and Multiplication, Division tables.
• Number 747567 is an odd Number.
• Number 747567 is a Prime Number
• Sum of all digits of 747567 is 36.
• Previous number of 747567 is 747566
• Next number of 747567 is 747568
## Square, Square Root, Cube, Cube Root of 747567
• Square Root of 747567 is 864.61956952176
• Cube Root of 747567 is 90.757677560045
• Square of 747567 is 558856419489
• Cube of 747567 is 417782616948133263
## Numeral System of Number 747567
• Binary Representation of 747567 is 10110110100000101111
• Octal Representation of 747567 is 2664057
• Hexadecimal Representation of 747567 is b682f
## Sin, Cos, Tan of Number 747567
• Sin of 747567 is -0.45399049973833
• Cos of 747567 is -0.89100652418899
• Tan of 747567 is 0.50952544949271
## Multiplication Table for 747567
• 747567 multiplied by 1 equals to 747,567
• 747567 multiplied by 2 equals to 1,495,134
• 747567 multiplied by 3 equals to 2,242,701
• 747567 multiplied by 4 equals to 2,990,268
• 747567 multiplied by 5 equals to 3,737,835
• 747567 multiplied by 6 equals to 4,485,402
• 747567 multiplied by 7 equals to 5,232,969
• 747567 multiplied by 8 equals to 5,980,536
• 747567 multiplied by 9 equals to 6,728,103
• 747567 multiplied by 10 equals to 7,475,670
• 747567 multiplied by 11 equals to 8,223,237
• 747567 multiplied by 12 equals to 8,970,804
## Division Table for 747567
• 747567 divided by 1 equals to 747567
• 747567 divided by 2 equals to 373783.5
• 747567 divided by 3 equals to 249189
• 747567 divided by 4 equals to 186891.75
• 747567 divided by 5 equals to 149513.4
• 747567 divided by 6 equals to 124594.5
• 747567 divided by 7 equals to 106795.28571429
• 747567 divided by 8 equals to 93445.875
• 747567 divided by 9 equals to 83063
• 747567 divided by 10 equals to 74756.7
• 747567 divided by 11 equals to 67960.636363636
• 747567 divided by 12 equals to 62297.25 | 748 | 2,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.789439 |
http://oeis.org/A121030 | 1,618,324,914,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072366.31/warc/CC-MAIN-20210413122252-20210413152252-00182.warc.gz | 67,340,445 | 3,689 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A121030 Multiples of 10 containing a 10 in their decimal representation. 35
10, 100, 110, 210, 310, 410, 510, 610, 710, 810, 910, 1000, 1010, 1020, 1030, 1040, 1050, 1060, 1070, 1080, 1090, 1100, 1110, 1210, 1310, 1410, 1510, 1610, 1710, 1810, 1910, 2010, 2100, 2110, 2210, 2310, 2410, 2510, 2610, 2710, 2810, 2910, 3010, 3100 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS R. Zumkeller, Table of n, a(n) for n = 1..10000 FORMULA a(n) ~ 10n. - Charles R Greathouse IV, Feb 12 2017 CROSSREFS Cf. A121041, A008592, A011531, A121022, A121023, A121024, A121025, A121026, A121027, A121028, A121029, A121031, A121032, A121033, A121034, A121035, A121036, A121037, A121038, A121039, A121040. Sequence in context: A169666 A169662 A124252 * A327786 A154810 A099820 Adjacent sequences: A121027 A121028 A121029 * A121031 A121032 A121033 KEYWORD nonn,base,easy AUTHOR Reinhard Zumkeller, Jul 21 2006 STATUS approved
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Last modified April 13 10:24 EDT 2021. Contains 342935 sequences. (Running on oeis4.) | 540 | 1,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-17 | latest | en | 0.543212 |
http://pdfmedfinder.com/e/elearning2.utcc.ac.th1.html | 1,534,865,431,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221218357.92/warc/CC-MAIN-20180821151743-20180821171743-00403.warc.gz | 303,324,796 | 5,742 | Où achat cialis sans ordonnance et acheter viagra en France.
## Elearning2.utcc.ac.th
CHAPTER 3: GRAPHICAL DESCRIPTIVE TECHNIQUES II
1. For what type of data is a histogram appropriate? 2. Twenty-five voters participating in a recent election exit poll in Alabama were asked to state their political party affiliation. Coding the data 1 for Republican, 2 for Democrat, and 3 for Independent, the
data collected were as follows: 3, 1, 2, 3, 1, 3, 3, 2, 1, 3, 3, 2, 1, 1, 3, 2, 3, 1, 3, 2, 3, 2, 1, 1, 3, 1, 2, 2,
1, and 3. Develop a frequency distribution and a relative frequency distribution for this data. What
does the data suggest about the strength of the political parties in Alabama?
NARRBEGIN: TeachersAges
Teachers Ages
The ages (in years) of a sample of 25 teachers are as follows:
47
3. {Teachers Ages Narrative} Draw a frequency histogram of this data which contains four classes. What 4. {Teachers Ages Narrative} Draw a frequency histogram of this data which contains six classes. What 5. {Teachers Ages Narrative} Draw a stem-and-leaf display of this data. What is the minimum and maximum age of the teachers in this data set? 6. {Teachers Ages Narrative} Construct an ogive for this data. Estimate the proportion of salespersons that are: 1) under 30 years of age; 2) 40 years of age or over; and 3) between 40 and 50 years of age.
NARRBEGIN: Test scores
Test scores
The scores on a calculus test for a random sample of 40 students are as follows:
63
7. {Test Grades Narrative} Construct a stem-and-leaf display for this data set. Describe the shape of the 8. {Test Grades Narrative} Construct frequency and relative frequency distributions for this data set using seven class intervals. Describe the shape of the data set. 9. {Test Grade Narrative} Construct a relative frequency histogram for this data set and discuss its shape. 10. {Test Grades Narrative} Describe the distribution of exam scores. 11. {Test Grades Narrative} Construct a cumulative frequency and a cumulative relative frequency distribution for this data. What proportion of the exam scores are less than 60? What proportion of the exam scores are 70 or more? 12. {Test Grades Narrative) Construct an ogive for this data set. Use the ogive to estimate the proportion of exam scores that are between 80 and 90. 13. Forty truck buyers were asked to indicate the car dealer they believed offered the best overall service. The four choices were A, B, C, and D as shown below: A Construct a table showing the frequencies and relative frequencies for this data set. What proportion of car buyers rated dealer D as the best? 14. A supermarket's monthly sales (in thousands of dollars) for the last year were as follows: Construct a relative frequency bar chart for this data set. How many observations are there in this data set? 15. Consider the following cumulative frequency distribution. Fill in the frequencies for each class in the above table. 16. The weights of a sample of 25 workers are given (in pounds): 164, 148, 137, 157, 173, 156, 177, 172, 169, 165, 145, 168, 163, 162, 174, 152, 156, 168, 154, 151, 174, 146, 134, 140, and 171. Construct an
ogive for the data. What proportion of the worker's weights are between 160 and 180 pounds; below
150 pounds; and at or above 140 pounds, respectively?
NARRBEGIN: Insurance Company
Insurance Company
A representative from a local insurance agency selected a random sample of insured homeowners and recorded the number of claims made in the last three years, with the following results: 17. {Insurance Company Narrative} How many homeowners are represented in the sample? 18. {Insurance Company Narrative} How many total claims are represented in the sample? 19. {Insurance Company Narrative} What proportion of homeowners had no claims in the last three 20. {Insurance Company Narrative} What number of claims was made by the highest proportion of homeowners?
NARRBEGIN: Electronics Company
Electronics Company
At a meeting of regional offices managers of a national electronics company, a survey was taken to
determine the number of employees the regional managers supervise in the operation of their
departments. The results of the survey are shown below.
21. {Electronics Company Narrative} How many regional offices are represented in the survey results? 22. {Electronics Company Narrative} Across all of the regional offices, how many total employees were supervised by those surveyed? ANS: (1 7) + (2 11) + (3 14) + (4 8) = (5 10) = 153 PTS: 1 23. {Electronics Company Narrative} What proportion of managers supervise 3 employees? 24. {Electronics Company Narrative} What is the cumulative relative frequency corresponding to 5 employees?
NARRBEGIN: Internet Classes
Internet Classes
A survey of 25 students was conducted to determine how they rate the quality of Internet classes.
Students were asked to rate the overall quality from 0 (no quality at all) to 100 (extremely good
quality). The stem-and-leaf display of the data is shown below.
25. {Internet Classes Narrative} What percentage of the students rated the overall quality of Internet 26. {Internet Classes Narrative} What percentage of the students rated the overall quality of Internet 27. {Internet Classes Narrative} What percentage of the students rated the overall quality of on-line classes as being between 50 and 75, inclusive? 28. {Internet Classes Narrative} What percentage of the students rated the overall quality of on-line 29. Explain the difference between a histogram and a line chart. 30. Briefly discuss the difference between cross-sectional data and time-series data. 31. Beef prices throughout the year (month 1 = January) are shown in the line chart below (per pound). Describe beef prices for this given year using this line chart. 32. An economics professor wants to study the relationship between income and education. A sample of 10 individuals is selected at random, and their income (in thousands of dollars) and education (in years) are shown below: Education Draw a scatter diagram for these data with the income on the vertical axis. Describe the relationship between income and education. 33. The number of houses sold in Miami Springs and the average monthly mortgage rates for 18 months randomly selected between January 2011 and April 2013 are shown in the following table. Draw a scatter diagram with the number of houses sold on the vertical axis. Describe the relationship between mortgage rate and number of houses sold. 34. Briefly explain the difference between correlation and causation in terms of a relationship between X 35. It is speculated that the number of police officers has a negative linear relationship with number of 36. What are the two most important characteristics to look for in a scatter diagram? 37. Can a scatter diagram be used to explore the relationship between two nominal variables? Explain why 36. In its 2000 report, a company presented the following data regarding its sales (in millions of dollars), and net income (in millions of dollars). Year The bar chart below was used to present these data.
Assume that you are an unscrupulous statistician and want to make the data appear more positive than
they really are. Redraw the graph by being selective about which years to include, getting the desired
effect.
The following bar chart shows the type of aspirin (if any) given to 100 cardiac patients within 15
minutes of their admission to the emergency room.
37. {Admission to ER Narrative} Notice that the Y-axis of the bar chart does not start at a frequency of zero. Create a new bar chart of the displayed data that accurately displays the frequency for each aspirin type. 38. {Admission to ER Narrative} In what way does the original bar chart distort the data as compared to a bar chart that starts at zero on the Y-axis (frequency axis)?
NARRBEGIN: Credit Hours
Credit Hours
The College of Business at The State University of California produced 3,400 credit hours during
Spring Semester, 2011he number and percentage of credit hours produced by each of the four
departments in the College of Business is shown below.
The following three-dimensional pie chart was constructed from the table above. 39. {Credit Hours Narrative} Construct a two-dimensional pie chart showing the percentages of credit 40. {Credit Hours Narrative} In what way does the original three-dimensional pie chart distort the data, compared to a two-dimensional pie chart? 41. Briefly explain why the histogram below is misleading. 42. Samantha recorded her amount of exercise time (in minutes) for 100 days. Both of the line charts below were created based on her same data set. Which line chart makes her exercise times look more variable and why?
NARRBEGIN: Home Care
Home Care
Data are collected on the number of house calls (x) made in a given week by each of 60 nurses in a
certain hospital. The same data set is displayed in each of the three histograms below.
43. {Home Care Narrative} How many nurses are there in this hospital? 44. {Home Care Narrative} Which histogram makes the differences in the house calls appear to be the 45. {Home Care Narrative} Which histogram makes the differences in the house calls appear to be the 46. {Home Care Narrative}Which graph do you think is the most appropriate display? Justify your 47. Below are two line charts where the percentage return for a stock is shown over time. The two graphs Explain why these two line charts look different. 48. An online dating service has a quick poll on its website showing the following results. Critique the graph portion of the table. How can it be improved?
Source: http://elearning2.utcc.ac.th/officialtcu/ECONTENT/SC123/SC123-Quiz/Quiz--Chapter%2003%20Graphical%20Descriptive%20Techniques%202%20SE.pdf
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Fusiones y adquisiciones en el sector eléctrico: Experiencia internacional en el análisis de casos CEER Centro de Estudios Económicos de la Regulación Universidad Argentina de la Empresa Lima 717, 1° piso C1073AAO BUENOS AIRES, ARGENTINA Teléfono: 54-11-43797693 Fax: 54-11-43797588 E-mail: ceer@uade.edu.ar http://www.uade.edu.ar/economia/ceer (Por favor, mire las últimas páginas de est
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4025_CH08_p398-443 01/02/04 12:34 PM Page 408 C H A P T E R 8 I n f e r e n c e s f r o m Tw o S a m p l e s Using Technology STATDISK Select Analysis from the main menu bar, then number of trials for Sample 1, in cell C1 enter the number of suc-select Hypothesis Testing, then Proportion-Two Samples. Enter cesses for Sample 2, and in cell D1 enter the number of trial | 2,522 | 10,662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-34 | longest | en | 0.851558 |
https://www.jipmagazine.nl/9752/work-power-and-energy.html | 1,591,081,730,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347423915.42/warc/CC-MAIN-20200602064854-20200602094854-00258.warc.gz | 751,089,985 | 5,899 | # Work Power And Energy
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What energy changes take place as this toy or object operat es q3hat form does the stored energy start out in q4hat form does the stored energy turn into q5hat form is the energy output in when it st ops q6hat made each object to move a cert ain disp lacement and what made.
Get Price | 1,956 | 8,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-24 | latest | en | 0.875832 |
https://squircleart.github.io/animation-nodes/charts-in-animation-nodes-grouped-bar-charts.html | 1,547,841,374,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660529.12/warc/CC-MAIN-20190118193139-20190118215139-00330.warc.gz | 664,884,565 | 6,159 | Grouped Bar Charts
Prerequisites:
The following is a list of all prerequisites in order to make the most out of the following tutorial.
In this second tutorial in the Charts In Animation Nodes tutorial series, we will create, automate and animate grouped bar charts.
We are going to use the same plane we created in the bar charts tutorial with some slight modification, so lets head to data parsing and grouped bars creation directly.
Data
Lets say we are comparing 5 GPUs in some game displaying their minimum, maximum and average frame rate, then our data may be like this:
GPU 1 : 147, 169, 28
GPU 2 : 154, 90, 58
GPU 3 : 165, 129, 82
GPU 4 : 158, 110, 30
GPU 5 : 175, 140, 69
The parsing process will also be similar to the way we parsed the data in the bar charts tutorial.
After we split the line using the separator :, we end up with two texts, one that include the name and one that include the fps separated by , so we split again using the separator , to get individual fps, then we parse them and append them to an output integer list. You will see why we didn’t classify them into separate list in a moment. You will also notice that I added an output parameter which is assigned to the length of the second split. For the example above, the parameter will be equal to 3 because we have minimum, maximum and average.
We now have the data parsed and ready to be used.
Grouped Bars
If you remember correctly, what we did in the previous tutorial is replicate a unity square along the x axis (using a range node) and then scale it according to its corresponding parsed value. In order to make a grouped bar chart, we have to replicate two times, one time to form each group (Which will contain 3 bars ) and another time to take each group and replicate it for each GPU. But if we replicated the group, we won’t be able to control the scale of each element of the group individually. So the answer is simple, instead of replicating the group itself, we will replicate its transformation matrices and then use those matrices to replicate the bars giving us the ability to control each bar independently, if you don’t understand this, no worries, you will understand in a moment.
The distribute matrices node can be used to create a certain amount of matrices arranged in a line, we will use it to locate bars within each group where the size controls the spaces between them:
The replicate matrices node can then be used to replicate those matrices along the x axis using the translation vector we created in the previous tutorial using the range node.
Those matrices can then be multiplied by scaling transformation matrices that contain the values we parsed in the y values where the x value is just some arbitrary variable that controles the width of the bars. That’s why we appended the parsed values without classification, because matrices are flat and not grouped or classified.
The resulted matrices can in turn be used to replicate the unity square to make the bars:
There is something we have to take care of though, lets say our y scale will range between 0 and 182 and the length of the y axis is 9. If we used the parsed data directly we will get very high bars, you probably know what to do by now. We normalize the fps by dividing by 182 and then multiply by the y axis length putting the bars in their right scale. And the result is already usable:
Now I am going to do something I haven’t told you about before and that is, giving different colors to our bars. I will add 3 materials which are red, blue and yellow, I want to assign the first material to the first bar of each group, second to second and third to third. Animation nodes let us define the material of each polygon easily using the Material indices input in the Mesh Object Output node . The material indices input is an integer list with length equal to the number of polygons in the mesh. The first integer is the index of the material of the first polygon, second integer is the index of the material of the second polygon and so on. So for our case, we need the indices to be $(0, 1, 2, 0, 1, 2, \cdots)$, this finite sequence can be easily achieved using the rule:
Where $n$ is the number of polygons in the mesh, resulting:
Giving a result:
And that’s it, we have our grouped bars chart, same automations we did in the previous tutorial applies here so no need to explain it.
Animation
If we used the same node trees we did in the previouse tutorial, we will get an already nice animation:
One might need to edit this animation such that all bars within each group rise at the same time and this can easily be achieved by using a transformation matrix instead of a translation vector in the replicate matrix node matrix where the y scale of the transformation matrix will be controlled using the same falloff you used before.
Giving the animation:
Challenge!
There is another method to achieve the previous animation using falloffs, can you find it?
I think that that’s it when it comes to bar charts, stay tuned for next type of charts we will be creating. | 1,096 | 5,084 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-04 | latest | en | 0.864059 |
https://forum.effectivealtruism.org/posts/GgPrbxdWhyaDjks2m/the-multiple-stage-fallacy | 1,702,185,389,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101195.85/warc/CC-MAIN-20231210025335-20231210055335-00209.warc.gz | 207,013,109 | 132,966 | # The Multiple Stage Fallacy
Mar 16 20164 min read 16
# 6
Frontpage
This was originally written by Eliezer Yudkowsky and posted on his Facebook wall. It is reposted here with permission from the author:
In August 2015, renowned statistician and predictor Nate Silver wrote "Trump's Six Stages of Doom" in which he gave Donald Trump a 2% chance of getting the Republican nomination (not the presidency, the nomination).
It's too late now to register an advance disagreement, but now that I've seen this article, I do think that Nate Silver's argument was a clear instance of something that I say in general you shouldn't do - what I used to call the Conjunction Fallacy Fallacy and have now renamed to the Multiple-Stage Fallacy (thanks to Graehl for pointing out the naming problem).
The Multiple-Stage Fallacy is when you list multiple 'stages' that need to happen on the way to some final outcome, assign probabilities to each 'stage', multiply the probabilities together, and end up with a small final answer. In other words, you take an arbitrary event, break it down into apparent stages, and say "But you should avoid the Conjunction Fallacy!" to make it seem very low-probability.
In his original writing, Nate listed 6 things that he thought Trump needed to do to get the nomination - "Trump's Six Stages of Doom" - and assigned 50% probability to each of them.
We're now past the first four stages Nate listed, and prediction markets give Trump a 74% chance of taking the nomination. On Nate's logic, that should have been a 25% probability. So while a low prior probability might not have been crazy and Trump came as a surprise to many of us, the specific logic that Nate used is definitely not holding up in light of current events.
On a probability-theoretic level, the three problems at work in the usual Multiple-Stage Fallacy are as follows:
1. First and foremost, you need to multiply *conditional* probabilities rather than the absolute probabilities. When you're considering a later stage, you need to assume that the world was such that every prior stage went through. Nate Silver was probably - though I here critique a man of some statistical sophistication - Nate Silver was probably trying to simulate his prior model of Trump accumulating enough delegates in March through June, not imagining his *updated* beliefs about Trump and the world after seeing Trump be victorious up to March.
1a. Even if you're aware in principle that you need to use conjunctive probabilities, it's hard to update far *enough* when you imagine the pure hypothetical possibility that Trump wins stages 1-4 for some reason - compared to how much you actually update when you actually see Trump winning! (Some sort of reverse hindsight bias or something? We don't realize how much we'd need to update our current model if we were already that surprised?)
2. Often, people neglect to consider disjunctive alternatives - there may be more than one way to reach a stage, so that not *all* the listed things need to happen. This doesn't appear to have played a critical role in Nate's prediction here, but I've often seen it in other cases of the Multiple-Stage Fallacy.
3. People have tendencies to assign middle-tending probabilities. So if you list enough stages, you can drive the apparent probability of anything down to zero, even if you solicit probabilities from the reader.
3a. If you're a motivated skeptic, you will be tempted to list more 'stages'.
Fallacy #3 is particularly dangerous for people who've read a lot about the dangers of overconfidence. Right now, we're down to two remaining stages in Nate's "six stages of doom" for Trump, accumulating the remaining delegates and winning the convention. The prediction markets assign 74% probability that Trump passes through both of them. So the conditional probabilities must look something like 90% probability that Trump accumulates enough delegates, and then 80% probability that Trump wins the convention given those delegates.
Imagine how overconfident this would sound without the prediction market! Oh, haven't you heard that what people assign 90% probability usually doesn't happen 9 times out of 10?
But if you're not willing to make "overconfident" probability assignments like those, then you can drive the apparent probability of anything down to zero by breaking it down into enough 'stages'. In fact, even if someone hasn't heard about overconfidence, people's probability assignments often trend toward the middle, so you can drive down their "personally assigned" probability of anything just by breaking it down into more stages.
For an absolutely ridiculous and egregious example of the Multiple-Stage Fallacy, see e.g. this page which commits both of the first two fallacies at great length and invites the third fallacy as much as possible: http://www.jefftk.com/p/breaking-down-cryonics-probabilities. To be sure, Nate Silver didn't do anything remotely on THAT order, but it does put all three subfallacies on clearer display than they appear in "Trump's Six Stages of Doom".
From beginning to end, I've never used this style of reasoning and I don't recommend that you do so either. Beware the Multiple-Stage Fallacy!
Edit (3/16): Reply from Jeff Kaufman below:
a) A more recent version of that post's approach with more people's estimates is http://www.jefftk.com/p/more-cryonics-probability-estimates
b) My estimates were careful to avoid (1) but I didn't make this clear enough in the writeup apparently. All the probabilities I'm using are conditional on the earlier things not happening.
c) In that post I explicitly excluded disjunctive paths because I really do think there's pretty much one path that's far more likely. All the candidates people have offered me for disjunctive paths for cryonics seem much less likely.
d) I'm pretty unhappy about the insinuation that my skepticism is motivated. I'd love it if cryonics was likely to work! I care about so many people that I would be devastated to lose. And when I initially wrote that post two friends (Jim, Michael) had me mostly convinced. Then thinking about why it still seemed unlikely to me I tried a Fermi style estimate and became way more skeptical.
e) I'd like to see more examples of people using this approach and whether it was useful before calling it a fallacy or a reasoning trap.
# Reactions
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To defend the critique of Jeff's reasoning a little, I do think Eliezer has a point with the 1a fallacy when he says that it's hard to properly condition on the fact that you've been surprised. For example, according to Jeff's estimates there's a ~3% chance that getting and keeping you frozen goes well. If this does happen you'd be hugely surprised at how well things go for cryonicists. There should be some explanation for that other than pure chance. (The problem is you can't search the space of explanations for each of the ~30 probabilities and adjust them appropriately. ) Here's one simple explanation: Cryonics gets big and successful. Perhaps that's unlikely a priori but given that something very weird happened it becomes plausible. This will strongly mess with the probabilities that determine if something goes wrong with reviving. The biggest one, 'The technology is never developed to extract the information', would certainly be lower. In fact, 9/10 probabilities would be go down. Sometimes they could also go up.
I doubt that Jeff managed to take all of these possibilities into account. Properly conditioning each of the ~30 events on each of the ones before it going well seems like a pretty daunting task. That doesn't mean Jeff is horrendously wrong but he probably does make mistake 1a because that's just hard to avoid with this type of reasoning.
The name "Multiple Stage Fallacy" seems to encourage equivocation: Is it a fallacy to analyze the probability of an event by breaking it down into multiple stages, or a fallacy to make the mistakes Eliezer points to?
For the Nate Silver example, Eliezer does aim to point out particular mistakes. But for Jeff, the criticism comes sort of between these two possibilities: There's a claim that Jeff makes these mistakes (which seems to be wrong - see Jeff's reply), but it's as if the mere fact of "Multiple Stages" means there's no need to actually make an argument.
Yes, I found the criticism-by-insinuation of Jeff's post unhelpful, because none of these errors were obvious. A more concrete discussion of disagreements might be interesting.
(For what it's worth Jeff's analysis still looks pretty plausible to me. My biggest disagreement is on the probabilities of something "other" going wrong, which look too modestly large to me after a decent attempt to think about what might fail. It's not clear that's even one of the kind of errors Eliezer is talking about.)
My biggest disagreement is on the probabilities of something "other" going wrong, which look too modestly large to me after a decent attempt to think about what might fail.
After a lot of discussion in the original post, I made a new model where I (a) removed steps I thought were very likely to succeed and (b) removed most of the mass from "other" since discussing with people had increased my confidence that we'd been exhaustive: http://lesswrong.com/lw/fz9
Sorry, if I'd done my homework I'd have linked to that and might have said you agreed with me!
I was pointing out the disagreement not to critique, but to highlight that in the piece Eliezer linked to as exhibiting the problems described, it seemed to me like the biggest issue was in fact a rather different problem.
I think this is dumb; I don't see any particular evidence that this happens very often, and I'm much more worried about people being overconfident about things based on tenuous, badly thought out, oversimplified models than I am about them being underconfident because of concerns like these.
According to Eliezer, in the cog sci literature people don't systematically undercorrect for biases like overconfidence when warned to avoid them. Rather, they're systematically miscalibrated about whether they're overcorrecting or undercorrecting. ("Yes, I know, I was surprised too.")
If our cure for moderate cases of overconfidence tends to produce extreme underconfidence, then we can be left worse off that we were originally, especially if there are community norms punishing people more for sounding overconfident than for sounding underconfident.
Hmm, though I agree with the idea that people tend to be overconfident, the critique of this style of reasoning is exactly that it leads to overconfidence. I think the argument "people tend to be underconfident, not overconfident" does not seem to bear a lot on the truth of this critique.
(e.g. underconfidence is having believes that tend towards the middling ranges, overconfidence is having extreme beliefs. Eliezer argues that this style of reasoning leads one to assign extremely low probabilities to events, which should be classified as overconfident)
But point 3 relies on underconfident estimates of the individual factors.
I'm not sure that addresses Buck's point. I just don't think you can reduce this to "people tend to be overconfident", even if it's a conclusion in a limited domain.
And the discussion on Jeff's FB post: https://www.facebook.com/jefftk/posts/775488981742.
"Back when I was doing predictions for the Good Judgement Project this is something that the top forecasters would use all the time. I don't recall it being thought inaccurate and the superforecasters were all pretty sharp cookies who were empirically good at making predictions."
I hate taking something like this and calling it the XXX fallacy. The unsubstantiated swipe at the end is an excellent example of why.
I think the unsubstantiated swipe at the end is a separate problem from calling it the XXX fallacy.
They're linked, though. Labelling it as a fallacy makes it easier to just point to an example of this kind of reasoning with the implication that it's fallacious (so apparently no need to substantiate).
Yeah that's a good point.
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· 4d ago · 9m read | 2,682 | 12,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-50 | longest | en | 0.960939 |
https://support.google.com/docs/answer/3094105 | 1,701,264,806,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100081.47/warc/CC-MAIN-20231129105306-20231129135306-00477.warc.gz | 626,267,205 | 310,770 | # STDEVP
Calculates the standard deviation based on an entire population.
STDEVP for BigQuery
Calculates the population standard deviation of a data column.
### Sample Usage
`STDEVP(table_name!price)`
## Syntax
`STDEVP(column)`
• `column` - The data column of the population.
Tip: Returning population standard deviation across multiple columns is not supported
### Sample Usage
`STDEVP(1,2,3,4,5,6,7,8,9,10)`
`STDEVP(A2:A100)`
### Syntax
`STDEVP(value1, [value2, ...])`
• `value1` - The first value or range of the population.
• `value2, ...` - Additional values or ranges to include in the population.
### Notes
• Although `STDEVP` is specified as taking a maximum of 30 arguments, Google Sheets supports an arbitrary number of arguments for this function.
• If the total number of values supplied as `value` arguments is not at least two, `STDEVP` will return the `#DIV/0!` error.
• `STDEVP` will return an error if any of the `value` arguments include text. To calculate standard deviation while interpreting text values as `0`, use `STDEVPA`.
• `STDEVP` calculates standard deviation for an entire population. To calculate standard deviation across a sample, use `STDEV`.
• `STDEVP` is equivalent to the square root of the variance, or `SQRT(VARP(...))` using the same dataset.
`VARPA`: Calculates the variance based on an entire population, setting text to the value `0`.
`VARP`: Calculates the variance based on an entire population.
`VARA`: Calculates the variance based on a sample, setting text to the value `0`.
`VAR`: Calculates the variance based on a sample.
`STDEVPA`: Calculates the standard deviation based on an entire population, setting text to the value `0`.
`STDEVA`: Calculates the standard deviation based on a sample, setting text to the value `0`.
`STDEV`: The STDEV function calculates the standard deviation based on a sample.
`SKEW`: Calculates the skewness of a dataset, which describes the symmetry of that dataset about the mean.
`KURT`: Calculates the kurtosis of a dataset, which describes the shape, and in particular the "peakedness" of that dataset.
`DVARP`: Returns the variance of an entire population selected from a database table-like array or range using a SQL-like query.
`DVAR`: Returns the variance of a population sample selected from a database table-like array or range using a SQL-like query.
`DSTDEVP`: Returns the standard deviation of an entire population selected from a database table-like array or range using a SQL-like query.
`DSTDEV`: Returns the standard deviation of a population sample selected from a database table-like array or range using a SQL-like query.
`DEVSQ`: Calculates the sum of squares of deviations based on a sample.
`AVEDEV`: Calculates the average of the magnitudes of deviations of data from a dataset's mean.
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Close search | 674 | 2,860 | {"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} | 2.625 | 3 | CC-MAIN-2023-50 | longest | en | 0.728425 |
http://mathvids.com/lesson/mathhelp/648-the-power-rule | 1,498,751,066,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128329372.0/warc/CC-MAIN-20170629154125-20170629174125-00691.warc.gz | 259,294,329 | 18,133 | ## The Power Rule
The Power Rule
• Currently 3.0/5 Stars.
3900 views, 5 ratings - 00:06:33
Taught by MrRutter
### Meets NCTM Standards:
Learn what the Power Rule says and how to use it in various contexts.
• What is the Power Rule in Calculus?
• What is the derivative of f(x) = a*x^n?
• How do you find the derivative of a function using the power rule?
• What is the derivative of 1/(2*sqrt(x)) or 1/2x^(-1/2)?
• What is the derivative of f(x) = 3x^2 + 9x - 3?
• What is the derivative of a constant?
• If f(x) is a number, why is the derivative equal to 0?
This lesson explains finding the derivative of functions using the power rule. This is one of the most basic and vital ideas for finding derivatives in Calculus. Some really great examples are done in this video.
• Currently 3.0/5 Stars.
Reviewed by MathVids Staff on April 18, 2009. | 245 | 846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.908755 |
https://262.run/how-much-weight-is-25-kg/ | 1,719,278,760,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00542.warc.gz | 67,852,004 | 51,644 | # How Much Weight Is 25 KG
How Much Weight Is 25 KG?
Weight is a fundamental measurement used worldwide to determine the heaviness of an object or a person. One commonly used metric unit to measure weight is kilograms (kg). In this article, we will explore how much weight is equivalent to 25 kg and address some frequently asked questions about this measurement.
The Conversion: Kilograms to Pounds
To understand the weight of 25 kg, it is helpful to convert it into pounds, a commonly used unit of weight in certain regions. One kilogram is equivalent to approximately 2.20462 pounds. Therefore, to convert 25 kg to pounds, we multiply the weight in kilograms by the conversion factor (2.20462):
25 kg * 2.20462 lb/kg = 55.1155 lb
Hence, 25 kilograms is approximately equal to 55.12 pounds.
Understanding the Weight of 25 KG
25 kg can represent various objects or entities, such as:
1. A Bag of Rice: A typical bag of rice, commonly found in grocery stores, weighs around 25 kg. This weight is manageable for most individuals.
2. Luggage: When traveling, airlines often impose weight restrictions on baggage. For instance, many airlines allow passengers to check in one or two pieces of luggage, each weighing up to 23 kg. Therefore, 25 kg would be considered slightly overweight for a single piece of luggage according to these regulations.
See also How Many Calories Are in a Kfc Chicken Pot Pie
3. Weights in Exercise: In the fitness world, weights are often used to build strength and endurance. Dumbbells, barbells, or weight plates are measured in kilograms. A set of dumbbells weighing 25 kg can be challenging for beginners but are commonly used for various exercises by experienced weightlifters.
4. Average Weight of a Child: The weight of a child can vary significantly depending on their age, height, and other factors. However, as a general reference, an average 5-year-old child weighs around 18-20 kg. Therefore, 25 kg would be considered heavier than average for a child of this age.
FAQs
Q: Is 25 kg heavy?
A: The perception of weight can vary from person to person. However, in most cases, 25 kg is considered moderately heavy and can be managed by a healthy adult.
Q: How can I convert kilograms to pounds?
A: To convert kilograms to pounds, multiply the weight in kilograms by 2.20462. This will give you the approximate weight in pounds.
Q: Is 25 kg too heavy for carry-on luggage?
A: Most airlines have specific weight restrictions for carry-on luggage. While it may vary, 25 kg is generally considered too heavy for carry-on luggage as it exceeds the weight limit imposed by many airlines.
Q: How much force is required to lift 25 kg?
A: The amount of force required to lift 25 kg depends on various factors, such as the individual’s strength and technique. In general, an average adult should be able to lift 25 kg with moderate effort. | 648 | 2,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-26 | latest | en | 0.93233 |
https://cs.stackexchange.com/questions/4840/proving-omegacf-omegaf/4844 | 1,600,440,647,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00515.warc.gz | 369,849,617 | 36,945 | # Proving $\Omega(cf) = \Omega(f)$
I'm trying to prove the following lemma:
$c$ is a positive real number and $f, g$ are functions from natural numbers to non-negative real numbers. I'm trying to prove rigorously that:
$\Omega(cf(n))$ = $\Omega(f(n))$.
I know that it is obvious but I'm trying to construct a proof that is as complete as possible. My current approach is like this:
This lemma is equivalent to saying that: $f(n) \in \Omega(cf(n))$ iff $cf(n) \in \Omega(f(n))$.
We can also restate that as:
1. If $t(n) \in \Omega(cf(n))$ then $t(n) \in \Omega(f(n))$.
2. If $t(n) \in \Omega(f(n))$ then $t(n) \in \Omega(cf(n))$.
For 1.,
$(*)$ $\exists d_1, d_2 \gt 0, \forall n \gt n_0, n_1, \forall n \in N$:
$t(n) \ge d_1cf(n)$ and $t(n) \ge d_2f(n)$
Now let's fix $d_1, d_2$ and $n_0, n_1$ to be any constants that fulfils $(*)$, such that: $n'=max\{n_0, n_1\}$ and $d_1c \ge d_2$, using this we can say that:
$t(n) \ge d_1cf(n) \ge d_2f(n)$ and hence the 1. is satisfied because $t(n) \in \Omega(cf(n)), \Omega(f(n))$. The proof of 2. is mutatis mutandis.
Do I have a mistake in my proof, is there a better/more elegant way to prove this lemma? Shortly how can I improve this?
• Please note that check-my-proof questions are typically boring for others if the the answer is a simple "yes" (or "no)", especially if the task is as simple as unfolding the definition twice. – Raphael Oct 2 '12 at 17:48
• yes you are right, it is obvious but it was a bit hard for me to prove it formally. – systemsfault Oct 2 '12 at 19:58
• My above statement is independent of the level of the question. (Nevertheless, in my experience questions of basic level -- meaning the asker is likely to have problems with the basics -- yield better results when discussed at a blackboard with a teacher.) – Raphael Oct 2 '12 at 20:00
• In fact that this was a question from an assignment of a course I took last semester and I got zero point with the proof I wrote above. In the beginning I was planning to write the proof in the assignment that phantom pointed out. But I found it too simple and instead wrote the one in the post. Now I was trying to figure out my mistake and find the correct solution as an exercise. The problem about the professor is that he is very busy person and it is very hard to find him outside his classes. – systemsfault Oct 3 '12 at 14:34
• Fair enough. But don't you have TAs, or fellow students who understood the material better than you? – Raphael Oct 3 '12 at 15:27
## 3 Answers
I think you're trying to take too many steps at once. Try to think about it one small step at a time.
Always think about:
• What are the definitions I have available
• What do I know already? What is proven?
## Proving sets equal
Since you have two sets, $\Omega(cf)$ and $\Omega(f)$ and you want to prove them equal, you first need to think about what it means for two sets to be equal:
$$A = B \Longleftrightarrow A \subseteq B \land A \supseteq B$$
This tells you, that you need to show two things:
1. $\Omega(cf) \subseteq \Omega(f)$
2. $\Omega(cf) \supseteq \Omega(f)$
## Proving $A$ to be a subset of $B$
For this, you need to apply the definition of $A \subseteq B$:
$$A \subseteq B \Longleftrightarrow \forall a \in A:(a \in B)$$
## Nature of our proof
As you can probably see by now, our proof will have the following form:
\begin{align*}"\subseteq":\\ & \text{Let } g \in \Omega(cf):\\ &\Rightarrow \dots\\ &\Rightarrow g \in \Omega(f) \end{align*}
The proof goes the same way for $"\supseteq"$.
## Filling in the dots
Keeping this structure in mind, the path you need to follow is already laid out.
Using this definition for \Omega: $\Omega(f) = \{ g |\, \exists\, n_0 \in \mathbb{N}, c > 0 \forall n \geq n_0: cf(n) \leq g(n) \}$
One direction of the proof could look as follows:
\begin{align*}"\subseteq":\\ & \text{Let } g \in \Omega(cf): \\ &\Rightarrow \exists\, n_0\in \mathbb{N}, c'>0: c'cf(n) \leq g(n)\ \ \forall n \geq n_0 &\text{definition of } \Omega\\ &\Rightarrow \exists\, d>0, n_0\in \mathbb{N}: d f(n) \leq g(n)\ \ \forall n \geq n_0 &\text{Let } d=c'c\\ &\Rightarrow g \in \Omega(f) &\text{definition of } \Omega \end{align*}
• This does not tell the OP where specifically the problems in their proposed proof are. – Raphael Oct 3 '12 at 11:57
• yeah thanks the explanation, the filling in the dots part was actually the second proof that I had in my mind :). At least you've approved my second proof, before I wrote it :). – systemsfault Oct 3 '12 at 14:16
• @Raphael Hmm yes, you have a good point. I might fill that in later this evening if I have the time. – phant0m Oct 3 '12 at 14:22
Why do such $d_1, d_2$ always exist?
You might not be able to increase $d_1$ indefinitely in order to fulfil $d_1c\geq d_2$ because $t(n) \geq d_1cf(n)$ might break along the way.
• hmm yes you are right. I'll update the post and I'm going to add my second proof. – systemsfault Oct 2 '12 at 19:52
My opinion is that the $=$ sign should be interpreted as $\subset$ in the asymptotic complexity context. Consider the case when we write $n^2=\Omega(n)$. This is mentioned in CLRS's "Introduction to Algorithms", page 49.
Another point is that considering the definition of $\Omega$, it naturally hides constant coefficients inside, so you can show the "equality" easily by stating $c_2=c\cdot c_1$, defined as follows:
$\Omega(f)=\{g|∃n_0∈N,c_2>0∀n≥n0:c_2f(n)≤g(n)\}$, $\Omega(cf)=\{g|∃n_0∈N,c_1>0∀n≥n0:c_1\cdot cf(n)≤g(n)\}$.
• No, the equals sign should be interpreted as equality. The question is equivalent to "Prove that $g\in\Omega(cf)$ if, and only if, $g\in\Omega(f)$." – David Richerby Jun 15 '16 at 11:00
• ok...why not...it depends on what does the OP want. I sometimes see things like $\Omega(n^2)=\Omega(n)$ in which $=$ means $\subset$ – Leo Jun 15 '16 at 12:09
• Well, in this case, the two sets are equal, so it certainly shouldn't mean $\subset$. And being asked to prove that $A\subseteq B$ is a very different thing to being asked if $A\subset B$ or $A\subseteq B$. – David Richerby Jun 15 '16 at 12:11 | 1,874 | 6,074 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-40 | latest | en | 0.912247 |
http://gmatclub.com/forum/in-the-figure-above-is-the-area-of-triangular-region-abc-134270.html?sort_by_oldest=true | 1,484,623,260,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279410.32/warc/CC-MAIN-20170116095119-00412-ip-10-171-10-70.ec2.internal.warc.gz | 124,213,959 | 69,394 | In the figure above, is the area of triangular region ABC : GMAT Data Sufficiency (DS)
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# In the figure above, is the area of triangular region ABC
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In the figure above, is the area of triangular region ABC [#permalink]
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09 Jun 2012, 23:28
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In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ?
(2) ∆ABC is isosceles.
Source: OG13 DS79
[Reveal] Spoiler: OA
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Re: Triangular region ABC equal to triangular region DBA? [#permalink]
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10 Jun 2012, 00:13
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I started with
Area = .5bh
Area(ABC) = .5*AC*CB
So we need to know whether AC*CB = AD*AB
This tells us the relationship between AC and AD but not enough to derive the full areas. We need to know about CB and AB.
Insufficient.
BCE
2) ∆ABC is isosceles.
This tells us that AC = CB, and furthermore, we then know AB, but nothing about AD.
Insufficient.
1+2) Here, we know the relationship between AC, AD, and CB explicitly. We can also determine AB.
Sufficient. Remember, we don't need to derive the area. Just determine whether or not we can determine if they will be equal.
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Re: Triangular region ABC equal to triangular region DBA? [#permalink]
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10 Jun 2012, 01:00
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After some more calculations I found the following solution, which is similar to yours:
Area of ABC = Area of DBA?
0.5 AC*CB = 0.5 = AD*AB ?
Use Pythagorean theorem to simplify further:
$$AB^2=AC^2+CB^2$$
$$AB=\sqrt{AC^2+CB^2}$$
Therefore, the question reduces to:
$$AC*AB = AD*\sqrt{AC^2+CB^2}?$$
(1) $$AC = \sqrt{2}*AD$$
So we can get rid of AC. Still need AD, AB and CB though:
$$AC*AB = AD*\sqrt{AC^2+CB^2}?$$
$$\sqrt{2}*AD*AB = AD*\sqrt{AC^2+CB^2}?$$
$$\sqrt{2}*AB =\sqrt{2*AD^2+CB^2}?$$
We cannot simplify or reduce further. Insufficient.
(2) ABC = isosceles means AC = CB
So we can replace AC:
$$AC*AB = AD*\sqrt{AC^2+CB^2}?$$
$$CB*AB = AD*\sqrt{2*CB^2}?$$
$$CB*AB = AD*\sqrt{2}*CB?$$
$$AB = AD*\sqrt{2}?$$
We cannot simplify or reduce further. Insufficient.
(1) and (2)
(1) actually answers the question we are left with in (2):
$$AB = AD*\sqrt{2}?$$
I am not sure if I could have done this in 2 mins.
Key take aways:
- Rephrase and simplify question as much as possible, given known formulas etc.
- To check C: use work you do for checking (1) also for checking (2) and vice versa.
- Don't get thrown off my complicated formulas
Hope it helps others. Please let me know if I made any mistake!
Thanks a lot!
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Re: In the figure above, is the area of triangular region ABC [#permalink]
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23 Oct 2012, 10:36
does anyone have any other alternative method to solve?
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Re: In the figure above, is the area of triangular region ABC [#permalink]
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23 Oct 2012, 21:39
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teal wrote:
does anyone have any other alternative method to solve?
You can solve sufficiency questions of geometry by drawing some diagrams too.
Attachment:
Ques3.jpg [ 4.81 KiB | Viewed 30800 times ]
We need to compare areas of ABC and DAB. Notice that given triangle ABC with a particular area, the length of AD is defined. If AD is very small, (shown by the dotted lines) the area of DAB will be very close to 0. If AD is very large, the area will be much larger than the area of ABC. So for only one value of AD, the area of DAB will be equal to the area of ABC.
Now look at the statements:
$$AD = AC/\sqrt{2}$$
The area of ABC is decided by AC and BC, not just AC. We can vary the length of BC to see the relation between AC and AD is not enough to say whether the areas will be the same (see diagram). So insufficient.
Attachment:
Ques4.jpg [ 7.15 KiB | Viewed 31075 times ]
(2) ∆ABC is isosceles.
Using both, ratio of sides of ABC are $$1:1:\sqrt{2}$$ = AC:BC : AB
Area of ABC = 1/2*1*1 = 1/2
Area of DAB = $$1/2*AD*AB = 1/2*1/\sqrt{2}*\sqrt{2} = 1/2$$
Areas of both the triangles is the same
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Joined: 22 Mar 2011 Posts: 612 WE: Science (Education) Followers: 99 Kudos [?]: 887 [24] , given: 43 Re: In the figure above, is the area of triangular region ABC [#permalink] ### Show Tags 24 Oct 2012, 01:04 24 This post received KUDOS 10 This post was BOOKMARKED jfk wrote: Attachment: OG13DS79v2.png In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ? (1) (AC)^2=2(AD)^2 (2) ∆ABC is isosceles. Source: OG13 DS79 The area of a right triangle can be easily expressed as half the product of the two legs. Area of triangle $$ABC$$ is $$0.5AC\cdot{BC}$$ and that of the triangle $$DBA$$ is $$0.5AD\cdot{AB}$$. The question in fact is "Is $$AC\cdot{BC} = AB\cdot{AD}$$?" (1) From $$AC^2=2AD^2$$, we deduce that $$AC=\sqrt{2}AD$$, which, if we plug into $$AC\cdot{BC} = AB\cdot{AD}$$, we get $$\sqrt{2}AD\cdot{BC}=AB\cdot{AD}$$, from which $$\sqrt{2}BC=AB$$. This means that triangle $$ABC$$ should necessarily be isosceles, which we don't know. Not sufficient. (2) Obviously not sufficient, we don't know anything about $$AD$$. (1) and (2): If $$AC=BC=x$$, then $$AB=x\sqrt{2}$$, and $$AD=\frac{x}{\sqrt{2}}$$. Then $$AC\cdot{BC}=AB\cdot{AD}=x^2$$. Sufficient. Answer C. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Intern Joined: 06 Apr 2011 Posts: 13 Followers: 0 Kudos [?]: 2 [0], given: 292 Re: In the figure above, is the area of triangular region ABC [#permalink] ### Show Tags 13 Mar 2013, 22:53 Nice and succinct explanation Eva... Intern Joined: 04 Sep 2012 Posts: 4 Followers: 0 Kudos [?]: 0 [0], given: 1 Re: In the figure above, is the area of triangular region ABC [#permalink] ### Show Tags 05 Jun 2013, 17:41 I still don't get it. To compare the Area of (ABC) and Area of (DBA): is 1/2(AC)(CB) = 1/2(AD)(AB) ? In the hint 1: (AC)^2=2(AD)^2 -> AC = sq(2)*(AD) -> AD > AC (1) Then look at the graph: (AB) is the hypotenuse of triangle (ABC) -> AB > CB (2) (1) and (2) : -> (AC)(CB) < (AB)(AD) -> the answer is NO, their areas are not equal Thus, the answer must be A. Can anyone please explain my mistake in here? Thanks Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7118 Location: Pune, India Followers: 2129 Kudos [?]: 13625 [1] , given: 222 Re: In the figure above, is the area of triangular region ABC [#permalink] ### Show Tags 05 Jun 2013, 23:07 1 This post received KUDOS Expert's post ltkenny wrote: I still don't get it. To compare the Area of (ABC) and Area of (DBA): is 1/2(AC)(CB) = 1/2(AD)(AB) ? In the hint 1: (AC)^2=2(AD)^2 -> AC = sq(2)*(AD) -> AD > AC (1) Then look at the graph: (AB) is the hypotenuse of triangle (ABC) -> AB > CB (2) (1) and (2) : -> (AC)(CB) < (AB)(AD) -> the answer is NO, their areas are not equal Thus, the answer must be A. Can anyone please explain my mistake in here? Thanks There's your error. $$AD = AC/\sqrt{2}$$ So AD < AC (not AD > AC) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: In the figure above, is the area of triangular region ABC [#permalink]
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06 Jun 2013, 16:55
Ah, I get it. Thank you for pointing my error.
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Re: In the figure above, is the area of triangular region ABC [#permalink]
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24 Aug 2013, 02:49
WAIT WAIT WAIT....
in my opinion statement 1) is sufficient.
AC is less than AD, and BC must be less than AB (leg vs hypotenuse) therefore Area cannot be equal. 1) sufficient.
choice A is correct?
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Re: In the figure above, is the area of triangular region ABC [#permalink]
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24 Aug 2013, 02:54
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leroyconje wrote:
WAIT WAIT WAIT....
in my opinion statement 1) is sufficient.
AC is less than AD, and BC must be less than AB (leg vs hypotenuse) therefore Area cannot be equal. 1) sufficient.
choice A is correct?
Refer to the posts above, AC>AD.
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Re: In the figure above, is the area of triangular region ABC [#permalink]
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24 Aug 2013, 02:55
mau5 wrote:
leroyconje wrote:
WAIT WAIT WAIT....
in my opinion statement 1) is sufficient.
AC is less than AD, and BC must be less than AB (leg vs hypotenuse) therefore Area cannot be equal. 1) sufficient.
choice A is correct?
Refer to the posts above, AC>AD.
MY GOOD what a distraction!
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Re: In the figure above, is the area of triangular region ABC [#permalink]
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18 Sep 2013, 16:14
EvaJager wrote:
jfk wrote:
Attachment:
OG13DS79v2.png
In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ?
(2) ∆ABC is isosceles.
Source: OG13 DS79
The area of a right triangle can be easily expressed as half the product of the two legs.
Area of triangle $$ABC$$ is $$0.5AC\cdot{BC}$$ and that of the triangle $$DBA$$ is $$0.5AD\cdot{AB}$$.
The question in fact is "Is $$AC\cdot{BC} = AB\cdot{AD}$$?"
(1) From $$AC^2=2AD^2$$, we deduce that $$AC=\sqrt{2}AD$$, which, if we plug into $$AC\cdot{BC} = AB\cdot{AD}$$, we get $$\sqrt{2}AD\cdot{BC}=AB\cdot{AD}$$, from which $$\sqrt{2}BC=AB$$. This means that triangle $$ABC$$ should necessarily be isosceles, which we don't know.
Not sufficient.
(2) Obviously not sufficient, we don't know anything about $$AD$$.
(1) and (2): If $$AC=BC=x$$, then $$AB=x\sqrt{2}$$, and $$AD=\frac{x}{\sqrt{2}}$$.
Then $$AC\cdot{BC}=AB\cdot{AD}=x^2$$.
Sufficient.
I'm a little confused with this method.
1) In statement 1, how can you deduce that the ABC needs to be an isosceles?
2) When you combine the statements, How do you know that $$AD=\frac{x}{\sqrt{2}}$$?
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Re: In the figure above, is the area of triangular region ABC [#permalink]
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18 Sep 2013, 17:49
russ9 wrote:
EvaJager wrote:
jfk wrote:
Attachment:
OG13DS79v2.png
In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ?
(2) ∆ABC is isosceles.
Source: OG13 DS79
The area of a right triangle can be easily expressed as half the product of the two legs.
Area of triangle $$ABC$$ is $$0.5AC\cdot{BC}$$ and that of the triangle $$DBA$$ is $$0.5AD\cdot{AB}$$.
The question in fact is "Is $$AC\cdot{BC} = AB\cdot{AD}$$?"
(1) From $$AC^2=2AD^2$$, we deduce that $$AC=\sqrt{2}AD$$, which, if we plug into $$AC\cdot{BC} = AB\cdot{AD}$$, we get $$\sqrt{2}AD\cdot{BC}=AB\cdot{AD}$$, from which $$\sqrt{2}BC=AB$$. This means that triangle $$ABC$$ should necessarily be isosceles, which we don't know.
Not sufficient.
(2) Obviously not sufficient, we don't know anything about $$AD$$.
(1) and (2): If $$AC=BC=x$$, then $$AB=x\sqrt{2}$$, and $$AD=\frac{x}{\sqrt{2}}$$.
Then $$AC\cdot{BC}=AB\cdot{AD}=x^2$$.
Sufficient.
I'm a little confused with this method.
1) In statement 1, how can you deduce that the ABC needs to be an isosceles?
2) When you combine the statements, How do you know that $$AD=\frac{x}{\sqrt{2}}$$?
In statement 1, the question boils down to: Is $$\sqrt{2}BC=AB$$?
or Is $$BC/AB=1/\sqrt{2}$$?
In a right triangle, the ratio of a leg and hypotenuse will be 1:\sqrt{2} if the third side is also 1 i.e. only if the triangle is isosceles. You can figure this from pythagorean theorem
$$1^2 + x^2 = \sqrt{2}^2$$
$$x = 1$$
On combining the statements, we know that ABC is isosceles so AC = BC. So ratio of sides $$AC:BC:AB = 1:1:\sqrt{2}$$i.e. the sides are $$x, x$$ and $$\sqrt{2}x$$
From statement 1 we know that $$AC = \sqrt{2}AD$$
So $$AC = x = \sqrt{2}AD$$
So $$AD = x/\sqrt{2}$$
This method is way too mechanical and prone to errors. Try to use the big picture approach.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 11 Aug 2013 Posts: 34 Followers: 0 Kudos [?]: 9 [0], given: 9 Re: In the figure above, is the area of triangular region ABC [#permalink] ### Show Tags 18 Nov 2013, 20:37 teal wrote: does anyone have any other alternative method to solve? Don't use numbers, use logic. Vandygrad above helped me break this down. Look up what an Isosceles triangle is: Which is a triangle that has 2 equal sides, and 2 equal angles, so with that information on hand, you know that it's a 45/45/90 , and look up how to find the area of a triangle A=Base*Height/2, or A=1/2(B*H). The question stem states that the area of ABC and DBA, is it the same? Yes/No? 1) No values are given for ABC/or or DBA. INSUF. So you have an equation for C, but no know lengths, so insufficient to solve for the area. Knowing the formula is irrelevant, values are important. Time Saver. 2) ABC=Isocolecs: so you know the triangle has 2 equal sides, and 2 equal angels. No lengths, to be able to determine the area of triangle ABC: Combined) C is solved for. Given: *****ABC = Isocolesces, you have two equal sides, that and knowing that you can reduce statement 1, then use the P. Theorem to solve: Stop here. Sufficient to solve. Intern Joined: 04 Jun 2014 Posts: 49 Followers: 0 Kudos [?]: 2 [0], given: 5 Re: In the figure above, is the area of triangular region ABC [#permalink] ### Show Tags 07 Aug 2014, 01:04 EvaJager wrote: jfk wrote: Attachment: OG13DS79v2.png In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ? (1) and (2): If $$AC=BC=x$$, then $$AB=x\sqrt{2}$$, and $$AD=\frac{x}{\sqrt{2}}$$. Then $$AC\cdot{BC}=AB\cdot{AD}=x^2$$. Sufficient. Answer C. Can anyone explane the last step? AC * BC = AB * AD = x^2 ? x * x = x√2 * x/√2 x^2 = x^2*√2/√2 ---> is this right to solve the equation like this? Math Expert Joined: 02 Sep 2009 Posts: 36520 Followers: 7067 Kudos [?]: 92926 [0], given: 10528 Re: In the figure above, is the area of triangular region ABC [#permalink] ### Show Tags 12 Aug 2014, 05:49 Expert's post 1 This post was BOOKMARKED lou34 wrote: EvaJager wrote: jfk wrote: Attachment: OG13DS79v2.png In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ? (1) and (2): If $$AC=BC=x$$, then $$AB=x\sqrt{2}$$, and $$AD=\frac{x}{\sqrt{2}}$$. Then $$AC\cdot{BC}=AB\cdot{AD}=x^2$$. Sufficient. Answer C. Can anyone explane the last step? AC * BC = AB * AD = x^2 ? x * x = x√2 * x/√2 x^2 = x^2*√2/√2 ---> is this right to solve the equation like this? Since AC = BC = x, then AC*BC = x^2. Since $$AB=x\sqrt{2}$$, and $$AD=\frac{x}{\sqrt{2}}$$, then $$AB * AD =x\sqrt{2}*\frac{x}{\sqrt{2}}=x^2$$. _________________ Intern Joined: 16 Mar 2014 Posts: 5 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: In the figure above, is the area of triangular region ABC [#permalink] ### Show Tags 12 Mar 2015, 07:58 I understand the mathematical/algebraic approach, but is it possible to answer this question using logic alone? For example: Statement 1 fixes the ratio of AC to AD, but point B is still free to move in space, so insufficient. Statement 2 fixes the ratio of AC to BC, but point D is still free to move in space, so insufficient. Together, the ratio of all the sides are fixed, so sufficient. Does the above approach make sense? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7118 Location: Pune, India Followers: 2129 Kudos [?]: 13625 [0], given: 222 Re: In the figure above, is the area of triangular region ABC [#permalink] ### Show Tags 12 Mar 2015, 18:53 swaggerer wrote: I understand the mathematical/algebraic approach, but is it possible to answer this question using logic alone? For example: Statement 1 fixes the ratio of AC to AD, but point B is still free to move in space, so insufficient. Statement 2 fixes the ratio of AC to BC, but point D is still free to move in space, so insufficient. Together, the ratio of all the sides are fixed, so sufficient. Does the above approach make sense? The two statements are fine but you need to think through the "using both" part clearly. I have discussed the logical approach here: in-the-figure-above-is-the-area-of-triangular-region-abc-134270.html#p1134780 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: In the figure above, is the area of triangular region ABC [#permalink] 12 Mar 2015, 18:53
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Display posts from previous: Sort by | 5,991 | 18,873 | {"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} | 3.609375 | 4 | CC-MAIN-2017-04 | latest | en | 0.860477 |
https://solsarin.com/how-to-solve-a-pyramid-rubiks-cube/ | 1,713,587,120,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817474.31/warc/CC-MAIN-20240420025340-20240420055340-00668.warc.gz | 482,365,744 | 26,389 | #### solsarin
the complate explain
# how to solve a pyramid rubik’s cube
## how to solve a pyramid rubik’s cube
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## How to Solve a Rubik’s Cube, Step by Step
### Before You Begin
Here are some things you should know about the Rubik’s Cube. Some of these points might strike you as trivial at first, but each affords some insight that will become clearer the more time you spend with the cube.
• The Rubik’s Cube has six faces.
• Each face is defined by its center. The face with the blue center will ultimately be blue when the cube is solved.
• Centers don’t move. White is typically opposite of yellow, blue is typically opposite green, and red is typically opposite orange.
• Corner pieces have three stickers and edge pieces have two stickers. When solving the cube, try to keep in mind that you are moving piecesnot stickers. Another way of thinking about this point is that a red sticker on a corner piece will never move to an edge position.
### Cube Notation
Solving the cube will require you to turn its faces. Each face is represented by a letter. The direction of a given rotation is denoted by the presence or absence of a prime (’) symbol.
Right face: R
Left face: L
Upward-pointing face: U
Downward-pointing face: D
Front face: F
Back face: B
R, L, U, D, F, or B means to turn the corresponding face 90 degrees clockwise
R’, L’, U’, D’, F’, or B’ means to turn the corresponding face 90 degrees counterclockwise.
R2, L2, U2, D2, F2, or B2 means to turn the corresponding face 180 degrees.
### : Make the Daisy
The goal of this step is to place four white edge stickers around the yellow center. When you are finished with this step.
Note: It doesn’t matter what color the grey squares are. Two things to keep in mind:
1. Once a white sticker is placed next to the yellow center, it does not need to be moved.
2. You can turn the top layer without disturbing anything next to the yellow center.
### Step Two: Create the White Cross
For each petal on the “daisy,” match the non-white sticker to the center piece of the same color. Once matched, turn the face with the matching center two times. Repeat this process three more times. When you are finished, the bottom face of the cube will have a white cross.
Note: For the rest of the solve, the white cross will be on the bottom. If you ever find the white cross somewhere else, something has gone wrong.
### Step Three: Solve the First Layer
Time to learn your first algorithms. The following “trigger moves” are the most basic of the bunch:
• Right Trigger = R U R’
• Left Trigger = L’ U’ L
Look for white stickers on the top layer that face the sides. (If you find a white sticker on the top face of the cube, or on the bottom layer of the cube pointing outward, we’ll deal with it later.) Each white sticker should be on a corner piece with three stickers. Rotate the top face of the cube so that the sticker beside the white sticker that is also outward facing (i.e., not the sticker on the top) diagonally matches the center of the same color.
Once you’ve paired them, face the color-matched stickers toward you. If the matched sticker in the top layer is right of the center, perform the Right Trigger. or If the matched sticker is left of center, perform the Left Trigger.
and If you have a white sticker facing the top, position the white sticker over something that is not white (because it will disrupt whatever is underneath), and, depending on if the piece is on the right or on the left, perform the following algorithm:
R, U, R’, R, U, R’
Or
L’, U, L, L’, U, L
If you have an outward facing white sticker in the bottom layer, face it toward you and position the cube so that it is either in the bottom left or bottom right corner of the side facing you, and perform either the left or right trigger, respectively, to relocate it to the top face of the cube.
### Step Four How to Solve a Rubik’s Cube: Solve the Middle Layer
Identify edge pieces on the top layer that do not have yellow stickers. (If it has a yellow sticker, it belongs on the top and not in the middle.) Once you find an edge without a yellow sticker, rotate the top face of the cube until the outward facing sticker on that edge piece is directly over the center piece of the same color.
Once it matches, look at the upward-facing sticker on that edge piece. That sticker will match the center on either the left or the right.
If it matches on the right, perform the following algorithm:
U + Right Trigger
Doing so will disturb the first layer. Fix the displaced white corner sticker as you did in step three.
If it matches on the left, perform the following algorithm:
U’ + Left Trigger
Doing so will disturb the first layer. Fix the displaced white corner sticker as you did in step three.
Occasionally you will find no edge pieces in the top layer without yellow stickers but the middle layer is not solved. In such cases, displace them is-matched middle-layer edge piece by performing the left or right trigger. There should now be an edge piece in the top layer without a yellow sticker. Solve for it as described above.
### Step Five How to Solve a Rubik’s Cube: Create the Yellow Cross
The goal of this step is to create a yellow cross on the upward-pointing face of the cube. This entire step hinges on the following algorithm:
F U R U’ R’ F’
If your top face has no yellow edge pieces, perform ***F U R U’ R’ F’. or If your top face has two yellow edge pieces such that they form a line with the center yellow piece, orient the cube such that the three yellow stickers form a vertical line and perform ***F U R U’ R’ F’. If your top face has two yellow edge pieces such that they form a backwards L, rotate the top face of the cube until the edge pieces are at the 12 and 9 positions of a clock and perform ***F U R U’ R’ F’. At this point, the top face of your cube should resemble a yellow cross. | 1,389 | 6,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-18 | latest | en | 0.909954 |
https://math.stackexchange.com/questions/755261/is-there-a-standard-proof-for-mathbb-psx-n-text-hits-a-text-before-b/762206 | 1,563,494,307,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525863.49/warc/CC-MAIN-20190718231656-20190719013656-00288.warc.gz | 471,170,049 | 38,544 | # Is there a standard proof for $\mathbb P(S^X_n\text{ hits }A\text{ before }B) >\mathbb P(S^Y_n\text{ hits }A\text{ before }B)$?
Let $X_i$ and $Y_i$ be two continuous random variables on $\mathbb{R}$ having distribution functions $F$ and $G$, respectively satisfying $G(y)>F(y)$ for all $y$. Let futhermore $S^X_n=\sum_{i=1}^n X_i$, $S^Y_n=\sum_{i=1}^n Y_i$, $A>0$, and $B<0$, Then, I wonder if there is a simple/standard proof for the following:
$$\mathbb P(S^X_n\text{ hits }A\text{ before }B) > \mathbb P(S^Y_n\text{ hits }A\text{ before }B)$$
Note: Both $X$ and $Y$ have negative means.
Thanks alot.
First remark : the events "$S_n^X$ hits $A$ before $B$" are negligible. Let's be rigorous. Let $\tau_A := \inf \{n \geq 0: S_n \geq A\}$, and let $\tau_B := \inf \{n \geq 0: S_n \leq B\}$, both with values in $\mathbb{N} \cup \{+\infty\}$. Then you wan to prove that:
$$\mathbb{P}_{(S_n^X)} (\tau_A < \tau_B) > \mathbb{P}_{(S_n^Y)} (\tau_A < \tau_B).$$
Essentially, I just want to point out that "hitting $A$" means "reaching $A$ or a higher value", and likewise for $B$. Note that $X$ and $Y$ are continuous, so that for each of them $\tau_A < +\infty$ a.s. or $\tau_B < +\infty$ a.s., so the event $\tau_A < \tau_B$ is well-defined up to a negligible subset (that wouldn't be the case if $X \equiv 0$, for instance).
Now, the idea is, as Did suggested, to use a coupling. The condition $G > F$ is equivalent to the fact that there exists a probability space $\Omega$ and a random variable $(X',Y')$ on $\Omega$ such that:
• $X' = X$ in distribution;
• $Y' = Y$ in distribution;
• $X' > Y'$ a.s.
(Edit, thanks to Einar Rødland) If there is such a coupling, it is easy to see that $G>F$ everywhere. The converse - which is what we are interested in - is more delicate. The idea is to take a random variable $U$ with a uniform distribution on $[0,1]$, and put $(X',Y') := (F^{-1} (U), G^{-1} (U))$, where $F^{-1}$ and $G^{-1}$ are the generalized inverses of $F$ and $G$ respectively. Now, since $F^{-1} > G^{-1}$ on $(0,1)$, we've won.
I can't find a good reference now, see e.g. the first slide here. I guess it's done in Villani's Optimal tranport: Old and new, but there must be a probability-oriented book somewhere with the minimum on stochastic ordering...
Now, let us take a sequence of i.i.d. random variables $(X_n',Y_n')_{n \geq 1}$ such that $(X',Y') = (X_1',Y_1')$ in distribution. Then $(X_n')_{n \geq 1} = (X_n)_{n \geq 1}$ in distribution, and $(Y_n')_{n \geq 1} = (Y_n)_{n \geq 1}$. Let $S_n^{X'} := \sum_{i=1}^n X_i'$, and likewise for $Y'$. Then:
• $(S_n^{X'})_{n \geq 1} = (S_n^X)_{n \geq 1}$ in distribution;
• $(S_n^{Y'})_{n \geq 1} = (S_n^Y)_{n \geq 1}$ in distribution.
In addition, $X' > Y'$ a.s., so $X_i' > Y_i'$ for all $i$ a.s.. Hence:
• $S_n^{X'} > S_n^{Y'}$ for all $n$ a.s..
Let $\tau_A^{X'} := \inf \{n \geq 0: S_n^{X'} \geq A\}$, and define $\tau_A^{Y'}$, $\tau_B^{X'}$, $\tau_B^{Y'}$ in the same way. Then $S_n^{Y'} \geq A$ implies that $S_n^{X'} \geq A$ for all $n$, so $\tau_A^{X'} \leq \tau_A^{Y'}$. In the same way, $\tau_B^{X'} \geq \tau_B^{Y'}$. Hence,
$$\mathbb{P}_{(S_n^X)} (\tau_A < \tau_B) = \mathbb{P} (\tau_A^{X'} < \tau_B^{X'}) \geq \mathbb{P} (\tau_A^{Y'} < \tau_B^{Y'}) = \mathbb{P}_{(S_n^Y)} (\tau_A < \tau_B).$$
This is the standard proof. Proving a strict inequality is trickier. We have to find an event of positive measure on which $\tau_A^{X'} < \tau_B^{X'}$ but $\tau_A^{Y'} > \tau_B^{Y'}$. This relies more on the specific data of the problem, and thus is less standard.
Note that $G > F$ implies that $\mathbb{P} (Y < t) > 0$ for all $t$, and $\mathbb{P} (X > t) > 0$ for all $t$.
Let $M \in (A, +\infty]$ be such that $\mathbb{P} (X_1' \geq M) = \mathbb{P} (Y_1' \geq A)$. Such an $M$ exists, since both $X$ and $Y$ are continuous random variables. If $X_1' \in (A,M)$ then $Y_1' < A$ (this comes from the specific construction of the optimal coupling). In addition,
$$\mathbb{P} (X_1' \in (A,M)) = F(M)-F(A) = G(A)-F(A) > 0.$$
Now, let us consider the event "$X_1' \in (A,M)$ and $Y_2' < B-A$". On this event, we have $\tau_A^{X'} = 1 < \tau_B^{X'}$ and $\tau_A^{Y'} > \tau_B^{Y'} = 2$. Finally, this event has probability $(G(A)-F(A))G(B-A)$, whence:
$$\mathbb{P}_{(S_n^X)} (\tau_A < \tau_B) - \mathbb{P}_{(S_n^Y)} (\tau_A < \tau_B) \geq (G(A)-F(A))G(B-A) > 0.$$
• Isn't the title of Villani's book: "Optimal transport: Old and New" and basically it is not really about coupling? – Arash Apr 20 '14 at 21:35
• Can this be useful: math.stackexchange.com/questions/503504/…? – Arash Apr 20 '14 at 21:35
• If you let $U\sim\text{Uniform}[0,1]$, let $X=F^{-1}(U)$ and $Y=G^{-1}(U)$, and you have $X$ and $Y$ with the desired distributions and $X<Y$. – Einar Rødland Apr 20 '14 at 21:38
• @Arash: thanks, that was a lapsus. If I remember well, there is a chapter (probably the first) on the basics of coupling, which (I guess) presents the optimal coupling on the real line. – D. Thomine Apr 20 '14 at 21:39
• @EinarRødland, We do not know if $F^{-1}$ exist or not so there are some subtleties here. Look at the end of my answer here: math.stackexchange.com/questions/503504/… – Arash Apr 20 '14 at 21:42 | 1,931 | 5,192 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-30 | latest | en | 0.797718 |
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# PHYS 1010 Pt1 - 3 rd law o For every force there’s an...
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PHYS 1010 Newton’s Laws—Laws to explain motion 1 st law—an object at rest will remain at rest; an object moving in a direction at a constant speed will remain at that speed and at that direction o Velocity Speed Direction o This law was actually first realized by Galileo He called it inertia; inertia has a Latin origin, and basically means inactive o Basically this law is just saying that objects will keep on doing what they are doing o It is force that is required to stop the motion; absence of force would just mean the object keeps moving per Newton’s force law. Force isn’t need to put something in motion, but needed to stop it. o Change in velocity=acceleration 2 nd law o Greater force greater acceleration o Greater mass smaller acceleration A * F A* 1/M o Mass= amount of matter in object o Weight=gravity acting on mass o Force=mass*acceleration o A=g (acceleration=acceleration of gravity) If there’s NO air resistance a=F/M
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Unformatted text preview: 3 rd law o For every force there’s an equal and opposite force Kepler’s law Ptolemy Retrograde motion Eccentricity—stretching the a round shape (Circle ellipse) Kepler’s First Law: The orbits of the planets are ellipses, with the Sun at one focus of the ellipse. Kepler’s Second Law: The line joining the planet to the Sun sweeps out equal areas in equal times as the planet travels around the ellipse Kepler’s Third Law: Newton’s Law of Gravitation Conservation of Angular Momentum Conservation of Energy Conservation of Momentum Energy—Quantity often understood as the ability to do work Forms of energy: Mechanical energy-1. Kinetic energy Definition of Simple Harmonic Motion (SHM): Periodic motion about an equilibrium position in a sinusoidal pattern. A body in SHM experiences an opposing or correcting force proportional to displacement and in opposite direction to motion....
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## This note was uploaded on 07/15/2011 for the course PHYS 1010 taught by Professor Heil during the Spring '08 term at UGA.
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PHYS 1010 Pt1 - 3 rd law o For every force there’s an...
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http://www.dreamincode.net/forums/topic/137892-java-programming-arrays/ | 1,440,989,719,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065488.33/warc/CC-MAIN-20150827025425-00223-ip-10-171-96-226.ec2.internal.warc.gz | 415,638,753 | 24,573 | # Java Programming---Arrays
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## 10 Replies - 1484 Views - Last Post: 09 November 2009 - 06:51 PMRate Topic: //<![CDATA[ rating = new ipb.rating( 'topic_rate_', { url: 'http://www.dreamincode.net/forums/index.php?app=forums&module=ajax§ion=topics&do=rateTopic&t=137892&s=bd218bcc945d0ca8a5ca2a218572e75a&md5check=' + ipb.vars['secure_hash'], cur_rating: 0, rated: 0, allow_rate: 0, multi_rate: 1, show_rate_text: true } ); //]]>
### #1 BabyKodou
Reputation: 0
• Posts: 12
• Joined: 02-November 09
# Java Programming---Arrays
Posted 09 November 2009 - 05:12 PM
How can i calculate the average of the given number of arrays, i've already calculated the sum but don't know how to get the average.
THIS IS THE CODE
import java.util.Random;
import java.io.*;
public class Average{
public static void main(String[]args){
int MyArray[]=new int[20];
int Largest=MyArray[0];
int sum=0;
Random Me=new Random();
for(int x=0;x<20;x++){
MyArray[x]=Me.nextInt(100);
System.out.print(MyArray[x]);
System.out.print(" ");
for(int j=0;j<20;j++){
if(Largest<MyArray[j]){
Largest=MyArray[j];
}
}
for(int m=0;m<20;m++){
sum+=MyArray[m];
}
}
System.out.println(": Largest number is: "+ Largest);
System.out.println("The sum of all the numbers is : " + sum);
}
}
Is This A Good Question/Topic? 0
## Replies To: Java Programming---Arrays
### #2 jimdandy75
• D.I.C Regular
Reputation: 37
• Posts: 311
• Joined: 30-June 08
## Re: Java Programming---Arrays
Posted 09 November 2009 - 05:29 PM
Well, it looks like you have 20 elements in your array. Wouldn't you just take the sum divided by 20 to get the average?
```System.out.println("The average of all the numbers is : " + sum/20);
```
This post has been edited by jimdandy75: 09 November 2009 - 05:32 PM
### #3 BabyKodou
Reputation: 0
• Posts: 12
• Joined: 02-November 09
## Re: Java Programming---Arrays
Posted 09 November 2009 - 05:47 PM
jimdandy75, on 9 Nov, 2009 - 04:29 PM, said:
Well, it looks like you have 20 elements in your array. Wouldn't you just take the sum divided by 20 to get the average?
```System.out.println("The average of all the numbers is : " + sum/20);
```
I've done that thank you, it worked!.
So can you help me show me how to search for a number in an array set of numbers. For example if a user wants to search if or if not a number is in the set of arrays.
### #4 theautokustomizer
• D.I.C Regular
Reputation: 16
• Posts: 250
• Joined: 20-September 09
## Re: Java Programming---Arrays
Posted 09 November 2009 - 05:50 PM
And as per your other problem, I would just use an
```if (MyArray[x] == userinput)
System.out.println("The number is in this array");
```
Obviously you would have to ask the question first and declare your input, but that is general idea of it....
This post has been edited by theautokustomizer: 09 November 2009 - 05:56 PM
### #5 WaHooCrazy7
Reputation: 3
• Posts: 19
• Joined: 09-November 09
## Re: Java Programming---Arrays
Posted 09 November 2009 - 05:51 PM
There is also another option(s) that would work, even if you changed the size of the array. Java has a length method for figuring out what the size of an array is. So you could do something like this:
```String[] foo;
foo = new String[20];
//While loop to populate foo
println "The average of Foo is: " +sum/foo.length + "\n";
```
That method will find the average size of any array that you give it to. One important note though is if you allocate 20 cells, but only fill 10 the length method is still going to return 20. (Should not be a problem in your case)
This post has been edited by WaHooCrazy7: 09 November 2009 - 06:02 PM
### #6 WaHooCrazy7
Reputation: 3
• Posts: 19
• Joined: 09-November 09
## Re: Java Programming---Arrays
Posted 09 November 2009 - 05:58 PM
BabyKodou, on 9 Nov, 2009 - 04:47 PM, said:
I've done that thank you, it worked!.
So can you help me show me how to search for a number in an array set of numbers. For example if a user wants to search if or if not a number is in the set of arrays.
That task can be accomplished by a function. If you simply want to know if a number is there or not, it can return a bool. So for example:
```//Assumes an array of Foo is already in existance
bool findNum(int x){
for (int i = 0; i < Foo.length; i++) {
if (Foo[i] == x){
return true; //Will break you out of the for loop
} else {
return false;
}
}
}
```
That will go through Foo and return True if it finds the value that you passed it. This can be modified to return the index of where it is in the array if that's what you need.
### #7 theautokustomizer
• D.I.C Regular
Reputation: 16
• Posts: 250
• Joined: 20-September 09
## Re: Java Programming---Arrays
Posted 09 November 2009 - 06:12 PM
WaHooCrazy7, on 9 Nov, 2009 - 04:58 PM, said:
BabyKodou, on 9 Nov, 2009 - 04:47 PM, said:
I've done that thank you, it worked!.
So can you help me show me how to search for a number in an array set of numbers. For example if a user wants to search if or if not a number is in the set of arrays.
That task can be accomplished by a function. If you simply want to know if a number is there or not, it can return a bool. So for example:
```//Assumes an array of Foo is already in existance
bool findNum(int x){
for (int i = 0; i < Foo.length; i++) {
if (Foo[i] == x){
return true; //Will break you out of the for loop
} else {
return false;
}
}
}
```
That will go through Foo and return True if it finds the value that you passed it. This can be modified to return the index of where it is in the array if that's what you need.
I LIKE IT!!!!
### #8 WaHooCrazy7
Reputation: 3
• Posts: 19
• Joined: 09-November 09
## Re: Java Programming---Arrays
Posted 09 November 2009 - 06:27 PM
theautokustomizer, on 9 Nov, 2009 - 05:12 PM, said:
WaHooCrazy7, on 9 Nov, 2009 - 04:58 PM, said:
BabyKodou, on 9 Nov, 2009 - 04:47 PM, said:
I've done that thank you, it worked!.
So can you help me show me how to search for a number in an array set of numbers. For example if a user wants to search if or if not a number is in the set of arrays.
That task can be accomplished by a function. If you simply want to know if a number is there or not, it can return a bool. So for example:
```//Assumes an array of Foo is already in existance
bool findNum(int x){
for (int i = 0; i < Foo.length; i++) {
if (Foo[i] == x){
return true; //Will break you out of the for loop
} else {
return false;
}
}
}
```
That will go through Foo and return True if it finds the value that you passed it. This can be modified to return the index of where it is in the array if that's what you need.
I LIKE IT!!!!
Actually I just realized my code is wrong...man I've been fried lately. Problem with it is that it will only check the first value in the array and if its not equal to x just return false. Basically I should have moved the else outside of the for loop. Here is the correct code:
```//Assumes an array of Foo is already in existance
bool findNum(int x){
for (int i = 0; i < Foo.length; i++) {
if (Foo[i] == x){
return true; //Will break you out of the for loop
}
}
//You wont get here if there is any value that would cause the if(Foo[i] == x) evaluate true
return false;
}
```
### #9 BabyKodou
Reputation: 0
• Posts: 12
• Joined: 02-November 09
## Re: Java Programming---Arrays
Posted 09 November 2009 - 06:36 PM
WaHooCrazy7, on 9 Nov, 2009 - 05:27 PM, said:
theautokustomizer, on 9 Nov, 2009 - 05:12 PM, said:
WaHooCrazy7, on 9 Nov, 2009 - 04:58 PM, said:
BabyKodou, on 9 Nov, 2009 - 04:47 PM, said:
I've done that thank you, it worked!.
So can you help me show me how to search for a number in an array set of numbers. For example if a user wants to search if or if not a number is in the set of arrays.
That task can be accomplished by a function. If you simply want to know if a number is there or not, it can return a bool. So for example:
```//Assumes an array of Foo is already in existance
bool findNum(int x){
for (int i = 0; i < Foo.length; i++) {
if (Foo[i] == x){
return true; //Will break you out of the for loop
} else {
return false;
}
}
}
```
That will go through Foo and return True if it finds the value that you passed it. This can be modified to return the index of where it is in the array if that's what you need.
I LIKE IT!!!!
Actually I just realized my code is wrong...man I've been fried lately. Problem with it is that it will only check the first value in the array and if its not equal to x just return false. Basically I should have moved the else outside of the for loop. Here is the correct code:
```//Assumes an array of Foo is already in existance
bool findNum(int x){
for (int i = 0; i < Foo.length; i++) {
if (Foo[i] == x){
return true; //Will break you out of the for loop
}
}
//You wont get here if there is any value that would cause the if(Foo[i] == x) evaluate true
return false;
}
```
What about the BinarySearch method using Java.util.Arrays class
### #10 WaHooCrazy7
Reputation: 3
• Posts: 19
• Joined: 09-November 09
## Re: Java Programming---Arrays
Posted 09 November 2009 - 06:46 PM
BabyKodou, on 9 Nov, 2009 - 05:36 PM, said:
What about the BinarySearch method using Java.util.Arrays class
Hmmm that would work, I just looked at the Javadoc's and I was not aware that the method was there. I'm currently taking a data structures class in C++ so I guess I'm just used to my professor making me write out this stuff. The binary search method is easy to use, just pass it the name of your array (e.g. Foo) and whatever value your looking for (e.g. 5). so something like....
```int location = binarySearch(Foo, 5);
```
Then if you want true / false just say if location is > 0 then the value exists in the array, or if the value is negative then the value is not in the array.
### #11 theautokustomizer
• D.I.C Regular
Reputation: 16
• Posts: 250
• Joined: 20-September 09
## Re: Java Programming---Arrays
Posted 09 November 2009 - 06:51 PM
We just got done with arrays, you are correct with the binarysearch, and I feel like an idiot for not thinking of it, lol.
This post has been edited by theautokustomizer: 09 November 2009 - 06:51 PM | 2,979 | 10,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2015-35 | longest | en | 0.667862 |
https://www.lessonup.com/en/lesson/bfB3HdFEPeuWKgr6L | 1,712,971,578,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00218.warc.gz | 817,563,795 | 37,658 | # Summary chapter 7: wordformulas 7.1 en 7.2
H7 Wordformulas
1 / 13
Slide 1: Slide
WiskundeMiddelbare schoolvwoLeerjaar 1
This lesson contains 13 slides, with text slides.
Lesson duration is: 30 min
## Items in this lesson
H7 Wordformulas
#### Slide 1 -Slide
7.1 From description to wordformulas
7.2 Graphs from formulas
7.3 Linear formules
7.4 Making formulas for graphs
7.5 Comparing graphs
#### Slide 2 -Slide
To make a wordformula from the setting on the left you must find regularity!
You can see at each table there are 4 chairs. So if you know the number of tables you have to multiply by 4. But, be aware of the fixed amount! At the end there are 2 more chairs
Number of tables times 4 plus 2 is the number of chairs
7.1
#### Slide 3 -Slide
And again: At this table always 6 chairs. So if you know the number of chairs you have to multiply with 6. But!!! At the end there are 2 more chairs so you must add 2 as well.
The number of tables times 6 plus 2 is the number of chairs.
Formula: number of tables x 6 + 2 = number of chairs
Fixed amount and 'per' number
#### Slide 4 -Slide
How to make a wordformula from the setting on the left you must find regularity!
You can see at each table there are 4 chairs. So if you know the number of tables you have to multiply by 4. But, be aware of the fixed amount! At the end there are 2 more chairs
Number of tables times 4 plus 2 is the number of chairs
#### Slide 5 -Slide
It doesn't matter how many cupcakes she makes, the 12 euros is just once. The 12 euros is what we call the fixed amount
Profit = .... - number of cupcakes x .....
#### Slide 6 -Slide
It doesn't matter how many cupcakes she makes, the 12 euros is just once. The 12 euros is what we call the fixed amount
Profit = 12 - number of cupcakes x 0,80
#### Slide 7 -Slide
7.2
Graphs from formulas: method-theory
#### Slide 8 -Slide
7.2
Graphs from formulas: method-theory
#### Slide 9 -Slide
In this formula time is in hours and the distance in km.
Calculate the distance in case you've cycled for 3 hours.
15 x time =distance
7.2
#### Slide 10 -Slide
In this formula time is in hours and the distance in km.
Calculate the distance in case you've cycled for 3 hours.
15 x time =distance
15 x 3 = distance
45 = distance
so 45 km cycling
7.2
#### Slide 11 -Slide
Formula: 15 x time =distance
Table:
7.2
time
0
1
2
3
distance
0
15
30
45
#### Slide 12 -Slide
Formula: 15 x time =distance
Table:
Graph:
7.2
Graphs from formulas: theory
time
0
1
2
3
distance
0
15
30
45 | 752 | 2,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.815982 |
http://easy-ciphers.com/gideoni | 1,566,763,087,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330800.17/warc/CC-MAIN-20190825194252-20190825220252-00089.warc.gz | 64,673,611 | 21,674 | Easy Ciphers Tools:
Caesar cipher
Caesar cipher, is one of the simplest and most widely known encryption techniques. The transformation can be represented by aligning two alphabets, the cipher alphabet is the plain alphabet rotated left or right by some number of positions.
When encrypting, a person looks up each letter of the message in the 'plain' line and writes down the corresponding letter in the 'cipher' line. Deciphering is done in reverse.
The encryption can also be represented using modular arithmetic by first transforming the letters into numbers, according to the scheme, A = 0, B = 1,..., Z = 25. Encryption of a letter x by a shift n can be described mathematically as
Plaintext: gideoni
cipher variations: hjefpoj ikfgqpk jlghrql kmhisrm lnijtsn mojkuto npklvup oqlmwvq prmnxwr qsnoyxs rtopzyt supqazu tvqrbav uwrscbw vxstdcx wytuedy xzuvfez yavwgfa zbwxhgb acxyihc bdyzjid cezakje dfablkf egbcmlg fhcdnmh
Decryption is performed similarly,
(There are different definitions for the modulo operation. In the above, the result is in the range 0...25. I.e., if x+n or x-n are not in the range 0...25, we have to subtract or add 26.)
Atbash Cipher
Atbash is an ancient encryption system created in the Middle East. It was originally used in the Hebrew language.
The Atbash cipher is a simple substitution cipher that relies on transposing all the letters in the alphabet such that the resulting alphabet is backwards.
The first letter is replaced with the last letter, the second with the second-last, and so on.
An example plaintext to ciphertext using Atbash:
Plain: gideoni Cipher: trwvlmr
Baconian Cipher
To encode a message, each letter of the plaintext is replaced by a group of five of the letters 'A' or 'B'. This replacement is done according to the alphabet of the Baconian cipher, shown below.
```a AAAAA g AABBA m ABABB s BAAAB y BABBA
b AAAAB h AABBB n ABBAA t BAABA z BABBB
c AAABA i ABAAA o ABBAB u BAABB
d AAABB j BBBAA p ABBBA v BBBAB
e AABAA k ABAAB q ABBBB w BABAA
f AABAB l ABABA r BAAAA x BABAB
```
Plain: gideoni Cipher: AABBA ABAAA AAABB AABAA ABBAB ABBAA ABAAA
Affine Cipher
In the affine cipher the letters of an alphabet of size m are first mapped to the integers in the range 0..m - 1. It then uses modular arithmetic to transform the integer that each plaintext letter corresponds to into another integer that correspond to a ciphertext letter. The encryption function for a single letter is
where modulus m is the size of the alphabet and a and b are the key of the cipher. The value a must be chosen such that a and m are coprime.
Considering the specific case of encrypting messages in English (i.e. m = 26), there are a total of 286 non-trivial affine ciphers, not counting the 26 trivial Caesar ciphers. This number comes from the fact there are 12 numbers that are coprime with 26 that are less than 26 (these are the possible values of a). Each value of a can have 26 different addition shifts (the b value) ; therefore, there are 12*26 or 312 possible keys.
Plaintext: gideoni
cipher variations:
hjefpoj
tzknroz
fpqvtop
rfwdvof
dvclxov
plitzol
nrujdor
zharfoh
lxgzhox
xnmhjon
jdsplod
vtyxnot
ikfgqpk
ualospa
gqrwupq
sgxewpg
ewdmypw
qmjuapm
osvkeps
aibsgpi
myhaipy
yonikpo
ketqmpe
wuzyopu
jlghrql
vbmptqb
hrsxvqr
thyfxqh
fxenzqx
rnkvbqn
ptwlfqt
bjcthqj
nzibjqz
zpojlqp
lfurnqf
xvazpqv
kmhisrm
wcnqurc
istywrs
uizgyri
gyfoary
solwcro
quxmgru
ckduirk
oajckra
aqpkmrq
mgvsorg
ywbaqrw
lnijtsn
xdorvsd
jtuzxst
vjahzsj
hzgpbsz
tpmxdsp
rvynhsv
dlevjsl
pbkdlsb
brqlnsr
nhwtpsh
zxcbrsx
mojkuto
yepswte
kuvaytu
wkbiatk
iahqcta
uqnyetq
swzoitw
emfwktm
qclemtc
csrmots
oixuqti
aydcsty
npklvup
zfqtxuf
lvwbzuv
xlcjbul
jbirdub
vrozfur
txapjux
fngxlun
rdmfnud
dtsnput
pjyvruj
bzedtuz
oqlmwvq
agruyvg
mwxcavw
ymdkcvm
kcjsevc
wspagvs
uybqkvy
gohymvo
sengove
eutoqvu
qkzwsvk
cafeuva
prmnxwr
bhsvzwh
nxydbwx
zneldwn
ldktfwd
xtqbhwt
vzcrlwz
hpiznwp
tfohpwf
fvuprwv
rlaxtwl
dbgfvwb
qsnoyxs
citwaxi
oyzecxy
aofmexo
melugxe
yurcixu
iqjaoxq
ugpiqxg
gwvqsxw
smbyuxm
echgwxc
rtopzyt
djuxbyj
pzafdyz
bpgnfyp
nfmvhyf
zvsdjyv
xbetnyb
jrkbpyr
vhqjryh
hxwrtyx
tnczvyn
fdihxyd
supqazu
ekvyczk
qabgeza
cqhogzq
ognwizg
awtekzw
ycfuozc
kslcqzs
wirkszi
iyxsuzy
uodawzo
gejiyze
tvqrbav
flwzdal
rbchfab
driphar
phoxjah
bxuflax
ltmdrat
xjsltaj
jzytvaz
vpebxap
hfkjzaf
uwrscbw
gmxaebm
scdigbc
esjqibs
qipykbi
cyvgmby
aehwqbe
munesbu
yktmubk
kazuwba
wqfcybq
iglkabg
vxstdcx
hnybfcn
tdejhcd
ftkrjct
rjqzlcj
dzwhncz
bfixrcf
nvoftcv
zlunvcl
lbavxcb
xrgdzcr
jhmlbch
wytuedy
iozcgdo
uefkide
gulskdu
skramdk
eaxioda
cgjysdg
owpgudw
amvowdm
mcbwydc
kinmcdi
xzuvfez
vfgljef
hvmtlev
tlsbnel
fbyjpeb
dhkzteh
pxqhvex
bnwpxen
ndcxzed
ztifbet
ljondej
yavwgfa
kqbeifq
wghmkfg
iwnumfw
umtcofm
gczkqfc
eilaufi
qyriwfy
coxqyfo
oedyafe
aujgcfu
mkpoefk
zbwxhgb
lrcfjgr
xhinlgh
jxovngx
vnudpgn
hdalrgd
fjmbvgj
rzsjxgz
dpyrzgp
pfezbgf
bvkhdgv
nlqpfgl
acxyihc
msdgkhs
yijomhi
kypwohy
woveqho
iebmshe
gkncwhk
satkyha
eqzsahq
qgfachg
cwliehw
omrqghm
bdyzjid
ntehlit
zjkpnij
lzqxpiz
xpwfrip
jfcntif
hlodxil
tbulzib
fratbir
rhgbdih
dxmjfix
pnsrhin
cezakje
oufimju
aklqojk
maryqja
yqxgsjq
kgdoujg
impeyjm
ucvmajc
gsbucjs
sihceji
eynkgjy
qotsijo
dfablkf
pvgjnkv
blmrpkl
nbszrkb
zryhtkr
lhepvkh
jnqfzkn
vdwnbkd
htcvdkt
tjidfkj
fzolhkz
rputjkp
egbcmlg
qwhkolw
cmnsqlm
octaslc
asziuls
mifqwli
korgalo
wexocle
iudwelu
ukjeglk
gapmila
sqvuklq
fhcdnmh
rxilpmx
dnotrmn
pdubtmd
btajvmt
njgrxmj
lpshbmp
xfypdmf
jvexfmv
vlkfhml
hbqnjmb
trwvlmr
gideoni
syjmqny
eopusno
qevcune
cubkwnu
okhsynk
mqticnq
ygzqeng
kwfygnw
wmlginm
icroknc
usxwmns
The decryption function is
where a - 1 is the modular multiplicative inverse of a modulo m. I.e., it satisfies the equation
The multiplicative inverse of a only exists if a and m are coprime. Hence without the restriction on a decryption might not be possible. It can be shown as follows that decryption function is the inverse of the encryption function,
ROT13 Cipher
Applying ROT13 to a piece of text merely requires examining its alphabetic characters and replacing each one by the letter 13 places further along in the alphabet, wrapping back to the beginning if necessary. A becomes N, B becomes O, and so on up to M, which becomes Z, then the sequence continues at the beginning of the alphabet: N becomes A, O becomes B, and so on to Z, which becomes M. Only those letters which occur in the English alphabet are affected; numbers, symbols, whitespace, and all other characters are left unchanged. Because there are 26 letters in the English alphabet and 26 = 2 * 13, the ROT13 function is its own inverse:
ROT13(ROT13(x)) = x for any basic Latin-alphabet text x
An example plaintext to ciphertext using ROT13:
Plain: gideoni Cipher: tvqrbav
Polybius Square
A Polybius Square is a table that allows someone to translate letters into numbers. To give a small level of encryption, this table can be randomized and shared with the recipient. In order to fit the 26 letters of the alphabet into the 25 spots created by the table, the letters i and j are usually combined.
1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z
Basic Form:
Plain: gideoni Cipher: 22424151433342
Extended Methods:
Method #1
Plaintext: gideoni
method variations: moiktso rtopyxt wytudcy bdyzihd
Method #2
Bifid cipher
The message is converted to its coordinates in the usual manner, but they are written vertically beneath:
```g i d e o n i
2 4 4 5 4 3 4
2 2 1 1 3 3 2 ```
They are then read out in rows:
24454342211332
Then divided up into pairs again, and the pairs turned back into letters using the square:
Plain: gideoni Cipher: ryoiblh
Method #3
Plaintext: gideoni
method variations: rrvqnsg rvqnsgr vqnsgrr qnsgrrv nsgrrvq sgrrvqn grrvqns
Read more ...[RUS] , [EN]
Permutation Cipher
In classical cryptography, a permutation cipher is a transposition cipher in which the key is a permutation. To apply a cipher, a random permutation of size E is generated (the larger the value of E the more secure the cipher). The plaintext is then broken into segments of size E and the letters within that segment are permuted according to this key.
In theory, any transposition cipher can be viewed as a permutation cipher where E is equal to the length of the plaintext; this is too cumbersome a generalisation to use in actual practice, however.
The idea behind a permutation cipher is to keep the plaintext characters unchanged, butalter their positions by rearrangement using a permutation
This cipher is defined as:
Let m be a positive integer, and K consist of all permutations of {1,...,m}
For a key (permutation) , define:
The encryption function
The decryption function
A small example, assuming m = 6, and the key is the permutation :
The first row is the value of i, and the second row is the corresponding value of (i)
The inverse permutation, is constructed by interchanging the two rows, andrearranging the columns so that the first row is in increasing order, Therefore, is:
Total variation formula:
e = 2,718281828 , n - plaintext length
Plaintext: gideoni
all 5040 cipher variations:
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Read more ...[1] , [2] , [3] | 24,115 | 49,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2019-35 | latest | en | 0.811491 |
http://freebooksdownloads.net/mathematics-for-classical-information-retrieval-roots-and-applications/ | 1,618,213,531,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00435.warc.gz | 37,728,726 | 10,834 | Home » Mathematics » Mathematics for Classical Information Retrieval: Roots and Applications
Mathematics for Classical Information Retrieval: Roots and Applications
• Category: Mathematics
• Author: Dariush Alimohammadi, Mary K. Bolin
• Pages: 86 pages
Read and download free eBook intituled Mathematics for Classical Information Retrieval: Roots and Applications in format – 86 pages created by Dariush Alimohammadi, Mary K. Bolin.
This book is about Information Retrieval (IR), particularly Classical Information Retrieval (CIR). It looks at these topics through their mathematical roots. The mathematical bases of CIR are briefly reviewed, followed by the most important and interesting models of CIR, including Boolean, Vector Space, and Probabilistic. The primary goal of book is to create a context for understanding the principles of CIR by discussing its mathematical bases.
Mathematics is a foundation and building block of all areas of knowledge. It particularly affects disciplines concerned with information organization, storage, retrieval, and exchange. Information is manipulated using computers, and computers have a mathematical basis. The word “computer” reveals this relationship. Students and practitioners of computer science, library and information science (LIS), and communications need a foundation in mathematics. IR, a subfield in all these disciplines, also needs mathematics as a common and formal language. Understanding CIR is not possible without basic mathematical knowledge.
The primary goal of book is to create a context for understanding the principles of CIR by discussing its mathematical bases. This book can be helpful for LIS students who are studying IR but have no knowledge of mathematics. Weakness in math impairs the ability to understand current issues in IR. While LIS students are the main target of this book, it may be of interest to computer science and communications students as well.
READ Lecture Notes for Linear Algebra | 379 | 1,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-17 | latest | en | 0.898203 |
https://www.physicsforums.com/threads/group-theory-why-transformations-of-hamiltonian-are-unitary.843116/ | 1,508,487,743,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823839.40/warc/CC-MAIN-20171020063725-20171020083725-00106.warc.gz | 985,263,395 | 14,260 | # Group Theory why transformations of Hamiltonian are unitary?
1. Nov 14, 2015
### applestrudle
This is what I have so far:
I'm trying to show that the matrix D has to be unitary. It is the matrix that transforms the wavefunction.
2. Nov 14, 2015
### HallsofIvy
Staff Emeritus
The matrix that transforms the wave function how? So that it preserves some property? Transformations do NOT have to be unitary unless you are trying to lengths of vectors.
3. Nov 14, 2015
### applestrudle
In lectures we were showing
Tψ(r) = ψ(Ur) = ΣDij ψ(r)
Dij has to be unitary and form a representation of T - I'm just trying to figure out the proof. Are you saying this is only try if you scale the position vector r? | 194 | 711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-43 | longest | en | 0.947292 |
https://codereview.stackexchange.com/questions/tagged/bitwise?tab=Frequent | 1,579,516,768,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598217.23/warc/CC-MAIN-20200120081337-20200120105337-00331.warc.gz | 382,862,397 | 32,999 | # Questions tagged [bitwise]
Bitwise refers to a bitwise operation which operates on one or more bit patterns or binary numerals at the level of their individual bits.
25 questions
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34k views
### Efficiently checking if a number is divisible by 3
I'm looking for code review, best practices and optimizations. ...
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### Printing longest sequence of zeroes
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### Find the binary gap of a number N
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### Testing for divisibility of numbers - follow-up
(Here is the follow up question) I had another go at trying to speed up my program described in my previous question. JS1's answer was particulary helpful and now my code is only about 20% slower ...
105 views
### Program to test for divisibility of numbers
(EDIT: here is the follow up qusetion) Using this program to test for the smallest number where its permutations or itself is divisible by 10 or less numbers is twice as slow as the fastest program I ...
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### Implementation of C Standard Library Function ntohl()
This is an implementation of ntohl() that I wrote as an exercise. ntohl() takes a uint32_t ...
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### Integer Log2 implemented using binary search
I saw a simple method to calculate Log2(N), where N is the power of 2: ...
229 views
### Array-like container for uints shorter than 8 bits
Follow-up: Array-like container for uints shorter than 8 bits (Rev 1) First of all: yes, I need this kind of array to generate data for a piece of hardware, and it must be stored and manipulated ...
2k views
### Diamond in the rough! Finding Gems in the rocks
Recently a question was asked that solved a hackerrank problem: Counting gems amongst the rocks The description of the challenge is: John has discovered various rocks. Each rock is composed of ...
917 views
As some of you know, I have been working with Gray codes and am trying to design a good gray_code class in modern C++. I will end up posting several parts of the ...
201 views
### Simulation of image transfer over noisy channel
Following this question, I've tried to rewrite some core methods to avoid using String for bit operations, and I ended up using ...
614 views
### Gray codes addition — Take 2
For a while, I have tried to find ways (known or new) to do stuff with Gray codes. As some of you probably know, I already tried to implement an algorithm to add two Gray codes without having to ...
384 views
### Array-like container for uints shorter than 8 bits (Rev 1)
This is a follow-up to Array-like container for uints shorter than 8 bits In short: The PackedBitfieldArray should be a kind of container for packed (unsigned) ...
2k views
### Custom integer class for calculating crypto algorithms
How can I improve upon this integer class that I wrote? I am/will be using this for some calculation intensive crypto algorithms, mainly POLY1305AES. I might even write RSA using this. I know that ...
357 views
### Rotating 3D bit board in C
I've got a 64 bit bitboard, but instead of interpreting as an 8 by 8 binary matrix, as is usually the case, I've taken it to the third dimension, so to speak. Now mirroring and even rotating by 180 ...
18k views
### Split a long integer into eight 4-bit values
It's an exercise from a book. There is a given long integer that I should convert into 8 4-bit values. The exercise is not very clear for me, but I think I understood the task correctly. I googled ...
444 views
### Packing and unpacking bits (2)
Part 1 here. I improved my code from my previous question (linked above). ...
4k views
### Codility binary gap solution using regex
Below is the code I implemented to solve the binary gap problem: Find longest sequence of zeros, bounded by ones, in binary representation of an integer. However, it seems to be different from ...
346 views
### Fastest sum of absolute values of 32 differences
A long should be treated as a list of 32 unsigned numbers, each of them just two bit long. So 0x1234F... represents ...
2k views
### Finding the fastest common prefix of 2 strings in Python
I'm comparing my algorithms, slice_prefix and bit_prefix, with existing ones to find the common prefix length of 2 strings as ...
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### Packing and unpacking bits
Part 2 here. I wrote this class for packing data. Would it benefit from being named BitPacker rather than BitStream (and change write/read to pack/unpack), or it works as is? How can I improve it? ...
2k views
### Hamming distance between numbers in JavaScript
In Leetcode it states that my runtime is only faster than 38% all of submitted JavaScript solutions. Is there anything I can change to make it more efficient? ...
155 views
### C bit mask calculator written using GTK+ 3
I've been studying C on my own using K&R 2nd Edition, and started exploring GTK+ to practice some of the concepts I've learned by making GUI applications. This is the first GUI application I've ... | 1,218 | 5,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.930756 |
https://www.physicsforums.com/threads/solving-integrals-int-frac-xdx-1-x-4.73887/ | 1,721,066,163,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00094.warc.gz | 820,253,406 | 15,399 | # Solving Integrals: \int\frac{xdx}{1+x^4}
• Echo 6 Sierra
In summary, an integral is a mathematical concept used to find the area under a curve on a graph. To solve an integral, one must use algebraic manipulation and integration techniques to find an antiderivative and evaluate it over a given range. The general process for solving integrals involves identifying the type, finding the antiderivative, applying techniques, and evaluating. The purpose of solving integrals is to find the total value of a function over a given range, which has many real-world applications. Common integration techniques include substitution, integration by parts, partial fractions, and trigonometric substitution.
Echo 6 Sierra
OK, for the bad mamma-jamma:
$$\int\frac{xdx}{1+x^4}$$ I picked $$u=x^2$$ and made du=2xdx which makes 1/2du=xdx.
Now I have: $$\displaystyle{\frac{1}{2}}$$$$\int\frac{du}{1+u^2}$$
If the almighty green-bottled lager from Holland has been good to me, I get:
$$\frac{1}{2} \arctan x^2+c$$
Is this: a) correct, b)good enough or c)can this be tweaked?
So help me, if I get at least a B in Calculus I will be near Spiritual Creaminess.
Yes it's correct.
First of all, great job on solving the integral! Your approach of using u-substitution and picking u=x^2 was very effective. Your final answer of 1/2 arctan x^2 + c is indeed correct and can be considered good enough. However, if you want to tweak it a bit, you can also express it as 1/2 arctan (1+x^2) + c. This is just a different way of writing the same answer and may be considered more simplified. Overall, your work shows a strong understanding of integration techniques and I have no doubt that you will do well in Calculus. Keep up the good work!
## What is an integral?
An integral is a mathematical concept that represents the area under a curve on a graph. It is used to find the total value of a function over a given range.
## How do you solve an integral?
To solve an integral, you need to use a combination of algebraic manipulation and integration techniques. This involves finding an antiderivative of the function and evaluating it over the given range.
## What is the general process for solving integrals?
The general process for solving integrals involves identifying the type of integral (definite or indefinite), finding an antiderivative, applying any necessary integration techniques, and evaluating the integral over the given range.
## What is the purpose of solving integrals?
Solving integrals allows us to find the total value of a function over a given range. This can be useful in many real-world applications, such as calculating areas, volumes, and rates of change.
## What are some common integration techniques used to solve integrals?
Some common integration techniques include substitution, integration by parts, partial fractions, and trigonometric substitution. These techniques can help simplify the integral and make it easier to solve.
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2K | 836 | 3,384 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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} | 3.953125 | 4 | CC-MAIN-2024-30 | latest | en | 0.8991 |
https://translate.academic.ru/Fourier%20series/lt/ | 1,586,167,791,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371620338.63/warc/CC-MAIN-20200406070848-20200406101348-00455.warc.gz | 640,577,884 | 14,302 | # Fourier series
• 1 daugiaserijinis
• 2 kelių dalių tęstinis pasakojimas
• 3 leisti atskiromis dalimis/tęstinį leidinį
• 4 paeiliui einantis
• 5 serialas
• 6 serialo statymas
• 7 serijinis
• 8 statyti serialą
• 9 susidedantis iš dalių
• 10 žudikas maniakas
### См. также в других словарях:
• Fourier series — Fourier transforms Continuous Fourier transform Fourier series Discrete Fourier transform Discrete time Fourier transform Related transforms … Wikipedia
• Fourier series — Fourier eilutė statusas T sritis automatika atitikmenys: angl. Fourier series vok. Fourier Reihe, f rus. ряд Фурье, m pranc. série de Fourier, f ryšiai: sinonimas – Furjė eilutė … Automatikos terminų žodynas
• Fourier series — [foor′ē ər] n. [formulated by FOURIER Baron Jean Baptiste Joseph] Math. the expansion of a periodic function into a series of sines and cosines … English World dictionary
• Fourier series — Furjė eilutė statusas T sritis fizika atitikmenys: angl. Fourier series vok. Fourier Reihe, f rus. ряд Фурье, m pranc. série de Fourier, f … Fizikos terminų žodynas
• Fourier series — Math. an infinite series that involves linear combinations of sines and cosines and approximates a given function on a specified domain. [1875 80; see FOURIER ANALYSIS] * * * In mathematics, an infinite series used to solve special types of… … Universalium
• Fourier series — noun Etymology: Baron J.B.J. Fourier Date: 1877 an infinite series in which the terms are constants multiplied by sine or cosine functions of integer multiples of the variable and which is used in the analysis of periodic functions … New Collegiate Dictionary
• Fourier series — noun Mathematics an infinite series of trigonometric functions used to represent a given periodic function. Origin named after the 19th cent. French mathematician Jean B. J. Fourier … English new terms dictionary
• Fourier series — Fou′rier se ries n. math. an infinite series that approximates a given function on a specified domain by using linear combinations of sines and cosines • Etymology: 1875–80; see Fourier analysis … From formal English to slang
• Fourier series — noun A series of cosine and sine functions or complex exponentials resulting from the decomposition of a periodic function … Wiktionary
• Fourier series — n. Math. an expansion of a periodic function as a series of trigonometric functions … Useful english dictionary
• Fourier series — Смотри закон (или число) Фурье … Энциклопедический словарь по металлургии
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https://www.hanspub.org/journal/PaperInformation.aspx?paperID=30495 | 1,590,851,372,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347409337.38/warc/CC-MAIN-20200530133926-20200530163926-00525.warc.gz | 739,464,640 | 26,747 | # 一种对缆索无损探伤微弱信号的处理方法A Weak Signal Process Method of Cable Nondestructive Inspection
DOI: 10.12677/HJCE.2019.83091, PDF, HTML, XML, 下载: 274 浏览: 399
Abstract: A method is designed to deal with the problem that the signal-to-noise ratio of the weak signal is small when processing the weak signal of nondestructive inspection of cables, which can’t clearly reflect the tiny defects inside the cables. This method collects the weak signal several times under the accurate timing control and accumulates the collected signal sequence strictly according to the time line. Because the distribution of signal noise in time domain belongs to normal distribution, i.e. uniform and symmetrical distribution on both sides of the real value will cancel each other after accumulating, while the real signal does not belong to normal distribution. After multiple ac-cumulations, the signal value will be amplified. By this method, the real signal can be amplified ef-fectively besides the noise, so that the weak signal generated by the weak internal defects can be clearly distinguished, and the accuracy of nondestructive inspection is greatly improved. This method is simple, easy to operate and has a wide range of applications, and has high popularization value.
1. 引言
2. 缆索无损探伤系统
Figure 1. Hardware diagram of cable NDT system
3. 信号处理
$p\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }\mathrm{exp}\left\{-\frac{{\left(x-\mu \right)}^{2}}{2{\sigma }^{2}}\right\}$ (1)
$F\left(x\right)=\frac{1}{\sqrt{2\text{π}}\sigma }{\int }_{-\infty }^{x}\mathrm{exp}\left\{-\frac{{\left(x-\mu \right)}^{2}}{2{\sigma }^{2}}\right\}\text{d}x$ (2)
$E\left(x\right)=\frac{1}{\sqrt{2\text{π}}\sigma }{\int }_{-\infty }^{+\infty }x\mathrm{exp}\left\{-\frac{{\left(x-\mu \right)}^{2}}{2{\sigma }^{2}}\right\}\text{d}x$ (3)
(4)
${S}_{\left[{T}_{1}\text{}{T}_{2}\text{}\cdots {T}_{n}\right]}=\left[\begin{array}{ccc}{F}_{\left(1\right){T}_{1}}& \cdots & {F}_{\left(1\right){T}_{n}}\\ ⋮& \ddots & ⋮\\ {F}_{\left(M\right){T}_{1}}& \cdots & {F}_{\left(M\right){T}_{n}}\end{array}\right]$ (5)
${S}_{\left[{T}_{1}\text{}{T}_{2}\text{}\cdots {T}_{n}\right]}=\left[M{\mu }_{1}M{\mu }_{2}\cdots M{\mu }_{n}\right]$ (6)
1) 由图1中的4产生时间线，作为整个流程的时序基准。
2) 在T1时刻，由图1中的2产生发射脉冲信号，并等待50 ms。
3) 等时间间隔，从图1中的3采集回波信号序列，保存为 $\left({F}_{1},{F}_{2},\cdots {F}_{n}\right)$
4) 重复2~3步骤M次，得到 ${\left({F}_{1},{F}_{2},\cdots {F}_{n}\right)}_{1}$ ~ ${\left({F}_{1},{F}_{2},\cdots {F}_{n}\right)}_{M}$ ，M组信号序列。
5) 将M组信号序列按照进行矩阵相加运算，得到结果序列。对结果序列进行归一化，得到最终结果序列数据。
6) 将结果序列通过图1中的5进行数据输出，显示成回波曲线，方便测试人员判断结果。
Figure 2. Signal process flow chart
4. 实验结果分析
Table 1. Signal processing results of coefficients
Figure 3. Original signal
Figure 4. Processed signal with coefficient 10
Figure 5. Processed signal with coefficient 50
Figure 6. Processed signal with coefficient 100
Figure 7. Processed signal with coefficient 1000
5. 结束语
[1] 邱宇, 朱承建. 钢丝绳无损探伤仪研究[J]. 电子世界, 2013(13): 128-130. [2] 宋静. 手持式超声波探伤仪应用软件设计[D]: [硕士学位论文]. 南京: 南京航空航天大学, 2009. [3] 董谦, 谢剑英. 超声波无损检测系统的噪声干扰分析与抑制[J]. 测控技术, 2001, 20(11): 50-52. [4] 赵文礼, 夏炜, 刘鹏, 等. 基于混沌理论的微弱信号放大原理与方法研究[J]. 物理学报, 2010, 59(5): 2962-2970. [5] 朱勇辉, 朱其猛. 超声波应力测试降噪技术[J]. 电焊机, 2018, 48(2): 105-107. [6] 彭瑾, 王黎, 高晓蓉, 等. 基于相位分析的超声波噪声信号抑制算法[J]. 铁道技术监督, 2009, 37(1): 34-37. [7] 徐灵活, 殷明, 章建程, 等. 正态分布法在噪声听觉损伤易感性评价中的应用探讨[J]. 海军医学杂志, 2007, 28(1): 5-7. [8] 王培容, 杨艺. 统计平均拟合正态分布曲线的迭代算法及其应用[J]. 重庆工商大学学报(自然科学版), 2001, 18(1): 37-40. [9] 金秋, 刘少岗. 线性不对称损失时非正态分布过程均值优化[J]. 运筹与管理, 2014(6): 23-28. [10] 辛士波. 基于双侧等截尾正态分布的均值-极差控制图[J]. 火力与指挥控制, 2012, 37(8): 180-183. | 1,542 | 3,539 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-24 | latest | en | 0.727571 |
https://learn.careers360.com/engineering/question-equation-img-altmathrma-x22-h-x-yb-y20-srchttpsentrancecorneroncodecogscomgiflatex5cmathrm7ba20x5e7b27dplus220h20x20yplusb20y5e7b27d3d07d/ | 1,723,513,779,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00130.warc.gz | 280,256,836 | 36,307 | #### Equation $\mathrm{a x^{2}+2 h x y+b y^{2}=0}$ represents a pairs of lines, combined equation of lines that can be obtained by taking the mirror of lines about the $\mathrm{x}$-axis isOption: 1 $\mathrm{a x^{2}+2 h x y+b y^{2}=0}$Option: 2 $\mathrm{b x^{2}+2 h x y+a y^{2}=0}$Option: 3 $\mathrm{b x^{2}-2 h x y+a y^{2}=0}$Option: 4 none of these
Let the lines represented by $\mathrm{a x^{2}+b y^{2}+2 h x y=0\, be \, y=m_{1} x\, and \, y=m_{2} x,then \: m_{1}+m_{2}=\frac{-2 \mathrm{~h}}{\mathrm{~b}}}$, $\mathrm{m_{1} m_{2}=a / b}$. If these lines reflected about the x-axis, there equation becomes $\mathrm{y+m_{1} x= 0\, and \, y+m_{2} x=0}$ and their combined equation is $\mathrm{\left(y+m_{1} x\right)\left(y+m_{2} x\right)=0}$
$\mathrm{\Rightarrow y^{2}+x y\left(m_{1}+m_{2}\right)+m_{1} m_{2} x^{2}=0}$
$\mathrm{\Rightarrow \mathrm{by}^{2}-2 \mathrm{hxy}+a \mathrm{x}^{2}=0}$ | 367 | 891 | {"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": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.559969 |
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In worksheet on word problems on linear equation in one variable student can practice different types of equation problems.
1. Convert the following statements into equations.
(a) 5 added to a number is 9.
(b) 3 subtracted from a number is equal to 12.
(c) 5 times a number decreased by 2 is 4.
(d) 2 times the sum of the number x and 7 is 13.
2. A number is 12 more than the other. Find the numbers if their sum is 48.
3. Twice the number decreased by 22 is 48. Find the number.
4. Seven times the number is 36 less than 10 times the number. Find the number.
5. 4/5 of a number is more than 3/4 of the number by 5 . Find the number.
6. The sum of two consecutive even numbers is 38. Find the numbers.
7. The sum of three consecutive odd numbers is 51. Find the numbers.
8. Rene is 6 years older than her younger sister. After 1 0 years, the sum of their ages will be 50 years. Find their present ages.
9. The length of a rectangle is 10 m more than its breadth. If the perimeter of rectangle is 80 m, find the dimensions of the rectangle.
10. A 300 m long wire is used to fence a rectangular plot whose length is twice its width. Find the length and breadth of the plot.
11. The denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and denominator is decreased by 1, the number obtained is 3/2, find the fraction.
12. A sum of $2700 is to be given in the form of 63 prizes. If the prize is of either$100 or $25, find the number of prizes of each type. 13. In a class of 42 students, the number of boys is 2/5 of the girls. Find the number of boys and girls in the class. 14. Among the two supplementary angles, the measure of the larger angle is 36° more than the measure of smaller. Find their measures. 15. My mother is 12 years more than twice my age. After 8 years, my mother’s age will be 20 years less than three times my age. Find my age and my mother’s age. 16. In an isosceles triangle, the base angles are equal and the vertex angle is 80°. Find the measure of the base angles. 17. Adman’s father is 49 years old. He is 5 years older than four times Adman’s age. What is Adman’s age? 18. The cost of a pencil is 25 cents more than the cost of a eraser. If the cost of 8 pencils and 10 erasers is$12.80, find the cost of each.
19. Divide 36 into two parts in such a way that 1/5 of one part is equal to 1/7 of the other.
20. The length of the rectangle exceeds its breadth by 3 cm. If the length and breadth are each increased by 2 cm, then the area of new rectangle will be 70 sq. cm more than that of the given rectangle. Find the length and breadth of the given rectangle.
Answers for worksheet on word problems on linear equation in one variable are given below so that student can check the exact answers of the equation problems.
1. (a) x + 5 = 9
(b) x - 3 = 12
(c) 5x - 2 = 4
(d) 2(x + 7) = 13
2. 18, 30
3. 35
4. 12
5. 100
6. 18, 20
7. 15, 17, 19
8. 12, 18 years
9. 15 cm, 25 cm
10. 50 m, 100 m
11. 13/21
12. 15, 48
13. 30 girls, 12 boys
14. 72°, 108°
15. 16 years, 44 years
16. 50°
17. 11 years
18. eraser – 60 cents, pencil – 85 cents
19. 15, 21
20. length = 18 m, breadth = 15 m
Equations
What is an Equation?
What is a Linear Equation?
How to Solve Linear Equations?
Solving Linear Equations
Problems on Linear Equations in One Variable
Word Problems on Linear Equations in One Variable
Practice Test on Linear Equations
Practice Test on Word Problems on Linear Equations
Equations - Worksheets
Worksheet on Linear Equations
Worksheet on Word Problems on Linear Equation
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The value of a digit in a given number depends on its place or position in the number. This value is called its place value. | 1,320 | 4,788 | {"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} | 3.734375 | 4 | CC-MAIN-2024-38 | latest | en | 0.929601 |
https://usmam.org/factors-of-60-that-add-up-to-11/ | 1,653,674,210,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662675072.99/warc/CC-MAIN-20220527174336-20220527204336-00241.warc.gz | 654,071,736 | 5,941 | Factors the 60 are integers that can be separated evenly right into 60. There are 12 determinants of 60 of i m sorry 60 chin is the best factor and its prime components are 2, 3 and 5 The sum of all determinants of 60 is 168.
You are watching: Factors of 60 that add up to 11
Factors the 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and also 60Negative determinants of 60: -1, -2, -3, -4, -5, -6, -10, -12, -15, -20, -30 and -60Prime determinants of 60: 2, 3, 5Prime administer of 60: 2 × 2 × 3 × 5 = 22 × 3 × 5Sum of determinants of 60: 168
1 What are the components of 60? 2 How come Calculate determinants of 60? 3 Factors of 60 by prime Factorization 4 Factors that 60 in Pairs 5 FAQs on components of 60
## What space the factors of 60?
The factors of 60 room all the number that offer the value 60 once multiplied. Together 60 is an also number, ithas more than one factor. To know why the is composite, let's remind the meaning of a composite number. A number is said to be composite if it has an ext than two factors.
60 ÷2 = 335 and also 12are determinants of 60.Factors the 60are1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.
## How to CalculateFactors the 60?
To calculate the determinants of 60, we have the right to use various methods prefer prime factorization and the department method. In element factorization, us express 60 as a product of its prime factors and in the division method, we check out what numbers divide 60 exactly.
Hence, the components of 60 are,1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and60.
Explore factors using illustrations and interactive examples.
## Factors of 60 by prime Factorization
Prime factorization is the process of creating a number as a product the its element factors. Allow uslearn just how to discover the factors of a number utilizing prime factorization. We start by in search of the the smallest prime numberwhich can divide 60leaving the remainder together 0.As 60 is an also number, 2 is one of its factors.
The number 60is split by the smallest prime number which divides 60exactly, i.e. It leaves a remainder 0. The quotient is then split by the the smallest or second smallest element number and the process continues it spins the quotient i do not care indivisible. Let united state divide 60by the element number 2⇒60 ÷2 = 30Now we need to divide the quotient 30by the next the very least prime number. Since 30is an also number, the isdivisible through 2⇒30 ÷2 =1515 willbe divisible by 3⇒ 15 ÷ 3= 55 is a element number and also hence not more divisible.
Prime administrate of 60=2×2 × 3× 5Therefore, the prime components of 60are 2, 3, and 5.
### Prime administrate by factor Tree
The other way of prime factorization is acquisition 60as the root and creatingbranches by dividing it by the the smallest prime number. This an approach is comparable to the department method above. The distinction lies in presenting the factorization. The figure listed below shows the element tree of 60.
Collect all the number in circles and also write 60as the product of this numbers.Hence, 60 = 2×2 × 3× 5
Pair components are the components of a number given in pairs, which once multiplied together, offer that original number.6 and also 10both are components of 60. Because that example, 6× 10= 60.The pair determinants of 60 would be the two numbers which, once multiplied together, result in 60.
The following table represents the pair factors of 60:
Product Pair factors 1× 60 (1, 60) 2× 30 (2, 30) 3 × 20 (3, 20) 4 × 15 (4, 15) 5 × 12 (5, 12) 6 × 10 (6, 10)
We deserve to have negative factors additionally for a given number.Thenegative pair determinants of 60are(-1, -60), (-2, -30), (-3, -20), (-4, -15), (-5, -12) and (-6, -10).
Important Notes:
The factors of 60 are1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.Every even number will have actually 2 as one of its factors.60 is a composite number as it has an ext factors various other than 1 and itself.
Challenging Questions:
What are the typical factors of 60 and 600?Are 2/3 and also 4/5 determinants of 600?
Example 1: Fifteenfriends have actually 60pizzasfrom a pizza shopand distributed among themselves equally. How plenty of pizzazwould each among them get?
Solution:
15 pizzasare to be divided among friends equally.This suggests we have to divide 60by 15.60 ÷ 15= 4Hence, each friend will get 4 pizzas.
Example 2:Sia wants to know if there are any kind of common factors of 60 and also 80. Aid her in finding the answer come it.
Solution:
Factors that 60 =1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and also 60Factors of 80=1, 2, 4, 5, 8, 10, 16, 20, 40,80The common factors that 60and 80= 1, 2, 4, 5, 10, and also 20Hence, the usual factors the 60and 80are1, 2, 4, 5, 10, and also 20.
Example 3: discover the product of every the prime factors of 60.
Solution:
Since, the prime factors of 60 space 2, 3, 5. Therefore, the product of prime factors = 2 × 3 × 5 = 30.
## FAQs on determinants of 60
### What room the determinants of 60?
The determinants of 60 room 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 and also its negative factors room -1, -2, -3, -4, -5, -6, -10, -12, -15, -20, -30, -60.
### What is the Greatest common Factor that 60 and also 21?
The factors of 60 and also 21 room 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 and 1, 3, 7, 21 respectively.Common determinants of 60 and also 21 room <1, 3>.Hence, the GCF the 60 and 21 is 3.
### What Numbers room the Prime components of 60?
The prime factors of 60 are 2, 3, 5.
See more: Ngk Bkr6Es-11 - Cross Reference, Platinum Tt
### What are the usual Factors the 60 and 38?
Since, the factors of 60 space 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 and also the components of 38 are 1, 2, 19, 38.Hence, <1, 2> room the usual factors that 60 and 38.
### What is the sum of all the determinants of 60?
Sum that all determinants of 60 = (22 + 1 - 1)/(2 - 1) × (31 + 1 - 1)/(3 - 1) × (51 + 1 - 1)/(5 - 1) = 168 | 1,923 | 5,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2022-21 | latest | en | 0.900925 |
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Depdendent Variable
Number of equations to solve: 23456789
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Dependent Variable
Number of inequalities to solve: 23456789
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worksheet on how to multiply mixed numbers
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nanagm
Registered: 15.04.2006
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Registered: 11.03.2004
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Registered: 05.12.2002
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Registered: 08.01.2002
From: Australia
Posted: Wednesday 03rd of Jan 14:02 Well, if it is as you say; then do visit http://www.algebra-equation.com/exponential-and-logarithmic-equations.html. This is a software that helps people excel their math. | 942 | 3,820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-09 | latest | en | 0.910745 |
http://www.convex.org/category/mathematics/arithmetic/ | 1,503,316,040,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108268.39/warc/CC-MAIN-20170821114342-20170821134342-00325.warc.gz | 516,639,036 | 11,354 | Vietnamese math puzzle
Alex Bellos shared the following math puzzle that was given to Vietnamese eight-year-olds. The challenge is to place the digits from 1 to 9 into the grid.
Preliminary remarks
1. Although it wasn’t stated explicitly, I assume that each digit should be used only once.
2. The colon signifies division.
3. There are many correct answers. I assume that students were meant to use enlightened trial and error to find a solution, but were not expected to find all solutions.
4. I am assuming the standard order of operations, but I am not sure if this was the original intention. Would an eight-year-old be expected to understand that multiplication and division have higher precedence than addition and subtraction?
5. Frankly, this puzzle is kind of boring for grown-ups.
Python code
Searching for solutions by hand did not appeal to me, so I wrote a Python program to search for all solutions. I hope that this example is helpful for people who are learning Python. (Perhaps it will inspire someone to write a better program than this.)
from itertools import permutations
standard_order = True
tolerance = 1e-7
target = 66
if standard_order:
expr = "? + 13 * ? / ? + ? + 12 * ? - ? - 11 + ? * ? / ? - 10"
else:
expr = "(((? + 13) * ? / ? + ? + 12) * ? - ? - 11 + ?) * ? / ? - 10"
expr = expr.replace("?", "%d")
for inputs in permutations(range(1,10)):
expression = expr % inputs
result = eval(expression)
if abs(result - target) < tolerance:
print ("%s = %s" % (expression, target))
Discussion
The itertools package provides many useful tools for iteration, including the permutation function, which generates all permutations of a list. Our script iterates through all 362880 permutations of the set of integers from 1 to 9.
Each permutation is substituted into the expression and evaluated. The program uses the eval function, which allows Python to run Python code within itself. This can be very dangerous, so use this function with care.
The division operations can lead to round-off error, so we only check that the result is very close to the target value. The variable tolerance sets the required precision.
Output
There are 136 solutions.
1 + 13 * 2 / 6 + 4 + 12 * 7 - 8 - 11 + 3 * 5 / 9 - 10 = 66
1 + 13 * 2 / 6 + 4 + 12 * 7 - 8 - 11 + 5 * 3 / 9 - 10 = 66
1 + 13 * 3 / 2 + 4 + 12 * 5 - 8 - 11 + 7 * 9 / 6 - 10 = 66
1 + 13 * 3 / 2 + 4 + 12 * 5 - 8 - 11 + 9 * 7 / 6 - 10 = 66
1 + 13 * 3 / 2 + 9 + 12 * 5 - 6 - 11 + 4 * 7 / 8 - 10 = 66
1 + 13 * 3 / 2 + 9 + 12 * 5 - 6 - 11 + 7 * 4 / 8 - 10 = 66
1 + 13 * 3 / 4 + 7 + 12 * 6 - 5 - 11 + 2 * 9 / 8 - 10 = 66
1 + 13 * 3 / 4 + 7 + 12 * 6 - 5 - 11 + 9 * 2 / 8 - 10 = 66
1 + 13 * 3 / 6 + 2 + 12 * 7 - 9 - 11 + 4 * 5 / 8 - 10 = 66
1 + 13 * 3 / 6 + 2 + 12 * 7 - 9 - 11 + 5 * 4 / 8 - 10 = 66
1 + 13 * 3 / 9 + 4 + 12 * 7 - 8 - 11 + 2 * 5 / 6 - 10 = 66
1 + 13 * 3 / 9 + 4 + 12 * 7 - 8 - 11 + 5 * 2 / 6 - 10 = 66
1 + 13 * 4 / 8 + 2 + 12 * 7 - 9 - 11 + 3 * 5 / 6 - 10 = 66
1 + 13 * 4 / 8 + 2 + 12 * 7 - 9 - 11 + 5 * 3 / 6 - 10 = 66
1 + 13 * 5 / 2 + 3 + 12 * 4 - 8 - 11 + 7 * 9 / 6 - 10 = 66
1 + 13 * 5 / 2 + 3 + 12 * 4 - 8 - 11 + 9 * 7 / 6 - 10 = 66
1 + 13 * 5 / 2 + 8 + 12 * 4 - 7 - 11 + 3 * 9 / 6 - 10 = 66
1 + 13 * 5 / 2 + 8 + 12 * 4 - 7 - 11 + 9 * 3 / 6 - 10 = 66
1 + 13 * 5 / 3 + 9 + 12 * 4 - 2 - 11 + 7 * 8 / 6 - 10 = 66
1 + 13 * 5 / 3 + 9 + 12 * 4 - 2 - 11 + 8 * 7 / 6 - 10 = 66
1 + 13 * 8 / 3 + 7 + 12 * 4 - 5 - 11 + 2 * 6 / 9 - 10 = 66
1 + 13 * 8 / 3 + 7 + 12 * 4 - 5 - 11 + 6 * 2 / 9 - 10 = 66
1 + 13 * 9 / 6 + 4 + 12 * 5 - 8 - 11 + 3 * 7 / 2 - 10 = 66
1 + 13 * 9 / 6 + 4 + 12 * 5 - 8 - 11 + 7 * 3 / 2 - 10 = 66
1 + 13 * 9 / 6 + 7 + 12 * 5 - 2 - 11 + 3 * 4 / 8 - 10 = 66
1 + 13 * 9 / 6 + 7 + 12 * 5 - 2 - 11 + 4 * 3 / 8 - 10 = 66
2 + 13 * 1 / 4 + 3 + 12 * 7 - 9 - 11 + 5 * 6 / 8 - 10 = 66
2 + 13 * 1 / 4 + 3 + 12 * 7 - 9 - 11 + 6 * 5 / 8 - 10 = 66
2 + 13 * 3 / 6 + 1 + 12 * 7 - 9 - 11 + 4 * 5 / 8 - 10 = 66
2 + 13 * 3 / 6 + 1 + 12 * 7 - 9 - 11 + 5 * 4 / 8 - 10 = 66
2 + 13 * 4 / 8 + 1 + 12 * 7 - 9 - 11 + 3 * 5 / 6 - 10 = 66
2 + 13 * 4 / 8 + 1 + 12 * 7 - 9 - 11 + 5 * 3 / 6 - 10 = 66
2 + 13 * 6 / 9 + 8 + 12 * 5 - 1 - 11 + 4 * 7 / 3 - 10 = 66
2 + 13 * 6 / 9 + 8 + 12 * 5 - 1 - 11 + 7 * 4 / 3 - 10 = 66
2 + 13 * 8 / 6 + 9 + 12 * 4 - 1 - 11 + 5 * 7 / 3 - 10 = 66
2 + 13 * 8 / 6 + 9 + 12 * 4 - 1 - 11 + 7 * 5 / 3 - 10 = 66
2 + 13 * 9 / 6 + 3 + 12 * 5 - 1 - 11 + 4 * 7 / 8 - 10 = 66
2 + 13 * 9 / 6 + 3 + 12 * 5 - 1 - 11 + 7 * 4 / 8 - 10 = 66
3 + 13 * 1 / 4 + 2 + 12 * 7 - 9 - 11 + 5 * 6 / 8 - 10 = 66
3 + 13 * 1 / 4 + 2 + 12 * 7 - 9 - 11 + 6 * 5 / 8 - 10 = 66
3 + 13 * 2 / 1 + 5 + 12 * 4 - 7 - 11 + 8 * 9 / 6 - 10 = 66
3 + 13 * 2 / 1 + 5 + 12 * 4 - 7 - 11 + 9 * 8 / 6 - 10 = 66
3 + 13 * 2 / 4 + 8 + 12 * 5 - 1 - 11 + 7 * 9 / 6 - 10 = 66
3 + 13 * 2 / 4 + 8 + 12 * 5 - 1 - 11 + 9 * 7 / 6 - 10 = 66
3 + 13 * 2 / 8 + 6 + 12 * 5 - 1 - 11 + 7 * 9 / 4 - 10 = 66
3 + 13 * 2 / 8 + 6 + 12 * 5 - 1 - 11 + 9 * 7 / 4 - 10 = 66
3 + 13 * 5 / 2 + 1 + 12 * 4 - 8 - 11 + 7 * 9 / 6 - 10 = 66
3 + 13 * 5 / 2 + 1 + 12 * 4 - 8 - 11 + 9 * 7 / 6 - 10 = 66
3 + 13 * 6 / 4 + 9 + 12 * 5 - 8 - 11 + 1 * 7 / 2 - 10 = 66
3 + 13 * 6 / 4 + 9 + 12 * 5 - 8 - 11 + 7 * 1 / 2 - 10 = 66
3 + 13 * 9 / 2 + 8 + 12 * 1 - 5 - 11 + 6 * 7 / 4 - 10 = 66
3 + 13 * 9 / 2 + 8 + 12 * 1 - 5 - 11 + 7 * 6 / 4 - 10 = 66
3 + 13 * 9 / 6 + 2 + 12 * 5 - 1 - 11 + 4 * 7 / 8 - 10 = 66
3 + 13 * 9 / 6 + 2 + 12 * 5 - 1 - 11 + 7 * 4 / 8 - 10 = 66
4 + 13 * 2 / 6 + 1 + 12 * 7 - 8 - 11 + 3 * 5 / 9 - 10 = 66
4 + 13 * 2 / 6 + 1 + 12 * 7 - 8 - 11 + 5 * 3 / 9 - 10 = 66
4 + 13 * 3 / 2 + 1 + 12 * 5 - 8 - 11 + 7 * 9 / 6 - 10 = 66
4 + 13 * 3 / 2 + 1 + 12 * 5 - 8 - 11 + 9 * 7 / 6 - 10 = 66
4 + 13 * 3 / 9 + 1 + 12 * 7 - 8 - 11 + 2 * 5 / 6 - 10 = 66
4 + 13 * 3 / 9 + 1 + 12 * 7 - 8 - 11 + 5 * 2 / 6 - 10 = 66
4 + 13 * 9 / 6 + 1 + 12 * 5 - 8 - 11 + 3 * 7 / 2 - 10 = 66
4 + 13 * 9 / 6 + 1 + 12 * 5 - 8 - 11 + 7 * 3 / 2 - 10 = 66
5 + 13 * 1 / 2 + 9 + 12 * 6 - 7 - 11 + 3 * 4 / 8 - 10 = 66
5 + 13 * 1 / 2 + 9 + 12 * 6 - 7 - 11 + 4 * 3 / 8 - 10 = 66
5 + 13 * 2 / 1 + 3 + 12 * 4 - 7 - 11 + 8 * 9 / 6 - 10 = 66
5 + 13 * 2 / 1 + 3 + 12 * 4 - 7 - 11 + 9 * 8 / 6 - 10 = 66
5 + 13 * 3 / 1 + 7 + 12 * 2 - 6 - 11 + 8 * 9 / 4 - 10 = 66
5 + 13 * 3 / 1 + 7 + 12 * 2 - 6 - 11 + 9 * 8 / 4 - 10 = 66
5 + 13 * 4 / 1 + 9 + 12 * 2 - 7 - 11 + 3 * 8 / 6 - 10 = 66
5 + 13 * 4 / 1 + 9 + 12 * 2 - 7 - 11 + 8 * 3 / 6 - 10 = 66
5 + 13 * 4 / 8 + 9 + 12 * 6 - 7 - 11 + 1 * 3 / 2 - 10 = 66
5 + 13 * 4 / 8 + 9 + 12 * 6 - 7 - 11 + 3 * 1 / 2 - 10 = 66
5 + 13 * 7 / 2 + 8 + 12 * 3 - 9 - 11 + 1 * 6 / 4 - 10 = 66
5 + 13 * 7 / 2 + 8 + 12 * 3 - 9 - 11 + 6 * 1 / 4 - 10 = 66
5 + 13 * 9 / 3 + 6 + 12 * 2 - 1 - 11 + 7 * 8 / 4 - 10 = 66
5 + 13 * 9 / 3 + 6 + 12 * 2 - 1 - 11 + 8 * 7 / 4 - 10 = 66
6 + 13 * 2 / 8 + 3 + 12 * 5 - 1 - 11 + 7 * 9 / 4 - 10 = 66
6 + 13 * 2 / 8 + 3 + 12 * 5 - 1 - 11 + 9 * 7 / 4 - 10 = 66
6 + 13 * 3 / 1 + 9 + 12 * 2 - 5 - 11 + 7 * 8 / 4 - 10 = 66
6 + 13 * 3 / 1 + 9 + 12 * 2 - 5 - 11 + 8 * 7 / 4 - 10 = 66
6 + 13 * 9 / 3 + 5 + 12 * 2 - 1 - 11 + 7 * 8 / 4 - 10 = 66
6 + 13 * 9 / 3 + 5 + 12 * 2 - 1 - 11 + 8 * 7 / 4 - 10 = 66
7 + 13 * 1 / 4 + 9 + 12 * 6 - 5 - 11 + 2 * 3 / 8 - 10 = 66
7 + 13 * 1 / 4 + 9 + 12 * 6 - 5 - 11 + 3 * 2 / 8 - 10 = 66
7 + 13 * 2 / 8 + 9 + 12 * 6 - 5 - 11 + 1 * 3 / 4 - 10 = 66
7 + 13 * 2 / 8 + 9 + 12 * 6 - 5 - 11 + 3 * 1 / 4 - 10 = 66
7 + 13 * 3 / 1 + 5 + 12 * 2 - 6 - 11 + 8 * 9 / 4 - 10 = 66
7 + 13 * 3 / 1 + 5 + 12 * 2 - 6 - 11 + 9 * 8 / 4 - 10 = 66
7 + 13 * 3 / 2 + 8 + 12 * 5 - 9 - 11 + 1 * 6 / 4 - 10 = 66
7 + 13 * 3 / 2 + 8 + 12 * 5 - 9 - 11 + 6 * 1 / 4 - 10 = 66
7 + 13 * 3 / 4 + 1 + 12 * 6 - 5 - 11 + 2 * 9 / 8 - 10 = 66
7 + 13 * 3 / 4 + 1 + 12 * 6 - 5 - 11 + 9 * 2 / 8 - 10 = 66
7 + 13 * 5 / 2 + 8 + 12 * 4 - 9 - 11 + 1 * 3 / 6 - 10 = 66
7 + 13 * 5 / 2 + 8 + 12 * 4 - 9 - 11 + 3 * 1 / 6 - 10 = 66
7 + 13 * 6 / 4 + 8 + 12 * 5 - 9 - 11 + 1 * 3 / 2 - 10 = 66
7 + 13 * 6 / 4 + 8 + 12 * 5 - 9 - 11 + 3 * 1 / 2 - 10 = 66
7 + 13 * 8 / 3 + 1 + 12 * 4 - 5 - 11 + 2 * 6 / 9 - 10 = 66
7 + 13 * 8 / 3 + 1 + 12 * 4 - 5 - 11 + 6 * 2 / 9 - 10 = 66
7 + 13 * 9 / 6 + 1 + 12 * 5 - 2 - 11 + 3 * 4 / 8 - 10 = 66
7 + 13 * 9 / 6 + 1 + 12 * 5 - 2 - 11 + 4 * 3 / 8 - 10 = 66
8 + 13 * 2 / 4 + 3 + 12 * 5 - 1 - 11 + 7 * 9 / 6 - 10 = 66
8 + 13 * 2 / 4 + 3 + 12 * 5 - 1 - 11 + 9 * 7 / 6 - 10 = 66
8 + 13 * 3 / 2 + 7 + 12 * 5 - 9 - 11 + 1 * 6 / 4 - 10 = 66
8 + 13 * 3 / 2 + 7 + 12 * 5 - 9 - 11 + 6 * 1 / 4 - 10 = 66
8 + 13 * 5 / 2 + 1 + 12 * 4 - 7 - 11 + 3 * 9 / 6 - 10 = 66
8 + 13 * 5 / 2 + 1 + 12 * 4 - 7 - 11 + 9 * 3 / 6 - 10 = 66
8 + 13 * 5 / 2 + 7 + 12 * 4 - 9 - 11 + 1 * 3 / 6 - 10 = 66
8 + 13 * 5 / 2 + 7 + 12 * 4 - 9 - 11 + 3 * 1 / 6 - 10 = 66
8 + 13 * 6 / 4 + 7 + 12 * 5 - 9 - 11 + 1 * 3 / 2 - 10 = 66
8 + 13 * 6 / 4 + 7 + 12 * 5 - 9 - 11 + 3 * 1 / 2 - 10 = 66
8 + 13 * 6 / 9 + 2 + 12 * 5 - 1 - 11 + 4 * 7 / 3 - 10 = 66
8 + 13 * 6 / 9 + 2 + 12 * 5 - 1 - 11 + 7 * 4 / 3 - 10 = 66
8 + 13 * 7 / 2 + 5 + 12 * 3 - 9 - 11 + 1 * 6 / 4 - 10 = 66
8 + 13 * 7 / 2 + 5 + 12 * 3 - 9 - 11 + 6 * 1 / 4 - 10 = 66
8 + 13 * 9 / 2 + 3 + 12 * 1 - 5 - 11 + 6 * 7 / 4 - 10 = 66
8 + 13 * 9 / 2 + 3 + 12 * 1 - 5 - 11 + 7 * 6 / 4 - 10 = 66
9 + 13 * 1 / 2 + 5 + 12 * 6 - 7 - 11 + 3 * 4 / 8 - 10 = 66
9 + 13 * 1 / 2 + 5 + 12 * 6 - 7 - 11 + 4 * 3 / 8 - 10 = 66
9 + 13 * 1 / 4 + 7 + 12 * 6 - 5 - 11 + 2 * 3 / 8 - 10 = 66
9 + 13 * 1 / 4 + 7 + 12 * 6 - 5 - 11 + 3 * 2 / 8 - 10 = 66
9 + 13 * 2 / 8 + 7 + 12 * 6 - 5 - 11 + 1 * 3 / 4 - 10 = 66
9 + 13 * 2 / 8 + 7 + 12 * 6 - 5 - 11 + 3 * 1 / 4 - 10 = 66
9 + 13 * 3 / 1 + 6 + 12 * 2 - 5 - 11 + 7 * 8 / 4 - 10 = 66
9 + 13 * 3 / 1 + 6 + 12 * 2 - 5 - 11 + 8 * 7 / 4 - 10 = 66
9 + 13 * 3 / 2 + 1 + 12 * 5 - 6 - 11 + 4 * 7 / 8 - 10 = 66
9 + 13 * 3 / 2 + 1 + 12 * 5 - 6 - 11 + 7 * 4 / 8 - 10 = 66
9 + 13 * 4 / 1 + 5 + 12 * 2 - 7 - 11 + 3 * 8 / 6 - 10 = 66
9 + 13 * 4 / 1 + 5 + 12 * 2 - 7 - 11 + 8 * 3 / 6 - 10 = 66
9 + 13 * 4 / 8 + 5 + 12 * 6 - 7 - 11 + 1 * 3 / 2 - 10 = 66
9 + 13 * 4 / 8 + 5 + 12 * 6 - 7 - 11 + 3 * 1 / 2 - 10 = 66
9 + 13 * 5 / 3 + 1 + 12 * 4 - 2 - 11 + 7 * 8 / 6 - 10 = 66
9 + 13 * 5 / 3 + 1 + 12 * 4 - 2 - 11 + 8 * 7 / 6 - 10 = 66
9 + 13 * 6 / 4 + 3 + 12 * 5 - 8 - 11 + 1 * 7 / 2 - 10 = 66
9 + 13 * 6 / 4 + 3 + 12 * 5 - 8 - 11 + 7 * 1 / 2 - 10 = 66
9 + 13 * 8 / 6 + 2 + 12 * 4 - 1 - 11 + 5 * 7 / 3 - 10 = 66
9 + 13 * 8 / 6 + 2 + 12 * 4 - 1 - 11 + 7 * 5 / 3 - 10 = 66
Russian peasant multiplication and loop invariants
Russian peasant multiplication is a method for multiplying numbers based on repeated doubling and halving. Nobody is quite sure why it is called “Russian peasant multiplication,” but the name has stuck.
Here is the procedure:
1. Write each number at the head of a column.
2. Double the number in the first column.
3. Halve the number in the second column. (If odd, ignore the remainder.)
4. Repeat (2) and (3) until the number in the second column is 1.
5. Cross out the rows with an even number in the second column.
6. Add up the remaining numbers in the first column.
Example: Multiply 43 by 18.
First number Second number
43 18
$86$ $9$
172 4
344 2
$688$ $1$
$\implies 43 \times 18 = 86 + 688 = 774.$
Here is a Python implementation of the algorithm.
def multiply(m, n):
p = 0
while n > 0:
if n % 2: # n is odd
p = p + m
m = m * 2
n = n // 2
return p
Correctness
Let’s see what happens to the values of $m$, $n$, and $p$ in one iteration. If $n$ is even, then $m$ is doubled and $n$ is halved, so the product $mn$ is unchanged. If $n$ is odd, then $m$ is doubled and $n$ is replaced with $(n-1)/2$, so the value of the product $mn$ is decreased.
$(2m) \cdot (n-1)/2 = mn – m$
To compensate for this decrease, the value of $p$ is increased by $m$ when $n$ is odd. Consequently, the value of $Q = mn + p$ does not change – it is a loop invariant. At the beginning of the loop we have $Q = mn$ (since $p=0$) and at the end we have $Q = p$ (since $n=0$). Therefore, the product $mn$ is equal to the final value of $p$.
A decimal variation
The Russian Peasant method is associated with binary (base two) arithmetic, since it involves multiplying and dividing by two. Is there a decimal (base ten) version? Consider the following:
def multiply_decimal(m, n):
p = 0
while n > 0:
r = n % 10
p = p + m * r
m = m * 10
n = n // 10
return p
Let’s analyze this algorithm. Each time through the loop, the values of $m$, $n$, and $p$ are replaced with new values $m’$, $n’$, and $p’$. The following relations hold between the new values and the old values.
\begin{align*} m’ &= 10m\\ n &= 10n’ + r\\ p’ &= p + mr \end{align*}
As before, the quantity $Q = mn+p$ is a loop invariant, since
$mn + p = m(10n’ + r) + p = (10m)n’ + (p + mr) = m’n’ + p’.$
Therefore, the function returns the product $mn$.
Example: Multiply 123 by 456.
$m$ $n$ $mr$
$123$ $456$ $123 \times 6 = 738$
$1230$ $45$ $1230 \times 5 = 6150$
$12300$ $4$ $12300 \times 4 = 49200$
$\implies 123 \times 456 = 738 + 6150 + 49200 = 56088.$
Note that this is essentially the standard algorithm for multiplying numbers.
123
× 456
—————
738
6150
49200
—————
56088
Fast exponentiation
We have seen that we multiply numbers via repeated doubling. In a similar way, we can raise a number to a power via repeated squaring. Here is a Python implementation:
def power(m, n):
p = 1
while n > 0:
if n % 2: # n is odd
p = p * m
m = m * m
n = n // 2
return p
The loop invariant $Q = m^n p$ can be used to prove that the function correctly returns the value of $m^n$. | 7,048 | 13,302 | {"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": 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} | 3.625 | 4 | CC-MAIN-2017-34 | longest | en | 0.907191 |
https://kr.mathworks.com/matlabcentral/cody/problems/42998-electrical-diode-current-calculation | 1,713,883,458,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818711.23/warc/CC-MAIN-20240423130552-20240423160552-00582.warc.gz | 308,666,137 | 22,873 | # Problem 42998. Electrical Diode Current Calculation
In engineering, there is not always a single equation that describes a phenomenon accurately enough to be applied in all instances of that phenomenon. Sometimes it is more useful to use one equation in a set of circumstances, and another equation in a different set of circumstances. One example of this is with electrical diodes. A simplification of the approximation of the electrical characteristics of a diode uses these two simple equations:
i = I_s * exp(v/V_T) for v > V_T
i = −I_s for v ≤ V_T
Where V_T = 0.026 Volts and I_S = 1*10-8 Amperes. Write a function with one input, the voltage v across this diode, and one output, the current i running through the diode. Use a conditional statement in writing this program.
### Solution Stats
49.47% Correct | 50.53% Incorrect
Last Solution submitted on Apr 23, 2024 | 210 | 879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-18 | latest | en | 0.871006 |
https://socratic.org/questions/jeff-reads-for-2-1-4-hours-then-he-watches-tv-for-1-2-3-long-how-long-did-jeff-r | 1,601,159,018,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00512.warc.gz | 598,099,798 | 5,956 | # Jeff reads for 2 1/4 hours. Then he watches TV for 1 2/3 long. How long did Jeff read and watch TV?
##### 1 Answer
Jun 21, 2018
"Total time spent on reading + watching TV " = color(brown)(3h : 55 min
#### Explanation:
color(crimson)(60 sec = 1 min, 60 min = 1 hour
$\text{Jeff read for } = 2 \left(\frac{1}{4}\right) h o u r s = 2 h : 15 \min$
$\text{Jeff watched TV for } = 1 \left(\frac{2}{3}\right) h o u r s = 1 h : 40 \min$
$\text{Total time spent on reading + watching TV } = 2 h : 15 \min + 1 h : 40 \min$
color(brown)(=> 3h : 55 min | 209 | 550 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-40 | latest | en | 0.786251 |
https://www.e-olymp.com/en/problems/1965 | 1,558,278,307,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254889.43/warc/CC-MAIN-20190519141556-20190519163556-00488.warc.gz | 761,804,761 | 2,831 | Problems
# Rank
Peter Vasichkin moved to another school. At the gym class he needs to find his place in the row...
#### Input
The first is the number n of students in the class. Then goes nonincreasing sequence of n numbers, meaning the growth of each person in line. After that given the growth x of Peter. All numbers are natural and do not exceed 200.
#### Output
Print the number under which Peter should stand in a line. If there are people with the same height as Peter has, then Peter should stand after them.
Time limit 1 second
Memory limit 128 MiB
Input example #1
```8
165 163 160 160 157 157 155 154
162
```
Output example #1
```3
```
Input example #2
```8
165 163 160 160 157 157 155 154
160
```
Output example #2
```5
```
Source Stage II Ukrainian School Olympiad 2011-2012, Berdichev | 225 | 805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-22 | latest | en | 0.860288 |
https://web2.0calc.com/questions/math-equation-problem | 1,558,571,034,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256980.46/warc/CC-MAIN-20190522223411-20190523005411-00250.warc.gz | 672,539,924 | 7,748 | +0
# Math Equation problem
0
289
3
This morning James had half as much money as his brother and 27 dollars less than his sister. Then his brother paid James and his sister each 12 dollars not to tell their parents what happened out in the garden last Wednesday. After the payoff, James's sister has twice as much money as James's brother. How much money does James have now? Thanks to whoever can help! Also, please give a detailed solution.
Mar 21, 2018
#1
+1
See 2 solutions here:
And here:
Mar 21, 2018
#2
+2338
+1
I think it is best to initialize variables first and foremost. A quick glance at the problem will reveal that the information mentions the amount of money that three individuals have at separate moments.
$$\text{Let }J=\text{James's Money}\\ \text{Let }B=\text{James's Brother's Money}\\ \text{Let }S=\text{James's Sister's Money}$$
Now, let's break down the information into its components and analyze them.
• If James had half as much money as his brother, then we can represent this situation via equation: $$J_1=\frac{B_1}{2}$$
• If James had 27 dollars less than his sister, then we can create another equation: $$J_1=S_1-27$$
After the secretive payoff, though, new information rolls out! The subscript of 1 on every variable is intentional; it is meant to denote the relationship of money prior to the shady payoff. I will use a subscript of 2 to represent amounts of money after the deal.
• If James's sister has twice as much money as James's brother, then we can create a third equation: $$S_2=2B_2$$
By utilizing subscripts, it is easy to notice that we cannot simply compare this information to the previous equations. However, we can use the information about the payoff to adjust the variables.
• If James's brother paid James and James's sister 12 dollars, then this has a three-way affect.
• $$J_1=J_2-12\\ S_1=S_2-12\\ B_1=B_2+12$$
Since the question specifically asks for the amount of money James has now (after the payoff), then we should adjust every variable such that it has a subscript of 2. Let's do that!
$$S_2=2B_2$$ $$J_1=\frac{B_1}{2}$$ $$J_1=S_1-27$$ Do the necessary substitution! $$J_2-12=\frac{B_2+12}{2}$$ $$J_2-12=S_2-12-27$$ Let's clean these equations up. $$\boxed{1}\hspace{1mm}S_2=2B_2$$ $$\boxed{2} \hspace{1mm} 2J_2=B_2+36$$ $$\boxed{3}\hspace{1mm}J_2=S_2-27$$ This creates a three-variable system of equations. Now, we must solve it!
Let's place the 1st and 3rd equations alongside each other. By adding the equations together (with the combination of a slight manipulation) it is possible to eliminate one variable from this system. This creates a fourth equation.
$$\boxed{1}\hspace{1mm}S_2=2B_2\Rightarrow \hspace{10mm}S_2\hspace{10mm}=2B_2\\ \boxed{3}\hspace{1mm}J_2=S_2-27\Rightarrow -S_2+J_2=-27\\ \boxed{4}\hspace{1mm}\hspace{46mm}J_2=2B_2-27$$
We can now introduce the 2nd equation into the mix, which will allow us to eliminate the other variable.
$$\boxed{4}\hspace{39mm}J_2=2B_2-27\\ \boxed{2} \hspace{1mm} 2J_2=B_2+36 \Rightarrow-4J_2=-2B_2-72\\ \hspace{39mm}-3J_2=-99\\ \hspace{47mm}J_2=33$$
According to this work, James has 33 dollars.
Mar 22, 2018
#3
+2338
+1
Upon further review, I have realized that I made a slight oversight on my end. Sorry about that! Unfortunately, I think everyone is prone to making such mistake. This is a learning experience--for you and me. Let's try to embrace it in the best way possible. The mistake occurs in the following segment:
• If James's brother paid James and James's sister 12 dollars, then this has a three-way affect.
• $$J_1=J_2-12\\ S_1=S_2-12\\ B_1=B_2+12$$
Did you spot the error yet? If not, that's OK! There is a lot of information to digest in this problem, so it is very easy to assume that I could not have blundered at this step. The problem lies at the $$B_1=B_2+12$$ part.
The original problem states that James's brother gives 12 dollars to both James and his sister. This means that James's brother loses $24 ($12 to James and $12 to the sister) in total--not$12 like I originally hinted at.
The working of the system of equation is mostly the same, though, and I get that James has \$41 right now! I have double and triple checked, so I think I will finally move on from this word problem.
TheXSquaredFactor Mar 22, 2018
edited by TheXSquaredFactor Mar 22, 2018 | 1,294 | 4,328 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2019-22 | latest | en | 0.952587 |
https://codehelppro.com/detail/c-programming/if-else/ | 1,642,876,339,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303868.98/warc/CC-MAIN-20220122164421-20220122194421-00551.warc.gz | 237,603,477 | 16,380 | # If Else
0 luist9776194511 November 13, 2020
If we have to execute a block of statements when a particular condition is met or not met. This is known as Decision making. For decision making, We have 4 types of control statements in C Programming.
• if statement
• if….else statement
• else…..if statement
• nested if….else statement
## 1. IF Statement:
If the statement is a powerful decision-making statement and is used to control the flow of execution of statements. It basically a two-way decision statement and is used together with an expression, i.e. test condition. The if statement evaluates the expression first and then, if the value the expression is true, it executes the statement within the block. Otherwise, it skips the statements within its block and continues from the first statement outside the if block. It takes the form,
``````if(condition)
{
statement-block;
}
statement-x;``````
### Flow Diagram
Let’s take an example to make clear concepts on IF-ELSE.
``````#include <stdio.h>
int main(){
int num = 10;
if(num % 2 == 0){
printf("Number is Even\n");
}
return 0;
}``````
The output of the above program is:
``Number is Even``
## 2. IF…ELSE Statement
Theif…else statement is an extension of the simpleifstatement. It is used when there are two possible actions – one when a condition is true, and the other when it is false. The general form is
``````if(condition)
{
true_block statement;
}
else
{
false_block statement;
}
statement-x;``````
### Flowchart
Let’s look upon one example to make clear concepts on if..else condition.
``````#include <stdio.h>
int main(){
int num = 10;
if(num % 2 == 0)
{
printf("Number is Even\n");
}
else
{
printf("Number is Odd\n");
}
return 0;
}``````
The output of the above program is:
``Number is Even``
## 3. ELSE…IF Statement
The else if the statement is used when there are more than two possible actions depending upon the outcome of the test. When an action is taken, no others can be executed or taken. In such a situation, the if…else if…else if….else statement is used. This structure takes the form,
``````if(condition-1)
{
statement-1;
}
else if(condition-2)
{
statement-2;
}
else if(condition-3)
{
statement-3;
}
else
{
default-statement;
}``````
Let’s look at one example to make clear concepts on else…if condition.
``````#include <stdio.h>
int main(){
int num = 15;
if(num % 2 == 0)
{
printf("Number is Even\n");
}
else if(num % 3 == 0)
{
printf("Number is Odd\n");
}
else
{
printf("Invalid Number\n");
}
return 0;
}``````
The output of the above program is:
``Number is Odd``
## 4. Nested IF…ELSE Statement
Nested IF…ELSE statement means there is anIF condition inside the if condition. The syntax looks like this.
``````if(condition)
{
if(condition-2)
{
statement-1;
}
else
{
statement-2;
}
}
else
{
statement-3;
}``````
Let’s look at examples to make clear concepts on nested if…else condition.
``````#include < stdio.h>
int main(){
int num = 15;
if(num % 2 == 0)
{
if(num > 5 == 0)
{
printf("Number is greater than 5 and Even\n");
}else{
printf("Number is lessa than 5 andEven\n");
}
}
else
{
printf("Number is Odd\n");
}
return 0;
}``````
The output of the above program is:
``Number is greater than 5 and Even`` | 876 | 3,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | longest | en | 0.779819 |
https://www.fit.vut.cz/study/course/268213/ | 1,713,802,122,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818312.80/warc/CC-MAIN-20240422144517-20240422174517-00816.warc.gz | 685,769,916 | 14,172 | Course details
Algorithms
IAL Acad. year 2023/2024 Winter semester 5 credits
Overview of fundamental data structures and their exploitation. Introduction to time complexity of algorithms. Specification of abstract data types (ADT). Specification and implementation of ADT's: lists, stack and its exploitation, queue, searching table. Algorithms upon the binary trees. Searching: sequential, in the ordered and in not ordered array, searching with the guard (sentinel), binary search, search tree, balanced trees (AVL), balanced multiway serach trees. Searching in hash-tables. Ordering (sorting), principles, sorting without the moving of items, sorting with multiple keys. Most common methods of sorting: Select-sort, Bubble-sort, Heap-sort, Insert-sort and its variants, Shell-sort, recursive and non-recursive notation of the Quick sort, Merge-sort, List-merge-sort, Radix-sort. Recursion and dynamic programming. Searching the patterns in the text. Graph algorithms. Proving of the correctness of programs.
5 ECTS credits represent approximately 125-150 hours of study workload:
• 39 hours of lectures
• 26 hours for two home assignments
• 35 hours of project work
• 20 hours of continual study
• 30 hours of study for the midterm and final examination
Guarantor
Course coordinator
Language of instruction
Czech, English
Completion
Credit+Examination (written)
Time span
• 39 hrs lectures
• 13 hrs projects
Assessment points
• 51 pts final exam (written part)
• 14 pts mid-term test (written part)
• 35 pts projects
Department
Lecturer
Instructor
Learning objectives
The student will learn the fundamentals of algorithm complexity and will be able to use gained knowledge in the design of programs. Student acquaints with basic abstract data types and learn how to implement nad use them. The student learns and commands recursive and non-recursive notation of basic algorithms and will be able to use gained knowledge in the design of programmes. Student understands the implementation and analysis of most of the used algorithms for searching and sorting. The student will understand basic text search algorithms and basic graph algorithms. The student will learn about recursive problem solving and the basics of dynamic programming. The student will learn the principles of methods of proving of the correctness of programs and will be able to use gained knowledge in the design of programmes.
• The student will learn the fundamentals of algorithm complexity and their intention.
• He/she acquaints with basic abstract data types and to commands its implementation and exploitation.
• He/she learns and commands recursive and non recursive notation of basic algorithms.
• Student will become familiar with different variants of search tables and understand their properties.
• Student overrules the implementation and analysis of most used algorithms for searching and sorting.
• The student learns terminology in Czech and English language.
• The student learns to participate on the small project as a member of a small team.
• Student learns to present and defend the results of the small project.
• The student will acquaint with the methods of proving of the correctness of programs and with the construction of proved programs and learn their significance.
Why is the course taught
Almost all computer programs need to store data during their execution. Students will make familiar with common data structures and understand their properties. It allows them later to choose the right data structure for a particular purpose. Students also learn algorithms associated with discussed data structures and will be able to analyse and evaluate them using the asymptotic time complexity method.
Recommended prerequisites
Prerequisite knowledge and skills
• Basic knowledge of the programming in a procedural programming language
• Knowledge of secondary school level mathematics
Study literature
• Mareš, M., Valla, T.: Průvodce labyrintem algoritmů, CZ.NIC, 2017, ISBN 978-80-88168-19-5, http://pruvodce.ucw.cz/
• Honzík, J., Hruška, T., Máčel, M.: Vybrané kapitoly z programovacích technik, Ed. stř. VUT Brno, 1991
• Cormen, T.H., Leiserson, Ch.E., Rivest, R.L.: Introduction to Algorithms, Cambridge: MIT Press, 2009
• Knuth, D.: The Art of Computer programming, Vol.1,2,3. Addison Wesley, 1968
• Wirth, N.: Alorithms+Data Structures=Programs, Prentice Hall, 1976
• Aho A.V., Hoppcroft J.E., Ullman J.D.: Data Structures and Algorithms, Addison Wesley, 1983
• Baase, S.: Computer Algorithms - Introduction to Design and Analysis. Addison Wesley, 1998
• Sedgewick,R.:Algoritmy v C, Addison Wesley 1998, Softpress 2003
Syllabus of lectures
1. Overview of data structures. Time complexity of algorithms.
2. Abstract data type and its specification. Specification, implementation and exploitation of the ADT list.
3. Specification, implementation and exploitation of ADT stack, queue, search table. Numeration of expressions with the use of the stack.
4. Algorithms upon the binary tree.
5. Searching, sequential, in the array, binary search, binary search trees, AVL tree.
6. Balanced multiway search trees. Hashing-tables.
7. Sorting, principles, without movement, multiple key.
8. Most common methods of sorting of arrays.
9. String-searching algorithms.
10. Recursion and the dynamic programming.
11. Graph algorithms.
12. Proving the correctness of algorithms.
Syllabus - others, projects and individual work of students
1. Two home assignments
2. Project with a mini-defence for a team of students.
Progress assessment
• Home assignments - 20 points
• Mid-term written examination - 14 point
• Evaluated project with the defence - 15 points
• Accreditation - a minimum of 20 points per semester is required for credit to be awarded.
• Final written examination - 51 points; The minimal number of points which can be obtained from the final written examination is 24. Otherwise, no points will be assigned to a student.
In case of illness or another serious obstacle, the student should inform the faculty about that and subsequently provide the evidence of such an obstacle. Then, it can be taken into account within evaluation:
• The student can ask the responsible teacher to extend the time for the home assignment.
• If a student cannot attend the mid-term exam, (s)he can ask to derive points from the evaluation of his/her first attempt of the final exam. To enter the final exam in this case, at least 14 points from home assignments and project are required.
• If a student cannot attend the defense of the project and the other team members agree with that (s)he can earn the same points from the project defence as for present members.
Exam prerequisites
• to earn min. 20 points within the semester
Plagiarism and not allowed cooperation will cause that involved students are not classified and disciplinary action can be initiated.
Schedule
DayTypeWeeksRoomStartEndCapacityLect.grpGroupsInfo
Tue exam 2024-01-09 A112 A113 D0206 D0207 D105 E104 E105 E112 G202 10:0011:50 řádná
Tue exam 2024-01-23 D0206 D0207 D105 E112 10:0011:50 1. termín
Tue exam 2023-10-17 E104 E105 E112 12:0013:00 Půlsemestrální zkouška - uterý 12:15
Tue lecture 1., 2., 3., 4., 5., 6., 7., 8., 9., 10., 11., 12. of lectures E104 E105 E112 12:0014:50320 2BIB 3BIT xx 20 - 29 Burgetová
Tue lecture 2023-12-12 E104 E105 E112 12:0014:50320 2BIB 3BIT xx 20 - 29 Křena
Tue other 2023-10-31 C231 12:0023:59 Zápočet
Tue exam 2023-10-17 E104 E105 E112 13:0014:30 Půlsemestrální zkouška - uterý 13:20
Wed other 2023-11-01 C231 00:0013:00 Zápočet
Wed exam 2023-10-18 E104 E105 E112 14:0015:00 Půlsemestrální zkouška - středa 14:15
Wed lecture 1., 2., 3., 4., 5., 6., 7., 8., 9., 10., 11., 12. of lectures E104 E105 E112 14:0016:50294 2BIA 3BIT xx 10 - 19 Hranický
Wed lecture 2023-12-13 E104 E105 E112 14:0016:50294 2BIA 3BIT xx 10 - 19 Křena
Wed exam 2023-10-18 E104 E105 E112 15:0016:30 Půlsemestrální zkouška - středa 15:20
Thu exam 2024-02-01 D105 10:0011:50 2. termín
Fri other 2023-12-08 C228 12:0012:15 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 12:1512:30 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 12:3012:45 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 12:4513:00 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 13:0013:15 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 13:1513:30 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 13:3013:45 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 13:4514:00 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 14:0014:15 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 14:1514:30 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 14:3014:45 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 14:4515:00 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 15:0015:15 Obhajoba náhradního projektu
Fri other 2023-12-08 C228 15:1515:30 Obhajoba náhradního projektu
Course inclusion in study plans
• Programme BIT, 2nd year of study, Compulsory
• Programme BIT (in English), 2nd year of study, Compulsory
• Programme IT-BC-3, field BIT, 2nd year of study, Compulsory | 2,463 | 9,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-18 | longest | en | 0.833266 |
http://forum.enjoysudoku.com/clueless-skyscraper-t5526.html | 1,516,572,963,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890893.58/warc/CC-MAIN-20180121214857-20180121234857-00206.warc.gz | 129,974,389 | 5,758 | ## Clueless Skyscraper
For fans of Killer Sudoku, Samurai Sudoku and other variants
### Clueless Skyscraper
Each row, column, and 3x3 box contains each digit 1-9. The digits and X's around the edges tell you how many skyscrapers are visible from that vantage point. Each X stands for a digit 1-9. The X's and the 3 above the top edge form a nonet, i.e. each digit 1-9 appears exactly once among these 9 characters.
Code: Select all
X X X 3 X X X X X
_ _ _ _ _ _ _ _ _
|_|_|_|_|_|_|_|_|_|
|_|_|_|_|_|_|_|_|_|
|_|_|_|_|_|_|_|_|_|
|_|_|_|_|_|_|_|_|_| 4
|_|_|_|_|_|_|_|_|_|
4 |_|_|_|_|_|_|_|_|_|
|_|_|_|_|_|_|_|_|_|
|_|_|_|_|_|_|_|_|_|
|_|_|_|_|_|_|_|_|_|
1
I'll leave it up to some of you geeks to improve the rendering of the above diagram (and to post your improvement, please!). Unless I have committed a serious error in my reasoning, the solution is unique.
Bill Smythe
Smythe Dakota
Posts: 536
Joined: 11 February 2006
Marvellous puzzle!
Here is a pictorial format:
(A,B,C,D,E,F,G,H each represents a distinct digit from 1,2,4,5,6,7,8,9.)
I believe the unique solution should be this one:
693741528
714852639
825963741
936174852
147285963
258396174
369417285
471528396
582639417
Unless both Bill and I have made a careless mistake in our logical reasoning.
Thanks again for the great neat puzzle, Billy!
udosuk
Posts: 2698
Joined: 17 July 2005
Yep, that was my solution too.
Thanks for the improved picture!
Bill Smythe
Smythe Dakota
Posts: 536
Joined: 11 February 2006
Bill, I've just posted this puzzle on another forum on behalf of you.
I know quite a few of skyscraper players in there, so hopefully more people would enjoy and appreciate your puzzle.
udosuk
Posts: 2698
Joined: 17 July 2005
Thanks for posting my puzzle over there, and for calling my attention to yet another nice forum (almost as good as this one).
Bill Smythe
Smythe Dakota
Posts: 536
Joined: 11 February 2006 | 587 | 1,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | longest | en | 0.848405 |
https://www.numbers-figures.com/is-125010-onehundredtwentyfivethousandandten-a-prime-number | 1,680,267,503,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00507.warc.gz | 1,018,377,471 | 4,612 | # Is 125010 a prime number? | Is onehundredtwentyfivethousandandten a prime number?
125010 is:
no prime number!
## What is a prime number?
A prime number is basically a number that is only by itself and one divisible. Further condition is that the number is greater than 1.
Since time immemorial, humans and computers calculate larger and larger primes. The current record is a number with 17425170 decimals (as of 2013).
Primes are the base for many other calculations in mathematics and are deeply manifested in the history of mankind. Primes have been discovered by the ancient Greeks. They are mainly used for cryptography. | 141 | 633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-14 | latest | en | 0.944196 |
https://mathoverflow.net/questions/285932/tail-bound-of-a-distribution/286216 | 1,571,285,623,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672548.33/warc/CC-MAIN-20191017022259-20191017045759-00222.warc.gz | 601,223,106 | 26,163 | # Tail bound of a distribution
Let $X_1, X_2, \ldots, X_n, Y_1, Y_2, \ldots, Y_n$ be independent binary random variables each being $1$ with probability $\frac{1}{k}$.
Let $Z = X_1(Y_1 + \cdots + Y_k) + X_2(Y_2 + \cdots + Y_{k+1}) + \cdots + X_n(Y_n + Y_1 + \cdots + Y_{k-1})$.
It is easy to see that the expected value of $Z$ is $\mu = \frac{n}{k}$.
Can someone suggest the best possible bound one can obtain for the probability $\Pr[Z \ge (1+\varepsilon) \mu]$?
The setting I am interested in where $n$ is large, and $k \approx \sqrt{n}$.
• Do you need really the precise asymptotics of that probability? Or a "reasonably good" bound (with fast decay in $n$) will do? – Serguei Popov Nov 13 '17 at 20:31
• I expect the decay to be exponential in $\sqrt{n}$ for $k \approx \sqrt{n}$. I need as good a bound as possible, but for now, even a reasonably good bound will be helpful. – user47772 Nov 13 '17 at 23:25
• The first idea I had in mind was the following: condition on $Y_1,\dots,Y_n$ to be reduced to a deviation inequality of independent sequence, in this case a weighted sum of i.i.d. However, we have to control the sum of variances, which does not seem easy. This can probably be done by having information about the eigenvalues of a symmetric real matrix, since it will be a quadratic form applied to $(y_1,\dots,y_n)$. – Davide Giraudo Nov 14 '17 at 14:27
## 1 Answer
Here are some ideas, which may be too long for a comment. Denote for fixed $n$ and $k$: $$Z_{n,k}:=\sum_{i=1}^n\left(X_i\sum_{j=1}^k Y_{i+j-1 \mod n}-\frac 1k\right).$$ Let $$A_{n,k}:=\sum_{i=1}^{n-k}\left(X_i\sum_{j=1}^k Y_{i+j-1} -\frac 1k\right)\mbox{ and }$$ $$B_{n,k}:=\sum_{i=n-k+1}^n\left(X_i\sum_{j=1}^k Y_{i+j-1 \mod n}-\frac 1k\right).$$ It seems that the contribution of $B_{n,k}$ in the setting you are interested in would not be too big. In order to bound that of $A_{n,k}$, let us denote $$I_l:=\left\{i\in\mathbb N, kl\leqslant i\leqslant\left(k+1\right)l-1\right\}$$ and $$C_{l}:=\sum_{i\in I_l}\left(X_i\sum_{j=1}^k Y_{i+j-1} -\frac 1k\right).$$ We can express $A_{n,k}$ in terms of a sum of $C_l$ plus a negligible term. Moreover the sequence $\left(C_{2l}\right)_{l\geqslant 1}$ is centered and independent, as well as $\left(C_{2l-1}\right)_{l\geqslant 1}$ hence we can apply an inequality by Fuk and Nagaev (1971). It remains to control the tail of $C_l$ and its variance. | 845 | 2,382 | {"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-2019-43 | latest | en | 0.845176 |
https://neophilosophical.blogspot.com/2017/07/inflation-of-hubble-constant.html | 1,532,326,576,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594954.59/warc/CC-MAIN-20180723051723-20180723071723-00245.warc.gz | 738,319,645 | 22,545 | ## Sunday, 16 July 2017
### Inflation of the Hubble Constant
I've been chewing on an old bone recently, metaphorically that is.
In A Little Expansion on the Lightness of Fine-Tuning, I wrote about how I visualised the way two spaceships might approach each other, with both of them travelling at half the speed of light (relative to an implied third observer) and yet have a closing velocity of less than the speed of light. The resultant model, for me at least, also managed to explain the spatial and temporal effects of special relativity.
A consequence of this model is that the universe is expanding at the speed of light and this expansion is time (see also On Time) - so that were you to be at rest in spatial terms, you would not be at rest in temporal terms, you would still be travelling "through" time by virtue of the universal expansion (at a rate equivalent to the speed of light).
The problem is that if the universe is expanding at the speed of light, per my model, then what about reports that the rate of expansion of the universe is increasing? The speed of light is invariant, so the rate of expansion of the universe should also be invariant - if my model is valid. What about inflation? Well, I'll get to inflation in a moment.
I have previously (and rhetorically) asked the question Is the Universe Expanding at the Speed of Light? My conclusion was that, if a single spatial Planck unit were added to the universe (in the direction we are looking) for every temporal Planck unit, then we would observe an expansion of the universe (today) at pretty much the rate that we observe the universe expanding at (today) - about 70 kilometres per second per megaparsec.
This was calculated, however, using a different model to that presented in A Little Expansion on the Lightness of Fine-Tuning. It was if I were looking at a segmented ruler, with each segment being a Planck length long and I was adding a unit of Planck length somewhere in the middle for every unit of Planck time. I then calculated the rate of the expansion of the ruler after 8.08x1060 units of Planck time (which is the age of the universe) and found that this matched the Hubble Constant.
So the question I have: is what happens if I use the onion like model to calculate the effect of the universe expanding at the speed of light? This is taken from A Little Expansion on the Lightness of Fine-Tuning:
I was trying to explain something a little different there, so it's a bit more cluttered than it might need to be. For our purposes at the moment, all we really need to consider is the difference between the value of the arc defined by xG at tE and its value at tG, noting the relevant angle θ. My contention is that the universe expands at c, so therefore Δt = c. Let us call the arc length x and refer to any change as Δx.
An arc length is calculated reasonably simply, x = θ.r (where θ is expressed in radians, and a full circle circumference is therefore given by x = 2π.r – see Hugging the World). We therefore know that the difference in x would be given by Δx = θ.Δr (and in this model Δx = θ.Δt). This gives us enough to work with.
Consider two moments in time:
x = θ.tnow
and
x + Δx = θ.(tnow + Δt)
And eliminate θ:
(x + Δx)/x = (tnow + Δt)/tnow
1 + Δx/x = 1 + Δt/tnow
Δx/x = Δt/tnow
Δx/Δt = x/tnow
So the rate of expansion of the universe is proportional to the age of the universe (tnow). The Hubble Constant is presented as expansion over a given distance, so:
Ho = Δx/Δt/x = 1/tnow
Phew, my model stands up - because the value of the Hubble Constant is actually the reciprocal of the age of the universe - noting that the age of the universe is not uniquely calculated from the reciprocal of the Hubble Constant, there are other methods (see strong priors a little further down the page). However, my model suggests that the universe is, in a sense, expanding at the speed of light - always has done so and always will do so - and the Hubble Constant should be decreasing with the age of the universe. Nevertheless, we have people telling us that the rate of expansion of the universe is increasing.
This sounds like a potential worry because if the Hubble Constant were increasing rather than decreasing, then its current value at the reciprocal of the age of the universe would be coincidental. This would be another, worrying example of fine-tuning that would have to be explained.
Fortunately, while there are observations that indicate that expansion of the universe might be increasing, the Hubble Constant is not. This might seem counter-intuitive. As Sean Carroll explains, the Hubble Constant gives us a scale by which to measure the velocity at which distant objects recede from us due to universal expansion, v = Ho.d, where d is the distance to the object receding away from us. But note that this is caveated with "due to universal expansion". Carroll is considering dark energy here. If, in addition to universal expansion, things are being pushed apart even only by smidgen, then d will be increasing at rate greater than v (where v is "due to universal expansion").
Note that there is some room to doubt whether there actually is this acceleration in the rate of universal expansion, I myself remain a bit dubious maybe in part because if dark energy is real then my model may be fatally flawed, despite explaining so many things so well.
---
But what about inflation, I hear you yell excitedly.
Well, there's two things. Firstly, we need to remember Hugging the World. The term tnow doesn't necessarily mean the time since any absolute beginning to the universe, it means time since some key event - what that event was may have been no more than a phase change from inflation to the current state of affairs. (Or from an earlier aeon to this aeon. This makes my model consistent with conformal cyclic cosmology, which can bypass inflation.)
Secondly, in my model, we currently have a nice, orderly, temporal expansion of the universe, with one layer (one moment) added "at a time". This isn't necessarily the way things have to be. Instead, there could have been a situation in which each Planck volume spawned a new Planck volume each unit of Planck time. This would lead to an exponential cascade of expansion - and to get to the lower limit of inflation, an increase in size by a factor of 1026, it is only necessary to have about 86 doublings … if it is assumed that every Planck volume splits during each doubling. However, there are about 2x1011 units of Planck time in the period during which inflation is thought to have occurred meaning that all that is required to achieve the minimum for inflation is that for each unit of Planck time, there would be an average one additional Planck volume for every existent 3x109 Planck volumes. This is what would happen about 3x109 units of Planck time in, or alternatively, at 1.62x10-34s – noting that inflation is believed to have occurred sometime about 10-33s in.
In other words, something very like inflation happens anyway with my model. | 1,618 | 7,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.966372 |
https://stats.stackexchange.com/questions/624403/in-the-monty-hall-problem-does-it-matter-that-the-host-knows-which-door-the-car | 1,721,480,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00339.warc.gz | 462,927,060 | 43,477 | In the Monty Hall problem, does it matter that the host knows which door the car is behind? If so, why?
If I'm thinking about this correctly, regardless of how the host chooses which door to open, there's a 1/3 chance the player initially picks the door with the car behind it, in which case they shouldn't switch doors and the door opened by the host will have been one of goat doors regardless of whether he was choosing at random or not (assuming he was choosing between the two doors not already chosen by the player). In the case that the player initially picks one of the goat doors, which has a 2/3 chance, if the host is picking a door randomly, there's a 50% he chooses a goat door, whereas there's a 100% chance of that of he knows where the car is and is purposely avoiding that door. But the way the problem is usually stated, we already know the door opened by the host was a goat door, so it seems that there should be 2/3 probability of the player choosing the car door if they switch to the unopened door, regardless of how the host decided which of the two unopened doors to open.
And yet, I've heard many people say that the key to understanding this problem is to realize that the host is intentionally opening a door with a goat and that, of he were instead choosing an unopened door at random, it really wouldn't matter if the player switches doors or not. Wikipedia seems to back this up:
For example, assume the contestant knows that Monty does not pick the second door randomly among all legal alternatives but instead, when given an opportunity to pick between two losing doors, Monty will open the one on the right. In this situation, the following two questions have different answers:
1. What is the probability of winning the car by always switching?
2. What is the probability of winning the car by switching given the player has picked door 1 and the host has opened door 3?
The answer to the first question is 2/3, as is correctly shown by the "simple" solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 given the host has opened door 3 (the door on the right) is 1/2. This is because Monty's preference for rightmost doors means that he opens door 3 if the car is behind door 1 (which it is originally with probability 1/3) or if the car is behind door 2 (also originally with probability 1/3).
This isn't quite the same distinction as whether the host is choosing randomly or not, but the core thing I'm not understanding is why it matters at all how he chooses which unopened door to open, since we already know which one it is. I'm trying to figure it out by relating it to other conditional probability scenarios where the prior probability is clearly relevant, such as the textbook example of a medical test with a high accuracy rate where the probability that a positive result is correct depends on the prior probability the patient had the disease. I understand Bayes' theorem and medical test examples and the like fine. But I'm not seeing a clear way to make an analogy to variations of the Monty Hall problem where the host isn't purposely never choosing the door with the car.
• If Monty opens a random door, does this mean he also has a chance to open the door with the car? What happens in that case? Commented Aug 20, 2023 at 10:58
• "why it matters at all how he chooses which unopened door to open, since we already know which one it is" what do you mean with 'we already know'? Who is 'we', the contestant that needs to make the choice? Commented Aug 20, 2023 at 11:09
• @SextusEmpiricus, yes the contestant already sees that a goat had been revealed behind the door opened by the host. Commented Aug 20, 2023 at 19:52
There are 3 golfers, named A, B, and C. One of them is a master, the others are amateurs. The master will always outgolf the amateurs. You want to guess which one is the master.
Scenario 1: you are told that all 3 of them played a round of golf, and B came last. A and C are clearly equally likely to be the master. This is equivalent to the modified Monty Hall problem where Monty opens one of the goat doors at random and happens to pick B.
Scenario 2: you are told that only B and C played a round of golf, and B came last. The probability that C is the master is 2/3. This is equivalent to the original Monty Hall problem where Monty is only allowed to pick a goat door from among B and C and picks B.
It is easier to picture with 100 doors.
Scenario 1: You pick 1 random door and Monty opens 98 doors that he knows have goats behind them.
Scenario 2: You pick 1 random door and Monty picks 1 random door. (Note that this is the same as him randomly picking 98 other doors to open first, since picking the first 98 doors or last 1 door amounts to the same thing.)
In Scenario 1 there is a 99% chance that the car is behind the doors you did not pick. By eliminating 98 goat doors non randomly that still leaves a 99% chance that the door he did not open has the car and a 1% chance your door has the car. There were always going to be at least 98 doors in the unchosen group that had goats. Opening those doors on purpose does not change the original 99% chance that a car was behind your unchosen 99 doors. You should always switch here since there is a 99% chance it is behind the door you did not choose.
In Scenario 2 there is a 2% chance that your door or Montys door have the car. If you open the other 98 doors and there randomly happened to be goats behind them then there is still an equal chance that your door or Montys door have the car. There is no benefit to switching in this scenario since each door now has a 50% chance of having the car.
One way to help further visualize this is to imagine that you always pick door 99, Monty always picks door 100, and the doors are always opened in order. After doors 1 through 98 are opened with goats you have the option to switch with Monty and take door 100. There was always an equal chance it would be behind 99 or 100, and there is now still an equal chance it will be behind 99 or 100.
Monty choosing randomly is absolutely different from Monty knowing and opening all the goat doors first on purpose.
The point about Monty's door choices is that it provides additional information, but also that the information interacts with the first choice by the contestant and the place of the true position of the price.
We can have alternative situations where one or both of the interactions are not present.
This question relates to the question: Monty Hall problem and causality where we saw the first diagram on the left below, and relating to this question we can add the adaptations that are made in the second and third diagram:
The images above are directed acyclic graphs that express causal relationships. At the start of the graph are three independent variables, u_door_with_the_prize, u_quizmaster and u_first_choice. The arrows indicate which variables depend on one and another. The first variation has the arrow between the 'u_door_with_the_prize' and the 'door opened by the quizmaster' eliminated. The second variation has an additional relationship between the 'second choice' and the 'u_door_with_the_prize'.
In the situation that you mentioned, Monty opens the doors independently from the the true position of the prize. That would be like the middle graph. Then, the strategy, to switch or not switch can not have any effect, because the final decision, however it is made, is causally independent from the actual position of the prize and therefore the strategy can't have a positive effect.
An exception would be when we allow for the strategy, and the second choice, to be in some way dependent on the position of the prize, and in that case we get the third graph. This would be the case when the strategy is not just to switch to the door that was left unopened by Monty, but when the strategy is also allowed to switch to other doors and in particular a door opened by Monty that might have the prize behind it.
This type of conditional strategy is similar to what is happening in the Wikipedia example where based on the two different type of situations (a plain strategy decided before knowing what Monty opens, or a strategy that relates to what Monty actually opens).
• I don't really understand how to interpret your diagrams. Would you mind clarifying? Commented Aug 20, 2023 at 23:12 | 1,896 | 8,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-30 | latest | en | 0.980222 |
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# MATH PROJECT - PowerPoint PPT Presentation
MATH PROJECT. DONE BY : LAILA HASSAN MOZA BADER HAMDA MUBARAK HOUR ESSA. TASK-1 PART-1. Write an algebraic expression for the charges under each option. Catering R us. All things catered 4U. 800+15x. 200+30x. PART-2.
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• DONE BY : LAILA HASSAN
• HAMDA MUBARAK
• HOUR ESSA
Write an algebraic expression for the charges under each option.
Catering R us
All things catered 4U
800+15x
200+30x
Recored the cost for each option for inviting 5,15,25,35,45 friends
R US - 800+15(5)=875
800+15(15)=1,025
800+15(25)=1175
800+15(35)=1325
800+15(45)=1475
4U- 200+30(5)=350
200+30(15)=650
200+30(25)=950
200+30(35)=1250
200+30(45)=1550
a) If Mohammed plans to invite 20 friends, which catering company should he choose? Show your work.
All thing catered 4U-:200+30(20)
=800AED
• catering R Us-: 800+15(20)
• =1,100AED
Mohammed should choose all thing catered 4U company.
b) If Mohammed plans to invite 50 friends, which catering company should he choose? Show your work.
Catering R Us -: 800+15(50)
=1.550AED
All things catered 4U-:200+30(50)
=1,700AED
Mohammed should choose all thing catered 4U company.
c) Find the exact number of persons that each of the two catering companies can accommodate for the same amount. How much will the amount be? Show your complete solution.
The amount of each company will be 1,400AED for 40 persons.
Ans: 800+15x=200+30x
800-200=30x-15x
600/15=15x/15
X=40
200+30(40)=1,400 800+15(40) =1,400
PART-2 catering companies can accommodate for the same amount. How much will the amount be? Show your complete solution.
Mohammed has 950 AED available to spend on his birthday party. How many friends can he invite under each of the catering options? Set up an equation for each option and solve it to find out how many friends (persons) that he can invite under each option.
Catering R Us :800+15x=950
15x=950-800
15x/15=150/15
X=10
10persons
All things catered 4U-:200+30x=950
30x=950-200
30x/30=750/30
X=25
25persons
PART-3 catering companies can accommodate for the same amount. How much will the amount be? Show your complete solution.
Mohammed hates to waste food and beverages. He does not like to be embarrassed in front of his friends either for not providing enough to eat and drink. He determines that 40 persons plus or minus 10% is a good number. Write and solve an absolute value equation to find the maximum and minimum number of friends that Mohammed should invite to his party.
The maximum number of friends is 44
The minimum number of friends is 36
| x-40|=4
X-40=+4or-4
X-40=4
X-40=-4
X=4+40
X=-4+40
X=44
X=36
TASK-4 catering companies can accommodate for the same amount. How much will the amount be? Show your complete solution.
PART-1
Three groups ,groups A,B,C of mohammed's friends decided to buy a watch as a memorable gift for his birthday contributed 1200AED in the ratio of 3:4:5.how much did each group contributed?
PART-2 catering companies can accommodate for the same amount. How much will the amount be? Show your complete solution.
A) what is the percentage is the highest contribution with the respect to the total?
• the highest contribution is C
• 500/1200x100=41.7%
B) what is the percentage is the lowest contribution with the respect to the total?
B) the lowest contribution is A
300/1200x100=25%
TASK-5 catering companies can accommodate for the same amount. How much will the amount be? Show your complete solution.
PART-1
Equations are used widely in our real-world life applications. Prepare a research about how solving equations helps us in different domains of our life.
The algebra activities on this page, make learning algebra easy and fun. Take the example in picture below.
BALANCED SCALED catering companies can accommodate for the same amount. How much will the amount be? Show your complete solution.
For example, if I add a certain amount of weight to one side of the scale, to make it stay balanced, I know I need to add the same weight to the other side of the scale.
Guess what happens if you don’t add the same amount of weight to one side as you do to the other side. Well, take a look at the picture below if you’re wondering.
UNBALANCED SCALED catering companies can accommodate for the same amount. How much will the amount be? Show your complete solution.
In Algebra, we always want the Scale to Stay Balanced. So in the above algebra example, too much weight was added to the right side of the scale. So the scale is not balanced anymore.
X+ catering companies can accommodate for the same amount. How much will the amount be? Show your complete solution.
=
So the equation is x+5=9
There are two blocks on the left, one has a 5 in it and the other is empty.
Write an algebra equation for the balance, and
Determine what the resulting value of the variable, x is.
two blocks on the left, side must equal 9, the block on the right side of the balance scale.
3X=12 ; X=4 catering companies can accommodate for the same amount. How much will the amount be? Show your complete solution.
6X=24; X=4
8X=64 ; X=8
THANK YOU catering companies can accommodate for the same amount. How much will the amount be? Show your complete solution.
AND WE HOPE THAT YOU LIKE OUR PRESENT . :) | 1,614 | 5,938 | {"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} | 3.9375 | 4 | CC-MAIN-2017-51 | latest | en | 0.738125 |
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# help
+1
86
2
Ben is 20 years older than Daniel. Ben and Daniel first met two years ago. Three years ago, Ben was 3 times as old as Daniel. How old is Daniel now?
Apr 21, 2020
#1
+711
+2
Hi guest!
This problem was a bit confusing for me - is "Ben and Daniel first met two years ago." extraneous information? It seems like it to me...
Well anyways, let's start.
We know that
$$B=20+D$$ (since Ben is 20 years older than Daniel)
$$B-3=3D-3$$ (since 3 years ago, Ben was thrice Daniel's age)
Now all we have to do it plug it in.
$$B=20+D$$
$$B=3D$$
$$20+D=3D$$
$$D=10$$
$$D+3= \text{his age now}$$
Daniel is now 13.
Apr 21, 2020
#2
+25994
+2
B = 20 + D
B-3 = 3(D-3) or -B + 3 = -3D+9 add to first equation
3 = -3D+9 + 20 + D
-26 = -2D D = 13
Apr 21, 2020 | 310 | 821 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-34 | latest | en | 0.916353 |
https://imathworks.com/physics/physics-question-about-physical-degree-of-freedom-in-maxwell-theory-why-coulomb-gauge-can-fix-all-redundant-degree-of-freedom/ | 1,723,434,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00789.warc.gz | 245,295,153 | 8,573 | # [Physics] Question about physical degree of freedom in Maxwell Theory: Why Coulomb gauge can fix all redundant degree of freedom
degrees of freedomelectromagnetismfield-theorygaugegauge-theory
Given $4$-potential $A^\mu(x)=(\phi(x),\mathbf{A}(x))$, the vacuum Maxwell equations:
$$\nabla^2\phi+\frac{\partial}{\partial t}(\nabla\cdot \mathbf{A} )=0$$
$$\nabla^2 \mathbf{A} -\frac{\partial^2 \mathbf{A} }{\partial t^2} – \nabla(\nabla\cdot\mathbf{A} + \frac{\partial \phi}{\partial t})=0$$
There is redundant degree of freedom(d.o.f) in $A^\mu(x)$:
$$A^\mu(x)\rightarrow A^\mu(x) +\partial^\mu \lambda(x)$$
In Coulomb gauge:
$$\nabla\cdot \mathbf{A}(x)=0 \tag{1}$$
Vacuum Maxweel equation becomes:
$$\nabla^2\phi =0 \tag{2}$$
$$\nabla^2 \mathbf{A} -\frac{\partial^2 \mathbf{A} }{\partial t^2} = \nabla( \frac{\partial \phi}{\partial t}) \tag{3}$$
Then we can always choose $\phi(x)=0$. So question becomes that physical degree of freedom is $A^\mu(x)= (0,\mathbf{A}(x))$ with one constraint $\nabla\cdot \mathbf{A}(x)=0$. Every textbook then says the physical degree of freedom is $2$. But it seems there are still redundant d.o.f , we can always make
$$\mathbf{A}(\mathbf{x},t)\rightarrow \mathbf{A}(\mathbf{x},t)+\nabla \Lambda(\mathbf{x})$$
such that
$$\nabla^2 \Lambda(\mathbf{x}) =0\tag{4}$$
But above equation is Laplace equation that has nontrivial solutions, harmonic function.
For example, $\Lambda(\mathbf{x}) = xyz$.
My questions
1. Using $\phi(x) =0$ and $\nabla\cdot \mathbf{A}(\mathbf{x},t)=0$, I have substract two redundant d.o.f. , why fixing $\Lambda(\mathbf{x})$ further can not substract more redundant d.o.f. ?
2. Many textbooks will argue that $A^\mu(x)$ should vanish at spacial infinity, so Laplace equation $(4)$ with zero boundary condition in infinity has only trivial solution. But why do we have to require $A^{\mu}$ vanish at spacial infinity? For example, a uniform magnetic field has $\mathbf{A}(x) = \mathbf{B}\times \mathbf{r}/2$ which does not vanish in infinity. If you require that $A^\mu$ vanish at spacial infinity, you even cannot get constant electrical or magnetic field solutions from Vacuum Maxwell equations. And even the elctromagnetic wave solution $e^{i k(t -x)}$ is also nonvanishing at spacial infinity. This question has some relation with Coulomb gauge fixing and “normalizability”
3. Why textbook says "Lorentz gauge is Lorentz invariant but cannot fix all redundant d.o.f. Coulomb gauge can fix all redundant d.o.f but is not Lorentz invariant. "? But it's obvious that only Coulomb gauge $(1)$ also cannot fix all redundant d.o.f. We can see that the gauge fixing $\phi =0$ is not a consequence of $(2)$. It's forced artificially and $\phi=0$ is independent from Coulomb gauge. For example, vacuum Maxwell equations $(2),(3)$ can have uniform electrical field solution $\phi(x)=-\mathbf{E}\cdot \mathbf{r}$, $\mathbf{A}=0$ satisfying only Coulomb gauge $(1)$ but $\phi(x)\neq 0$. If we requires $\phi=0$ and $\nabla \cdot \mathbf{A}=0$, the solution becomes $\phi=0$ and $\mathbf{A}= \mathbf{E} t$.
1. The main thing one has to be careful about here is the boundary conditions on the gauge field $A_\mu(x)$ and the gauge parameter $\Lambda(x)$. In particular, the set of gauge transformations that one is allowed to "gauge away" must vanish at "infinity". By infinity, here one often means both spatial infinity as well as null infinity. You cannot gauge fix "large gauge transformations" also known as global gauge transformations since those correspond to actual physical symmetries of the theory with physical consequences on the system (deduced via conservation laws). The simplest such example is when $\Lambda(x) = \lambda =$ constant. These gauge symmetries give rise to charge conservation and are very important. They cannot and should not be removed. Having said this, we go back to the equation in Coulomb gauge. The residual gauge transformations are generated by functions $\Lambda(x)$ satisfying $$\nabla^2 \Lambda(x) = 0 \, , \qquad \Lambda(x) \to 0 ~~\text{at infinity.}$$ You can now convince yourself that the only solution to these equations is $$\Lambda(x) = 0\, .$$ This answers your first question.
2. The answer to your second question about the "why" of these boundary conditions is the requirement of finite energy flux through the boundaries of the system. The idea is that we restrict to solutions which have the property that if finite energy is input into the system through a certain boundary, then finite energy should be released from the system. Here, I am not talking about the total energy which will always be fixed due to energy conservation. What we are talking about here is the local energy flux through each boundary. We are only interested in solutions to Maxwell's equations where even the local energy density is finite at all points and no singularities are created. All such solutions need to have the property that the "vanish" in a suitable way near a suitable boundary. This requirement, as you correctly point out, excludes constant electric or magnetic fields whose total energy is proportional to the volume of the system. The wave solutions do not die of at spatial infinity but that's OK since spatial infinity is not a suitable boundary for these solutions. Wave-like solutions travel outward and reach null infinity ${\mathscr I}^+$ as opposed to spatial infinity. In other words, even though wave-solutions are non-vanishing on spatial infinity, they do not contribute to the energy density there. Therefore, we need that the energy flux on ${\mathscr I}^+$ due to wave-solutions be finite and you can check that this is indeed the case.
3. $\phi(x) \neq 0$ does not mean that it is a degree of freedom. A degree of freedom is defined as a "part" or component of the field $A_\mu(x)$ that is not determined by the equations of motion, and it is therefore, completely free to choose. To state this more precisely, a degree of freedom is the data that must be prescribed (completely freely) on a Cauchy surface so as to ensure unique time evolution into the past and future. It is the piece of information about the field that completely determines every other aspect of it. Thus, $\phi$ is never a degree of freedom. In any gauge you choose, it is easy to see that it is determined entirely in terms of $A_i(x)$. | 1,664 | 6,334 | {"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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-33 | latest | en | 0.726834 |
https://slideplayer.com/slide/2386324/ | 1,531,975,182,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590493.28/warc/CC-MAIN-20180719031742-20180719051742-00241.warc.gz | 779,614,026 | 20,472 | # Planning a Picnic Division of a Whole Number by a Fraction: 5.NF.7b.
## Presentation on theme: "Planning a Picnic Division of a Whole Number by a Fraction: 5.NF.7b."— Presentation transcript:
Planning a Picnic Division of a Whole Number by a Fraction: 5.NF.7b
End of the Year Picnic There will be an end of the school year picnic. Each group has been assigned a particular food or beverage to bring. Your job is to determine how many servings your food or beverage will provide.
Steps: Draw a picture, graphic or number line to represent your food or beverage item(s) Divide it by the given fraction and show your work on the drawing. Write a brief explanation of how the number of servings was determined. Answer the follow-up questions. Use the student check sheet along the way.
Group 1: Watermelon There are 6 watermelons. If a serving of watermelon is 2/7 of a watermelon, how many people will these 6 watermelons serve?
Fruit Juice There are 12 bottles of fruit juice. A serving of fruit juice is 2/9 of a bottle. How many servings are there in all?
Ground Beef Hamburgers will be served at the picnic. It takes 2/5 lb of ground beef to make one hamburger patty. If there are 8 lbs of ground beef, how many hamburgers can be made?
Brownies There are 6 pounds of brownies. A serving size for a brownie is 3/10 of a pound. How many servings of brownies are there?
Closure: Patterns Think about the picnic problems you completed. (See board for reference). Do you notice any patterns to how the quotient is determined?
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Induction
“Because all the many flames observed before have been hot is a good reason to believe this flame will be hot” is an example of an inductive argument, and a rational one.
An inductive argument is an argument from contingent (not logically necessary) premises which are, or could have been, observed, to a contingent conclusion about something that has not been, and may not be able to be, observed. An inductive argument must also have its conclusion say about the unobserved something like what the premises says about the observed.
In classical, frequentist statistics inductive arguments are forbidden—not frowned upon, but disallowed. Even some Bayesians have adopted the skeptical belief that inductive arguments are “ungrounded”, or that there is a “problem” with induction. This is not the time to enter into a discussion of why these thoughts are false: David Stove’s masterpiece “The Rationality of Induction” can be consulted for particulars.
Anyway, only academics pretend to be mystified by induction, and only in writing. They never act as if induction is irrational. For example, I’ve never met a skeptical philosopher willing to jump off a tall building. I assume inductive arguments to be rational and unproblematic.
There are deductive arguments and non-deductive arguments; not all non-deductive arguments are inductive ones, though it is a common mistake to say so (perhaps because Carnap often made this slip). Logical probability can be used for any type of argument. Frequentist probability is meant to represent a substitute for deductive arguments in non-deductive contexts, somewhat in line with Popper’s ideas on falsification. We will talk about that concept in a different post.
Confidence Intervals
In frequentist statistics, a confidence interval (CI) is function of the data. Custom dictates a “95%” CI; though the size is irrelevant to its interpretation. Often, at least two data points must be in hand to perform the CI calculation. This, incidentally, is another limitation of frequentist theory.
The CI says something about the value of an unobservable parameter or parameters of a probability model. It does not say anything about observables, theories, or hypotheses. It merely presents an interval (usually contiguous) that relates to a parameter.
Its interpretation: If the “experiment” in which you collected your data were to be repeated a number of times that approached the limit of infinity, and in each of those experiments you calculated a CI, then 95% of the resulting (infinite) set of CIs will “cover” the true value of the parameter.
Problems
Problem one: The “experiment” must be, but almost never is, defined rigorously. But even when it is, the idea that you could recreate an experiment that is identical to the milieu in which you collected your original data an infinite number of times is ludicrous. That milieu must be the same in each re-running—except that it must be “randomly” different. That “randomly” is allowed to remain vague and undefined.
Problem two, and the big one. Even if you can satisfy yourself that an infinite number of trials is possible, you are still confronted with the following fact. The CI you collected on your data has only one interpretation: either the true value of the parameter lies within it or it does not. Pause here. That is all you are ever allowed to say.
The italicized statement—a tautology—is so important, so crucial, so basic to the critique of frequentist theory that few can keep it in focus, yet nothing can be as simple.
Whatever interval you construct, no matter how wide or how small, the only thing you are allowed to say is that the true value of the parameter lies within it or it does not. And since any interval whatsoever also meets this tautological test—the interval [1, 2] for example—then the CI we have actually calculated in our problem means nothing.
Yet everybody thinks their CI means something. Further, everybody knows that as more data collected, the calculated CIs grow narrower, which seems to indicate that our confidence about where the true value of the parameter lies grows stronger.
This is false. Strictly false, in frequentist theory.
Making any definite statement about a CI other than the above-mentioned tautology is a mistake. The most common is to say that “there is a 95% chance that the parameter lies within the CI.” That interpretation is a Bayesian one.
The other, just mentioned, is to say that narrower CIs are more certain about the value of the parameter than are wider CIs. That is an inductive argument which attempts to bypass the implications of the tautology.
Neyman
The gentleman that invented confidence intervals, Dzerzij (Jerzy) Neyman knew about the interpretational problems of confidence intervals, and was concerned. But he was more concerned about inductive arguments, which he thought had no business in statistics.
Neyman tried to take refuge in arguments like this: “Well, you cannot say that there is a 95% chance that the actual value of the parameter is in your interval; but if statisticians everywhere were to use confidence intervals, then in the long run, 95% of their intervals will contain their actual values.”
The flaw in that workaround argument is obvious (make sure you see it). And so, with nowhere else to turn, in 1937 Neyman resorted to a dodge and said this: “ The statistician…may be recommended…to state that the value of the parameter…is within [the just calculated interval]” merely by an act of will.
Since that time, statisticians having been illegally willing CIs to mean more than they do.
Induction and CIs
You will often read of “numerical simulation experiments” in which a statistician tries out his new method of estimating a parameter. He will simulate a, necessarily finite, run of CIs where the true value of the parameter is known and note the percentage of simulated CIs that cover the parameter.
If the percentage is close to 95%, then the statistician will state that his procedure is good. He will convince himself that his method is giving proper results: that is—this is crucial—he will convince himself that his estimation method/theory is likely to be true.
Just think: he will use an inductive argument from his observed experimental data to infer that future CIs will be well behaved. But this is forbidden in classical statistics. You are nowhere allowed to infer the probability that a theory is true or false: nowhere.
Any such inference is the result of using induction.
Of course, classical statisticians everywhere use induction, especially when interpreting the results of studies. We just never seem to remember that the frequentist theory of probability forbids such things. Challenge two: find one study whose conclusions do not contain inductive arguments.
Logical Probability, Bayesian
Any theory of probability should be all-encompassing. It shouldn’t just work for the technical apparatus inside a probability model, and not work for events outside that limited framework. A proper theory should apply to its technical apparatus, its conclusions and the extrapolations made from them. The Bayesian and logical theories of probability, of course, are general and apply to statements of any kind.
Somehow, frequentists can use one form of probability for their models and then another for their interpretations of their models. This inconsistency is rarely noted; perhaps because it is more than an inconsistency: it is a fatal to the frequentist position.
Now, if you have ever had any experience with CIs, you know that they often “work.” That is, we can interpret them as if there were a 95% chance that the true value of the parameter lies withing them, etc.
This is only an artifact caused by the close association of Bayesian and classical theory, where the Bayesian procedure opts for “non-informative” priors. This is coincidental, of course, because the association fails to obtain in complex situations.
There are those who would reconcile frequentist and Bayesian theories. They say, “What we want are Bayesian procedures that have good frequentist properties.” In other words, they want Bayesian theory to operate at the individual problem level, but they want the compilation, or grouping, of those cases to exhibit “long-run” stability.
But this is merely Neyman’s mistake restated. If each individual problem is ideal or optimal (in whatever sense), then the group of them, considered as a group, is also ideal or optimal. Plus, you do not want to sacrifice optimality for your study for the tenuous goal of making groups of studies amicable to frequentist desire.
Recommendation
As always: we should cease immediately teaching frequentist mathematical theory to all but PhD students in probability. In particular, no undergraduates should hear of it; nor should casual users.
Next stop: p-values.
1. Bernie says:
Did my last message get stuck in a filter somewhere?
Matt:
Are there any current statisticians who explicitly champion the frequentist position?
Are there any significant practical implications of the differences in the two positions? I thought about doctors telling patients about the risks associated with a procedure. In such cases many of the immediately obvious moderating factors can be measured. However, it seems pretty transparent that no doctor could withstand much of a cross-examination as to the basis of his or her risk assessment.
I would bet many insurance companies base their risk assessments on a frequentist position. I initially thought that the insurance company that is advertising accident forgiveness may have adopted a Bayesian approach – but on further reflections they may have simply looked at the frequency with which policy holders have multiple accidents.
2. Briggs says:
Bernie,
(You accidentally posted it on the Ernie Harwell thread.)
No: there are only a few statisticians left who exclusively champion frequentism. Most are compromisers. They concentrate—rightly, for the most part—of getting answers to practical questions. The philosophical debates don’t interest them.
Because of this we have tremendous inertia. All our class notes and textbooks are in the frequentist tradition. Who has time to change them?
I’d say we are at a point at which the introduction of new methods is perfectly timed. Nobody can ever remember to use confidence intervals and p-values properly—despite the hundreds of reminders that are written perennially. Too, our software has reached the point where implementations of modern methods are trivial.
What the majority of students (of all ages, and at most levels) want are procedures that are easy to understand and to implement. Procedures which are logically coherent and are natural and easy to interpret.
Bayesian (logical probability) procedures are simple: once you learn how to do it in one problem, it’s the same in every other problem. Just as in frequentism, there are always models to choose from, but there are no tests to choose. Testing, as we know, leads to cheating and overconfidence.
The strongest point, besides internal coherence, is that Bayesian procedures can be used to talk directly about observables. We need to move away from this odd reliance on unobservable parameters for our inferences. Once more, we are always too sure of ourselves when speaking of parameters.
And why do it wrong when doing it right is so much easier? Bayes is easier to teach and to retain.
A huge problem we have as teachers is that we can’t make up our minds whether we’re teaching a mathematical or a empirico-philosophical classes. Mathematics at least has the advantage of being difficult. And it does form the guts of what we do. But that doesn’t mean we should teach it that way.
3. JJD says:
OMG, this sounds serious. Do “frequentists” patrol the streets in uniforms and hit you in the face with a rifle butt if you don’t calculate your p-values in the approved way? (I am not talking about university professors and journal editors — I can show you my scars — but actual working people who use statistics).
It might be less aggravating to argue against “1930’s statistical methods” instead of personalizing the issue. Have a beer and cool off there, Matt. It is 95% certain that there is no vast conspiracy of perverted and irrational “frequentists.” Even old dinosaurs such as myself who (without meaning any harm) are occasional users of confidence intervals and hypothesis tests would welcome clear expositions of alternatives offered by modern statistics.
I think the prevalence of 1930’s methodology in the workplace is partly explained by ignorance of what has been developed since then. You are right — Stats 101 needs a serious overhaul, and the outdated textbooks need to be replaced. The same kind of problem exists in other fields too.
4. Briggs says:
JJD,
You’re right: it’s my natural combative nature. Probably causes more troubles than solves problems.
On the other hand, I’d say we’re in an epidemic of over certainty caused by poor practices. If it were constrained to academia, I wouldn’t care so much. But it’s reached the government and taken a nasty turn towards absolute assurance.
5. Bernie says:
Matt:
I am assuming that ensemble runs of computer models with subsequent predictions and CIs are fruits of this poisoned tree?
6. John says:
Bayesian statistics doesn’t really ever allow inference either. It tells you how to change your existing (prior) beliefs, but it’s never clear what those prior beliefs should be, at least, for the purposes of publication. Maybe you have some prior belief, but how many of your readers would agree? There’s no way of differentiating a reasonable prior (noninformative perhaps?) from a prior that says “I am highly confident that my pet theory for which I have no empirical evidence.”
Conversely, in many actual experimental settings, the logic of frequentism makes a lot of sense. “I did an experiment and found that two things were different. I have done frequentist statistics and determined that somebody else would also find them different 95% of the time if they did my same experiment over again.” The inference comes afterwards. Since it seems that this experiment will produce this difference reliably, does this interfere with theories that wouldn’t predict such an effect?
This might sound unprincipled, but actually Bayesian statistics only seems more principled because it builds some (though of course not all) of the unprincipled aspects of science (specifically, the prior and inferential model) into the statistics. I can see why some mathematically-inclined individuals might prefer this way of doing things, but I don’t see how it’s any better in principle.
7. Briggs says:
John,
Ah, but there is the field of objective Bayesian statistics, otherwise known as logical probability, in which evidence can be used to specify the model and its parameters (PDF). This is not so with frequentist statistics, in which the model is always unspecified; always, that is, ad hoc, subjective and prone to abuse.
There is no universal solution to uncertainty, though. No matter frequentist, subjective, or objective Bayesian, all inferential statements are conditional on the given model. Which is of far greater importance than priors set on parameters.
You’ll find the Bayes can be split in two: subjective and objective. The former is more prone to insist on precise answers for all questions. Oddly, the objective branch does not and allows imprecision in its answers.
Also, as the standard demonstration goes, the influence of the prior on the posterior parameter distribution is small to disappearing as your evidence increases. Even stronger, the priors’ influence on the posterior observable (predictive) distribution is nearly zip. And don’t neglect the correspondence between many frequentist procedures and Bayesian results assuming “flat” priors.
And how about situations in which no data is possible; say, counterfactuals? There is no frequency of something that did not happen. But this is no bar to forming a logical probability on counterfactual.
Your example is false. You say, “I did an experiment and found that two things were different. I have done frequentist statistics and determined that somebody else would also find them different 95% of the time if they did my same experiment over again.”
You may not make that kind of claim in frequentist theory. You may, of course, in Bayesian theory; what you have said is an inductive inference.
In frequentist theory, you may make statements conditional on the model’s parameters taking certain specified values (usually zero; but subjectively chosen that way to make it zero). But these are odd statements to make and do not relate directly to observables.
Bernie,
Well, we always have the “approximate” (inductive) inference available to us. Assuming the experiments are well done, and the (subjectively chosen) models used are reasonable, then the answers spit out can be interpreted in a (loose) inductive way.
Think of it this way: nobody, even champions of frequentism, believes that they results they produce are exact. Everybody builds into the results a bit of fuzz. But this is not allowed in frequentist theory, because doing so means engaging in yet another inductive inference. Of the type, “I’ve seen many models before and none of them are perfect; they are all off by a little; therefore this one will be too.” Again, can’t say that in frequentism.
And once more we have the embedding problem…which we’ll skip!
8. JH says:
Admittedly, I haven’t given careful thought (lazy me) to this post, so my comment below might seem a bit unorganized.
Yes, you are correct. Some people simply are not interest; and those who are/was interested in the philosophy know that Bayesian vs. frequentist is an old debate. Moreover, I would say that most statisticians know of various concepts of probability.
[T]he influence of the prior on the posterior parameter distribution is small to disappearing as your evidence increases.
You mean, as your sample size increases… Hmm, I though Bayesian doesn’t use the large-sample theory. ^_^
You may say there is such a thing called objective Bayesian, to me, the word “objective†is misleading.
This is not so with frequentist statistics, in which the model is always unspecified; always, that is, ad hoc, subjective and prone to abuse.
I am not sure what you meant here. The likelihood function is the essential component of statistical modeling, irrespective of the inference methods. Once you have modeled a theoretically, practically appropriate likelihood function, one may choose Bayesian, frequentist, or likekihood-ist, or nonparametric or whatever inference method. So why would the model be always unspecified in frequentist statistics?
Many statisticians have chosen to use more than one method and then carefully compare the results. And there are happy coincidences where Bayesian and frequentist CIs look the same.
As a whole, the basic idea the Bayesian is simple but it can get complicated computationally. We know that in practice people blindly run whatever models without checking model assumptions. Off the top my head, here is one worrisome problem about Bayesian analysis. It seems to me that people would conveniently employ conjugate priors. And out of numerous (infinite?) possible distributions, normal, gamma, beta etc just turn out to be the right choice??!! So, I am not sure whether frequentist or Bayeisan statistics are more prone to abuse.
Well, I don’t wish to pick on Bayesian or frequentist I see them as different statistical tools. As far as I know, every methodology has its faults and merits.
And this comment is too long!
9. Mike B says:
Why does everyone forget Fisher in these Frequentist/Bayesian debates? The greatest scientific statistician of the 20th Century was neither a Frequentist nor a Bayesian.
Bradley Efron neatly delineates them in this lecture, summarized by a triangle, with one apex each for Frequentist, Bayesian, and Fisherian.
http://www.uweb.ucsb.edu/~utungd00/fisher.pdf
10. Briggs says:
Mike B,
Excellent point. Fisher’s views on induction were somewhat inconstant. Unlike the school that followed him, he allowed the usefulness of induction, but only up to a point. He accepted its use in developing mathematical theorems, for example (not for their proofs, of course). But he did not allow it to be formally quantified probabilistically. At this point, possibly recognizing the paradox and attempting to reconcile it, he developed his idea of fiducial probability. He wanted to avoid explicit use of evidence—that is to say, induction—and especially disliked the idea of “prior” distributions (I’ve never loved that name; it is misleading.)
Of course, fiducial inference was never a success. It gave rise to many counterexamples, and did not provide actual probability measures. Fisher himself, in a departure from his usual self-confidence, never loved his own theory. He let it lie fallow.
Fisher, too, disliked confidence intervals. Many of the standard complaints against them originated with him.
And let’s not forget that Fisher’s p-values were built in such a way to approximate Popper’s falsification standard for probabilistic arguments, for which no falsification is possible. Fisher acknowledged this, but Popper never did.
Unfortunately, we all know the problems of p-values. I plan on discussing these in another article. I’ll point out an little-known argument against their use.
11. Bernie says:
Matt:
Can I suggest that you use concrete examples to illustrate the differences in the different positions. After a while, and without spending significant time expanding my technical vocabulary, the practical significance of the discussion gets obscured. After reading about the work of Godel, Heisenberg, Arrow and Briggs ( ;>) ), I am convinced that too many people are too certain of too many things … what’s next? | 4,519 | 22,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-13 | latest | en | 0.928444 |
https://richardvigilantebooks.com/how-much-can-you-notch-out-of-a-joist/ | 1,716,743,838,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00102.warc.gz | 426,848,089 | 11,128 | # How much can you notch out of a joist?
## How much can you notch out of a joist?
The maximum depth of a notch at the end of a joist (where it rests on a wall or beam) can’t exceed one-quarter of the joist depth. Maximum notch depth in the outer third of a joist is one-sixth of the joist depth. Limit the length of notches to one-third of the joist depth. No notching in the middle third of a joist.
Can I cut a notch in a floor joist?
You can cut a notch at the end of the joist to 1/4 of the joist’s depth (maximum). Along the outer third of a joist, you can cut a notch a maximum of 1/6 the joist depth and 1/3 the joist length without compromising its strength.
How far can a 2×12 floor joist span without support?
In general terms, joists spaced 16 inches on center can span 1.5 times in feet their depth in inches. A 2×8 up to 12 feet; 2×10 to 15 feet and 2×12 to 18 feet.
### Can you notch an LVL?
Notching of LVL beams should be avoided whenever possible, especially on the tension side of a member. Tension- side notching of LVL beams is not recommended except at end bearings and then only under specific conditions.
Can I drill holes in LVL beams?
Drilling holes in IB (International Beams) LVL (laminated veneer lumber) is generally discouraged. However, small diameter round holes, appropriately located in the wide face of the LVL, are permitted as indicated in the table and notes below.
Is LVL stronger than dimensional lumber?
LVL offers several advantages over typical milled lumber: Made in a factory under controlled specifications, it is stronger, straighter, and more uniform. Due to its composite nature, it is much less likely than conventional lumber to warp, twist, bow, or shrink.
#### How far will a 12 inch LVL span?
A double 2×12 beam can span 12 feet; a (2) 2×10 can span 10 feet and so on. In respect to this, how far can you span LVL? Measure your total span between members and ensure that it is not greater than 60 feet. Because of transport limitations, the maximum standard length for manufactured beams is sixty feet.
Which is stronger LVL or PSL?
PSL is a little more durable than LVL because of densification under high pressure and different applied techniques. However, in LVL densification to increase its strength is very limited. Due to this, more bonding pressure also ensures a strong adhesiveness for bonding veneer strands in PSL.
Can a 2×12 span 20 feet?
In general terms, joists spaced 16 inches on center can span 1.5 times in feet their depth in inches. A 2×8 up to 12 feet; 2×10 to 15 feet and 2×12 to 18 feet. The larger the deck, the larger the joists.
## How far can I span a triple 2×12?
Joist SpansSouthern Pine3-2X1012′-9″3-2X1215′-0″Douglas Fir-Larch, Hem-Fir, Spruce-Pine-Fir, Redwood, Cedars, Ponderosa Pine, Red Pine3X6 OR 2-2X65′-2″3X8 OR 2-2X86′-7″17
What size LVL Do I need to span 20 feet?
In that case, you need something like a 12-16″ GLULAM or LVL to span the 20′ and can use simple 2×8-10 dimensional lumber 16″OC as floor joists.
How far can a 4 ply 2×12 beam span?
You will see that a 4-ply, 2 X 12 can span a maximum of 13′ while supporting 12′ joists on either side of it.
### How far can a 6×6 beam span without support?
What is the maximum span for a 6×6 beam? Each 6×6 beam only spans 12 feet, of which 2 of those feet are supported by a knee brace. So technically the beams only span 10ft.
How far can a 4×4 span without support?
A 2×8 up to 12 feet; 2×10 to 15 feet and 2×12 to 18 feet. The larger the deck, the larger the joists. For best results, refer to our wood and composite deck joist span table.
What size lumber can span 20 feet?
In that case, you need something like a 12-16″ GLULAM or LVL to span the 20′ and can use simple 2×8-10 dimensional lumber 16″OC as floor joists….What size rafters do I need to span 20 feet?Maximum Span (ft – in)Nominal Size (inches)Rafter Spacing, Center to Center (inches)Lumber Grade2 x 102420′ – 9”2 •
#### What size lumber can span 24 feet?
2x6s are generally used for ceiling joists, esp since you said 24″ oc and reallllly short spans only (5 or 6 feet ). Most floor loads would be a minimum of 2×8 @ 16″ oc and most preferbaly 2×10.
What size beam do I need for a 15 foot span?
How far can a 2×10 ceiling joist span without support?
Joist spacing of 24 inches is allowed for spans between 16 to 20 feet using 2-inch by 10-inch lumber of these three grades.
## Can a 2×6 span 12 feet?
Spaced at 12 inches, the joist may only span 16 feet 8 inches….How far can you span a 2×6 floor joist?Maximum Span (ft – in)Nominal Size (inches)Joist Spacing Center to Center (inches)Lumber Grade2 x 6249′ – 7″2 x 81215′ – 10″1 more row• | 1,344 | 4,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.924588 |
https://iq.opengenus.org/sum-of-squares-of-first-n-numbers/ | 1,701,757,158,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100545.7/warc/CC-MAIN-20231205041842-20231205071842-00187.warc.gz | 375,748,434 | 22,069 | ×
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# Sum of squares of first N numbers ( Σ n² )
#### Algorithms List of Mathematical Algorithms
Reading time: 30 minutes | Coding time: 2 minutes
Our focus is to find the sum of the quares of the first N numbers that is from 1 to N. The simplest approach is to square each number and add it which will take linear time (N) if we assume multiplication takes constant time. With an insightful equation, we can solve this in constant time O(1).
Example:
If N = 5, then the sum F_2(5) is:
Sum = 1^2 + 2^2 + 3^2 + 4^2 + 5^2
Sum = 1 + 4 + 9 + 16 + 25
Sum = 55
Let F_2(N) be the function denoting the sum of squares of the first N numbers.
The insightful equation is:
$$F_2(N) = \sum_{x=1}^N x^2 = N * (N+1) * (2*N + 1) / 6$$
If we apply if for N = 5, we get:
F_2(5) = 5 * 6 * 11 / 6 = 5 * 11 = 55
There are two approaches to arrive at the solution:
• Induction method
• Binomial method
In Induction method, we will assume a equation of a degree as a solution and used known results to find the actual parameters of the equation. Provided you started with the right degree polynomial, this method will give you the right answer.
In Binomial method, we will start with an expansion of an equation and use it and solution of sum of first N integers to arrive at our current solution. This depends to previous results but is an elegant result.
## Induction method
Let us assume that the polynomial of degree 3 captures the sum. So, our equation will look like:
F_2(X) = A * X^3 + B * X^2 + C * X + D
As there are 4 parameters (A, B, C and D), we will need 4 pre-known solutions to solve this. Following are the four solutions:
F_2(0) = 0
F_2(1) = 1 = 1^2
F_2(2) = 5 = 1^2 + 2^2
F_2(3) = 14 = 1^2 + 2^2 + 3^2
We have choosen 4 values of X = 0, 1, 2 and 3.
When we place this in our initial equation, we get the following:
F_2(0) = D = 0 ... EQ(1.1)
F_2(1) = A + B + C + D = 1 ... EQ(1.2)
F_2(2) = 8A + 4B + 2C + D = 5 ... EQ(1.3)
F_2(3) = 27A + 9B + 3C + D = 14 ... EQ(1.4)
With EQ(1.1), we get the value of D as 0. With this the rest of the three equations are modified as:
F_2(1) = A + B + C = 1 ... EQ(1.21)
F_2(2) = 8A + 4B + 2C = 5 ... EQ(1.31)
F_2(3) = 27A + 9B + 3C = 14 ... EQ(1.41)
To solve for A, B and C, we need to solve the above equations.
EQ(2.1) = EQ(1.31) - 4 * EQ(1.21)
EQ(2.2) = EQ(1.41) - 9 * EQ(1.21)
We get EQ(2.1) and EQ(2.2) as follows:
EQ(2.1) = 4A - 2C = 1
EQ(2.2) = 18A - 6C = 5
Now, if we do the following, we will get the value of A.
EQ(2.2) - 3 * EQ(2.1)
6A - 0 = 2
A = 1/3
With A = 1/3 in EQ(2.1), we will get the value of C as follows:
4*(1/3) - 2C = 1
2C = 1/3
C = 1/6
Now, with A = 1/3 and C = 1/6 in EQ(1.21), we get the following:
A + B + C = 1 ... EQ(1.21)
1/3 + B + 1/6 = 1
B = 1 - 1/2
B = 1/2
Hence, the value of coefficients are:
A = 1/3
B = 1/2
C = 1/6
D = 0
With this, our final equation is:
F_2(X) = (1/3) * X^3 + (1/2) * X^2 + (1/6) * X
F_2(X) = (2 * X^3 + 3 * X^2 + X) / 6
F_2(X) = X * (2 * X^2 + 3 * X + 1) / 6
F_2(X) = X * (X+1) * (2X-1) / 6
Hence, we have arrive at the equation. Use it and enjoy its power.
## Binomial method
To arrive at the solution, we will expand this expression (X-1)^3. Our focus is to find F_2(x) (Let us denote this as EQ1):
$$F_2(x) = 1^2 + 2^2 + ... + x^2$$
We know that (EQ2):
$$F_0(x) = \sum_{i=1}^N i^0 = 1^0 + 2^0 + ... + x^0 = x$$
$$F_1(x) = \sum_{i=1}^N i^1 = 1^1 + 2^1 + ... + x^1 = x * (x+1) / 2$$
Another relation, we will need is the following (EQ3):
$$\sum_{x=1}^N x^3 - \sum_{x=1}^N (x-1)^3 = N^3$$
This is because all terms are cancelled except the last one. Follow this to get the idea:
Σ x^3 - Σ (x-1)^3
= Σ (x^3 - (x-1)^3)
= (1^3 - 0^3 ) + (2^3 - 1^3) + ... + (N^3 - (N-1)^3)
// move negative terms to the end
= 1^3 + 2^3 + ... + N^3 - 0^3 - 1^3 - 2^3 - ... - (N-1)^3
// terms are cancelled except N^3
= N^3
Now, as we have prepared everything, we shall move to the final step. Let us expand (x-1)^3.
(X-1)^3 = (X-1)(X^2 +1-2X) = X^3 - 3*X^2 + 3X - 1
If we place summation in this above expansion, we will get the following (EQ4):
Σ (x-1)^3 = Σ x^3 - 3 * Σ x^2 + 3 * Σ x - Σ 1
=> Σ x^3 - Σ (x-1)^3 = 3 * Σ x^2 - 3 * Σ x + Σ 1
Now, if we use EQ1, EQ2 and EQ3 on EQ4, we get the following:
$$\sum_{x=1}^N x^3 - \sum_{x=1}^N (x-1)^3 = 3 * \sum_{x=1}^N x^2 - 3 * \sum_{x=1}^N x + \sum_{x=1}^N 1$$
$$N^3 = 3 * F_2(N) - 3 * F_1(N) + N$$
$$N^3 = 3 * F_2(N) - 3 * N * (N+1) / 2 + N$$
$$F_2(N) = ( N^3 + 3 * N * (N+1) / 2 - N ) / 3$$
$$F_2(N) = (2 * N^3 + 3 * N^2 + N ) / 6$$
$$F_2(N) = N * (2 * N^2 + 3 * N + 1 ) / 6$$
$$F_2(N) = N * (N+1) * (2N+1) / 6$$
Hence, we have arrived at our solution.
With this, we can find the sum of squares of first N numbers in constant time.
## Implementation
Following is the implementation of the above technique:
class opengenus
{
static int sum(int N)
{
return N * (N+1) * (2*N+1) / 6;
}
public static void main (String[] args)
{
int N = 10;
}
}
Output:
385
With this, you have the complete knowledge of this domain. Enjoy.
#### OpenGenus Tech Review Team
The official account of OpenGenus's Technical Review Team. This team review all technical articles and incorporates peer feedback. The team consist of experts in the leading domains of Computing.
Improved & Reviewed by:
Sum of squares of first N numbers ( Σ n² ) | 2,207 | 5,392 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2023-50 | longest | en | 0.869919 |
https://questions.examside.com/past-years/medical/question/two-waves-having-equation-x1-asinomega-t-k-neet-physics-waves-apl4crjw8iidukiy | 1,716,159,872,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00296.warc.gz | 434,410,004 | 40,674 | 1
AIPMT 2001
MCQ (Single Correct Answer)
+4
-1
Two waves having equation x1 = $$a$$sin($$\omega$$t $$-$$ kx + $$\phi$$1), x2 = asin($$\omega$$t $$-$$kx + $$\phi$$2). If in the resultant wave the frequency and amplitude remain equal to amplitude of superimposing waves, the phase difference between them is
A
$${\pi \over 6}$$
B
$${{2\pi } \over 3}$$
C
$${\pi \over 4}$$
D
$${\pi \over 3}$$
2
AIPMT 2001
MCQ (Single Correct Answer)
+4
-1
The equation of a wave is represented by
y $$=$$ 10$$-$$4 sin(100t $$-$$ $${x \over {10}}$$) m. then the velocity of wave will be
A
100 m/s
B
4 m/s
C
1000 m/s
D
10 m/s
3
AIPMT 2001
MCQ (Single Correct Answer)
+4
-1
If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved then its fundamental frequency will become
A
$${\pi \over 4}$$
B
$$\sqrt 2 n$$
C
n
D
$${n \over {\sqrt 2 }}$$
4
AIPMT 2000
MCQ (Single Correct Answer)
+4
-1
The equations of two waves acting in perpendicular directions are given as
x = $$a$$cos($$\omega$$t +$$\delta$$) and y = $$a$$cos($$\omega$$t + $$\alpha$$), where $$\delta$$ = $$\alpha$$ + $${\pi \over 2}$$, the resultant wave represents
A
a parabola
B
a circle
C
an ellipse
D
a straight line
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A particle executing shm of amplitude 4 cm and T=4sec The time taken by it to move from positive extreme position to half the amplitude is
8 years ago
Rohith Gandhi
24 Points
Dear Pallavi,
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best Pallavi !!!Regards,Askiitians ExpertsRohith Gandhi P
8 years ago
Yati Raj
24 Points
At half distance from the extreme positions, K.E = PE1/2mv2=1/2kx2$1/2 m (A\omega cos\omega t + \phi )^{2} = 1/2 (\omega^{2}m)(A/2)^{^{2}} \Rightarrow cos^{2} \omega t = 1/4 \Rightarrow (cos 2\pi /T) t = cos\pi /3 \Rightarrow (2\pi /4) t = \pi /3 \Rightarrow t = (2/3) seconds.$ Please try to understand the answer. I felt it as the easy way and so I am sharing it with you. as the typing of equations took me time in my computer, I made it really short. If you are going to appear for the boards, you can include a lot more steps.
11 months ago
Think You Can Provide A Better Answer ?
## Other Related Questions on Mechanics
View all Questions »
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• View Details | 484 | 1,646 | {"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": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.85355 |
https://java.tutorialink.com/how-to-find-smallest-number-based-on-user-input-java/ | 1,721,678,492,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517915.15/warc/CC-MAIN-20240722190551-20240722220551-00475.warc.gz | 280,131,278 | 12,630 | # How to Find Smallest Number Based On User Input – Java
I’m trying to find the smallest number from the inputs (5 inputs) of users. However, every time I click run and write down the numbers, the smallest number will always be “0”. Here is my code:
```import java.util.Scanner;
public class Q2_9 {
public static void main(String[] args) {
// TODO Auto-generated method stub
int counter = 1;
double number;
double smallest = 0;
Scanner s = new Scanner(System.in);
while (counter < 6) {
System.out.println("Enter the number: ");
number = s.nextDouble();
if (smallest > number) {
smallest = number;
}
counter++;
}
System.out.println("The smallest number is " + smallest);
s.close();
}
}
```
Algorithm:
1. Input the first number and assume it to be the smallest number.
2. Input the rest of the numbers and in this process replace the smallest number if you the input is smaller than it.
Demo:
```import java.util.Scanner;
public class Main {
public static void main(String[] args) {
final int LIMIT = 5;
int counter = 1;
double number;
Scanner s = new Scanner(System.in);
// Input the first number
System.out.print("Enter the number: ");
number = s.nextDouble();
double smallest = number;
// Input the rest of the numbers
while (counter <= LIMIT - 1) {
System.out.print("Enter the number: ");
number = s.nextDouble();
if (smallest > number) {
smallest = number;
}
counter++;
}
System.out.println("The smallest number is " + smallest);
}
}
```
A sample run:
```Enter the number: -4040404
Enter the number: 808080
Enter the number: -8080808
Enter the number: 8989898
Enter the number: -8989898
The smallest number is -8989898.0
```
Note: Do not close a `Scanner(System.in)` as it also closes `System.in` and there is no way to open it again.
User contributions licensed under: CC BY-SA
3 People found this is helpful | 453 | 1,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.656192 |
https://harbourfronts.com/time-weighted-average-price/ | 1,709,075,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00719.warc.gz | 309,876,706 | 25,990 | Time Weighted Average Price
Almost every security trader will rather go for a trading strategy that minimizes risks and increases chances of achieving success. And, the time-weighted average price (TWAP) is one such trade execution strategy.
TWAP reduces the impact of large orders on the security market, making it a perfect choice for high-volume traders. Furthermore, calculating the time-weighted average price uses an easy-to-understand formula. And, in this article, we will help you understand the time-weighted average price better, including how to calculate it.
What is the Time Weighted Average Price?
Time-weighted average price (TWAP) tells you the average price of a security over a predetermined period. TWAP is a relevant tool for effective trading as it offers traders an alternative strategy to executing large orders. Using the TWAP, you can break up a large order into smaller chunks to minimize the impact of large orders on the market.
The ultimate goal is to improve the price of security via executing trades evenly over a specified period. For instance, if you want to purchase 10,000 of a company’s shares, applying the TWAP strategy, you could break the trade down, purchasing 1,000 shares every 15 mins.
The Time-weighted average price is not so different from the Volume -weighted average price (VWAP). While the VWAP considers volume in its calculation, TWAP does not. Normally, VWAP is an ideal order execution algorithm, but not when you foresee unfavorable market price momentum.
How to Calculate Time Weighted Average Price
Calculating TWAP does not demand any complex mathematical procedure, you only need to master the formula. You can calculate TWAP by taking the average of the “typical price” – Open, High, Low, and Close price of each bar, and then calculating their averages for ‘n’ number of periods.
The formula for calculating TWAP is:
TP=(Open+Low+Close+High)/4
TWAP=TP1+TP2+…+TPnn
In application, assuming you want TWAP value for 5 periods, then:
• Take the Average of Open, High, Low, and Close values of each 5 bars – a1,a2,a3,a4,a5
• Take an average of a1 to a5. TWAP= (a1+a2+a3………..+a5)/5
However, if you are faced with daily prices with several after-hours movements, you should use the open, high, low, and close when factoring in the typical price. In cases where the open and close are similar, such as intraday prices on liquid stocks, using the open, high, and low is advisable.
When you have gotten the TWAP, it is also important to detect whether the security is overvalued or undervalued. To do this you have to compare the order price. An order price that is below TWAP indicates that the security is undervalued, and if above the TWAP, it is an indicator that the security is overvalued.
Conclusion
The time-weighted average price plays a major role in trading especially when you are dealing with large orders. By breaking down your order into smaller chunks and executing them within a specific period you could reduce the impact of large orders and improve price.
Further questions
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http://math.stackexchange.com/questions/4814/what-are-the-names-of-graphs-with-2-in-outbound-edges | 1,469,350,901,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823989.0/warc/CC-MAIN-20160723071023-00216-ip-10-185-27-174.ec2.internal.warc.gz | 156,633,828 | 16,957 | # What are the names of graphs with 2 in/outbound edges?
I know that many types graphs have unique names, ie. a directed graph where each node has exactly one outbound edge is known as a functional graph. Do the graphs with exactly 2 inbound/outbound edges have a name?
Bonus: If the 2 inbound/outbound graph is named, is the nth inbound/outbound graph named?
-
This is known as a $d$-regular graph (at least if the graph is undirected). If your graph is a sub-graph of a bigger graph, then it is a $d$-factor. Finally, for $d=2$ it's also called a cycle cover, since it's a collection of cycles (if it's finite). For $d=1$ it's called a matching. For $d=3$ it is known as a cubic graph. | 182 | 691 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2016-30 | latest | en | 0.985419 |
http://www.jiskha.com/display.cgi?id=1363802001 | 1,495,710,277,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608058.57/warc/CC-MAIN-20170525102240-20170525122240-00285.warc.gz | 545,529,282 | 3,838 | # Pre-Algebra
posted by on .
Find the mean of the following data set:
25, 30, 20, 20, 25, 35, 35, 10
a. 15
b. 20
c. 25
d. 30
b. 20
• Pre-Algebra - ,
http://www.mathsisfun.com/mean.html
Did you add them up and divide by 8?
• Pre-Algebra - ,
YES SO THE ANSWER IS 30
• Pre-Algebra - ,
25 + 25 = 50
35 + 35 = 70
30 + 20 + 20 + 10 = 80
50 + 70 + 80 = ??
Divide that number by 8.
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https://www.aqua-calc.com/calculate/weight-to-volume/substance/cyanogen-blank-iodide | 1,716,126,104,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00045.warc.gz | 590,327,946 | 8,227 | # Volume of Cyanogen iodide
## cyanogen iodide: convert weight to volume
### Volume of 100 grams of Cyanogen iodide
centimeter³ 35.21 milliliter 35.21 foot³ 0 oil barrel 0 Imperial gallon 0.01 US cup 0.15 inch³ 2.15 US fluid ounce 1.19 liter 0.04 US gallon 0.01 meter³ 3.52 × 10-5 US pint 0.07 metric cup 0.14 US quart 0.04 metric tablespoon 2.35 US tablespoon 2.38 metric teaspoon 7.04 US teaspoon 7.14
### The entered weight of Cyanogen iodide in various units of weight
carat 500 ounce 3.53 gram 100 pound 0.22 kilogram 0.1 tonne 0 milligram 100 000
#### How many moles in 100 grams of Cyanogen iodide?
There are 653.93 millimoles in 100 grams of Cyanogen iodide
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Topic: Of Interest
Replies: 38 Last Post: Nov 5, 2012 7:48 PM
Messages: [ Previous | Next ]
Robert Hansen Posts: 11,345 From: Florida Registered: 6/22/09
Re: Of Interest
Posted: Oct 30, 2012 6:21 AM
att1.html (5.1 K)
On Oct 30, 2012, at 12:51 AM, Louis Talman <talmanl@gmail.com> wrote:
> 1. You seem to be suggesting that the Proceedings of the National Academy of Science is a third-rate journal.
Because it publishes research that can be refuted?
>
> 2. You seem to be suggesting that you can "fry" someone else's conclusions without showing that they aren't repeatable.
Results are repeatable (or not repeatable). Conclusions are checked for reasonableness.
I fried it on reasonableness.
If I wrote a lengthy paper with experimental results etc. and my conclusion at the end of all that is that the sky can't be blue, would that be reasonable? If anything, it would mean simply that I don't understand how the sky gets its color.
If every living creature has some measure of quantitive sense, yet only a very small fraction (a subset of only one species, humans) develop mathematical sense, what does that say for the correlation of quantitive sense and mathematical sense?
These researchers are attempting now to correlate the level of early quantitive sense to the level of later mathematical sense, supposedly to show that quantitive sense somehow acts as the foundation for later mathematical sense. I have two issues with this, the nature of the experiment and the nature of the conclusion.
If I was testing quantitive sense I would use piles of jelly beans or toothpicks. The researchers are using rather intricate images. They may be correlating a higher and more rational spatial sense with mathematical sense, not quantitive sense with mathematical sense. But I am not at odds with a correlation between quantitive and mathematical sense. It is the cause and effect part that concerns me. Why test kids quantitively when they are young and mathematical when they are older? To give credence to a cause and effect relationship between quantitive and mathematical senses. If we had tested the kids mathematically, both when they were young and older, it would spoil the plot. Our conclusion would be that kids that are mathematically advanced when they are older were mathematically advanced when they were younger. Big whoop.
If I were to present a mathematical proof with even 1/10th of the fallacy found in some of these "scientific" papers on education, you would rip me to shreds. Why do you hate it so when I challenge this stuff? I support the quest more than you think. But it is a quest at this point. A fledgling and tedious endeavor. They need to stop trying to win this thing with one race.
This is what I wrote on another thread...
"The researcher is allowed to suppose but those suppositions must be accompanied by caveats and counter explanations. When doubt exists a true scientist never puts the responsibility of that doubt on the reader, ever. The researcher doesn't conclude unless there is an irrefutable chain of evidence to back up those conclusions. Leaps of faith do not count (except in quantum physics). Education research is fledgling, but science is a tedious business and the truth comes grudgingly."
Bob Hansen
Date Subject Author
10/29/12 Louis Talman
10/29/12 Robert Hansen
10/30/12 Louis Talman
10/30/12 Robert Hansen
10/30/12 Paul A. Tanner III
10/30/12 Louis Talman
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10/31/12 Louis Talman
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11/5/12 Robert Hansen
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11/5/12 Gary Tupper
11/1/12 Wayne Bishop
11/3/12 Robert Hansen
11/3/12 Robert Hansen
11/3/12 Louis Talman
11/3/12 Robert Hansen
10/29/12 Robert Hansen | 1,102 | 4,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | longest | en | 0.944362 |
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Equations/23480: how do u solve a slope equation1 solutions Answer 12265 by rapaljer(4667) on 2006-01-04 15:40:49 (Show Source): You can put this solution on YOUR website!I'm not sure what you are asking here. If you are given the equation of a line in STANDARD FORM (Ax+By = C), then in order find the slope, you must solve for y in terms of x. Then the coefficient of x is the slope. R^2 at SCC
Exponential-and-logarithmic-functions/23477: HOw would you solve this problem and please show the steps? 1 solutions Answer 12264 by rapaljer(4667) on 2006-01-04 15:38:07 (Show Source): You can put this solution on YOUR website!First of all congratulations on writing the equation box for this particular problem!! Next, remove parentheses in the denominator. Remember that when you raise a power to a power you multiply exponents!! Next, you can divide out the in the numerator with in the denominator, leaving in the denominator (subtract exponents!!). You can divide out the in the numerator with the in the denominator, leaving a in the denominator. The 2 stays in the numerator, and the 5 stays in the denominator. The final result is: R^2 at SCC
Equations/22003: 3(2*)-3*+4=2*+5 x=1 solutions Answer 12238 by rapaljer(4667) on 2006-01-04 00:36:51 (Show Source): You can put this solution on YOUR website!I think you have an error in the statement of this problem. The symbol * means multiply, and you have to have TWO numbers to have a multiplication!! R^2 at SCC
Expressions-with-variables/23451: This is regarding Simplification involving Addition & Subtraction: Problem: (3f-f^2+fg)-(f-3f^2-2fg) My answer is: 2f+2f^2+3fg (Is this correct?) 1 solutions Answer 12237 by rapaljer(4667) on 2006-01-04 00:29:27 (Show Source): You can put this solution on YOUR website!Yes, I agree!!! R^2 at SCC
percentage/23445: find the percent of change, rounded to the nearest whole percent. Describe the change as an increase or decrease. \$22,000 to \$22,001.551 solutions Answer 12236 by rapaljer(4667) on 2006-01-04 00:06:21 (Show Source): You can put this solution on YOUR website!The amount of increase is 22,001.55-22,000 = 1.55 The percent of increase (always make a ratio of the amount of decrease to the original amount) = which is 0.0000775 -- This is VERY close to zero. The nearest whole percent is 0%. R^2 at SCC
Linear-equations/23444: how do you graph this linear equation 5x + 2y = 101 solutions Answer 12235 by rapaljer(4667) on 2006-01-03 23:37:34 (Show Source): You can put this solution on YOUR website!Because this is in standard form, it's probably easiest to find the two intercepts: 5x +2y = 10 If x = 0, then 2y=10, so y = 5 If y = 0, then 5x = 10, so x = 2. Graph these two points, and connect the points with a line. It should look like this: R^2 at SCC
percentage/23446: choose the equation you would use to answer the question: What percent of 175 is 35. a.175+35=n b.175n=35 c.35n=175 d. n=175/35 e.175(35)=n1 solutions Answer 12234 by rapaljer(4667) on 2006-01-03 23:22:34 (Show Source): You can put this solution on YOUR website!Let n = the unknown ("What percent" means n!) "of" usually means "times" "is" means "=" What percent of 175 is 35 This looks like the multiple choice b. 175n = 35 To eliminate the d. answer, you can divide both sides by 175, which gives you , which is NOT d! Final answer = b. 175n = 35 R^2 at SCC
Expressions-with-variables/23447: This is regarding Simplification involving Addition & Subtraction: The Problem is: -(4-5a)-(-2+3a) My answer I get is: -6-8a (Is this correct?) It isn't working for me. Thanks1 solutions Answer 12233 by rapaljer(4667) on 2006-01-03 23:15:53 (Show Source): You can put this solution on YOUR website!-(4-5a)-(-2+3a) -4 + 5a + 2 - 3a -2 + 2a R^2 at SCC
Inverses/23442: -2(2c-4)=1/3(-12c+24) I need your help with this question. Please.1 solutions Answer 12232 by rapaljer(4667) on 2006-01-03 23:13:53 (Show Source): You can put this solution on YOUR website! The first step for me is to clear the fraction, by multiplying both sides by the denominator which is 3. This is an IDENTITY, so the solution is all real values of c. R^2 at SCC
Expressions-with-variables/23443: solve this pair of equations by substitution 4x+y=-19 7x-y=81 solutions Answer 12229 by rapaljer(4667) on 2006-01-03 22:11:26 (Show Source): You can put this solution on YOUR website!To solve by substitution, you have to solve for one of the variables in terms of the other. Probably the first equation is the easiest to solve for y. 4x+y=-19 y = -4x - 19 Substitute this into the second equation for y: 7x-y=8 7x - (______) = 8 7x - (-4x - 19) = 8 7x + 4x + 19 = 8 11x + 19 = 8 11x + 19 - 19 = 8-19 11x = -11 x= -1 Now, substitute this value of x into the y = equation to solve for y: y = -4x - 19 y = -4(-1) - 19 y = 4 -19 y=-15 Check by substituting into the OTHER equation: 7x - y = 8 7(-1)-(-15) = 8 -7+15 = 8 8=8 It checks! R^2 at SCC
Inverses/23441: 4/3u-6=7/3u+8 Can You help me figure this problem out?1 solutions Answer 12227 by rapaljer(4667) on 2006-01-03 22:05:14 (Show Source): You can put this solution on YOUR website!This is not clear exactly what you mean. Do you mean: ? If so, then remember this one? means that Likewise: means that Multiply the parentheses: Get all the variables on the right side by subtracting 12u from each side: Add +42 to each side: Divide both sides by 9 That's a pretty ugly answer, but that's life sometimes!! It's a hard problem to check, and the check on one like this is probably harder than just solving it again. If anyone finds an error, please send me an Email so I can fix it!! R^2 at SCC
Geometry_Word_Problems/21471: A piece of cloth measures 2 yards, 1 foot, and 4 inches in length. What is its length in feet? 1 solutions Answer 12223 by rapaljer(4667) on 2006-01-03 21:45:36 (Show Source): You can put this solution on YOUR website!Change everything to feet. 2 yards = 6 feet, 1 foot = 1 foot, and 4 inches = 4/12 = 1/3 foot. The total is 6+ 1 + 1/3 = 7 1/3 feet or 22/7 feet, or 7.333. . . feet. R^2 at SCC
Sequences-and-series/23417: Find the first five (5) terms of the sequence (begin with n = 1). A=n!/n^2 I still do not understand how to do this one. 1 solutions Answer 12221 by rapaljer(4667) on 2006-01-03 21:38:57 (Show Source): You can put this solution on YOUR website!First, hopefully you know what n! means. For examples: 1! = 1 2! = 2*1 = 2 3!= 3*2*1=6 4!=4*3*2*1 = 24 5! = 5*4*3*2*1 = 120 etc. n! = the product of the integers beginning with n and counting down to 1. Now, Beginning with n=1: n=1, so n=2, so n=3, so n=4, so n=5, so R^2 at SCC
Probability-and-statistics/23421: what is the answer to 5c/8m-7m/c?1 solutions Answer 12219 by rapaljer(4667) on 2006-01-03 21:30:15 (Show Source): You can put this solution on YOUR website!What did you want to do, find a common denominator and subtract the fractions? The LCD = 8mc Multiply the first fraction by and the second fraction by . R^2 at SCC
Rational-functions/23428: (x/x+y)+(x/x-y)1 solutions Answer 12216 by rapaljer(4667) on 2006-01-03 21:06:15 (Show Source): You can put this solution on YOUR website! Because this is an addition problem, it calls for a common denominator, which is (x+y)(x-y). Now, multiply the first fraction by and the second fraction by It should look like this: + R^2 at SCC
real-numbers/23431: Please help me with this problem. Where do I begin Mary traveled 12mph to Joes house and returned by car at 36mph. How far from Mary does Joe live?1 solutions Answer 12213 by rapaljer(4667) on 2006-01-03 20:54:45 (Show Source): You can put this solution on YOUR website!I don't think there is enough information to solve this one. R^2 at SCC
Graphs/23438: when is there one answer? when are there two answers? and when is there no answer?1 solutions Answer 12212 by rapaljer(4667) on 2006-01-03 20:53:00 (Show Source): You can put this solution on YOUR website!I'm not sure what kind of problems you are talking about. If you have two straight line graphs, then the lines USUALLY intersect in ONE point. That is, USUALLY, except for TWO possibilities. If the lines are parallel, then the lines do NOT intersect, so there will be NO SOLUTION. The other possibility is that the lines could actually be the SAME LINE. In this case, the entire line is the solution, so there are infinitely many solutions. Two straight lines CANNOT intersect in TWO points. However, if you had a quadratic function (i.e., a parabola), then it could intersect a straight line in TWO points (or in ONE point, or no points which would be no solution). Is this what you had in mind? R^2 at SCC
Linear-equations/23436: how do you solve y-2=1/3(y+6)?1 solutions Answer 12210 by rapaljer(4667) on 2006-01-03 20:47:42 (Show Source): You can put this solution on YOUR website!I would multiply both sides of the equation by the denominator which is 3. That clears the fractions and makes life a little easier! Now, get all the y terms to the left side by adding -y to each side. Add +6 to both sides: Check: Substituet y = 6 back into the original equation: It checks!! R^2 at SCC
Graphs/23416: ok im doin checking for solutions and graphing inequallities. i kinda kno what im doin but when it comes 2 graphimg im lost plz help 4x+2y_<61 solutions Answer 12208 by rapaljer(4667) on 2006-01-03 20:39:32 (Show Source): You can put this solution on YOUR website!Graph the line . Since the equation is in standard form, I would find the two intercepts: . Let x = 0, then 2y = 6, so y = 3 Let y = 0, then 4x = 6, so y = 6/4 or 3/2 Graph the points and connect the dots with a line that should look like this: Now, because the intequality to graph is , since the inequality is a LESS THAN OR EQUAL TO you need to graph a SOLID LINE and SHADE BELOW THE LINE. I can't do this in algebra.com, so you will have to do this. Remember, a SOLID LINE and SHADE BELOW THE LINE. R^2 at SCC
Graphs/23394: When would you want to use the slope-intercept versus the point-slope form of a line?1 solutions Answer 12190 by rapaljer(4667) on 2006-01-03 15:20:04 (Show Source): You can put this solution on YOUR website!I would always use the slope-intercept method. For decades it seems I taught to find the equation of a line using the point-slope formula. I thought this was the #3 formula in all of mathematics, behind the quadratic formula and the Theorem of Pythagoras. Then someone showed me that the slope intercept formula y = mx+b was so much easier! So it probably depends upon who you ask!! R^2 at SCC
Quadratic_Equations/23385: I need to find the product. Help me please?? (2x + 1/3)(5x – 2/3) 1 solutions Answer 12175 by rapaljer(4667) on 2006-01-03 09:15:24 (Show Source): You can put this solution on YOUR website! I recommend that you FOIL it out. Do you know about FOIL? F= First times First O= Outer times Outer I = Inner times Inner L = Last times Last F = O = I = L= Total: Combine like terms: Total: R^2 at SCC
Rational-functions/23383: solve and graph the inequality. Can someone help me answer this? x+3/x less than or equal to -2 1 solutions Answer 12174 by rapaljer(4667) on 2006-01-03 09:00:29 (Show Source): You can put this solution on YOUR website!This one would be The first step is probably to set this inequality to zero by adding +2 to each side. WARNING: DO NOT MULTIPLY BOTH SIDES OF THE INEQUALITY BY A VARIABLE x!! As I said, do NOT multiply both sides of the inequality by x, but you CAN find a common denominator of x, and express the left side as a single fraction! That is allowed, and it will help solve the problem. Now draw the graph. The numerator does not factor, and probably does NOT have any real roots. The denominator will equal zero at x=0, so this is an ASYMPTOTE. With this done, now draw the graph, either with a graphing calculator or with the calculator in algebra.com. For this, you can use the original form of the inequality after you set it equal to zero, : In this problem since it is a LESS THAN OR EQUAL TO problem, you will be looking for all values ON OR BELOW the x-axis. This will be the values of x to the left of the y-axis. Because the problem says "less than or EQUAL to", you would normally be INCLUDING THE ENDPOINTS. However, you NEVER include DENOMINATOR endpoints, since division by zero is NEVER allowed. So the final answer is x<0. Interval notation is (-inf, 0). R^2 at SCC
Quadratic_Equations/23384: I am to find the product. Can someone help?? 4a^3 (3ab^3 – 2ab^3) 1 solutions Answer 12173 by rapaljer(4667) on 2006-01-03 08:45:33 (Show Source): You can put this solution on YOUR website! This is a special problem in that you can just combine like terms in the parentheses: Remember that R^2 at SCC
Rational-functions/23382: Solve the rational inequality. State and graph the solution set. Can someone help? x/x+2>-1 1 solutions Answer 12172 by rapaljer(4667) on 2006-01-03 08:01:34 (Show Source): You can put this solution on YOUR website!I'm assuming that you mean . If so, then I would begin by setting the inequality to zero. DO NOT MULTIPLY BOTH SIDES OF THE INEQUALITY BY (X+2)!! . Next, find a common denominator, which is x+2: Now, you have two choices for method of solving the inequality. You can use some algebra explanations to solve it, or you can use a graphing calculator (or the algebra.com calculator!!). I will choose the latter method. You need to graph . Before drawing the graph, notice that there is one place that the numerator equals zero, and that is at x= -1. There is one value of x that would make the denominator zero, which is NOT allowed, and that would be x= -2. This tells me that there is a ROOT at x = -1 and an ASYMPTOTE at x= -2. Now, draw the graph with this in mind: Now, since the problem is to solve , which is solving a problem that is "GREATER THAN", you need to find all values of x for which the graph is ABOVE the x-axis. And do NOT include the endpoints. Notice that the graph is above the x-axis from -infinity to -2, and also from -1 to infinity. This means that the solution is x<-2 or x>1. In interval notation this will be:(-inf, -2) U (-1, inf). R^2 at SCC
Inequalities/23373: 6x-(3-2x)>3x+2 how do you do this??????1 solutions Answer 12161 by rapaljer(4667) on 2006-01-02 23:43:22 (Show Source): You can put this solution on YOUR website!6x-(3-2x)>3x+2 Distribute the -: 6x - 3 + 2x > 3x +2 8x - 3 > 3x +2 Subtract 3x from each side: 8x - 3 -3x > 3x +2 -3x 5x -3 > 2 Add +3 to each side: 5x-3 + 3 > 2+ 3 5x > 5 Divide both sides by 5: x>1 Graph on a numberline by going 1 unit to the right, and shade everything to the right of 1, with an arrow on the right end of the numberline, since x>1 means x is any number LARGER than 1, and since it goes forever in the right direction. To indicate that the endpoint at x=1 is NOT included, make an open circle at the point x=1. In higher math, we use an open parentheses on the point x=1, opening to the right. R^2 at SCC
Coordinate-system/23374: please solve y=4*2+4. i have to write an equation is slope intercept form and the two points on my line are (2,4) and (3,8). the slope is 4. please write the equation in slope intercept and slove it! i need this like asap!!!! thank you ssoo much!!1 solutions Answer 12160 by rapaljer(4667) on 2006-01-02 23:35:56 (Show Source): You can put this solution on YOUR website!If you know that the slope is 4 , and it passes through the point (2,4), then your first step is to use y=mx+b to find b which is the y-intercept. Substitute everything you know (m=4, x= 2 and y = 4) into that formula: y=mx+b 4 = 4*2 + b 4=8 +b b=-4 So graph the line by starting at the y-intercept going DOWN 4 units on the y-axis. Put a point there. Then with your pencil on that point, rise UP 4, and go RIGHT 1, and put another point. Connect the points, and that should give you a line that goes through the points you mentioned in the problem. The equation of the line is , and the graph should look like this: R^2 at SCC
Numbers_Word_Problems/23371: Okay , I am in 8th grade , and i am in Algebra 1 Honors , and for homework we have 13 Krypto problems , if some of you dont know what that is , its where u use 5 numbers to get to a certain number. So one of my problems is 5 , 2 , 7 , 11 , 13 = 17 so you have to use those numbers to get to 17 . Please Help 1 solutions Answer 12158 by rapaljer(4667) on 2006-01-02 21:48:52 (Show Source): You can put this solution on YOUR website!How about this: 5 + 2 * 7 + 11 -13 = 17 I've never heard of Krypto numbers. Can you change the order of the numbers? Do you have to use them all exactly once?? Interesting! R^2 at SCC
Inequalities/23356: whats the answer please? 2y-4_>121 solutions Answer 12147 by rapaljer(4667) on 2006-01-02 21:04:12 (Show Source): You can put this solution on YOUR website! Add +4 to each side: Divide by 2: Locate 8 on the right side of the numberline, and the solution is all values to the right of the 8 and INCLUDING the 8. You can indicate this either by drawing a SOLID circle on 8 and shade to the right of 8 with an arrow on the end of the line to indicate that it continues forever. Also, instead of the solid circle, in higher math, you would place a BRACKET [ on the 8, and opening to the right, and shade to the right. R^2 at SCC
real-numbers/23365: please help me solve this problem the area of a rectangle is 22cm2 the width is w the length is 3w+5 I got as far as: w(3w+5)=22cm2 can you help me solve this?1 solutions Answer 12146 by rapaljer(4667) on 2006-01-02 20:59:43 (Show Source): You can put this solution on YOUR website!w(3w+5)=22 This is going to be quadratic, so multiply out the parentheses, and set it equal to zero: 3w^2 + 5w -22 = 0 (3x ______)(w ______)=0 Factoring this will probably be the hard part of the problem. Here we go: Two numbers whose product is 22 would probably be 2 * 11. Try the 11 in the first slot and the 2 in the last slot, and the middle term should come out to -6w +11w= 5w: (3w+11)(w-2) = 0 Two solutions: 3w+11= 0 3w=-11 w=-11/3 Reject! The width cannot be negative! w-2=0 w=2 cm. This one works! Now, find the length: L = 3w+5 L =3*2+5 = 6+5 =11 cm. Check: Area = LW = 11*2 = 22 cm^2. R^2 at SCC
Graphs/23366: graph the inequality: y<-x-4 this is very difficult please help.1 solutions Answer 12144 by rapaljer(4667) on 2006-01-02 20:51:50 (Show Source): You can put this solution on YOUR website!Before you can graph , first graph y = -x-4, which is a straight line with y intercept -4 and slope -1. This means to plot the y-intercept by going DOWN 4 units on the y axis. The the slope of -1 means . With your pencil on the y-intercept at -4, go DOWN 1 and RIGHT 1 unit and put another point. Graph this line as a DOTTED line because the inequality with the symbol "<" means DO NOT INCLUDE THE LINE. Also shade BELOW the line. In algebra.com I can only draw the line for you. You will need to make it a DOTTED LINE, and you will have to shade BELOW the line. Dont' forget: DOTTED LINE, SHADE BELOW THE LINE!! R^2 at SCC
Numeric_Fractions/23358: ```Hi, can you help me solve 3x-2 2 ————— - ————— x+2 x-2```1 solutions Answer 12141 by rapaljer(4667) on 2006-01-02 20:43:54 (Show Source): You can put this solution on YOUR website! Because this is a subtraction problem, you must first find a common denominator, and in this case the LCD = (x+2)(x-2) You'll have to multiply the first fraction by and the second fraction by . It should look like this: - Place this all over the common denominator and multiply out the binomials in the numerators. HOWEVER, be CAREFUL to keep the negative between the numerators. This can be factored, but it does NOT reduce the fraction, so factoring is optional: R^2 at SCC | 6,989 | 21,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2013-20 | latest | en | 0.206385 |
https://www.esaral.com/q/if-pt-is-a-tangent-to-a-circle-with-centre-o-and-pq-is-a-chord-of-the-circle-such-that-qpt-70-37463 | 1,726,481,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00307.warc.gz | 678,730,400 | 11,814 | # If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70
Question:
If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70 then find the measure of ∠POQ
Solution:
We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90
Now, ∠OPQ = ∠OPT − ∠TPQ = 90 − 70 = 20
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 20 (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180 (Angle sum property of a triangle)
⇒ ∠POQ = 180 − 20 − 20 = 140 | 220 | 605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.879031 |
https://nl.mathworks.com/help/matlab/ref/triangulation.incenter.html | 1,726,647,131,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00630.warc.gz | 380,422,783 | 18,289 | # incenter
Incenter of triangulation elements
## Syntax
``C = incenter(TR)``
``C = incenter(TR,ID)``
``````[C,r] = incenter(___)``````
## Description
````C = incenter(TR)` returns the coordinates of the incenters of each triangle or tetrahedron in the triangulation `TR`.```
example
````C = incenter(TR,ID)` returns the coordinates of the incenter of each triangle or tetrahedron specified by `ID`. The identification numbers of the triangles or tetrahedra in `TR` are the corresponding row numbers of the property `TR.ConnectivityList`.```
example
``````[C,r] = incenter(___)``` also returns the radii of the inscribed circles or spheres.```
example
## Examples
collapse all
Create a 2-D Delaunay triangulation.
```x = [0 1 1 0 0.5]'; y = [0 0 1 1 0.5]'; TR = delaunayTriangulation(x,y);```
Compute the incenters of the triangles.
`C = incenter(TR);`
Plot the triangles and incenters.
```triplot(TR) axis equal axis([-0.2 1.2 -0.2 1.2]) hold on plot(C(:,1),C(:,2),'*r') ```
`load tetmesh`
Calculate the incenter coordinates of the first five tetrahedra in the triangulation, in addition to the radii of their inscribed spheres.
```TR = triangulation(tet,X); [C,r] = incenter(TR,[1:5]')```
```C = 5×3 -6.1083 -31.0234 8.1439 -2.1439 -31.0283 5.8742 -1.9555 -31.9463 7.4112 -4.3019 -30.8460 10.5169 -3.1596 -29.3642 6.1851 ```
```r = 5×1 0.7528 0.9125 0.8430 0.6997 0.7558 ```
## Input Arguments
collapse all
Triangulation representation, specified as a scalar `triangulation` or `delaunayTriangulation` object.
Data Types: `triangulation` | `delaunayTriangulation`
Triangle or tetrahedron IDs, specified as a scalar or a column vector whose elements each correspond to a single triangle or tetrahedron in the triangulation object. The identification number of each triangle or tetrahedron is the corresponding row number of the `ConnectivityList` property.
Data Types: `double`
## Output Arguments
collapse all
Incenters, returned as a matrix whose rows contain the coordinates of an incenter.
Data Types: `double`
Radii of the inscribed circles or spheres, returned as a vector.
Data Types: `double`
## Version History
Introduced in R2013a | 660 | 2,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.675278 |
https://sandialabs.github.io/Prove-It/packages/proveit/numbers/summation/__pv_it/axioms/811b651e86314814e3ce87463a2b65b55b6a76d90/expr.html | 1,713,220,222,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00209.warc.gz | 455,707,932 | 3,828 | # from the theory of proveit.numbers.summation¶
In [1]:
import proveit
# Automation is not needed when building an expression:
proveit.defaults.automation = False # This will speed things up.
proveit.defaults.inline_pngs = False # Makes files smaller.
# import Expression classes needed to build the expression
from proveit import b, c
from proveit.numbers import LessEq
In [2]:
# build up the expression from sub-expressions
expr = LessEq(b, c)
expr:
In [3]:
# check that the built expression is the same as the stored expression
assert expr == stored_expr
assert expr._style_id == stored_expr._style_id
print("Passed sanity check: expr matches stored_expr")
Passed sanity check: expr matches stored_expr
In [4]:
# Show the LaTeX representation of the expression for convenience if you need it.
print(stored_expr.latex())
b \leq c
In [5]:
stored_expr.style_options()
namedescriptiondefaultcurrent valuerelated methods
operation'infix' or 'function' style formattinginfixinfix
wrap_positionsposition(s) at which wrapping is to occur; '2 n - 1' is after the nth operand, '2 n' is after the nth operation.()()('with_wrapping_at', 'with_wrap_before_operator', 'with_wrap_after_operator', 'without_wrapping', 'wrap_positions')
justificationif any wrap positions are set, justify to the 'left', 'center', or 'right'centercenter('with_justification',)
directionDirection of the relation (normal or reversed)normalnormal('with_direction_reversed', 'is_reversed')
In [6]:
# display the expression information
stored_expr.expr_info()
core typesub-expressionsexpression
0Operationoperator: 1
operands: 2
1Literal
2ExprTuple3, 4
3Variable
4Variable | 406 | 1,646 | {"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} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.494292 |
https://www.experts-exchange.com/questions/21141694/Building-Array-Compile-error.html | 1,524,689,743,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947957.81/warc/CC-MAIN-20180425193720-20180425213720-00027.warc.gz | 765,150,025 | 22,283 | Solved
# Building Array....Compile error.
Posted on 2004-09-22
Medium Priority
224 Views
--------------------------------------------------------------------------------------------------
Code begin
--------------------------------------------------------------------------------------------------
/*
instance variables
-----------------------------------------
private double[][] M array representing the Matrix
private final int MAX_SIZE constant representing the maximum size of a Matrix;
it should be set to 100
-----------------------------------------
The Code:
public methods
-----------------------------------------
Matrix(int sizeN) A constructor that creates a Matrix of
size N x N filled with zeros
N must be in the range [0, MAX_SIZE].
If N is negative, create a Matrix of size 0 x 0.
If N is larger than MAX_SIZE, create a Matrix
of size MAX_SIZE x MAX_SIZE.
Matrix(int sizeN, double[][] initValues) A constructor that creates a Matrix of size N x N
and fills it with the values found in the array initValues.
N must be in the range [0, MAX_SIZE]. If N is negative,
create a Matrix of size 0 x 0. If N is larger than MAX_SIZE,
create a Matrix of size MAX_SIZE x MAX_SIZE.
-----------------------------------------
*/
public class Matrix
{
//initialize array M
private double[][] M;
//variable to set max size of array
private final int MAX_SIZE = 100;
//Constructor One
//Create a new Matrix, initializing it to zeros
//The size must be a value in the range [0, MAX_SIZE]
public Matrix(int sizeN)
{
if (sizeN > MAX_SIZE)
{
System.out.println("The maximum size of a matrix is " +
MAX_SIZE);
sizeN = MAX_SIZE;
}
else if (sizeN < 0)
{
System.out.println("The size of a matrix cannot be negative");
sizeN = 0;
}
//build array M with sizeN x sizeN initalized to 0
M = new double[sizeN][sizeN];
for (int i = 0; i < sizeN; i++)
for (int j = 0; j < sizeN; j++)
M[i][j] = 0.0;
}
//Constructor Two
//Create a new Matrix, initializing it with the above parameters
//The dimensions must be in the range [1, MAX_SIZE]
public Matrix(int sizeN, double[][] initValues)
{
initValues = new double[sizeN][sizeN];
double fillArray;
for (int i = 1; i < MAX_SIZE; i++)
fillArray = 0;
for (int j = 1; j < MAX_SIZE; j++)
fillArray += initValues [i][j];
}
--------------------------------------------------------------------------------------------------
code end
--------------------------------------------------------------------------------------------------
Problem: (Compile error)
--------------------------------------------------------------------------------------------------
cannot resolve symbol
symbol : variable i
location: class Matrix
fillArray += initValues [ i ] [ j ];
^
What's the problem? Up to "//Constructor Two" the code compiles without error.
As you probably know I'm writing a "matrix" class.
Last do you see any flaws with the rest of the build either in code itself or my approach?
Thanks.
0
Question by:smkJackstraw
• 5
LVL 3
Accepted Solution
CI-Ia0s earned 400 total points
ID: 12129653
Try putting brackets around the body of that last for loop...
0
Author Comment
ID: 12129705
You know....syntax errors will be the death of me. It worked! Before I award the points, would you answer this last question:
Why didn't I get the syntax error with the previous loop: //build array M with sizeN x sizeN initalized to 0
There nearly identical. Which is more 'proper' syntax? (with reguards to Java AND the idea that I'm learning Java as a stepping stone to C++)
0
LVL 3
Expert Comment
ID: 12129771
Unfortunately, the Java compiler often doesn't catch errors with brackets and loops/if-statements (I've had similar issues when I've forgotten brackets, though with an if-statement). I always use brackets and I have never seen it done without them. The reasons for this are:
1. No chance of having the compiler miss it and give you some other error/get a run time error.
2. It's easier to follow. If you can see where every loop begins and ends with two clearly defined markers it makes it easier to skim the code (an important skill when you start writing long code). It's also easier to follow for others who may read your code. :)
0
LVL 92
Expert Comment
ID: 12129779
> Why didn't I get the syntax error with the previous loop: //build array M with sizeN x sizeN initalized to 0
Because there was only a single statement in the loop, so brackets were optional.
0
LVL 3
Expert Comment
ID: 12129780
P.S. I'd add brackets to the previous "//build array M with sizeN x sizeN initalized to 0" to make sure there's no confusion. Even if it runs correctly now, you may end up doing some editing to find it not working later.
0
LVL 3
Expert Comment
ID: 12129788
>>Because there was only a single statement in the loop, so brackets were optional.
i.e. The compiler sees this:
for (int i = 1; i < MAX_SIZE; i++){
fillArray = 0;
}
for (int j = 1; j < MAX_SIZE; j++){
fillArray += initValues [i][j];
}
0
LVL 3
Expert Comment
ID: 12129824
:)
0
## Featured Post
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Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.
Have a better answer? Share it in a comment. | 1,273 | 5,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-17 | latest | en | 0.510938 |
https://en.wikipedia.org/wiki/Quasi-separated | 1,495,798,547,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608659.43/warc/CC-MAIN-20170526105726-20170526125726-00223.warc.gz | 909,957,486 | 9,759 | # Quasi-separated morphism
(Redirected from Quasi-separated)
In algebraic geometry, a morphism of schemes f from X to Y is called quasi-separated if the diagonal map from X to X×YX is quasi-compact (meaning that the inverse image of any quasi-compact open set is quasi compact). A scheme X is called quasi-separated if the morphism to Spec Z is quasi-separated. Quasi-separated algebraic spaces and algebraic stacks and morphisms between them are defined in a similar way, though some authors include the condition that X is quasi-separated as part of the definition of an algebraic space or algebraic stack X. Quasi-separated morphisms were introduced by Grothendieck (1964, 1.2.1) as a generalization of separated morphisms.
All separated morphisms (and all morphisms of Noetherian schemes) are automatically quasi-separated. Quasi-separated morphisms are important for algebraic spaces and algebraic stacks, where many natural morphisms are quasi-separated but not separated.
The condition that a morphism is quasi-separated often occurs together with the condition that it is quasi-compact.
## Examples
• If X is a locally Noetherian scheme then any morphism from X to any scheme is quasi-separated, and in particular X is a quasi-separated scheme.
• Any separated scheme or morphism is quasi-separated.
• The line with two origins over a field is quasi-separated over the field but not separated.
• If X is an "infinite dimensional vector space with two origins" over a field K then the morphism from X to spec K is not quasi-separated. More precisely X consists of two copies of Spec K[x1,x2,....] glued together by identifying the nonzero points in each copy.
• The quotient of an algebraic space by an infinite discrete group acting freely is often not quasi-separated. For example if K is a field of characteristic 0 then the quotient of the affine line by the group Z of integers is an algebraic space that is not quasi-separated. This algebraic space is also an example of a group object in the category of algebraic spaces that is not a scheme; quasi-separated algebraic spaces that are group objects are always group schemes. There are similar examples given by taking the quotient of the group scheme Gm by an infinite subgroup, or the quotient of the complex numbers by a lattice. | 490 | 2,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.939249 |
https://www.mo4tech.com/%E3%80%90-%E5%85%B8-%E5%9E%8B-%E8%8C%83-%E4%BE%8B-%E3%80%91-a-string-representing-a-numeric-value.html | 1,670,057,266,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00167.warc.gz | 957,480,951 | 5,550 | # 【 典 型 范 例 】 A string representing a numeric value
Posted on Dec. 2, 2022, 5:52 p.m. by Inaaya Barman
Category: The back-end
This is the second day of my participation in the More text Challenge. For more details, see more text Challenge
## The title
Implement a function that determines whether a string represents a numeric value (both integer and decimal).
The values (in order) can be divided into the following sections:
1. A number of Spaces
2. A decimal or an integer
3. (Optional) an 'e' or 'e' followed by an integer
4. A number of Spaces
Decimals (in order) can be divided into the following parts:
1. (Optional) a symbol character ('+' or '-')
2. One of the following formats:
• At least one digit followed by a dot '.'
• At least one digit followed by a dot '.' followed by at least one digit
• A dot '.' followed by at least one digit
Integers (in order) can be divided into the following parts:
1. (Optional) a symbol character ('+' or '-')
2. At least one digit
Some values are listed as follows:
[" + 100 ", "5 e2", "123", "3.1416", "1 e - 16", "0123"] part non-numerical listed below:
["12e", "1a3.14", "1.2.3", "+-5", "12e+5.4"]
Source: LeetCode
## Train of thought
There are a variety of solutions to this problem: "regular", "DFA", "simulation", here with a more easy to understand the string simulation method
1. This problem is almost exactly the same as that in Leetcode 65, except for the addition of a numeric part (several Spaces), so initially you need to call trim() to remove the whitespace before and after
The string is divided into two parts: e(e),e(e),e(e),e(e),e(e),e(e),e(e),e(e),e(e),e(e),e(e),e(e),e(e
3. Implement an isChecked method to verify that the string is valid
• + or - can only be in the first position
• .only once
• Numbers appear at least once
`````` public boolean isNumber(String s) {
s = s.trim();// Remove the Spaces before and after
char[] cs = s.toCharArray();
int n = cs.length;
int idx = -1;
//1. Find the first e or the position of e
for (int i = 0; i cs.length; i++) {
if (cs[i] == 'e' || cs[i] == 'E') {
if(idx ! = -1) {
return false; // Note that e(e) appears twice
} else{ idx = i; }}}boolean ans = true;
// In this e or e as the center as the partition
if (idx == -1) { // If idx=-1, there is no E and E. E (E) must be followed by an integer. It can be preceded by either a decimal or an integer
ans = isChecked(cs, 0, n - 1.false);
} else {
ans = isChecked(cs, 0, idx - 1.false);
ans = isChecked(cs, idx + 1, n - 1.true);
}
return ans;
}
public boolean isChecked(char[] cs, int start, int end, boolean mustInt) {
if (start end) return false;
boolean point = false;
boolean hasNum = false;
//1. If the first position is +- sign
if (cs[start] == '+' || cs[start] == The '-') start++;
for (int i = start; i = end; i++) {
if (cs[i] = '0' cs[i] = '9') {
hasNum = true;
} else if (cs[i] == '. ' !point) {
if (mustInt) return false; // No decimal point if it must be an integer
point = true;
} else {
return false; }}return hasNum;
}
Copy the code``````
O(N)
## Extra space complexity
O(N) uses cs array storage, of course if you use the charAt method you can reduce the extra space complexity to O(1)
Search | 934 | 3,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.798605 |
https://encyclopedia2.thefreedictionary.com/Local+uniform+convergence | 1,638,168,259,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358688.35/warc/CC-MAIN-20211129044311-20211129074311-00357.warc.gz | 290,836,348 | 11,518 | # Uniform Convergence
(redirected from Local uniform convergence)
## uniform convergence
[′yü·nə‚fȯrm kən′vər·jəns]
(mathematics)
A sequence of functions {ƒn (x)} converges uniformly on E to ƒ(x) if given ε > 0 there is an N such that |ƒn (x) - ƒ(x)| < ε="" for="" all="">x in E provided n > N.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
## Uniform Convergence
an important special case of convergence. A sequence of functions fn (x) (n = 1, 2, 3,…) is said to converge uniformly on a given set to the limit function f(x) if, for every ∊ > 0, there exists a number N = N(∊) such that, when η > N, ǀ f(x) – fn (x)ǀ < ∊ for all points x in the set.
For example, the sequence of functions fn(x) = xn converges uniformly on the closed interval [0, 1 /2] to the limit f(x) = 0. To show that this is true, let us take n > In (1/∊)/In 2. It then follows that ǀf(x)–fn (x)ǀ ≤ (1/2)n < ∊ for all x, 0 ≤ x ≤ 1/2. This sequence of functions, however, does not converge uniformly on the interval [0, 1]. Here, the limit function is f(x) = 0 for 0 ≤ x< 1 and f(1) = 1. The reason for the failure to converge uniformly is that for arbitarily large η there exist points η that satisfy the inequalities and for which ǀf (η) - fn(η)ǀ = ηn > 1/2.
The notion of uniform convergence admits of a simple geometric interpretation. The uniform convergence of a sequence of functions fn(x) on some closed interval to the function f(x) means that, for any ∊ > 0, all curves y = fn (x) with large enough n will be located within a strip that is 2∊ in width and is bounded by the curves y = f(x) ± ∊ for any x in the interval (see Figure 1).
Figure 1
Uniformly converging sequences of functions have a number of important properties. For example, the limit of a uniformly converging sequence of continuous functions is also continuous. On the other hand, the example given above shows that the limit of a sequence of functions that does not converge uniformly may be discontinuous. An important role is played in mathematical analysis by Weierstrass’ theorem, which states that every function continuous on a closed interval can be represented as the limit of a uniformly converging sequence of polynomials. (SeeAPPROXIMATION AND INTERPOLATION OF FUNCTIONS.) | 671 | 2,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-49 | latest | en | 0.891705 |
http://en.academic.ru/dic.nsf/cide/134032/Plane | 1,495,736,182,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608120.92/warc/CC-MAIN-20170525180025-20170525200025-00037.warc.gz | 116,333,129 | 13,050 | Plane table
Plane Plane, a. [L. planus: cf. F. plan. See {Plan}, a.] Without elevations or depressions; even; level; flat; lying in, or constituting, a plane; as, a plane surface. [1913 Webster]
Note: In science, this word (instead of plain) is almost exclusively used to designate a flat or level surface. [1913 Webster]
{Plane angle}, the angle included between two straight lines in a plane.
{Plane chart}, {Plane curve}. See under {Chart} and {Curve}.
{Plane figure}, a figure all points of which lie in the same plane. If bounded by straight lines it is a rectilinear plane figure, if by curved lines it is a curvilinear plane figure.
{Plane geometry}, that part of geometry which treats of the relations and properties of plane figures.
{Plane problem}, a problem which can be solved geometrically by the aid of the right line and circle only.
{Plane sailing} (Naut.), the method of computing a ship's place and course on the supposition that the earth's surface is a plane.
{Plane scale} (Naut.), a scale for the use of navigators, on which are graduated chords, sines, tangents, secants, rhumbs, geographical miles, etc.
{Plane surveying}, surveying in which the curvature of the earth is disregarded; ordinary field and topographical surveying of tracts of moderate extent.
{Plane table}, an instrument used for plotting the lines of a survey on paper in the field.
{Plane trigonometry}, the branch of trigonometry in which its principles are applied to plane triangles. [1913 Webster]
The Collaborative International Dictionary of English. 2000.
### Look at other dictionaries:
• Plane table — Plane ta ble See under {Plane}, a. [1913 Webster] … The Collaborative International Dictionary of English
• plane table — n. a surveying device for plotting maps in the field: it consists of a drawing board mounted on a tripod with an alidade that can be moved about over the table to establish a reference point … English World dictionary
• Plane table — A plane table is a device used in surveying and related disciplines to provide a solid and level surface on which to make field drawings, charts and maps.HistoryThe first known instrument for directly producing a drawing of a site was developed… … Wikipedia
• plane-table — ˈ ̷ ̷ˌ ̷ ̷ ̷ ̷ verb Etymology: plane table intransitive verb : to make use of a plane table transitive verb : to plot with a plane table … Useful english dictionary
• plane-table — [c]/ˈpleɪn teɪbəl/ (say playn taybuhl) verb (plane tabled, plane tabling) –verb (t) 1. to survey (a tract of land) with a plane table. –verb (i) 2. to carry out surveying using a plane table … Australian English dictionary
• plane-table — /playn tay beuhl/, v.t., v.i., plane tabled, plane tabling. to survey with a plane table. [1870 75] * * * … Universalium
• plane table — plane′ ta ble n. sur a drawing board mounted on a tripod, used in the field, with an alidade, for surveying tracts of land • Etymology: 1600–10 … From formal English to slang
• plane table — drafting table, table used for drawing blueprints … English contemporary dictionary
• plane table — Survey. a drawing board mounted on a tripod, used in the field, with an alidade, for surveying tracts of land. Also, plain table. [1600 10] * * * … Universalium
• plane table — noun Date: 1607 an instrument consisting essentially of a drawing board on a tripod with a ruler pointed at the object observed and used for plotting the lines of a survey directly from observation … New Collegiate Dictionary | 871 | 3,527 | {"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} | 3.265625 | 3 | CC-MAIN-2017-22 | longest | en | 0.869679 |
http://stackoverflow.com/questions/tagged/sum?sort=unanswered | 1,394,917,521,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678699570/warc/CC-MAIN-20140313024459-00091-ip-10-183-142-35.ec2.internal.warc.gz | 144,576,241 | 26,804 | # Tagged Questions
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https://www.definitions.net/definition/chaotic%20attractor | 1,679,606,454,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945183.40/warc/CC-MAIN-20230323194025-20230323224025-00615.warc.gz | 835,608,567 | 17,391 | # Definitions for chaotic attractorchaot·ic attrac·tor
### Princeton's WordNetRate this definition:0.0 / 0 votes
1. strange attractor, chaotic attractornoun
an attractor for which the approach to its final point in phase space is chaotic
### WikipediaRate this definition:0.0 / 0 votes
1. chaotic attractor
In the mathematical field of dynamical systems, an attractor is a set of states toward which a system tends to evolve, for a wide variety of starting conditions of the system. System values that get close enough to the attractor values remain close even if slightly disturbed. In finite-dimensional systems, the evolving variable may be represented algebraically as an n-dimensional vector. The attractor is a region in n-dimensional space. In physical systems, the n dimensions may be, for example, two or three positional coordinates for each of one or more physical entities; in economic systems, they may be separate variables such as the inflation rate and the unemployment rate.If the evolving variable is two- or three-dimensional, the attractor of the dynamic process can be represented geometrically in two or three dimensions, (as for example in the three-dimensional case depicted to the right). An attractor can be a point, a finite set of points, a curve, a manifold, or even a complicated set with a fractal structure known as a strange attractor (see strange attractor below). If the variable is a scalar, the attractor is a subset of the real number line. Describing the attractors of chaotic dynamical systems has been one of the achievements of chaos theory. A trajectory of the dynamical system in the attractor does not have to satisfy any special constraints except for remaining on the attractor, forward in time. The trajectory may be periodic or chaotic. If a set of points is periodic or chaotic, but the flow in the neighborhood is away from the set, the set is not an attractor, but instead is called a repeller (or repellor).
### Numerology
1. Chaldean Numerology
The numerical value of chaotic attractor in Chaldean Numerology is: 7
2. Pythagorean Numerology
The numerical value of chaotic attractor in Pythagorean Numerology is: 4
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##### a person who fells trees
• A. lumberman
• B. muddle
• C. recital
• D. nitrile | 996 | 3,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.906386 |
http://www.xpmath.com/forums/showthread.php?p=35618 | 1,540,158,345,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514355.90/warc/CC-MAIN-20181021203102-20181021224602-00009.warc.gz | 583,721,332 | 12,271 | 7th grade quiz part 1 - XP Math - Forums
XP Math - Forums 7th grade quiz part 1
01-08-2012 #1
Dragonballful23
Points: 12,950, Level: 74
Activity: 33.3%
Last Achievements
Join Date: Dec 2011
Posts: 386
Q=question
Q.1)
Evaluate for x = 11: x + 4
A.
13.
B.
14.
C.
15.
D.
16.
Q.2) Evaluate for n = 0: 2 (4 + n) - 5
A.
0.
B.
1.
C.
2.
D.
3.
Q.3) Evaluate for x = 8 and y = 10: 3x + 2y
A.
40.
B.
41.
C.
42.
D.
43.
Q.4) Write an algebraic expression for: 77 more than the product of 2 and u
A.
77 more than 2u.
B.
2u - 77.
C.
77 + 2u.
D.
2u = 77.
Q.5) What is the absolute value of -10 minus the absolute value of 4 plus 1.
A.
15.
B.
-15.
C.
5.
D.
-5.
Q.6) Which is bigger? -14 _____ 0
A.
-14 < 0
B.
-14 > 0
C.
-14 = 0
Q.7) What is: 6 + (-4)
A.
2.
B.
10.
C.
-2.
D.
-10.
Q.8) Evaluate for m = -5: m + 7
A.
-2.
B.
12.
C.
2.
D.
-12.
Q.9) What is: 8 - (-4)
A.
2.
B.
6.
C.
-12.
D.
12.
Q.10)
Evaluate for m = -8: 9 - m
A.
-17.
B.
17.
C.
1.
D.
-1.
__________________
Finally finished all my tests!!!
I am ekko main now ^_^
(ign: supersaiyan2363)
It's not how much time you have, it's how you use it.
-Ekko
01-22-2012 #2
Jonathan W
Guest
Posts: n/a
Quote:
Originally Posted by Dragonballful23 Q=question Q.1) Evaluate for x = 11: x + 4 A. 13. B. 14. C. 15. D. 16. Q.2) Evaluate for n = 0: 2 (4 + n) - 5 A. 0. B. 1. C. 2. D. 3. Q.3) Evaluate for x = 8 and y = 10: 3x + 2y A. 40. B. 41. C. 42. D. 43. Q.4) Write an algebraic expression for: 77 more than the product of 2 and u A. 77 more than 2u. B. 2u - 77. C. 77 + 2u. D. 2u = 77. Q.5) What is the absolute value of -10 minus the absolute value of 4 plus 1. A. 15. B. -15. C. 5. D. -5. Q.6) Which is bigger? -14 _____ 0 A. -14 < 0 B. -14 > 0 C. -14 = 0 Q.7) What is: 6 + (-4) A. 2. B. 10. C. -2. D. -10. Q.8) Evaluate for m = -5: m + 7 A. -2. B. 12. C. 2. D. -12. Q.9) What is: 8 - (-4) A. 2. B. 6. C. -12. D. 12. Q.10) Evaluate for m = -8: 9 - m A. -17. B. 17. C. 1. D. -1.
1.) C
2.) D
3.) None, the answer is 44.
4.) C
5.) None, the answer is 6.
6.) A
7.) A
8.) C
9.) D
10.) B
01-22-2012 #3
JosePereira11
Points: 5,846, Level: 49
Activity: 0%
Last Achievements
Join Date: May 2011
Posts: 124
Quote:
Originally Posted by Jonathan W 1.) C 2.) D 3.) None, the answer is 44. 4.) C 5.) None, the answer is 6. 6.) A 7.) A 8.) C 9.) D 10.) B
actualy the 5) is 7. /-10/ -4 = 6 6+1 = 7 however, if the question is badly written it can be /-10/ - (4+1) = 5 and it would be the answer c)
i didnt see the others, only 3 and 5 cause jonathan said that they were wrong
__________________
You need only 2 objectives in life:
when you look at the past and you say that you coudn't have done any better than that is the 1º one.
The 2º one is to do better than that
Then, you must learn 2 rules in your life:
Rule number one : respect yourself and the others
Rule number two: first learn rule number one
Last edited by JosePereira11; 01-22-2012 at 02:39 PM..
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Contact Us - XP Math - Forums - Archive - Privacy Statement - Top | 1,399 | 3,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-43 | latest | en | 0.642695 |
http://www.expertsmind.com/questions/simplify-the-expression-301141848.aspx | 1,627,871,232,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154302.46/warc/CC-MAIN-20210802012641-20210802042641-00518.warc.gz | 63,721,497 | 12,049 | ## Simplify the expression, Algebra
Assignment Help:
-(1-5n)-7n
#### Math, 10000000004*56464684654654
10000000004*56464684654654
#### Evaluting, how do you simplify using order of operations
how do you simplify using order of operations
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The title of this section is perhaps a little misleading. The title appears to imply that we're going to look at equations which involve any radicals. However, we are going to li
#### Math, i need help with my math homework
i need help with my math homework
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#### Solve out the equations using simpler method, Example Solve out each of ...
Example Solve out each of the following equations. 7 x = 9 Solution Okay, although we say above that if we contained a logarithm in fron
#### Joy, Kevin randomly selected 1 card from a Standard deck of 52 cards. what ...
Kevin randomly selected 1 card from a Standard deck of 52 cards. what is the probabilty that he will get the king of hearts?? Multiple choice .... 4/13.... 17/52...... 13/4..... 5
#### Linear programing, Trees in urban areas help keep air fresh by absorbing ca...
Trees in urban areas help keep air fresh by absorbing carbon dioxide. A city has \$2100 to spend on planting spruce and maple trees. The land available for planting is 45,000 ft2. H
#### Change of base formula - logarithms, The last topic that we have to discuss...
The last topic that we have to discuss in this section is the change of base formula. Most of the calculators these days are able of evaluating common logarithms & natural logar
#### Ged prep, i dont want the answer but how i shouuld find the answer plz :A r...
i dont want the answer but how i shouuld find the answer plz :A rectangular display case houses a square pyramid. Both have the same square base measuring x inches on a side. The t | 516 | 2,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.890712 |
http://www.chegg.com/homework-help/questions-and-answers/014-kg-block-wooden-table-given-sharp-blow-hammer-block-slides-across-table-stop-coefficie-q1616226 | 1,435,759,233,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375094931.19/warc/CC-MAIN-20150627031814-00137-ip-10-179-60-89.ec2.internal.warc.gz | 334,946,677 | 11,821 | A 0.14- kg block on a wooden table is given a sharp blow with a hammer. The block then slides across the table to a stop. The coefficient of friction between the block and table is 0.28. The hammer's force on the block as a function of time is well approximated by an isosceles triangle with it's base on the horizontal time axis. The length of the base is 0.0061 s and the maximum force is 49.5 N. How far does the block go? Use a graphical technique. | 116 | 452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | latest | en | 0.896422 |
http://thermospokenhere.com/wp/01_tsh/A9855___sled_mass/sled_mass.html | 1,712,993,662,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00369.warc.gz | 31,165,083 | 3,150 | THERMO Spoken Here! ~ J. Pohl © TOC NEXT ~ 61
Mass and Sled
The sketch (right) is from a physics text. According to the drawing, there is one horizontal force that acts with magnitude, F, directed to the right. The implication is acceleration of masses m and M occurs "at the very instance" of observation.
Calculate: The greatest force such that NO sliding of the mass on the sled will occur.
♦ For the event, since the sled and mass have no relative motion, the Second Law is written the masses taken together, (m + M). The force is the single entity having magnitude |F| and direction as I:
(1)The direction of force is known. We take the dot product of the equation with "I".The unknown terms are: “F” and “d²X/dt²”.
In Eqn (1) the acceleration is not known and the force magnitude is not known. The event occurs in the "X"-direction so the Second Law yields only one equation. Therefore we one equation and two unknowns ~ insufficient information to solve. We must analyze a smaller system.
For the condition of "no sliding," the acceleration of the mass M is the same as the sled m. Thus the Second Law for the mass M alone is:
(2)The unknown terms are: “F” and “d²X/dt²”.
In Equation (2), the friction force is written as f. It is common for the friction force to have values less than its maximum during an event. The maximum value of the friction force is written to the right in Equation (2)
The maximum force to the right is associated with the maximum friction force. Equations (1) and (2) become:
(3)The unknown terms are: “F” and “d²X/dt²”.
When these equations are solved simultaneously, one obtains:
(4)The unknown terms are: “F” and “d²X/dt²”.
♦ Sometimes a question regarding a system with its dynamics "in progress," can be answered without any understanding whatsoever of the initiation of that "progress." Skilled persons can answer these questions. Whoever wrote this physics problem did not see the event get started.
sled_Mass_2.png
In classical mechanics (some manner of) initiation can be prescribed. For beginners, to be able to construct how the event began (or might have begun) is helpful. The sketch is an "expanded version" of our "physics problem."
I show one possible "initial condition (above right). All matter of that "initial condition" has zero acceleration; everything moves at SOME constant velocity... the easiest constant velocity is ZERO velocity; everything is at rest relative to Earth. In this sketch I post pone addressing friction by representing it as a "shear pin."
Explanation (to be provided later) is missing.
For times of the event for the sled (m) and mass (M) have no inclination for relative motion), their accelerations are equal:
Mass and Sled
The sketch is from a physics text. According to the drawing, there is one horizontal force that acts with magnitude, F, directed to the right. The implication is acceleration of masses m and M occurs "at the very instance" of observation.
Calculate: The greatest force such that NO sliding of the mass on the sled will occur.
Premise presently unwritted! | 714 | 3,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-18 | latest | en | 0.945935 |
https://socratic.org/questions/what-is-the-frequency-of-f-theta-sin-6-t-cos-32-t | 1,618,210,479,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00099.warc.gz | 620,726,643 | 5,795 | # What is the frequency of f(theta)= sin 6 t - cos 32 t ?
Dec 2, 2016
$\pi$
#### Explanation:
Frequency of $\sin 6 t - \to \frac{2 \pi}{6} = \frac{\pi}{3}$
Frequency of $\cos 32 t - \to \frac{2 \pi}{32} = \frac{\pi}{16}$
Find least common multiple of $\frac{\pi}{3} \mathmr{and} \frac{\pi}{16}$
$\frac{\pi}{3}$ .....x (3) ... -->$\pi$
$\frac{\pi}{16}$ ....x (16) ... --> $\pi$
Frequency of f(t) -->$\pi$ | 159 | 408 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-17 | latest | en | 0.632964 |
https://the-algorithms.com/algorithm/gaussian-elimination | 1,716,141,550,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057819.74/warc/CC-MAIN-20240519162917-20240519192917-00107.warc.gz | 512,130,396 | 22,065 | #### Gaussian Elimination
p
P
```"""
Gaussian elimination method for solving a system of linear equations.
Gaussian elimination - https://en.wikipedia.org/wiki/Gaussian_elimination
"""
import numpy as np
from numpy import float64
from numpy.typing import NDArray
def retroactive_resolution(
coefficients: NDArray[float64], vector: NDArray[float64]
) -> NDArray[float64]:
"""
This function performs a retroactive linear system resolution
for triangular matrix
Examples:
2x1 + 2x2 - 1x3 = 5 2x1 + 2x2 = -1
0x1 - 2x2 - 1x3 = -7 0x1 - 2x2 = -1
0x1 + 0x2 + 5x3 = 15
>>> gaussian_elimination([[2, 2, -1], [0, -2, -1], [0, 0, 5]], [[5], [-7], [15]])
array([[2.],
[2.],
[3.]])
>>> gaussian_elimination([[2, 2], [0, -2]], [[-1], [-1]])
array([[-1. ],
[ 0.5]])
"""
rows, columns = np.shape(coefficients)
x: NDArray[float64] = np.zeros((rows, 1), dtype=float)
for row in reversed(range(rows)):
total = np.dot(coefficients[row, row + 1 :], x[row + 1 :])
x[row, 0] = (vector[row][0] - total[0]) / coefficients[row, row]
return x
def gaussian_elimination(
coefficients: NDArray[float64], vector: NDArray[float64]
) -> NDArray[float64]:
"""
This function performs Gaussian elimination method
Examples:
1x1 - 4x2 - 2x3 = -2 1x1 + 2x2 = 5
5x1 + 2x2 - 2x3 = -3 5x1 + 2x2 = 5
1x1 - 1x2 + 0x3 = 4
>>> gaussian_elimination([[1, -4, -2], [5, 2, -2], [1, -1, 0]], [[-2], [-3], [4]])
array([[ 2.3 ],
[-1.7 ],
[ 5.55]])
>>> gaussian_elimination([[1, 2], [5, 2]], [[5], [5]])
array([[0. ],
[2.5]])
"""
# coefficients must to be a square matrix so we need to check first
rows, columns = np.shape(coefficients)
if rows != columns:
return np.array((), dtype=float)
# augmented matrix
augmented_mat: NDArray[float64] = np.concatenate((coefficients, vector), axis=1)
augmented_mat = augmented_mat.astype("float64")
# scale the matrix leaving it triangular
for row in range(rows - 1):
pivot = augmented_mat[row, row]
for col in range(row + 1, columns):
factor = augmented_mat[col, row] / pivot
augmented_mat[col, :] -= factor * augmented_mat[row, :]
x = retroactive_resolution(
augmented_mat[:, 0:columns], augmented_mat[:, columns : columns + 1]
)
return x
if __name__ == "__main__":
import doctest
doctest.testmod()
``` | 740 | 2,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.475539 |
http://slidermath.com/rpoly/Lincurv3.shtml | 1,679,430,442,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00271.warc.gz | 54,083,841 | 2,016 | Math Help - Name a Parabola Game - Use Y-intercept of Graph - Tips:
- A ball thrown upwards in still air follows an approximate parabolic arc.
- The curve y=-4x2+3 is a parabola that opens downwards and has a y-intercept of 3.
- The curve y=4x2+5 is a parabola that opens upwards and intersects the y-axis at 5.
- The second degree (or quadratic) relation y=ax2+b represents a set of ordered pairs (x,y)
which form a parabolic shape that passes through the point (0,b).
When 'a' is positive, the parabola opens upwards. When 'a' is negative, the parabola opens downwards.
Increasing the size of 'a', narrows the width of the mouth of the parabola.
When 'a' is zero, the parabola y=ax2+b reduces to the horizontal line y=b.
- Your Game Score is reduced by the number of fish hits.
- To slow the game speed repeat tap/click on the word Slider.
- To increase the game speed repeat tap/click on the word Math.
- Speed can also be adjusted with a keyboard's - and + keys.
- Refresh/Reload the web page to restart the game. | 269 | 1,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-14 | latest | en | 0.888486 |
https://quizizz.com/en/comparing-fractions-worksheets-class-2?page=1 | 1,722,997,538,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640667712.32/warc/CC-MAIN-20240807015121-20240807045121-00420.warc.gz | 390,486,591 | 25,344 | ## Recommended Topics for you
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## Explore printable Comparing Fractions worksheets for 2nd Class
Comparing Fractions worksheets for Class 2 are an essential tool for teachers looking to help their students build a strong foundation in math, specifically in the area of fractions. These worksheets provide a variety of exercises and activities designed to engage young learners and make the process of understanding and comparing fractions both fun and accessible. With a range of difficulty levels, these worksheets cater to the diverse needs of Class 2 students, allowing teachers to differentiate instruction and ensure that all students are challenged and supported in their learning. By incorporating Comparing Fractions worksheets for Class 2 into their lesson plans, teachers can effectively reinforce key concepts and help students develop the skills they need to succeed in math.
In addition to Comparing Fractions worksheets for Class 2, Quizizz offers a comprehensive platform that allows teachers to create interactive quizzes, games, and other engaging activities to supplement their students' learning. With Quizizz, teachers can easily customize content to align with their curriculum and address specific learning objectives. The platform also provides valuable data and insights on student performance, enabling teachers to identify areas where students may need additional support or practice. By integrating Quizizz into their instruction, teachers can enhance the learning experience for their Class 2 students and make math, including fractions, an enjoyable and rewarding subject. With a combination of high-quality worksheets and the interactive features of Quizizz, teachers can ensure that their students are well-equipped to tackle the challenges of math in Class 2 and beyond. | 648 | 2,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.927237 |
https://www.coursehero.com/file/6451476/Pre-Calc-Homework-Solutions-147/ | 1,496,045,506,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612036.99/warc/CC-MAIN-20170529072631-20170529092631-00230.warc.gz | 1,082,121,382 | 34,798 | Pre-Calc Homework Solutions 147
# Pre-Calc Homework Solutions 147 - Section 4.3 Find the...
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Find the zeros of y 0 : 5 0 2 x 4 2 27 x 2 1 54 5 0 x 2 55 } 27 6 4 3 ˇ 3 w 3 w } x 56 ! } 2 § 7 § 2 § 4 3 § ˇ § 3 w § 3 w } § < 6 1.56 Note that we do not use x ! } 2 § 7 § 1 § 4 3 § ˇ § 3 w § 3 w } § < 6 3.33, because these values are outside of the domain. (a) [ 2 3, 2 ˇ 6 w ] and [0, ˇ 6 w ] or, < [ 2 3, 2 2.45] and [0, 2.45] (b) [ 2 ˇ 6 w , 0] and [ ˇ 6 w , 3] or, < [ 2 2.45, 0] and [2.45, 3] (c) Approximately ( 2 1.56, 1.56) (d) Approximately ( 2 3, 2 1.56) and (1.56, 3) (e) Local maxima: ( 6 ˇ 6 w ,6 ˇ 3 w ) < ( 6 2.45, 10.39); local minima: (0, 0) and ( 6 3, 0) (f) < ( 6 1.56, 6.25) 23. y 95 } 1 1 1 x 2 } Since y 9. 0 for all x , y is always increasing. y 05 } d d x } (1 1 x 2 ) 2 1 52 (1 1 x 2 ) 2 2 (2 x ) 5 } (1 2 1 2 x x 2 ) 2 } Graphical support: [ 2 4, 4] by [ 2 2, 2]
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Fall '08 term at University of Florida.
Ask a homework question - tutors are online | 587 | 1,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-22 | longest | en | 0.765992 |
https://stats.stackexchange.com/questions/453157/variance-between-two-coefficients-in-linear-regression-model | 1,701,994,849,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100705.19/warc/CC-MAIN-20231207221604-20231208011604-00230.warc.gz | 608,360,613 | 40,549 | # Variance between two coefficients in linear regression model
I am trying to figure out the variance of the difference between two coefficients in a linear regression model. If I am given the design matrix (X^T X) and the value of sigma, how do I go about solving: var(beta.1.hat - beta.2.hat)? My thought process is finding the variance for each part using the formula var(beta.j.hat) = sigma^2((X^T X)^-1 subscript jj. Then var(beta.1.hat - beta.2.hat) should be equal to: var(beta.1.hat) + var(beta.2.hat) - 2 Cov(beta.1.hat,beta.2.hat)
Is my logic correct? Please give me some suggestions or assistance thanks!
• If you know the variance of $Y$, then you could multiply $\beta$ by a matrix $[1, -1]$ and apply the same to the least squares estimator and calculate the variance as such.
– user275978
Mar 8, 2020 at 0:21
Assuming you have 2 variables, write $$\beta = (\beta_1, \beta_2)^T$$ and let $$c = (1, -1)^T$$. Recall that in ordinary least squares, $$Var(\hat\beta) = \sigma^2 (X^TX)^{-1}$$. \begin{align*} Var(\hat \beta_1 - \hat \beta_2) &= Var(c^T\hat\beta)\\ &= c^TVar(\hat\beta)c\\ &=\sigma^2 c^T(X^TX)^{-1}c \end{align*} This reduces to what you have, so your logic is correct.
If you don't remember the variance (or other parts of least squars. Posit model $$Y=X\beta + \epsilon$$ where $$\epsilon \sim (0, \sigma^2I)$$. Minimize the following objective function: \begin{align*} f(\beta) &= (Y-X\beta)^T(Y-X\beta)\\ f'(\beta) &= -2X^T(Y-X\beta) = 0\\ \hat\beta &= (X^TX)^{-1}X^TY\\ Var(\hat\beta) &= Var((X^TX)^{-1}X^TY)\\ &= (X^TX)^{-1}X^T Var(Y) X(X^TX)^{-1}\\ &= (X^TX)^{-1}X^T Var(X\beta + \epsilon) X(X^TX)^{-1}\\ &= (X^TX)^{-1}X^T Var(\epsilon) X(X^TX)^{-1}\\ &= (X^TX)^{-1}X^T \cdot \sigma^2I \cdot X(X^TX)^{-1}\\ &= \sigma^2(X^TX)^{-1} \end{align*} Note: normality assumption is not required unless you need to estimate $$\sigma^2$$ and/or do inference. | 655 | 1,883 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-50 | longest | en | 0.628953 |
https://jvejournals.com/article/21039 | 1,579,955,943,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672440.80/warc/CC-MAIN-20200125101544-20200125130544-00263.warc.gz | 496,830,560 | 15,426 | ## Design and analysis of magnetic circuit of permanent magnet eddy current brake
Yumeng Fan1 , Guolai Yang2
1, 2School of Mechanical Engineering, Nanjing University of Science and Technology, Nanjing, China
1Corresponding author
Vibroengineering PROCEDIA, Vol. 28, 2019, p. 111-117. https://doi.org/10.21595/vp.2019.21039
Received 19 September 2019; accepted 26 September 2019; published 19 October 2019
Copyright © 2019 Yumeng Fan, et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract.
The eddy current brake has the advantages of no frictional contact or hydraulic fluid, high structural reliability, etc. The existing linear eddy current brakes are mostly flat type. A cylindrical permanent magnet eddy current brake is proposed in this paper, whose air gap magnetic field is generated by a series of ring-shaped permanent magnets and guided by iron pole to conductor layers, and can achieve higher air gap magnetic flux density. This paper introduces its basic structure and working principle. In order to obtain the analytical model of magnetic circuit design, the equivalent magnetic circuit method is used to analyze and derive the magnetic circuit, and it is verified by the axisymmetric finite element model. To derive the braking force generated by the eddy current brake, the layer theory approach is applied. The influence of electromagnetic parameters on the force characteristic is obtained by finite element numerical calculation, which provides a theoretical basis for the optimal design of the eddy current brake.
Keywords: eddy current brake, magnetic circuit design, analytical model, finite element analysis.
#### 1. Introduction
Compared with the traditional brake, the eddy current brake has the advantages of simple working principle, high structural reliability and no friction [1]. It makes the eddy current brake being widely applied, such as high-speed train track braking system and electromagnetic aircraft ejection system braking [2-4]. In order to solve the problem of short-distance braking of a large impulse machinery, a new type of eddy current brake is designed.
The fundamental mechanism of the eddy current brakes has been studied by many scholars [5], and some scholars have analyzed the performances of it with different structures [6]. Jia Qiang has designed a linear eddy current buffer by using Halbach array [7]. Kou Baoquan et al has designed two different kinds of magnetic circuit hybrid eddy current brake [1, 8], which can produce a greater braking force.
For the braking force calculation of the eddy current brake, Romanian scholar Boldea et al. equivalent the primary excitation winding to an infinitely thin current layer, and then derived Expression [9] according to the layer theory. There are also many other different derivation methods for the braking force models of different structures of eddy current buffers [10, 11].
Most of the existing linear eddy current brakes are flat-plate structures or hybrid excitation circuits. This paper designs and analyzes a brake model with a cylindrical structure using only permanent magnet. It has the following advantages over other types of eddy current brakes. First of all, the structure is simple, the installation is convenient, and it is not easy to suffer loss and error. Secondly, a large braking force can be generated, which makes it possible to brake a high-speed moving object in a short time and a short distance.
In order to carry out in-depth research, analytical mode were carried out for magnetic circuit design analysis, and finite element simulation and model analysis were also carried out for eddy current and braking force.
#### 2. Cylindrical permanent magnet eddy current brake model
Fig. 1(a) illustrates the partial three-dimensional structural schematic of the studied cylindrical linear EMB, which mainly consists of two parts: the primary part, which consists of a moving rod combined with a sequence of ring-shaped, axially magnetized permanent magnets separated by pure iron poles, and the secondary part, which consists of an outer tube and an inner tube. High-velocity relative movement between the primary and secondary parts will occur under the intensive impact load. Due to the electromagnetic induction, circumferential eddy currents are generated in the secondary. They are sequentially arranged along the iron pole interval, and is opposite to the adjacent eddy currents. At the same time, the circumferential eddy currents induce a new magnetic field interacting with the primary magnetic flux to generate braking force, which hinders the relative motion of the primary and secondary according to Lenz’s law.
Fig. 1(b) exhibits the layout of the permanent magnets which are oriented with magnetic poles of the same polarity facing each other. The iron pole is located between the two permanent magnets to increase the amount of magnetic flux entering the secondary. Therefore, the decrease of the quiescent operation point caused by the flux leakage in the cylinder hole of permanent magnet and excessive air-gap reluctance is reduced. The thick bold arrows show the direction of field line.
As shown in Fig. 1(b), ${\tau }_{m}$, $\tau$, ${r}_{i}$, ${r}_{o}$, ${r}_{d}$, ${R}_{i}$, ${R}_{o}$ are the thickness of the permanent magnet; the pole pitch; the internal radius of the inner cylinder; the external radius of the inner cylinder (the internal radius of the outer cylinder), the external radius of the EMB, the internal radius of the permanent magnet, the external radius of the permanent magnet, respectively.
Fig. 1. a) Permanent magnet eddy current brake and (b) permanent magnet distribution map
a)
b)
#### 3.1. Analytical model of no-load magnetic field
The no-load state of the eddy current brake refers to a state in which the relative velocity of the primary and secondary is zero, that is, the eddy current in the conductor plate is zero. Analyze under no-load condition can rapidly provide verify whether the performance and size are compatible with the envisioned application without involving comprehensive eddy current formulation. Fig. 2(a) shows the magnetic circuit distribution and magnetic flux flow direction of the eddy current brake in the no-load state. This paper did not neglect the leakage flux, adding it into the equivalent magnetic circuit, which improve the accuracy of the analytical model.
The calculation formula for each part of the magnetic circuit is as follows.
The magnetomotive force generated by the permanent magnet is:
(1)
${F}_{c}={H}_{c}{\tau }_{m},$
where ${H}_{c}$ is the permanent magnet coercivity.
The reluctance of the air gap is as follows, and the reluctance of the inner cylinder, the iron pole can be obtained similarly:
(2)
${R}_{\delta }=\frac{\mathrm{l}\mathrm{n}\left({r}_{i}/{r}_{1}\right)}{2{\mu }_{0}\pi \left(\tau -{\tau }_{m}\right)},$
where: ${r}_{1}$ is the external radius of the iron pole.
Fig. 2. a) Magnetic path and flux flow direction and b) its equivalent magnetic circuit
a)
b)
The magnetic circuit 1 and 4 and the permanent magnet have a ring structure. The reluctance of permanent magnet is as Eq. (3), and the reluctance of the 1 and 4 part can be obtained similarly:
(3)
${R}_{m}=\frac{{\tau }_{m}}{{\mu }_{0}{\mu }_{m}\pi \left({R}_{o}^{2}-{R}_{i}^{2}\right)},$
where: ${\mu }_{m}$ is the relative magnetic permeability of the permanent magnet.
The magnetic permeance of the 6 part of the magnetic circuit can be expressed as:
(4)
${G}_{6}={\int }_{0}^{2\pi }{\int }_{{R}_{i}-\frac{\tau }{2}}^{{R}_{i}-{\tau }_{m}}\frac{{\mu }_{0}r}{\pi \left({R}_{i}-r\right)}drd\theta .$
Fig. 3. The magnetic circuit of the 6 part
Fig. 4. Leakage flux between iron poles
The integral path of the leakage reluctance at the inner and outer ends of the permanent magnet is shown in Fig. 4. Although the relative magnetic permeability of the iron pole is larger than that of the permanent magnet, the axial length of the permanent magnet is longer, thus the magnetically leakage cannot be ignored.
Therefore, the expression of the leakage flux in the hole of the permanent magnet is:
(5)
${G}_{mm1}={\int }_{0}^{2\pi }{\int }_{\tau m/4}^{\tau m/2}\frac{{\mu }_{0}\left({R}_{i}-r\right)}{\pi r}drd\theta .$
The outer leakage flux is divided into two parts: the part through conductor and part only in the air gap:
(6)
${G}_{mm2}={\int }_{0}^{2\pi }{\int }_{\left({r}_{i}-{R}_{0}\right)/2}^{{r}_{i}-{R}_{0}}\frac{{\mu }_{0}\left({R}_{0}+r\right)}{\pi r}drd\theta ,$
(7)
${G}_{mm3}={\int }_{0}^{2\pi }{\int }_{0}^{{l}_{0}}\frac{{\mu }_{0}\left({r}_{0}-{R}_{0}\right)/\sqrt{2}}{\pi \left({r}_{0}-{R}_{0}\right)}+\frac{{\mu }_{s}{r}_{0}}{{\tau }_{m}-2r}drd\theta ,$
(8)
${l}_{0}=\frac{{\tau }_{m}}{2}-2\left({r}_{i}-{R}_{0}\right).$
The total leakage reluctance of the permanent magnet can be obtained by paralleling the three-part’s leakage reluctance:
(9)
${R}_{mm}=\frac{{R}_{mm1}{R}_{mm2}{R}_{mm3}}{{R}_{mm1}{R}_{mm2}+{R}_{mm1}{R}_{mm3}+{R}_{mm2}{R}_{mm3}}.$
According to the equivalent magnetic circuit of the electromagnetic buffer and the Kirchhoff magnetic potential difference law, we can get:
(10)
make:
(11)
$\left\{\begin{array}{l}{t}_{1}={R}_{m}/2{t}_{2}={R}_{mm}/2{t}_{3}={R}_{1}/2,\\ {t}_{4}={R}_{5}+{R}_{6}/2{t}_{5}={R}_{2}+{R}_{3}+{R}_{4}/2+{R}_{c}+{R}_{\delta },\end{array}\right\$
By solving the above equations, we get:
(12)
${\Phi }_{\delta }=\frac{{t}_{2}{t}_{4}{t}_{5}{F}_{C}}{{t}_{5}\left({t}_{4}+{t}_{5}\right)\left({t}_{1}{t}_{2}+{t}_{1}{t}_{3}+{t}_{2}{t}_{3}\right)+{t}_{4}{t}_{5}^{2}\left({t}_{1}+{t}_{2}\right)}.$
Therefore, the gap magnetic density at the air gap place r under no-load condition is:
(13)
${B}_{\delta }=\frac{{t}_{2}{t}_{4}{t}_{5}{F}_{C}}{2\pi r\left(\tau -{\tau }_{m}\right)\left({t}_{5}\left({t}_{4}+{t}_{5}\right)\left({t}_{1}{t}_{2}+{t}_{1}{t}_{3}+{t}_{2}{t}_{3}\right)+{t}_{4}{t}_{5}^{2}\left({t}_{1}+{t}_{2}\right)\right)}.$
The design parameters of the eddy current brake are shown in Table 1.
Table 1. Design parameters of eddy current brake
Parameters / mm Value Parameters / mm Value The internal radius of the permanent magnet ${R}_{i}$ 20 The external radius of the inner cylinder ${r}_{0}$ 64 The external radius of the permanent magnet ${R}_{0}$ 61 The external radius of the outer cylinder ${r}_{d}$ 72 The external radius of the iron pole ${r}_{1}$ 62.5 The thickness of the pole pitch τ 30 The internal radius of the inner cylinder ${r}_{i}$ 63 The thickness of the permanent magnet ${\tau }_{m}$ 20
#### 3.2. Analysis model of braking force
In this chapter, the layer method approach is used to derive the braking force of the inner tube conductor layer. The permanent magnet is equivalent to an infinitely thin current layer so that the magnetic motive force generated by the current layer is equal to the magnetic motive force generated by the actual winding and the permanent magnet. According to the electromagnetic field theory, the corresponding differential equations and their general solutions in different regions can be obtained.
Fig. 5. Finite element and analysis results of air gap flux density
For the air gap region, because the electrical conductivity of the air gap is zero, so the equation is as follows. And the region of the outer tube is similar to it:
(14)
$\frac{{\partial }^{2}{A}_{1}}{\partial {z}^{2}}+\frac{{\partial }^{2}{A}_{1}}{\partial {r}^{2}}=0.$
For the inner tube conductor layer, the eddy current is induced in the layer when the brake is working. And the differential equation in the inner tube layer can be derived:
(15)
where ${\sigma }_{c}$ is the conductivity (S/m) of the inner cylinder, and $v$ is the relative velocity between the primary and secondary.
The expression of the eddy current in the inner tube layer can be derived from the differential equations. And the expression of the eddy current loss and braking force can be obtained:
(16)
where $p$ is the polar logarithm; $\overline{{J}_{e}}$ is the conjugate complex number of ${J}_{e}$.
#### 4. Finite element analysis
Using a set of design data as Table 1 to establish the finite element model, the magnetic flux density of the device at the speed of 5, 10 and 1 5m/s are as Fig. 6. It is observed that the reaction field of eddy currents tilts the field line coming into the secondary and differs in degree at different speed.
Fig. 7 are force-speed characteristic curves obtained by using different air gap thickness and inner tube thickness, respectively. It can be seen from Fig. 7(a) that the thickness of the air gap has little effect on the braking force when the speed is small, but the larger the thickness of the air gap at a high speed, the smaller the braking force generated, because the air gap flux density is reduced. It can be seen from Fig. 7(b) that both the peak braking force and the critical speed increase as the thickness of the inner tube increases. At the same time, the braking force increases as the thickness of inner tube increases when the speed is less than 9 m/s, but the opposite when the speed is greater than 9 m/s.
In addition, the calculation results show that the peak braking force of the permanent magnet eddy current brake can reach 1.5e5N, and the critical speed is about 8 m/s, which is larger than the ordinary eddy current brake. That is, at high speeds, the brakes can reach a sufficiently large braking force requirement.
Fig. 6. Magnetic flux density ($T$) at a) 5 m/s, b) 10 m/s, and c) 15 m/s
a)
b)
c)
Fig. 7. Force characteristic curves for different a) air gap thicknesses and b) inner tube thicknesses
a)
b)
#### 5. Conclusions
In this paper, a new type of cylindrical permanent magnet eddy current brake is proposed, and its basic structure and working principle are introduced. In order to obtain the analytical model of the magnetic circuit, the equivalent circuit method and layer theory approach are used to analyze and derive the magnetic circuit. The model is also calculated by the axisymmetric finite element model, and the analytical model is verified through comparing. The influence of the thickness of the air gap and the thickness of the inner tube on the braking force characteristic curve is obtained using the finite element model, which provides a theoretical basis for the future optimal design of this eddy current brake.
There are still errors of the analytical model proposed this paper with practical device, demagnetization effect of permanent magnet and armature reaction would be added into consideration in future study.
#### References
1. Kou B., Jin Y., Zhang H., et al. Characteristic analysis of hybrid excitation linear electromagnetic dampers. Proceedings of the CSEE, Vol. 33, Issue 24, 2013, p. 143-151. [CrossRef]
2. Wang Z., Hua X., Chen Z. Experimental study on vibration control of a model footbridge by a tiny eddy current tuned mass damper with permanent magnets. Journal of Vibration and Shock, Vol. 33, Issue 20, 2014, p. 129-132. [CrossRef]
3. Yan Guobin Study of Hybrid Excitation Rail Eddy Current Brake System of High-Speed Train. Zhejiang University, 2010. [CrossRef]
4. Xiao Yao, Wu Jun The analysis and design of permanent-magnet eddy current brake for EMALS. Small and Special Electrical Machines, Vol. 41, Issue 8, 2013. [CrossRef]
5. Bae J. S., Hwang J. H., Park J. S., et al. Modeling and experiments on eddy current damping caused by a permanent magnet in a conductive tube. Journal of Mechanical Science and Technology, Vol. 23, Issue 11, 2009, p. 3024-3035. [Publisher]
6. Jang S. M., Lee S. H. Comparison of three types of permanent magnet linear eddy-current brakes according to magnetization pattern. IEEE Transactions on Magnetics, Vol. 39, Issue 5, 2003, p. 3004-3006. [Publisher]
7. Jia Q., Gao Y., Tong Y., et al. Design of linear eddy current buffer. Journal of Engineering Design, Vol. 18, Issue 3, 2011, p. 209-213. [CrossRef]
8. Kou B., Jin Y., Zhang H., et al. Analysis and design of hybrid excitation linear eddy current brake. IEEE Transactions on Energy Conversion, Vol. 29, Issue 2, 2014, p. 496-506. [Publisher]
9. Boldea I., Babescu M. Multilayer theory of D.C. linear brakes with solid-iron secondary. Proceedings of the Institution of Electrical Engineers, Vol. 123, Issue 3, 1976, p. 220. [Publisher]
10. Yin Xiangrui Research on Long Stroke Secondary Moving Permanent Magnetic Linear Eddy Current Brake. Harbin Institute of Technology, 2017. [CrossRef]
11. Zhao Xiaobo Study on Electromagnetic Properties and Braking Performance of Permanent Magnet Type Eddy Current Retarder. Nanjing Agriculture University, 2009. [CrossRef] | 4,226 | 16,792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 36, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.869914 |
https://www.askiitians.com/forums/Analytical-Geometry/when-should-we-use-family-of-straight-lines_110552.htm | 1,713,635,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00349.warc.gz | 590,483,643 | 43,882 | # when should we use family of straight lines?
Grade:10
## 2 Answers
Shobhit Varshney IIT Roorkee
askIITians Faculty 33 Points
9 years ago
Hi,
The family of straight lines is used when we need to represent large number of lines representing a particular condition.
thanks............................................................................................................................................................................................................................................
S. Agarwal
33 Points
9 years ago
Find the equation of the line which passes through the point of intersection of L1:2x – 3y + 5 = 0 and L2: 3x – 2y + 5 = 0, and also passes through the origin.
Here’s one way to solve the problem.
(i) Find the point of intersection L1 and L2: This point will be (-1, 1)
(ii) Use the two point form to find the required equation: (y – 0)=[(1-0)/(-1-0)](x – 0) or x + y = 0.
But that’s a boring method. I’m interested in family of lines. So here’s another method.
As discussed before, equation of any line passing through the point of intersection of the lines L1=0 and L2 =0 will be of the form L1+λL2 =0, where λ is a parameter.
That is, the required line will be of the form (2x – 3y + 5)+λ(3x – 2y + 5) = 0. That’s half the work done! But what about λ?
Another condition is given to us: The line passes through the origin as well. We can therefore substitute the coordinates to obtain the value of λ.
We have,
2(0)-3(0)+5 + λ(3(0)-2(0)+5) = 0,
or λ=-1.
And we’re done!
The required line is (2x – 3y + 5)+(-1)(3x – 2y + 5) = 0, or x + y = 0.
We didn’t even have to find the point of intersection of the given lines. Saved some effort.
So remember, any line passing through the point of intersection of two given lines will be given by L1+λL2 =0, and the value of the parameter λ will be determined by another given condition
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https://www.ebookee.net/Vibration-Analysis-and-Structural-Dynamics-for-Civil-Engineers-Essentials-and-Group-Theoretic-For_3983612.html | 1,579,714,685,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607314.32/warc/CC-MAIN-20200122161553-20200122190553-00194.warc.gz | 847,060,653 | 7,550 | # Vibration Analysis and Structural Dynamics for Civil Engineers Essentials and Group-Theoretic For...
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Posted on 2019-03-18, by voska89.
Description
Vibration Analysis and Structural Dynamics for Civil Engineers: Essentials and Group-Theoretic Formulations by Alphose Zingoni
English | 2014 | ISBN: 0415522560, 0415522552 | 276 pages | PDF | 17 MB
Appeals to the Student and the Seasoned Professional
While the analysis of a civil-engineering structure typically seeks to quantify static effects (stresses and strains), there are some aspects that require considerations of vibration and dynamic behavior. Vibration Analysis and Structural Dynamics for Civil Engineers: Essentials and Group-Theoretic Formulations is relevant to instances that involve significant time-varying effects, including impact and sudden movement. It explains the basic theory to undergraduate and graduate students taking courses on vibration and dynamics, and also presents an original approach for the vibration analysis of symmetric systems, for both researchers and practicing engineers. Divided into two parts, it first covers the fundamentals of the vibration of engineering systems, and later addresses how symmetry affects vibration behavior.
Part I treats the modeling of discrete single and multi-degree-of-freedom systems, as well as mathematical formulations for continuous systems, both analytical and numerical. It also features some worked examples and tutorial problems. Part II introduces the mathematical concepts of group theory and symmetry groups, and applies these to the vibration of a diverse range of problems in structural mechanics. It reveals the computational benefits of the group-theoretic approach, and sheds new insights on complex vibration phenomena.
The book consists of 11 chapters with topics that include:
* The vibration of discrete systems or lumped parameter models
* The free and forced response of single degree-of-freedom systems
* The vibration of systems with multiple degrees of freedom
* The vibration of continuous systems (strings, rods and beams)
* The essentials of finite-element vibration modelling
* Symmetry considerations and an outline of group and representation theories
* Applications of group theory to the vibration of linear mechanical systems
* Applications of group theory to the vibration of structural grids and cable nets
* Group-theoretic finite-element and finite-difference formulations
Vibration Analysis and Structural Dynamics for Civil Engineers: Essentials and Group-Theoretic Formulations acquaints students with the fundamentals of vibration theory, informs experienced structural practitioners on simple and effective techniques for vibration modelling, and provides researchers with new directions for the development of computational vibration procedures.
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Vibration Analysis and Structural Dynamics for Civil Engineers Essentials and Group-Theoretic For...
Related Books | 679 | 3,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-05 | longest | en | 0.898954 |
https://www.convertunits.com/from/arcminute/to/degree | 1,571,772,109,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987823061.83/warc/CC-MAIN-20191022182744-20191022210244-00363.warc.gz | 842,310,330 | 7,177 | ## ››Convert arcminute to degree
arcminute degree
How many arcminute in 1 degree? The answer is 60.
We assume you are converting between arcminute and degree.
You can view more details on each measurement unit:
arcminute or degree
The SI derived unit for angle is the radian.
1 radian is equal to 3437.7467707849 arcminute, or 57.295779513082 degree.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between arcminutes and degrees.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of arcminute to degree
1 arcminute to degree = 0.01667 degree
10 arcminute to degree = 0.16667 degree
20 arcminute to degree = 0.33333 degree
30 arcminute to degree = 0.5 degree
40 arcminute to degree = 0.66667 degree
50 arcminute to degree = 0.83333 degree
100 arcminute to degree = 1.66667 degree
200 arcminute to degree = 3.33333 degree
## ››Want other units?
You can do the reverse unit conversion from degree to arcminute, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Minute
A minute is: * a unit of time equal to 1/60th of an hour and to 60 seconds. (Some rare minutes have 59 or 61 seconds; see leap second.)
## ››Definition: Degree
A degree (or in full degree of arc), usually symbolized by the symbol °, is a measurement of plane angles, or of a location along a great circle of a sphere (such as the Earth or the celestial sphere), representing 1/360 of a full rotation.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 512 | 2,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-43 | latest | en | 0.871723 |
https://www.fractalforums.com/index.php?action=gallery;sa=view;id=19706 | 1,709,318,729,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475422.71/warc/CC-MAIN-20240301161412-20240301191412-00111.warc.gz | 771,929,953 | 11,405 | The All New FractalForums is now in Public Beta Testing! Visit FractalForums.org and check it out!
It's a strange world
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Description: Fragmentarium-Asurf DE-Kn8
#info Amazing Surface by Kali - Based on Tglad's Amazing Box
#define providesInit
#include "DE-Kn8.frag"
#define KN_VOLUMETRIC
#define USE_IQ_CLOUDS
#include "MathUtils.frag"
#define USE_INF_NORM
#group AmazingSurface
uniform int Iterations; slider[0,17,300]
uniform int ColorIterations; slider[0,3,300]
uniform float Scale; slider[-3,1.5,3.0]
uniform int FoldType; slider[1,1,3]
uniform vec3 PreTranslation; slider[(-5,-5,-5),(0,0,0),(5,5,5)]
uniform vec2 FoldValues; slider[(0,0),(1,1),(5,5)]
uniform bool Julia; checkbox[false]
uniform vec3 JuliaValues; slider[(-5,-5,-5),(-1,-1,-1),(5,5,5)]
uniform vec3 RotVector; slider[(-1,-1,-1),(1,1,1),(1,1,1)]
uniform float RotAngle; slider[-180,0,180]
mat3 rot;
void init() {
rot = rotationMatrix3(normalize(RotVector), RotAngle);
}
vec4 scale = vec4(Scale, Scale, Scale, abs(Scale))/MinRad2;
float absScalem1 = abs(1-Scale);
float AbsScaleRaisedTo1mIters = pow(abs(Scale), float(1-Iterations));
float DE(vec3 pos) {
vec4 p = vec4(pos,1);
vec3 c = Julia?JuliaValues:p.xyz;
for (int i=0; i<Iterations; i++) {
if (FoldType==1) p.xy=abs(p.xy+FoldValues)-abs(p.xy-FoldValues)-p.xy;
if (FoldType==2) p.xy=FoldValues-abs(abs(p.xy)-FoldValues);
if (FoldType==3) p.xy=abs(p.xy+FoldValues);
p.xyz+=PreTranslation;
float r2 = dot(p.xyz, p.xyz);
if (i<ColorIterations) orbitTrap = min(orbitTrap, abs(vec4(p.xyz,r2)));
p = p*scale;
p.xyz+=c;
p.xyz*=rot;
}
return ((length(p.xyz) - absScalem1) / p.w - AbsScaleRaisedTo1mIters);
}
#preset timemit1
FOV = 0.65625
Eye = -0.967282,1.658636,-0.1424379
Target = 3.104564,-40.57953,0.015433
Up = -0.0475652,-0.0008506,0.9992288
Gamma = 0.8999
ToneMapping = 5
Exposure = 0.833217
Brightness = 0.9803922
Contrast = 0.9470752
Saturation = 0.9733894
GaussianWeight = 1
AntiAliasScale = 0.0995733
Detail = -3.62453
FudgeFactor = 1
Dither = 0.43043
NormalBackStep = 5.957
AO = 0,0,0,1
CamLight = 0.811765,0.909804,0.937255,1.230226
CamLightMin = 0.5628492
Glow = 1,0.721569,0.490196,0.31962
GlowMax = 74
BaseColor = 1,1,1
OrbitStrength = 0.6269285
X = 0.105882,0.129412,0.129412,0.9710345
Y = 0.45098,0.364706,0.219608,1
Z = 0.827451,0.764706,0.772549,0.9241379
R = 0.87451,0.858824,0.67451,0.1235335
BackgroundColor = 0.921569,0.811765,0.168627
CycleColors = true
Cycles = 2.622887
EnableFloor = true NotLocked
FloorNormal = 0.0174703,-0.0356394,-1
FloorHeight = 0.1988835
FloorColor = 0.7254902,1,0.4666667
FocalPlane = 2.54299
Aperture = 0.07814
InFocusAWidth = 1
ApertureNbrSides = 5 NotLocked
ApertureRot = 136.325
ApStarShaped = false NotLocked
ReflectionsNumber = 1 NotLocked
Bloom = true
BloomIntensity = 0.2117982
BloomPow = 1.832865
BloomTaps = 17
DetailAO = -1.544693
MaxRaySteps = 921
MaxDistance = 567.12
AoCorrect = 0.016632
Specular = 0.1102688
SpecularExp = 16
Reflection = 0.2313725,0.172549,0.145098
SpotGlow = true
SpotLight = 1,0.631373,0.109804,8.94961
LightPos = 1.220527,5.686547,1.48405
LightSize = 0.2186207
LightFallOff = 0.289492
LightGlowExp = 0.5777311
HF_Fallof = 2.021722
HF_Const = 0.215917
HF_Intensity = 0.2363636
HF_Dir = -0.3145604,-0.0824176,0.3928571
HF_Offset = -3.393855
HF_Color = 0.4784314,0.6666667,0.8823529,1.118644
HF_Scatter = 1.6635
HF_Anisotropy = 0,0,0
HF_FogIter = 8
CloudScale = 0.57362
CloudFlatness = 0.12432
CloudTops = -7.8968
CloudBase = -8.9474
CloudDensity = 0.09519
CloudRoughness = 0.5894
CloudContrast = 1.7731
CloudColor = 0.905882,0.886275,0.588235
SunLightColor = 0.698039,0.203922,0.105882
Iterations = 84
ColorIterations = 14
Scale = 1.728778
FoldType = 3
PreTranslation = -2.772242,-3.476868,0.13879
FoldValues = 0.6026629,0.6236861
Julia = true
JuliaValues = -4.050633,-5,-0.281294
RotVector = 0.0434783,1,0.3576438
RotAngle = -40.11283
#endpreset
Stats:
Total Favorities: 0 View Who Favorited
Filesize: 542.29kB
Height: 1080 Width: 1920
Discussion Topic: View Topic
Keywords: timemit fragmentarium amazingsurf
Posted by: Tim Emit October 14, 2016, 11:53:40 AM
Rating: by 2 members. | 1,681 | 4,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.403063 |
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