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https://math.stackexchange.com/questions/232309/how-to-interpret-material-conditional-and-explain-it-to-freshmen | 1,721,436,918,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514972.64/warc/CC-MAIN-20240719231533-20240720021533-00747.warc.gz | 343,535,751 | 49,743 | # How to interpret material conditional and explain it to freshmen?
After studying mathematics for some time, I am still confused.
The material conditional “$\rightarrow$” is a logical connective in classical logic. In mathematical texts one often encounters the symbol “$\Rightarrow$”, which is read as “implies” or “if … then ….” It is customary and reasonable to treat “$\Rightarrow$” as the material conditional, i. e. as equivalent to saying “the antecedent is false or the consequence is true (or both)”.
Now there is some controversy about whether the material conditional really captures conditional statements because it doesn't really say anything about a causal connection between the antecedent and the consequence. This is quite often illustrated by the means of statements in natural languages such as “the moon is made out of cheese $\Rightarrow$ all hamsters are green” – since the moon isn't made out of cheese, is this statement true? This remained problematic to me.
While I came to accept the material conditional as a good way of describing implications and conditionals, I'm having a hard time to explain this usage to freshmen whenever I get asked.
My questions are: How can we best justify the interpretation of “$\Rightarrow$” as a material conditional? Why is it so well-suited for mathematics? How can we interpret or read it to understand it better? Can my confusion about it be led back to some kind of misunderstandig or misinterpretation of something?
I have yet very poor background in mathematical logic (I sometimes browse wikipedia articles about it), but I'd have no problem with a technical answer to this question if it clarifies the situation.
• see this recent post. There, Rick Decker gives a great example to use to convey the "logic" of material implication. There's also a reference to John Corcoran's explication of the "Meanings of Implication", which gives greater justice to how "implies" is used and what those uses means. Also, If you search the site using the tag logic and the word "implies" and/or the words "if then", you'll turn up a LOT of ways to explicate the notion of material implication! Commented Nov 7, 2012 at 18:51
• See also here and here. Commented Nov 7, 2012 at 19:03
• ...and here. Commented Nov 7, 2012 at 19:09
• @amwhy I stumbled upon some of these postings. And incidentally just now upon the last one you refered to. It really has a potential to clear things up for me. Many thanks! Still, it would be nice to have some other interpretation of the material implication, e. g. way to read it, which captures its truth-functional character better. I guess this is the main part of my question. Commented Nov 7, 2012 at 19:22
• Are there other kinds of impliations in math? I seen somthing called a "formal implication" but this donst seem well suited. Am I right? Commented Nov 1, 2015 at 14:11
The original question asked "Why is the material conditional so well-suited for mathematics?" Here's a central consideration which others have not touched on.
One thing mathematicians need to be very clear about is the use of statements of generality and especially statements of multiple generality – you know the kind of thing, e.g. the definition of continuity that starts for any $\epsilon$ ... there is a $\delta$ ... And the quantifier-variable notation serves mathematicians brilliantly to regiment statements of multiple generality and make them utterly unambiguous and transparent. (It is when we come to arguments involving generality that borrowing notation from logic to use in our mathematical English becomes really helpful.)
Quantifiers matter to mathematicians, then: that should be entirely uncontentious. OK, so now think about restricted quantifiers that talk about only some of a domain (e.g. talk not about all numbers but just about all the even ones). How might we render Goldbach's Conjecture, say? As a first step, we might write
$\forall n$(if $n$ is even and greater than 2, then $n$ is the sum of two primes)
Note then, we restrict the universal quantifier by using a conditional. So now think about the embedded conditional here.
What if $n$ is odd, so the antecedent of the conditional is false. If we say this instance of the conditional lacks a truth-value, or may be false, then the quantification would have non-true instances and so would not be true! But of course we can't refute Goldbach's Conjecture by looking at odd numbers!! So in these cases, if the quantified conditional is indeed to come out true when Goldbach is right, then we'll have to say that the irrelevant instances of the conditional with a false antecedent come out true by default. Come out "vacuously" true, if you like. In other words, the embedded conditional will have to be treated as a material conditional which is true when the antecedent is false.
So: to put it a bit tendentiously and over-briefly, if mathematicians are to deal nicely with expressions of generality using the quantifier-variable notation they have come to know and love, they will have to get used to using material conditionals too.
• Peter, my mistake. What you point out in your comment is correct: you referred to "a" central consideration, not to "the" central consideration i.e. you asserted "$\exists$... vs. $(\exists!)$... central consideration!". I'll delete my comment accordingly and add that I found your post to be "a central" and enriching answer to the post. Commented Nov 8, 2012 at 14:42
• @amWhy No problem -- so I'll delete my comment too! :-) Commented Nov 8, 2012 at 16:30
• I think, this is indeed a key point. Therefore I accept this answer, even though I feel there is more to it. I also find the answer given here, as also mentioned by amWhy, quite enlightening. Commented Nov 15, 2012 at 7:47
• I agree to answer give by Peter Smith which is accepted as correct answer. I just want to mention that, It took me a while to understand, interpret and comprehend what Peter Smith has to say in his answer. If you really want to get the gist of correct answer then I would recommend to go through following link - gowers.wordpress.com/2011/09/28/basic-logic-connectives-implies If you first read the above given link and then come back and re-read the correct answer by Peter, you will be certainly enlightened :). Commented Sep 22, 2013 at 8:00
Perhaps this will help to capture the truth-functional character of material implication:
The truth-value of an inclusion (subset) relation between sets corresponds to the truth-value of an implication relation, where $\subseteq$ corresponds to the $\rightarrow$ relation.
E.g., suppose $A\subseteq B$. Then if it is true that $x\in A$, then it must be true that $x\in B$, since $B$ contains $A$. However, if $x\notin A$ (if it is false that $x \in A$), it does not mean that then $x\notin B$, since if $A\subseteq B$, then $B$ may very well contain elements that $A$ does not contain.
Similarly, suppose we have that $p\rightarrow q$. If $p$ is true, then it must be the case that $q$ is true. But if $p$ is false, that does not necessarily mean then that $q$ is necessarily false. (For all we know, perhaps $q$ is true regardless of whether or not $p$ is true.) So $q$ can be true, while $p$ is false.
I don't know if this analogy helps or not. But it was the above analogy (correspondence) that helped me to firmly grasp the logic of material implication.
Here's a more down-to-earth example you may have already stumbled upon:
CLAIM: "If (it rains), then (I'll take an umbrella)":
I'd be lying (my assertion would be false) if (it rains = true), and I do not (take an umbrella).
But perhaps it's cloudy out, and I decide I'll take an umbrella , just in case it rains. In this case:
If it doesn't rain (it rains = false), but I took my umbrella (true), my claim above would not be a lie (it would not be false).
An intuitive way to understand the material conditional is as a promise. If you hold up your end of the deal (so the antecedent is true), then I must hold up mine (the conclusion is true). But if you break the promise, then I can do what ever I want without me breaking the promise. And if I am going to hold up my end of the bargain no matter what, it doesn't particularly matter if you hold up yours.
And as far as why it works so well in mathematics, it might be because it is truth functional and mimics some part of what implication means in a non-mathematical context. It doesn't have anything to do with causality, which is often how it is used out of math; it only relies on the truth values of the constituent statements.
The issue seems to be with the behaviour of $p\to q$ when $p$ is false. If $p\to q$ were false when $p$ is false, then you could conclude $p$ from $p\to q$ without any extra premises, and therefore guarantee $q$ as well. That betrays the idea of this being a conditional.
If (you stole the cookie) then (you're a horrible person)
Just from this, which most people with a sweet tooth would accept, I could then conclude that you stole the cookie and are in fact a horrible person. I don't need to assert that you stole the cookie separately, I can just conclude it from the truth of this statement. This behaviour loses the important "if" part of the conditional. To capture
if $P$ (happened), then $Q$ (would happen)
we need to allow for vacuous truths. As a positive example:
If (you won the bet) then (I would have paid you)
Is still true whether you actually win the bet or not.
The material conditional P => Q expresses an ordering relationship among two statements such that Q is "not less true" than P. It is only concerned with comparing truth values and not with what P and Q mean nor how they are related. Its use is intended to prevent us from starting with true assumptions and reaching false conclusions.
How we know that Q is at least as true as P is a different matter. The bare statement P => Q says nothing about how we know it is true; whether it is a bit of useless trivia, a useful working assumption, or a derived conclusion.
The material conditional is often simply defined as follows:
$$A \implies B ~~\equiv ~~ \neg (A \land \neg B)$$
This "definition" turns out to be a theorem of classical propositional logic that can be derived from what might be called "first principles" using a form of natural deduction.
First, we need to prove that $$(A\implies B) \implies \neg (A \land \neg B)$$
1. $$A\implies B~~~~$$ (Assume)
2. $$A ~\land ~\neg B~~~~$$ (Assume, to obtain a contradiction)
3. $$A~~~~$$ (Elim $$\land$$, 2a)
4. $$\neg B~~~~$$ (Elim $$\land$$, 2b)
5. $$B~~~~$$ (Elim $$\implies$$, 1, 3)
6. $$B \land \neg B ~~~~$$ (Intro $$\land$$, 5, 4)
7. $$\neg (A ~\land ~\neg B)~~~~$$ (Intro $$\neg$$, 2, 6)
8. $$(A\implies B) \implies \neg (A \land \neg B)~~~~$$ (Intro $$\implies$$, 1, 7)
Now, we need to prove that $$\neg (A \land \neg B)\implies(A\implies B)$$
1. $$\neg (A \land \neg B)~~~~$$ (Assume)
2. $$A~~~~$$ (Assume)
3. $$\neg B~~~~$$ (Assume, to obtain a contradiction)
4. $$A \land \neg B~~~~$$ (Intro $$\land$$, 2, 3)
5. $$(A \land \neg B) \land \neg (A \land \neg B)~~~~$$ (Intro $$\land$$, 4, 1)
6. $$\neg \neg B~~~~$$ (Intro $$\neg$$, 3, 5)
7. $$B~~~~$$ (Elim $$\neg$$, 6)
8. $$A \implies B~~~~$$ (Intro $$\implies$$, 2,7)
9. $$\neg (A \land \neg B)\implies(A\implies B)~~~~$$ (Intro $$\implies$$, 1, 8)
Combining these two results, we have as required:
$$A \implies B ~~\equiv ~~ \neg (A \land \neg B)$$
Note that the only properties of the implication operator ($$\implies$$) that are used here are:
1. Introduction of the $$\implies$$ operator by means of conditional (or direct) proof resulting in the discharging of an active premise.
2. Elimination of the $$\implies$$ operator by detachment (modus ponens)
• Yes. And also thank you for the comment on this you wrote a couple of months ago. I noted and thought about it. This is an extremely valuable answer, adding something new to the discussion. This motivation may be a tad too formal for freshmen, but the fact that the truth table of material implication is already forced in classical logic by the deduction rules we expect of it is in my opinion indeed the most powerful explanation I’ve seen. I mean, it just explains it. Commented Dec 23, 2022 at 1:31
I would like to show that the truth table of the " IF...THEN" operator is not so far from usual natural language conventions.
Remark. Here by " if ...then" I only talk about material implication ( not about logical/strict implication, that is " necessarily (if A then B) " ).
Let us adopt ( if you please) the following principles :
(1) "I cannot have said something wrong about a thing/ subject I did not talk about."
(2) When I talk in a declarative manner , either what I say is wrong/false, or what I say is "ok", "right", "true".
Now suppose I say " IF the winner of the next US Presidential Elections is a democrat, THEN this winner will be a woman."
What did I talk about? I only talked about the case in which the winner is a democrat.
And what did I say about this (possible) situation? I said that this situation will not be accompanied with a situation in which the winner is not a woman.
So the only case in which others are entitled to tell me that I said something wrong is the case in which (1) the winner is a democrat (2) but is not a woman.
In case the winner is a democrat and is a woman: I talked about this situation, and what I said is true.
In case the winner is not a democrat: I did not talk about this case, so, according to our two initial principles, what I said cannot be false relatively to this case, and consequently, my conditional assertion is true.
Remark. - I do not say that the corresponding logical implication is true in cases TT, TF and FF. Though I am right as long as it is not the case that the winner is a democrat and is not a woman, that does not mean that I would be right in saying : if the winner is a democrat, necessarily it will be a woman. This would amount to saying that the case " democrat + not a woman" is impossible. A much stronger claim than the one I have actually made
Sure you can have a valid "implication relationship" between two errant pieces of nonsense.
If we have M=>H between moon and hamster populations then the truth table is M=>H == not(M and not H). In modal logic (many-world) terms this means we allow all worlds to be possible EXCEPT the one that has a cheesy moon and at least one non-green hamster.
This is the main effect of the implication relationship, it acts as a prohibition on one case, indeed this is the essence of an asymmetric dependency. M=>H means M depends on H in the sense that we cannot have M without H. If H is false then M cannot be true. Asymmetric dependency relationships are very common, hence the utility of this connector.
In the other direction it matters not, there are at least two worlds that have a normal moon, and in only one of them the hamsters are all green. You are NOT forced to live in that world!
The Wason selection task is a good illustration of utility of material conditional support for deductive reasoning: | 3,713 | 15,141 | {"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": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-30 | latest | en | 0.951356 |
http://universaldependencies.org/docs/en/dep/nummod.html | 1,643,217,552,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00261.warc.gz | 74,558,218 | 4,174 | ## `nummod`: numeric modifier
A numeric modifier of a noun is any number phrase that serves to modify the meaning of the noun with a quantity.
``````Sam ate 3 sheep
nummod(sheep, 3)
``````
``````Sam spent forty dollars
nummod(dollars, forty)
``````
``````Sam spent \$ 40
nummod(\$, 40)
``````
## Treebank Statistics (UD_English)
This relation is universal.
3092 nodes (1%) are attached to their parents as `nummod`.
2147 instances of `nummod` (69%) are right-to-left (child precedes parent). Average distance between parent and child is 1.49676584734799.
The following 26 pairs of parts of speech are connected with `nummod`: en-pos/NOUN-en-pos/NUM (1966; 64% instances), en-pos/PROPN-en-pos/NUM (531; 17% instances), en-pos/SYM-en-pos/NUM (301; 10% instances), en-pos/NUM-en-pos/NUM (91; 3% instances), en-pos/VERB-en-pos/X (56; 2% instances), en-pos/PROPN-en-pos/NOUN (34; 1% instances), en-pos/PUNCT-en-pos/X (32; 1% instances), en-pos/NOUN-en-pos/NOUN (16; 1% instances), en-pos/ADJ-en-pos/X (13; 0% instances), en-pos/NOUN-en-pos/X (11; 0% instances), en-pos/NOUN-en-pos/DET (8; 0% instances), en-pos/NOUN-en-pos/ADV (6; 0% instances), en-pos/PROPN-en-pos/X (6; 0% instances), en-pos/ADJ-en-pos/NUM (4; 0% instances), en-pos/NOUN-en-pos/ADJ (4; 0% instances), en-pos/PROPN-en-pos/PROPN (2; 0% instances), en-pos/X-en-pos/X (2; 0% instances), en-pos/ADJ-en-pos/NOUN (1; 0% instances), en-pos/ADV-en-pos/X (1; 0% instances), en-pos/NOUN-en-pos/SYM (1; 0% instances), en-pos/NUM-en-pos/ADV (1; 0% instances), en-pos/PRON-en-pos/NUM (1; 0% instances), en-pos/SYM-en-pos/ADJ (1; 0% instances), en-pos/SYM-en-pos/SYM (1; 0% instances), en-pos/VERB-en-pos/NUM (1; 0% instances), en-pos/X-en-pos/NOUN (1; 0% instances).
``````# visual-style 1 bgColor:blue
# visual-style 1 fgColor:white
# visual-style 2 bgColor:blue
# visual-style 2 fgColor:white
# visual-style 2 1 nummod color:blue
1 Four four NUM CD NumType=Card 2 nummod _ _
2 months month NOUN NNS Number=Plur 3 nmod:npmod _ _
4 , , PUNCT , _ 7 punct _ _
5 we we PRON PRP Case=Nom|Number=Plur|Person=1|PronType=Prs 7 nsubjpass _ _
6 were be AUX VBD Mood=Ind|Tense=Past|VerbForm=Fin 7 auxpass _ _
7 married marry VERB VBN Tense=Past|VerbForm=Part|Voice=Pass 0 root _ SpaceAfter=No
8 . . PUNCT . _ 7 punct _ _
``````
``````# visual-style 8 bgColor:blue
# visual-style 8 fgColor:white
# visual-style 7 bgColor:blue
# visual-style 7 fgColor:white
# visual-style 7 8 nummod color:blue
1 Darin Darin PROPN NNP Number=Sing 2 name _ _
2 Fisher Fisher PROPN NNP Number=Sing 3 nsubj _ _
3 wrote write VERB VBD Mood=Ind|Tense=Past|VerbForm=Fin 0 root _ _
4 this this DET DT Number=Sing|PronType=Dem 5 det _ _
5 response response NOUN NN Number=Sing 3 dobj _ _
6 on on ADP IN _ 7 case _ _
7 January January PROPN NNP Number=Sing 3 nmod _ _
8 25 25 NUM CD NumType=Card 7 nummod _ SpaceAfter=No
9 , , PUNCT , _ 7 punct _ _
10 2005 2005 NUM CD NumType=Card 7 nummod _ SpaceAfter=No
11 : : PUNCT : _ 3 punct _ _
``````
``````# visual-style 2 bgColor:blue
# visual-style 2 fgColor:white
# visual-style 1 bgColor:blue
# visual-style 1 fgColor:white
# visual-style 1 2 nummod color:blue
1 \$ \$ SYM \$ _ 0 root _ SpaceAfter=No
2 5250 5250 NUM CD NumType=Card 1 nummod _ _
3 Deposited deposit VERB VBN Tense=Past|VerbForm=Part 1 advcl _ _ | 1,289 | 3,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | longest | en | 0.73965 |
http://www.jiskha.com/display.cgi?id=1283705778 | 1,498,271,309,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320209.66/warc/CC-MAIN-20170624013626-20170624033626-00192.warc.gz | 555,657,635 | 4,218 | # Math
posted by .
Jane is 2mi offshore in a boat and wishes to reach a coastal village 6mi down a straight shoreline from the point nearest the boat. She can row 2mph and can walk 5mph. Where should she land her boat to reach the village in the least amount of time?
I'm totally stumped by this question. So far I have the picture of the problem, and that's about it. Help please?!
Thank you
• Math -
This is a good Calculus question.
A - position of boat
B - point on shore closest to A
C- point on shore where he should row to (our answer)
V - position of the village
Clearly ABC is a right angle,
let BC = x, then CV = 6-x
AB = 2
AC = √(x^2 + 4)
Time in water = √(x^2 + 4)/2
time along shore = (6-x)/5
Time = (1/2)(x^2 + 4)^(1/2) + 6/5 - x/5
dTime/dx = (1/4)(x^2+4)^(-1/2)(2x) - 1/5
set this equal to zero and solve for x
let me know what you got.
• Math -
4/sqrt(21)? or 0.87287 | 296 | 898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-26 | latest | en | 0.93788 |
https://www.usacoverage.com/life-insurance/pretax-return-on-revenue.html | 1,701,436,204,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00061.warc.gz | 1,175,065,720 | 19,291 | # Pretax Return on Revenue
A return on revenue (ROR) is a means of measuring the company’s profitability from one year to another in order to determine if the company is doing well in the business industry by using a simple formula. This is calculated by means of dividing the net income without interest to the revenue. In terms of insurance figures of pretax return on revenue, this is done by dividing the operating earnings without interest and taxes by the net premiums earned. The only difference between the revenue and the net income is the expenses in such a way that the increase in ROR will describe that there are less expense to a greater net income. Through this, it is possible to identify and recognize the expenses made by the company in the business. This method is usually used by moist companies in order to determine the changes when it comes to the company’s profitability from period to period such as a yearly analysis.
The main goal of many companies is to increase to its heights the profits from one period to another. While taking into account that the net income itself provides the idea if the goals are being achieved though it does not give the whole picture of it. By the resulting quotient of the net income and the revenue, it is now possible to view the expenses obtained during that certain period and also determine how these expenses created impact on the income. In short, this method is being used by the company to determine if income return is increased, stayed the same as the usual or decreased from the previous time compared to last year and even to certain period of like 7 years ago.
What company owners want to see is the flow of the business which will indicate of at least a small increase in the revenue’s return. If the figure will show an increase more than the previous period, it clearly shows that the company’s expenses are facilitated efficiently and is producing outstanding net profit. If in event that the figure will show that there is a decrease on the return on revenue, this is a concrete sign that the expenses made by the company are not being managed well as time past. These pieces of information may provide the company a greater picture of the expenses that needs to be minimized without a stake of damaging the good quality of the products or the decline of meeting the demands of the customers in terms of production level.
Most of the companies usually compare its return on revue from one year to another. In some businesses, this kind of comparison may be done on semi-annual basis to allow the business see its status from six months to another 6 months. Depending on the company’s choice on the period, return on revenue is such an important factor to determine if business is taking its step forward and if there are certain aspect in the operation of the business that should be addressed before the total drop down occurs. | 560 | 2,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.969321 |
https://www.kodytools.com/units/power/from/btuph/to/btupmin | 1,726,645,557,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00827.warc.gz | 771,811,387 | 17,312 | # BTU/Hour to BTU/Minute Converter
1 BTU/Hour = 0.016666666666667 BTU/Minute
## One BTU/Hour is Equal to How Many BTU/Minute?
The answer is one BTU/Hour is equal to 0.016666666666667 BTU/Minute and that means we can also write it as 1 BTU/Hour = 0.016666666666667 BTU/Minute. Feel free to use our online unit conversion calculator to convert the unit from BTU/Hour to BTU/Minute. Just simply enter value 1 in BTU/Hour and see the result in BTU/Minute.
Manually converting BTU/Hour to BTU/Minute can be time-consuming,especially when you don’t have enough knowledge about Power units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online BTU/Hour to BTU/Minute converter tool to get the job done as soon as possible.
We have so many online tools available to convert BTU/Hour to BTU/Minute, but not every online tool gives an accurate result and that is why we have created this online BTU/Hour to BTU/Minute converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert BTU/Hour to BTU/Minute (btu/h to btu/min)
By using our BTU/Hour to BTU/Minute conversion tool, you know that one BTU/Hour is equivalent to 0.016666666666667 BTU/Minute. Hence, to convert BTU/Hour to BTU/Minute, we just need to multiply the number by 0.016666666666667. We are going to use very simple BTU/Hour to BTU/Minute conversion formula for that. Pleas see the calculation example given below.
$$\text{1 BTU/Hour} = 1 \times 0.016666666666667 = \text{0.016666666666667 BTU/Minute}$$
## What Unit of Measure is BTU/Hour?
BTU/hour or BTU per hour is a unit of measurement for power. One btu/hour is equal to 0.2930711 watts.
## What is the Symbol of BTU/Hour?
The symbol of BTU/Hour is btu/h. This means you can also write one BTU/Hour as 1 btu/h.
## What Unit of Measure is BTU/Minute?
BTU/minute or BTU per minute is a unit of measurement for power. One btu/minute is equal to 60 btu/hour.
## What is the Symbol of BTU/Minute?
The symbol of BTU/Minute is btu/min. This means you can also write one BTU/Minute as 1 btu/min.
## How to Use BTU/Hour to BTU/Minute Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select BTU/Hour and in the first input field, enter a value.
• From the second dropdown, select BTU/Minute.
• Instantly, the tool will convert the value from BTU/Hour to BTU/Minute and display the result in the second input field.
## Example of BTU/Hour to BTU/Minute Converter Tool
BTU/Hour
1
BTU/Minute
0.016666666666667
# BTU/Hour to BTU/Minute Conversion Table
BTU/Hour [btu/h]BTU/Minute [btu/min]Description
1 BTU/Hour0.016666666666667 BTU/Minute1 BTU/Hour = 0.016666666666667 BTU/Minute
2 BTU/Hour0.033333333333333 BTU/Minute2 BTU/Hour = 0.033333333333333 BTU/Minute
3 BTU/Hour0.05 BTU/Minute3 BTU/Hour = 0.05 BTU/Minute
4 BTU/Hour0.066666666666667 BTU/Minute4 BTU/Hour = 0.066666666666667 BTU/Minute
5 BTU/Hour0.083333333333333 BTU/Minute5 BTU/Hour = 0.083333333333333 BTU/Minute
6 BTU/Hour0.1 BTU/Minute6 BTU/Hour = 0.1 BTU/Minute
7 BTU/Hour0.11666666666667 BTU/Minute7 BTU/Hour = 0.11666666666667 BTU/Minute
8 BTU/Hour0.13333333333333 BTU/Minute8 BTU/Hour = 0.13333333333333 BTU/Minute
9 BTU/Hour0.15 BTU/Minute9 BTU/Hour = 0.15 BTU/Minute
10 BTU/Hour0.16666666666667 BTU/Minute10 BTU/Hour = 0.16666666666667 BTU/Minute
100 BTU/Hour1.67 BTU/Minute100 BTU/Hour = 1.67 BTU/Minute
1000 BTU/Hour16.67 BTU/Minute1000 BTU/Hour = 16.67 BTU/Minute
# BTU/Hour to Other Units Conversion Table
ConversionDescription
1 BTU/Hour = 2.9307107017222e-10 Gigawatt1 BTU/Hour in Gigawatt is equal to 2.9307107017222e-10
1 BTU/Hour = 0.00039301477894522 Horsepower1 BTU/Hour in Horsepower is equal to 0.00039301477894522
1 BTU/Hour = 0.00029307107017222 Kilowatt1 BTU/Hour in Kilowatt is equal to 0.00029307107017222
1 BTU/Hour = 2.9307107017222e-7 Megawatt1 BTU/Hour in Megawatt is equal to 2.9307107017222e-7
1 BTU/Hour = 2.9307107017222e-16 Petawatt1 BTU/Hour in Petawatt is equal to 2.9307107017222e-16
1 BTU/Hour = 2.9307107017222e-13 Terawatt1 BTU/Hour in Terawatt is equal to 2.9307107017222e-13
1 BTU/Hour = 0.29307107017222 Watt1 BTU/Hour in Watt is equal to 0.29307107017222
1 BTU/Hour = 293071070172220000 Attowatt1 BTU/Hour in Attowatt is equal to 293071070172220000
1 BTU/Hour = 0.016666666666667 BTU/Minute1 BTU/Hour in BTU/Minute is equal to 0.016666666666667
1 BTU/Hour = 0.00027777777777778 BTU/Second1 BTU/Hour in BTU/Second is equal to 0.00027777777777778
1 BTU/Hour = 252 Calorie/Hour1 BTU/Hour in Calorie/Hour is equal to 252
1 BTU/Hour = 4.2 Calorie/Minute1 BTU/Hour in Calorie/Minute is equal to 4.2
1 BTU/Hour = 0.069998822530864 Calorie/Second1 BTU/Hour in Calorie/Second is equal to 0.069998822530864
1 BTU/Hour = 29.31 Centiwatt1 BTU/Hour in Centiwatt is equal to 29.31
1 BTU/Hour = 0.00039846576241254 Cheval Vapeur1 BTU/Hour in Cheval Vapeur is equal to 0.00039846576241254
1 BTU/Hour = 225439.28 Clusec1 BTU/Hour in Clusec is equal to 225439.28
1 BTU/Hour = 2.93 Deciwatt1 BTU/Hour in Deciwatt is equal to 2.93
1 BTU/Hour = 0.029307107017222 Dekawatt1 BTU/Hour in Dekawatt is equal to 0.029307107017222
1 BTU/Hour = 10550558526.2 Dyne Centimeter/Hour1 BTU/Hour in Dyne Centimeter/Hour is equal to 10550558526.2
1 BTU/Hour = 175842642.1 Dyne Centimeter/Minute1 BTU/Hour in Dyne Centimeter/Minute is equal to 175842642.1
1 BTU/Hour = 2930710.7 Dyne Centimeter/Second1 BTU/Hour in Dyne Centimeter/Second is equal to 2930710.7
1 BTU/Hour = 10550558526.2 Erg/Hour1 BTU/Hour in Erg/Hour is equal to 10550558526.2
1 BTU/Hour = 175842642.1 Erg/Minute1 BTU/Hour in Erg/Minute is equal to 175842642.1
1 BTU/Hour = 2930710.7 Erg/Second1 BTU/Hour in Erg/Second is equal to 2930710.7
1 BTU/Hour = 2.9307107017222e-19 Exawatt1 BTU/Hour in Exawatt is equal to 2.9307107017222e-19
1 BTU/Hour = 293071070172220 Femtowatt1 BTU/Hour in Femtowatt is equal to 293071070172220
1 BTU/Hour = 778.17 Foot Pound/Hour1 BTU/Hour in Foot Pound/Hour is equal to 778.17
1 BTU/Hour = 12.97 Foot Pound/Minute1 BTU/Hour in Foot Pound/Minute is equal to 12.97
1 BTU/Hour = 0.21615812932898 Foot Pound/Second1 BTU/Hour in Foot Pound/Second is equal to 0.21615812932898
1 BTU/Hour = 25036.86 Foot Poundal/Hour1 BTU/Hour in Foot Poundal/Hour is equal to 25036.86
1 BTU/Hour = 417.28 Foot Poundal/Minute1 BTU/Hour in Foot Poundal/Minute is equal to 417.28
1 BTU/Hour = 6.95 Foot Poundal/Second1 BTU/Hour in Foot Poundal/Second is equal to 6.95
1 BTU/Hour = 10758575.59 Gram Force Centimeter/Hour1 BTU/Hour in Gram Force Centimeter/Hour is equal to 10758575.59
1 BTU/Hour = 179309.59 Gram Force Centimeter/Minute1 BTU/Hour in Gram Force Centimeter/Minute is equal to 179309.59
1 BTU/Hour = 2988.49 Gram Force Centimeter/Second1 BTU/Hour in Gram Force Centimeter/Second is equal to 2988.49
1 BTU/Hour = 0.0029307107017222 Hectowatt1 BTU/Hour in Hectowatt is equal to 0.0029307107017222
1 BTU/Hour = 0.00039285666242925 Horsepower [Electrical]1 BTU/Hour in Horsepower [Electrical] is equal to 0.00039285666242925
1 BTU/Hour = 0.00039846576241254 Horsepower [Metric]1 BTU/Hour in Horsepower [Metric] is equal to 0.00039846576241254
1 BTU/Hour = 0.0003928340191815 Horsepower [Water]1 BTU/Hour in Horsepower [Water] is equal to 0.0003928340191815
1 BTU/Hour = 396.31 Inch Ounce-Force Revolution/Minute1 BTU/Hour in Inch Ounce-Force Revolution/Minute is equal to 396.31
1 BTU/Hour = 1055.06 Joule/Hour1 BTU/Hour in Joule/Hour is equal to 1055.06
1 BTU/Hour = 17.58 Joule/Minute1 BTU/Hour in Joule/Minute is equal to 17.58
1 BTU/Hour = 0.29307107017222 Joule/Second1 BTU/Hour in Joule/Second is equal to 0.29307107017222
1 BTU/Hour = 0.25199576111111 Kilocalorie/Hour1 BTU/Hour in Kilocalorie/Hour is equal to 0.25199576111111
1 BTU/Hour = 0.0041999293518519 Kilocalorie/Minute1 BTU/Hour in Kilocalorie/Minute is equal to 0.0041999293518519
1 BTU/Hour = 0.000069998822530864 Kilocalorie/Second1 BTU/Hour in Kilocalorie/Second is equal to 0.000069998822530864
1 BTU/Hour = 107.59 Kilogram Force Meter/Hour1 BTU/Hour in Kilogram Force Meter/Hour is equal to 107.59
1 BTU/Hour = 1.79 Kilogram Force Meter/Minute1 BTU/Hour in Kilogram Force Meter/Minute is equal to 1.79
1 BTU/Hour = 0.029884932180941 Kilogram Force Meter/Second1 BTU/Hour in Kilogram Force Meter/Second is equal to 0.029884932180941
1 BTU/Hour = 107.59 Kilopond Meter/Hour1 BTU/Hour in Kilopond Meter/Hour is equal to 107.59
1 BTU/Hour = 1.79 Kilopond Meter/Minute1 BTU/Hour in Kilopond Meter/Minute is equal to 1.79
1 BTU/Hour = 0.029884932180941 Kilopond Meter/Second1 BTU/Hour in Kilopond Meter/Second is equal to 0.029884932180941
1 BTU/Hour = 0.000001 MMBTU/Hour1 BTU/Hour in MMBTU/Hour is equal to 0.000001
1 BTU/Hour = 293071.07 Microwatt1 BTU/Hour in Microwatt is equal to 293071.07
1 BTU/Hour = 293.07 Milliwatt1 BTU/Hour in Milliwatt is equal to 293.07
1 BTU/Hour = 293071070.17 Nanowatt1 BTU/Hour in Nanowatt is equal to 293071070.17
1 BTU/Hour = 1055.06 Newton Meter/Hour1 BTU/Hour in Newton Meter/Hour is equal to 1055.06
1 BTU/Hour = 17.58 Newton Meter/Minute1 BTU/Hour in Newton Meter/Minute is equal to 17.58
1 BTU/Hour = 0.29307107017222 Newton Meter/Second1 BTU/Hour in Newton Meter/Second is equal to 0.29307107017222
1 BTU/Hour = 0.00039846576241254 Pferdestarke1 BTU/Hour in Pferdestarke is equal to 0.00039846576241254
1 BTU/Hour = 293071070172.22 Picowatt1 BTU/Hour in Picowatt is equal to 293071070172.22
1 BTU/Hour = 0.00029884932180941 Poncelet1 BTU/Hour in Poncelet is equal to 0.00029884932180941
1 BTU/Hour = 6.95 Pound Square Foot/Cubic Second1 BTU/Hour in Pound Square Foot/Cubic Second is equal to 6.95
1 BTU/Hour = 0.000083333333382304 Ton of Refrigeration1 BTU/Hour in Ton of Refrigeration is equal to 0.000083333333382304
1 BTU/Hour = 0.29307107017222 Volt Ampere1 BTU/Hour in Volt Ampere is equal to 0.29307107017222
1 BTU/Hour = 2.9307107017222e+23 Yoctowatt1 BTU/Hour in Yoctowatt is equal to 2.9307107017222e+23
1 BTU/Hour = 2.9307107017222e-25 Yottawatt1 BTU/Hour in Yottawatt is equal to 2.9307107017222e-25
1 BTU/Hour = 293071070172220000000 Zeptowatt1 BTU/Hour in Zeptowatt is equal to 293071070172220000000
1 BTU/Hour = 2.9307107017222e-22 Zettawatt1 BTU/Hour in Zettawatt is equal to 2.9307107017222e-22
1 BTU/Hour = 0.001 Mega BTU/Hour1 BTU/Hour in Mega BTU/Hour is equal to 0.001
1 BTU/Hour = 1 BTU/Hour [TH]1 BTU/Hour in BTU/Hour [TH] is equal to 1
1 BTU/Hour = 0.016677819705968 BTU/Minute [TH]1 BTU/Hour in BTU/Minute [TH] is equal to 0.016677819705968
1 BTU/Hour = 0.00027796366176613 BTU/Second [TH]1 BTU/Hour in BTU/Second [TH] is equal to 0.00027796366176613
1 BTU/Hour = 252.16 Calorie/Hour [TH]1 BTU/Hour in Calorie/Hour [TH] is equal to 252.16
1 BTU/Hour = 4.2 Calorie/Minute [TH]1 BTU/Hour in Calorie/Minute [TH] is equal to 4.2
1 BTU/Hour = 0.070045666867166 Calorie/Second [TH]1 BTU/Hour in Calorie/Second [TH] is equal to 0.070045666867166
1 BTU/Hour = 0.00029307107017222 Kilovolt Ampere1 BTU/Hour in Kilovolt Ampere is equal to 0.00029307107017222
1 BTU/Hour = 1.05505585262e-15 Exajoule/Hour1 BTU/Hour in Exajoule/Hour is equal to 1.05505585262e-15
1 BTU/Hour = 1.7584264210333e-17 Exajoule/Minute1 BTU/Hour in Exajoule/Minute is equal to 1.7584264210333e-17
1 BTU/Hour = 2.9307107017222e-19 Exajoule/Second1 BTU/Hour in Exajoule/Second is equal to 2.9307107017222e-19
1 BTU/Hour = 1.05505585262e-12 Petajoule/Hour1 BTU/Hour in Petajoule/Hour is equal to 1.05505585262e-12
1 BTU/Hour = 1.7584264210333e-14 Petajoule/Minute1 BTU/Hour in Petajoule/Minute is equal to 1.7584264210333e-14
1 BTU/Hour = 2.9307107017222e-16 Petajoule/Second1 BTU/Hour in Petajoule/Second is equal to 2.9307107017222e-16
1 BTU/Hour = 1.05505585262e-9 Terajoule/Hour1 BTU/Hour in Terajoule/Hour is equal to 1.05505585262e-9
1 BTU/Hour = 1.7584264210333e-11 Terajoule/Minute1 BTU/Hour in Terajoule/Minute is equal to 1.7584264210333e-11
1 BTU/Hour = 2.9307107017222e-13 Terajoule/Second1 BTU/Hour in Terajoule/Second is equal to 2.9307107017222e-13
1 BTU/Hour = 0.00000105505585262 Gigajoule/Hour1 BTU/Hour in Gigajoule/Hour is equal to 0.00000105505585262
1 BTU/Hour = 1.7584264210333e-8 Gigajoule/Minute1 BTU/Hour in Gigajoule/Minute is equal to 1.7584264210333e-8
1 BTU/Hour = 2.9307107017222e-10 Gigajoule/Second1 BTU/Hour in Gigajoule/Second is equal to 2.9307107017222e-10
1 BTU/Hour = 0.00105505585262 Megajoule/Hour1 BTU/Hour in Megajoule/Hour is equal to 0.00105505585262
1 BTU/Hour = 0.000017584264210333 Megajoule/Minute1 BTU/Hour in Megajoule/Minute is equal to 0.000017584264210333
1 BTU/Hour = 2.9307107017222e-7 Megajoule/Second1 BTU/Hour in Megajoule/Second is equal to 2.9307107017222e-7
1 BTU/Hour = 1.06 Kilojoule/Hour1 BTU/Hour in Kilojoule/Hour is equal to 1.06
1 BTU/Hour = 0.017584264210333 Kilojoule/Minute1 BTU/Hour in Kilojoule/Minute is equal to 0.017584264210333
1 BTU/Hour = 0.00029307107017222 Kilojoule/Second1 BTU/Hour in Kilojoule/Second is equal to 0.00029307107017222
1 BTU/Hour = 10.55 Hectojoule/Hour1 BTU/Hour in Hectojoule/Hour is equal to 10.55
1 BTU/Hour = 0.17584264210333 Hectojoule/Minute1 BTU/Hour in Hectojoule/Minute is equal to 0.17584264210333
1 BTU/Hour = 0.0029307107017222 Hectojoule/Second1 BTU/Hour in Hectojoule/Second is equal to 0.0029307107017222
1 BTU/Hour = 105.51 Dekajoule/Hour1 BTU/Hour in Dekajoule/Hour is equal to 105.51
1 BTU/Hour = 1.76 Dekajoule/Minute1 BTU/Hour in Dekajoule/Minute is equal to 1.76
1 BTU/Hour = 0.029307107017222 Dekajoule/Second1 BTU/Hour in Dekajoule/Second is equal to 0.029307107017222
1 BTU/Hour = 10550.56 Decijoule/Hour1 BTU/Hour in Decijoule/Hour is equal to 10550.56
1 BTU/Hour = 175.84 Decijoule/Minute1 BTU/Hour in Decijoule/Minute is equal to 175.84
1 BTU/Hour = 2.93 Decijoule/Second1 BTU/Hour in Decijoule/Second is equal to 2.93
1 BTU/Hour = 105505.59 Centijoule/Hour1 BTU/Hour in Centijoule/Hour is equal to 105505.59
1 BTU/Hour = 1758.43 Centijoule/Minute1 BTU/Hour in Centijoule/Minute is equal to 1758.43
1 BTU/Hour = 29.31 Centijoule/Second1 BTU/Hour in Centijoule/Second is equal to 29.31
1 BTU/Hour = 1055055.85 Millijoule/Hour1 BTU/Hour in Millijoule/Hour is equal to 1055055.85
1 BTU/Hour = 17584.26 Millijoule/Minute1 BTU/Hour in Millijoule/Minute is equal to 17584.26
1 BTU/Hour = 293.07 Millijoule/Second1 BTU/Hour in Millijoule/Second is equal to 293.07
1 BTU/Hour = 1055055852.62 Microjoule/Hour1 BTU/Hour in Microjoule/Hour is equal to 1055055852.62
1 BTU/Hour = 17584264.21 Microjoule/Minute1 BTU/Hour in Microjoule/Minute is equal to 17584264.21
1 BTU/Hour = 293071.07 Microjoule/Second1 BTU/Hour in Microjoule/Second is equal to 293071.07
1 BTU/Hour = 1055055852620 Nanojoule/Hour1 BTU/Hour in Nanojoule/Hour is equal to 1055055852620
1 BTU/Hour = 17584264210.33 Nanojoule/Minute1 BTU/Hour in Nanojoule/Minute is equal to 17584264210.33
1 BTU/Hour = 293071070.17 Nanojoule/Second1 BTU/Hour in Nanojoule/Second is equal to 293071070.17
1 BTU/Hour = 1055055852620000 Picojoule/Hour1 BTU/Hour in Picojoule/Hour is equal to 1055055852620000
1 BTU/Hour = 17584264210333 Picojoule/Minute1 BTU/Hour in Picojoule/Minute is equal to 17584264210333
1 BTU/Hour = 293071070172.22 Picojoule/Second1 BTU/Hour in Picojoule/Second is equal to 293071070172.22
1 BTU/Hour = 1055055852620000000 Femtojoule/Hour1 BTU/Hour in Femtojoule/Hour is equal to 1055055852620000000
1 BTU/Hour = 17584264210333000 Femtojoule/Minute1 BTU/Hour in Femtojoule/Minute is equal to 17584264210333000
1 BTU/Hour = 293071070172220 Femtojoule/Second1 BTU/Hour in Femtojoule/Second is equal to 293071070172220
1 BTU/Hour = 1.05505585262e+21 Attojoule/Hour1 BTU/Hour in Attojoule/Hour is equal to 1.05505585262e+21
1 BTU/Hour = 17584264210333000000 Attojoule/Minute1 BTU/Hour in Attojoule/Minute is equal to 17584264210333000000
1 BTU/Hour = 293071070172220000 Attojoule/Second1 BTU/Hour in Attojoule/Second is equal to 293071070172220000
1 BTU/Hour = 0.00002987624957156 Horsepower [Boiler]1 BTU/Hour in Horsepower [Boiler] is equal to 0.00002987624957156 | 6,294 | 16,019 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.826155 |
https://betselection.cc/roulette-forum/kimo-li-roulette-foundation-numbers/?prev_next=next | 1,571,067,620,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653876.31/warc/CC-MAIN-20191014150930-20191014174430-00476.warc.gz | 420,340,940 | 14,805 | 08 Use Math/Statistics to beat Roulette/Baccarat Part 2.
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### Topic: Use Math/Statistics to beat Roulette/Baccarat Part 2. (Read 16550 times)
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#### Nickmsi
• Sr. Member
• Posts: 283
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #45 on: August 19, 2018, 12:28:49 am »
• Reply
• Hi Ozon,
I tested the Triplets with progression on 10,000 Single Zero spins and on 10,000 No Zero Spins, results per attached.
While both made a profit, the Single Zero was erratic and below profit for 3,000 spins or so.
Perhaps a different progression might help.
Cheers\\Nick
#### Sputnik
• Hero Member
• Posts: 1317
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #46 on: August 20, 2018, 09:26:57 am »
• Reply
• Hello Ozon - Holloway made a progression for bets that have higher strike ratio than 50% for even money bets.
Cheers
#### Sputnik
• Hero Member
• Posts: 1317
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #47 on: August 20, 2018, 02:50:30 pm »
• Reply
• Hello Nickmsi - I will put together one explanation about this selection based upon my opinion and understanding.
What you do is to play against patterns not repeat and the probability or the odds increase for each losing group.
For example, you have Red and play Black twice, the triplet and continue doing so on a rolling basis.
So each random march will end with a losing sequence and that combination is the principal of 1/3 because you bet against triplets.
So when you look at two loses you have one series of three and if you lose again you have another series of three and if you lose again you have a series of three.
They can come in any combination, and the march is to bet twice against triplets.
When you look at the sequence with the vertical perspective you see three patterns of the same formation repeating three times in a row.
And when you look at the sequence with the horizontal perspective you see the triplet that made you lose.
RRB
RRB
RRB
RBR
RBR
RBR
RBB
RBB
RBB
RRR
RRR
RRR
BRB
BRB
BRB
BBR
BBR
BBR
BRR
BRR
BRR
BBB
BBB
BBB
Now two patterns XOO XOX create a direct win and one pattern is a tie OOX and one losing pattern, the triplet XXX
If you continue betting after six loses the probability and the odds increase to 1 in 16 where you get groups of four or patterns with four in a row repeating three times in a row.
RBBR
RBBR
RBBR
And so it continues to grow with equilibrium and probability and odds for each new triplet.
I would say that you should stop after six losses and wait for a fictive win and start over.
The odds and probability is very small to get three repeaters back to back.
Here is also the possibility to experiment with triggers.
Cheers
#### ozon
• Full Member
• Posts: 153
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #48 on: August 21, 2018, 04:34:25 pm »
• Reply
• Hi Sputnik
I have this progression
Probably even slight modifications made by Bayes.
It has 46 steps.
I've never really simulated it.
I wonder if he is able to bring profit even after losing these 46 steps, sometimes.
#### Nickmsi
• Sr. Member
• Posts: 283
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #49 on: August 21, 2018, 09:49:12 pm »
• Reply
• Hi Patrik and Ozon,
Thanks for your insights Patrik and Ozon if you have the 46 Step Holloway progression could you post it here or send it to my
email nickmsi@aol.com.
I am finishing up my vacation with the grandsons and would love to test it with the Triplets. I am sure I have it at home but it would save me some time.
Cheers
Nick
#### BEAT-THE-WHEEL
• Hero Member
• Posts: 514
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #50 on: August 22, 2018, 12:40:14 am »
• Reply
• Sputnik,
Wait for the extreme variance to pass, as your idea, that...wait for 6losses, then a fictive win, and start bet...
#### Sputnik
• Hero Member
• Posts: 1317
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #51 on: August 22, 2018, 07:27:00 am »
• Reply
• Here is Holloway's progression, raise, and fall ...
1
1
1
1
1
2
2
2
2
3
3
3
4
4
5
5
6
7
8
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12
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16
18
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150
#### Nickmsi
• Sr. Member
• Posts: 283
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #52 on: August 22, 2018, 09:35:00 am »
• Reply
• Thanks Patrick, that is just what I needed.
Cheers
Nick
#### Albalaha
• Hero Member
• Posts: 1962
• Gender:
• Learn about randomness before trying to fight that
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #53 on: August 25, 2018, 03:07:00 am »
• Reply
• Hello jsintl
The progression we are using is as follows:
Flat bet for first 100 spins
If in profit after 100 spins, keep flat betting. Always flat bet when in profit.
After 100 spins if the total Losses greater than the total wins, increase bet +1 unit.
Keep bet the same for next 11 spins, then if in profit flat bet, else raise bet + 1 unit
Recheck for new bet every 11 spins.
We are currently winning 97/100 sessions.
The largest bet so far is 29 units.
The biggest drawdown so far is -304.
Total spins played = 68,117
Total Profit = 1,507 units
Average Profit/Session=15 units
Profit/spin=.022
Hope this helps.
Cheers
Nick
Isn't it only a matter of luck that you are in a net profit still? Your progression is slow but is it enough to ward off big losses? Will it not end with a huge loss taking away whatever you have won so far as any other progression does?
Email: earnsumit@gmail.com - VIsit my blog: http://albalaha.lefora.com
#### alrelax
• B&M Player since 1980
• Administrator
• Posts: 3213
• Gender:
• 'Caring for Kids' Nonprofit Children's Assistance
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #54 on: August 25, 2018, 02:52:52 pm »
• Reply
• Isn't it only a matter of luck that you are in a net profit still? Your progression is slow but is it enough to ward off big losses? Will it not end with a huge loss taking away whatever you have won so far as any other progression does?
And what you just said, "Will it not end with a huge loss taking away whatever you have won so far as any other progression does?", is basically what happens almost every single time, everywhere I ever play. Even with the better skilled players.
There are very easy defense tactics for the player, but most if not all, will believe way too much in themselves and be governed only by their own misconception of the game.
My Blog within BetSelection Board: https://betselection.cc/alrelax's-blog/
Played well over 29,555 shoes of baccarat since I started playing at B&M USA casinos.
"Don't say it's a winning hand until you are getting paid for it".
Played numerous properties in Las Vegas, Reno, Southern California, Atlantic City, Connecticut, South Florida, The South/Southeast as well as most areas of The Midwest.
Baccarat, actually a mixture of Watergate, attacking the Gotti Family and the famous ear biting Tyson fight leading to disqualification and a near riot. Bac has all that more.
EMAIL: Betselectionboard@Gmail.Com
#### Nickmsi
• Sr. Member
• Posts: 283
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #55 on: August 25, 2018, 10:49:24 pm »
• Reply
• Hello Sumit, good to hear from you.
And yes, both you and Glen are correct.
It is a negative progression that we first started with and that will fail in long run but what we want to see if what defenses were necessary, ie, what stop loss would be the worse, what was the maximum bets, could a Trailing Profit Stop be employed, etc.
Now we are testing a positive progression as this system wins more than loses and we are still seeing how the defense is doing.
Cheers,
Nick
#### Albalaha
• Hero Member
• Posts: 1962
• Gender:
• Learn about randomness before trying to fight that
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #56 on: August 26, 2018, 03:45:24 am »
• Reply
• And what you just said, "Will it not end with a huge loss taking away whatever you have won so far as any other progression does?", is basically what happens almost every single time, everywhere I ever play. Even with the better skilled players.
There are very easy defense tactics for the player, but most if not all, will believe way too much in themselves and be governed only by their own misconception of the game.
It happens with everybody because everybody uses same erroneous approach of increasing bets without any safeguard. The worst moments wait for the highest bets to get you the inevitable total loss. An empty belief that I won't get that worst moment will keep killing those stereotype progressions.
Sadly, the topic headline doesn't match with what we are discussing herein. We aren't using any kind of maths/stats to improve from where we started. Unless a progression can successfully pass through the virtually possible worst patches without killing itself, it is not playable, in my humble opinion.
Email: earnsumit@gmail.com - VIsit my blog: http://albalaha.lefora.com
#### Johno-Egalite
• Full Member
• Posts: 146
• I used to be known as "MarkTeruya" FYI
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #57 on: September 04, 2018, 11:24:53 pm »
• Reply
• This video very quickly confirms the advantage of the bet selection OLD. If we equate the finding to Baccarat, TH (in essence a chop) is will occur twice as frequent as any double HH, which of course s confirmed by the Zumma stats (50% of all hands are chops v's 25% of all hands being a single repeater).
Just like to add, any and all losing patterns can be defined, we as system players can pick and choose our nemesis pattern, hence we should avoid those methods that fail against the most common occurring patterns, which is why I thought VDW was not a practical option to take to the gaming tables.
Apologies if my response is dated.
Maths is great like that. Once it's been proven that no method exists to do what you claim, it's not necessary to go through the details of your system to prove that it doesn't work. You claim that it does something which can be proven impossible, therefore your claim is false. The details don't matter. I use the names Junket, Junket King, Lugi, Mark Teruya, Rolex, Relex, Rolex Watch, Mark, Eaglite, JohnO & More depending on what day it is and whom I am attempting to be!
#### Sputnik
• Hero Member
• Posts: 1317
##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #58 on: September 06, 2018, 01:19:52 pm »
• Reply
• Nick, I send you a personal message!
Cheers
#### alrelax
• B&M Player since 1980
• Administrator
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• Gender:
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##### Re: Use Math/Statistics to beat Roulette/Baccarat Part 2.
« Reply #59 on: March 05, 2019, 03:29:56 pm »
• Reply
• A lot has been posted by Alrelax and Gizmotron about how they win using their years of experience, knowledge of the game, situational awareness etc.
Others like Xander and Mike have posted that you need an edge.
I am not here to debate the merits of Alrelax and Gizmotron’s methods, but I would like to explore Xander’s and Mikes position of having an EDGE.
What is an EDGE? Can you actually get an EDGE? To me an EDGE would be something mathematical, statistical or physical that would turn the odds or probability in your favor.
The above is a quote from the OP. Yes, you are absolutely correct about wins and losses. You can never be on all sides with an Edge. It is impossible, so--IMO, why attempt to do that. However, trying it leaves room to improve your knowledge of reality the game produces. Problem is, most of us are not in reality and refuse to see it.
It is also like someone saying, that restaurant specializes in seafood and is well known for having the best seafood in town. There is no way they could have a great Asian dish, it is impossible. But, maybe they really do? Maybe their sole Asian dish is horrible. Maybe is great on the days a certain chef is working and off on the other days when that chef is not present?
Same in baccarat. But there is no edge that can give you consistent wins on anything that will redundantly appear.
People attempt to convert computer statistics to live gaming and those statistics were not derived from the amount of hands that you will be playing in front of.
The real edges come from other avenues within the game of baccarat. I don't know about the game of roulette, perhaps so or not? But in baccarat the edges are not definable by mathematical numbers. Disclosure: IMO.
But there are edges.
My Blog within BetSelection Board: https://betselection.cc/alrelax's-blog/
Played well over 29,555 shoes of baccarat since I started playing at B&M USA casinos.
"Don't say it's a winning hand until you are getting paid for it".
Played numerous properties in Las Vegas, Reno, Southern California, Atlantic City, Connecticut, South Florida, The South/Southeast as well as most areas of The Midwest.
Baccarat, actually a mixture of Watergate, attacking the Gotti Family and the famous ear biting Tyson fight leading to disqualification and a near riot. Bac has all that more.
EMAIL: Betselectionboard@Gmail.Com
imPulse2 © Bloc | 3,500 | 13,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2019-43 | latest | en | 0.871099 |
https://www.infoarena.ro/job_detail/12901?action=view-source | 1,611,030,852,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517966.39/warc/CC-MAIN-20210119042046-20210119072046-00296.warc.gz | 852,807,457 | 4,489 | Cod sursa(job #12901)
Utilizator Data 5 februarie 2007 10:50:41 Jocul Flip 20 fpc done Arhiva de probleme 2.54 kb
``````var fi,fo:text;
n,m,i,j:integer;
v,aux:array[1..17,1..17] of longint;
l1,l2,max,peak:longint;
a,b:array[0..35000,1..15] of byte;
function suma:longint;
var i,j:integer;
sum:longint;
begin
sum:=0;
for i:=1 to n do
for j:=1 to m do
sum:=sum+aux[i,j];
suma:=sum;
end;
procedure invcol(x:integer);
var i:integer;
begin
for i:=1 to n do
aux[i,x]:=-aux[i,x];
end;
procedure invlin(x:integer);
var j:integer;
begin
for j:=1 to m do
aux[x,j]:=-aux[x,j];
end;
procedure gosolve;
var i,j,k:longint;
begin
for i:=0 to l1 do
for j:=0 to l2 do
begin
aux:=v;
for k:=1 to n do
if a[i,k]=1 then invlin(k);
for k:=1 to m do
if b[j,k]=1 then invcol(k);
peak:=suma;
if peak>max then max:=peak;
end;
end;
procedure b2I(x:longint);
var int,i,j:longint;
begin
for i:=0 to x do
begin
int:=i;
j:=1;
while int<>0 do
begin
a[i,j]:=int mod 2;
int:=int div 2;
inc(j);
end;
end;
end;
procedure b2II(x:longint);
var int,i,j:longint;
begin
for i:=0 to x do
begin
int:=i;
j:=1;
while int<>0 do
begin
b[i,j]:=int mod 2;
int:=int div 2;
inc(j);
end;
end;
end;
begin
assign(fi,'flip.in'); reset(fi);
assign(fo,'flip.out'); rewrite(fo);
for i:=1 to n do
for j:=1 to m do
case n of
1: l1:=1;
2: l1:=3;
3: l1:=7;
4: l1:=15;
5: l1:=31;
6: l1:=63;
7: l1:=127;
8: l1:=255;
9: l1:=511;
10: l1:=1023;
11: l1:=2047;
12: l1:=4095;
13: l1:=8191;
14: l1:=16383;
15: l1:=32767;
end;
b2I(l1);
case m of
1: l2:=1;
2: l2:=3;
3: l2:=7;
4: l2:=15;
5: l2:=31;
6: l2:=63;
7: l2:=127;
8: l2:=255;
9: l2:=511;
10: l2:=1023;
11: l2:=2047;
12: l2:=4095;
13: l2:=8191;
14: l2:=16383;
15: l2:=32767;
end;
b2II(l2);
{ for i:=0 to l1 do
begin
for j:=1 to n do
write(fo,a[i,j],' ');
writeln(fo);
end;
for i:=0 to l2 do
begin
for j:=1 to m do
write(fo,b[i,j],' ');
writeln(fo);
end; }
gosolve;
writeln(fo,max);
close(fi);
close(fo);
end.
`````` | 838 | 1,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-04 | latest | en | 0.28513 |
http://www.centerspace.net/examples/nmath/csharp/analysis/constrained-least-squares-example.php | 1,638,981,297,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363515.28/warc/CC-MAIN-20211208144647-20211208174647-00303.warc.gz | 96,917,723 | 5,426 | # C# Constrained Least Squares Example
← All NMath Code Examples
```using System;
using CenterSpace.NMath.Core;
using CenterSpace.NMath.Analysis;
namespace CenterSpace.NMath.Analysis.Examples.CSharp
{
/// <summary>
/// A .NET example in C# showing how to use ConstrainedLeastSquares class to
/// solve the constrained least squares problem
/// Cx = d, subject to the constraints
/// Ax < b
/// </summary>
public class ConstrainedLeastSquaresExample
{
static void Main( string[] args )
{
// Solve, in the least squares sense, Cx = d, subect to Ax <= b
// and -0.1 <= x[i] <= 2.0
var C = new DoubleMatrix( "5x4 [0.9501 0.7620 0.6153 0.4057 " +
"0.2311 0.4564 0.7919 0.9354 " +
"0.6068 0.0185 0.9218 0.9169 " +
"0.4859 0.8214 0.7382 0.4102 " +
"0.8912 0.4447 0.1762 0.8936]" );
var d = new DoubleVector( 0.0578, 0.3528, 0.8131, 0.0098, 0.1388 );
// Constraint coefficient matrix
var A = new DoubleMatrix( "3x4[0.2027 0.2721 0.7467 0.4659 " +
"0.1987 0.1988 0.4450 0.4186 " +
"0.6037 0.0152 0.9318 0.8462]" );
// Constraints right hand sides.
var b = new DoubleVector( 0.5251, 0.2026, 0.6721 );
// Create the constrained least squares problem for minimizing
// || Cx - d||^2 subject to Ax <= b and -0.1 <= x[i] <= 2.0
// We first construct the problem object from the matrix C and the
// vector d, we then add the constraints.
var problem = new ConstrainedLeastSquaresProblem( C, d );
// Add the inequality constraints Ax <= b using a constraint tolerance
// of 0.00001. This allows for small violations of the constraints.
// Specifically the constraints will be considered satisfied for a
// vector x if
// Ax <= b + 0.00001
double constraintTolerance = 0.00001;
for ( int i = 0; i < A.Rows; i++ )
{
problem.AddUpperBoundConstraint( A.Row( i ), b[i], constraintTolerance );
}
// All variable values for the solution must satisfy the bounds
// -0.1 <= x[i] <= 2.0
var lb = new DoubleVector( problem.NumVariables, -0.10 );
var ub = new DoubleVector( problem.NumVariables, 2.0 );
for ( int i = 0; i < problem.NumVariables; i++ )
{
problem.AddBounds( i, lb[i], ub[i], .00001 );
}
// Create the solver instance.
var solver = new ConstrainedLeastSquares();
// The ConstrainedLeastSquares solver uses a QP (Quadratic Programming) solver
// to solve the constrained least squares problem.
// The current default QP solver is the NMath active set quadratic programming
// solver with default options.
bool success = solver.Solve( problem );
Console.WriteLine( "Default solver success = " + success );
Console.WriteLine( "Default solver solution x = " + solver.X.ToString( "0.0000" ) );
Console.WriteLine( "Default solver residual norm = " + solver.ResidualNorm.ToString( "0.0000" ) );
Console.WriteLine( "Default solver performed {0} iterations", solver.Iterations );
// You can pass in an instance of a quadratic programming solver for the
// constrained least squares class to use. This allows you to set
// option on the QP solver and inspect results of the QP
// solver.
var interiorPointQp = new InteriorPointQPSolver();
var solverParams = new InteriorPointQPSolverParams
{
MaxIterations = 10000,
PresolveLevel = InteriorPointQPSolverParams.PresolveLevelOption.None
};
solver.Solve( problem, interiorPointQp, solverParams );
Console.WriteLine( "\nInterior point solver success = " + success );
Console.WriteLine( "Interior point QP result = " + interiorPointQp.Result );
Console.WriteLine( "Interior point QP solver iteration count = " + solver.Iterations );
Console.WriteLine( "Interior point solver solution x = " + solver.X.ToString( "0.0000" ) );
Console.WriteLine( "Interior point solver residual norm = " + solver.ResidualNorm.ToString( "0.0000" ) );
// If you use the active set QP solver you can determine which constraints
// are active in the solution by accessing the active set QP solvers
// Lagrange multiplier property. A constraint is active if its corresponding
// Lagrange multiplier is non-zero.
var activeSetQP = new ActiveSetQPSolver();
success = solver.Solve( problem, activeSetQP );
Console.WriteLine( "\nActive set solver success = " + success );
Console.WriteLine( "Active set solver solution x = " + solver.X.ToString( "0.0000" ) );
Console.WriteLine( "Active set solver residual norm = " + solver.ResidualNorm.ToString( "0.0000" ) );
// Print out the active constraints.
for ( int i = 0; i < activeSetQP.LagrangeMultiplier.Length; i++ )
{
if ( activeSetQP.LagrangeMultiplier[i] != 0.0 )
{
Console.WriteLine( "Constraint {0} = {1} is active", i, problem.Constraints[i].ToString() );
}
}
Console.WriteLine();
Console.WriteLine( "Press Enter Key" ); | 1,325 | 4,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-49 | latest | en | 0.540043 |
https://la.mathworks.com/matlabcentral/cody/problems/68-kaprekar-steps/solutions/551096 | 1,606,418,683,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188899.42/warc/CC-MAIN-20201126171830-20201126201830-00187.warc.gz | 365,820,003 | 17,346 | Cody
# Problem 68. Kaprekar Steps
Solution 551096
Submitted on 31 Dec 2014 by Jon
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 3276; y_correct = 5; assert(isequal(KaprekarSteps(x),y_correct))
2 Pass
%% x = 3; y_correct = 6; assert(isequal(KaprekarSteps(x),y_correct))
3 Pass
%% x = 691; y_correct = 7; assert(isequal(KaprekarSteps(x),y_correct))
4 Pass
%% x = 3333; y_correct = Inf; assert(isequal(KaprekarSteps(x),y_correct))
5 Pass
%% x = 1; y_correct = 5; assert(isequal(KaprekarSteps(x),y_correct))
6 Pass
%% x = 6174; y_correct = 0; assert(isequal(KaprekarSteps(x),y_correct))
7 Pass
%% x = 1234; y_correct = 3; assert(isequal(KaprekarSteps(x),y_correct))
8 Pass
%% x = 3141; y_correct = 5; assert(isequal(KaprekarSteps(x),y_correct))
9 Pass
%% x = 8080; y_correct = 6; assert(isequal(KaprekarSteps(x),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 362 | 1,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-50 | latest | en | 0.447487 |
https://en.m.wikipedia.org/wiki/Great_retrosnub_icosidodecahedron | 1,603,421,647,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880519.12/warc/CC-MAIN-20201023014545-20201023044545-00077.warc.gz | 312,343,247 | 12,627 | # Great retrosnub icosidodecahedron
Great retrosnub icosidodecahedron
Type Uniform star polyhedron
Elements F = 92, E = 150
V = 60 (χ = 2)
Faces by sides (20+60){3}+12{5/2}
Wythoff symbol | 2 3/2 5/3
Symmetry group I, [5,3]+, 532
Index references U74, C90, W117
Dual polyhedron Great pentagrammic hexecontahedron
Vertex figure
(34.5/2)/2
Bowers acronym Girsid
In geometry, the great retrosnub icosidodecahedron or great inverted retrosnub icosidodecahedron is a nonconvex uniform polyhedron, indexed as U74. It has 92 faces (80 triangles and 12 pentagrams), 150 edges, and 60 vertices.[1] It is given a Schläfli symbol sr{3/2,5/3}.
3D model of a great retrosnub icosidodecahedron
## Cartesian coordinates
Cartesian coordinates for the vertices of a great retrosnub icosidodecahedron are all the even permutations of
(±2α, ±2, ±2β),
(±(α−βτ−1/τ), ±(α/τ+β−τ), ±(−ατ−β/τ−1)),
(±(ατ−β/τ+1), ±(−α−βτ+1/τ), ±(−α/τ+β+τ)),
(±(ατ−β/τ−1), ±(α+βτ+1/τ), ±(−α/τ+β−τ)) and
(±(α−βτ+1/τ), ±(−α/τ−β−τ), ±(−ατ−β/τ+1)),
with an even number of plus signs, where
α = ξ−1/ξ
and
β = −ξ/τ+1/τ2−1/(ξτ),
where τ = (1+5)/2 is the golden mean and ξ is the smaller positive real root of ξ3−2ξ=−1/τ, namely
${\displaystyle \xi ={\frac {\left(1+i{\sqrt {3}}\right)\left({\frac {1}{2\tau }}+{\sqrt {{\frac {\tau ^{-2}}{4}}-{\frac {8}{27}}}}\right)^{\frac {1}{3}}+\left(1-i{\sqrt {3}}\right)\left({\frac {1}{2\tau }}-{\sqrt {{\frac {\tau ^{-2}}{4}}-{\frac {8}{27}}}}\right)^{\frac {1}{3}}}{2}}}$
or approximately 0.3264046. Taking the odd permutations of the above coordinates with an odd number of plus signs gives another form, the enantiomorph of the other one. Taking the odd permutations with an even number of plus signs or vice versa results in the same two figures rotated by 90 degrees.
The circumradius for unit edge length is
${\displaystyle R={\frac {1}{2}}{\sqrt {\frac {2-x}{1-x}}}=0.580002\dots }$
where ${\displaystyle x}$ is the appropriate root of ${\displaystyle x^{3}+2x^{2}={\Big (}{\tfrac {1\pm {\sqrt {5}}}{2}}{\Big )}^{2}}$ . The four positive real roots of the sextic in ${\displaystyle R^{2},}$
${\displaystyle 4096R^{12}-27648R^{10}+47104R^{8}-35776R^{6}+13872R^{4}-2696R^{2}+209=0}$
are the circumradii of the snub dodecahedron (U29), great snub icosidodecahedron (U57), great inverted snub icosidodecahedron (U69), and great retrosnub icosidodecahedron (U74). | 931 | 2,374 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-45 | latest | en | 0.70938 |
https://lawakjenaka.com/qa/question-what-is-the-difference-between-z-score-and-t-score.html | 1,627,209,685,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00111.warc.gz | 368,636,799 | 7,867 | # Question: What Is The Difference Between Z Score And T Score?
## What is the difference between z score and t statistic?
The major difference between using a Z score and a T statistic is that you have to estimate the population standard deviation.
The T test is also used if you have a small sample size (less than 30)..
## What does the Z test tell you?
A z-test is a statistical test to determine whether two population means are different when the variances are known and the sample size is large. It can be used to test hypotheses in which the z-test follows a normal distribution. A z-statistic, or z-score, is a number representing the result from the z-test.
## What is the T score for severe osteoporosis?
A T-score of −2.5 or lower indicates that you have osteoporosis. The greater the negative number, the more severe the osteoporosis. Bone density is within 1 SD (+1 or −1) of the young adult mean. Bone density is between 1 and 2.5 SD below the young adult mean (−1 to −2.5 SD).
## What is the z value?
The Z-value is a test statistic for Z-tests that measures the difference between an observed statistic and its hypothesized population parameter in units of the standard deviation. … Converting an observation to a Z-value is called standardization.
## What is the main difference between z score and T score quizlet?
Terms in this set (35) The main difference between a z-score and t-test is that the z-score assumes you do/don’t know the actual value for the population standard deviation, whereas the t-test assumes you do/don’t know the actual value for the population standard deviation.
## What does the T score tell you?
The t-value measures the size of the difference relative to the variation in your sample data. Put another way, T is simply the calculated difference represented in units of standard error. The greater the magnitude of T, the greater the evidence against the null hypothesis.
## Is a higher Z score better?
The higher Z-score indicates that Jane is further above the Mean than John. fairly small while others are quite large, but the method of ranking is the same. An 80 Percentile means that 80% of the data elements are below that point.
## What is an advantage of T scores over z scores?
For example, a t score is a type of standard score that is computed by multiplying the z score by 10 and adding 50. One advantage of this type of score is that you rarely have a negative t score. As with z scores, t scores allow you to compare standard scores from different distributions.
## What is a normal T score?
A T-score of -1.0 or above is normal bone density. Examples are 0.9, 0 and -0.9. A T-score between -1.0 and -2.5 means you have low bone density or osteopenia. Examples are T-scores of -1.1, -1.6 and -2.4.
## What is a good T stat?
Thus, the t-statistic measures how many standard errors the coefficient is away from zero. Generally, any t-value greater than +2 or less than – 2 is acceptable. The higher the t-value, the greater the confidence we have in the coefficient as a predictor.
## What is the most common standard for statistical significance?
Significance levels show you how likely a pattern in your data is due to chance. The most common level, used to mean something is good enough to be believed, is . 95. This means that the finding has a 95% chance of being true.
## What is the difference between z test and t test?
Z-tests are statistical calculations that can be used to compare population means to a sample’s. T-tests are calculations used to test a hypothesis, but they are most useful when we need to determine if there is a statistically significant difference between two independent sample groups.
## What are Z and T scores?
Z score is the subtraction of the population mean from raw score and then divides the result with population standard deviation. T score is a conversion of raw data to the standard score when the conversion is based on the sample mean and sample standard deviation.
## Do I use Z or t test?
So when we should perform the Z test and when we should perform t-Test? … For a large sample size, Sample Variance will be a better estimate of Population variance so even if population variance is unknown, we can use the Z test using sample variance. | 943 | 4,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-31 | latest | en | 0.914949 |
https://shulchanaruchharav.com/halacha/how-many-strings-must-one-tie/ | 1,716,912,063,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00520.warc.gz | 434,971,433 | 60,847 | # How many strings must one tie
How many strings must one tie?
The Tzitzis of each corner must consist of four strings.[2] The four strings have to be entered into the corner hole of the Tallis and folded in half, for a total of eight strings.[3]
Used more than 4 strings?[4] If one ties more than four strings to the corner, even if he adds only one more string, he transgresses the Biblical prohibition of Baal Tosif.[5] In such a case, the fringe which contains the extra string is invalid until the extra strings are removed.[6]
Tied two sets of Tzitzis to one corner: See Halacha 7C!
May one enter one long string and then cut it?[7] One may form the eight strings through using a single long string. This can be accomplished through taking a single long string, folding it into four, and then entering it into the corner hole. When it is folded in half after being entered into the corner hole the string has eight sides. One then cuts the top of each string, making it into eight valid strings. If one forgot and began braiding the Tzitzis prior to cutting the top of the strings, it is invalid. This applies even if one only made one loop, and on top of it a single [set of a double[8]] knot, and then remembered.[9] [One must undo the Tzitzis and retie it to that hole.] If however one cut the top of the strings immediately after making the loop, prior to making even a single knot on top of the loop, the Tzitzis remains valid [and one may continue with the remainder of the braid].[10] Certainly if one cut the strings immediately after the first knot, prior to making any loops, the Tzitzis remains valid. Nevertheless, when using a single long string that is folded to four, it is proper to initially cut the string prior to inserting it into the hole, and thus enter four separate strings into the hole.[11]
Q&A
In the event that one used five strings to make the Tzitzis braid, how is that fringe to be validated?
It suffices to cut one string off from the Tzitzis, and it is not required to untie the entire fringe, remove the string, and then retie it from the start.[12] Some[13] however write that one is to undo the entire fringe, remove the extra string, and then retie it.
If one added the extra string with intent that it is for mere beauty and not for the Mitzvah, does one transgress Baal Tosif?
Some Poskim[14] rule that in such a case one does not transgress Baal Tosif. However some rule that the Tzitzis nevertheless remains invalid, as explained next.
If a string or thread accidently became tangled within the four strings, is it valid?
Some Poskim[15] rule one may not have any extra item mixed into the Tzitzis strings while they are tied and braided, and if the Tzitzis contains such an item it is possibly invalid.[16] This matter however has been omitted from all other codifiers, hence implying that we do not hold like this opinion. According to all, if the string became entangled in the Tzitzis after it was already braided, it does not invalidate the Tzitzis.[17]
What is the law if the strings of two corners became tied to each other?
The Tzitzis remains valid so long as one is able to undo the knots.[18] One is to untangle the Tzitzis prior to the blessing, as explained in chapter 2 Halacha 6!
[2] The reason: In times that Techeiles was available, the Mitzvah of Tzitzis was to take two strings which are dyed with the color called Techeiles and two non-dyed strings which are called the white strings [Chutei Lavan]. However today that we no longer have any more Techeiles, one must take four white strings, as Tzitzis are invalid if they do not contain four strings. This is learned from the verse which states “Gedilim Taaseh Lecha”, and a single Gedil [which is a braid] consist of no less than two strings which are braded over each other. Now, the verses states Gedilim in plural, hence teaching that it consists of two braids, which is a total of four strings. [Admur ibid; Taz 11/13; Menachos 39b]
[3] Admur 11/5; Michaber 11/12; Menachos 42a
The reason: This is learned from the verse “Pesil Techeiles”, which teaches that each of the strings must be folded like a wick. [Admur ibid] Rashi in end of Shlach explains the 8 strings correspond to the 8 days between the exodus from Egypt and the singing by the Yam Suf. The Baal Haturim explains that the 8 strings correspond to the 8 limbs of a person which can sin, which are the ears, eyes, mouth, nose, hands, legs, the Erva, and the heart. Thus, the 8 strings remind a person not to sin.
[5] Other opinions: Some rule that if one adds an extra string, the prohibition is due to Chatzitza and not due to Baal Tosif. [Raavad 1/15, brought in Tehila Ledavid 11/1]
Does one transgress Baal Tosif even if he does not wear the Tallis: lt is implied that one transgresses Baal Tosif even prior to wearing the Tallis in this form, and rather the mere tying is already a transgression. Vetzaruch Iyun!
[6] The reason: Since one transgressed Baal Tosif, therefore the entire set is invalid. [Admur ibid; M”B 11/60] The Gemara explains that this is because it is considered as if the braids and knots were not made at all. [Menachos 40b]
[7] 11/23; Michaber 11/13
[8] M”B 11/64; See Admur 11/28
[9] The reason why even one loop and knot invalidates: As Biblically it suffices for one to make one loop [i.e. Chulya] and one knot, as the verse simply states “Make braids” and does not state how many loops it is to contain. Therefore, as soon as one made a single loop and knot the Tzitzis is considered to have been made with invalidation, as at the time of the braiding it did not have four whole strings, but rather one string which was folded into eight parts. Therefore, even if one now cuts the heads of the strings, it is of no benefit, being that the actual braid was already made with an invalidation, and the Torah states “Taaseh Velo Min Hasuiy”. [ibid]
[10] Admur ibid; M”A, Levush and Taz; See M”B 11/64
The reason: As since one did not yet make the entire Biblically valid Tzitzis with an invalidation prior to cutting the tops, therefore it is not invalidated due to Taaseh Velo Min Hasuiy. [ibid]
Other opinions: Some Poskim hold that one is to undo the Tzitzis even if a knot was not made on top of the loop, being that each part of the Tzitzis must be made in a Kosher fashion. [Olas Tamid 11/8 and 10; Tosafus Menachos 39 in name of Rashi;; See Biur Halacha 11/10 “Venitku”] Some Poskim rule one must be stringent like this opinion. Chayeh Adam 11/18; Aruch Hashulchan 11/17; Chazon Ish 2/5; 3/11; M”B ibid rules to be initially stringent like this opinion]
[11] The reason: This is to be done in order to ensure there is no step of the making of the Tzitzis that contains invalidation. [ibid] From here we see that even the insertion of the strings into the hole is to be done with validity. [Ketzos Hashulchan 6 footnote 40]
[12] P”M 11 M”Z 13; Biur Halacha 11/12 “Hosif” based on Rama 10/6; Admur 10/14; Tehila Ledavid 11/1 that so applies according to those Poskim that do not rule like the Ravaad ibid
The reason: This does not involve Taaseh Velo Min Hasuy because the Gemara ibid states that so long as one transgresses Baal Tosif it is as if he has not done anything yet, and hence when the extra string is removed it is now considered that he tied the strings there. [ibid; See Admur ibid for the same explanation regarding one who tied two pairs of fringes to a corner]
How much of the extra string must be cut? Some Poskim question whether the entire extra string must be removed or it suffices to cut the strings that extend past the braids [Gedil]. [P”M ibid; Biur Halacha ibid] Some Poskim however rule that if the string remains in the braid, it is completely invalid. [Chazon Ish 3/16] Vetzaruch Iyun if according to all it suffices to simply cut the extra string from the bottom, in its area of attachment. [See Piskeiy Teshuvos 11 footnote 184]
[13] Chazon Ish 3/16; Piskeiy Teshuvos 11/28 based on Michaber 10/6
[14] P”M in introduction to Tzitzis; Biur Halacha 11/3 “Venaasu” in name of Peri Megadim; Vetzaruch Iyun from Biur Halacha 34/2 “Bealma” in name of many Achronim who argued on the Bach regarding one who added an extra Bayis to the Shel Rosh without intent for the Mitzvah.
[15] Tehila Ledavid 11/1 based on Raavad 1/15
[16] The reason: As according to the Raavad a Chatzitza invalidates the fringe. [ibid]
[17] See Piskeiy Teshuvos 11/28 footnote 189-190
[18] Admur 23/3 “Some are accustomed upon entering into a cemetery to tie the Tzitzis of two corners to each other in order to nullify the Mitzvah of Tzitzis from them [and hence allow one to enter the cemetery with the Tzitzis revealed]. However, in truth this does not help at all, as although the Tzitzis are tied to each other, nevertheless the Mitzvah of Tzitzis is not nullified through doing so, as he intends to return and untie them. [Admur 23/3; M”A 23/1] | 2,417 | 8,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-22 | latest | en | 0.91958 |
http://www.ck12.org/geometry/Distance-Formula-in-the-Coordinate-Plane/?difficulty=at+grade&by=community | 1,448,612,186,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398448389.58/warc/CC-MAIN-20151124205408-00110-ip-10-71-132-137.ec2.internal.warc.gz | 352,919,207 | 15,607 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Distance Formula in the Coordinate Plane
## Length between two points using a right triangle.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Distance Formula in the Coordinate Plane
This concept teaches students how to find the distance between two points using the distance formula.
0 | 183 | 929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2015-48 | longest | en | 0.861779 |
https://leetcode.ca/2020-04-19-1602-Find-Nearest-Right-Node-in-Binary-Tree/ | 1,720,895,098,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00101.warc.gz | 319,519,368 | 8,098 | # 1602. Find Nearest Right Node in Binary Tree
## Description
Given the root of a binary tree and a node u in the tree, return the nearest node on the same level that is to the right of u, or return null if u is the rightmost node in its level.
Example 1:
Input: root = [1,2,3,null,4,5,6], u = 4
Output: 5
Explanation: The nearest node on the same level to the right of node 4 is node 5.
Example 2:
Input: root = [3,null,4,2], u = 2
Output: null
Explanation: There are no nodes to the right of 2.
Constraints:
• The number of nodes in the tree is in the range [1, 105].
• 1 <= Node.val <= 105
• All values in the tree are distinct.
• u is a node in the binary tree rooted at root.
## Solutions
BFS or DFS.
• /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode findNearestRightNode(TreeNode root, TreeNode u) {
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
while (!q.isEmpty()) {
for (int i = q.size(); i > 0; --i) {
root = q.pollFirst();
if (root == u) {
return i > 1 ? q.peekFirst() : null;
}
if (root.left != null) {
q.offer(root.left);
}
if (root.right != null) {
q.offer(root.right);
}
}
}
return null;
}
}
• /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* findNearestRightNode(TreeNode* root, TreeNode* u) {
queue<TreeNode*> q{ {root} };
while (q.size()) {
for (int i = q.size(); i; --i) {
root = q.front();
q.pop();
if (root == u) return i > 1 ? q.front() : nullptr;
if (root->left) q.push(root->left);
if (root->right) q.push(root->right);
}
}
return nullptr;
}
};
• # Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def findNearestRightNode(self, root: TreeNode, u: TreeNode) -> Optional[TreeNode]:
q = deque([root])
while q:
for i in range(len(q) - 1, -1, -1):
root = q.popleft()
if root == u:
return q[0] if i else None
if root.left:
q.append(root.left)
if root.right:
q.append(root.right)
• /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func findNearestRightNode(root *TreeNode, u *TreeNode) *TreeNode {
q := []*TreeNode{root}
for len(q) > 0 {
for i := len(q); i > 0; i-- {
root = q[0]
q = q[1:]
if root == u {
if i > 1 {
return q[0]
}
return nil
}
if root.Left != nil {
q = append(q, root.Left)
}
if root.Right != nil {
q = append(q, root.Right)
}
}
}
return nil
}
• /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {TreeNode} u
* @return {TreeNode}
*/
var findNearestRightNode = function (root, u) {
const q = [root];
while (q.length) {
for (let i = q.length; i; --i) {
root = q.shift();
if (root == u) {
return i > 1 ? q[0] : null;
}
if (root.left) {
q.push(root.left);
}
if (root.right) {
q.push(root.right);
}
}
}
return null;
}; | 1,093 | 3,701 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-30 | latest | en | 0.450061 |
https://www.delftstack.com/howto/python-pandas/kde-plot-visualization-with-pandas-and-seaborn/ | 1,670,310,058,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00559.warc.gz | 771,008,579 | 12,548 | # KDE Plot Visualization With Pandas and Seaborn
Samreena Aslam Jan-30, 2022 Jan-24, 2022 Pandas Pandas DataFrame
KDE is `Kernel Density Estimate`, used to visualize the probability density of continuous and non-parametric data variables. When you want to visualize the multiple distributions, the `KDE` function produces a less cluttered plot that is more interpretable.
Using `KDE`, we can visualize multiple data samples using a single graph plot, which is a more efficient method in data visualization.
`Seaborn` is a python library like `matplotlib`. Seaborn can be integrated with `pandas` and `numpy` for data representations.
Data scientists use this library to create informative and beautiful statistical charts and graphs. Using these presentations, you can understand the clear concepts and flow of information within different modules.
We can plot univariate and bivariate graphs using the KDE function, Seaborn, and Pandas.
We will learn about the KDE plot visualization with pandas and seaborn. This article will use a few samples of the `mtcars` dataset to show the KDE plot visualization.
Before starting with the details, you need to install or add the `seaborn` and `sklearn` libraries using the pip command.
``````pip install seaborn
pip install sklearn
``````
## Data Visualization Using Normal KDE Plot and Seaborn in Python
We can plot the data using the normal KDE plot function with the Seaborn library.
In the following example, we have created 1000 data samples using the random library then arranged them in the array of `numpy` because the Seaborn library only works well with `numpy` and Pandas `dataframes`.
Example Code:
``````import seaborn as sn
import matplotlib.pyplot as plt
import numpy as np
data = np.random.randn(1000)
# KDE Plot with seaborn
plt.show()
``````
Output:
We can also visualize the above data sample vertically or revert the above plot using the KDE and Seaborn library. We used the plot property `vertical=True` to revert the above plot.
Example Code:
``````import seaborn as sn
import matplotlib.pyplot as plt
import numpy as np
data = np.random.randn(1000)
# KDE Plot with seaborn
res = sn.kdeplot(data, color='green', vertical=True, shade='True')
plt.show()
``````
Output:
## One-Dimensional KDE Plot Using Pandas and Seaborn in Python
We can visualize the probability distribution for a single target or continuous attribute using the KDE plot. In the following example, we have read a CSV file of the `mtcars` dataset.
There are more than 350 entries in our dataset, and we will visualize the univariate distribution along the x-axis.
Example Code:
``````import seaborn as sn
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
# read CSV file of dataset using pandas
# kde plot using seaborn
plt.show()
``````
Output:
You can also flip the plot by visualizing the data variable along the y-axis.
Example Code:
``````import seaborn as sn
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
# Read CSV file using pandas
# KDE plotting using seaborn
plt.show()
``````
Output:
We can visualize the probability distribution of multiple target values in a single plot.
Example Code:
``````import seaborn as sn
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
# Read CSV file using pandas
# KDE plotting using seaborn
plt.show()
``````
Output:
## Two-Dimensional or Bivariate KDE Plot Using Pandas and Seaborn in Python
We can visualize data in two-dimensional or bivariate KDE plots using the seaborn and pandas library.
In this way, we can visualize the probability distribution of a given sample against multiple continuous attributes. We visualized the data along the x and y-axis.
Example Code:
``````import seaborn as sn
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
# Read CSV file using pandas
# KDE plotting using seaborn
plt.show()
``````
Output:
Similarly, we can plot the probability distribution of multiple samples using a single KDE plot.
Example Code:
``````import seaborn as sn
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
# Read CSV file using pandas
# KDE plotting using seaborn
plt.show()
``````
Output:
## Conclusion
We demonstrated in this tutorial using the KDE plot visualization using Pandas and Seaborn library. We have seen how to visualize the probability distribution of single and multiple samples in a one-dimensional KDE plot.
We discussed how to use the KDE plot with Seaborn and Pandas to visualize the two-dimensional data.
## Related Article - Pandas DataFrame
• Get Pandas DataFrame Column Headers as a List
• Delete Pandas DataFrame Column
• Convert Pandas Column to Datetime
• Convert a Float to an Integer in Pandas DataFrame
• Sort Pandas DataFrame by One Column's Values
• Get the Aggregate of Pandas Group-By and Sum | 1,085 | 4,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-49 | longest | en | 0.795623 |
https://www.proprofs.com/discuss/q/605517/2000w-tennis-ball-machine-releases-0028kg-balls-speed-100-02 | 1,582,849,492,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146907.86/warc/CC-MAIN-20200227221724-20200228011724-00119.warc.gz | 773,394,224 | 66,747 | A 2000 W tennis ball machine releases 0.028 kg balls at a speed of - ProProfs Discuss
# A 2000-W tennis ball machine releases 0.028-kg balls at a speed of 100 m/s in 0.20 s. The efficiency of the machine is
A. 70%
B. 14%
C. 28%
D. 1.4%
E. 7.0%
This question is part of thermal physics IB
Asked by Songj, Last updated: Feb 17, 2020
Efficiency = output / input 100, so in this case power output / power input
Power output = W / t, where the work done is the equal to the change in kinetic energy of a ball
Pout = (0.5mv^2 - 0) / t = 0.5 0.028 100^2 / 0.2 = 700 W
efficiency = 700 / 2000 100 = 35%
1
1
#### kimeuegene
Kimeuegene
Replied on May 30, 2018
Yes, this is correct. It should be fixed by now
#### Thisuntheekshana
Thisuntheekshana
Distance =100/0.2=500 work done=500×0.028=14
Power=14J/0.2s=70
Efficiency=70/2000×100%=3.5%
John Smith | 318 | 857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.799674 |
https://zeligstudio.com/how-to-play-dominoes-5/ | 1,696,446,877,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511406.34/warc/CC-MAIN-20231004184208-20231004214208-00646.warc.gz | 1,177,412,409 | 18,566 | # How to Play Dominoes
Dominoes are flat, thumb-sized rectangular blocks bearing from one to six spots or pips (plus blanks) on each end. Each domino belongs to a suit.
The most common domino set contains 28 tiles; larger sets are available. These larger sets are typically used for games with multiple players or to play long, strategic games.
## Rules
There are many different types of domino games that can be played. Each has its own rules and dynamics, but they all share some basic principles. The player who successfully places all of their dominoes wins the game. They then earn the sum of all the other players’ points.
The starting player begins by placing a tile in the center of the table. The next player must match one end of their tile to the open end of the first domino. This is called the line of play and there are specific instructions on how to do this for each game. If a double is played, it must be placed perpendicular to the line of play and both ends must touch.
In some games, part of the score is obtained by counting the pips on the ends of the lines of play. For example, if a 5-5 is played and it is not a spinner, the count would be 10. The number of pips on each side of the domino must also be taken into account when making a score.
## Variations
There are many variations of domino. Students can use a scratch paper domino mat or whiteboard with an expo marker to practice decimal rounding. They may also choose to play a subtraction game. Each player starts with seven dominoes. When they can’t play a domino, they must pass their turn. The next player then picks a domino from the boneyard to add to their own personal train.
When a player begins to string together dominoes on their personal train, they must place a marker on the domino so that other players cannot add to it. This rule makes it more difficult to pass your turn, but is a good way to practice your strategy.
Some games allow players to buy tiles from the boneyard, which means that some of their dominoes remain in the stock after winning a round. These tiles are then added to the winner’s score. In addition, some games have a set number of pips that must be reached to win.
## Materials
Over the centuries, dominoes have been made of various materials. In addition to plastic, they have been made of bone and ivory, dark hardwoods such as ebony, with contrasting black or white pips; stone (e.g., marble or granite); metals such as brass or pewter; ceramic clay and even frosted glass. Some sets also feature a felt surface to keep the dominoes from scratching the table.
Dominoes are typically twice as long as they are wide, which makes it easy to stack them. The ends of the dominoes are marked with values ranging from six to zero. If a domino has two same values inked on its ends, it is called a double; otherwise, it is a single.
Although most domino games are played with more than one player, some are solitaire. This form of the game is popular with people who are not comfortable playing cards, as it circumvents religious proscriptions. These games tend to be more mathematically challenging than other types of domino, and require a greater degree of concentration.
## Scoring
A player scores points in domino by laying tiles end to end. If the exposed ends of the first double (called a spinner) match (ones touch one’s and two’s touch two’s), each side of that tile counts. If all sides of the tile count, the total is rounded up to the nearest multiple of five.
In games that involve more than one player, the winner is determined when a player has no more dominoes left in his hand. This is sometimes called a block. If the game becomes blocked, the players count up the pips on their remaining tiles and the player with the lowest value wins that round.
Some games are scored in a more complicated manner. For example, in the popular Mexican Train variant, a player scores by completing a chain of three or more matching sets of tiles. In this case, the number of pips in a set is tallied and rounded up to a multiple of five. | 906 | 4,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-40 | latest | en | 0.970166 |
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# In economics, the concept of “propensity to consume” refers to the pro
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In economics, the concept of “propensity to consume” refers to the pro [#permalink]
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Updated on: 26 Sep 2019, 00:10
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In economics, the concept of “propensity to consume” refers to the proportion of disposable income – income after taxes and transfers – that individuals spend on consumption. Marginal propensity to consume (MPC) is a related metric that quantifies induced consumption, the idea that as income increases a consumer’s consumption will also increase. MPC is the proportion of that additional income that an individual consumes; for example, if a household earns one extra dollar of disposable income, and the marginal propensity to consume is 0.65, then of that dollar, the household will spend 65 cents and save 35 cents.
In a standard Keynesian model, the MPC will be less than the average propensity to consume (APC) because in the short-run some (autonomous) consumption does not change with income. Short-term decreases in income do not lead to reductions in consumption, because people reduce savings to stabilize consumption. Over the long-run, as wealth and income rise, consumption also rises; the marginal propensity to consume out of long-run income is closer to the average propensity to consume.
Economists often distinguish between the marginal propensity to consume out of permanent income and the marginal propensity to consume out of temporary income, because if consumers expect a change in income to be permanent, then they have a greater incentive to increase their consumption. This implies that the Keynesian multiplier – the measure of that consumption’s impact on additional consumption in the marketplace – should be larger in response to permanent changes in income than it is in response to temporary changes in income (though the earliest Keynesian analyses ignored these subtleties). However, the distinction between permanent and temporary changes in income is often subtle in practice, and it is often quite difficult to designate a particular change in income as being permanent or temporary. What is more, the marginal propensity to consume should also be affected by factors such as the prevailing interest rate and the general level of consumer surplus that can be derived from purchasing.
Spoiler: :: OA&OE
C. In this Inference question, it can be inferred from the passage (in paragraph 2) that short-term decreases in income force people to increase their consumption relative to earnings, as you’re told that consumption does not decrease in the short-run. This, then would mean that the ratio of consumption to earnings goes up, therefore increasing MPC.
Question ID: 09960
1. According to the passage, it can be inferred that:
(A) When a household’s income increases, its marginal propensity to consume decreases.
(B) Most households cannot accurately delineate between permanent and temporary changes in income.
(C) Decreases in income generally lead to short-run increases in marginal propensity to consume.
(D) Early Keynesian analyses did not allow for a Keynesian multiplier for income changes with regard to marginal propensity to consume.
(E) In the short run, it is impossible for a household to have a negative marginal propensity to consume.
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Originally posted by hazelnut on 10 Oct 2017, 04:51.
Last edited by SajjadAhmad on 26 Sep 2019, 00:10, edited 1 time in total.
Updated - Complete topic (675).
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25 Oct 2017, 10:46
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nairaneesh wrote:
I am confused because MPC is applicable only when additional income comes in to play.
Read this line in the second paragraph - "Short-term decreases in income do not lead to reductions in consumption, because people reduce savings to stabilize consumption."
Earlier, if Income was \$1000 and MPC was 0.72 (consumption 1000* 0.72 = \$720), and now if the income falls to \$800, the consumption remains same, then the MPC increases to 0.9 (720/800).
Hope this helps!
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25 Oct 2017, 10:30
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The passage talks about MPC, APC and consumption and how they are related.
In between B and C. I kept B as an option as I read through it. Passage says it is difficult to differentiate but it doesn't mention that it is for household...so kept on hold
Option C.. It says when income reduces in short run consumption remains same..that means savings are reduced and a greater proportion of income is dedicated to consumption. Hence option C.
Kudos if you like the explanation.
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03 Nov 2017, 22:53
Sprincejindal wrote:
nairaneesh wrote:
I am confused because MPC is applicable only when additional income comes in to play.
Read this line in the second paragraph - "Short-term decreases in income do not lead to reductions in consumption, because people reduce savings to stabilize consumption."
Earlier, if Income was \$1000 and MPC was 0.72 (consumption 1000* 0.72 = \$720), and now if the income falls to \$800, the consumption remains same, then the MPC increases to 0.9 (720/800).
Hope this helps!
I agree with what you say but the very definition of MPC is that additional income that the family consumes. So if there is a decrease in income ,the MPC will be negative no ?
I think if the answer choice were to be rephrased, Decreases in income generally lead to short-run increases in propensity to consume. Choice c would be right.
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04 Nov 2017, 03:20
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goforgmat wrote:
Sprincejindal wrote:
nairaneesh wrote:
I am confused because MPC is applicable only when additional income comes in to play.
Read this line in the second paragraph - "Short-term decreases in income do not lead to reductions in consumption, because people reduce savings to stabilize consumption."
Earlier, if Income was \$1000 and MPC was 0.72 (consumption 1000* 0.72 = \$720), and now if the income falls to \$800, the consumption remains same, then the MPC increases to 0.9 (720/800).
Hope this helps!
I agree with what you say but the very definition of MPC is that additional income that the family consumes. So if there is a decrease in income ,the MPC will be negative no ?
I think if the answer choice were to be rephrased, Decreases in income generally lead to short-run increases in propensity to consume. Choice c would be right.
MPC is not the additional income that a family consumes, but it is the proportion of that additional income that a family consumes. It is the ratio of changes in consumtion to changes in income. So when the income (Denominator) reduces, but the consumption (Numerator) remains the same, then the MPC (Ratio) increases.
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28 Dec 2018, 06:51
nairaneesh wrote:
I am confused because MPC is applicable only when additional income comes in to play.
Read this line in the second paragraph - "Short-term decreases in income do not lead to reductions in consumption, because people reduce savings to stabilize consumption."
Earlier, if Income was \$1000 and MPC was 0.72 (consumption 1000* 0.72 = \$720), and now if the income falls to \$800, the consumption remains same, then the MPC increases to 0.9 (720/800).
Hope this helps![/quote]
I agree with what you say but the very definition of MPC is that additional income that the family consumes. So if there is a decrease in income ,the MPC will be negative no ?
I think if the answer choice were to be rephrased, Decreases in income generally lead to short-run increases in propensity to consume. Choice c would be right.
[/quote]
But inst the MPC calculated on the additional income and not the total income. So if the current income is say, €1000, what can be calculated is the APC. If the income increases by say, €200, then, MPC would be calculable only in the €200 and not €1200. This is what I am inferring.
I understand this is not the learning from this GMAT question, but only curios to understand the solution.
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22 Jan 2020, 21:27
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23 Jan 2020, 17:55
I am confused among options B, C and D
Quote:
Read this line in the second paragraph - "Short-term decreases in income do not lead to reductions in consumption, because people reduce savings to stabilize consumption."
Earlier, if Income was \$1000 and MPC was 0.72 (consumption 1000* 0.72 = \$720), and now if the income falls to \$800, the consumption remains same, then the MPC increases to 0.9 (720/800).
As said by Sprincejindal above, I understand why option C is correct.
Quote:
However, the distinction between permanent and temporary changes in income is often subtle in practice, and it is often quite difficult to designate a particular change in income as being permanent or temporary.
From the quote above, in option B, the passage suggests that it is quite difficult to designate a particular change, but doesn't say that it is the household that are unable to designate a change in income. Hence option B is also ruled out.
Quote:
This implies that the Keynesian multiplier – the measure of that consumption’s impact on additional consumption in the marketplace – should be larger in response to permanent changes in income than it is in response to temporary changes in income (though the earliest Keynesian analyses ignored these subtleties).
. But when it comes to option D, according to the sentence above, we see that the earlier Keynesian analyses did not account for differences in permanent changes in income and temporary changes in income. And these income changes affect the MPC. Hence shouldn't option D be right?
Re: In economics, the concept of “propensity to consume” refers to the pro [#permalink] 23 Jan 2020, 17:55
Display posts from previous: Sort by | 2,798 | 11,938 | {"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} | 3.03125 | 3 | CC-MAIN-2020-16 | longest | en | 0.922349 |
https://forums.getpaint.net/topic/108577-tiling-effect-perspective-woes/ | 1,716,073,021,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057516.1/warc/CC-MAIN-20240518214304-20240519004304-00189.warc.gz | 223,422,149 | 40,692 | # Tiling Effect Perspective woes
## Recommended Posts
I'm trying to keep things in perspective but they keep being blown out of all proportion!
I've written tiling effect that I'm pleased with. It uses MJW's super-sampling ideas and some ideas I had for brick like and reflective tiling and yields good results apart from the tilt. It really needs perspective tilting and I cannot solve the problem without creating unwanted curved perspective lines?
If the perspective problem is solvable, I would then (hopefully) do the calculations in Render so that it could benefit from super-sampling at the 'furthest' distance (Capped to a maximum of say 100 samples). 'Simple' backwards tilting would be ideal (not around the cross hairs, as enlarging below the axis never looks good IMO).
The dropbox link below contains the codelab code, the .dll and an image I've been using for testing.
I've been driving myself nuts trying to get the tilting to work correctly so any help will be greatly appreciated.
Btw I chose the name Wat Tyler as a play on words and as a hero to revolting peasants like myself
https://www.dropbox.com/sh/grkg1wvp4ioyve8/AABAf4OAJ1D9RGLnTM4hq0F3a?dl=0
Red ochre Plugin pack.............. Diabolical Drawings ................Real Paintings
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I believe the only way to get the perspective correct is to use homogeneous coordinates. That may sound complicated, but it just involves addition, multiplication, and division.
This explanation probably won't be enough to write code from, but it may make it clearer how it works.
You start of with the image coordinates, (x, y). There's an assumed "w" coordinate of 1, so it can be considered to be (x, y, 1). (I call it "w", but others might call it "z". It represents the distance from the viewer.)
The coordinates are transformed to other coordinates using a transformation:
x' = Mxx * x + Mxy * y + Mxw
y' = Myx * x + Myy * y + Myw
w' = Mwx * x + Mwy * y + Mww
The nine values, Mxx, Mxy, Mxw, etc., depend on how the image is transformed; that is, how it's rotated, translated, and scaled.
This is usually thought of as a matrix multiplication, written:
/ x' \ / Mxx Mxy Mxw \ / x \
| y' | = | Myx Myy Myw | | y |
\ w' / \ Mwx Mwy Mww / \ 1 /
Now this has three components, x', y', w', yet you can only display two. The two coordinates are produced by dividing x' and y' by w':
x" = x'/w'
y" = y'/w'
The division is called the "perspective division," and it's what causes the size of objects to decrease as they get farther away.
Here's the tricky part. In PDN plugins, you need to solve the opposite problem. You start with (x", y"), which are the coordinates of the destination pixel, and you need to find the (x, y) coordinates in the source image that will transform to them. Fortunately, it's not too difficult, because the whole process is reversible. The matrix can be "inverted" to produce a new matrix such that:
/ x* \ / Mxx' Mxy' Mxw' \ / x" \
| y* | = | Myx' Myy' Myw' | | y" |
\ w* / \ Mwx' Mwy' Mww' / \ 1 /
or,
x* = Mxx' * x" + Mxy' * y" + Mxw"
y* = Myx' * x" + Myy' * y" + Myw"
w* = Mwx' * x" + Mwy' * y" + Mww"
Then the points that are used in src.GetBilinearSampleClamped(x, y) are:
x = x*/w*
y = y*/w*
(if w* is less than or equal to 0, the point is behind the viewer, so it's invisible.)
So, once the values for Mxx', Mxy', Mxw', Myx', Myy', Myw', Mwx', Mwy', Mww' are known, it's just a matter of some multiplications, additions, and divisions.
I don't understand what you're doing in the plugin well enough yet to tell you how to find the values. I downloaded the code, so I can probably figure it out, but it would help if you'd explain it a little.
(In case it isn't clear, the primes (' and ") and star (*) are just used to distinguish between different versions of the coordinates.)
(Also, one small technical point: because the x and y values are divided by w, all the matrix entries can be scaled by the same amount without affecting the result. Therefore, the transposed co-factor matrix can be used instead of the inverted matrix, which simplifies the computation a bit. Also, the total transformation is usually a combination of easily invertible operations, such as "rotate about the x axis," followed by "translate in the x and y direction," etc. The total inverse is just the individual inverses in reverse order. (Don't worry if that's not clear. I'm just trying to show that the computations involved are usually not particularly difficult.))
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Thank you MJW for the explanation and the P.M. Unfortunately my knowledge of matrices is terrible!
I downloaded the code, so I can probably figure it out, but it would help if you'd explain it a little.
Sorry, there are explanations in the codelab code.
Basically, it calculates the X & Y distances from each src pixel to a moveable centre point. These values are then multiplied for zooming and/or rotation and fed through a method which does the tiling.
Tiling is achieved basically by using the remainder when divided by width or height to access the src image. This part of the effect works perfectly.
The problem is the tilting back into the Z direction.
I was hoping for a simpler 'trigonometric' algorithm for tilt based on the src Y coordinate but fear this is not possible.
Well, it should be possible but I'm clearly going wrong somewhere.'Close but no cigar'.
If the two systems are compatible it would be useful to have a choice of tiling pattern combined with high quality (super-sampled) perspective tilting.
Thank you again - I will be in touch via P.M. when I've studied your post further (I better read those text books from school in the 1970's !)
Red ochre Plugin pack.............. Diabolical Drawings ................Real Paintings
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I've had a chance to look at the code, and I'll try to fix the perspective transformation soon (sometime today, I hope), then post the modified version or put it somewhere you can retrieve it.
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If you can get a result without the perspective lines curving at higher tilt angles that would be brilliant.
Hopefully I could then adapt the existing sub-pixel sampling method to include it.
Then add 'prop rules', icon, sample image etc.
Red ochre Plugin pack.............. Diabolical Drawings ................Real Paintings
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The sub-pixel sampling should be no problem. Just as it currently does, it will take a destination pixel location and transform it into a source pixel location, which can then accessed with the bilinear sample routines. The destination locations can be for sub-pixels as well as pixels.
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I used these transformation routines in my (unfinished) faux landscape generator.
```
// Rotational transformation.
// Rotates the layer rectangle into a diamond, with top-left corner rotated to the top-middle (i.e. by 45 degrees clockwise).
// float x1 = (0.5f * (Width + x - y));
// float y1 = (0.5f * (x + y));
// However, I'm not rotating...
float x1 = (float)x;
float y1 = (float)y;
// Project flat 2d point into 3d.
// Simple 3d projection to convert x, y & height into perspective on a 2d screen.
// The basic idea behind any perspective projection is to divide horizonal and vertical position by depth so
// that objects further away appear smaller and vice-versa.
float x2 = HalfWidth;
float y2 = HeightMod;
float z2 = HalfWidth - cVN + (y1 * (float)Amount10); // last float is the apparent view height. Larger = higher view, smaller = sea level view.
float x3 = (x1 - HalfWidth) * (float)Amount11; // last is the view width. Larger = wider field of view.
float y3 = (Width - y1) * (float)(Amount12/10) + 1.0f; // small float is the distance. Larger = further away, smaller = closer
x1 = x2 + x3 / y3;
x1 = (x1<0) ? 0 : (x1>=Width) ? Width-1 : x1;
y1 = y2 + z2 / y3;
y1 = (y1<0) ? 0 : (y1>=Height) ? Height-1 : y1;
dst[(int)x1,(int)y1] = srcBgra;
```
cVN is a Value Noise integer. Basically a heightmap value.
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So I don't keep you waiting, here is a version that seems to work. I'm still trying to reconcile some differences in this code and Rotate/Zoom in how tilt angles are treated. The Rotate/Zoom angles have to be a little larger to produce the same image. I added an extra control, purely for experimental purposes. It's called Perspective Scale, and amounts to how far away the viewer is from the plane. The larger the value, the less rapidly the lines will converge. I'm not sure the way I handle this is the most sensible, and I don't currency understand how the equivalent feature is handled by Rotate/Zoom. (It's constant in Rotate/Zoom, but I don't know how that constant is determined.)
To produce the same distortion as Rotate/Zoom with a 75 degree tilt, I used a 72.3 degree tilt, and a Perspective Scale of 3.43. I don't yet fully understand the computations in Rotate/Zoom. They're kind of confusing, and of course, the disassembly has no comments.
As you can see, the changes are pretty minor. Just computing two values in Render, and replacing the perspective code in the transformation. Note I compute W in the perspective code, but don't divide by it till the rotate and zoom are complete. W is unaffected by those transformations, and doesn't affect them.
Hidden Content:
```// Name: Wat Tyler
// Author: Red ochre (John Robbins)
// URL: http://www.getpaint.net/redirect/plugins.html
// Title: Wat Tyler Bad Tilt Beta 9/5/16 Red Ochre
#region UICode
Pair<double, double> Amount1 = Pair.Create( 0.0 , 0.0 ); // Centre
byte Amount2 = 1; // Tiling options|reflect|repeat|reflect brick|repeat brick|clamp|none
bool Amount3 = true; // [0,1] Limit to integers
double Amount4 = 1; // [1,20] Horizontal(X) zoom out
double Amount5 = 1; // [1,20] Vertical (Y) zoom out
bool Amount6 = true; // [0,1] Link X & Y zoom
double Amount7 = 0; // [0,90] Tilt back angle BAD!
double Amount8 = 0; // [-180,180] Rotation
bool Amount9 = true; // [0,1] Limit rotation (15 degree steps)
bool Amount10 = false; // [0,1] Faster (lower quality)
double Amount11 = 3.0; //[1,20] Perspective Scale (for demonstation purposes)
#endregion
double PI = Math.PI;
double PI2 = Math.PI/2;
// Methods here
private ColorBgra move (Surface src,float X,float Y,float xoffset,float yoffset,float Xzoom,float Yzoom,float cosZ,float sinZ, double Tangle)
{
Rectangle sel = this.EnvironmentParameters.GetSelection(src.Bounds).GetBoundsInt();
int sL = sel.Left;
int sR = sel.Right;
int sT = sel.Top;
int sB = sel.Bottom;
int H = sel.Height;
int W = sel.Width;
float hW = (float)(W)/2;
float hH = (float)(H)/2;
ColorBgra Np;
//let C be the centre of zoom and rotation at a moveable but static position.
//1. subtract selection top or left
//2. work out distance to (moveable but static) crosshairs
//3. distort new xdist & ydist for rotation.
//4. distort for zoom out
//5. currently distort for tilt back VERY WRONG!!!
// Ideally this would be calculated in Render to change the number of samples made.
//6. feed new x & y through tesselation methods.
//7. add back selection top and left and access pixel values to feed back to SSpix method,
// where average taken and fed back to Render
float Cx = xoffset + hW;
float Cy = yoffset + hH;
float xdist = X - Cx;//selection.Left not in X (only in x)
float ydist = Y - Cy;
//try tilt here to get maths correct--------------------
//DOES NOT WORK
// I have tried every possible calculation I can think of here:
// Tanh, powers etc etc etc.
// I always end up with curved perspective lines?
// It really hurts me to give up but I just don't have the understanding of mathematics to do this correctly
//
//float xwarpT = xdist;
//float ywarpT = ydist;
//float Hrat = (H -Y)/H;
//float iHrat = 1 - Hrat;
//xwarpT = (float)((xdist + iHrat) + (Hrat * Math.Tan(Tangle) * xdist));//curved
//ywarpT = (float)((ydist + iHrat) + (Hrat * Math.Tan(Tangle) * ydist * 8));//all rubbish
float ywarpT = yTiltScale * ydist;
float w = wTiltScale * ydist + 1.0f;
//----------------------------Ends Tilt -------------------
float xwarpR = (cosZ * xdist) - (sinZ * ywarpT);//Rotation
float ywarpR = (cosZ * ywarpT) + (sinZ * xdist);
float xwarpZ = xwarpR * Xzoom;//Zoom out
float ywarpZ = ywarpR * Yzoom;
if (w <= 0.0f)
{
Np = ColorBgra.Transparent;
}
else
{
float recipW = 1.0f / w;
xwarpZ *= recipW;
ywarpZ *= recipW;
float nxP = xwarpZ + hW;
float nyP = ywarpZ + hH;
bool Oob = false;//out of bounds for case 5 only
float repx = (int)Math.Abs(nxP/W);// NOW FLOAT! (really int but saves loads of boxing/casting)
float repy = (int)Math.Abs(nyP/H);
//Syntax,value of this = condition ? value if true:value if false
//REMEMBER selection irrelevant here!!! use W, H & zero only.
switch(Amount2)
{
case 0://reflect - default
if(nxP < 0 ){nxP = - nxP;}
if(nxP >= W ){nxP = repx%2 >= 1 ? W - nxP%W: nxP%W;}
if(nyP < 0 ){nyP = - nyP;}
if(nyP >= H ){nyP = repy%2 >= 1 ? H - nyP%H: nyP%H;}
break;
case 1://repeat
if(nxP < 0){nxP = W - Math.Abs(nxP%W);}
if(nxP >= W){nxP = nxP%W;}
if(nyP < 0){nyP = H - Math.Abs(nyP%H);}
if(nyP >= H){nyP = nyP%H;}
break;
case 2://brick reflect
if(nyP >= 0){repy += 1;}//must be first
if(repy%2 == 0 ){nxP = hW + nxP;}
repy = (int)Math.Abs(nyP/H);//must be re-calculated as changed values
repx = (int)Math.Abs(nxP/W);
if(nxP < 0){nxP = - nxP;}
if(nxP >= W){nxP = repx%2 >= 1 ? W - nxP%W: nxP%W;}
if(nyP < 0){nyP = - nyP;}
if(nyP >= H){nyP = repy%2 >= 1 ? H - nyP%H: nyP%H ;}
break;
case 3://brick repeat
nyP = nyP + 0.0000001f;//bodge!!!! - no idea where the real bug is?
if(nyP > 0){repy += 1;}
if(repy%2 == 0){nxP = hW + nxP;}
if(nxP < 0){nxP = W - Math.Abs(nxP%W);}
nxP = nxP%W;
if(nyP < 0){nyP = H - Math.Abs(nyP%H);}
nyP = nyP%H;
break;
case 4://clamp
if(nxP < sL){nxP = sL;}
if(nxP > sR){nxP = sR;}
if(nyP < sT){nyP = sT;}
if(nyP > sB){nyP = sB;}
break;
case 5://none
if(nxP < sL || nxP > sR || nyP < sT || nyP > sB){Oob = true;}
break;
}
nxP = nxP + sL;//add selection top and left back in
nyP = nyP + sT;
ColorBgra klear = ColorBgra.Transparent;
Np = src.GetBilinearSampleClamped(nxP,nyP);
if(Oob){Np = klear;}
}
return Np;
}
private ColorBgra SSpix(Surface src,int x, int y,float Xstep,float Ystep, int xsamples, int ysamples,float xoffset,float yoffset,float Xzoom,float Yzoom,float cosZ,float sinZ,double Tangle)
{ //Sub-pixel Sampling method
ColorBgra SStemp = ColorBgra.Aquamarine;
int SSno = xsamples * ysamples;
int B = 0;int G = 0;int R = 0;int A = 0;
float tempB = 0;float tempG = 0;float tempR = 0;float tempA = 0;
float X = 0;float Y = 0;
double Trat = 0;
Rectangle sel = EnvironmentParameters.GetSelection(src.Bounds).GetBoundsInt();
double H = sel.Height;
for (int Ys = 0; Ys < ysamples;Ys++)
{
Y = (y - sel.Top) + (Ys * Ystep);
for (int Xs = 0; Xs < xsamples;Xs++)
{
X = (x - sel.Left) + (Xs * Xstep);//note: now float
SStemp = move(src, X, Y, xoffset, yoffset, Xzoom, Yzoom, cosZ, sinZ, Tangle);
tempB += SStemp.B;
tempG += SStemp.G;
tempR += SStemp.R;
tempA += SStemp.A;// sum values
}
}
B = (int)(tempB/SSno);//end of SPS loop... divide by number of pixels sampled to find average
G = (int)(tempG/SSno);
R = (int)(tempR/SSno);
A = (int)(tempA/SSno);
return ColorBgra.FromBgra(Int32Util.ClampToByte(, Int32Util.ClampToByte(G), Int32Util.ClampToByte(R), Int32Util.ClampToByte(A));
}
// There's nothing wrong with using "global" variables (which are actually class variables)
// to communicate with subroutines, provided that the values never change during the
// rendering process. The values are state, not arguments. It seems to me to be so much
// neater than passing a huge number of unchanging arguments.
float yTiltScale;
float wTiltScale;
void Render(Surface dst, Surface src, Rectangle rect)
{
Rectangle sel = EnvironmentParameters.GetSelection(src.Bounds).GetBoundsInt();
float W = (float) sel.Width;
float hW = (float)(sel.Width/2);
float hH = (float)(sel.Height/2);
float H = (float)sel.Height;
int B = 0;
int G = 0;
int R = 0;
int A = 0;
int tempB = 0;
int tempG = 0;
int tempR = 0;
int tempA = 0;
ColorBgra cp = ColorBgra.Aquamarine; //just to declare a value
float xoffset = (float)(Amount1.First * hW);
float yoffset = (float)(Amount1.Second * hH);
float Xzoom = (float)Amount4;
float Yzoom = (float)Amount5;if(Amount6){Yzoom = (float)Amount4;}
if(Amount3){Xzoom = (int)(Xzoom);Yzoom = (int)(Yzoom);}
int xsamples = (int)Xzoom;
int ysamples = (int)Yzoom;
float Xstep = (float)(1.0/xsamples);
float Ystep = (float)(1.0/ysamples);
double Zrot = PI * Amount8/180;//all in radians now
if(Amount9){int uaf = (int)(Amount8/15);Zrot = PI * (uaf * 15)/180;}//confine to multiples of 7.5 degrees
float cosZ = (float)Math.Cos(-Zrot);
float sinZ = (float)Math.Sin(-Zrot);//just prefer things rotating as the slider does
double Tangle = PI * Amount7/180;//-ve halfPI to +ve PI
double tanT = Math.Tan(Tangle), cosT = Math.Cos(Tangle);
yTiltScale = (float)(1.0 / cosT);
wTiltScale = (float)(tanT) / ((float)Amount11 * hH);
for (int y = rect.Top; y < rect.Bottom; y++)
{
for (int x = rect.Left; x < rect.Right; x++)
{
double Trat = (H - (y - sel.Top))/H;
if(Amount10){xsamples = ysamples = 1;}
cp = SSpix(src,x,y,Xstep,Ystep,xsamples,ysamples,xoffset,yoffset,Xzoom,Yzoom,cosZ,sinZ,Tangle);//call Sub-pixel sampling method here
dst[x,y] = cp;
}
}
}```
EDIT: I'm convinced I'm not doing the perspective transformation correctly. The foreshortening is wrong, so the image stretches in Y as the tilt increases. I have an idea why, but I'll need to do a little math. I tried to derive it using 3x3 matrices instead of the usual 4x4, and I think it led me astray.
EDIT 2: What a silly mistake! The posted code seems just fine. I was experimenting with something in my version, and managed to use the wrong version of the Y coordinate at one step in the process. Arghh!
Edited by MJW
##### Share on other sites
Excellent initial results. Straight perspective lines! Hurray!
I will get back to you when I've studied it a bit more but in the meantime many, many thanks! :star: :star:
Red ochre Plugin pack.............. Diabolical Drawings ................Real Paintings
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× | 5,407 | 18,527 | {"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} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.907229 |
https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_19&diff=prev&oldid=136093 | 1,627,368,599,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152236.64/warc/CC-MAIN-20210727041254-20210727071254-00487.warc.gz | 114,797,463 | 11,603 | # Difference between revisions of "2017 AMC 8 Problems/Problem 19"
## Problem 19
For any positive integer $M$, the notation $M!$ denotes the product of the integers $1$ through $M$. What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$ ?
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$
## Solution 1
Factoring out $98!+99!+100!$, we have $98!(1+99+99*100)$ which is $98!(10000)$ Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$. The $19$ is because of all the multiples of $5$. Now $10,000$ has $4$ factors of $5$, so there are a total of $22 + 4 = \boxed{\textbf{(D)}\ 26}$ factors of $5$.
## Solution 2
The number of $5$'s in the factorization of $98! + 99! + 100!$ is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by $5$, until you can't divide by $5$ anymore. Factorizing $98! + 99! + 100!$, you get $98!(1+99+9900)=98!(10000)$. To find the number of trailing zeroes in 98!, we do $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22$. Now since $10000$ has 4 zeroes, we add $22 + 4$ to get $\boxed{\textbf{(D)}\ 26}$ factors of $5$. | 481 | 1,313 | {"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": 32, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2021-31 | latest | en | 0.774426 |
https://www.coursehero.com/file/6770097/Hw3Fl05Alg/ | 1,495,866,188,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608870.18/warc/CC-MAIN-20170527055922-20170527075922-00116.warc.gz | 1,095,040,368 | 95,276 | Hw3Fl05Alg
# Hw3Fl05Alg - up characteristic equation T(N = T(N-1 T(N-2...
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CSE 5211/4081 Algorithm Analysis HomeWork 3 Due: 9/29/05 Points: 15/25 .1. Write the recurrence equation for time complexity of the following algorithm and solve it using Master Theorem. [5] Algorithm Useless (Global array A, index first , index last ) .(1) if first = = last .(2) return Else { .(3) for int i := first through last do .(4) for int j := last through first do .(5) global array B[j] := A[i]*A[j]; .(6) int mid1 := first+(last – first)/3; .(7) int mid2 := first+ 2*(last – first)/3; .(8) Useless(A, first, mid1); .(9) Useless(A, mid1 +1, mid2); .(10) Useless(A, mid2 +1, last); }; End if; End algorithm. [For lines 3 through 5 you may use the final asymptotic complexity function rather than showing the calculation via the summation series. Ignore the constant-time steps, e.g., line 6 and 7, in setting up the recurrence equation. If you are playing with any particular problem size, then use n=3^k for some integer k, e.g., n=27.] .2. Solve the following homogeneous recurrence equation for general solution by setting
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Unformatted text preview: up characteristic equation: T(N) = T(N-1) + T(N-2) [ Note that this is the Fibonacci number and not the complexity of calculating it.] The roots of the quadratic equation ax 2 + bx + c = 0 are (-b + sqareroot(b 2 – 4ac))/2a and (-b - sqareroot(b 2 – 4ac))/2a. [5] GRAD QUSETION 2b. Set up the characteristic equation for: T(N) = T(N-1) + T(N-2) +1 [5] UNDERGRAD QUESTION: 3. Solve the following recurrence equation for general solution by setting up its characteristic equation, and then mention the order (theta) function for that solution. T(N) = 2T(N-1) - T(N-2) [5] GRAD QUESTION 3. Solve the following recurrence equation for general solution by setting up its characteristic equation, and then mention the order (theta) function for that solution. T(N) = 2T(N-1) - T(N-2) +1 [5] .4. GRAD QUESTION: Analyze average case complexity of QuickSelect algorithm by setting up its recurrence equation and then solving it by telescopic method. [5]...
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## This note was uploaded on 02/10/2012 for the course CSE 5211 taught by Professor Dmitra during the Spring '12 term at FIT.
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Hw3Fl05Alg - up characteristic equation T(N = T(N-1 T(N-2...
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Ask a homework question - tutors are online | 765 | 2,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2017-22 | longest | en | 0.829055 |
https://neutronstars.utk.edu/code/o2scl/html/inte.html | 1,679,442,621,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00356.warc.gz | 504,903,839 | 18,429 | # Integration¶
O2scl
## One-dimensional integration based on GSL, CERNLIB, and Boost¶
Several classes integrate arbitrary one-dimensional functions:
• Integration over a finite interval:
• Integration from $$a$$ to $$\infty$$:
• Integration from $$-\infty$$ to $$b$$:
• Integration from $$-\infty$$ to $$\infty$$:
• Integration over a finite interval for a function with singularities:
• Cauchy principal value integration over a finite interval:
• Integration over a function weighted by cos(x) or sin(x):
• Fourier integrals:
• Integration over a weight function
$W(x)=(x-a)^{\alpha}(b-x)^{\beta}\log^{\mu}(x-a)\log^{\nu}(b-x)$
is performed by inte_qaws_gsl.
Note that some of the integrators support multiprecision, see Multiprecision Support.
There are two competing factors that can slow down an adaptive integration algorithm: (1) each evaluation of the integrand can be numerically expensive, depending on how the function is defined, and (2) the process of subdividing regions and recalculating values is almost always numerically expensive in its own right. For integrands that are very smooth (e.g., analytic functions), a high-order Gauss-Kronrod rule (e.g., 61-point) will achieve the desired error tolerance with relatively few subdivisions. For integrands with discontinuities or singular derivatives, a low-order rule (e.g., 15-point) is often more efficient.
## GSL-based integration details¶
For the GSL-based integration routines, the variables o2scl::inte::tol_abs and o2scl::inte::tol_rel have the same role as the quantities usually denoted in the GSL integration routines by epsabs and epsrel. In particular, the integration classes attempt to ensure that
$|\mathrm{result}-I| \leq \mathrm{Max}(\mathrm{tol\_abs}, \mathrm{tol\_rel}|I|)$
and returns an error to attempt to ensure that
$|\mathrm{result}-I| \leq \mathrm{abserr} \leq \mathrm{Max}(\mathrm{tol\_abs},\mathrm{tol\_rel}|I|)$
where I is the integral to be evaluated. Even when the corresponding descendant of o2scl::inte::integ() returns success, these inequalities may fail for sufficiently difficult functions. All of the GSL integration routines except for inte_qng_gsl use a workspace given in inte_workspace_gsl which holds the results of the various subdivisions of the original interval.
The GSL routines were originally based on QUADPACK, which is available at http://www.netlib.org/quadpack .
For adaptive GSL integration classes, the type of Gauss-Kronrod quadrature rule that is used to approximate the integral and estimate the error of a subinterval is set by o2scl::inte_kronrod_gsl::set_rule().
The number of subdivisions of the integration region is limited by the size of the workspace, set in o2scl::inte_kronrod_gsl::set_limit(). The number of subdivisions required for the most recent call to o2scl::inte::integ() or o2scl::inte::integ_err() is given in o2scl::inte::last_iter. This number will always be less than or equal to the workspace size.
Note
The GSL integration routines can sometimes lose precision if the integrand is everywhere much smaller than unity. Some rescaling may be required in these cases.
## GSL-based integration error messages¶
The error messages given by the adaptive GSL integration routines tend to follow a standard form and are documented here. There are several error messages which indicate improper usage and cause the error handler to be called regardless of the value of o2scl::inte::err_nonconv:
There are also convergence errors which will call the error handler unless o2scl::inte::err_nonconv is false. See What is an error? for more discussion on convergence errors versus fatal errors:
• Cannot reach tolerance because of roundoff error on first attempt in class::function(). [ exc_eround ] Each integration attempt tests for round-off errors by comparing the computed integral with that of the integrand’s absolute value (i.e., $$L^1$$ -norm). A highly oscillatory integrand may cause this error.
• A maximum of 1 iteration was insufficient in class::function(). [ exc_emaxiter ] This occurs if the workspace is allocated for one interval and a single Gauss-Kronrod integration does not yield the accuracy demanded by o2scl::inte::tol_abs and o2scl::inte::tol_rel.
• Bad integrand behavior in class::function(). [ exc_esing ] This occurs if the integrand is (effectively) singular in a region, causing the subdivided intervals to become too small for floating-point precision.
• Maximum number of subdivisions 'value' reached in class::function(). [ exc_emaxiter ] This occurs if the refinement algorithm runs out of allocated workspace. The number of iterations required for the most recent call to o2scl::inte::integ() or o2scl::inte::integ_err() is given in o2scl::inte::last_iter. This number will always be less than or equal to the workspace size.
• Roundoff error prevents tolerance from being achieved in class::function(). [ exc_eround ] The refinement procedure counts round-off errors as they occur and terminates if too many such errors accumulate.
• Roundoff error detected in extrapolation table in inte_singular_gsl::qags(). [ exc_eround ] This occurs when error-terms from the $$\varepsilon$$ -algorithm are are monitored and compared with the error-terms from the refinement procedure. The algorithm terminates if these sequences differ by too many orders of magnitude. See o2scl::inte_singular_gsl::qelg().
• Integral is divergent or slowly convergent in inte_singular_gsl::qags(). [ exc_ediverge ] This occurs if the approximations produced by the refinement algorithm and the extrapolation algorithm differ by too many orders of magnitude.
• Exceeded limit of trigonometric table in inte_qawo_gsl_sin()::qawo(). [ exc_etable ] This occurs if the maximum level of the table of Chebyshev moments is reached.
## One-dimensional integration example¶
This example computes the integral $$\int_{-\infty}^{\infty} e^{-x^2} ~dx$$ with inte_qagi_gsl, the integral $$\int_0^{\infty} e^{-x^2} ~dx$$ with inte_qagiu_gsl, the integral $$\int_{-\infty}^{0} e^{-x^2} ~dx$$ with inte_qagil_gsl, and the integral $$\int_0^1 \left[ \sin (2 x) + \frac{1}{2} \right]~dx$$ with both inte_qag_gsl and inte_adapt_cern, and compares the computed results with the exact results.
/* Example: ex_inte.cpp
-------------------------------------------------------------------
An example to demonstrate numerical integration.
*/
#include <cmath>
#include <o2scl/test_mgr.h>
#include <o2scl/constants.h>
#include <o2scl/funct.h>
#include <o2scl/inte_qag_gsl.h>
#include <o2scl/inte_qagi_gsl.h>
#include <o2scl/inte_qagiu_gsl.h>
#include <o2scl/inte_qagil_gsl.h>
using namespace std;
using namespace o2scl;
using namespace o2scl_const;
class cl {
public:
// We'll use this to count the number of function
// evaulations required by the integration routines
int nf;
// A function to be integrated
double integrand(double x) {
nf++;
return exp(-x*x);
}
// Another function to be integrated
double integrand2(double x) {
nf++;
return sin(2.0*x)+0.5;
}
};
int main(void) {
cl acl;
test_mgr t;
t.set_output_level(1);
funct f1=std::bind(std::mem_fn<double(double)>
(&cl::integrand),&acl,std::placeholders::_1);
funct f2=std::bind(std::mem_fn<double(double)>
(&cl::integrand2),&acl,std::placeholders::_1);
// We don't need to specify the function type in the integration
// objects, because we're using the default function type (type
// funct).
inte_qag_gsl<> g;
inte_qagi_gsl<> gi;
inte_qagiu_gsl<> gu;
inte_qagil_gsl<> gl;
// The result and the uncertainty
double res, err;
// An integral from -infinity to +infinity (the limits are ignored)
acl.nf=0;
int ret1=gi.integ_err(f1,0.0,0.0,res,err);
cout << "inte_qagi_gsl: " << endl;
cout << "Return value: " << ret1 << endl;
cout << "Result: " << res << " Uncertainty: " << err << endl;
cout << "Number of iterations: " << gi.last_iter << endl;
cout << "Number of function evaluations: " << acl.nf << endl;
cout << endl;
t.test_rel(res,sqrt(pi),1.0e-8,"inte 1");
// An integral from 0 to +infinity (the second limit argument is
// ignored in the line below)
acl.nf=0;
gu.integ_err(f1,0.0,0.0,res,err);
cout << "inte_qagiu_gsl: " << endl;
cout << "Return value: " << ret1 << endl;
cout << "Result: " << res << " Uncertainty: " << err << endl;
cout << "Number of iterations: " << gu.last_iter << endl;
cout << "Number of function evaluations: " << acl.nf << endl;
cout << endl;
t.test_rel(res,sqrt(pi)/2.0,1.0e-8,"inte 2");
// An integral from -infinity to zero (the first limit argument is
// ignored in the line below)
acl.nf=0;
gl.integ_err(f1,0.0,0.0,res,err);
cout << "inte_qagil_gsl: " << endl;
cout << "Return value: " << ret1 << endl;
cout << "Result: " << res << " Uncertainty: " << err << endl;
cout << "Number of iterations: " << gl.last_iter << endl;
cout << "Number of function evaluations: " << acl.nf << endl;
cout << endl;
t.test_rel(res,sqrt(pi)/2.0,1.0e-8,"inte 3");
// An integral from 0 to 1 with the GSL integrator
acl.nf=0;
g.integ_err(f2,0.0,1.0,res,err);
cout << "inte_qag_gsl: " << endl;
cout << "Return value: " << ret1 << endl;
cout << "Result: " << res << " Uncertainty: " << err << endl;
cout << "Number of iterations: " << g.last_iter << endl;
cout << "Number of function evaluations: " << acl.nf << endl;
cout << endl;
t.test_rel(res,0.5+sin(1.0)*sin(1.0),1.0e-8,"inte 4");
// The same integral with the CERNLIB integrator
acl.nf=0;
ca.integ_err(f2,0.0,1.0,res,err);
cout << "inte_adapt_cern: " << endl;
cout << "Return value: " << ret1 << endl;
cout << "Result: " << res << " Uncertainty: " << err << endl;
cout << "Number of iterations: " << ca.last_iter << endl;
cout << "Number of function evaluations: " << acl.nf << endl;
cout << endl;
t.test_rel(res,0.5+sin(1.0)*sin(1.0),1.0e-8,"inte 5");
t.report();
return 0;
}
## Gauss-Kronrod integration coefficients¶
Top
namespace o2scl_inte_gk_coeffs
A namespace for the GSL adaptive Gauss-Kronrod integration coefficients.
origin permits storing only those abscissae in the interval
$$[0, 1).$$ Similiarly for the weights. The odd-indexed abscissae xgk[1], xgk[3],… belong to the low-order Gauss rule. For the full Gauss-Kronrod rule, the array wgk[] holds the weights corresponding to the abscissae xgk[]. For the low-order rule, wg[i] is the weight correpsonding to xgk[2*i+1].
Documentation from GSL
:
Gauss quadrature weights and kronrod quadrature abscissae and weights as evaluated with 80 decimal digit arithmetic by L. W. Fullerton, Bell Labs, Nov. 1981.
Variables
static const double qk15_xgk[8] = {0.991455371120812639206854697526329, 0.949107912342758524526189684047851, 0.864864423359769072789712788640926, 0.741531185599394439863864773280788, 0.586087235467691130294144838258730, 0.405845151377397166906606412076961, 0.207784955007898467600689403773245, 0.000000000000000000000000000000000}
Abscissae of the 15-point Kronrod rule.
static const double qk15_wg[4] = {0.129484966168869693270611432679082, 0.279705391489276667901467771423780, 0.381830050505118944950369775488975, 0.417959183673469387755102040816327}
Weights of the 7-point Gauss rule.
static const double qk15_wgk[8] = {0.022935322010529224963732008058970, 0.063092092629978553290700663189204, 0.104790010322250183839876322541518, 0.140653259715525918745189590510238, 0.169004726639267902826583426598550, 0.190350578064785409913256402421014, 0.204432940075298892414161999234649, 0.209482141084727828012999174891714}
Weights of the 15-point Kronrod rule.
static const double qk21_xgk[11] = {0.995657163025808080735527280689003, 0.973906528517171720077964012084452, 0.930157491355708226001207180059508, 0.865063366688984510732096688423493, 0.780817726586416897063717578345042, 0.679409568299024406234327365114874, 0.562757134668604683339000099272694, 0.433395394129247190799265943165784, 0.294392862701460198131126603103866, 0.148874338981631210884826001129720, 0.000000000000000000000000000000000}
Abscissae of the 21-point Kronrod rule.
static const double qk21_wg[5] = {0.066671344308688137593568809893332, 0.149451349150580593145776339657697, 0.219086362515982043995534934228163, 0.269266719309996355091226921569469, 0.295524224714752870173892994651338}
Weights of the 10-point Gauss rule.
static const double qk21_wgk[11] = {0.011694638867371874278064396062192, 0.032558162307964727478818972459390, 0.054755896574351996031381300244580, 0.075039674810919952767043140916190, 0.093125454583697605535065465083366, 0.109387158802297641899210590325805, 0.123491976262065851077958109831074, 0.134709217311473325928054001771707, 0.142775938577060080797094273138717, 0.147739104901338491374841515972068, 0.149445554002916905664936468389821}
Weights of the 21-point Kronrod rule.
static const double qk31_xgk[16] = {0.998002298693397060285172840152271, 0.987992518020485428489565718586613, 0.967739075679139134257347978784337, 0.937273392400705904307758947710209, 0.897264532344081900882509656454496, 0.848206583410427216200648320774217, 0.790418501442465932967649294817947, 0.724417731360170047416186054613938, 0.650996741297416970533735895313275, 0.570972172608538847537226737253911, 0.485081863640239680693655740232351, 0.394151347077563369897207370981045, 0.299180007153168812166780024266389, 0.201194093997434522300628303394596, 0.101142066918717499027074231447392, 0.000000000000000000000000000000000}
Abscissae of the 31-point Kronrod rule.
static const double qk31_wg[8] = {0.030753241996117268354628393577204, 0.070366047488108124709267416450667, 0.107159220467171935011869546685869, 0.139570677926154314447804794511028, 0.166269205816993933553200860481209, 0.186161000015562211026800561866423, 0.198431485327111576456118326443839, 0.202578241925561272880620199967519}
Weights of the 15-point Gauss rule.
static const double qk31_wgk[16] = {0.005377479872923348987792051430128, 0.015007947329316122538374763075807, 0.025460847326715320186874001019653, 0.035346360791375846222037948478360, 0.044589751324764876608227299373280, 0.053481524690928087265343147239430, 0.062009567800670640285139230960803, 0.069854121318728258709520077099147, 0.076849680757720378894432777482659, 0.083080502823133021038289247286104, 0.088564443056211770647275443693774, 0.093126598170825321225486872747346, 0.096642726983623678505179907627589, 0.099173598721791959332393173484603, 0.100769845523875595044946662617570, 0.101330007014791549017374792767493}
Weights of the 31-point Kronrod rule.
static const double qk41_xgk[21] = {0.998859031588277663838315576545863, 0.993128599185094924786122388471320, 0.981507877450250259193342994720217, 0.963971927277913791267666131197277, 0.940822633831754753519982722212443, 0.912234428251325905867752441203298, 0.878276811252281976077442995113078, 0.839116971822218823394529061701521, 0.795041428837551198350638833272788, 0.746331906460150792614305070355642, 0.693237656334751384805490711845932, 0.636053680726515025452836696226286, 0.575140446819710315342946036586425, 0.510867001950827098004364050955251, 0.443593175238725103199992213492640, 0.373706088715419560672548177024927, 0.301627868114913004320555356858592, 0.227785851141645078080496195368575, 0.152605465240922675505220241022678, 0.076526521133497333754640409398838, 0.000000000000000000000000000000000}
Abscissae of the 41-point Kronrod rule.
static const double qk41_wg[11] = {0.017614007139152118311861962351853, 0.040601429800386941331039952274932, 0.062672048334109063569506535187042, 0.083276741576704748724758143222046, 0.101930119817240435036750135480350, 0.118194531961518417312377377711382, 0.131688638449176626898494499748163, 0.142096109318382051329298325067165, 0.149172986472603746787828737001969, 0.152753387130725850698084331955098}
Weights of the 20-point Gauss rule.
static const double qk41_wgk[21] = {0.003073583718520531501218293246031, 0.008600269855642942198661787950102, 0.014626169256971252983787960308868, 0.020388373461266523598010231432755, 0.025882133604951158834505067096153, 0.031287306777032798958543119323801, 0.036600169758200798030557240707211, 0.041668873327973686263788305936895, 0.046434821867497674720231880926108, 0.050944573923728691932707670050345, 0.055195105348285994744832372419777, 0.059111400880639572374967220648594, 0.062653237554781168025870122174255, 0.065834597133618422111563556969398, 0.068648672928521619345623411885368, 0.071054423553444068305790361723210, 0.073030690332786667495189417658913, 0.074582875400499188986581418362488, 0.075704497684556674659542775376617, 0.076377867672080736705502835038061, 0.076600711917999656445049901530102}
Weights of the 41-point Kronrod rule.
static const double qk51_xgk[26] = {0.999262104992609834193457486540341, 0.995556969790498097908784946893902, 0.988035794534077247637331014577406, 0.976663921459517511498315386479594, 0.961614986425842512418130033660167, 0.942974571228974339414011169658471, 0.920747115281701561746346084546331, 0.894991997878275368851042006782805, 0.865847065293275595448996969588340, 0.833442628760834001421021108693570, 0.797873797998500059410410904994307, 0.759259263037357630577282865204361, 0.717766406813084388186654079773298, 0.673566368473468364485120633247622, 0.626810099010317412788122681624518, 0.577662930241222967723689841612654, 0.526325284334719182599623778158010, 0.473002731445714960522182115009192, 0.417885382193037748851814394594572, 0.361172305809387837735821730127641, 0.303089538931107830167478909980339, 0.243866883720988432045190362797452, 0.183718939421048892015969888759528, 0.122864692610710396387359818808037, 0.061544483005685078886546392366797, 0.000000000000000000000000000000000}
Abscissae of the 51-point Kronrod rule.
static const double qk51_wg[13] = {0.011393798501026287947902964113235, 0.026354986615032137261901815295299, 0.040939156701306312655623487711646, 0.054904695975835191925936891540473, 0.068038333812356917207187185656708, 0.080140700335001018013234959669111, 0.091028261982963649811497220702892, 0.100535949067050644202206890392686, 0.108519624474263653116093957050117, 0.114858259145711648339325545869556, 0.119455763535784772228178126512901, 0.122242442990310041688959518945852, 0.123176053726715451203902873079050}
Weights of the 25-point Gauss rule.
static const double qk51_wgk[26] = {0.001987383892330315926507851882843, 0.005561932135356713758040236901066, 0.009473973386174151607207710523655, 0.013236229195571674813656405846976, 0.016847817709128298231516667536336, 0.020435371145882835456568292235939, 0.024009945606953216220092489164881, 0.027475317587851737802948455517811, 0.030792300167387488891109020215229, 0.034002130274329337836748795229551, 0.037116271483415543560330625367620, 0.040083825504032382074839284467076, 0.042872845020170049476895792439495, 0.045502913049921788909870584752660, 0.047982537138836713906392255756915, 0.050277679080715671963325259433440, 0.052362885806407475864366712137873, 0.054251129888545490144543370459876, 0.055950811220412317308240686382747, 0.057437116361567832853582693939506, 0.058689680022394207961974175856788, 0.059720340324174059979099291932562, 0.060539455376045862945360267517565, 0.061128509717053048305859030416293, 0.061471189871425316661544131965264, 0.061580818067832935078759824240066}
Weights of the 51-point Kronrod rule.
static const double qk61_xgk[31]
Abscissae of the 61-point Kronrod rule.
static const double qk61_wg[15] = {0.007968192496166605615465883474674, 0.018466468311090959142302131912047, 0.028784707883323369349719179611292, 0.038799192569627049596801936446348, 0.048402672830594052902938140422808, 0.057493156217619066481721689402056, 0.065974229882180495128128515115962, 0.073755974737705206268243850022191, 0.080755895229420215354694938460530, 0.086899787201082979802387530715126, 0.092122522237786128717632707087619, 0.096368737174644259639468626351810, 0.099593420586795267062780282103569, 0.101762389748405504596428952168554, 0.102852652893558840341285636705415}
Weights of the 30-point Gauss rule.
static const double qk61_wgk[31]
Weights of the 61-point Kronrod rule.
Top
namespace o2scl_inte_qng_coeffs
Documentation from GSL
:
Gauss-Kronrod-Patterson quadrature coefficients for use in quadpack routine qng. These coefficients were calculated with 101 decimal digit arithmetic by L. W. Fullerton, Bell Labs, Nov
Variables
static const double x1[5] = {0.973906528517171720077964012084452, 0.865063366688984510732096688423493, 0.679409568299024406234327365114874, 0.433395394129247190799265943165784, 0.148874338981631210884826001129720}
Abcissae common to the 10-, 21-, 43- and 87-point rule
static const double w10[5] = {0.066671344308688137593568809893332, 0.149451349150580593145776339657697, 0.219086362515982043995534934228163, 0.269266719309996355091226921569469, 0.295524224714752870173892994651338}
Weights of the 10-point formula
static const double x2[5] = {0.995657163025808080735527280689003, 0.930157491355708226001207180059508, 0.780817726586416897063717578345042, 0.562757134668604683339000099272694, 0.294392862701460198131126603103866}
Abcissae common to the 21-, 43- and 87-point rule
static const double w21a[5] = {0.032558162307964727478818972459390, 0.075039674810919952767043140916190, 0.109387158802297641899210590325805, 0.134709217311473325928054001771707, 0.147739104901338491374841515972068}
Weights of the 21-point formula for abcissae x1
static const double w21b[6] = {0.011694638867371874278064396062192, 0.054755896574351996031381300244580, 0.093125454583697605535065465083366, 0.123491976262065851077958109831074, 0.142775938577060080797094273138717, 0.149445554002916905664936468389821}
Weights of the 21-point formula for abcissae x2
static const double x3[11] = {0.999333360901932081394099323919911, 0.987433402908088869795961478381209, 0.954807934814266299257919200290473, 0.900148695748328293625099494069092, 0.825198314983114150847066732588520, 0.732148388989304982612354848755461, 0.622847970537725238641159120344323, 0.499479574071056499952214885499755, 0.364901661346580768043989548502644, 0.222254919776601296498260928066212, 0.074650617461383322043914435796506}
Abscissae common to the 43- and 87-point rule
static const double w43a[10] = {0.016296734289666564924281974617663, 0.037522876120869501461613795898115, 0.054694902058255442147212685465005, 0.067355414609478086075553166302174, 0.073870199632393953432140695251367, 0.005768556059769796184184327908655, 0.027371890593248842081276069289151, 0.046560826910428830743339154433824, 0.061744995201442564496240336030883, 0.071387267268693397768559114425516}
Weights of the 43-point formula for abscissae x1, x3
static const double w43b[12] = {0.001844477640212414100389106552965, 0.010798689585891651740465406741293, 0.021895363867795428102523123075149, 0.032597463975345689443882222526137, 0.042163137935191811847627924327955, 0.050741939600184577780189020092084, 0.058379395542619248375475369330206, 0.064746404951445885544689259517511, 0.069566197912356484528633315038405, 0.072824441471833208150939535192842, 0.074507751014175118273571813842889, 0.074722147517403005594425168280423}
Weights of the 43-point formula for abscissae x3
static const double x4[22] = {0.999902977262729234490529830591582, 0.997989895986678745427496322365960, 0.992175497860687222808523352251425, 0.981358163572712773571916941623894, 0.965057623858384619128284110607926, 0.943167613133670596816416634507426, 0.915806414685507209591826430720050, 0.883221657771316501372117548744163, 0.845710748462415666605902011504855, 0.803557658035230982788739474980964, 0.757005730685495558328942793432020, 0.706273209787321819824094274740840, 0.651589466501177922534422205016736, 0.593223374057961088875273770349144, 0.531493605970831932285268948562671, 0.466763623042022844871966781659270, 0.399424847859218804732101665817923, 0.329874877106188288265053371824597, 0.258503559202161551802280975429025, 0.185695396568346652015917141167606, 0.111842213179907468172398359241362, 0.037352123394619870814998165437704}
Abscissae of the 87-point rule
static const double w87a[21] = {0.008148377384149172900002878448190, 0.018761438201562822243935059003794, 0.027347451050052286161582829741283, 0.033677707311637930046581056957588, 0.036935099820427907614589586742499, 0.002884872430211530501334156248695, 0.013685946022712701888950035273128, 0.023280413502888311123409291030404, 0.030872497611713358675466394126442, 0.035693633639418770719351355457044, 0.000915283345202241360843392549948, 0.005399280219300471367738743391053, 0.010947679601118931134327826856808, 0.016298731696787335262665703223280, 0.021081568889203835112433060188190, 0.025370969769253827243467999831710, 0.029189697756475752501446154084920, 0.032373202467202789685788194889595, 0.034783098950365142750781997949596, 0.036412220731351787562801163687577, 0.037253875503047708539592001191226}
Weights of the 87-point formula for abscissae x1, x2, x3
static const double w87b[23] = {0.000274145563762072350016527092881, 0.001807124155057942948341311753254, 0.004096869282759164864458070683480, 0.006758290051847378699816577897424, 0.009549957672201646536053581325377, 0.012329447652244853694626639963780, 0.015010447346388952376697286041943, 0.017548967986243191099665352925900, 0.019938037786440888202278192730714, 0.022194935961012286796332102959499, 0.024339147126000805470360647041454, 0.026374505414839207241503786552615, 0.028286910788771200659968002987960, 0.030052581128092695322521110347341, 0.031646751371439929404586051078883, 0.033050413419978503290785944862689, 0.034255099704226061787082821046821, 0.035262412660156681033782717998428, 0.036076989622888701185500318003895, 0.036698604498456094498018047441094, 0.037120549269832576114119958413599, 0.037334228751935040321235449094698, 0.037361073762679023410321241766599}
Weights of the 87-point formula for abscissae x4 | 8,721 | 25,712 | {"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} | 2.78125 | 3 | CC-MAIN-2023-14 | longest | en | 0.789223 |
http://www.examguruadda.in/2015/10/reasoning-quiz_14.html | 1,488,143,599,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172077.66/warc/CC-MAIN-20170219104612-00614-ip-10-171-10-108.ec2.internal.warc.gz | 399,015,662 | 35,615 | # Reasoning Quiz
Directions (Q. 1–5): Study the following information carefully to answer the given question :
A, B, C, D, E, F, G and H are sitting around a circular table but not facing the centre. C is second to the right of B. B is fourth to the left of E. F and A are not immediate neighbors of B. F is third to the left of A. H is not an immediate neighbor of A. G is not on the immediate left of A.
1. Which of the following is true about D ?
1) D is second to the left of F.
2) D is on the immediate left of A.
3) D is opposite of C.
4) D is third to the left of H.
5) None of these
2. What is the position of A with respect to B ?
1) Immediate left
2) Fifth to the left
3) Second to the right
4) Third to the right
5) None of these
3. Who is third to the left of C ?
1) G
2) A
3) D
4) B
5) E
4. Which of the following pairs are neighbor of G ?
1) H, B
2) D, A
3) A, B
4) Can't be determined
5) None of these
5. Who sits on the immediate left of E ?
1) C
2) F
3) H
4) D
5) None of these
1. 2
2. 5 Second to left of B.
3. 1
4. 3
5. 2 | 348 | 1,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-09 | longest | en | 0.893343 |
https://studysoup.com/tsg/statistics/378/contemporary-mathematics/chapter/18064/chapter-20 | 1,606,239,874,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176922.14/warc/CC-MAIN-20201124170142-20201124200142-00315.warc.gz | 508,971,498 | 10,776 | ×
×
# Solutions for Chapter CHAPTER 20 : INVESTMENTS
## Full solutions for Contemporary Mathematics | 6th Edition
ISBN: 9780538481267
Solutions for Chapter CHAPTER 20 : INVESTMENTS
Solutions for Chapter CHAPTER 20
4 5 0 388 Reviews
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##### ISBN: 9780538481267
Contemporary Mathematics was written by and is associated to the ISBN: 9780538481267. Since 50 problems in chapter CHAPTER 20 : INVESTMENTS have been answered, more than 6080 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Contemporary Mathematics, edition: 6. Chapter CHAPTER 20 : INVESTMENTS includes 50 full step-by-step solutions.
Key Statistics Terms and definitions covered in this textbook
• Alternative hypothesis
In statistical hypothesis testing, this is a hypothesis other than the one that is being tested. The alternative hypothesis contains feasible conditions, whereas the null hypothesis speciies conditions that are under test
• Asymptotic relative eficiency (ARE)
Used to compare hypothesis tests. The ARE of one test relative to another is the limiting ratio of the sample sizes necessary to obtain identical error probabilities for the two procedures.
• Bias
An effect that systematically distorts a statistical result or estimate, preventing it from representing the true quantity of interest.
• Bimodal distribution.
A distribution with two modes
• Binomial random variable
A discrete random variable that equals the number of successes in a ixed number of Bernoulli trials.
• Cause-and-effect diagram
A chart used to organize the various potential causes of a problem. Also called a ishbone diagram.
• Center line
A horizontal line on a control chart at the value that estimates the mean of the statistic plotted on the chart. See Control chart.
• Comparative experiment
An experiment in which the treatments (experimental conditions) that are to be studied are included in the experiment. The data from the experiment are used to evaluate the treatments.
• Conditional probability density function
The probability density function of the conditional probability distribution of a continuous random variable.
• Conidence level
Another term for the conidence coeficient.
• Consistent estimator
An estimator that converges in probability to the true value of the estimated parameter as the sample size increases.
• Contour plot
A two-dimensional graphic used for a bivariate probability density function that displays curves for which the probability density function is constant.
• Cumulative normal distribution function
The cumulative distribution of the standard normal distribution, often denoted as ?( ) x and tabulated in Appendix Table II.
• Defect concentration diagram
A quality tool that graphically shows the location of defects on a part or in a process.
• Discrete distribution
A probability distribution for a discrete random variable
• Extra sum of squares method
A method used in regression analysis to conduct a hypothesis test for the additional contribution of one or more variables to a model.
• Forward selection
A method of variable selection in regression, where variables are inserted one at a time into the model until no other variables that contribute signiicantly to the model can be found.
• Fraction defective
In statistical quality control, that portion of a number of units or the output of a process that is defective.
• Fractional factorial experiment
A type of factorial experiment in which not all possible treatment combinations are run. This is usually done to reduce the size of an experiment with several factors.
• Goodness of fit
In general, the agreement of a set of observed values and a set of theoretical values that depend on some hypothesis. The term is often used in itting a theoretical distribution to a set of observations.
× | 776 | 3,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-50 | latest | en | 0.87964 |
https://www.doubtnut.com/qna/391602674 | 1,717,088,057,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00059.warc.gz | 631,642,988 | 35,360 | # Two liquid drops of radii 1 mm and 2 mm merge in vacuum isothermally. Radius of resulting drop is
A
3mm
B
31/3mm
C
32/3mm
D
6mm
Video Solution
Text Solution
Verified by Experts
## R3=r31+r32
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## Two soap bubbles of radii 1 mm and 2 mm merge isothermally . Then radius of the new bubble formed would be
A3 mm
B2/3 mm
C3/2 mm
D5 mm
• Question 1 - Select One
## Two soap bubbles having radii 3 cm and 4 cm in vacuum, coalesce under isothermal conditions. The radius of the new bubble is
A3
B4
C5
D7
• Question 1 - Select One
## Two spherical soap bubbles of radii a and b in vacuum coalesce under isothermal conditions. The resulting bubble has a radius given by
A(a+b)2
Baba+b
Ca2+b2
Da+b
• Question 1 - Select One
## Two spherical soap bubbles of a radii 1 cm and 2 cm vacuum coalesce under isothermal conditions . The resultant bubble has a radius of
A2/5 cm
B5 cm
C3 cm
D6 cm
• Question 1 - Select One
## Two spherical soap bubbles of radii r1 and r2 in vacuume collapse under isothermal condition. The resulting bubble has radius R such that
AR=r1+r2
BR=(r1+r2)12
CR=r21+r22
DR=[(r2r2(r1+r2)]
• Question 1 - Select One
## Two liquid drops have their diameters as 1 mm and 2 mm. The ratio of excess pressures in them is
A1:2
B2:1
C4:1
D1:4
• Question 1 - Select One
## Two soap bubble of radii 3 mm and 4 mm are in contact radius of curvature of interface between those two bubbles is
A1mm
B7mm
C12mm
D127mm
• Question 1 - Select One
## Choose the most appropriate option. Two liquid drops of equal radii are falling through air with the terminal velocity v. If these two drops coalesce to form a single drop, its terminal velocity will be
A2v
B2v
C34v
D32v
### Similar Questions
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 809 | 2,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-22 | latest | en | 0.837199 |
http://www.purplemath.com/learning/viewtopic.php?f=8&t=1906&p=5666 | 1,527,073,032,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00460.warc.gz | 448,176,049 | 6,341 | ## How do you go from a quadratic function in...
Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
Absolutely
Posts: 2
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Contact:
### How do you go from a quadratic function in...
generic form $ax^2+bx+c$ to the standard parabola form $a(x-h)^2+k$ by completing the square?
I know that the vertex of a parabola (h, k) is $h=\frac{-b}{2a}$ and $k=f(h)$ but I can't figure out how the first form gets to the second. I just want to know how the standard form is derived form the generic form by completing the square in a general way (meaning without numbers, just the letters to represent the constants); step-by-step please!
Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:
### Re: How do you go from a quadratic function in...
generic form $ax^2+bx+c$ to the standard parabola form $a(x-h)^2+k$ by completing the square? I know that the vertex of a parabola (h, k) is $h=\frac{-b}{2a}$ and $k=f(h)$ but I can't figure out how the first form gets to the second. I just want to know how the standard form is derived form the generic form by completing the square in a general way (meaning without numbers, just the letters to represent the constants); step-by-step please!
assume $a\neq0$
$ax^2+bx+c$
$=a(x^2+\frac{b}{a}x)+c$
$=a\left(x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2\right)+c$
$=a\left(x+\frac{b}{2a}\right)^2-a\left(\frac{b}{2a}\right)^2+c$
$=a\left(x-\frac{-b}{2a}\right)^2-a\left(\frac{b}{2a}\right)^2+c$
$=a\left(x-\frac{-b}{2a}\right)^2+c-\frac{b^2}{4a}$
$=a\left(x-h\right)^2+k$
Absolutely
Posts: 2
Joined: Mon Apr 04, 2011 8:57 pm
Contact:
### Re: How do you go from a quadratic function in...
Thanks! Seeing the algebraic manipulations behind these derivations makes the formulas more sensible to me. | 592 | 1,879 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 16, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-22 | latest | en | 0.787433 |
https://scipy-lectures.org/intro/summary-exercises/optimize-fit.html | 1,713,686,435,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.87/warc/CC-MAIN-20240421071342-20240421101342-00105.warc.gz | 478,508,776 | 7,636 | # 1.6.11.2. Non linear least squares curve fitting: application to point extraction in topographical lidar data¶
The goal of this exercise is to fit a model to some data. The data used in this tutorial are lidar data and are described in details in the following introductory paragraph. If you’re impatient and want to practice now, please skip it and go directly to Loading and visualization.
## Introduction¶
Lidars systems are optical rangefinders that analyze property of scattered light to measure distances. Most of them emit a short light impulsion towards a target and record the reflected signal. This signal is then processed to extract the distance between the lidar system and the target.
Topographical lidar systems are such systems embedded in airborne platforms. They measure distances between the platform and the Earth, so as to deliver information on the Earth’s topography (see [1] for more details).
[1] Mallet, C. and Bretar, F. Full-Waveform Topographic Lidar: State-of-the-Art. ISPRS Journal of Photogrammetry and Remote Sensing 64(1), pp.1-16, January 2009 http://dx.doi.org/10.1016/j.isprsjprs.2008.09.007
In this tutorial, the goal is to analyze the waveform recorded by the lidar system [2]. Such a signal contains peaks whose center and amplitude permit to compute the position and some characteristics of the hit target. When the footprint of the laser beam is around 1m on the Earth surface, the beam can hit multiple targets during the two-way propagation (for example the ground and the top of a tree or building). The sum of the contributions of each target hit by the laser beam then produces a complex signal with multiple peaks, each one containing information about one target.
One state of the art method to extract information from these data is to decompose them in a sum of Gaussian functions where each function represents the contribution of a target hit by the laser beam.
Therefore, we use the scipy.optimize module to fit a waveform to one or a sum of Gaussian functions.
>>> import numpy as np
and visualize it:
>>> import matplotlib.pyplot as plt
>>> t = np.arange(len(waveform_1))
>>> plt.plot(t, waveform_1)
[<matplotlib.lines.Line2D object at ...>]
>>> plt.show()
As shown below, this waveform is a 80-bin-length signal with a single peak with an amplitude of approximately 30 in the 15 nanosecond bin. Additionally, the base level of noise is approximately 3. These values can be used in the initial solution.
## Fitting a waveform with a simple Gaussian model¶
The signal is very simple and can be modeled as a single Gaussian function and an offset corresponding to the background noise. To fit the signal with the function, we must:
• define the model
• propose an initial solution
• call scipy.optimize.leastsq
### Model¶
A Gaussian function defined by
can be defined in python by:
>>> def model(t, coeffs):
... return coeffs[0] + coeffs[1] * np.exp( - ((t-coeffs[2])/coeffs[3])**2 )
where
• coeffs[0] is (noise)
• coeffs[1] is (amplitude)
• coeffs[2] is (center)
• coeffs[3] is (width)
### Initial solution¶
One possible initial solution that we determine by inspection is:
>>> x0 = np.array([3, 30, 15, 1], dtype=float)
### Fit¶
scipy.optimize.leastsq minimizes the sum of squares of the function given as an argument. Basically, the function to minimize is the residuals (the difference between the data and the model):
>>> def residuals(coeffs, y, t):
... return y - model(t, coeffs)
So let’s get our solution by calling scipy.optimize.leastsq() with the following arguments:
• the function to minimize
• an initial solution
• the additional arguments to pass to the function
>>> from scipy.optimize import leastsq
>>> t = np.arange(len(waveform_1))
>>> x, flag = leastsq(residuals, x0, args=(waveform_1, t))
>>> print(x)
[ 2.70363341 27.82020742 15.47924562 3.05636228]
And visualize the solution:
fig, ax = plt.subplots(figsize=(8, 6))
plt.plot(t, waveform_1, t, model(t, x))
plt.xlabel('Time [ns]')
plt.ylabel('Amplitude [bins]')
plt.legend(['Waveform', 'Model'])
plt.show()
Remark: from scipy v0.8 and above, you should rather use scipy.optimize.curve_fit() which takes the model and the data as arguments, so you don’t need to define the residuals any more.
## Going further¶
• Try with a more complex waveform (for instance waveform_2.npy) that contains three significant peaks. You must adapt the model which is now a sum of Gaussian functions instead of only one Gaussian peak.
• In some cases, writing an explicit function to compute the Jacobian is faster than letting leastsq estimate it numerically. Create a function to compute the Jacobian of the residuals and use it as an input for leastsq.
• When we want to detect very small peaks in the signal, or when the initial guess is too far from a good solution, the result given by the algorithm is often not satisfying. Adding constraints to the parameters of the model enables to overcome such limitations. An example of a priori knowledge we can add is the sign of our variables (which are all positive).
• See the solution.
• Further exercise: compare the result of scipy.optimize.leastsq() and what you can get with scipy.optimize.fmin_slsqp() when adding boundary constraints.
[2] The data used for this tutorial are part of the demonstration data available for the FullAnalyze software and were kindly provided by the GIS DRAIX. | 1,257 | 5,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-18 | latest | en | 0.887269 |
https://www.clutchprep.com/chemistry/practice-problems/136440/an-illustration-of-an-ethanol-fuel-cell-is-given-below-the-half-reactions-writte-1 | 1,638,804,069,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00150.warc.gz | 756,068,794 | 36,466 | Galvanic Cell Video Lessons
Concept
# Problem: An illustration of an ethanol fuel cell is given below:The half reactions (written as reductions) for the fuel cell at 25 °C are given below.O2 (g) + 4 H3O+ (aq) + 4 e- → 6 H2O (l) ε°red = 1.23 VCH3COOH (aq) + 4 H3O+ (aq) + 4 e- → CH3CH2OH (aq) + 5 H2O (l) ε°red = ??? acetic acid ethanolUsing the (appropriate) information below, calculate the overall potential for the fuel cell (Δε°) at 25 °C. Please circle your answer.
###### FREE Expert Solution
We are asked to calculate the overall potential for the fuel cell (ε°) at 25 °C.
Recall:
where:
n = number of moles electrons
F = Faraday’s constant, 96485 C/mol e-
ε° = standard reduction potential
We can use the following equation to solve for ΔG˚rxn:
We go through the following steps to solve the problem:
Step 1: Write the overall chemical reaction
Step 2: Calculate ΔG°rxn.
Step 3: Calculate the overall potential for the fuel cell
Step 1: Write the overall chemical reaction
Recall the mnemonics LEO GER.
Lose Gain
Electron Electrons
Oxidation Reduction
86% (227 ratings)
###### Problem Details
An illustration of an ethanol fuel cell is given below:
The half reactions (written as reductions) for the fuel cell at 25 °C are given below.
O2 (g) + 4 H3O+ (aq) + 4 e- → 6 H2O (l) ε°red = 1.23 V
CH3COOH (aq) + 4 H3O+ (aq) + 4 e- → CH3CH2OH (aq) + 5 H2O (l) ε°red = ???
acetic acid ethanol
Using the (appropriate) information below, calculate the overall potential for the fuel cell (Δε°) at 25 °C. Please circle your answer. | 513 | 1,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-49 | latest | en | 0.635433 |
https://brainly.in/question/205544 | 1,485,081,924,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281421.33/warc/CC-MAIN-20170116095121-00240-ip-10-171-10-70.ec2.internal.warc.gz | 787,253,779 | 10,193 | # A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
2
by KmvidhimaC
2015-10-14T17:18:34+05:30
Volume of old cube =a³
⇒(12)³=1728 cm³
No.of cubes =8
Volume of old cube = 8* volume of 1 new cube
1728=8*a³
1728/8=a³
12/2=a
6 cm=a
Surface area of old cube/ surface area of new cube=6*(12)²/6*(6)²
=144/36=4/1=4:1
surface area of a cube = 6a square, then the surface area of older cube = 6 * 12square =>864 sq.cm and the surface area of new cube = 6* 6 square =>216
arre you have cancelled the 6 on both numerator and denominator . I am REALLY SORRY. Nice job
2015-10-14T18:02:10+05:30
Side of a large cube(a) = 12cm
volume of the cube = (12³)cm³
=> 1728 cm³
eight small cubes of equal volumes are cut from the larger one
∴volume of larger cube = 8* volume of one small cube.
=>1728 = 8 * volume of small cube
=>1728/8 = volume of small cube
∴volume of one small cube = 216.
then side of the smaller cube = ∛216 = 6cm
The ratio of surface area of larger cube and small cube = 864:216 = 4:1 | 386 | 1,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-04 | latest | en | 0.814205 |
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# ics103-major01-062 - King Fahd University of Petroleum and...
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King Fahd University of Petroleum and Minerals Information and Computer Science Department ICS 103: Computer Programming in C Spring Semester 2006-2007 (Term-062) Major Exam-I Time:100 minutes Wednesday, March 28, 2007 Name: ID#: PLEASE CIRCLE YOUR SECTION BELOW: Section 01 02 03 04 05 06 7 Time SM-8-9 SM 9-10 SM 11-12 UT 9-10 UT 11:12 UT 8-9 UT 10-11 Note: Copying or Discussion will result in zero grade for all the students involved. Attempt all questions. Question # Maximum Marks Obtained Marks 1 7 x 2 = 14 2 10 3 10 4 10 5 6 6 6 7 8 8 8 9 14 10 14 Total 100 1
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Question 1: (14 points) Find the values of the following expressions. expression Value -3<=-2<=-1 5>3>=2+1 6+3<5&&!5-3>6 -3+3*-5%4+(5-2)/2.0 11.5-(2+3*(7-4)/4) Question 2 (10 points ) Write the corresponding mathematical or C expression C EXPRESSION Mathematical Expression y x z x y + - 2 pow(sqrt(x)+y,0.5) 2 7 3 1 + - y x xy 1+exp(y+x)/y-sqrt(z) 2
Q uestion 3: 10 points Show the output of the following program in the space provided below it. Each square corresponds to one space.
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Ask a homework question - tutors are online | 539 | 1,784 | {"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} | 3.46875 | 3 | CC-MAIN-2018-09 | latest | en | 0.68346 |
https://www.unitsconverters.com/en/Semicircle-To-Sign/Unittounit-7753-3670 | 1,708,707,845,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474440.42/warc/CC-MAIN-20240223153350-20240223183350-00443.warc.gz | 1,069,004,013 | 35,496 | Formula Used
1 Semicircle = 6.00000000008563 Sign
Semicircles to Signs Conversion
semicircle stands for semicircles and sign stands for signs. The formula used in semicircles to signs conversion is 1 Semicircle = 6.00000000008563 Sign. In other words, 1 semicircle is 7 times bigger than a sign. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
Convert Semicircle to Sign
How to convert semicircle to sign? In the angle measurement, first choose semicircle from the left dropdown and sign from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from sign to semicircle? You can check our sign to semicircle converter.
How to convert Semicircle to Sign?
The formula to convert Semicircle to Sign is 1 Semicircle = 6.00000000008563 Sign. Semicircle is 6 times Bigger than Sign. Enter the value of Semicircle and hit Convert to get value in Sign. Check our Semicircle to Sign converter. Need a reverse calculation from Sign to Semicircle? You can check our Sign to Semicircle Converter.
How many Radian is 1 Semicircle?
1 Semicircle is equal to 6 Radian. 1 Semicircle is 6 times Bigger than 1 Radian.
How many Degree is 1 Semicircle?
1 Semicircle is equal to 6 Degree. 1 Semicircle is 6 times Bigger than 1 Degree.
How many Gradian is 1 Semicircle?
1 Semicircle is equal to 6 Gradian. 1 Semicircle is 6 times Bigger than 1 Gradian.
How many Minute is 1 Semicircle?
1 Semicircle is equal to 6 Minute. 1 Semicircle is 6 times Bigger than 1 Minute.
Semicircles to Signs Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like angle finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like semicircle to sign through multiplicative conversion factors. When you are converting angle, you need a Semicircles to Signs converter that is elaborate and still easy to use. Converting Semicircle to Sign is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Semicircle to Sign conversion along with a table representing the entire conversion.
Let Others Know | 628 | 2,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-10 | latest | en | 0.903602 |
https://buy-essay.com/the-wildhorse-company-sells-sports-decals-that-can-be-personalized-with-a-players-name-a-team-name-and-a-jersey-numbe/ | 1,679,579,703,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945168.36/warc/CC-MAIN-20230323132026-20230323162026-00218.warc.gz | 183,451,246 | 12,981 | The Wildhorse Company sells sports decals that can be personalized with a player’s name, a team name, and a jersey numbe
The Wildhorse Company sells sports decals that can be personalized with a player’s name, a team name, and a jersey number for \$7.00 each. Wildhorse buys the decals from a supplier for \$2.60 each and spends an additional \$0.70 in variable operating costs per decal. The results of last month’s operations are as follows:
Sales revenue \$14,000 Cost of goods sold 5,200 Gross profit 8,800 Operating expenses 3,100 Operating income \$5,700
1. Calculate contribution margin per unit. (Round answer to 2 decimal places, e.g. 0.38.)
2. What is Wildhorse’s monthly breakeven point in units? In dollars? (Use your answer of breakeven units to calculate the breakeven point in dollars. Round Breakeven units and point in dollar to 0 decimal places, e.g. 25,000.)
3. What is Wildhorse’s margin of safety? (Round answers to 0 decimal places, e.g. 25,000.)
1 Sales revenue 7 Less: Variable costs Cost of goods sold 2.6 Operating expenses 0.7 Contribution margin per unit 3.7 2 Break even point in units = Fixed costs/ Unit contribution margin = 1700/3.7= 459 Break even point in sales = 459*7= 3213 3 margin of safety = Actual sales-Break even sales = 14000-3213=
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At this stage, our editor will go through your essay and make sure your writer did meet all the instructions. | 730 | 3,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-14 | latest | en | 0.869206 |
https://alumniagri.in/task/a-wire-is-7x-3-metres-long-a-length-of-3x-4-metres-is-cut-17961877 | 1,708,603,992,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00721.warc.gz | 93,998,563 | 4,792 | Math, asked by roychampa, 9 days ago
A wire is 7x – 3 metres long. A length of 3x + 4 metres is cut out of it for use. How much wire is left?
2
4x + 1 metre
Step-by-step explanation:
7x - 3metres - 3x + 4metres
Collect like terms
7x - 3x - 3metres + 4metres
4x+ 1 metre
4
Total Length = 7x-3
Length cut out = 3x + 4
Wire left = 7x-3 -(3x+4)
=7x-3-3x-4
=7x-3x-3-4 | 167 | 377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-10 | latest | en | 0.860668 |
https://www.airmilescalculator.com/distance/waw-to-lis/ | 1,670,034,907,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710918.58/warc/CC-MAIN-20221203011523-20221203041523-00800.warc.gz | 667,910,463 | 91,307 | # How far is Lisbon from Warsaw?
Distance between Warsaw (Warsaw Chopin Airport) and Lisbon (Lisbon Airport) is 1711 miles / 2754 kilometers / 1487 nautical miles. Estimated flight time is 3 hours 44 minutes.
Driving distance from Warsaw (WAW) to Lisbon (LIS) is 2059 miles / 3313 kilometers and travel time by car is about 33 hours 36 minutes.
1711
Miles
2754
Kilometers
1487
Nautical miles
3 h 44 min
193 kg
## Distance from Warsaw to Lisbon
There are several ways to calculate distance from Warsaw to Lisbon. Here are two common methods:
Vincenty's formula (applied above)
• 1711.279 miles
• 2754.036 kilometers
• 1487.061 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1708.127 miles
• 2748.964 kilometers
• 1484.322 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Warsaw to Lisbon?
Estimated flight time from Warsaw Chopin Airport to Lisbon Airport is 3 hours 44 minutes.
## What is the time difference between Warsaw and Lisbon?
The time difference between Warsaw and Lisbon is 1 hour. Lisbon is 1 hour behind Warsaw.
Warsaw time to Lisbon time converter
## Flight carbon footprint between Warsaw Chopin Airport (WAW) and Lisbon Airport (LIS)
On average flying from Warsaw to Lisbon generates about 193 kg of CO2 per passenger, 193 kilograms is equal to 427 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Warsaw to Lisbon
Shortest flight path between Warsaw Chopin Airport (WAW) and Lisbon Airport (LIS).
## Airport information
Origin Warsaw Chopin Airport
City: Warsaw
Country: Poland
IATA Code: WAW
ICAO Code: EPWA
Coordinates: 52°9′56″N, 20°58′1″E
Destination Lisbon Airport
City: Lisbon
Country: Portugal
IATA Code: LIS
ICAO Code: LPPT
Coordinates: 38°46′52″N, 9°8′9″W | 536 | 2,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-49 | latest | en | 0.868591 |
https://gyankosh.net/exceltricks/index-match-in-excel/amp/ | 1,721,266,586,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514816.43/warc/CC-MAIN-20240718003641-20240718033641-00252.warc.gz | 237,845,948 | 25,938 | # EXCEL TRICKS-USING INDEX-MATCH
## INTRODUCTION
INDEX-MATCH is the combination of two great functions INDEX and MATCH which are very useful in the practical usage of Excel. This combination helps us to find out the values, corresponding to any lookup value from the same or another table within a fraction of seconds.
It becomes very important to learn these tricks in order to be proficient in EXCEL.
Let us first review both of these functions individually, so that we can understand how they would work in conjunction to make our lookup better and faster.
INDEX FUNCTION:
INDEX FUNCTION returns the value of a cell or array in a specified relative location given by row no. and column no. in a selected range.
A relative position is a position with respect to the select LOOKUP_RANGE.
For example, suppose
We have five COLORS and five CODES in which we need to find out a value on a particular position.[ Data at a particular location specified by row and column ].
So we can use the INDEX function for this scenario and it’ll find out the item to be searched at a particular position.
Suppose there is a table of 4 columns and 4 rows. If we want to search that what element is present in a row no. 3 and column no. 4, we’ll take the help of INDEX FUNCTION.
If you want to have an in-depth understanding of INDEX FUNCTION by solving the examples. HERE IS THE LINK.
MATCH FUNCTION:
MATCH FUNCTION looks up VALUE_TO_BE_FOUND in the given range and returns its relative position.
Relative position is the position with respect to the selected LOOKUP_RANGE.
e.g.
We have five COLORS and five CODES and we need to find out at which position, any specified color is present.
The lookup value will be the color code, and the lookup range will be the range of colors.
The result will be the relative position of the color with respect to the selection.
If you want to have an in-depth understanding of MATCH FUNCTION by solving the examples. HERE IS THE LINK.
Now if you notice, both the functions are almost opposite to each other. So it is possible to combine both of them to get better use of both the functions to lookup and retrieve the value from a big pool of data. Let us see, how we can use them.
## COMBINING INDEX AND MATCH
Index and Match functions are opposite to each other.
Index function finds out the value of an item at a particular location in an array whereas Match function returns the relative position of an item after matching it from a given list.
So we can combine both the functions to retrieve the value from any column, against a lookup value which will be searched by the MATCH function for us. Let us try to put this in the form of a function.
Syntax is given in the next Section.
## PURPOSE OF INDEX-MATCH IN EXCEL
The combined INDEX-MATCH functions can be used to pull out any value which is corresponding to the another lookup value from another set of data which may or may not be on the same sheet.
## SYNTAX: INDEX-MATCH
The INDEX-MATCH FUNCTION will be used in the following format when used for lookup.
=INDEX( Array/Range from where the value is finally needed, MATCH(LOOKUP VALUE, Array in which the value will be found, match mode ), relative column number )
Array/Range from where the value is finally needed
This is the table from which the value will be extracted finally .
MATCH Lookup Value The unique value on the basis of which, we’ll find the desired value.
Array in which the value will be found This is the column where the value will be found / Column to be searched for.
Match Mode Different match modes of MATCH FUNCTION
Relative Column Number The column number of the INDEX (first argument) which will be returned for the value after the matching has been done.
## EXAMPLE 1:INDEX MATCH SIMPLE LOOKUP
We have a data sample with the students having different marks in different subjects. We’ll try a simple lookup of the data through index match first.
The data given is shown in the picture.
We’ll try to find the student with 83 marks in LANGUAGE.
After that, we’ll find the person who got 67 marks too.
EXPLANATION:
We put the formula as
=INDEX(G7:G12,MATCH(83,K7:K12,0),1)
G7:G12 is the table from which we need the result. Right now we selected just the column from where we need the name.
Match started as it’d return us to the index.
83 is the value to be found.
k7:k12 IS THE ARRAY WHERE 67 WILL BE COMPARED WITH.
0 IS THE EXACT MATCH
The third argument of the INDEX function is the column number to return which is 1 as we have selected a single column.
The result appears as JOE which is correct.
After this, we change the lookup value to 67 and see that the result changes to David.
## EXAMPLE 2:INDEX MATCH- COMPLICATED LOOKUP
### DATA SAMPLE
Find out the students who have the highest marks in Science. The data is given as the following table
Let us try to find out the highest marks in Science.
### EXPLANATION
We have to find the highest marks in Science. For that, we need to lookup for the highest marks first which we can do with the help of LARGE FUNCTION.
So we put it inside the Match Function. The function used to solve the problem is
=INDEX(G26:G31,MATCH(LARGE(I26:I31,1),I26:I31,0),1)
G26:G31 is the array from which we need the output.
Match function is started.
We’ll lookup the value which is highest for which we’ll use the LARGE function.
I26:I31 is the array from which the value is to be matched.
0 is for the exact match.
1 in the index function is for returning the value of the first column.
The answer is shown as DAVID which is correct.
## CONFUSION CLARIFICATIONS
### THE RELATIVE ROW AND COLUMN NUMBER
A cell is filled with YELLOW COLOR which we want to find. Now let us see in how many ways we can get this value.
The row and column number of the INDEX FUNCTION is relative to the RANGE SELECTED.
If we select Table 1, the value to be found will be at the row no. 4 and column no. 3.
If we select Table 2, the value to be found will be at the row no. 4 and column no. 3 again.
If we select Table 3, the value to be found will be at the row no. 3 and column no. 2.
So all these three different notations result in the same result. So always take care of the range selected and row and column no. selected. | 1,419 | 6,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-30 | latest | en | 0.885275 |
https://www.unleashcodingskills.com/2018/04/toandfro-to-and-fro-solution-spoj.html | 1,606,709,254,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141205147.57/warc/CC-MAIN-20201130035203-20201130065203-00669.warc.gz | 898,516,987 | 36,007 | TOANDFRO - To and Fro Solution SPOJ - UCS - Unleash-Coding-Skills
## Tuesday, 17 April 2018
TOANDFRO - To and Fro
Link for Problem TOANDFRO - To and Fro problem:
Problem:
### Input
There will be multiple input sets. Input for each set will consist of two lines. The first line will contain an integer in the range 2...20 indicating the number of columns used. The next line is a string of up to 200 lower case letters. The last input set is followed by a line containing a single 0, indicating end of input.
### Output
Each input set should generate one line of output, giving the original plaintext message, with no spaces.
```Input:
5
toioynnkpheleaigshareconhtomesnlewx
3
ttyohhieneesiaabss
0
Output:
theresnoplacelikehomeonasnowynightx
thisistheeasyoneab```
The CPP solution for this problem is
```#include <iostream>
#include <string>
using namespace std;
int main() {
int i,j,k,t,n,e;
string s;
while(1)
{
cin>>n;
if(n==0)
break;
cin>>s;
long int len=s.length();
char a[len/n][205];
e=0,k=0;
for(i=0;i<len;i++)
{
if(e==0)
{
for(j=0;j<n;j++)
{
a[k][j]=s[i];
i++;
}
k++;
e=1;
}
else
{
for(j=n-1;j>=0;j--)
{
a[k][j]=s[i];
i++;
}
k++;
e=0;
}
i--;
}
for(i=0;i<n;i++)
{
for(j=0;j<(len/n);j++)
cout<<a[j][i];
}
cout<<endl;
}
return 0;
}
```
Happy Coding..... | 412 | 1,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-50 | longest | en | 0.43872 |
http://gmatclub.com/forum/need-help-improving-verbal-from-94900.html | 1,485,186,478,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282932.75/warc/CC-MAIN-20170116095122-00034-ip-10-171-10-70.ec2.internal.warc.gz | 125,011,487 | 52,268 | need help improving verbal from 37 : GMAT Verbal Section
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# need help improving verbal from 37
Author Message
Intern
Joined: 17 Mar 2010
Posts: 39
Followers: 2
Kudos [?]: 64 [0], given: 26
need help improving verbal from 37 [#permalink]
### Show Tags
27 May 2010, 10:12
It appears that I am stuck at v 37...in the last gmatpre i had 37 (with about 6 repeats from OG) !! and in Mgmat i consistently score 37.
I don't know what to do...I have read and taken notes from Powerscore CR, Powerscore SC and Manhattan SC...my hit rate in the OG has been 80% in CR and SC and 70% in RC yet in tests my rate is about 50-60%.
I am practicing daily with LSAT questions for CR and RC and I have about 75% accuracy. I just don't know what to do anymore...
_________________
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Manager
Joined: 30 Jun 2004
Posts: 177
Location: Singapore
Followers: 1
Kudos [?]: 24 [0], given: 5
Re: need help improving verbal from 37 [#permalink]
### Show Tags
28 May 2010, 20:19
Sometimes it's something other than concepts and practice that makes the difference.
1. Try to relax in the test.
2. Focus on physical fitness.
3. Do some activity that improves concentration.
Manager
Joined: 30 Mar 2010
Posts: 86
Followers: 0
Kudos [?]: 59 [0], given: 45
Re: need help improving verbal from 37 [#permalink]
### Show Tags
05 Jun 2010, 09:52
Try 1000 SC, CR, RC..
_________________
Click on kudos, if you like my post!
Senior Manager
Joined: 12 Apr 2010
Posts: 446
GMAT 1: Q V
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Re: need help improving verbal from 37 [#permalink]
### Show Tags
09 Jun 2010, 07:06
leilak wrote:
It appears that I am stuck at v 37...in the last gmatpre i had 37 (with about 6 repeats from OG) !! and in Mgmat i consistently score 37.
I don't know what to do...I have read and taken notes from Powerscore CR, Powerscore SC and Manhattan SC...my hit rate in the OG has been 80% in CR and SC and 70% in RC yet in tests my rate is about 50-60%.
I am practicing daily with LSAT questions for CR and RC and I have about 75% accuracy. I just don't know what to do anymore...
You mentioned that your having 80% in SC, to improve yourself dont work on an entire test, try to work on the topics which you are constantly making errors, example i made more errors in modifiers, tense so i try to improve in those area.
_________________
Re: need help improving verbal from 37 [#permalink] 09 Jun 2010, 07:06
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Display posts from previous: Sort by | 1,022 | 3,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.908314 |
https://www.uaudio.com/blog/total-harmonic-distortion/?___store=default&___from_store=uaudiofr&setsite=fr&setsite=en | 1,659,974,157,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570868.47/warc/CC-MAIN-20220808152744-20220808182744-00433.warc.gz | 945,586,963 | 21,925 | Total Harmonic Distortion
What is THD and how is it measured?
THD stands for Total Harmonic Distortion and can be used to estimate the degree to which a system is nonlinear. A THD measurement can be made by applying a sine wave as an input to a system, and measuring the total energy which appears at the output of the system at harmonics of the input frequency. A sinusoidal input is used because sinusoids contain energy at only a single frequency; output energy that appears at any other frequency is thus the result of nonlinearities or time-varying system behavior. Energy that appears exactly at harmonics of the input frequency is almost certainly generated by nonlinearities in the system. Other periodic test signals or broadband signals are not ideal for THD measurements, because energy contained in the input signal can mask energy created by nonlinearities in the system. It is therefore important to have a pure input signal, i.e., one that has almost all of its energy contained at a single frequency.
In order to perform a true THD measurement, energy must be measured separately at harmonics of the input signal. Amplitudes at each harmonic frequency are squared, and then summed. The square root of the sum is the value of THD. Sometimes, this value is presented as a fraction (percentage) of the input signal's amplitude. In order to make the THD measurement possible, only a finite number of harmonics can be considered in the measurement. For mild nonlinearities, it is usually the case that energy falls off at higher harmonics, so that most of the THD energy is contained in the low harmonics. For severe nonlinearities, many harmonics must be included in order to obtain an accurate measurement.
In most cases, the amount of THD will depend on the amplitude of the input signal. For systems that saturate, it is to be expected that THD levels will increase with increasing signal amplitude. For systems with crossover distortion, it is possible that relative THD levels will fall with increasing input amplitude. Figure 1 shows input and output signals for a nonlinear system. The input signal is a pure sinusoid, while the output signal has several low-order distortion components totaling about 31% THD. On the right-hand side, spectra for the input and output signals are plotted. This system has no additive noise or modulation mechanism, so all energy at the output is contained at harmonics of the input frequency.
Figure 1: Block diagram for flanger and phaser
Many times, an estimate for THD is obtained by making what is called a Total Harmonic Distortion plus Noise, or THD+N measurement. THD+N figures are obtained by measuring the total energy present in the output, excluding energy at the input frequency. Usually, this measurement is accomplished by passing the system output through a notch filter, which eliminates energy at the input frequency, and measuring the total energy at the output of the notch filter. If it is expected that nonlinearities are the major sources of out-of-band energy, there will only be significant energy at harmonics of the input frequency. Therefore, the notch filter will be adequate if it has significant rejection at the input frequency, and is narrow enough to have nearly unity gain at the second harmonic of the input frequency.
In order to perform a true THD measurement, energy must be measured separately at harmonics of the input signal.
THD+N measurements can be made more simply than THD measurements, because energy at multiple frequencies does not have to be calculated separately. Because THD+N measurements include a continuous spectrum rather than discrete (harmonic) spectral points, THD+N figures reflect system noise, crosstalk, and interference at the outputs, as well as nonlinearities. Because of this, THD+N can be used as an overall figure of merit for a system.
For THD+N figures dominated by noise, relative measured values can be expected to decrease as signal amplitude grows. For measurements dominated by nonlinearities, relative measured values will increase with signal amplitude. For systems that produce both noise and nonlinear distortion, a minimum THD+N figure will be obtained at an input amplitude that produces an output comfortably above the noise floor, but below the levels at which significant distortion occurs.
Figure 2 shows input and output waveforms for a nonlinear system with an additive noise component. The upper right-hand plot shows the log-spectrum of the output waveform. The noise floor for the system can be seen along the bottom of the plot. The lower right-hand plot shows figures for THD and THD+N. The THD figure is not really affected by the presence of noise, having a value similar to that shown in Figure 1. However, the THD+N shows a higher value due to the broadband energy contained in the noise.
Figure 2: Transfer function for flanger's delay element
A related measure of distortion is the IMD, or Intermodulation Distortion measurement. IMD is measured by using the sum of two or more sinusoids as an input signal. Typically, the frequencies combined in the input are not harmonically related. With multiple frequencies present at the input, system nonlinearities produce distortion products at sums and differences of multiples of the input frequencies. Thus, rather than being harmonically related, IMD components are separated by the lower of the two input frequencies. This is convenient, because it allows for examination of many distortion components in a narrow bandwidth. With THD techniques, it is necessary to use low frequency input signals to create multiple distortion components in a narrow band. If the system being measured does not support low frequencies, THD measurements then become difficult.
See Audio Precision's "Audio Measurement Handbook" by Bob Metzler, available for download, for more information on THD and THD+N.
— Dave Berners | 1,175 | 5,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2022-33 | latest | en | 0.944452 |
https://math.stackexchange.com/questions/2924934/maximal-subgroups-that-force-solvability | 1,702,090,476,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100781.60/warc/CC-MAIN-20231209004202-20231209034202-00082.warc.gz | 418,412,342 | 36,405 | # Maximal subgroups that force solvability.
For which finite groups $$M$$ is it the case that every finite group $$G$$ with $$M$$ as a maximal subgroup solvable?
If $$M$$ satisfies this condition then $$M$$ is solvable. Also, if $$M$$ is abelian then $$M$$ satisfies this condition. Futhermore, I believe that if $$M$$ is nilpotent and if all 2-subgroups of $$M$$ are normal subgroups of $$M$$ (if Sylow 2-subgroups of $$M$$ are abelian or quaternion, for example) then $$M$$ satisfies this condition (proof below).
More specific questions:
1) Is there a non-nilpotent group that satisfies this condition?
2) Which 2-groups satisfy this condition?
Apparently, the dihedral group of order 8 satisfies this condition (see Mikko Korhonen's comment on this post). Also, if $$M\times N$$ satisfies this condition then $$M$$ and $$N$$ both satisfy this condition.
(This proof is adapted from j.p.'s answer to the linked question). Let $$G$$ be minimal such that $$G$$ is not solvable and such that $$G$$ contains a maximal subgroup $$M$$ that is nilpotent and whose 2-subgroups are normal. If $$M$$ contains a nontrivial normal subgroup $$N$$ of $$G$$ then $$G/N$$ contradicts the minimality of $$G$$. Thus, $$M$$ does not contain nontrivial normal subgroups of $$G$$. In particular, $$N_G(P)=M$$ for all Sylow $$p$$-subgroups $$P$$ of $$M$$. Then $$P$$ is a Sylow $$p$$-subgroup of $$N_G(P)$$ so $$P$$ is a Sylow $$p$$-subgroup of $$G$$. This shows that $$M$$ is a Hall subgroup of $$G$$.
If $$P$$ is a Sylow $$p$$-subgroup of $$M$$ and if $$Q$$ is a nontrivial normal subgroup of $$P$$ then $$N_G(Q)=M$$ which has a normal $$p$$-complement. For $$p=2$$, Frobenius' normal $$p$$-complement theorem gives that $$G$$ has a normal $$p$$-complement. For $$p\geq3$$, Thompson's normal $$p$$-complement theorem or Glauberman's normal $$p$$-complement theorem gives that $$G$$ has a normal $$p$$-complement (since you only have to consider characteristic $$p$$-subgroups).
Thus, for each prime $$p$$ dividing the order of $$M$$, $$G$$ has a normal $$p$$-complement. Then $$M$$ has a normal complement $$N$$ in $$G$$. Since $$M$$ is solvable but $$G$$ is not solvable, $$N$$ is not solvable. In particular, $$N$$ does not admit a fixed-point-free automorphism of prime order. If $$m\in Z(M)$$ has prime order then $$C_N(m)$$ is nontrivial. Then $$C_N(m)M$$ is a subgroup of $$G$$ that properly contains $$M$$ so $$C_N(m)M=G$$ by the maximality of $$M$$. Comparing cardinalities shows that $$C_N(m)=N$$ so $$m\in Z(G)$$. Then $$\langle m\rangle$$ is a nontrivial normal subgroup of $$G$$ contained in $$M$$ which is a contradiction.
• A famous example here is when $M$ is a $p$-group for an odd prime $p$. Sep 21, 2018 at 6:41
• I wonder if there is a reduction to the case where $G$ is almost simple? That is, could we show that $M$ "forces solvability as a maximal subgroup" if and only if $M$ is solvable and $M$ is not a maximal subgroup of any almost simple group? If the answer is yes, then your questions are answered by a paper of Li and Zhang from 2011 (Proceedings of the LMS), who classify the solvable maximal subgroups of almost simple groups. Sep 27, 2018 at 1:50
• About 2), just to give an example: if $p = 2^n - 1$ is a Mersenne prime ($n > 3$), then $PSL(2,p)$ has the dihedral group of order $2^{n-1}$ as a maximal subgroup. Sep 27, 2018 at 1:52
• Interesting. I computed maximal subgroups of a number of simple groups a while ago. It looked like all small nontrivial semi-direct products of two cyclic groups occurred with the exception of the dihedral group of order 8. I still don't know whether the dihedral group of order 8 forces solvability. Sep 27, 2018 at 2:21
• @ThomasBrowning: In "A condition for the solvability of a finite group" (1961) Deskins shows that if $G$ has a nilpotent maximal subgroup of nilpotency class $\leq 2$, then $G$ is solvable. In particular, the dihedral group of order $8$ forces solvability as a maximal subgroup Sep 27, 2018 at 3:46 | 1,174 | 3,985 | {"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": 78, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-50 | latest | en | 0.842289 |
https://search.r-project.org/CRAN/refmans/bcp/html/makeAdjGrid.html | 1,623,861,236,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487625967.33/warc/CC-MAIN-20210616155529-20210616185529-00569.warc.gz | 459,878,272 | 2,149 | ## Creating the adjacency structure for grid graphs
### Description
`makeAdjGrid()` produces a sparse representation of the adjacency structure for grid graphs, useful as the `adj` argument in `bcp()`.
### Usage
```makeAdjGrid(n, m = NULL, k = 8)
```
### Arguments
`n` the number of rows of vertices in the graph data. `m` (optional) the number of column of vertices in the graph data. If not given, we assume `m = n`. `k` (optional) the number of neighbors assumed for a typical vertex (see details below), either 4 or 8. Default number of neighbors is assumed to be 8.
### Details
`makeAdjGrid()` produces a list representation of the adjacency structure for grid graphs. The i-th entry in the list gives a vector of neighbor ids for the i-th node. Note that neighbor ids are offset by 1 because indexing starts at 0 in C++. If `k = 8`, then we assume each node is joined via edges to its 8 neighbors in the (top left, top middle, top right, left, right, bottom left, bottom middle, and bottom right) directions, where applicable. If `k = 4`, then we assume each node is joined via edges to its 4 neighbors in the (top, right, bottom, left) directions, where applicable.
### Author(s)
Xiaofei Wang
`bcp` for performing Bayesian change point analysis.
### Examples
```# generates an adjacency list for a 10 node by 5 node grid, assuming a maximum of 8 neighbors
# generates an adjacency list for a 10 node by 5 node grid, assuming a maximum of 4 neighbors
### show a grid example
## Not run:
set.seed(5)
z <- rep(c(0, 2), each=200)
y <- z + rnorm(400, sd=1)
if (require("ggplot2")) {
df <- data.frame(mean=z, data = y, post.means = out\$posterior.mean[,1],
post.probs = out\$posterior.prob,
i = rep(1:20, each=20), j = rep(1:20, times=20))
# visualize the data
g <- ggplot(df, aes(i,j)) +
geom_tile(aes(fill = data), color='white') +
ggtitle("Observed Data")
print(g)
# visualize the means
g <- ggplot(df, aes(i,j)) +
geom_tile(aes(fill = mean), color='white') +
ggtitle("True Means")
print(g)
# visualize the posterior means/probs
g <- ggplot(df, aes(i,j)) +
geom_tile(aes(fill = post.means), color='white') +
ggtitle("Posterior Means")
print(g)
g <- ggplot(df, aes(i,j)) +
geom_tile(aes(fill = post.probs), color='white') + | 620 | 2,247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.708054 |
https://www.hameroha.com/category/mathematics/ | 1,561,571,487,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000414.26/warc/CC-MAIN-20190626174622-20190626200622-00380.warc.gz | 785,274,429 | 26,034 | # Category: Mathematics
## Cross Product of two Vectors with Examples
Share This Topic: Course Details: Course Name: Applied Mathematics I Chapter Title: Vectors, Lines and Planes Subtitle: Cross Product of two Vectors Cross Product of two Vectors with Examples You can also read Scalar Multiplication Vs Scalar Product of Vectors
## Projection and Resolution of Vectors
Share This Topic: Course Details: Course Name: Applied Mathematics I Chapter Title: Vectors, Lines and Planes Subtitle: Projection and Resolution of Vectors Projection and Resolution of Vectors
## Scalar Multiplication Vs Scalar Product
Share This Topic: Scalar Multiplication Vs Scalar Product Product described below in the pdf clearly defines the difference between scalar multiplication and scalar product of vectors. You can also download the pdf file for your offline reading. Course Details: Course Name: Applied Mathematics I Chapter Title: Vectors, Lines and Planes Subtitle: Scalar Multiplication Vs Scalar Product
## Equality of two Vectors
Share This Topic:1 1ShareIn this post Equality of two Vectors in Applied Mathematics will be explained in detail with definition. Course Details: Course Name: Applied Mathematics I Chapter Title: Vectors, Lines and Planes Subtitle: Equality of two Vectors Two non-zero vectors and are said to be equal, if and only if they have the same direction | 281 | 1,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-26 | longest | en | 0.822448 |
http://clay6.com/qa/33361/find-the-equation-of-the-plane-that-contains-the-point-1-1-2-and-is-perp-to | 1,480,868,968,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541324.73/warc/CC-MAIN-20161202170901-00390-ip-10-31-129-80.ec2.internal.warc.gz | 57,970,604 | 27,770 | Browse Questions
# Find the equation of the plane that contains the point $(1,-1,2)$ and is $\perp$ to the planes $2x+3y-2z=5$ and $x+2y-3z=8$. Hence find the distance of the point $(-2,5,5)$ from the plane obtained above.
Toolbox:
• Distance of the point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$ is given by $\bigg|\large\frac{ax_1+by_1+cz_1+d}{\sqrt {a^2+b^2+c^2}}\bigg|$
To form the equation of a plane we need a point on it and normal vector $(\overrightarrow n)$
Step1
Given that the required plane contains the point $(1,-1,2)$
Let the normal to the required plane be $\overrightarrow n$
Also it is given that the required plane is $\perp$ to the planes
$2x+3y-2z-5=0$..............(i) and
$x+2y-3z-8$..........(ii)
Normal to the plane is $\overrightarrow n_1=(2,3,-2)$
Normal to the plane (ii) is $\overrightarrow n_2=(1,2,-3)$
Since the required plane is $\perp$ to (i) and (ii),
$\overrightarrow n=\overrightarrow n_1\times\overrightarrow n_2$
$\overrightarrow n=\left |\begin {array} \hat i & \hat j & \hat k\\2 & 3 & -2\\1 & 2 & -3\end {array}\right|=\hat i(-9+4)-\hat j(-6+2)+\hat k(4-3)$
$i.e., \overrightarrow n=(-5,4,1)$
Step 2
$\therefore$ Equation of the required plane is
$-5x+4y+z+d=0$
Given that this plane contains the point $(1,-1,2)$
$\Rightarrow\:$ It satisfies the equation of the plane.
$\Rightarrow\:-5\times 1+4\times (-1)+2+d=0$
$\Rightarrow\:d=7$
$\Rightarrow\:$ Equation of the required plane is $5x-4y-z-7=0$
Step 3
We know that distance of the point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$ is given by
$\bigg|\large\frac{ax_1+by_1+cz_1+d}{\sqrt {a^2+b^2+c^2}}\bigg|$
The distance of the point $(-2,5,5)$ from the plane $5x-4y-z-7=0$ is given by
$\bigg|\large\frac{5\times (-2)+(-4)\times 5+(-1)\times 5-7}{\sqrt {5^2+(-4)^2+(-1)^2}}\bigg|$
$=\bigg|\large\frac{-10-20-5-7}{\sqrt 42}\bigg|$
$=\large\frac{42}{\sqrt {42}}$$=\sqrt {42}$ units. | 733 | 1,885 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2016-50 | longest | en | 0.664929 |
https://jp.mathworks.com/matlabcentral/answers/445682-surface-plot-form-1d-data?s_tid=prof_contriblnk | 1,579,776,849,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250610004.56/warc/CC-MAIN-20200123101110-20200123130110-00477.warc.gz | 491,590,739 | 19,591 | # Surface plot form 1d data
17 ビュー (過去 30 日間)
mk_ballav 2019 年 2 月 18 日
Answered: Star Strider 2019 年 2 月 18 日
I want to make a mesh plot from 1D equation. I have an one-dimensional equation. I want to make a mesh plot with such equation. I am expecting the diagonal of the matrix resulting solution of equation.
%% 1D line plot
x = linspace(0,1000,1000);
C = x.*(1-2/3*x);
figure(1):plot(x,C)
%% create 2D mesh from x-cordinate;
[X,Y] = meshgrid(x,x);
%%% how to Convert C to 2D matrix
surf(X,Y,CC)
#### 0 件のコメント
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### 件の回答 (1)
Star Strider 2019 年 2 月 18 日
We have absolutely no idea what result you are expecting.
Try this:
%% 1D line plot
x = linspace(0,1000,1000);
C = @(x) x.*(1-2/3*x);
figure(1)
plot(x,C(x))
%% create 2D mesh from x-cordinate;
[X,Y] = meshgrid(x);
%%% how to Convert C to 2D matrix
surf(X,Y,C(X+Y), 'EdgeColor','none')
#### 0 件のコメント
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サインイン してこの質問に回答します。
Translated by | 332 | 933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-05 | latest | en | 0.450024 |
https://support.certara.com/forums/topic/1261-using-crossover-for-tmax-test/?setlanguage=1&langid=1&forcePrint=1 | 1,638,037,776,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00129.warc.gz | 635,752,183 | 12,982 | # using Crossover for Tmax test
Tmax Crossover sign test rank sum
### #1 0521
0521
• Members
• 33 posts
Posted 08 November 2018 - 03:18 AM
Dear all，
When using Crossover for Tmax testing, how can I judge whether it is equivalent by the result?
I read the "The nonparametric methods proposed by Koch (1972)" document, and I think he is doing the following steps to judge:
1. Determine if the P value of the sequence is greater than 0.05,
2. If the P value of the sequence is less than 0.05, the test of "Treatment" cannot be performed.
3. If the P value of the sequence is greater than 0.05, it is judged whether the P value of the treatment is greater than 0.05.
4. If the P value of the Treatment is greater than the equivalent, otherwise it is not equivalent.
5. If the P value of the sequence and the P value of the Period are both greater than 0.05, a sign test can be used instead of the rank sum test used in the above steps.
Assuming α = 0.05, all of the above P values are the results in the "Effects" table.
Is the above judgment process correct?
If not, what is the correct judgment process?
Best，
0521
• Members
• 33 posts
Posted 09 November 2018 - 09:39 AM
Dear 0521,
Please refer to the link below, and there is a detail explanation bu Helmut.
https://support.cert...ossover-design/
https://forum.bebac....asc=DESC#p14841
Best,
Chandramouli
### Also tagged with one or more of these keywords: Tmax, Crossover, sign test, rank sum
#### 0 user(s) are reading this topic
0 members, 0 guests, 0 anonymous users | 417 | 1,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.853014 |
https://math.stackexchange.com/questions/752105/prove-that-chi-v-1-otimes-v-2-g-chi-v-1-g-cdot-chi-v-2-g | 1,580,094,410,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694176.67/warc/CC-MAIN-20200127020458-20200127050458-00040.warc.gz | 549,573,572 | 31,647 | # Prove that $\chi_{V_1 \otimes V_2} (g) = \chi_{V_1} (g) \cdot \chi_{V_2} (g).$
Here, $\chi$ is the character of the sub-representation, i.e., Given
$\rho : G \to GL(V)$ is a representation, then the function $\chi_{\rho}: G \to \mathbb{C}: \chi_{\rho}(g) \to Tr(\rho_g)$.
I do not fully understand the notation here, as it was just sort of thrown at me without any explanation. Above is what I want to prove, that is:
$\chi_{V_1 \otimes V_2} (g) = \chi_{V_1} (g) \cdot \chi_{V_2} (g)$.
I know that since the trace map is invariant for similar matrices, the choice off basis does not affect the answer. If anyone would be so kind as to point me to how to prove this, it would be nice. I'm not asking for the full proof, but at least a starting point.
My thinking is that we need to choose some sort of matrix representation for $\rho$ and look at what the tensor product does to two randomly chosen matrix representations $V_1$ and $V_2$.
Thanks!
• Read about Kronecker Products of Matrices – Geoff Robinson Apr 13 '14 at 17:52
• Really want you want to show is that $Tr(A\otimes B) = Tr(A)Tr(B)$ as this does all of the work for you. – Cameron Williams Apr 13 '14 at 17:52
One thing I have noticed is that representation theorists love to be loose with notation and definitions (e.g. calling a $G$-invariant subspace a subrepresentation and the restriction of a representation to a $G$-invariant subspace a subrepresentation). It bothers me a lot. This is such an instance, where they have replaced $\rho_1\otimes\rho_2$ with $V_1\otimes V_2$. This does not match with the notational definition you've been given even though a representation theorist would likely argue that it's clear what is meant. I'm of the opinion that notation should be unambiguous and completely consistent throughout.
Suppose you have two representations $\rho_1:G\rightarrow GL(V_1)$ and $\rho_2:G\rightarrow GL(V_2)$. Better notation is perhaps $\chi_{\rho_1\otimes\rho_2}$ in place of $\chi_{V_1\otimes V_2}$. Piggybacking off of my comment, what this then says is that you want to show that the following is true:
$$\chi_{\rho_1\otimes\rho_2}(g) = \chi_{\rho_1}(g)\chi_{\rho_2}(g).$$
Or equivalently:
$$\operatorname{Tr}((\rho_1\otimes\rho_2)(g)) = \operatorname{Tr}(\rho_1(g))\operatorname{Tr}(\rho_2(g)).$$
• This makes is much more clear. Thank you. I completely agree that people that use representation theory tend to not care about the notation, and that makes it extremely difficult to follow. – Calculus08 Apr 13 '14 at 18:09
• Yeah I had plenty of issues with this. It's really troubling because it's a really nice branch of mathematics but there are some major miscues here and there (from my perspective).. – Cameron Williams Apr 13 '14 at 18:10
• In my experience the notation is only confusing in the beginning. Anyway, the claimed fact is not actually a fact of representation theory but one of linear algebra; ${\rm tr}(A\otimes B)={\rm tr}(A){\rm tr}(B)$ holds for any matices. This can be done by determining an eigenbasis for $A\otimes B$ via eigenbases of $A$ and $B$ (even if our scalars are not algebraically closed, we can extend them to accomplish this task). – anon Apr 13 '14 at 18:14 | 923 | 3,205 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.140625 | 3 | CC-MAIN-2020-05 | latest | en | 0.922968 |
https://oeis.org/A308463 | 1,716,227,104,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00426.warc.gz | 384,289,114 | 4,217 | The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
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A308463 a(n) = Sum_{k=1..n, gcd(n,k) = 1} Stirling2(n,k). 1
1, 1, 4, 7, 51, 16, 876, 2045, 15475, 15256, 678569, 2006863, 27644436, 46065293, 669225640, 5235101739, 82864869803, 234937438645, 5832742205056, 25117329128165, 235703526149476, 1886712616836675, 44152005855084345, 102153081219673712, 3428690854204959151 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 LINKS Table of n, a(n) for n=1..25. MATHEMATICA a[n_] := Sum[If[GCD[n, k] == 1, StirlingS2[n, k] , 0], {k, 1, n}]; Table[a[n], {n, 1, 25}] PROG (PARI) a(n) = sum(k=1, n, if (gcd(n, k)==1, stirling(n, k, 2))); \\ Michel Marcus, May 28 2019 CROSSREFS Cf. A000110, A008277, A056188, A096314, A308037. Sequence in context: A094609 A249936 A049191 * A261672 A013467 A330450 Adjacent sequences: A308460 A308461 A308462 * A308464 A308465 A308466 KEYWORD nonn AUTHOR Ilya Gutkovskiy, May 28 2019 STATUS approved
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Last modified May 20 13:41 EDT 2024. Contains 372715 sequences. (Running on oeis4.) | 545 | 1,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.610797 |
http://quant.stackexchange.com/questions/tagged/heston?sort=unanswered&pageSize=15 | 1,469,455,457,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824230.71/warc/CC-MAIN-20160723071024-00034-ip-10-185-27-174.ec2.internal.warc.gz | 193,324,201 | 17,045 | # Tagged Questions
A type of stochastic volatility model developed by associate finance professor Steven Heston in 1993 for analyzing bond and currency options. The Heston model is a closed-form solution for pricing options that seeks to overcome the shortcomings in the Black-Scholes option pricing model related to ...
128 views
140 views
### Simulation of Heston process
I am currently working on implementing Heston model in matlab for option pricing (in this case I am trying to price a European call) and I wanted to compare the results I obtain from using the exact ...
104 views
### Calibration of Heston version of CIR
I'd like to calibrate a variant of Heston model for interest rates which is describe by this couple of SDE \begin{aligned}dr_t&=a(b-r_t)+\sqrt{r_t}\sigma_t dW_t^1 \\ d\sigma_t&=k(\theta-\...
53 views
### Euler discretization bias, heston model
I am performing option pricing using Heston model and Euler discretization. I'm getting the following result: ...
98 views
### Heston model - Andersen scheme implementation
I would like to implement Andersen scheme for Heston simulation. On the following snipped is my code for generating asset path: ...
77 views
### Initial values for Heston Model calibration
I'm doing a Heston model in Matlab using simple Monte Carlo simulations (5.000 paths and 2 steps per day, simulating 360 days). When I try to calibrate the Heston parameters using fminsearch it takes ...
147 views
### Reference request about stochastic volatility model
I'm fiddling with estimation of stochastic volatility models and have build up a somewhat flexible framework using indirect inference. I would like to try and throw a lot of different continuous ...
100 views
### Problems with exact Heston simulations
I am just wondering if there is any problem with the so-called "exact" Heston simulations? So far what I have seen are the good things about it, what are the disadvantages? Because if it is so perfect,...
142 views
### Maximum Likelihood Estimation Heston Model using Matlab
My question is based on the MLE of the Heston model discussed in this paper URL: http://www.princeton.edu/~yacine/stochvol.pdf with Matlab code: http://www.princeton.edu/~yacine/closedformmle.htm ... | 514 | 2,249 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2016-30 | latest | en | 0.877014 |
https://sublinear.info/index.php?title=Open_Problems:78 | 1,656,670,626,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103940327.51/warc/CC-MAIN-20220701095156-20220701125156-00322.warc.gz | 589,384,907 | 5,797 | # Problem 78: Linear Sketching Over $F_2$
For a function $f:\{0,1\}^n\rightarrow\{0,1\}$, we define its deterministic linear sketch complexity $D^\text{lin}(f)$ as a the smallest number $k$ such that there exist $k$ sets $S_1,\ldots,S_k \subseteq [n]$ such that for any $x\in \{0,1\}^n$, we can compute $f(x)$ using $\sum_{i\in S_1} x_i, \ldots, \sum_{i\in S_k} x_i$, where the sum is mod $2$. For randomized linear sketch complexity, which is denoted by $R^\text{lin}(f)$, the $k$ sets are chosen in advance from a joint distribution and are available for recovering $f(x)$. Please see the paper by Kannan, Mossel, Sanyal and Yaroslavtsev [KannanMSY-18] for more details.
Given $f$, we also define $f^+:\{0,1\}^n\times \{0,1\}^n \rightarrow\{0,1\}$ as $f^+ (x,y) = f(x\oplus y)$ for all $x,y\in\{0,1\}^n$, where $\oplus$ denotes bitwise XOR. It is known that $D^\text{lin}(f) = D^\rightarrow(f^+)$ [MontanaroO-09], where $D^\rightarrow$ denotes one-way communication complexity (Alice sends one message to Bob).
Prove (or disprove) the following conjecture: $R^\text{lin}(f) = \tilde{\Theta}(R^\rightarrow(f^+))$. | 392 | 1,115 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-27 | latest | en | 0.745798 |
https://mycalcu.com/1.3-atm-to-bar | 1,713,148,597,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00067.warc.gz | 380,408,663 | 13,116 | # 1.3 atm to bar
1.3 Atm To Bar calculator converts 1.3 atm into Bar and vice versa.
## How many bars are there in 1.3 Atm (standard atmosphere)?
You can quickly calculate the answer by multiplying the 1.3 atm by 1.013.
## What is the value of 1.3 standard atmosphere in bar?
1.3 Atm is equal to 1.3169 Bar.
## 1.3 Standard Atmosphere Other Conversion
Bar 1.3169 Megapascal 0.131726 PSI 19.1048 Pascal 131722 Kilopascal 131.3 Torr 988
## Relevant Calculators
1.3 atm To Bar calculator converts 1.3 atm into bars quickly. It also converts bar to atm as well. Furthermore, it converts 1.3 atm into other units such as torr, pascal, kilopascal, and more.
atm bar
1.300 atm 1.317225 bar
1.301 atm 1.318238 bar
1.302 atm 1.319252 bar
1.303 atm 1.320265 bar
1.304 atm 1.321278 bar
1.305 atm 1.322291 bar
1.306 atm 1.323305 bar
1.307 atm 1.324318 bar
1.308 atm 1.325331 bar
1.309 atm 1.326344 bar
1.310 atm 1.327358 bar
1.311 atm 1.328371 bar
1.312 atm 1.329384 bar
1.313 atm 1.330397 bar
1.314 atm 1.331411 bar
1.315 atm 1.332424 bar
1.316 atm 1.333437 bar
1.317 atm 1.33445 bar
1.318 atm 1.335464 bar
1.319 atm 1.336477 bar
1.320 atm 1.33749 bar
1.321 atm 1.338503 bar
1.322 atm 1.339517 bar
1.323 atm 1.34053 bar
1.324 atm 1.341543 bar
1.325 atm 1.342556 bar
1.326 atm 1.34357 bar
1.327 atm 1.344583 bar
1.328 atm 1.345596 bar
1.329 atm 1.346609 bar
1.330 atm 1.347623 bar
1.331 atm 1.348636 bar
1.332 atm 1.349649 bar
1.333 atm 1.350662 bar
1.334 atm 1.351676 bar
1.335 atm 1.352689 bar
1.336 atm 1.353702 bar
1.337 atm 1.354715 bar
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1.340 atm 1.357755 bar
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1.342 atm 1.359782 bar
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1.344 atm 1.361808 bar
1.345 atm 1.362821 bar
1.346 atm 1.363835 bar
1.347 atm 1.364848 bar
1.348 atm 1.365861 bar
1.349 atm 1.366874 bar
1.350 atm 1.367888 bar
1.351 atm 1.368901 bar
1.352 atm 1.369914 bar
1.353 atm 1.370927 bar
1.354 atm 1.371941 bar
1.355 atm 1.372954 bar
1.356 atm 1.373967 bar
1.357 atm 1.37498 bar
1.358 atm 1.375994 bar
1.359 atm 1.377007 bar
1.360 atm 1.37802 bar
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1.362 atm 1.380047 bar
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1.365 atm 1.383086 bar
1.366 atm 1.3841 bar
1.367 atm 1.385113 bar
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1.370 atm 1.388153 bar
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1.373 atm 1.391192 bar
1.374 atm 1.392206 bar
1.375 atm 1.393219 bar
1.376 atm 1.394232 bar
1.377 atm 1.395245 bar
1.378 atm 1.396259 bar
1.379 atm 1.397272 bar
1.380 atm 1.398285 bar
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1.382 atm 1.400312 bar
1.383 atm 1.401325 bar
1.384 atm 1.402338 bar
1.385 atm 1.403351 bar
1.386 atm 1.404365 bar
1.387 atm 1.405378 bar
1.388 atm 1.406391 bar
1.389 atm 1.407404 bar
1.390 atm 1.408418 bar
1.391 atm 1.409431 bar
1.392 atm 1.410444 bar
1.393 atm 1.411457 bar
1.394 atm 1.412471 bar
1.395 atm 1.413484 bar
1.396 atm 1.414497 bar
1.397 atm 1.41551 bar
1.398 atm 1.416524 bar
1.399 atm 1.417537 bar
1.400 atm 1.41855 bar
1.401 atm 1.419563 bar
1.402 atm 1.420577 bar
1.403 atm 1.42159 bar
1.404 atm 1.422603 bar
1.405 atm 1.423616 bar
1.406 atm 1.42463 bar
1.407 atm 1.425643 bar
1.408 atm 1.426656 bar
1.409 atm 1.427669 bar
1.410 atm 1.428683 bar
1.411 atm 1.429696 bar
1.412 atm 1.430709 bar
1.413 atm 1.431722 bar
1.414 atm 1.432736 bar
1.415 atm 1.433749 bar
1.416 atm 1.434762 bar
1.417 atm 1.435775 bar
1.418 atm 1.436789 bar
1.419 atm 1.437802 bar
1.420 atm 1.438815 bar
1.421 atm 1.439828 bar
1.422 atm 1.440842 bar
1.423 atm 1.441855 bar
1.424 atm 1.442868 bar
1.425 atm 1.443881 bar
1.426 atm 1.444895 bar
1.427 atm 1.445908 bar
1.428 atm 1.446921 bar
1.429 atm 1.447934 bar
1.430 atm 1.448948 bar
1.431 atm 1.449961 bar
1.432 atm 1.450974 bar
1.433 atm 1.451987 bar
1.434 atm 1.453001 bar
1.435 atm 1.454014 bar
1.436 atm 1.455027 bar
1.437 atm 1.45604 bar
1.438 atm 1.457054 bar
1.439 atm 1.458067 bar
1.440 atm 1.45908 bar
1.441 atm 1.460093 bar
1.442 atm 1.461107 bar
1.443 atm 1.46212 bar
1.444 atm 1.463133 bar
1.445 atm 1.464146 bar
1.446 atm 1.46516 bar
1.447 atm 1.466173 bar
1.448 atm 1.467186 bar
1.449 atm 1.468199 bar
1.450 atm 1.469213 bar
1.451 atm 1.470226 bar
1.452 atm 1.471239 bar
1.453 atm 1.472252 bar
1.454 atm 1.473266 bar
1.455 atm 1.474279 bar
1.456 atm 1.475292 bar
1.457 atm 1.476305 bar
1.458 atm 1.477319 bar
1.459 atm 1.478332 bar
1.460 atm 1.479345 bar
1.461 atm 1.480358 bar
1.462 atm 1.481372 bar
1.463 atm 1.482385 bar
1.464 atm 1.483398 bar
1.465 atm 1.484411 bar
1.466 atm 1.485425 bar
1.467 atm 1.486438 bar
1.468 atm 1.487451 bar
1.469 atm 1.488464 bar
1.470 atm 1.489478 bar
1.471 atm 1.490491 bar
1.472 atm 1.491504 bar
1.473 atm 1.492517 bar
1.474 atm 1.493531 bar
1.475 atm 1.494544 bar
1.476 atm 1.495557 bar
1.477 atm 1.49657 bar
1.478 atm 1.497584 bar
1.479 atm 1.498597 bar
1.480 atm 1.49961 bar
1.481 atm 1.500623 bar
1.482 atm 1.501637 bar
1.483 atm 1.50265 bar
1.484 atm 1.503663 bar
1.485 atm 1.504676 bar
1.486 atm 1.50569 bar
1.487 atm 1.506703 bar
1.488 atm 1.507716 bar
1.489 atm 1.508729 bar
1.490 atm 1.509743 bar
1.491 atm 1.510756 bar
1.492 atm 1.511769 bar
1.493 atm 1.512782 bar
1.494 atm 1.513796 bar
1.495 atm 1.514809 bar
1.496 atm 1.515822 bar
1.497 atm 1.516835 bar
1.498 atm 1.517849 bar
1.499 atm 1.518862 bar
1.500 atm 1.519875 bar
1.501 atm 1.520888 bar
1.502 atm 1.521902 bar
1.503 atm 1.522915 bar
1.504 atm 1.523928 bar
1.505 atm 1.524941 bar
1.506 atm 1.525955 bar
1.507 atm 1.526968 bar
1.508 atm 1.527981 bar
1.509 atm 1.528994 bar
1.510 atm 1.530008 bar
1.511 atm 1.531021 bar
1.512 atm 1.532034 bar
1.513 atm 1.533047 bar
1.514 atm 1.534061 bar
1.515 atm 1.535074 bar
1.516 atm 1.536087 bar
1.517 atm 1.5371 bar
1.518 atm 1.538114 bar
1.519 atm 1.539127 bar
1.520 atm 1.54014 bar
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1.522 atm 1.542167 bar
1.523 atm 1.54318 bar
1.524 atm 1.544193 bar
1.525 atm 1.545206 bar
1.526 atm 1.54622 bar
1.527 atm 1.547233 bar
1.528 atm 1.548246 bar
1.529 atm 1.549259 bar
1.530 atm 1.550273 bar
1.531 atm 1.551286 bar
1.532 atm 1.552299 bar
1.533 atm 1.553312 bar
1.534 atm 1.554326 bar
1.535 atm 1.555339 bar
1.536 atm 1.556352 bar
1.537 atm 1.557365 bar
1.538 atm 1.558379 bar
1.539 atm 1.559392 bar
1.540 atm 1.560405 bar
1.541 atm 1.561418 bar
1.542 atm 1.562432 bar
1.543 atm 1.563445 bar
1.544 atm 1.564458 bar
1.545 atm 1.565471 bar
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1.547 atm 1.567498 bar
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1.550 atm 1.570538 bar
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1.567 atm 1.587763 bar
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1.569 atm 1.589789 bar
1.570 atm 1.590803 bar
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1.580 atm 1.600935 bar
1.581 atm 1.601948 bar
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1.584 atm 1.604988 bar
1.585 atm 1.606001 bar
1.586 atm 1.607015 bar
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1.592 atm 1.613094 bar
1.593 atm 1.614107 bar
1.594 atm 1.615121 bar
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1.600 atm 1.6212 bar
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1.603 atm 1.62424 bar
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1.606 atm 1.62728 bar
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1.610 atm 1.631333 bar
1.611 atm 1.632346 bar
1.612 atm 1.633359 bar
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1.619 atm 1.640452 bar
1.620 atm 1.641465 bar
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1.624 atm 1.645518 bar
1.625 atm 1.646531 bar
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1.628 atm 1.649571 bar
1.629 atm 1.650584 bar
1.630 atm 1.651598 bar
1.631 atm 1.652611 bar
1.632 atm 1.653624 bar
1.633 atm 1.654637 bar
1.634 atm 1.655651 bar
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1.686 atm 1.70834 bar
1.687 atm 1.709353 bar
1.688 atm 1.710366 bar
1.689 atm 1.711379 bar
1.690 atm 1.712393 bar
1.691 atm 1.713406 bar
1.692 atm 1.714419 bar
1.693 atm 1.715432 bar
1.694 atm 1.716446 bar
1.695 atm 1.717459 bar
1.696 atm 1.718472 bar
1.697 atm 1.719485 bar
1.698 atm 1.720499 bar
1.699 atm 1.721512 bar
1.700 atm 1.722525 bar
1.701 atm 1.723538 bar
1.702 atm 1.724552 bar
1.703 atm 1.725565 bar
1.704 atm 1.726578 bar
1.705 atm 1.727591 bar
1.706 atm 1.728605 bar
1.707 atm 1.729618 bar
1.708 atm 1.730631 bar
1.709 atm 1.731644 bar
1.710 atm 1.732658 bar
1.711 atm 1.733671 bar
1.712 atm 1.734684 bar
1.713 atm 1.735697 bar
1.714 atm 1.736711 bar
1.715 atm 1.737724 bar
1.716 atm 1.738737 bar
1.717 atm 1.73975 bar
1.718 atm 1.740764 bar
1.719 atm 1.741777 bar
1.720 atm 1.74279 bar
1.721 atm 1.743803 bar
1.722 atm 1.744817 bar
1.723 atm 1.74583 bar
1.724 atm 1.746843 bar
1.725 atm 1.747856 bar
1.726 atm 1.74887 bar
1.727 atm 1.749883 bar
1.728 atm 1.750896 bar
1.729 atm 1.751909 bar
1.730 atm 1.752923 bar
1.731 atm 1.753936 bar
1.732 atm 1.754949 bar
1.733 atm 1.755962 bar
1.734 atm 1.756976 bar
1.735 atm 1.757989 bar
1.736 atm 1.759002 bar
1.737 atm 1.760015 bar
1.738 atm 1.761029 bar
1.739 atm 1.762042 bar
1.740 atm 1.763055 bar
1.741 atm 1.764068 bar
1.742 atm 1.765082 bar
1.743 atm 1.766095 bar
1.744 atm 1.767108 bar
1.745 atm 1.768121 bar
1.746 atm 1.769135 bar
1.747 atm 1.770148 bar
1.748 atm 1.771161 bar
1.749 atm 1.772174 bar
1.750 atm 1.773188 bar
1.751 atm 1.774201 bar
1.752 atm 1.775214 bar
1.753 atm 1.776227 bar
1.754 atm 1.777241 bar
1.755 atm 1.778254 bar
1.756 atm 1.779267 bar
1.757 atm 1.78028 bar
1.758 atm 1.781294 bar
1.759 atm 1.782307 bar
1.760 atm 1.78332 bar
1.761 atm 1.784333 bar
1.762 atm 1.785347 bar
1.763 atm 1.78636 bar
1.764 atm 1.787373 bar
1.765 atm 1.788386 bar
1.766 atm 1.7894 bar
1.767 atm 1.790413 bar
1.768 atm 1.791426 bar
1.769 atm 1.792439 bar
1.770 atm 1.793453 bar
1.771 atm 1.794466 bar
1.772 atm 1.795479 bar
1.773 atm 1.796492 bar
1.774 atm 1.797506 bar
1.775 atm 1.798519 bar
1.776 atm 1.799532 bar
1.777 atm 1.800545 bar
1.778 atm 1.801559 bar
1.779 atm 1.802572 bar
1.780 atm 1.803585 bar
1.781 atm 1.804598 bar
1.782 atm 1.805612 bar
1.783 atm 1.806625 bar
1.784 atm 1.807638 bar
1.785 atm 1.808651 bar
1.786 atm 1.809665 bar
1.787 atm 1.810678 bar
1.788 atm 1.811691 bar
1.789 atm 1.812704 bar
1.790 atm 1.813718 bar
1.791 atm 1.814731 bar
1.792 atm 1.815744 bar
1.793 atm 1.816757 bar
1.794 atm 1.817771 bar
1.795 atm 1.818784 bar
1.796 atm 1.819797 bar
1.797 atm 1.82081 bar
1.798 atm 1.821824 bar
1.799 atm 1.822837 bar
1.800 atm 1.82385 bar
1.801 atm 1.824863 bar
1.802 atm 1.825877 bar
1.803 atm 1.82689 bar
1.804 atm 1.827903 bar
1.805 atm 1.828916 bar
1.806 atm 1.82993 bar
1.807 atm 1.830943 bar
1.808 atm 1.831956 bar
1.809 atm 1.832969 bar
1.810 atm 1.833983 bar
1.811 atm 1.834996 bar
1.812 atm 1.836009 bar
1.813 atm 1.837022 bar
1.814 atm 1.838036 bar
1.815 atm 1.839049 bar
1.816 atm 1.840062 bar
1.817 atm 1.841075 bar
1.818 atm 1.842089 bar
1.819 atm 1.843102 bar
1.820 atm 1.844115 bar
1.821 atm 1.845128 bar
1.822 atm 1.846142 bar
1.823 atm 1.847155 bar
1.824 atm 1.848168 bar
1.825 atm 1.849181 bar
1.826 atm 1.850195 bar
1.827 atm 1.851208 bar
1.828 atm 1.852221 bar
1.829 atm 1.853234 bar
1.830 atm 1.854248 bar
1.831 atm 1.855261 bar
1.832 atm 1.856274 bar
1.833 atm 1.857287 bar
1.834 atm 1.858301 bar
1.835 atm 1.859314 bar
1.836 atm 1.860327 bar
1.837 atm 1.86134 bar
1.838 atm 1.862354 bar
1.839 atm 1.863367 bar
1.840 atm 1.86438 bar
1.841 atm 1.865393 bar
1.842 atm 1.866407 bar
1.843 atm 1.86742 bar
1.844 atm 1.868433 bar
1.845 atm 1.869446 bar
1.846 atm 1.87046 bar
1.847 atm 1.871473 bar
1.848 atm 1.872486 bar
1.849 atm 1.873499 bar
1.850 atm 1.874513 bar
1.851 atm 1.875526 bar
1.852 atm 1.876539 bar
1.853 atm 1.877552 bar
1.854 atm 1.878566 bar
1.855 atm 1.879579 bar
1.856 atm 1.880592 bar
1.857 atm 1.881605 bar
1.858 atm 1.882619 bar
1.859 atm 1.883632 bar
1.860 atm 1.884645 bar
1.861 atm 1.885658 bar
1.862 atm 1.886672 bar
1.863 atm 1.887685 bar
1.864 atm 1.888698 bar
1.865 atm 1.889711 bar
1.866 atm 1.890725 bar
1.867 atm 1.891738 bar
1.868 atm 1.892751 bar
1.869 atm 1.893764 bar
1.870 atm 1.894778 bar
1.871 atm 1.895791 bar
1.872 atm 1.896804 bar
1.873 atm 1.897817 bar
1.874 atm 1.898831 bar
1.875 atm 1.899844 bar
1.876 atm 1.900857 bar
1.877 atm 1.90187 bar
1.878 atm 1.902884 bar
1.879 atm 1.903897 bar
1.880 atm 1.90491 bar
1.881 atm 1.905923 bar
1.882 atm 1.906937 bar
1.883 atm 1.90795 bar
1.884 atm 1.908963 bar
1.885 atm 1.909976 bar
1.886 atm 1.91099 bar
1.887 atm 1.912003 bar
1.888 atm 1.913016 bar
1.889 atm 1.914029 bar
1.890 atm 1.915043 bar
1.891 atm 1.916056 bar
1.892 atm 1.917069 bar
1.893 atm 1.918082 bar
1.894 atm 1.919096 bar
1.895 atm 1.920109 bar
1.896 atm 1.921122 bar
1.897 atm 1.922135 bar
1.898 atm 1.923149 bar
1.899 atm 1.924162 bar
1.900 atm 1.925175 bar
1.901 atm 1.926188 bar
1.902 atm 1.927202 bar
1.903 atm 1.928215 bar
1.904 atm 1.929228 bar
1.905 atm 1.930241 bar
1.906 atm 1.931255 bar
1.907 atm 1.932268 bar
1.908 atm 1.933281 bar
1.909 atm 1.934294 bar
1.910 atm 1.935308 bar
1.911 atm 1.936321 bar
1.912 atm 1.937334 bar
1.913 atm 1.938347 bar
1.914 atm 1.939361 bar
1.915 atm 1.940374 bar
1.916 atm 1.941387 bar
1.917 atm 1.9424 bar
1.918 atm 1.943414 bar
1.919 atm 1.944427 bar
1.920 atm 1.94544 bar
1.921 atm 1.946453 bar
1.922 atm 1.947467 bar
1.923 atm 1.94848 bar
1.924 atm 1.949493 bar
1.925 atm 1.950506 bar
1.926 atm 1.95152 bar
1.927 atm 1.952533 bar
1.928 atm 1.953546 bar
1.929 atm 1.954559 bar
1.930 atm 1.955573 bar
1.931 atm 1.956586 bar
1.932 atm 1.957599 bar
1.933 atm 1.958612 bar
1.934 atm 1.959626 bar
1.935 atm 1.960639 bar
1.936 atm 1.961652 bar
1.937 atm 1.962665 bar
1.938 atm 1.963679 bar
1.939 atm 1.964692 bar
1.940 atm 1.965705 bar
1.941 atm 1.966718 bar
1.942 atm 1.967732 bar
1.943 atm 1.968745 bar
1.944 atm 1.969758 bar
1.945 atm 1.970771 bar
1.946 atm 1.971785 bar
1.947 atm 1.972798 bar
1.948 atm 1.973811 bar
1.949 atm 1.974824 bar
1.950 atm 1.975838 bar
1.951 atm 1.976851 bar
1.952 atm 1.977864 bar
1.953 atm 1.978877 bar
1.954 atm 1.979891 bar
1.955 atm 1.980904 bar
1.956 atm 1.981917 bar
1.957 atm 1.98293 bar
1.958 atm 1.983944 bar
1.959 atm 1.984957 bar
1.960 atm 1.98597 bar
1.961 atm 1.986983 bar
1.962 atm 1.987997 bar
1.963 atm 1.98901 bar
1.964 atm 1.990023 bar
1.965 atm 1.991036 bar
1.966 atm 1.99205 bar
1.967 atm 1.993063 bar
1.968 atm 1.994076 bar
1.969 atm 1.995089 bar
1.970 atm 1.996103 bar
1.971 atm 1.997116 bar
1.972 atm 1.998129 bar
1.973 atm 1.999142 bar
1.974 atm 2.000156 bar
1.975 atm 2.001169 bar
1.976 atm 2.002182 bar
1.977 atm 2.003195 bar
1.978 atm 2.004209 bar
1.979 atm 2.005222 bar
1.980 atm 2.006235 bar
1.981 atm 2.007248 bar
1.982 atm 2.008262 bar
1.983 atm 2.009275 bar
1.984 atm 2.010288 bar
1.985 atm 2.011301 bar
1.986 atm 2.012315 bar
1.987 atm 2.013328 bar
1.988 atm 2.014341 bar
1.989 atm 2.015354 bar
1.990 atm 2.016368 bar
1.991 atm 2.017381 bar
1.992 atm 2.018394 bar
1.993 atm 2.019407 bar
1.994 atm 2.020421 bar
1.995 atm 2.021434 bar
1.996 atm 2.022447 bar
1.997 atm 2.02346 bar
1.998 atm 2.024474 bar
1.999 atm 2.025487 bar
## More Calculations
1 atm to bar 1.1 atm to bar 1.2 atm to bar 1.4 atm to bar 1.5 atm to bar 1.6 atm to bar 1.7 atm to bar 1.8 atm to bar 1.9 atm to bar | 7,974 | 16,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-18 | latest | en | 0.237628 |
http://www.nelson.com/nelson/school/elementary/mathK8/math7/quizzes/math7quizzes/gr7_ch1_les6-edited.htm | 1,516,297,825,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887535.40/warc/CC-MAIN-20180118171050-20180118191050-00495.warc.gz | 524,553,153 | 5,837 | Name: Try It Out -- Chapter 1, Lesson 6: Square Roots
Numeric Response Use mental math to calculate the square root. 1. 36 2. 64 3. 1600 Calculate the square root. 4. 361 5. 10 000 6. 484 7. 3364 8. 1225 9. 8836 10. 4489 | 92 | 225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-05 | latest | en | 0.301713 |
http://coding.derkeiler.com/Archive/Fortran/comp.lang.fortran/2012-07/msg00105.html | 1,368,864,448,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696381249/warc/CC-MAIN-20130516092621-00066-ip-10-60-113-184.ec2.internal.warc.gz | 52,080,494 | 4,658 | Best way to initialize 2D matrix in place
• From: "Nasser M. Abbasi" <nma@xxxxxxxxx>
• Date: Thu, 12 Jul 2012 01:45:05 -0500
I am trying to come up with the best method to declare
and initialize a 2D matrix in Fortran. But I want
to do that in one place. i.e. Not declare it, then do
the assignment later.
After a bit of experimentation, I found this method,
and wanted to ask the experts if there is an
improvement to it that can be made.
This initializes a 2 by 3 matrix (2 rows, 3 columns)
-------------------
program foo5
implicit none
INTEGER, dimension(2,3)::A = &
transpose(reshape([2,3,4 , &
4,0,7], [3,2]))
print *,shape(A)
end program foo5
-----------------------------
gfortran -std=f2003 -Wextra -Wall -pedantic -Wsurprising -Wconversion foo5.f90
./a.out
2 3
I wanted the initialization to show as the matrix
would appear in the text book.
The above does that. I do not like that I had
to use [3,2] for SHAPE in there, while the matrix A
is actually a (2,3). But found no way around this,
since I need to do a transpose, I had to do this. If
I do not do transpose, the matrix will be stored inside
as
2 4 0
3 4 7
instead of what I want, which is, what I typed
2 3 4
4 0 7
Again, the rules of the game: all must be done in
one place. Which means, one must use reshape() and
shape() as above, right? I did not see another way
so far. I am using gfortran 4.6
thanks,
--Nasser
.
Relevant Pages
• Re: Best way to initialize 2D matrix in place
... and initialize a 2D matrix in Fortran. ... end program foo5 ... I do not do transpose, the matrix will be stored inside ... Don't do this often, but look the optional order argument to the reshape function, ...
(comp.lang.fortran)
• Re: Setting pointer to null!
... I think this debugging fetaure emerged originally ... developer forgets to initialize a variable and it accidentally ... to declare a bunch that would be "initialized" in relatively distant code. ... because you prevent the compiler from initializing them ...
(microsoft.public.vc.language)
• Re: "no variable or argument declarations are necessary."
... the compilor will allert you ... If I forget to declare several variables in C, ... >> forget to initialize several variables in Python, ... > runtime error per "forgot to initialize". ...
(comp.lang.python)
• Re: Replicating results
... gradient calculation used in a large optimization routine. ... It will make sure that if you do something like declare a variable called "x1" and use a variable called "xl" the compiler will tell you that xl was undeclared. ... Make sure that you initialize all of your variables, ...
(comp.lang.fortran)
• Re: question on scope of a variable
... G'day "Chip Orange", ... If I have a std code module and declare at the top of it, ... >to be able to initialize it, or have access to it's value if it's ... >While this compiles for me, I'm getting a runtime error that indicates the ...
(microsoft.public.word.vba.general) | 777 | 2,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2013-20 | latest | en | 0.883411 |
http://forum.allaboutcircuits.com/threads/current-flow-sensor-switch.115065/ | 1,485,251,440,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284376.37/warc/CC-MAIN-20170116095124-00349-ip-10-171-10-70.ec2.internal.warc.gz | 102,068,143 | 18,005 | Current flow sensor switch
Discussion in 'General Electronics Chat' started by pafitis, Sep 2, 2015.
1. pafitis Thread Starter New Member
May 8, 2015
16
0
I'am new here to this wonderfull site and im glad that i find it. Now i believe each and every one should be creative and and must be happy to spread their knowledge freely every were.
What im asking its simple circuit consisting of a small current transformer that will detect an ac current (from 220V water pump),where the live wire is passing thru its hole.
As we all know that will produce an ac voltage of around (200-300mV) at its leads.
As long as there such voltage, can anyone design a circuit that will drive a relay of 12v and then by using
its contacts to connect an 220V hour meter for example.
2. Alec_t AAC Fanatic!
Sep 17, 2013
5,980
1,138
Welcome to AAC!
Actually, we don't all know that . The voltage generated will depend on a) the turns ratio of the current transformer, b) the value of the burden resistor across the transformer output and c) the current through the input. Can you post the spec/datasheet of the transformer? Do you know the current draw of the pump?
3. pafitis Thread Starter New Member
May 8, 2015
16
0
Thank you for your reply........i know that because i had my experiment....i took that small current transformer i run the live wire thru its hole,then i switch on the pump (6 amps load) and it gaves me
220mV !!!
Now, do you have a circuit for driving a 12V relay???
4. dl324 Distinguished Member
Mar 30, 2015
3,391
653
Welcome to AAC!
For this example, you could simply connect the hour meter across the power leads to the well pump.
5. pafitis Thread Starter New Member
May 8, 2015
16
0
Thank you for your reply...If it is so simple as what you said, i won't to ask for help .
Cannot have access.
6. dl324 Distinguished Member
Mar 30, 2015
3,391
653
In that case, you could take the signal from the current transformer, amplify, rectify, filter, and use the presence of voltage on the filter cap to control the relay driver.
7. Alec_t AAC Fanatic!
Sep 17, 2013
5,980
1,138
What value burden resistor? What is the transformer rating?
If you can increase the resistor value without exceeding that rating then you should get a high enough output voltage from the transformer to switch a transistor without the amplification step that dl324 mentioned.
8. AnalogKid Distinguished Member
Aug 1, 2013
4,709
1,301
220 mV isn't much to work with. And, I wonder where that number comes from. If it is an AC DVM reading of an open circuit CT secondary, that's not very much. Still some information missing.
OP - please post a sketch of the AC power source, pump wiring, and current transformer. Also, what did you use to measure thee CT secondary?
ak
Oct 2, 2009
5,451
1,066
10. pafitis Thread Starter New Member
May 8, 2015
16
0
Thank you for your reply.......all i can see its sensors,sensors,sensors, sensors,sensors, sensors.....a complete and customade design/circuit for the purpose i need, cannot see nowhere. ??????
11. pafitis Thread Starter New Member
May 8, 2015
16
0
I believe your miles away from analog circuits.........you don't need a special meter to measure it, this reading came out from the secondary of this current transformer. Its similar to power transfomers you apply for example 220V onto primary and you get 12V onto secondary. In that case you run the wire thru its core and you get voltage which depends on how much amps, and how many turns are winding the core.
12. blocco a spirale AAC Fanatic!
Jun 18, 2008
1,463
372
Will someone please provide him with a complete custom made design before he demands a full refund and takes his business elsewhere?
sirch2 likes this.
13. AnalogKid Distinguished Member
Aug 1, 2013
4,709
1,301
I believe you are miles away from understanding what a current transformer is. It is *not* "similar to power transformers." You do not "get voltage which depends on how much amps". You get *current* based on how much amps. Its output resembles a current source more than a voltage source. Because of this, the input impedance of the measurement instrument (scope, meter, etc.) has a direct effect on the reading and its accuracy. What did you use to measure the output of the current transformer? Is there a terminating or burden resistor on the output? If you don't understand the difference between a voltage source and a current source, none of the answers you get here will make any sense.
ak
14. BillB3857 Senior Member
Feb 28, 2009
2,402
348
I scavenged the coil from an old GFCI, tied an LED and properly positioned diode to protect the LED to it and made my own Opto Isolator using a photo transistor. The GFCI coil is monitoring my sump pump and the photo transistor is triggering a counter to tell me how many times the pump cycles. A similar setup could be used to trigger a relay.
15. pafitis Thread Starter New Member
May 8, 2015
16
0
If you don't have something usefull to say or suggest, will you please remain silinced ??
16. pafitis Thread Starter New Member
May 8, 2015
16
0
Many thanks for your usefull reply .....this is first positive posted answer from this member. Iam now close to what im try to achieve,if i get more like this i believe at the end of this week i will be able to made this circuit happen, thanks again.!!!!
17. AnalogKid Distinguished Member
Aug 1, 2013
4,709
1,301
With the coil driving an opto, isn't that two isolation barriers in series? Excellent safety, but you needed the system to work with lower input current you could have the coil drive a 2N4401 directly. or something like that.
ak
18. ScottWang Moderator
Aug 23, 2012
4,934
777
Will you draw a circuit or block diagram and labeling the V/I and parameters, it will make members have the same target to discuss with you and you can get the answer more fast as you wish?
19. sirch2 Well-Known Member
Jan 21, 2013
1,008
351
blocco a spirale likes this.
20. blocco a spirale AAC Fanatic!
Jun 18, 2008
1,463
372
ScottWang likes this. | 1,541 | 6,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.925911 |
http://www.nottingham.ac.uk/xpert/scoreresults.php?keywords=Film%20in%20history/hi&start=13220&end=13240 | 1,369,178,870,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700871976/warc/CC-MAIN-20130516104111-00094-ip-10-60-113-184.ec2.internal.warc.gz | 629,647,169 | 24,734 | The History of American Literature-Melville & Dickinson
This program explores the famous literary works of Melville & Dickinson. Expert discusses these authors and the meaning of their works.
Author(s): No creator set
Nature by Numbers
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The instructor uses an electronic chalkboard to demonstrate how to bisect a line segment. One example is modeled using astep by step approach to walk the learner through the process of bisecting the line segment by using a compass and ruler.
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Author(s): No creator set | 1,291 | 6,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.898595 |
https://au.mathworks.com/matlabcentral/cody/problems/44311-number-of-even-elements-in-fibonacci-sequence/solutions/1317305 | 1,582,984,401,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875149238.97/warc/CC-MAIN-20200229114448-20200229144448-00477.warc.gz | 278,374,218 | 15,853 | Cody
Problem 44311. Number of Even Elements in Fibonacci Sequence
Solution 1317305
Submitted on 27 Oct 2017 by cokakola
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
d = 14; y_correct = 4; assert(isequal(evenFibo(d),y_correct))
2 Pass
d = 20; y_correct = 6; assert(isequal(evenFibo(d),y_correct))
3 Pass
d = 50; y_correct = 16; assert(isequal(evenFibo(d),y_correct))
4 Pass
d = 100; y_correct = 33; assert(isequal(evenFibo(d),y_correct))
5 Pass
d = 150; y_correct = 50; assert(isequal(evenFibo(d),y_correct))
6 Pass
d = 200; y_correct = 66; assert(isequal(evenFibo(d),y_correct))
7 Pass
d = 500; y_correct = 166; assert(isequal(evenFibo(d),y_correct))
8 Pass
d = 1000; y_correct = 333; assert(isequal(evenFibo(d),y_correct))
9 Pass
d = 1e4; y_correct = 3333; assert(isequal(evenFibo(d),y_correct))
10 Pass
d = 2e4; y_correct = 6666; assert(isequal(evenFibo(d),y_correct))
11 Pass
d = 3e5; y_correct = 1e5; assert(isequal(evenFibo(d),y_correct))
12 Pass
d = 6e6; y_correct = 2e6; assert(isequal(evenFibo(d),y_correct)) % % %% % d = 9223372036854775807; % y_correct = 3074457345618258432; % assert(isequal(evenFibo(d),y_correct)) | 463 | 1,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-10 | latest | en | 0.403062 |
https://archived.hpcalc.org/museumforum/thread-125830-post-125933.html | 1,723,042,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694594.35/warc/CC-MAIN-20240807143134-20240807173134-00004.warc.gz | 80,641,347 | 13,627 | Trivia quiz and mini-challenge: Classic HP35 « Next Oldest | Next Newest »
▼ Allen Unregistered Posts: 562 Threads: 27 Joined: Oct 2006 10-07-2007, 11:57 AM Two quick quizzes for your Sunday pleasure: 1. Who bought the first HP 35 when it was first released in 1972? 2. Sometimes you will see a calculator display read: ` 123456789 ` as proof that the keybord is working. But there is more than one way to get the calculator to display this. What keystroke sequence (on the HP35 or HP35s) provides the FEWEST number of keys used. Obviously there are two easy answers: (1 key) Press Sigma+ 123 Million times, or (9 keys) Press the numbers in 1-9 is there another way?? ▼ Jake Schwartz Unregistered Posts: 349 Threads: 66 Joined: Apr 2007 10-07-2007, 05:10 PM Quote: 1. Who bought the first HP 35 when it was first released in 1972? Hint: Check HP Key Notes, Volume 6 Number 1 (February 1982) Page 14 in the lefthand column for the answer. Jake Schwartz Paul Dale Unregistered Posts: 3,229 Threads: 42 Joined: Jul 2006 10-07-2007, 05:19 PM I'm not sure of the sequence that requires the fewest keystrokes, I've not thought up anything shorter than keying in the digits. What might be more interesting would be a practical method (i.e. at most a couple of hundred keystrokes) that uses a minimum of *different* keys to get the number 123456789 on the display. I.e. what minimal functionality is required in order to get this display. Off the top of my head, I can think of a simple algorithm that requires four different functions / five different keys (need that shift). To get the '1' digit do something like: ``` CLx orange shift 10^x ENTER ENTER + ENTER + ENTER + + orange shift 10^x ``` Repeat similar sequences for the other digits and add them together. I think I could replace the shift 10^x sequence with a CLx, COS, +, ENTER, y^x series but this doesn't reduce the number of different keys which have to function and would increase the length significantly. I'm sure somebody here can do better... - Pauli ▼ Gerson W. Barbosa Unregistered Posts: 2,761 Threads: 100 Joined: Jul 2005 10-07-2007, 05:57 PM On the HP-35, the best I was able to come up with was this 10-key sequence: ```10 ENTER xy 91 - 81 / ``` Simpling typing 123456789 is still better... Gerson. ▼ Paul Dale Unregistered Posts: 3,229 Threads: 42 Joined: Jul 2006 10-07-2007, 06:07 PM You mean yx instead of xy I hope :-) This sequence is one operation shorter but the same number of keystrokes: ``` 10 10x 91 - 81 / ``` - Pauli ▼ Gerson W. Barbosa Unregistered Posts: 2,761 Threads: 100 Joined: Jul 2005 10-07-2007, 06:14 PM Quote: You mean yx instead of xy I hope :-) No, on the classic HP-35, that's the way it is! Unfortunately it also lacks 10x. Anyway, your sequence is great on the newer HP-35S. Regards, Gerson. ▼ Paul Dale Unregistered Posts: 3,229 Threads: 42 Joined: Jul 2006 10-07-2007, 06:24 PM Oops, somewhere between between reading the question and attempting to solve it, I slipped from a 35 to a 35s :-( My earlier algorithm is also wrong for a 35. - Pauli Allen Unregistered Posts: 562 Threads: 27 Joined: Oct 2006 10-07-2007, 08:07 PM Quote: I'm not sure of the sequence that requires the fewest keystrokes.. Not necessarily looking for the fewest keystrokes, but the fewest number of working keys. E.g. ```81 + 91 + ``` would be 4 different keys: +,1,8 and 9. ▼ Paul Dale Unregistered Posts: 3,229 Threads: 42 Joined: Jul 2006 10-07-2007, 08:26 PM huh? That sequence doesn't seem like it would work properly... Does it? However, here is a sequence that only requires three working keys and 83 keystrokes: ``` 1 ENTER + 1 + ENTER + 1 + ENTER + ENTER + 1 + ENTER + ENTER + 1 + ENTER + 1 + ENTER + ENTER + 1 + ENTER + 1 + ENTER + 1 + ENTER + 1 + ENTER + ENTER + ENTER + 1 + ENTER + 1 + ENTER + ENTER + 1 + ENTER + ENTER + ENTER + ENTER + 1 + ENTER + ENTER + 1 + ENTER + ENTER + 1 + ``` You can get rid of the single numeric (1) keystrokes and use CLx, COS and extra ENTERs appropriately. To understand what I did, take the binary expansion of 123456789 and read the bits left to right. - Pauli ▼ Gerson W. Barbosa Unregistered Posts: 2,761 Threads: 100 Joined: Jul 2005 10-07-2007, 09:25 PM ```11111.11111 ENTER * 1.1 - ``` 17 keystrokes but five keys. You're the winner! Gerson. ▼ Paul Dale Unregistered Posts: 3,229 Threads: 42 Joined: Jul 2006 10-07-2007, 09:34 PM What about this slight modification: ```9 1/x EEX 5 * ENTER * 1.1 - ``` 11 key strokes, 9 different keys. - Pauli Edited: 7 Oct 2007, 9:35 p.m. Namir Unregistered Posts: 2,247 Threads: 200 Joined: Jun 2005 10-07-2007, 09:00 PM How about: 987654321 9 - 8 / I started with a number that is the reverse of 123456789. Also, 1111111111 10 - 9 / Namir Edited: 7 Oct 2007, 9:05 p.m. Paul Dale Unregistered Posts: 3,229 Threads: 42 Joined: Jul 2006 10-07-2007, 09:30 PM Another almost solution at 7 commands: ``` 5 10x 9 / x2 1 - ``` To avoid the "10x", use: "EEX 5 ENTER" instead of the first two lines. Likewise, replace the "x2" with "ENTER *" Eight keystrokes on my 15c, no idea on the 35. Looks okay in FIX 0 mode. - Pauli ▼ Paul Dale Unregistered Posts: 3,229 Threads: 42 Joined: Jul 2006 10-07-2007, 09:32 PM Change the final '1' to '1.1' to remove the dependency on the display setting at the cost of two additional keystrokes. This puts us more than typing the number in directly :-( - Pauli ▼ designnut Unregistered Posts: 264 Threads: 65 Joined: Sep 2007 10-08-2007, 12:27 AM I just key in .0081 1/x Sam ▼ Namir Unregistered Posts: 2,247 Threads: 200 Joined: Jun 2005 10-08-2007, 01:16 AM You get: 123.45679 Notice that there is an 8 missing ...... so it's back to the drawing board my friend!! Namir PS: I saw this same mistake on another web site. Hmmmmmmmm .. did you copy your answer from that site? Edited: 8 Oct 2007, 11:57 a.m. Alex L Unregistered Posts: 43 Threads: 2 Joined: Mar 2007 10-08-2007, 01:57 PM Quote: What keystroke sequence (on the HP35 or HP35s) provides the FEWEST number of keys used. Obviously there are two easy answers: 1. (1 key) Press Sigma+ 123 Million times, or 2. (9 keys) Press the numbers in 1-9 is there another way?? 3 keys, but lots of keystrokes: 9 ENTER ENTER ENTER + (13717421 times) This has the added "advantage" of allowing one to truthfully declare: "Tested basic math operation and numeric keyboard." Stefan Vorkoetter Unregistered Posts: 217 Threads: 21 Joined: Aug 2007 10-08-2007, 01:59 PM How about: 3 ENTER x 3803 x 3607 x Thirteen keystrokes, seven different keys, only five working digit keys. Stefan ▼ Stefan Vorkoetter Unregistered Posts: 217 Threads: 21 Joined: Aug 2007 10-08-2007, 02:02 PM Or the slightly shorter: 9 ENTER 3803 x 3607 x Twelve keystrokes, but eight different keys of which six are digit keys. Namir Unregistered Posts: 2,247 Threads: 200 Joined: Jun 2005 10-08-2007, 03:26 PM Enter the number as the hexadecimal: 75BCD15h And the convert it to the decimal. Still not less than 9 keys. And another method. Store 1E9 in register A and then repeat: RCL A ENTER RAND * + You are generating random numbers between 1E9 and 2E9. A lucky random number will give you 123456789. The process uses 7 keys. These keystrokes test how touch the keyboard is!! ▼ Stefan Vorkoetter Unregistered Posts: 217 Threads: 21 Joined: Aug 2007 10-08-2007, 03:33 PM I don't think the Classic HP35 has a RAND key. ▼ Namir Unregistered Posts: 2,247 Threads: 200 Joined: Jun 2005 10-08-2007, 04:48 PM We are having fun with this problem. Not adhering 100% to the rules is part of the fun. If we want to get serious, this challenge is downright meaningless!! But like I said, we are having fun. Namir Edited: 8 Oct 2007, 4:50 p.m. Paul Dale Unregistered Posts: 3,229 Threads: 42 Joined: Jul 2006 10-08-2007, 03:46 PM Finally got one that is less than nine keystrokes on a 15c. Not applicable to a 35 though. A total of 7 commands, 8 keystrokes and 7 working keys required: ``` 9 ENTER 10x delta% 9 0 / ``` There are plenty of other ways to produce the 90 if we're in degrees mode (e.g. 0 COS-1; 1 SIN-1 or ENTER TAN-1). - Pauli ▼ Namir Unregistered Posts: 2,247 Threads: 200 Joined: Jun 2005 10-08-2007, 04:52 PM You solution also works with the HP-41C/CV/CX. Good show man!! Namir Allen Unregistered Posts: 562 Threads: 27 Joined: Oct 2006 10-08-2007, 05:12 PM Wow! Excellent submissions!! Jake definitely got the award for the first part. He found the source of the trivia. HP Keynotes reports that a Professor R.J. Donnelly of the University of Oregon purchased the first production HP35, which as of 1982 is not in use, rather kept in a Bank box for safe keeping. The part 2 award goes to Pauli for both the number of solutions and for the solution with the fewest KEYS needed to create the display: ` 123456789 ` Namir's submission was the closest to my original solution requiring 4 keys, but Pauli managed to find a solution with only 3 keys!!! EXCELLENT! My original 4-key solution could be done a few different ways, but the general formula, starting from a cleared stack (this can be done without pressing any buttons) is: ```1111111111 - - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 9 / ``` Four keys: 1,9,[-] and [/]. This could also be done with (4444444444-40)/6/6 using 4 keys:4,6,[-],and [/]. Paul Dale Unregistered Posts: 3,229 Threads: 42 Joined: Jul 2006 10-08-2007, 06:00 PM And of course if we're after a display of 1234567890 we can use this sequence which is one keystroke shorter but again not good for a 35 (6 commands, 7 keystrokes, 6 working keys on the 15c): ``` 9 ENTER 10x delta% 9 / ``` If we're happy with 12345.6789 displayed FIX 4 there is this one again for the 15c (6 commands, 10 keystrokes, 9 working keys): ``` 9 ENTER TANH CHS delta% x2 ``` - Pauli
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Author: Amelie Arnulf
Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
Top | 1,843 | 9,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-40 | latest | en | 0.928943 |
http://slideplayer.com/slide/3866800/ | 1,571,517,230,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986697760.44/warc/CC-MAIN-20191019191828-20191019215328-00213.warc.gz | 182,113,987 | 22,550 | # Waves!.
## Presentation on theme: "Waves!."— Presentation transcript:
Waves!
What is a wave? A wave is an oscillation that travels from one place to another. If you poke a floating ball, it oscillates up and down. The oscillation spreads outward from where it started.
Waves When you drop a ball into water, some of the water is pushed aside and raised by the ball.
Waves Waves are rhythmic disturbances that transfer energy from one place to another (caused by vibrations. i.e. water after the ball drop). Waves travel at the same speed – the speed of light. (3 x 10^8 m/s or 186,000 miles/second) Speed of sound – 340 m/s
Parts of a wave You can think of a wave as a moving series of high points and low points. A crest is the high point of the wave. A trough is the low point.
Parts of a wave The frequency of a wave is the rate at which every point on the wave moves up and down every second Frequency means “how often”. Unit – Hertz (Hz) means vibrations per second
Let’s try this! If the coil of a slinky makes two vibrations every second, what is the Frequency? Answer: 2 Hz! If a woodpecker pecks at a tree 15 times a second, what is the Frequency? Answer: 15 Hz!
Parts of a wave The amplitude of a water wave is the maximum height the wave rises above the level surface.
Parts of a wave Wavelength is the distance from any point on a wave to the same point on the next cycle of the wave. The distance between one crest and the next crest is a wavelength.
A wave moves one wavelength in each cycle
Transverse Wave Vibration moves at right angles to movement of wave
Height of wave depends on amount of energy Tidal waves carry a lot of energy
Longitudinal Wave Direction wave travels is the same direction in which the source vibrates.
Longitudinal Transverse
5. Wave behavior: a. Reflection - the bouncing back of a wave. 1) Sound echoes 2) Light images in mirrors
b. Refraction - the bending of a wave caused by a change in speed as the wave moves from one medium to another.
c. Diffraction - the bending of a wave around the edge of an object.
1) Water waves bending around islands 2) Water waves passing through a slit and spreading out
3) Diffraction depends on the size of the obstacle or opening compared to the wavelength of the wave. Less occurs if wavelength is smaller than the object. More occurs if wavelength is larger than the object.
4) AM radio waves are longer and can diffract around large buildings and mountains; FM can’t.
d. Interference - two or more waves overlapping to form a new wave.
Sound waves that constructively interfere are louder
1) Constructive (in phase) Sound waves that constructively interfere are louder
Sound waves that destructively interfere are not as loud
2) Destructive (out of phase) Sound waves that destructively interfere are not as loud
e. Standing wave - a wave pattern that occurs when two waves equal in wavelength and frequency meet from opposite directions and continuously interfere with each other. node antinode
f. Resonance - the ability of an object to vibrate by absorbing energy at its natural frequency. | 703 | 3,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-43 | latest | en | 0.905801 |
https://brilliant.org/problems/simple-yet-challenging-integration-problem/ | 1,524,173,587,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937045.30/warc/CC-MAIN-20180419204415-20180419224415-00557.warc.gz | 581,246,801 | 16,825 | Simple Yet Challenging Integration Problem
Calculus Level 4
Evaluate the Definite Integral: $$\int _{ 0 }^{ \pi /4 }{ \ln { (\tan { x+1) dx } } } .$$ The answer must be a decimal in terms of $$\pi$$. For example, if the answer is 2$$\pi,$$ then the answer should be 2.
× | 88 | 273 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-17 | latest | en | 0.575478 |
https://haesleinhuepf.github.io/BioImageAnalysisNotebooks/41_descriptive_statistics/Bland_altman_analysis.html | 1,721,786,965,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00641.warc.gz | 250,458,583 | 17,314 | # Method comparison#
Assume for a specific type of measurement, there are two methods for performing it. A common question in this context is if both methods could replace each other. Therefore, similarity of measurements is investigated. One method for this is Bland-Altman analysis called after Martin Bland and Douglas Altman.
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
Before we dive into Bland-Altman analysis, we have a look at straight-forward methods for comparing methods. An important assumption is that we worked with paired data. That means we can apply the two measurement methods to the same sample without destroying it and without the two methods harming each other.
# make up some data
measurement_A = [1, 9, 7, 1, 2, 8, 9, 2, 1, 7, 8]
measurement_B = [4, 5, 5, 7, 4, 5, 4, 6, 6, 5, 4]
# show measurements as table
pd.DataFrame([measurement_A, measurement_B], ["A", "B"]).transpose()
A B
0 1 4
1 9 5
2 7 5
3 1 7
4 2 4
5 8 5
6 9 4
7 2 6
8 1 6
9 7 5
10 8 4
## Comparison of means#
A very simple method for comparing arrays of measurements is comparing their means.
print("Mean(A) = " + str(np.mean(measurement_A)))
print("Mean(B) = " + str(np.mean(measurement_B)))
Mean(A) = 5.0
Mean(B) = 5.0
Using this method one could conclude that both methods deliver similar measurements because their mean is equal. However, this might be misleading.
## Scatter plots#
A more visual method for method comparison is drawing scatter plots. In these plots measurements of the one method are plotted against the other method.
plt.plot(measurement_A, measurement_B, "*")
plt.plot([0, 10], [0, 10])
plt.axis([0, 10, 0, 10])
plt.xlabel('measurement A')
plt.ylabel('measurement B')
plt.show()
Obviously A and B lead to quite different results. If the blue points would lie on the orange line, we would conclude that the measurements are related.
## Histograms#
As we concluded already that both measurements lie in different ranges, we should take a look at the distribution. Histograms are a good plot of choice. To make sure histograms for both measurements are visualized the same, e.g. with the same range on the x-axis, we can write our own little draw_histogram function:
def draw_histogram(data):
counts, bins = np.histogram(data, bins=10, range=(0,10))
plt.hist(bins[:-1], bins, weights=counts)
plt.axis([0, 10, 0, 4])
plt.show()
draw_histogram(measurement_A)
draw_histogram(measurement_B)
# Correlation#
For measuring the relationship between two measurements, we can take Pearson’s definition of a correlation coefficient
The data for the following expriment is taken from Altman & Bland, The Statistician 32, 1983, Fig. 1.
# new measurements
measurement_1 = [130, 132, 138, 145, 148, 150, 155, 160, 161, 170, 175, 178, 182, 182, 188, 195, 195, 200, 200, 204, 210, 210, 215, 220, 200]
measurement_2 = [122, 130, 135, 132, 140, 151, 145, 150, 160, 150, 160, 179, 168, 175, 187, 170, 182, 179, 195, 190, 180, 195, 210, 190, 200]
# scatter plot
plt.plot(measurement_1, measurement_2, "o")
plt.plot([120, 220], [120, 220])
plt.axis([120, 220, 120, 220])
plt.show()
# Determining Pearson's correlation coefficient r with a for-loop
import numpy as np
# get the mean of the measurements
mean_1 = np.mean(measurement_1)
mean_2 = np.mean(measurement_2)
# get the number of measurements
n = len(measurement_1)
# get the standard deviation of the measurements
std_dev_1 = np.std(measurement_1)
std_dev_2 = np.std(measurement_2)
# sum the expectation of
sum = 0
for m_1, m_2 in zip(measurement_1, measurement_2):
sum = sum + (m_1 - mean_1) * (m_2 - mean_2) / n
r = sum / (std_dev_1 * std_dev_2)
print ("r = " + str(r))
r = 0.9435300113035253
# Determine Pearson's r using scipy
from scipy import stats
stats.pearsonr(measurement_1, measurement_2)[0]
0.9435300113035257
# Bland-Altman plots#
Bland-Altman plots are a way to visualize differences between paired measurements specifically. When googling for python code that draws such plots, one can end up with this solution:
# A function for drawing Bland-Altman plots
# source https://stackoverflow.com/questions/16399279/bland-altman-plot-in-python
import matplotlib.pyplot as plt
import numpy as np
def bland_altman_plot(data1, data2, *args, **kwargs):
data1 = np.asarray(data1)
data2 = np.asarray(data2)
mean = np.mean([data1, data2], axis=0)
diff = data1 - data2 # Difference between data1 and data2
md = np.mean(diff) # Mean of the difference
sd = np.std(diff, axis=0) # Standard deviation of the difference
plt.scatter(mean, diff, *args, **kwargs)
plt.axhline(md, color='gray', linestyle='--')
plt.axhline(md + 1.96*sd, color='gray', linestyle='--')
plt.axhline(md - 1.96*sd, color='gray', linestyle='--')
plt.xlabel("Average")
plt.ylabel("Difference")
# draw a Bland-Altman plot
bland_altman_plot(measurement_1, measurement_2)
plt.show()
## Exercise#
Process the banana dataset again, e.g. using a for-loop that goes through the folder ../data/banana/, and processes all the images. Measure the size of the banana slices using the scikit-image thresholding methods threshold_otsu and threshold_yen. Compare both methods using the techniques you learned above. | 1,506 | 5,279 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-30 | latest | en | 0.905399 |
http://mathoverflow.net/questions/25042?sort=newest | 1,369,029,367,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698411148/warc/CC-MAIN-20130516100011-00014-ip-10-60-113-184.ec2.internal.warc.gz | 170,845,653 | 12,021 | ## Packing twelve spherical caps to maximize tangencies
Suppose that $v_i$, for $i \in \{1, 2, \ldots 11, 12\}$, are twelve unit length vectors based at the origin in $R^3$. Suppose that $|v_i - v_j| \geq 1$ for all $i \neq j$. What arrangement of the $v_i$ maximizes the number of pairs $\{i,j\}$ so that $|v_i - v_j| = 1$?
If C is a cube of sidelength $\sqrt{2}$ centered at the origin then we can place the $v_i$ at the midpoints of the twelve edges. Taking the convex hull of the $v_i$ gives a cube-octahedron of edge-length one. See here for a picture. If you cut the cubeoctahedron along a hexagonal equator and rotate the top half by sixty degrees you get another polyhedron. Both of these have 24 edges. Are these the unique maximal solutions to the above problem?
Notice that if you place the $v_i$ at the arguably nicer vertices of a icosahedron then the $v_i$ become too widely separated. It is easy to check this by making a physical model!
I spent some time thinking about areas of spherical polygons and restrictions on the graph of edges (and its dual graph) coming from the Euler characteristic. However, I don't think I got very far - in particular ruling out pentagons seems to be a crucial point that I couldn't deal with. Finally, to explain the problem title: instead of thinking of unit vectors with spacing restrictions, consider the (equivalent) problem of placing twelve identical spherical caps, of radius $\pi/12$, on the unit sphere with disjoint interiors in such a way as to maximize the number of points of tangency.
This question was asked of me by an applied mathematician. It comes from a problem involving packing balls in three-space, minimizing some quantity that is computed by knowing pairwise distances. The solution to the kissing problem thus justifies the "twelve" appearing in the problem statement. The projection of surrounding balls to a central one gives the spherical caps.
-
Interesting question. I can find answer using my program, which was made for solving Tammes problem for 13 points. But I need some time for answer.
UPD: I wrote program. Result: 24 is a maximal number of edges. I did in three steps. First, I enumerated planar graphs with 12 vertices with at least 25 edges, at most 5 edges in a vertex and at most hexagonal faces. Total number of suc graphs is 67497.
Second, I eliminated by linear programming by considering values of face angles as variables. My constrains was: 1. angle in triangle is ~1.2310 2. each angle no less than 1.2310 3. sum of angles around vertex is 2*pi 4. opposite angles of rectangle are equal 5. sum of non-opposite angles in rectangle between 3.607 and 3.8213
I solve feasibility of this LP problem (with some tolerance) After this step all graph were eliminated.
-
Are there other solutions with 24 edges other than the ones given by Sam Nead? – jc May 18 2010 at 15:42
I see, by staring at the dihedral angle of a tetrahedron, that the vertex degrees are either 1, 2, 3, 4, or 5. A "sliding" argument rules out degrees equal to 1. But I don't see why the face degrees are less than seven. (In fact, the original poser showed me another sliding argument to convince me that all faces are disks, ie rule out annular or worse faces.) Also, how do you get the upper and lower bounds on the sum of a pair of adjacent angles of a rombus? (Is the maximum really realized at $2\pi - 2 \arccos(1/3)$ or does it just "look" that way?) – Sam Nead May 18 2010 at 17:03
I was wrong then exclude septagons. I did it for Tammes problem because where is lemma that septagons are impossible for irreducible graphs. But here they are not irreducible. I will recalculate with all faces tomorrow (as well question for 24 edges). About rhombus, from spherical Pythagorean theorem where is constrain : $cot(a/2)*cot(b/2) = cos d$, where d - length of the side, a,b - angles. Maximum sum of angles applies for regular quadrilateral, so yes maximal summ is $2 pi - 2 \arccos(1/3)$ – Alexey Tarasov May 18 2010 at 17:46
I recalculated for all good graphs. Total amount of graphs with 24 edges and more is 221501. By elimination program only two graph were survived. These graphs are described in the question. Does anybody know about existing non-computer proof of this fact? – Alexey Tarasov May 19 2010 at 16:22
I was fast because I already have working approach. Graphs are generated by program plantri and filtered by my program. Total times about 15 minutes. Program for solving Tammes problem for N=13 is quite complicated. It is written on the perl it's size 1800 lines. Part used here more simple. Program will available soon on my webpage with dcs.isa.ru/taras/tammes13 – Alexey Tarasov May 21 2010 at 9:03
show 1 more comment | 1,213 | 4,718 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2013-20 | latest | en | 0.934387 |
https://ncatlab.org/nlab/show/bar%20construction | 1,720,783,548,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2024-30/segments/1720763514387.30/warc/CC-MAIN-20240712094214-20240712124214-00723.warc.gz | 268,576,599 | 22,351 | # nLab bar construction
Contents
### Context
#### Higher algebra
higher algebra
universal algebra
# Contents
## Idea
In categorical algebra, the bar construction takes an algebra $A$ of a monad and systematically “puffs it up”, replacing it with a simplicial object $Bar_T(A)$ in which all equations in the original algebra are replaced by 1-simplices, all equations between equations (called syzygies) are replaced by 2-simplices, and so on.
More precisely, the bar construction takes an algebra $(A, \rho)$ of a monad $(T, \mu, \epsilon)$ on a category $C$ to an augmented simplicial object $Bar_T(A)$ in the Eilenberg-Moore category $C^T$ of that monad. The face and degeneracy maps of this augmented simplicial object are given by the structure maps of the monad and its action on the algebra $A$:
$Bar_T(A) \coloneqq \left( \cdots \stackrel{\longrightarrow}{\stackrel{\longrightarrow}{\longrightarrow}} T T A \stackrel{\stackrel{\mu \cdot Id_A}{\longrightarrow}}{\stackrel{\phantom{A} T \cdot \rho \phantom{A}}{\longrightarrow}} T A \stackrel{\rho}{\longrightarrow} A \right) \,.$
If we apply the forgetful functor $U : C^T \to C$ to $Bar_T(A)$, we get an augmented simplicial object in $C$ called the bar resolution of $A$:
$U Bar_T(A) = \left( \cdots \stackrel{\longrightarrow}{\stackrel{\longrightarrow}{\longrightarrow}} U T T A \stackrel{\stackrel{U (\mu \cdot Id_A)}{\longrightarrow}}{\stackrel{\phantom{A} U(T \cdot \rho) \phantom{A}}{\longrightarrow}} U T A \stackrel{U \rho}{\longrightarrow} U A \right) \,.$
This is called a resolution of $A$ because it is equipped with a simplicial homotopy equivalence to the underlying $C$-object $U A \in C$, in a sense clarified by the “acyclic structure” of Definition . Moreover the bar resolution of $A$ has a universal property: it is initial among resolutions of $A$, as explained in Theorem .
Here we distinguish between the bar construction and the bar resolution, although the terms are often used interchangeably. It is worth noting that the bar construction $Bar_T(A)$ is not, in general, simplicially homotopy equivalent to $A$ in the category of simplicial objects in $C^T$. Only after applying the forgetful functor $U \colon C^T \to C$ do we obtain a key feature of the bar resolution: the simplicial homotopy equivalence $U Bar_T(A) \simeq U A$.
(Above we are identifying objects in a category, either $A \in C^T$ or $U A \in C$, with simplicial objects in that category that are “discrete”.)
## Definition
Let $\mathbf{E}$ be a category and let $(T, m: T T \to T, u: 1_{\mathbf{E}} \to T)$ be a monad on $\mathbf{E}$. We let $\mathbf{E}^T$ denote the category of $T$-algebras, and $U: \mathbf{E}^T \to \mathbf{E}$ the forgetful functor which is monadic, with left adjoint $F$.
Recall that the augmented simplex category $\Delta_a$, viz. the category consisting of finite ordinals1 and order-preserving maps, is the “walking monoid”, i.e., is initial among strict monoidal categories equipped with a monoid object. The monoidal product on $\Delta_a$ is ordinal addition $[m]+[n] = [m+n]$. If $[n]$ is the $n$-element ordinal, then the terminal object $[1]$ carries a unique monoid structure and represents the “generic monoid”2.
Similarly $\Delta_a^{op}$ is the walking comonoid. Since the comonad $F U$ on $\mathbf{E}^T$ can be regarded as a comonoid in the strict monoidal category of endofunctors $[\mathbf{E}^T, \mathbf{E}^T]$ (with endofunctor composition as monoidal product), there is a strong monoidal functor (or in fact a unique strict monoidal functor)
$\Delta_a^{op} \stackrel{Bar_T}{\to} [\mathbf{E}^T, \mathbf{E}^T]$
that takes the generic monoid $[1]$ to $F U$ and generally $[n]$ to $(F U)^{\circ n}$.
If furthermore $A$ is a $T$-algebra, there is an evaluation functor
$[\mathbf{E}^T, \mathbf{E}^T] \stackrel{eval_A}{\to} \mathbf{E}^T$
and we have the following definition:
###### Definition
The bar construction $Bar_T(A)$ is the simplicial $T$-algebra given by the composite functor
$\Delta_a^{op} \stackrel{Bar_T}{\to} [\mathbf{E}^T, \mathbf{E}^T] \stackrel{eval_A}{\to} \mathbf{E}^T.$
The composite
$\Delta_a^{op} \stackrel{Bar_T(A)}{\to} \mathbf{E}^T \stackrel{U}{\to} \mathbf{E}$
will here be called the bar resolution of $A$.
In the notation of two-sided bar constructions, the bar construction would be written as $Bar_T(A) = B(F, T, A)$, and the bar resolution as $B(T, T, A)$.
## Décalage and resolutions
### Décalage
To explain the sense in which $U Bar_T(A)$ is a resolution of (the constant simplicial object) $A$, we recall the fundamental décalage construction. Very simply, putting
$D = [1] + (-): \Delta_a^{op} \to \Delta_a^{op}$
the décalage functor on simplicial objects $C^{\Delta_a^{op}}$ (valued in a category $\mathbf{C}$) is the functor
$P: \mathbf{C}^{\Delta_a^{op}} \stackrel{\mathbf{C}^D}{\to} \mathbf{C}^{\Delta_a^{op}}.$
Note that $D$ has a comonad structure (inherited from the comonoid structure on $[1]$ in $\Delta_a^{op}$, see also at décalage – comonad structure), and therefore $P$ also carries a comonad structure. Notice also that there is a comonad map $D \to [1]\circ !$ (where $[1]: 1 \to \Delta_a^{op}$ is left adjoint to $!: \Delta_a^{op} \to 1$ since $[1]$ is initial in $\Delta_a^{op}$), induced by the evident natural inclusion $[1]+i: [1]+[0] \to [1]+[m]$ in $\Delta_a$. This in turn induces a comonad map $P X \to {|X|}$ where ${|-|}$ is the composite (“discretization”):
$C^{\Delta_a^{op}} \stackrel{ev_{[1]}}{\to} C \stackrel{diag}{\to} C^{\Delta_a^{op}}.$
The notation $P$ is chosen because décalage is essentially a kind of path space construction, i.e., in the case $\mathbf{C} = Set$ it is a simplicial sets analogue of a topological pullback
$\array{ P X & \to & X^I & \stackrel{ev_1}{\to} X \\ \downarrow & & \downarrow_\mathrlap{ev_0} & \\ {|X|} & \underset{id}{\to} & X }$
where $id: {|X|} \to X$ is the identity inclusion of the underlying set with the discrete topology. $P X$ is essentially a sum of spaces of based paths $(\alpha: (I, 0) \to (X, x_0)$ over all possible choices of basepoint $x_0$, fibered over $X$ by taking $\alpha$ to $\alpha(1)$. Each space of based paths is contractible and therefore $P X$ is acyclic.
The following definition names a nonce expression; this author (Todd Trimble) does not know how this is (or might be) referred to in the literature:
###### Definition
An acyclic structure on a simplicial object $X: \Delta_a^{op} \to C$ is a $P$-coalgebra structure $X \to P X$.
Here a $P$-coalgebra structure on $X$ is the same as a right $D$-coalgebra (or $D$-comodule) structure, given by a simplicial map $h: X \to X \circ D$ satisfying evident equations. In more nuts-and-bolts terms, it consists of a series of maps $h_n: X([n]) \to X([n+1])$ satisfying suitable equations.
The map $h: X \to X D$ may be viewed as a homotopy. Again, turning to the topological analogue for intuition, the corresponding $h: X \to P X$ is a homotopy (or rather, the composite $X \to P X \to X^I$ can be turned into a homotopy $I \times X \to X$). The coalgebra structure $h: X \to P X$ has a retraction given by the counit $\varepsilon: P X \to X$, so $X$ becomes a retract of an acyclic space, hence acyclic itself.
###### Remark
Definition gives an absolute notion of acyclicity, in the sense that if $X: \Delta_a^{op} \to \mathbf{C}$ carries an acyclic structure $h: X \to X D$ and $G: \mathbf{C} \to \mathbf{C}'$ is any functor, then $G X$ automatically carries an acyclic structure $G h: G X \to G X D$. (For example, $G X$ becomes acyclic in a standard model category sense under any functor $G: \mathbf{C} \to Set$.)
### Resolutions
Returning now to the bar resolution $U Bar_T(A)$: there is a canonical natural isomorphism $T \circ U Bar_T \cong U Bar_T \circ D$ obtained as the following 2-cell pasting (where $U Bar_T$ abbreviates the top and bottom horizontal composites)
(1)$\array{ \Delta_a^{op} & \stackrel{Bar_T}{\to} & [\mathbf{E}^T, \mathbf{E}^T] & \stackrel{[id, U]}{\to} & [\mathbf{E}^T, \mathbf{E}] \\ _\mathllap{D = [1] + (-)} \downarrow & \cong & _\mathllap{[id, F U]} \downarrow & \cong & \downarrow_\mathrlap{[id, U F] = [id, T]} \\ \Delta_a^{op} & \underset{Bar_T}{\to} & [\mathbf{E}^T, \mathbf{E}^T] & \underset{[id, U]}{\to} & [\mathbf{E}^T, \mathbf{E}], }$
whence there is a homotopy
$h = (U Bar_T \stackrel{u U Bar_T}{\to} T U Bar_T \cong U Bar_T D).$
###### Proposition
The map $h$ is an acyclic structure, def. , i.e., a right $D$-coalgebra structure.
###### Proof
We verify the coassociativity condition for the coaction $h: U Bar_T \to U Bar_T D$; the counit condition is checked along similar lines. The comultiplication of the comonad $F U$ is $\delta \coloneqq F u U$, and putting $\eta = u U: U \to U F U$ for a right $F U$-coaction, the coassociativity of $\eta$ follows from a naturality square
$\array{ U & \stackrel{\eta}{\to} & U F U \\ _\mathllap{\eta} \downarrow & & \downarrow_\mathrlap{U\delta} \\ U F U & \underset{\eta F U}{\to} & U F U F U. }$
Apply $[id_{\mathbf{E}^T}, -]$ to this coassociativity square to get another coassociativity, this time for the comonad $K \coloneqq [id_{\mathbf{E}^T}, F U]$ on $[\mathbf{E}^T, \mathbf{E}^T]$ (with comultiplication denoted $\delta_K$) and coaction $H \coloneqq [id, \eta]: [id, U] \to [id, U] \circ K$. Thus there is an equalizing diagram
$[id, U] \stackrel{H}{\to} [id, U]K \stackrel{\overset{[id, U]\delta_K}{\to}}{\underset{H K}{\to}} [id, U]K K.$
Because $Bar_T: \Delta_a^{op} \to [\mathbf{E}^T, \mathbf{E}^T]$ is a strong monoidal functor (see the left isomorphism in (1)), the squares in
$\array{ [id, U] Bar_T & \stackrel{H Bar_T}{\to} & [id, U]K Bar_T & \stackrel{\overset{[id, U]\delta_K Bar_T}{\to}}{\underset{H K Bar_T}{\to}} & [id, U]K K Bar_T \\ & _\mathllap{h}{\searrow} & \downarrow_\mathrlap{\cong} & & \downarrow_\mathrlap{\cong} \\ & & [id, U] Bar_T D & \stackrel{\overset{[id, U]Bar_T \delta_D}{\to}}{\underset{h D}{\to}} & [id, U]Bar_T D D }$
commute serially, with the triangle commuting by definition of $h$. This completes the verification.
By Remark , it follows that $U Bar_T(A)$, obtained by applying evaluation at a $T$-algebra $A$, carries an acyclic structure as well. In this sense we may say that $U Bar_T(A)$ (which has $A$ as its augmented component in dimension $-1$) is an acyclic resolution of the constant simplicial $T$-algebra at $A$ that carries a $T$-algebra structure.
## Properties
### Universal property
We now state and prove a universal property of the bar construction $Bar_T(A)$.
###### Definition
Let $(T, m: T T \to T, u: 1 \to T)$ be a monad on a category $\mathbf{E}$. A $T$-algebra resolution is a simplicial object $Y: \Delta_a^{op} \to \mathbf{E}^T$ together with an acyclic structure on $U Y: \Delta_a^{op} \to \mathbf{E}$. A morphism between $T$-algebra resolutions is a natural transformation $\phi: Y \to Y'$ such that $U\phi: U Y \to U Y'$ is a $P$-coalgebra map.
Let $AlgRes_T$ be the category of $T$-algebra resolutions. There is a forgetful functor
$G: AlgRes_T \to \mathbf{E}^T$
that takes an algebra resolution $Y$ to its augmentation component $Y[0]$.
###### Theorem
The functor $\hom_{\mathbf{E}^T}(A, G-): AlgRes_T \to Set$ is represented by $Bar_T(A)$; i.e., $Bar_T(-): \mathbf{E}^T \to AlgRes_T$ is left adjoint to $G$.
The proof is distributed over two lemmas.
###### Lemma
Given a $T$-algebra resolution $Y$ and a $T$-algebra map $f: A \to Y[0]$, there is at most one $T$-algebra resolution map $\phi: Bar_T(A) \to Y$ such that $\phi[0] = f$.
###### Proof
The $P$-coalgebra structure $h: U Bar_T(A) \to U Bar_T(A) \circ D$ is defined on components $U Bar_T(A)[n] = T^n A$ by $h[n] = u T^n(A): T^n(A) \to T^{n+1}(A)$. Thus in order that $U\phi$ be a $P$-coalgebra map, we must have that the diagram
$\array{ T^n A & \stackrel{u T^n A}{\to} & U F T^n(A) \\ _\mathllap{U\phi[n]} \downarrow & & \downarrow_\mathrlap{U\phi[n+1]} \\ U Y[n] & \underset{h_Y[n]}{\to} & U Y[n+1] }$
commutes. Here $\phi[n]: T^n (A) \to Y[n]$ determines a unique $T$-algebra map $g: F T^n(A) \to Y[n+1]$ such that
$U(g) \circ u T^n(A) = h_Y[n] \circ U\phi[n]$
since $F$ is left adjoint to $U$. Thus, starting with $\phi[0] = f$ as given, each algebra map $\phi[n]$ uniquely determines its successor $\phi[n+1] = g$.
###### Remark
The preceding proof does not show that the $\phi[n]$ fit together to form a map $\phi$ of simplicial $T$-algebras (i.e., to respect faces and degeneracies); it merely shows at most one such $T$-algebra resolution map is possible. But once we show that $\phi$ respects faces and degeneracies, the proof of Theorem will be complete.
###### Lemma
Given a $T$-algebra resolution $Y$ and an algebra map $f: X \to Y[0]$, there is at least one $T$-algebra resolution map $\phi: Bar_T(X) \to Y$ with $\phi[0] = f$.
###### Proof
It is enough to produce such a map $\phi: Bar_T(Y[0]) \to Y$ in the case $f = 1_{Y[0]}$, since the case for general $f: X \to Y[0]$ is then given by a composite
$Bar_T(X) \stackrel{Bar_T(f)}{\to} Bar_T(Y[0]) \stackrel{\phi}{\to} Y.$
We will do something slightly more general. For any category $\mathbf{C}$, the endofunctor category $[\mathbf{C}^{\Delta_a^{op}}, \mathbf{C}^{\Delta_a^{op}}]$ has a comonoid object $P = - \circ D$, so that there is an induced strong monoidal functor
$\Delta_a^{op} \stackrel{\tilde{P}}{\to} [\mathbf{C}^{\Delta_a^{op}}, \mathbf{C}^{\Delta_a^{op}}]$
which, upon evaluating at an object $Y$ of $\mathbf{C}^{\Delta_a^{op}}$, gives a functor
$B(Y, D, D) \coloneqq eval_Y \circ \tilde{P}: \Delta_a^{op} \to \mathbf{C}^{\Delta_a^{op}}$
with $B(Y, D, D)[n] = Y D^n$, so that $B(Y, D, D)$ is a double simplicial object. Taking $\mathbf{C} = \mathbf{E}^T$ and taking $Y$ to be a $T$-algebra resolution with acyclic structure $h: Y \to Y D$, we will produce a (double) simplicial map
$\Phi: B(T, T, Y) \to B(Y, D, D)$
where $\Phi[n]: T^n Y \to Y D^n$ is defined recursively as in the proof of Lemma , by setting $\Phi[0] = 1_Y$ and taking $\Phi[n+1]: T^{n+1} Y \to Y D^{n+1}$ the unique simplicial $T$-algebra map such that
$\array{ T^n Y & \stackrel{u T^n Y}{\to} & T^{n+1} Y \\ _\mathllap{\Phi[n]} \downarrow & & \downarrow_\mathrlap{\Phi[n+1]} \\ Y D^n & \underset{h D^n}{\to} & Y D^{n+1} }$
commutes for all $n$. Once we verify the claim that $\Phi$ respects faces and degeneracies, the same will be true for $\phi[n] = \Phi[n][0]: (T^n Y)[0] = T^n(Y[0]) \to (Y D^n)[0] = Y[n]$, whence the proof will be complete by Remark .
The claim is proved by induction on $n$. Let $\epsilon: D \to 1_{\Delta_a^{op}}$ be the counit and $\delta: D \to D D$ be the comultiplication. We have face maps
$T^j m T^{n-j-1} Y: T^{n+1} Y \to T^n Y, \qquad Y D^j \epsilon D^{n-j}: Y D^{n+1} \to Y D^n$
for $j = 0$ to $n$, under the special convention that $m T^{-1}Y: T Y \to Y$ denotes the action $\alpha: T Y \to Y$. We also have degeneracy maps
$T^j u T^{n-j} Y: T^n Y \to T^{n+1} Y, \qquad Y D^{j-1} \delta D^{n-j}: Y D^n \to Y D^{n+1}$
for $j = 1$ to $n$. We proceed as follows.
• To check preservation of face maps, we treat separately the cases where $j = 0$ and $j \geq 1$.
• For $j = 0$, we must check commutativity of the square in
$\array{ T^n Y & \stackrel{u T^n Y}{\to} & T^{n+1} Y & \stackrel{\Phi[n+1]}{\to} & Y D^{n+1} \\ & _\mathllap{id} \searrow & \downarrow_\mathrlap{m T^{n-1} Y} & & \downarrow_\mathrlap{Y\epsilon D^n} \\ & & T^n Y & \underset{\Phi[n]}{\to} & Y D^n. }$
Since all the maps are algebra maps and $u T^n Y: T^n Y \to T^{n+1} Y$ exhibits $T^{n+1} Y$ as the free algebra on $T^n Y$, it suffices to check commutativity around the perimeter. (N.B.: the triangle commutes, even in the case where $n=0$ which we need to start the induction.) By definition of $\Phi[n+1]$, commutativity of the perimeter boils down to commutativity of
$\array{ T^n Y & \stackrel{u T^n Y}{\to} & T^{n+1} Y & \stackrel{\Phi[n+1]}{\to} & Y D^{n+1} \\ & _\mathllap{\Phi[n]} \searrow & & \nearrow_\mathrlap{h D^n} & \downarrow_\mathrlap{Y\epsilon D^n} \\ & & Y D^n & \underset{id}{\to} & Y D^n }$
where the triangle commutes by one of the acyclic structure equations.
• For $j \geq 1$, the commutativity of the right square in
$\array{ T^n Y & \stackrel{u T^n Y}{\to} & T^{n+1} Y & \stackrel{\PhiY[n+1]}{\to} & Y D^{n+1} \\ _\mathllap{T^{j-1}m T^{n-j-1} Y} \downarrow & nat. & \downarrow_\mathrlap{T^j m T^{n-j-1} Y} & & \downarrow_\mathrlap{Y D^j \epsilon D^{n-j}} \\ T^{n-1} Y & \underset{u T^{n-1} Y}{\to} & T^n Y & \underset{\Phi[n]}{\to} & Y D^n \\ & _\mathllap{\Phi[n-1]} \searrow & & \nearrow_\mathrlap{h D^{n-1}} & \\ & & Y D^{n-1} & & }$
is again by appeal to a freeness argument where we just need to check commutativity of the perimeter, noting commutativity of the left square by naturality and that of the bottom quadrilateral by the recursive definition of $\Phi[n]$. But the perimeter commutes by examining the diagram
$\array{ T^n Y & \stackrel{u T^n Y}{\to} & T^{n+1} Y & \stackrel{\Phi[n+1]}{\to} & Y D^{n+1} \\ _\mathllap{T^{j-1}m T^{n-j-1} Y} \downarrow & _\mathllap{\Phi[n]} \searrow & & \nearrow_\mathrlap{h D^n} & \downarrow_\mathrlap{Y D^j \epsilon D^{n-j}} \\ T^{n-1} Y & ind. & Y D^n & nat. & Y D^n \\ & _\mathllap{\Phi[n-1]} \searrow & \downarrow & \nearrow_\mathrlap{h D^{n-1}} & \\ & & Y D^{n-1} & & }$
(where the middle vertical arrow is $Y D^{j-1} \epsilon D^{n-j}$) using the inductive hypothesis in the bottom left parallelogram.
• To check preservation of degeneracy maps, we treat separately the cases $j=1$ and $j \geq 2$.
• For $j = 1$, the commutativity of the top right square in
$\array{ T^{n-1} Y & \stackrel{u T^{n-1} Y}{\to} & T^n Y & \stackrel{\Phi[n]}{\to} & Y D^n \\ _\mathllap{u T^{n-1} Y} \downarrow & nat. & \downarrow _\mathrlap{T u T^{n-1} Y} & & \downarrow_\mathrlap{Y \delta D^{n-1}} \\ T^n Y & \underset{u T^n Y}{\to} & T^{n+1} Y & \underset{\Phi[n+1]}{\to} & Y D^{n+1} \\ & _\mathllap{\Phi[n]} \searrow & & \nearrow_\mathrlap{h D^n} & \\ & & Y D^n & & }$
is by appeal to a freeness argument where we just need to check commutativity of the perimeter (the special case $n=1$ being used to start the induction). But this boils down to commutativity of the diagram
$\array{ T^{n-1} Y & \stackrel{u T^{n-1} Y}{\to} & T^n Y & \stackrel{\Phi[n]}{\to} & Y D^n \\ _\mathllap{u T^{n-1} Y} \downarrow & \searrow_\mathrlap{\Phi[n-1]} & & _\mathllap{h D^{n-1}} \nearrow & \downarrow_\mathrlap{Y \delta D^{n-1}} \\ T^n Y & & Y D^{n-1} & & Y D^{n+1} \\ & _\mathllap{\Phi[n]} \searrow & \downarrow_\mathrlap{h D^{n-1}} & \nearrow_\mathrlap{h D^n} & \\ & & Y D^n & & }$
where the bottom right quadrilateral commutes by one of the acyclic structure equations.
• For $j \geq 2$, the commutativity of the top right square in
$\array{ T^{n-1} Y & \stackrel{u T^{n-1} Y}{\to} & T^n Y & \stackrel{\Phi[n]}{\to} & Y D^n \\ _\mathllap{T^{j-1} u T^{n-j} Y} \downarrow & nat. & \downarrow _\mathrlap{T^j u T^{n-j} Y} & & \downarrow_\mathrlap{Y D^{j-1} \delta D^{n-j}} \\ T^n Y & \underset{u T^n Y}{\to} & T^{n+1} Y & \underset{\Phi[n+1]}{\to} & Y D^{n+1} \\ & _\mathllap{\Phi[n]} \searrow & & \nearrow_\mathrlap{h D^n} & \\ & & Y D^n & & }$
is once again by appeal to a freeness argument where we just need to check commutativity of the perimeter. Here it boils down to commutativity of
$\array{ T^{n-1} Y & \stackrel{u T^{n-1} Y}{\to} & T^n Y & \stackrel{\Phi[n]}{\to} & Y D^n \\ _\mathllap{T^{j-1} u T^{n-j} Y} \downarrow & \searrow_\mathrlap{\Phi[n-1]} & & _\mathllap{h D^{n-1}} \nearrow & \downarrow_\mathrlap{Y D^{j-1} \delta D^{n-j}} \\ T^n Y & ind. & Y D^{n-1} & nat. & Y D^{n+1} \\ & _\mathllap{\Phi[n]} \searrow & \downarrow & \nearrow_\mathrlap{h D^n} & \\ & & Y D^n & & }$
where the middle vertical arrow is $Y D^{j-2} \delta D^{n-j}$.
This completes the proof.
## Special cases
### The classifying space and universal bundle of a group
There is a monad $T$ on $Set$ whose algebras are left $G$-sets, given by
$T X \,=\, G \times X \,.$
The terminal $G$-set $1$, namely the singleton set equipped with the trivial action of $G$, is an algebra for this monad $T$.
Now recalling that the bar construction takes any algebra of any monad and turns it into a simplicial object in the category of algebras of that monad, write $Bar_T(1)$ for the augmented simplicial set obtained by applying this to $1 \in G Set$.
By Theorem , its underlying simplicial set $U Bar_T(1)$ is contractible. This simplicial set is often denoted $E G$ (and as such known as the universal simplicial $G$-principal bundle) or $W G$ (see here at simplicial classifying space).
Working through the details, we see that the set of $n$-simplices of $E G$ is $G^{n+1}$, and all the face maps $\partial_i \colon G^{n+1} \to G^n$ are given by multiplying successive entries in the $(n+1)$-tuple, except for the last face map, which encodes the trivial action of $G$ on $1$:
$\partial_i (g_0, g_1, \dots, g_n) \;=\; (g_0, \dots, g_i g_{i+1}, \dots, g_n)$
for $i = 0, \dots, n-1$, while
$\partial_n (g_0, g_1, \dots, g_n) \;=\; (g_0, g_1, \dots, g_{n-1}) \,.$
The quotient $B G \coloneqq (E G)/G$ is known as the simplicial classifying space of $G$ (also denoted $\overline{W}G$), and its topological realization is a topological classifying space for $G$-principal bundles. Finally the topological realization of the quotient map $E G \to B G$ is the “universal $G$-principal bundle”.
The simplicial set $B G$ is also the simplicial nerve of the delooping groupoid of $G$.
### For modules over an algebra
Let $A$ be a commutative associative algebras over some ring $k$. Write $A Mod$ for the category of right modules over $A$.
For $M$ a right module, also $M \otimes_k A$ is canonically a right module. This construction extends to a functor
$T = (-) \otimes_k A : A Mod \to A Mod \,.$
The monoid structure on $A$ makes $T$ into a monad on $A Mod$: the monad product and unit are given by the product and unit in $A$.
For any right $A$-module $M$, the action $\rho \colon M \otimes A \to N$ makes $M$ into an algebra of the monad $T$. The bar construction $Bar_T(A)$ is then the simplicial $A$-module
$\cdots \stackrel{\longrightarrow}{\stackrel{\longrightarrow}{\longrightarrow}} N \otimes_k A \otimes_k A \stackrel{\overset{Id \otimes \mu}{\longrightarrow}}{\underset{\rho \otimes Id}{\longrightarrow}} N \otimes_k A \,.$
Under the Moore complex functor of the Dold-Kan correspondence, this simplicial $A$-module is identified with a chain complex of $A$-modules whose differential is given by the alternating sums of the face maps indicated above.
This chain complex is what originally was called the bar complex in homological algebra. It got its name because the first authors denoted its elements using a notation involving vertical bars (Ginzburg).
This chain complex provides a free resolution of $A$, which can be used to compute the Tor group
$Tor(M, A \times A) \,.$
This gives the Hochschild homology of $A$. See there for more details.
See (Fresse).
## References and Literature
The original reference for bar constructions in the generality of monads is
• Roger Godement, Topologie algébrique et theorie des faisceaux, Actualités Sci. Ind. 1252, Hermann, Paris (1958) [webpage, pdf]
A general discussion of bar construction for monads is at
Textbook accounts can be found at:
The bar complex of a bimodule is reviewed for instance in
around page 16.
The bar complex for E-infinity algebras is discussed in
• Benoit Fresse, The bar complex of an E-infinity algebra, Advances in Mathematics Volume 223, Issue 6, 1 April 2010, Pages 2049-2096
The compositional structure of the bar construction of several monads, as well as its interpretation in terms of partial evaluations is studied in
1. N.B.: including the empty ordinal.
2. If $X: \Delta_a^{op} \to \mathbf{C}$ is a simplicial object, then $X([n])$ is what is usually denoted $X_{n-1}$, the object of cells in dimension $n-1$. Note that $X([0]) = X_{-1}$ is the augmented component. The $n$ can be thought of as the number of vertices of a simplex of dimension $n-1$. We choose the index $n$ over the geometric dimension $n-1$ as it is more convenient for our purposes.
Last revised on June 12, 2024 at 16:46:11. See the history of this page for a list of all contributions to it. | 8,274 | 24,271 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 326, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.775425 |
https://established1962.wordpress.com/2013/11/16/magic-squares-2-creating-a-4x4-magic-square/ | 1,675,304,405,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499954.21/warc/CC-MAIN-20230202003408-20230202033408-00844.warc.gz | 262,370,759 | 28,332 | # Magic Squares 2 – Creating a 4×4 Magic Square
Rather surprisingly I find 4×4 magic squares offer more opportunity in the classroom than 3×3.
To start with, though I can never remember the details for a 3×3 square, it’s easy to create a 4×4 square in just a few moments.
Start by writing the numbers 1 to 16 into a 4×4 square in the natural way.
Then exchange the numbers in the two yellow positions, and do the same with the numbers in the other two corners.
You now have this arrangement.
Now exchange the two numbers in the two red positions, and do the same with the other two numbers in the central square.
That’s it! You now have a 4×4 magic square in which every row, every column, and the two diagonals add up to 34.
.
.
### 2 responses
1. […] Magic Squares 2 – Creating a 4×4 Magic Square (established1962.wordpress.com) Before reading this post I could not make a 4 x 4 Magic Square without reading specific instructions each time. Now I can do in from memory. Great post! […]
2. […] Magic Squares 2 – Creating a 4×4 Magic Square (established1962.wordpress.com) Before reading this post I could not make a 4 x 4 Magic Square without reading specific instructions each time. Now I can do in from memory. Great post! […] | 314 | 1,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2023-06 | latest | en | 0.894954 |
https://www.teacherspayteachers.com/Product/Functions-Review-Game-8th-Grade-1988819 | 1,484,639,149,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279489.14/warc/CC-MAIN-20170116095119-00056-ip-10-171-10-70.ec2.internal.warc.gz | 1,003,347,077 | 46,871 | Teachers Pay Teachers
# Functions Review Game (8th Grade)
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How to unzip files.
3.64 MB | 23 pages
### PRODUCT DESCRIPTION
During this functions BINGO review activity, your students will work in cooperative groups to practice, determining output of a function given the input, comparing rate of change of functions represented in different ways, determining whether a function is linear or nonlinear, writing a rule to model a function, analyzing and sketching graphs, and more! The PowerPoint slides are editable so you can adjust to the needs of your students.
Included in this product:
-18 review questions aligned to the CCSS (with answers)
-BINGO board
-Number Cards 1-24 (directions for use included)
-Teacher's Guide with objectives, materials, and procedure
To play, group students into teams of 4-5. Pass out mini-whiteboards to each student, if available. Each group gets a BINGO board, which they should fill in with the numbers 1-24 in random order. Each student in a group is assigned a number 1-5.
For each slide, students collaborate quietly to ensure all students understand and are able to arrive at the correct answer. Once an appropriate amount of time has been provided to answer the question, call out one of the numbers you assigned at the beginning of class (1, 2, 3, 4, 5). The student in each group with that number must hold up their whiteboard. If they are correct, their group gets credit for the question. If they are incorrect, their group does not get credit for the question even if other students in the group had the correct answer. Encourage students to help each other out!
To determine how students who received credit for the question will cross off a square on the BINGO board, draw a card from the pile of the 1-24 cards (included in this product). Repeat for all other slides or as time allows.
This product is a zip file. The instructions, BINGO board, and numbers are in a PDF file. The PowerPoint game is editable to adjust to the needs of your own students.
Get all 8th Grade BINGO Review Games here:
8th Grade Math Common Core-Aligned Standardized Prep BINGO Games
Get all functions resources here: Functions Bundle
Teach 8th grade math? Consider this incredible deal: 8th Grade Math MEGA (Growing) Bundle
You might also like:
Input and Output: Evaluating and Identifying Functions
Old Math Guy: Matching Input and Output of Functions
Growth Rates: Comparing Rate of Change in Non-Proportional Relationships
Math War: A Game to Practice Comparing Rate of Change Non-Proportional Edition
Is it Linear? Comparing Linear and Nonlinear Functions
Sort Activity: Classifying Families of Functions
Road Trip! Modeling and Analyzing Linear Relationships
Task Cards: Determining the Rate of Change and Initial Value of Linear Functions
Up, Down, and Around: Analyzing Graphs
Ratios and Proportional Relationships BINGO Review Game
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22 ratings | 935 | 4,189 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.92841 |
https://puzzling.stackexchange.com/questions/14556/identify-the-incorrect-letter-in-this-unusual-sequence | 1,718,236,889,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00272.warc.gz | 449,309,516 | 38,849 | # Identify the incorrect letter in this unusual sequence?
Here's a curious sequence of letters:
H H H E L I P O A
L S C S E C S A C
But one of the letters is wrong! Can you figure out which letter is wrong, and what the sequence represents?
(note: two lines used for sequence are simply for the sake of keeping it neat, the newline is not part of the puzzle)
• Looked like the periodic table elements :( Commented May 8, 2015 at 15:25
P should be B.
The periodic table overlayed with the Fibonacci sequence. H = 1, H = 1, He = 2, Li = 3, B = 5, O = 8, Al = 13... | 162 | 568 | {"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} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.927498 |
http://www.math.utah.edu/~gustafso/index2280S2009.html | 1,563,383,440,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525355.54/warc/CC-MAIN-20190717161703-20190717183703-00501.warc.gz | 239,432,802 | 3,633 | # Grant Gustafson at math.utah.edu
2280 8:35am Spring 2009
Updated: Monday May 18: 18:11PM, 2009 Today: Wednesday July 17: 11:10AM, 2019
# Intro Differential Equations 2280 S2009
### Announcements
• Web records now available, 9am on 15 May 2009. Send email if you find an error that affects your final letter grade or rank in class.
• Exams outside the door 113jwb on 12 May. Final exam solutions key, S2009. Top score solutions by Sara S., David D., Simon W.
• Sample Final Exam [20 problems] is ready to download 30 April. Revised with answers and solutions on 2 May. Please advise of any errors found, and the correction will be posted here. Bart, Josh, Kelli, Dan, Chelsie and Jennifer discovered a wrong answer for problems 7 and 18a. The pdf below has been corrected. Sample final exam problems, S2009.
Solutions to pass one of exam 3 on 22 April are available here:
Exam 3 answers, 22 April 2009, Version 1
Wood, Harris and Young discovered a wrong answer in (4b), correct answer is x(t)=exp(2t)-exp(t), y(t)=0.
## 2280 Due Dates, Maple Projects and Extra Credit
1. 2280 [8:35am] due dates for submitted work
Due dates updated daily
3. 2280 Extra Credit. Cancel zeros and fifty-fives! More maple extra credit appears in certain chapters. Extra credit problems
## 2280 Syllabus and Dailies Spring 2009
1. Gustafson's 2280 course syllabus Spring 2009
pdf postscript syllabus (90k (119K pdf)
Differential Equations, Edwards-Penney 4/E 2008
2. Gustafson's 2280 gradesheet Spring 2009
A form for keeping your own records ( 98K pdf)
3. 2280 format suggestions for reports
How to improve your written work. Writing reports. ( 25K pdf)
4. How to do maple graphics using a modem or dsl
Low speed internet and maple graphics (3.8K txt)
## 2280 Slides and Documents
1. Slides projected in class from the notebook computer are duplicated on the web here:
Directory from classroom notebook computer
## 2280 Midterm and Final Exams
Click here for all S2009 exam materials, including samples, old exam keys with solutions, keys to this semester's exams, and related study guides for exams.
## Maple Tutorials and Maplesoft offer
3. A rookie maple tutorial for the impatient from Indiana University Click Here
4. Angie Gardiner tutorial source. Save to local disk and launch it in maple. Save file (.mws) [usually right click]
5. Details for maple 12 under unix, windows, OS/X How to use maple 12 (2k text)
6. Offer from Waterloo Maple to buy a copy of maple 12 for your OS/X, Linux or MS-Windows home computer or laptop [\$50.00].
## 2280 Problem Notes and Frequently Asked Questions
1. Presently, there are none, but the companion book Differential Equations and Linear Algebra by Edwards-Penney is used for math 2250, and there are numerous notes on those problems. Beware: the chapters correspond, but have a different order, e.g., Laplace is earlier in our textbook. Also, some extra chapters appear in our text, and they will not have 2250 problem notes.
2. If I get time to sort this out, then I will, and make available problem notes keyed to our textbook.
3. 2250 Problem Notes. Extra info on dailies. Problem notes and FAQ
# FLASH Animations of applications
• The physics animations in the link below [85 animations] were written by David M. Harrison, Dept. of Physics, Univ. of Toronto. They are Copyright 2002 - 2006 David M. Harrison.
Harrison's physics animations
The areas covered in the 85 animations appear below. A high percentage of them touch the subject matter of this course.
1. Chaos
2. Classical Mechanics
3. Electricity and Magnetism
4. Micrometer Caliper
5. Miscellaneous
6. Nuclear
7. Optics
8. Oscilloscope
9. Quantum Mechanics
10. Relativity
11. Sound Waves
12. Vectors
13. Waves
``` Grant B. Gustafson | 991 | 3,750 | {"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} | 2.640625 | 3 | CC-MAIN-2019-30 | latest | en | 0.829461 |
https://www.thestudentroom.co.uk/showthread.php?t=4518454&page=2 | 1,527,438,568,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794869272.81/warc/CC-MAIN-20180527151021-20180527171021-00061.warc.gz | 843,546,494 | 40,880 | You are Here: Home >< Chat >< Chat
# How many people could be fed with £90000.? watch
Announcements
1. (Original post by john2054)
You would be much better off donating to a genuine orphanage, to help the thousands of orphans, and their carers, who have to get by, day by day...
Correct and preferably one that's local to you.
2. Well there are 50,000 grans of rice in a kg.
1kg at Asda is 42p meaning with 9k you could buy 21428kgs of rice.
Thus, if you fed each person one grain of rice you could feed (21,428 x 50,000) = 1,071,400,000 people.
3. (Original post by King_786)
It would cost around £40 billion to end world poverty.
I don't think so.
Quick google pointed me here which seems sensible enough
To end extreme poverty worldwide in 20 years, Sachs calculated that the total cost per year would be about \$175 billion. This represents less than one percent of the combined income of the richest countries in the world.
4. You could feed people for a bit and then they would be back to where they were quickly.
You can't just throw money at it. 40bil would provide a temporary solution but within a year, poverty would be back in full force.
You could pour all the money you want into Africa but it will stay poor because of the political structures and the culture. Unless those things change, nothing will improve. It is very difficult to change those things from Europe. It is for the people of those countries to improve things for themselves.
5. (Original post by Alfissti)
Correct and preferably one that's local to you.
No i wouldn't stand to gain from it, but i know of one in Kenya that i can put you in touch with, if you are still interested?
6. (Original post by Sternumator)
You could feed people for a bit and then they would be back to where they were quickly.
You can't just throw money at it. 40bil would provide a temporary solution but within a year, poverty would be back in full force.
You could pour all the money you want into Africa but it will stay poor because of the political structures and the culture. Unless those things change, nothing will improve. It is very difficult to change those things from Europe. It is for the people of those countries to improve things for themselves.
I agree with this. No amount of money can end world poverty, because it is to do with the social and political structures in place. And just pumping money in to them, won't change a thing.
Just look at the millions we spend on the poor via red nose day each year. But there are still poor. So answer me this, why is that!??
7. Depends on the food and country
90k would feed a lot more people in third world countries than it would here
Tinned food and bread would feed a lot more than if you opted for organic food
Either way, you'll still feed a good percentage of hungry families
Quality over quantity remember, if even it only feeds 100 people then you've still fed 100 people
8. (Original post by john2054)
No i wouldn't stand to gain from it, but i know of one in Kenya that i can put you in touch with, if you are still interested?
Sorry, I live in Norway and come to England for business and classes, therefore I generally do not believe in doing anything other than in these 2 countries,
Personally believe in charities that improve local communities either by means of environmental or social improvement only.
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Course: 2024/2025
Process Control
(14273)
Master in Industrial Engineering (Plan: 170 - Estudio: 226)
EPI
Coordinating teacher: GARRIDO BULLON, LUIS SANTIAGO
Department assigned to the subject: Systems Engineering and Automation Department
Type: Compulsory
ECTS Credits: 3.0 ECTS
Course:
Semester:
Requirements (Subjects that are assumed to be known)
Previous knowledge of Ordinary Differential Equations, Matrix Calculus and Automatic Control is assumed.
Objectives
Ability to design and project automated production systems and advanced process control
Skills and learning outcomes
Description of contents: programme
1 Modeling and Analysis of Systems in State Space. 1.1 Introduction to the concept of state and state space. 1.2 Dynamic systems. 1.3 Linearity and invariance. 1.4 Linearization. 1.5 Representation of systems in state space. 1.6 Interconnection of systems. 1.7 Obtaining the state model. 1.8 Linear transformations. 1.9 Obtaining the transfer function from the state model. 2 Solution of the equations of state. 2.1 Transition matrix. 2.2 Calculation of the transition matrix. Properties. 2.3 Solution of the equations of state in discrete time systems. 3 Status feedback control. 3.1 Controllability and observability. 3.2 Complete controllability of the state of a system. 3.3 Complete controllability of a system's output. 3.4 Complete observability of the state of a system. 3.5 Invariance of controllability and observability before transformations. 3.6 Status feedback control: pole positioning method. 3.7 Adjusting the positions of the poles in a closed chain. 3.8 Gain adjustment. 3.9 Modification of the type of a system. 4 Design of state observers. 4.1 Concept of state observer. 4.2 Conditions for observing the state. 4.3 Complete order status observer. 4.4 Error dynamics in the complete order observer. 4.5 Design of the matrix of gains of the observer feedback. 4.6 Closed-loop dynamics of the system with state feedback and state observer.
Learning activities and methodology
The training activities include: - Lectures, where knowledge that students should acquire will be presented. To facilitate their development students receive class notes and have basic reference texts that facilitates follow lessons and develop further work. - Resolution of exercises by the student self-assessment and will serve to acquire the necessary skills. - Classes of problems, which are developed and discuss the problems that are proposed to students. - Lab, where students experimentally verify the theoretical concepts and results seen in class. - Lab in computer room where computer are resolved proposed problems. - Lectures, classes resolution of questions in small groups, student presentations, individual tutorials and personal work, including research, tests and exams aimed at the acquisition of theoretical knowledge will involve 1.5 ECTS credits. - Laboratory practices and kinds of problems in small groups, individual tutorials and personal work, including research, tests and exams aimed at the acquisition of practical skills related to the program will involve subjects 1.5 ECTS credits.
Assessment System
• % end-of-term-examination 50
• % of continuous assessment (assigments, laboratory, practicals...) 50
Calendar of Continuous assessment
Basic Bibliography
• Tripathi. Modern Control Systems: An Introduction. Jones & Bartlett Learning. 2010 | 762 | 3,422 | {"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} | 2.671875 | 3 | CC-MAIN-2024-30 | latest | en | 0.826163 |
https://www.zweigmedia.com/RealWorld/quiz/Calcchap2TF.html | 1,652,842,606,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00774.warc.gz | 1,273,956,515 | 2,593 | ## True/False Quizzes for Applied Calculus Topic: Introduction to the Derivative (Chapter 3)
Chapter 2 quiz
Chapter 4 quiz
Quiz Index
Summary of This Topic
Review Exercises
On-Line Utilities
On-Line Tutorials
Everything for Calculus
Everything for Finite Math & Calculus
Note! If your screen is not partitioned into three frames, press here to bring up the frames that will enable you to use these quizzes.
1. f'(2) is the slope of the secant line to the curve y = f(x) at the point (2, f(2)).
2. f'(2) is the slope of the tangent line to the curve y = f(x) at the point (2, f(2)).
3. If f'(x) = 0, then the tangent line to the curve y = f(x) at (x, f(x)) is horizontal.
4. If f'(x) 0, then the tangent line to the curve y = f(x) at (x, f(x)) is not horizontal.
5. If f(x) = x3(x2 1), then f'(x) = 3x2(2x 0).
6. If the position s of a particle is given by s = 2t (t = time in seconds), then the particle has a speed of 2 units per second
7. The average velocity of a particle over the time interval [a, b] is given by the slope of the tangent of the position vs. time graph through the point (a, b).
8. The average velocity of a particle over the time interval [a, b] is given by the slope of the secant of the position vs. time graph through the points with time coordinates a and b.
9. The difference quotient of f is the slope of the secant line through the points (x, f(x)) and (x+h, f(x+h)).
10. The derivative is obtained from the difference quotient by setting h=0
11. The difference quotient is the limit of the derivative as h0.
12. If f(x)3 as x4, then f(3) = 4.
13.
If x3_ f(x) = 4, and x3 f(x) exists, then x3+ f(x) = 4.
14.
x3 x3 - xx2 + 1 = x3 x3 x2 = 3
15.
x 1x3 = 0
16. If a soccer ball is at rest and s represents its position, then ds/dt = 0.
17. If the marginal revenue is zero at some value of q, then the revenue is increasing rapidly as q increases.
18. If R(q) is a company's annual revenue (in dollars) as a function of the number of items it sells, then R'(1,000) measures the number of items it sells per year at a production level of 1,000.
19. Suppose that H(n) is the height of an individual (in feet) at age n years. If H'(12) = 0.3, this means that the height of the individual is increasing at a rate of 0.3 feet per year on his or her twelfth birthday.
20. If R(q) is a company's revenue (in dollars) as a function of the number of items it sells, then the units of R'(q) are dollars per item.
Last Updated: April, 1997 | 733 | 2,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-21 | latest | en | 0.873929 |
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# Newsom
Spring 2006
## Post Hoc Tests
Familywise Error
Also known as alpha inflation or cumulative Type I error. Familywise error (FWE) represents the
probability that any one of a set of comparisons or significance tests is a Type I error. As more tests are
conducted, the likelihood that one or more are significant just due to chance (Type I error) increases.
One can estimate familywise error with the following formula:
FWE 1 (1 EC )c
where FWE is the familywise error rate,EC is the alpha rate for an individual test (almost always
considered to be .05), and c is the number of comparisons. c as used in the formula is an exponent, so
the parenthetical value is raised to the cth power.
Bonferroni
The Bonferroni simply calculates a new pairwise alpha to keep the familywise alpha value at .05
(or another specified value). The formula for doing this is as follows:
FWE
c
where B is the new alpha based on the Bonferroni test that should be used to evaluate each comparison
or significance test, FWE is the familywise error rate as computed in the first formula, and c is the number
of comparisons (statistical tests).
The Bonferroni is probably the most commonly used post hoc test, because it is highly flexible, very
simple to compute, and can be used with any type of statistical test (e.g., correlations)not just post hoc
tests with ANOVA. The traditional Bonferroni, however, tends to lack power. The loss of power occurs
for several reasons: (1) the familywise error calculation depends on the assumption that, for all tests, the
null hypothesis is true. This is unlikely to be the case, especially after a significant omnibus test; (2) all
tests are assumed to be orthogonal (i.e., independent or nonoverlapping) when calculating the
familywise error test, and this is usually not the case when all pairwise comparisons are made; (3) the
test does not take into account whether the findings are consistent with theory and past research. If
consistent with previous findings and theory, an individual result should be less likely to be a Type I error;
and (4) Type II error rates are too high for individual tests. In other words, the Bonferroni overcorrects for
Type I error.
Modified Bonferroni Approaches
Several alternatives to the traditional Bonferroni have been developed, including those developed
by Holm, Holland and Copenhaver, Hommel, Rom, and others (see Olejnik, Li, Supattathum, & Huberty,
1997 for a review). These tests have greater power than the Bonferroni while retaining its flexible
approach that allows for use with any set of statistical tests (e.g., t-tests, correlations, chi-squares).
Sidak-Bonferroni. Sidak (1967) suggested a relatively simple modification of the Bonferroni formula that
would have less of an impact on statistical power but retain much of the flexibility of the Bonferroni
method (Keppel & Wickens, 2004 discuss this testing approach). Instead of dividing by the number of
comparisons, there is a slightly more complicated formula:
S B 1 1 FWE
1/ c
where S-B is the Sidak-Bonferroni alpha level used to determine significance (something less than .05),
FWE is the computed familywise error according to the formula at the top of the first page, and c is the
number of comparisons or statistical tests conducted in the family. The p-values obtained from the
Newsom
USP 534 Data Analysis I
Spring 2006
computer printout must be smaller than S-B to be considered significant. One can also extend this test to
other statistical tests, such as correlations. In the case of correlations, one could replace dfA with the
number of variables that are used in the group of correlations tests. c would represent the number of
correlations in the correlation matrix. This approach is convenient and easy to do but has not received
any systematic study, and it is likely that a single, simple correction will not result in the most efficient
balance of Type I and Type II errors.
Hochbergs Sequential Method. This test uses a specific sequential method called a step-up approach
as a more powerful alternative to the Bonferroni procedure. Sequential methods use a series of steps in
the correction, depending on the result of each prior step. Contrasts are initially conducted and then
ordered according to p-values (from smallest to largest in the step-up approach). Each step corrects
for the previous number of tests rather than all the tests in the set. This test is a good, high power
alternative to the other modified Bonferroni approaches as long as confidence intervals are not needed.
Approaches for Pairwise Comparisons with ANOVA Designs
Dunn. Identical to the Bonferroni correction.
Scheffe. The Scheffe test computes a new critical value for an F test conducted when comparing two
groups from the larger ANOVA (i.e., a correction for a standard t-test). The formula simply modifies the Fcritical value by taking into account the number of groups being compared: (a 1) Fcrit. The new critical
value represents the critical value for the maximum possible familywise error rate. As you might
suppose, this also results in a higher than desired Type II error rate, by imposing a severe correction.
Fisher LSD. The Fisher LSD test stands for the Least Significant Difference test (rather than what you
might have guessed). The LSD test is simply the rationale that if an omnibus test is conducted and is
significant, the null hypothesis is incorrect. (If the omnibus test is nonsignificant, no post hoc tests are
conducted.) The reasoning is based on the assumption that if the null hypothesis is incorrect, as
indicated by a significant omnibus F-test, Type I errors are not really possible (or less likely), because
they only occur when the null is true. So, by conducting an omnibus test first, one is screening out group
differences that exist due to sampling error, and thus reducing the likelihood that a Type I error is present
among the means. Fishers LSD test has been criticized for not sufficiently controlling for Type I error.
Still, the Fisher LSD is sometimes found in the literature.
Dunnet. The Dunnet test is similar to the Tukey test (described below) but is used only if a set of
comparisons are being made to one particular group. For instance, we might have several treatment
groups that are compared to one control group. Since this is rarely the of interest, and the Tukey serves
a much more general purpose, I recommend the Tukey test.
Tukey a (also known as Tukeys HSD for honest significant difference). Tukeys test calculates a new
critical value that can be used to evaluate whether differences between any two pairs of means are
significant. The critical value is a little different because it involves the mean difference that has to be
exceeded to achieve significance. So one simply calculates one critcal value and then the difference
between all possible pairs of means. Each difference is then compared to the Tukey critical value. If the
difference is larger than the Tukey value, the comparison is significant. The formula for the critical value
is as follows:
d T qT
MS s / A
n
qT is the studentized range statistic (similar to the t-critcal values, but different), which one finds in a table
(Table C.9 in the Myers & Well text), MSs/A is the mean square error from the overall F-test, and n is the
sample size for each group. Error df referred to in the table is the dfs/A used in the ANOVA test. FWE is
the desired familywise error rate. This is the test I usually recommend, because studies show it has
Newsom
USP 534 Data Analysis I
Spring 2006
greater power than the other tests under most circumstances and it is readily available in computer
packages. The Tukey-Kramer test is used by SPSS when the group sizes are unequal. It is important to
note that the power advantage of the Tukey test depends on the assumption that all possible pairwise
comparisons are being made. Although this is usually what is desired when post hoc tests are
conducted, in circumstances where not all possible comparisons are needed, other tests, such as the
Dunnett or a modified Bonferroni method should be considered because they may have power
Games-Howell. This test is used with variances are unequal (see Unequal Variances below) and also
takes into account unequal group sizes. Severely unequal variances can lead to increased Type I error,
and, with smaller sample sizes, more moderate differences in group variance can lead to increases in
Type I error. The Games-Howell test, which is designed for unequal variances, is based on Welchs
correction to df with the t-test and uses the studentized range statistic. This test appears to do better
than the Tukey HSD if variances are very unequal (or moderately so in combination with small sample
size) or can be used if the sample size per cell is very small (e.g., <6).
I have included only a subset of all the possible post hoc corrections for familywise error. And,
believe it or not, familywise error correction procedures currently available in most statistical packages
(only some of which I have focused on here) represent only a subset of the approaches which have been
proposed and studied.
Klockars, Hancock, and McAweeney (1995) discuss many of the post hoc ANOVA procedures, some of
which seem to advantages over the traditional approaches such as the Tukey currently available in
statistical software packages. The alternative procedures are often classified as sequential vs.
simultaneous, weighted vs. unweighted, and step-up vs. step-down, and they involve elaborate
computational procedures which are inconvenient to do by hand especially for a large number of
comparisons. Modified Bonferroni procedures have been designed for a broader array of statistical
circumstances beyond post hoc ANOVA tests (e.g., correlations or chi-square tests). Olejnik and
colleagues (1997) review the modified Bonferroni procedures and their computations. They conclude that
most of the modified Bonferroni procedures have clear advantages over the traditional Bonferroni
procedure, but small differences among the alternatives in the amount of power or control of Type I error.
Their results suggest that Roms (1990) procedure has the most power (not currently available in SPSS).
Other authors have reviewed post hoc tests with additional attention to unequal error variances (e.g.,
Kromrey & La Rocca, 1995; Seaman, Levin, & Serlin,1991). How heterogeneous (i.e., unequal) the
error variances must be in order to cause problems is difficult to discern, because their impact is greater
with lower sample sizes. Unfortunately, tests such as the Levine tests for unequal variances have lower
power when sample size is smaller, so they may be least likely to indicate a problem with unequal
variances when it is most likely to affect Type I errors. In terms of post ANOVA tests, the Games-Howell
is good if there are large differences in variances between groups.
Keppel, G., & Wickens, T.D. (2004). Design and analysis: A researchers handbook (4rd Edition). Upper Saddle River, NJ: Pearson.
Klockars, A.J.,Hancock, G.R., & McAweeney, M.J. (1995). Power of unweighted and weighted versions of simultaneous and sequential multiplecomparison procedures. Psychological Bulletin, 118, 300-307.
Kromrey, J.D., & La Rocca, M.A. (1995). Power and Type I error rates of new pairwise multiple comparison procedures under heterogeneous
variances. Journal of Experimental Education, 63, 343-362.
Olejnik,S., Li, J., Supattathum, S., and Huberty, C.J. (1997). Multiple testing and statistical power with modified Bonferroni procedures. Journal
of educational and behavioral statistics, 22, 389-406.
Seaman, M.A., Levin, J.R., & Serlin, R.C. (1991). New developments in pairwise multiple comparisons: Some powerful and practicable
procedures. Psychological Bulletin, 110, 577-586. | 2,701 | 11,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-24 | latest | en | 0.910565 |
france-pastry.com | 1,601,301,473,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401601278.97/warc/CC-MAIN-20200928135709-20200928165709-00328.warc.gz | 41,045,352 | 7,536 | # Top Parentheses in Mathematics Tips!
The end result of dividing two rational numbers is another rational number as soon as the divisor isn't 0. When using more elaborate equations, which might combine several terms and utilize several operations, grouping the terms together will help organize the equation. We still must demonstrate the averaging property. All these exponents of powers require to get defined.
It's much less good once you need both superscripts and subscripts on the identical expression. Note that you're permitted to put both partial sentences and total sentences within parentheses. essaysource.com As a result, since we might remove parentheses, we might also place them. Parentheses and brackets are extremely typical in mathematical formulas. These fall below the OoO as parenthesis.
## The Secret to Parentheses in Mathematics
We are going to be worried about the minus in a subsequent step. Another individual might choose to produce the problem a little simpler by multiplying first. Aside from that, it's only a matter of multiplying every one of those steps and adding everything together. Your program needs to be in a position to read and process numerous lines of input. There aren't any terms which could be combined, this predicament is complete.
## The Nuiances of Parentheses in Mathematics
The body of the text of these things may be put in italics to be able to distinguish them from the majority of the text of the write-up. Inside this example we had the ability to combine two of the terms to simplify the last answer. A typical mathematics paper comprises lots of the elements of a normal academic article, including an abstract, an introduction, and reference section. Otherwise, this punctation mark wouldn't be utilized in academic writing. You can do something similar with commas, like I did in the previous sentence and am doing now, but if you've got a number of commas in the sentence for different purposes, things can find a bit out of hand, like they may be here. By way of example, take one of my favourite authors.
## Definitions of Parentheses in Mathematics
Make sure students can determine the sections of the expression before continuing. They isolate themselves into small groups for protection as opposed to relying on a far larger society. Q. I mean exactly what this fitness is about. Students in such lessons are learning how to stick to a rote procedure to arrive at a solution.
## The Little-Known Secrets to Parentheses in Mathematics
In modern mathematical provisions, this is called extending the system. Intervals are chunks of the actual line. One of the most frequent techniques is to find rid of a term on a single side by subtracting it from either side. Every real number has a distinctive location at stake.
You distribute the complete 2x2 just enjoy the numbers in the prior examples. Also, it's simple to demonstrate that the associative law is valid, and thus the parentheses in (2) can be omitted with no ambiguities. Operators which are shown on the identical row have the identical precedence. However, it's additionally a huge topic due to the presence of so much mathematical notation. In various disciplines, languages or in region brackets may carry unique meanings, but the majority of the time depending on the context of the application.
Thus, let's talk about that which we can actually do. It could be convenient to have a wonderful scratchpad handy. BEDMAS or PEDMAS is among them.
## The Tried and True Method for Parentheses in Mathematics in Step by Step Detail
It's also advisable to write a brief program to check your solution. She's also continually ranked as one of the greatest instructors in California. Our classes are comparatively small especially at the top division level. Teaching for conceptual understanding has many advantages.
She doesn't hide her entire name but there's no demand for her to ensure it is obvious within her work. A couple of the names below are somewhat weird. I can consider a few distinct words such as this. So we definitely want some manner of turning things around. To begin with, I'd like to say this is in no way to question the manner somebody is writing. Employing an analogy might be a terrific approach to explain a new idea, to draw conclusions in a new instance, or a means to convince a person to change their mind about a specific matter.
## Finding Parentheses in Mathematics
But in case you have more than 5 fields then utilizing a simple for loop would be perfect. As for the way the compiler determines when to take care of parentheses in every single way, the rules are quite intricate and there are in fact a few circumstances where the compiler isn't going to parse your expression in the intended way. Be aware that the SQL should end with semi-colon if you've got multiple queries in the query window. Employing the various grouping symbols is merely a wonderful way of assisting the user keep an eye on the different pairs of symbols. If you wish to typeset equations beyond the very simplest ones, you'll want to use the amsmath package, among the many packages that it is possible to use with LaTeX. In your parser, when you encounter a word, you've got in order to tell whether it is a variable or a normal function.
## What to Expect From Parentheses in Mathematics?
This standard is crucial to simplifying and solving different algebra issues. It is very important to make sure you've got a strong foundation before you move on. In the end, it's important to grasp the math behind the 'acronym'. Abstract In several of the quizzes and exams for this training course, you're going to be typing mathematics.
Last modification: Thu 12 Sep 2019 | 1,139 | 5,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-40 | latest | en | 0.944064 |
https://plainmath.net/83714/a-district-in-the-ashanti-region-town-ha | 1,660,055,423,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570977.50/warc/CC-MAIN-20220809124724-20220809154724-00687.warc.gz | 424,417,234 | 14,657 | # A district in the Ashanti region town has 102 towns and villages. If you interviewed everyone living in one village, would you be interviewing a population or a sample from the village?
A district in the Ashanti region town has 102 towns and villages. If you interviewed everyone living in one village, would you be interviewing a population or a sample from the village?
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frisiao
It is given that,
A district in the Ashanti region town has 102 towns and villages. It means we can divide the district into towns and villages.
We know that, if a population is divided into various group or cluster and complete sampling is done over on just a few groups or cluster, then this type of sampling scheme is said to be cluster sampling.
The above-given situation is exactly similar to that in cluster sampling.
Thus, If you interviewed everyone living in one village, we would be interviewing a Population from the village. | 237 | 1,129 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 35, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-33 | latest | en | 0.965401 |
https://www.physicsforums.com/threads/electric-and-magnetic-fields.804767/ | 1,508,734,915,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825575.93/warc/CC-MAIN-20171023035656-20171023055656-00593.warc.gz | 989,930,010 | 16,872 | # Electric and magnetic fields
1. Mar 23, 2015
### vysero
1. The problem statement, all variables and given/known data
The below region of space shown has uniform electric and/or magnetic fields, dependent on the conditions listed for cases A-G. Identical positively charged particles enters these regions with the same velocity.
Rank the magnitude of the force on the particle just after it enters these regions for cases A-G from greatest to least.
2. Relevant equations
F = qE + qv x B
F = qE
Fb = qvBsin(*)
3. The attempt at a solution
A) Since the particle is positive(+) I said there would be a force F namely ( ---> ) on the particle.
B) When only B is present there is no force on the particle, Fb = qvBsin(0) = 0. This is because both B and V are parallel.
C) Same as A: ( ---> )
D) Same as A: ( ---> )
E) Here the direction of the force changes but its magnitude remains the same so: ( <--- )
F) There is an upwards force in this case. However, its magnitude is the same as was in A.
G) I have this as being two forces on the particle so G is the largest.
Overall I ranked them as follows: G>A=C=D=E=F>B
Not sure if I did these correctly any help would be appreciated.
2. Mar 23, 2015
### Dick
Sounds ok to me.
3. Mar 23, 2015
### vysero
Wait... you mean I got it right.. like on my own ._. I think I will go have a drink!
4. Mar 23, 2015
### Dick
Sure, have a drink. I can't see anything wrong with your logic. | 392 | 1,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-43 | longest | en | 0.939175 |
https://www.got-it.ai/solutions/excel-chat/excel-problem/11035-if-function | 1,702,283,907,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00711.warc.gz | 857,487,759 | 13,750 | # Excel - IF Function Problem - Expert Solution
Question description:
This user has given permission to use the problem statement for this blog.
I need help with an IF statement in Excel. I work for a youth hockey organization and I have the scores of the recent games. A win of more than 5 goals or a loss of more than 5 goals is called a 'blowout' game. How can I write a formula to say if a value in a cell is >5 or <5, return the value of 1 so that I can simply count the number of 'blowout' games? If the answer does not meet criteria return the value of 0.
Solved by B. A. in 20 mins
This is the chat thread from the real Excelchat help session. It contains no private user information.
Excelchat Expert 20/09/2017 - 08:29
Welcome, Thanks for choosing Got It Pro-Excel!
Excelchat Expert 20/09/2017 - 08:29
How may I help you today!
User 20/09/2017 - 08:29
I can write the formula for the >5 and the <5 but I cannot seem to combine into one formula.
Excelchat Expert 20/09/2017 - 08:30
No worries I'll write it for yo
Excelchat Expert 20/09/2017 - 08:30
*you
User 20/09/2017 - 08:30
THANK YOU!
Excelchat Expert 20/09/2017 - 08:30
would you please me give a moment to solve your problem?
User 20/09/2017 - 08:30
sure
Excelchat Expert 20/09/2017 - 08:30
I’ll be back to you soon Thanks.
Excelchat Expert 20/09/2017 - 08:31
You need formula if value is greater than or less than 5 it should return 1?
User 20/09/2017 - 08:31
yes
Excelchat Expert 20/09/2017 - 08:32
OK
User 20/09/2017 - 08:32
otherwise return 0
Excelchat Expert 20/09/2017 - 08:32
Otherwise means ?
User 20/09/2017 - 08:32
if not >5 or <5 return 0 in the formula
Excelchat Expert 20/09/2017 - 08:33
Greater than 5 are all values greater than five & less than means 0-5, than what may be other value?
User 20/09/2017 - 08:34
correct
User 20/09/2017 - 08:34
0,1,2,3,4 and 5 would return 0
Excelchat Expert 20/09/2017 - 08:34
Please explain than?
Excelchat Expert 20/09/2017 - 08:34
OK
User 20/09/2017 - 08:34
a score of 9-2 is a difference of 7
User 20/09/2017 - 08:34
I call that delta of 7 a blowout game because it's >5
User 20/09/2017 - 08:34
a score of 2-1 is not, delta is only 1
Excelchat Expert 20/09/2017 - 08:37
=IF(A1>5,1,IF(A1=5,0,IF(A1<5,0)))
Excelchat Expert 20/09/2017 - 08:37
I think this is you need
Excelchat Expert 20/09/2017 - 08:37
Please verify or clarify?
User 20/09/2017 - 08:37
let me try it on a sample set of scores
User 20/09/2017 - 08:37
one second
Excelchat Expert 20/09/2017 - 08:37
Yes please
User 20/09/2017 - 08:41
a value of -9 did not work properly
User 20/09/2017 - 08:41
came back as 0
Excelchat Expert 20/09/2017 - 08:41
OK
Excelchat Expert 20/09/2017 - 08:41
You need that -9 should give 1?
User 20/09/2017 - 08:41
yes
Excelchat Expert 20/09/2017 - 08:42
Others are good?
User 20/09/2017 - 08:42
I think I fixed the end of the formula
User 20/09/2017 - 08:42
let me see what the number 1 brings back
Excelchat Expert 20/09/2017 - 08:42
Yes please
User 20/09/2017 - 08:43
=IF(A1>5,1,IF(A1=5,0,IF(A1<5,0))) the value of 2 does not come back as 0
Excelchat Expert 20/09/2017 - 08:43
One minute
User 20/09/2017 - 08:43
0.1.2.3.4 and 5 should all come back as 0
Excelchat Expert 20/09/2017 - 08:43
Here its giving 0
Excelchat Expert 20/09/2017 - 08:44
For my sheet its giving 0 for all 0,1,2,3,4,5
Excelchat Expert 20/09/2017 - 08:44
Can you see here on google sheets?
User 20/09/2017 - 08:44
send me the formula one more time please
User 20/09/2017 - 08:44
It might be me
Excelchat Expert 20/09/2017 - 08:45
Have you seen please?
Excelchat Expert 20/09/2017 - 08:45
=IF(A1>5,1,IF(A1=5,0,IF(A1<5,0)))
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# Lesson 8 - 1 - PowerPoint PPT Presentation
## Lesson 8 - 1
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##### Presentation Transcript
1. Lesson 8 - 1 Discrete Distribution Binomial Taken from http://www.pendragoncove.info/statistics/statistics.htm And modified slightly
2. Knowledge Objectives • Describe the conditions that need to be present to have a binomial setting. • Define a binomial distribution. • Explain when it might be all right to assume a binomial setting even though the independence condition is not satisfied. • Explain what is meant by the sampling distribution of a count. • State the mathematical expression that gives the value of a binomial coefficient. Explain how to find the value of that expression. • State the mathematical expression used to calculate the value of binomial probability.
3. Construction Objectives • Evaluate a binomial probability by using the mathematical formula for P(X = k). • Explain the difference between binompdf(n, p, X) and binomcdf(n, p, X). • Use your calculator to help evaluate a binomial probability. • If X is B(n, p), find µx and x (that is, calculate the mean and variance of a binomial distribution). • Use a Normal approximation for a binomial distribution to solve questions involving binomial probability
4. Vocabulary • Binomial Setting – random variable meets binomial conditions • Trial – each repetition of an experiment • Success – one assigned result of a binomial experiment • Failure – the other result of a binomial experiment • PDF – probability distribution function; assigns a probability to each value of X • CDF – cumulative (probability) distribution function; assigns the sum of probabilities less than or equal to X • Binomial Coefficient – combination of k success in n trials • Factorial – n! is n (n-1) (n-2) … 2 1
5. Binomial Mean and Std Dev A binomial experiment with n independent trials and probability of success p has Mean μx = np Standard Deviation σx = √np(1-p)
6. Example 1 Find the mean and standard deviation of a binomial distribution with n = 10 and p = 0.1 Mean: μx = np = 10(0.1) = 1 Standard Deviation: σx = √np(1-p) = 10(0.1)(0.9) = 0.9 = 0.9487
7. Using Normal Apx to Binomials As binominal’s number of trials increases the formula for a binomial becomes unworkable (a situation alleviated with statistical software). So statisticians developed a procedure to use a continuous distribution, the normal, to estimate a discrete distribution. This procedure is used later with proportions.
8. Example 2 Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide random sample of 2500 adults if shopping was often frustrating and time-consuming. Assume that 60% of all US adults would agree if asked the same question, what is the probability that 1520 or more of the sample would agree? P(X ≥ 1520) = 1 – P(X ≤ 1519) = 1 – binomcdf(2500, 0.6, 1519) = 0.2131 𝜇x = np = 2500 (0.6) = 1500 σx= √np(1-p) = √ 2500(0.6)(0.4) = 600 = 24.49 P(X ≥ 1520) = using normCDF(1520, E99, 1500, 24.49) = 0.2071 a difference of 0.0061 or less than 0.6%
9. Example 2 cont Histogram showing normal apx to binomial
10. Simulating Binomial Events • To simulate a binomial event, we must know: • how random variable X and “success” is defined • probability of success • number of trials • Calculator has a randbin(1,p,n) function that will generate results with “1”s as successes • We can store the results in lists and do statistics on the lists as usual
11. Example 3 Each entry in a table of random digits like Table B in our book has a probability of 0.1 of being a zero. • Find probability of finding exactly 4 zeros in a line 40 digits long. • What is the probability that a group of five digits from the table will contain at least 1 zero? P(X = 4) = binompdf(40, 0.1, 4) = 0.206 P(X ≥ 1) = 1 – P(X=0) = 1 ‒ binompdf (5, 0.1, 0) = .4095 = 1 – P(no zeros)5 = 0.4095 (before binomials) = 1 – (0.9)5 = 0.4095
12. Example 4 A university claims that 80% of its basketball players get their degree. An investigation examines the fates of a random sample of 24 players who entered the program over a period of several years. Of these players, 12 graduated and 12 are no longer in school. If the university's claim is true, • What is the probability that at most 12 out of 24 players graduate? b) What is the probability that all 24 graduate? P (x < 12) = binomcdf (24, .8, 12) = .000978 binompdf (24, .8, 24) = .0047
13. Summary and Homework • Summary • Binomial experiments have 4 specific criteria that must be met • Means and Variance for a Binomial • E(X) = np and V(X) = np(1-p) • Normal distribution (continuous) can approximate a Binomial (discrete) • Calculator has a random binomial generator | 1,350 | 4,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-51 | latest | en | 0.712887 |
https://www.omnicalculator.com/sports/tennis-balls | 1,695,872,817,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00152.warc.gz | 994,782,691 | 65,136 | Tennis Ball Usage Calculator
Created by Piotr Małek and Álvaro Díez
Reviewed by Steven Wooding
Last updated: Jun 05, 2023
Are you ready for some ballsy statistics? The Australian Open is heating up, and the players are complaining about the balls not being up to par, but replacing them is not an easy task. Don't believe us? At Omni Calculator, we've done the math for you.
Our tennis ball usage calculator takes into account the number of players, matches, games, and sets, as well as balls lost to the crowd, to give you an accurate estimate of the number of tennis balls used during a tournament. But wait, there's more! Not only is it a fun tool with insight into the lesser-known tennis stats, but it can also help tournament organizers plan for their next event. So grab your racket, and let's dive into the numbers!
When playing tennis, balls are disposable. Not only can they break or get "lost" in the crowd, but they also degrade over time. One would think that it's not a big problem, you can stock a few balls per match, and everything will be fine. But nothing could be further from the truth! In fact, for major tournaments with over a hundred players, balls represent a significant purchase, with tens of thousands of them used in the span of a week. How is that possible? Let's take a look at it.
According to ATP and WTA standards, a new set of 6 balls is put into the game after the first 7 games of any given match, and then every next 9 games, they're replaced with another 6 balls. For a simple 6-0, 6-0 sweep in the Women's match, only 16 balls will be used. On the other hand, if two Men's tournament participants go into a 5-set clash with 12 games on average in each set, 36 balls will be used. On top of those numbers, hundreds of balls "magically" disappear after landing in the crowds. Who, after all, doesn't want to have that kind of souvenir?
Keep in mind that a lot of events happen during each Grand Slam tournament which accounts for countless balls used. Those are the most popular singles and doubles for both Men and Women. Also, mixed couples play the tournament, along with juniors, disabled players as well as plenty of participants in invitation tournaments.
Are you interested in sports and calculations? We have a wide assortment of sports calculators for you, from marathon pace calculator and hiking calculator to cricket follow-on calculator!
How can I determine number of balls used in a tournament?
The formula to determine the number of balls in a tournament is:
Balls in a tournament = Balls per match × Number of matches
So all you have to do is:
1. Note the number of balls used per match;
2. Multiply them by the total number of matches; and
3. The result is the number of balls used in a tournament.
FAQ
How many times should I use a tennis ball?
If you are a casual tennis player, you may need 2-3 balls per match.
But a professional player may need a new ball for every set.
Is it OK to reuse a tennis ball?
You may reuse tennis balls as long as they do not affect the gameplay. One ball lasts around 3-4 regular game sessions.
Can I keep tennis balls for long?
Tennis balls will last you around 2-3 weeks with moderate game plays.
You can keep tennis balls for a long time provided that you keep them in their pressurized container and do not use them, but even if you do not use them, they expire in 2 years.
How many balls are needed in 20 games?
The total number of balls used in a tournament that lasts 20 games is 31.
The assumption for this result is six players in the tournament, and 1% of balls are lost in the crowd.
Piotr Małek and Álvaro Díez
Select a tournament
Australian Open
Who is playing?
All
Games per set
Sets per match
Balls lost to crowd
%
Results
The tournament will use
10,408
balls
Environmental impact
13,309
lb
CO₂-eq
Plastic waste
1,285
lb
🌳 ~3251 trees are needed to absorb the CO₂ emissions.
This amount of CO₂eq is also equivalent to:
• 📱 772690 smartphones charged
• 2571 usgal of gasoline consumed
• 🚗 24093.2 mi by an average passenger vehicle
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Ideal egg boiling
Quantum physicist's take on boiling the perfect egg. Includes times for quarter and half-boiled eggs. | 1,087 | 4,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.968796 |
https://www.scribd.com/document/204571044/Simple-Harmonic-Oscillator-Solution | 1,582,435,225,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145746.24/warc/CC-MAIN-20200223032129-20200223062129-00241.warc.gz | 911,163,113 | 64,220 | You are on page 1of 6
# Simple Harmonic Oscillator Solution!
## (for the curious)
Hi guys! Well I see some of you just cant accept a formula given without any proof, so I tried to come with one. Warning: you should know basics on complex numbers and exponential derivatives. So here it is: The differential equation (DE) we want to solve is:
x is the second derivative of x with respect to time, or acceleration. x is just the position. Since we have derivative of x with respect to time, then x must be a function of t (if not, the derivative with respect to time would be zero).
So lets continue! Lets assume our solution has the form (unknown, just for now ;) ). Plug it in our formula. Remember:
. ( )
## We see either one of both terms in the multiplication has to be 0: or ( Notice )
never equals 0, so you can take it out: you wont get a solution from it. Were left with: ( )
Again, notice no real value of r will satisfy the quadratic equation. We must introduce the concept of complex numbers. For those of you who are not familiar with complex numbers, just to make it simple, believe in me on this: We call the imaginary unit. So now, a number squared can give you a negative number! You dont see that in real numbers. Now you can express the quadratic as the multiplication of two terms: ( ) ( )( )
If you expand it and use the facts I mentioned above on , you can verify theyre equal.
## ? Now we have r. Actually, we have 2, so we have 2
What? But we need ONE solution! That cant be right. Which one do we choose? I want you to see this video first: https://www.khanacademy.org/math/differential-equations/second-orderdifferential-equations/linear-homogeneous-2nd-order/v/2nd-order-linear-homogeneousdifferential-equations-1
I assume youve already watched the video. So from the video you see two important facts: Our solutions on x are incomplete. We can actually multiply our solutions on x1 and x2 each by any constant, and each of them separately will also suffice the DE (Differential Equation). Each of our solutions is a partial solution. We obtain the general solution by summing all our partial solutions.
So, our more complete partial solutions are (with the constants):
And our general solution will be the sum of all our partial solutions, x1 and x2:
Now, again for those who dont know much on complex numbers, we have this beautiful equation called the Eulers Formula:
I wont show the proof here, but again, you have to believe me. Notice we have similar expressions on our solution for x. We can expand our terms using Eulers Formula (see, our x would be wt in our case):
## Now use some basic trigonometry. Remember: and Thus:
Collect terms:
See since a and b are constants, their sum or difference will be other constants as well. If we multiply them, we get constants too! Our new constants will be:
Replacing:
Both
and
## Well use Eulers Formula again. From Eulers Formula:
See is the real part of , and is the imaginary part of . (The real part is that part without the , and the imaginary part, the one that has the ). In more formal notation, we say:
Since
## (Since sinx doesnt have the i).
Lets now return to our original function, x. We can plug these relations up there. ( )
Now youll have to believe me again and treat the Real operator as if it were as other functions, so we can factor it out (thats why I used the second expression of sine, in terms of the Real operator, so we can factor them out!). Its kind of intuitive too, and you can even prove it ;) Factor the common terms: [ ]
For those who dont know complex numbers, a complex number has the form:
## And it can be expressed also as:
Which is called the polar form. Theres a relation between a, b, c and d, so complex numbers can be converted from polar to the common numerical form and backwards. However it wont be shown here. Just see if we have two constants (a and b) on the common form, we also have constants on the other form, which will be different. So we use the polar coordinates to convert that term: c-id
The rules from exponentials also follow with complex exponentials (sum of exponents) ( So we want the real part of that expression. ) | 963 | 4,188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2020-10 | latest | en | 0.91514 |
https://www.mbamission.com/blog/2008/10/02/the-quest-for-700-weekly-gmat-challenge-answer-11/ | 1,611,557,073,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00737.warc.gz | 862,461,579 | 10,966 | ## Blog
### The Quest for 700: Weekly GMAT Challenge (Answer)
Yesterday, Integrated Learning posted a 700 level GMAT question on our blog. Today, they have followed up with the answer:
If a number has remainder c when it is divided by 7 it will have either remainder c or remainder c+7 when divided by 14 (it is of the form 7k+c, and the remainder depends on whether k is even or odd). Thus (1st statement) – we could have a remainder that is less than 5 (eg. 15), and we could have a remainder greater than 5 (eg. 13). The statement is not sufficient.On the other hand (2nd statement) – the remainder when dividing by 14 will always be greater or equal to the remainder when dividing by 7. Therefore statement 2 is sufficient.
Integrated Learning provides professional, experienced GMAT tutors throughout the United States.
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http://www.cfd-online.com/Forums/fluent/31272-help-needed-gambit-problem-2-d-print.html | 1,474,847,940,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660436.26/warc/CC-MAIN-20160924173740-00111-ip-10-143-35-109.ec2.internal.warc.gz | 381,148,181 | 2,299 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
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- - Help Needed: Gambit Problem (2-D) (http://www.cfd-online.com/Forums/fluent/31272-help-needed-gambit-problem-2-d.html)
Huawei April 13, 2003 12:25
Help Needed: Gambit Problem (2-D)
Hi, My Friends:
I have a problem to plot my calculation domain(2-D). One part of its form is the function of y=tanh(x). Could I find this function in Gambit, and directly use it to plot the form of domain?
(By the way, at present, I used a not good method to plot that part domain(y=tanh(x)). I input many points of this function into Gambit, then connected these points and smooth the curve to build the domain).
If you could take the time and trouble to give me some advice, I would much appreciate it.
Alex Munoz April 13, 2003 15:06
Re: Help Needed: Gambit Problem (2-D)
Hi Collin
I guess what you should do is to generate your geometry on Autocad and then export it to gambit.
Warning: perhap you will find some edges are not connected therefore you must to fix your geometry before meshing.
Check on this web side how to import files from autocad to gambit
http://www.udel.edu/topics/software/...uide/ugtoc.htm
I don't think you will find a person in this discussion forum that will be able to address that problem. Therefore, I suggest that you contact Fluent user support, perhaps they can give you a better solution
Regards
Alex Munoz
Huawei April 13, 2003 18:16
Re: Help Needed: Gambit Problem (2-D)
Thank you very much, Alex!
Alex Munoz April 13, 2003 20:45
Re: Help Needed: Gambit Problem (2-D)
Hi
I forgot to tell you that you can write temperature profile in a surface
Best regards
Alex Munoz
All times are GMT -4. The time now is 19:59. | 477 | 1,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | longest | en | 0.884818 |
https://fr.maplesoft.com/support/help/addons/view.aspx?path=MultiSet&L=F | 1,701,367,710,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100229.44/warc/CC-MAIN-20231130161920-20231130191920-00095.warc.gz | 294,189,844 | 23,146 | MultiSet - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
MultiSet
overview of MultiSets
Calling Sequence MultiSet( element_spec, ... ) MultiSet['generalized']( element_spec, ... )
Parameters
element_spec - specifies the MultiSet elements and their multiplicities generalized - literal index which indicates that a generalized MultiSet should be constructed
Description
• A MultiSet is a data structure which stores and manipulates an unordered collection of elements which may be repeated. It is implemented as a Maple object. The procedure exports of a MultiSet are used to create, update, query and otherwise interact with one or more MultiSet objects.
• A MultiSet can be constructed from another MultiSet, from a set or list of elements, or from an expression sequence of elements with their multiplicities:
– If M is a MultiSet, then MultiSet(M) produces a new MultiSet with the same elements and multiplicities.
– If L is a Maple list or set of two-element lists, where the second element of each list is a non-negative integer (normal case) or real number (generalized case), then MultiSet(L) is a MultiSet whose elements are the first elements of each list with multiplicities given by the corresponding second elements of each list.
– If L is any other Maple list, then MultiSet(L) is a MultiSet of the same elements, with multiplicities the same as in L.
– If S is any other Maple set, then MultiSet(S) is a MultiSet of the same elements, each with multiplicity 1.
– Each of the expressions M( a, b=2, c=3 ) and M( a, [b, 2], [c, 3] ) constructs a MultiSet in which the element a has multiplicity 1, b has multiplicity 2 and c has multiplicity 3. Here a, b, and c can be any Maple expressions.
– Multiplicities must be non-negative integers, unless the generalized index is provided on the constructor, in which case arbitrary (real) numeric multiplicities are also permitted. An element with multiplicity 0 is removed from its MultiSet.
• MultiSets are displayed using a set-of-lists-of-pairs notation, but MultiSets are not sets in the usual Maple sense. The convert command can be used to realize a MultiSet in a variety of different alternate formats.
• To test whether an expression is a MultiSet, use type(..., MultiSet).
• To iterate over a MultiSet, see MultiSet Iteration.
• For commands which operate on more than one MultiSet, for example intersect, at least one operand must be a MultiSet. Other operands can be MultiSets, sets or lists; a non-MultiSet operand will be converted to a MultiSet before the operation is carried out.
• Generalized and standard (non-generalized) MultiSets cannot be combined in commands which operate on more than one MultiSet. Note that this means that if a generalized MultiSet appears in an operation, for example, union, with another argument which is not a MultiSet (for example, a list or set), then that other argument will be converted to a generalized MultiSet before proceeding with the operation.
• The command IsGeneralized(M) can be used to determine if a MultiSet is generalized or not.
List of MultiSet Object Commands
• The following is a list of the commands which work with MultiSet objects.
Examples
> $M≔\mathrm{MultiSet}\left(a=2,\left[b,4\right],c\right)$
${M}{≔}\left\{\left[{a}{,}{2}\right]{,}\left[{b}{,}{4}\right]{,}\left[{c}{,}{1}\right]\right\}$ (1)
> $N≔{\mathrm{MultiSet}}_{\mathrm{generalized}}\left(\left[\left[x,4\right],\left[y,\frac{3}{2}\right],\left[z,-2\right]\right]\right)$
${N}{≔}\left\{\left[{x}{,}{4}\right]{,}\left[{y}{,}\frac{{3}}{{2}}\right]{,}\left[{z}{,}{-2}\right]\right\}$ (2)
> $\mathrm{evalb}\left(M=\mathrm{MultiSet}\left(\mathrm{Entries}\left(M\right)\right)\right)$
${\mathrm{true}}$ (3)
> $\mathrm{Entries}\left(M\right)$
$\left\{\left[{a}{,}{2}\right]{,}\left[{b}{,}{4}\right]{,}\left[{c}{,}{1}\right]\right\}$ (4)
Compatibility
• The MultiSet object was introduced in Maple 2016.
• For more information on Maple 2016 changes, see Updates in Maple 2016. | 1,082 | 4,069 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.743197 |
https://www.intimatebritneyspears.us/lottery-prediction-software-can-it-work/ | 1,624,055,694,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487641593.43/warc/CC-MAIN-20210618200114-20210618230114-00227.warc.gz | 744,757,692 | 6,067 | # Intimate
Blog
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In every area of your life, everything is more organised if you find a plan upon how to do points. The same thing is applicable to enjoying the lotto. In order to possess better chances of successful the lotto, you will need to study its prior results, determine the probable results, and come up with a achievable way with how to make these kinds of results work for you. With endurance, you may possibly find yourself to be a winner sooner than you imagine. | 692 | 3,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.970401 |
https://da.wikipedia.org/wiki/Skabelon:Intorient | 1,606,772,005,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00677.warc.gz | 247,443,303 | 14,018 | # Skabelon:Intorient
Spring til navigation Spring til søgning
Skabelondokumentation[vis] [redigér] [historik] [opfrisk]
This template is used to include the oriented integrals around closed surfaces (or hypersurfaces in higher dimensions), usually in a mathematical formula. They are additional symbols to \oiint and \oiiint which are not yet rendered on wikipedia.
## Arguments
• preintegral the text or formula immediately before the integral
• symbol the integral symbol,
Select one of... Arrow up, integrals over a closed Arrow down, integrals over a closed
1-surface 2-surface 3-surface 1-surface 2-surface 3-surface
Clockwise
orientation
oint= oiint= oiiint= varoint= varoiint= varoiiint=
Counterclockwise
orientation
ointctr= oiintctr= oiiintctr= varointctr= varoiintctr= varoiiintctr=
The default is
• intsubscpt the subscript below the integral
• integrand the text or formula immediately after the formula
All parameters are optional.
## Examples
• The work done in a thermodynamic cycle on an indicator diagram: ${\displaystyle W=}$${\displaystyle {\scriptstyle \Gamma }}$${\displaystyle p{\rm {d}}V}$
{{intorient
| preintegral=$W=$
| symbol = varoint
| intsubscpt = ${\scriptstyle \Gamma}$
| integrand = $p{\rm d}V$
}}
• In complex analysis for contour integrals: ${\displaystyle {\scriptstyle \Gamma }}$${\displaystyle {\frac {{\rm {d}}z}{(z+a)^{3}z^{1/2}}}}$
{{intorient|
| preintegral =
|symbol=varoint
| intsubscpt = ${\scriptstyle \Gamma}$
| integrand = $\frac{{\rm d}z}{(z+a)^3z^{1/2}}$
}}
• Line integrals of vector fields: ${\displaystyle {\scriptstyle \partial S}}$${\displaystyle \mathbf {F} \cdot {\rm {d}}\mathbf {r} =-}$${\displaystyle {\scriptstyle \partial S}}$${\displaystyle \mathbf {F} \cdot {\rm {d}}\mathbf {r} }$
{{intorient|
| preintegral = {{intorient|
| preintegral =
|symbol=oint
| intsubscpt = ${\scriptstyle \partial S}$
| integrand = $\mathbf{F}\cdot{\rm d}\mathbf{r}=-$
}}
|symbol=ointctr
| intsubscpt = ${\scriptstyle \partial S}$
| integrand = $\mathbf{F}\cdot{\rm d}\mathbf{r}$
}}
• Other examples: ${\displaystyle {\scriptstyle \Sigma }}$${\displaystyle (E+H\wedge T){\rm {d}}^{2}\Sigma }$
{{Intorient|
| preintegral =
|symbol=oiiintctr
| intsubscpt = ${\scriptstyle \Sigma}$
| integrand = $(E+H\wedge T) {\rm d}^2 \Sigma$
}}
${\displaystyle {\scriptstyle \Omega }}$${\displaystyle (E+H\wedge T){\rm {d}}^{4}\Omega }$
{{Intorient|
| preintegral =
|symbol=varoiiintctr
| intsubscpt = ${\scriptstyle \Omega}$
| integrand = $(E+H\wedge T) {\rm d}^4 \Omega$
}} | 807 | 2,514 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-50 | latest | en | 0.418771 |
https://immersivelabz.com/plant-population-frequency-by-quadrat-method/ | 1,726,321,378,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.38/warc/CC-MAIN-20240914125424-20240914155424-00804.warc.gz | 281,647,000 | 42,312 | Biology Practicals Class 12
# Plant Population Frequency by Quadrat Method
• Perform science experiments at your fingertips
• Learn anytime and from anywhere
• 3D gamified way of advanced learning
• Accessible through Mobile, Laptop, Desktop, and Tablet
• Gain a competitive edge in IIT and NEET exams
• This biology class 12 practical helps learners to understand plant population frequency by quadrat method.
• This biology experiment explains the difference between often confused terms such as population, species, and individuals in a simple way.
• Learners will understand how to choose the investigation area for calculating the plant population frequency.
• This biology experiment helps learners understand how to plot the quadrat through simulation.
• The formula to calculate plant population frequency is explained through this biology class 12 practical in a lucid manner.
• This biology practical also helps learners to prepare the complex observation table easily.
### Simulation Details
Duration – 30 Minutes
Easily Accessible
Language – English
Platforms – Android & Windows
Description
Population is a term that refers to a group of individuals from the same species that live in the same place at the same time. Plant population is the number of plants per unit area.
Species is a collection of closely related organisms that are genetically similar. The individuals of species can generally interbreed and produce fertile offspring.
The frequency of the population is the uniform presence of individuals of a species within a community. To calculate the population frequency, we consider the number of species occurrences in the selected area and divide the number by the number of samples we’ve studied.
Quadrat method is used to calculate the plant population density. Quadrat is a square of a specific size, usually of 1-meter square. These squares are chosen randomly and generally marked with wire outlines.
Typically, a succession of quadrats is sampled to allow comparisons between phenomena, such as the distribution of a specific plant in two distinct places.
The plant population frequency by quadrat method is calculated as:
Frequency = Number of sampling units (quadrats) in which the species occurs / Total number of sampling units (quadrats) employed for the study.
Check all science experiments. | 444 | 2,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.910908 |
http://www.jiskha.com/display.cgi?id=1229404233 | 1,498,202,992,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320023.23/warc/CC-MAIN-20170623063716-20170623083716-00586.warc.gz | 559,565,089 | 4,329 | # chemistry desperately
posted by on .
i am completely lost, i confused myself.
i need to find the concentration of HC2H3O2
HC2H3O2 + NaOH ==> NaC2H3O2 + HOH
NaOH
C= 0.3 M
v= 0.25 L
n= 0.075 moles
M= 40.00g/mol
m= 3.00 g
NaC2H3O2
n= 0.075 moles
m= 0.4973 g(ammount of acetic acid in
solution)
HC2H302
v= 0.020 L
M= 60.06g/mol
thats all the information i have. i don't know how to calculate it. my friend said c1v1=c2v2 but i don't know what numbers id use
• chemistry desperately - ,
You are confused because you have too much information piled in together. You need to think this through. And you don't say what unit the concentration should be in.
M(NaOH) x L(NaOH) = mols NaOH.
mols NaOH = mols acetic acid since the equation is 1 mol NaOH to 1 mol acetic acid.
Now M(acid) x L(acid) = mols. You know mols, you know L acid (is that the 0.020 L), calculate M acid. That will be the molarity of the acetic acid in that 20 mL sample. By the way, I can't believe you used 0.25 L of the NaOH. That's 250 mL which is a huge volume to be titrated with. That means you would have had to refill a 50 mL buret five times. (Of course this could be a problem.)
I'm getting ready to leave for the night so if there is a quick question, please type it in before I leave.
• chemistry desperately - ,
alright thank you i think i should be able to figure it out. and i know we only did 3 trials though | 447 | 1,400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.937024 |
https://leanprover-community.github.io/mathlib_docs/ring_theory/ideal/basic.html | 1,702,194,259,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00238.warc.gz | 402,804,876 | 53,436 | # mathlib3documentation
ring_theory.ideal.basic
# Ideals over a ring #
THIS FILE IS SYNCHRONIZED WITH MATHLIB4. Any changes to this file require a corresponding PR to mathlib4.
This file defines ideal R, the type of (left) ideals over a ring R. Note that over commutative rings, left ideals and two-sided ideals are equivalent.
## Implementation notes #
ideal R is implemented using submodule R R, where • is interpreted as *.
## TODO #
Support right ideals, and two-sided ideals over non-commutative rings.
@[reducible]
def ideal (R : Type u) [semiring R] :
A (left) ideal in a semiring R is an additive submonoid s such that a * b ∈ s whenever b ∈ s. If R is a ring, then s is an additive subgroup.
Equations
• = R
@[protected]
theorem ideal.zero_mem {α : Type u} [semiring α] (I : ideal α) :
0 I
@[protected]
theorem ideal.add_mem {α : Type u} [semiring α] (I : ideal α) {a b : α} :
a I b I a + b I
theorem ideal.mul_mem_left {α : Type u} [semiring α] (I : ideal α) (a : α) {b : α} :
b I a * b I
@[ext]
theorem ideal.ext {α : Type u} [semiring α] {I J : ideal α} (h : (x : α), x I x J) :
I = J
theorem ideal.sum_mem {α : Type u} [semiring α] (I : ideal α) {ι : Type u_1} {t : finset ι} {f : ι α} :
( (c : ι), c t f c I) t.sum (λ (i : ι), f i) I
theorem ideal.eq_top_of_unit_mem {α : Type u} [semiring α] (I : ideal α) (x y : α) (hx : x I) (h : y * x = 1) :
I =
theorem ideal.eq_top_of_is_unit_mem {α : Type u} [semiring α] (I : ideal α) {x : α} (hx : x I) (h : is_unit x) :
I =
theorem ideal.eq_top_iff_one {α : Type u} [semiring α] (I : ideal α) :
I = 1 I
theorem ideal.ne_top_iff_one {α : Type u} [semiring α] (I : ideal α) :
I 1 I
@[simp]
theorem ideal.unit_mul_mem_iff_mem {α : Type u} [semiring α] (I : ideal α) {x y : α} (hy : is_unit y) :
y * x I x I
def ideal.span {α : Type u} [semiring α] (s : set α) :
The ideal generated by a subset of a ring
Equations
Instances for ideal.span
@[simp]
theorem ideal.submodule_span_eq {α : Type u} [semiring α] {s : set α} :
@[simp]
theorem ideal.span_empty {α : Type u} [semiring α] :
@[simp]
theorem ideal.span_univ {α : Type u} [semiring α] :
theorem ideal.span_union {α : Type u} [semiring α] (s t : set α) :
theorem ideal.span_Union {α : Type u} [semiring α] {ι : Sort u_1} (s : ι set α) :
ideal.span ( (i : ι), s i) = (i : ι), ideal.span (s i)
theorem ideal.mem_span {α : Type u} [semiring α] {s : set α} (x : α) :
x (p : ideal α), s p x p
theorem ideal.subset_span {α : Type u} [semiring α] {s : set α} :
theorem ideal.span_le {α : Type u} [semiring α] {s : set α} {I : ideal α} :
I s I
theorem ideal.span_mono {α : Type u} [semiring α] {s t : set α} :
s t
@[simp]
theorem ideal.span_eq {α : Type u} [semiring α] (I : ideal α) :
= I
@[simp]
theorem ideal.span_singleton_one {α : Type u} [semiring α] :
theorem ideal.mem_span_insert {α : Type u} [semiring α] {s : set α} {x y : α} :
x (a z : α) (H : z , x = a * y + z
theorem ideal.mem_span_singleton' {α : Type u} [semiring α] {x y : α} :
x ideal.span {y} (a : α), a * y = x
theorem ideal.span_singleton_le_iff_mem {α : Type u} [semiring α] (I : ideal α) {x : α} :
ideal.span {x} I x I
theorem ideal.span_singleton_mul_left_unit {α : Type u} [semiring α] {a : α} (h2 : is_unit a) (x : α) :
theorem ideal.span_insert {α : Type u} [semiring α] (x : α) (s : set α) :
theorem ideal.span_eq_bot {α : Type u} [semiring α] {s : set α} :
(x : α), x s x = 0
@[simp]
theorem ideal.span_singleton_eq_bot {α : Type u} [semiring α] {x : α} :
ideal.span {x} = x = 0
theorem ideal.span_singleton_ne_top {α : Type u_1} {x : α} (hx : ¬) :
@[simp]
theorem ideal.span_zero {α : Type u} [semiring α] :
@[simp]
theorem ideal.span_one {α : Type u} [semiring α] :
theorem ideal.span_eq_top_iff_finite {α : Type u} [semiring α] (s : set α) :
(s' : finset α), s' s
theorem ideal.mem_span_singleton_sup {S : Type u_1} {x y : S} {I : ideal S} :
x ideal.span {y} I (a b : S) (H : b I), a * y + b = x
def ideal.of_rel {α : Type u} [semiring α] (r : α α Prop) :
The ideal generated by an arbitrary binary relation.
Equations
• = {x : α | (a b : α) (h : r a b), x + b = a}
@[class]
structure ideal.is_prime {α : Type u} [semiring α] (I : ideal α) :
Prop
An ideal P of a ring R is prime if P ≠ R and xy ∈ P → x ∈ P ∨ y ∈ P
Instances of this typeclass
theorem ideal.is_prime_iff {α : Type u} [semiring α] {I : ideal α} :
I.is_prime I {x y : α}, x * y I x I y I
theorem ideal.is_prime.ne_top {α : Type u} [semiring α] {I : ideal α} (hI : I.is_prime) :
theorem ideal.is_prime.mem_or_mem {α : Type u} [semiring α] {I : ideal α} (hI : I.is_prime) {x y : α} :
x * y I x I y I
theorem ideal.is_prime.mem_or_mem_of_mul_eq_zero {α : Type u} [semiring α] {I : ideal α} (hI : I.is_prime) {x y : α} (h : x * y = 0) :
x I y I
theorem ideal.is_prime.mem_of_pow_mem {α : Type u} [semiring α] {I : ideal α} (hI : I.is_prime) {r : α} (n : ) (H : r ^ n I) :
r I
theorem ideal.not_is_prime_iff {α : Type u} [semiring α] {I : ideal α} :
¬I.is_prime I = (x : α) (H : x I) (y : α) (H : y I), x * y I
theorem ideal.zero_ne_one_of_proper {α : Type u} [semiring α] {I : ideal α} (h : I ) :
0 1
theorem ideal.bot_prime {R : Type u_1} [ring R] [is_domain R] :
@[class]
structure ideal.is_maximal {α : Type u} [semiring α] (I : ideal α) :
Prop
• out :
An ideal is maximal if it is maximal in the collection of proper ideals.
Instances of this typeclass
theorem ideal.is_maximal_def {α : Type u} [semiring α] {I : ideal α} :
theorem ideal.is_maximal.ne_top {α : Type u} [semiring α] {I : ideal α} (h : I.is_maximal) :
theorem ideal.is_maximal_iff {α : Type u} [semiring α] {I : ideal α} :
I.is_maximal 1 I (J : ideal α) (x : α), I J x I x J 1 J
theorem ideal.is_maximal.eq_of_le {α : Type u} [semiring α] {I J : ideal α} (hI : I.is_maximal) (hJ : J ) (IJ : I J) :
I = J
@[protected, instance]
def ideal.is_coatomic {α : Type u} [semiring α] :
theorem ideal.is_maximal.coprime_of_ne {α : Type u} [semiring α] {M M' : ideal α} (hM : M.is_maximal) (hM' : M'.is_maximal) (hne : M M') :
M M' =
theorem ideal.exists_le_maximal {α : Type u} [semiring α] (I : ideal α) (hI : I ) :
(M : ideal α), M.is_maximal I M
Krull's theorem: if I is an ideal that is not the whole ring, then it is included in some maximal ideal.
theorem ideal.exists_maximal (α : Type u) [semiring α] [nontrivial α] :
(M : ideal α), M.is_maximal
Krull's theorem: a nontrivial ring has a maximal ideal.
@[protected, instance]
def ideal.nontrivial {α : Type u} [semiring α] [nontrivial α] :
theorem ideal.maximal_of_no_maximal {R : Type u} [semiring R] {P : ideal R} (hmax : (m : ideal R), P < m ¬m.is_maximal) (J : ideal R) (hPJ : P < J) :
J =
If P is not properly contained in any maximal ideal then it is not properly contained in any proper ideal
theorem ideal.span_pair_comm {α : Type u} [semiring α] {x y : α} :
ideal.span {x, y} = ideal.span {y, x}
theorem ideal.mem_span_pair {α : Type u} [semiring α] {x y z : α} :
z ideal.span {x, y} (a b : α), a * x + b * y = z
@[simp]
theorem ideal.span_pair_add_mul_left {R : Type u} [comm_ring R] {x y : R} (z : R) :
ideal.span {x + y * z, y} = ideal.span {x, y}
@[simp]
theorem ideal.span_pair_add_mul_right {R : Type u} [comm_ring R] {x y : R} (z : R) :
ideal.span {x, y + x * z} = ideal.span {x, y}
theorem ideal.is_maximal.exists_inv {α : Type u} [semiring α] {I : ideal α} (hI : I.is_maximal) {x : α} (hx : x I) :
(y i : α) (H : i I), y * x + i = 1
theorem ideal.mem_sup_left {R : Type u} [semiring R] {S T : ideal R} {x : R} :
x S x S T
theorem ideal.mem_sup_right {R : Type u} [semiring R] {S T : ideal R} {x : R} :
x T x S T
theorem ideal.mem_supr_of_mem {R : Type u} [semiring R] {ι : Sort u_1} {S : ι } (i : ι) {x : R} :
x S i x supr S
theorem ideal.mem_Sup_of_mem {R : Type u} [semiring R] {S : set (ideal R)} {s : ideal R} (hs : s S) {x : R} :
x s x
theorem ideal.mem_Inf {R : Type u} [semiring R] {s : set (ideal R)} {x : R} :
x ⦃I : ⦄, I s x I
@[simp]
theorem ideal.mem_inf {R : Type u} [semiring R] {I J : ideal R} {x : R} :
x I J x I x J
@[simp]
theorem ideal.mem_infi {R : Type u} [semiring R] {ι : Sort u_1} {I : ι } {x : R} :
x infi I (i : ι), x I i
@[simp]
theorem ideal.mem_bot {R : Type u} [semiring R] {x : R} :
x x = 0
def ideal.pi {α : Type u} [semiring α] (I : ideal α) (ι : Type v) :
ideal α)
I^n as an ideal of R^n.
Equations
theorem ideal.mem_pi {α : Type u} [semiring α] (I : ideal α) (ι : Type v) (x : ι α) :
x I.pi ι (i : ι), x i I
theorem ideal.Inf_is_prime_of_is_chain {α : Type u} [semiring α] {s : set (ideal α)} (hs : s.nonempty) (hs' : s) (H : (p : ideal α), p s p.is_prime) :
@[simp]
theorem ideal.mul_unit_mem_iff_mem {α : Type u} (I : ideal α) {x y : α} (hy : is_unit y) :
x * y I x I
theorem ideal.mem_span_singleton {α : Type u} {x y : α} :
x ideal.span {y} y x
theorem ideal.mem_span_singleton_self {α : Type u} (x : α) :
theorem ideal.span_singleton_le_span_singleton {α : Type u} {x y : α} :
theorem ideal.span_singleton_eq_span_singleton {α : Type u} [comm_ring α] [is_domain α] {x y : α} :
theorem ideal.span_singleton_mul_right_unit {α : Type u} {a : α} (h2 : is_unit a) (x : α) :
theorem ideal.span_singleton_eq_top {α : Type u} {x : α} :
theorem ideal.span_singleton_prime {α : Type u} {p : α} (hp : p 0) :
theorem ideal.is_maximal.is_prime {α : Type u} {I : ideal α} (H : I.is_maximal) :
@[protected, instance]
def ideal.is_maximal.is_prime' {α : Type u} (I : ideal α) [H : I.is_maximal] :
theorem ideal.span_singleton_lt_span_singleton {β : Type v} [comm_ring β] [is_domain β] {x y : β} :
theorem ideal.factors_decreasing {β : Type v} [comm_ring β] [is_domain β] (b₁ b₂ : β) (h₁ : b₁ 0) (h₂ : ¬is_unit b₂) :
ideal.span {b₁ * b₂} < ideal.span {b₁}
theorem ideal.mul_mem_right {α : Type u} {a : α} (b : α) (I : ideal α) (h : a I) :
a * b I
theorem ideal.pow_mem_of_mem {α : Type u} {a : α} (I : ideal α) (ha : a I) (n : ) (hn : 0 < n) :
a ^ n I
theorem ideal.is_prime.mul_mem_iff_mem_or_mem {α : Type u} {I : ideal α} (hI : I.is_prime) {x y : α} :
x * y I x I y I
theorem ideal.is_prime.pow_mem_iff_mem {α : Type u} {I : ideal α} (hI : I.is_prime) {r : α} (n : ) (hn : 0 < n) :
r ^ n I r I
theorem ideal.pow_multiset_sum_mem_span_pow {α : Type u} (s : multiset α) (n : ) :
s.sum ^ * n + 1) ideal.span ((multiset.map (λ (x : α), x ^ (n + 1)) s).to_finset)
theorem ideal.sum_pow_mem_span_pow {α : Type u} {ι : Type u_1} (s : finset ι) (f : ι α) (n : ) :
s.sum (λ (i : ι), f i) ^ (s.card * n + 1) ideal.span ((λ (i : ι), f i ^ (n + 1)) '' s)
theorem ideal.span_pow_eq_top {α : Type u} (s : set α) (hs : = ) (n : ) :
ideal.span ((λ (x : α), x ^ n) '' s) =
@[protected]
theorem ideal.neg_mem_iff {α : Type u} [ring α] (I : ideal α) {a : α} :
-a I a I
@[protected]
theorem ideal.add_mem_iff_left {α : Type u} [ring α] (I : ideal α) {a b : α} :
b I (a + b I a I)
@[protected]
theorem ideal.add_mem_iff_right {α : Type u} [ring α] (I : ideal α) {a b : α} :
a I (a + b I b I)
@[protected]
theorem ideal.sub_mem {α : Type u} [ring α] (I : ideal α) {a b : α} :
a I b I a - b I
theorem ideal.mem_span_insert' {α : Type u} [ring α] {s : set α} {x y : α} :
x (a : α), x + a * y
@[simp]
theorem ideal.span_singleton_neg {α : Type u} [ring α] (x : α) :
theorem ideal.eq_bot_or_top {K : Type u} (I : ideal K) :
I = I =
All ideals in a division (semi)ring are trivial.
@[protected, instance]
def ideal.is_simple_order {K : Type u} :
Ideals of a division_semiring are a simple order. Thanks to the way abbreviations work, this automatically gives a is_simple_module K instance.
theorem ideal.eq_bot_of_prime {K : Type u} (I : ideal K) [h : I.is_prime] :
I =
theorem ideal.bot_is_maximal {K : Type u} :
theorem ideal.mul_sub_mul_mem {R : Type u_1} [comm_ring R] (I : ideal R) {a b c d : R} (h1 : a - b I) (h2 : c - d I) :
a * c - b * d I
theorem ring.exists_not_is_unit_of_not_is_field {R : Type u_1} [nontrivial R] (hf : ¬) :
(x : R) (H : x 0), ¬
Also see ideal.is_simple_order for the forward direction as an instance when R is a division (semi)ring.
This result actually holds for all division semirings, but we lack the predicate to state it.
theorem ring.ne_bot_of_is_maximal_of_not_is_field {R : Type u_1} [nontrivial R] {M : ideal R} (max : M.is_maximal) (not_field : ¬) :
When a ring is not a field, the maximal ideals are nontrivial.
theorem ideal.bot_lt_of_maximal {R : Type u} [comm_ring R] [nontrivial R] (M : ideal R) [hm : M.is_maximal] (non_field : ¬) :
< M
def nonunits (α : Type u) [monoid α] :
set α
The set of non-invertible elements of a monoid.
Equations
@[simp]
theorem mem_nonunits_iff {α : Type u} {a : α} [monoid α] :
theorem mul_mem_nonunits_right {α : Type u} {a b : α} [comm_monoid α] :
b a * b
theorem mul_mem_nonunits_left {α : Type u} {a b : α} [comm_monoid α] :
a a * b
theorem zero_mem_nonunits {α : Type u} [semiring α] :
0 0 1
@[simp]
theorem one_not_mem_nonunits {α : Type u} [monoid α] :
1
theorem coe_subset_nonunits {α : Type u} [semiring α] {I : ideal α} (h : I ) :
theorem exists_max_ideal_of_mem_nonunits {α : Type u} {a : α} (h : a ) :
(I : ideal α), I.is_maximal a I | 4,796 | 12,935 | {"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} | 2.890625 | 3 | CC-MAIN-2023-50 | latest | en | 0.623474 |
http://www.geeksforgeeks.org/regular-languages-and-finite-automata-gq/ | 1,506,099,624,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689028.2/warc/CC-MAIN-20170922164513-20170922184513-00689.warc.gz | 458,146,302 | 23,146 | ## Regular languages and finite automata
Question 1
Consider the languages L1 = and L2 = {a}. Which one of the following represents L1 L2* U L1*
A A B B C C D D
GATE CS 2013 Regular languages and finite automata
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Question 1 Explanation:
L1 L2* U L1* Result of L1 L2* is [Tex]\phi[/Tex]. {[Tex]\phi[/Tex]} indicates an empty language. Concatenation of [Tex]\phi[/Tex] with any other language is [Tex]\phi[/Tex]. It works as 0 in multiplication. L1* = [Tex]\phi[/Tex]* which is {[Tex]\epsilon[/Tex]}. Union of [Tex]\phi[/Tex] and {[Tex]\epsilon[/Tex]} is {[Tex]\epsilon[/Tex]}
Question 2
Consider the DFA given. Which of the following are FALSE?
1. Complement of L(A) is context-free.
2. L(A) = L((11*0+0)(0 + 1)*0*1*)
3. For the language accepted by A, A is the minimal DFA.
4. A accepts all strings over {0, 1} of length at least 2.
A 1 and 3 only B 2 and 4 only C 2 and 3 only D 3 and 4 only
GATE CS 2013 Regular languages and finite automata
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Question 2 Explanation:
1 is true. L(A) is regular, its complement would also be regular. A regular language is also context free. 2 is true. 3 is false, the DFA can be minimized to two states. Where the second state is final state and we reach second state after a 0. 4 is clearly false as the DFA accepts a single 0.
Question 3
W hat is the complement of the language accepted by the NFA shown below?
A A B B C C D D
GATE CS 2012 Regular languages and finite automata
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Question 3 Explanation:
The given alphabet contains only one symbol {a} and the given NFA accepts all strings with any number of occurrences of ‘a’. In other words, the NFA accepts a+. Therefore complement of the language accepted by automata is empty string.
Question 4
Given the language L = {ab, aa, baa}, which of the following strings are in L*?
1) abaabaaabaa
2) aaaabaaaa
3) baaaaabaaaab
4) baaaaabaa
A 1, 2 and 3 B 2, 3 and 4 C 1, 2 and 4 D 1, 3 and 4
GATE CS 2012 Regular languages and finite automata
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Question 4 Explanation:
Question 5
Consider the set of strings on {0,1} in which, every substring of 3 symbols has at most two zeros. For example, 001110 and 011001 are in the language, but 100010 is not. All strings of length less than 3 are also in the language. A partially completed DFA that accepts this language is shown below. The missing arcs in the DFA are
A A B B C C D D
GATE CS 2012 Regular languages and finite automata
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Question 5 Explanation:
Question 6
Definition of a language L with alphabet {a} is given as following.
L={| k>0, and n is a positive integer constant}
What is the minimum number of states needed in DFA to recognize L?
A k+1 B n+1 C 2^(n+1) D 2^(k+1)
GATE CS 2011 Regular languages and finite automata
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Question 6 Explanation:
Question 7
A deterministic finite automation (DFA)D with alphabet {a,b} is given below Which of the following finite state machines is a valid minimal DFA which accepts the same language as D?
A A B B C C D D
GATE CS 2011 Regular languages and finite automata
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Question 7 Explanation:
Options (B) and (C) are invalid because they both accept ‘b’ as a string which is not accepted by give DFA. (D) is invalid because it accepts "bba" which are not accepted by given DFA.
Question 8
Let w be any string of length n is {0,1}*. Let L be the set of all substrings of w. What is the minimum number of states in a non-deterministic finite automaton that accepts L?
A n-1 B n C n+1 D 2n-1
GATE CS 2010 Regular languages and finite automata
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Question 8 Explanation:
We need minimum n+1 states to build NFA that accepts all substrings of a binary string. For example, following NFA accepts all substrings of “010″ and it has 4 states.
Question 9
Which one of the following languages over the alphabet {0,1} is described by the regular expression: (0+1)*0(0+1)*0(0+1)*?
A The set of all strings containing the substring 00. B The set of all strings containing at most two 0’s. C The set of all strings containing at least two 0’s. D The set of all strings that begin and end with either 0 or 1.
GATE-CS-2009 Regular languages and finite automata
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Question 9 Explanation:
The regular expression has two 0′s surrounded by (0+1)* which means accepted strings must have at least 2 0′s.
Question 10
Which one of the following is FALSE?
A There is unique minimal DFA for every regular language B Every NFA can be converted to an equivalent PDA. C Complement of every context-free language is recursive. D Every nondeterministic PDA can be converted to an equivalent deterministic PDA.
GATE-CS-2009 Regular languages and finite automata
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Question 10 Explanation:
Deterministic PDA cannot handle languages or grammars with ambiguity, but NDPDA can handle languages with ambiguity and any context-free grammar. So every nondeterministic PDA can not be converted to an equivalent deterministic PDA.
There are 80 questions to complete. | 1,382 | 4,977 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-39 | latest | en | 0.854229 |
https://www.cfd-online.com/Forums/main/73579-upwind-solvers-1d-compressible-euler-eqns.html | 1,503,344,797,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109525.95/warc/CC-MAIN-20170821191703-20170821211703-00671.warc.gz | 899,691,186 | 16,023 | # Upwind solvers for 1D compressible Euler eqns
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March 11, 2010, 14:39 Upwind solvers for 1D compressible Euler eqns #1 New Member Join Date: Mar 2010 Posts: 2 Rep Power: 0 Sponsored Links I am an introductory student in CFD. I ma solving 1D Euler equations in non-conservative form. I am able to calculate the eigenvalues and eigenvector matrix. I get eigenvalues of u, u+c, and u-c. I need to find F+ and F-. I am trying to understand the Steger-Warming and Van Leer flux splitting methods. I am having trouble figuring out how to do the Steger-Warming and Van Leer. There is no info on these schemes in Wiki. Can someone explain these simply or provide an example of them in use?
March 11, 2010, 18:31 Something like this? #2 Member Join Date: Mar 2009 Posts: 33 Rep Power: 10 Here describes Van Leer's splitting. http://chimeracfd.com/programming/gr...uxvanleer.html The Steger-Warming splitting is simple. For the Euler equations, we have F = AU where F is the flux, U is the conservative state vector, and A is the Jacobian (dF/dU)! Yes, that's true for the Euler. So, if we decompose A into a positive part and negative part: A = RDL = R(Dp + Dm)L = RDpL + RDmL = Ap + Am, where R is the right-eigenvector matrix, D is the diagonal matrix with eigenvalues in the diagonal, L is the left-eigenvector (inverse of R), and Dp is D with positive eigenvalues only, Dm is D with only negative eigenvalues, THEN we can write the flux vector as F = Fp + Fm = ApU + AmU So, the positive flux is ApU and the negative flux is AmU! There is a source code for Van Leer's flux and the Steger-Warming flux at http://www.ossanworld.com/cfdbooks/cfdcodes.html Good luck. gory
May 1, 2010, 04:19 #3 New Member erwin Join Date: Apr 2010 Posts: 2 Rep Power: 0 hi all.i need computer programme for solving 1D euler EQUATİONS van leer and roe scheme.plz help me
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https://testbook.com/question-answer/for-what-values-of-k-is-the-system-of-equations-2k--619737db65b907cc997ce230 | 1,642,580,468,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301264.36/warc/CC-MAIN-20220119064554-20220119094554-00573.warc.gz | 585,103,912 | 30,795 | # For what values of k is the system of equations 2k2x + 3y - 1 = 0, 7x - 2y + 3 = 0, 6kx + y + 1 = 0 consistent?
1. $$\rm \frac{3±\sqrt{11}}{10}$$
2. $$\rm \frac{21±\sqrt{161}}{10}$$
3. $$\rm \frac{3±\sqrt{7}}{10}$$
4. $$\rm \frac{4±\sqrt{11}}{10}$$
Option 2 : $$\rm \frac{21±\sqrt{161}}{10}$$
## Detailed Solution
Concept:
Consider three linear eqaution in two variable:
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
a3x + b3y + c3 = 0
Condition for the consistency of three simultaneous linear equations in 2 variables:
$$\left| {\begin{array}{*{20}{c}} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3 \end{array}} \right|=0$$
ax2 + bx + c = 0
x = $$\rm \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$
Calculation:
2k2x + 3y - 1 = 0 ....(1)
7x - 2y + 3 = 0 ....(2)
6kx + y + 1 = 0 ....(3)
For consistency of given simultaneous equation,
$$\left| {\begin{array}{*{20}{c}} 2k^2&3&-1\\ 7&-2&3\\ 6k&1&1 \end{array}} \right|=0$$
⇒ 2k2(-2- 3) - 3(7 - 18k) - 1(7 + 12k) = 0
⇒ -10k2 - 21 + 54k - 7 - 12k = 0
⇒ -10k2 + 42k - 28 = 0
5k2 - 21k + 14 = 0
By using the formula,
$$x=\rm \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$
$$\Rightarrow k=\rm \frac{-(-21) \pm \sqrt{(-21)^{2} - 4(5)(14)}}{2\times 5}$$
$$\therefore k=\rm \frac{21 \pm \sqrt{161}}{10}$$ | 642 | 1,264 | {"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} | 4.4375 | 4 | CC-MAIN-2022-05 | latest | en | 0.2999 |
https://jp.mathworks.com/matlabcentral/cody/problems/322-number-of-circles-in-a-number/solutions/1216109 | 1,590,798,929,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347406785.66/warc/CC-MAIN-20200529214634-20200530004634-00097.warc.gz | 403,116,330 | 16,767 | Cody
# Problem 322. Number of Circles in a Number
Solution 1216109
Submitted on 19 Jun 2017 by Gregory
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
assert(isequal(ncircles(50237),1))
y = 1
2 Pass
assert(isequal(ncircles(56774),1))
y = 1
3 Pass
assert(isequal(ncircles(18828),6))
y = 6
4 Pass
assert(isequal(ncircles(32420),1))
y = 1
5 Pass
assert(isequal(ncircles(71604),2))
y = 2
6 Pass
assert(isequal(ncircles(55293),1))
y = 1
7 Pass
assert(isequal(ncircles(14229),1))
y = 1
8 Pass
assert(isequal(ncircles(38037),3))
y = 3
9 Pass
assert(isequal(ncircles(39657),2))
y = 2
10 Pass
assert(isequal(ncircles(57675),1))
y = 1
11 Pass
assert(isequal(ncircles(1941),1))
y = 1
12 Pass
assert(isequal(ncircles(57759),1))
y = 1
13 Pass
assert(isequal(ncircles(93220),2))
y = 2
14 Pass
assert(isequal(ncircles(10688),6))
y = 6
15 Pass
assert(isequal(ncircles(73215),0))
y = 0
16 Pass
assert(isequal(ncircles(97053),2))
y = 2
17 Pass
assert(isequal(ncircles(60889),7))
y = 7
18 Pass
assert(isequal(ncircles(71967),2))
y = 2
19 Pass
assert(isequal(ncircles(30276),2))
y = 2
20 Pass
assert(isequal(ncircles(45903),2))
y = 2
21 Pass
assert(isequal(ncircles(4803),3))
y = 3
22 Pass
assert(isequal(ncircles(38536),3))
y = 3
23 Pass
assert(isequal(ncircles(36172),1))
y = 1
24 Pass
assert(isequal(ncircles(28759),3))
y = 3
25 Pass
assert(isequal(ncircles(81672),3))
y = 3
26 Pass
assert(isequal(ncircles(45053),1))
y = 1
27 Pass
assert(isequal(ncircles(80664),5))
y = 5
28 Pass
assert(isequal(ncircles(79018),4))
y = 4
29 Pass
assert(isequal(ncircles(28296),4))
y = 4
30 Pass
assert(isequal(ncircles(6831),3))
y = 3
31 Pass
assert(isequal(ncircles(5494),1))
y = 1
32 Pass
assert(isequal(ncircles(63753),1))
y = 1
33 Pass
assert(isequal(ncircles(42429),1))
y = 1
34 Pass
assert(isequal(ncircles(90554),2))
y = 2
35 Pass
assert(isequal(ncircles(41733),0))
y = 0
36 Pass
assert(isequal(ncircles(15406),2))
y = 2
37 Pass
assert(isequal(ncircles(54000),3))
y = 3
38 Pass
assert(isequal(ncircles(93710),2))
y = 2
39 Pass
assert(isequal(ncircles(66096),5))
y = 5
40 Pass
assert(isequal(ncircles(39466),3))
y = 3
41 Pass
assert(isequal(ncircles(25900),3))
y = 3
42 Pass
assert(isequal(ncircles(84792),3))
y = 3
43 Pass
assert(isequal(ncircles(94506),3))
y = 3
44 Pass
assert(isequal(ncircles(37700),2))
y = 2
45 Pass
assert(isequal(ncircles(6729),2))
y = 2
46 Pass
assert(isequal(ncircles(18159),3))
y = 3
47 Pass
assert(isequal(ncircles(57575),0))
y = 0
48 Pass
assert(isequal(ncircles(18589),5))
y = 5
49 Pass
assert(isequal(ncircles(29145),1))
y = 1
50 Pass
assert(isequal(ncircles(46167),2))
y = 2 | 1,117 | 2,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-24 | latest | en | 0.418726 |
https://raweb.inria.fr/rapportsactivite/RA2009/apics/uid12.html | 1,679,640,748,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00644.warc.gz | 546,933,864 | 13,613 | ## Section: Scientific Foundations
### Identification and approximation
Identification typically consists in approximating experimental data by the prediction of a model belonging to some model class. It consists therefore of two steps, namely the choice of a suitable model class and the determination of a model in the class that fits best with the data. The ability to solve this approximation problem, often non-trivial and ill-posed, impinges on the effectiveness of a method.
Particular attention is payed within the team to the class of stable linear time-invariant systems, in particular resonant ones, and in isotropically diffusive systems, with techniques that dwell on functional and harmonic analysis. In fact one often restricts to a smaller class —e.g. rational models of suitable degree (resonant systems, see section 4.3 ) or other structural constraints— and this leads us to split the identification problem in two consecutive steps:
1. Seek a stable but infinite (numerically: high) dimensional model to fit the data. Mathematically speaking, this step consists in reconstructing a function analytic in the right half-plane or in the unit disk (the transfer function), from its values on an interval of the imaginary axis or of the unit circle (the band-width). We will embed this classical ill-posed issue (i.e. the inverse Cauchy problem for the Laplace equation) into a family of well-posed extremal problems, that may be viewed as a regularization scheme of Tikhonov-type. These problems are infinite-dimensional but convex (see section 3.1.1 ).
2. Approximate the above model by a lower order one reflecting further known properties of the physical system. This step aims at reducing the complexity while bringing physical significance to the design parameters. It typically consists of a rational or meromorphic approximation procedure with prescribed number of poles in certain classes of analytic functions. Rational approximation in the complex domain is a classical but difficult non-convex problem, for which few effective methods exist. In relation to system theory, two specific difficulties superimpose on the classical situation, namely one must control the region where the poles of the approximants lie in order to ensure the stability of the model, and one has to handle matrix-valued functions when the system has several inputs and outputs, in which case the number of poles must be replaced by the McMillan degree (see section 3.1.2 ).
When identifying elliptic (Laplace, Beltrami) partial differential equations from boundary data, point 1. above can be recast as an inverse boundary-value problem with (overdetermined Dirichlet-Neumann) data on part of the boundary of a plane domain (recover a function, analytic in a domain, from incomplete boundary data). As such, it arises naturally in higher dimensions when analytic functions get replaced by gradients of harmonic functions (see section 4.2 ). Motivated by free boundary problems in plasma control and questions of source recovery arising in magneto/electro-encephalography, we aim at generalizing this approach to the real Beltrami equation in dimension 2 (section 6.3.3 ) and to the Laplace equation in dimension 3 (section 6.3.1 ).
Step 2. above, i.e., meromorphic approximation with prescribed number of poles—is used to approach other inverse problems beyond harmonic identification. In fact, the way the singularities of the approximant (i.e. its poles) relate to the singularities of the approximated function is an all-pervasive theme in approximation theory: for appropriate classes of functions, the location of the poles of the approximant can be used as an estimator of the singularities of the approximated function (see section 6.3.2 ).
We provide further details on the two steps mentioned above in the sub-paragraphs to come.
#### Analytic approximation of incomplete boundary data
Keywords : extremal problems, inverse problems, Hardy spaces, harmonic functions, Beltrami equations.
Participants : Laurent Baratchart, Slah Chaabi, Yannick Fischer, Juliette Leblond, Jean-Paul Marmorat, Jonathan Partington, Stéphane Rigat [ Univ. Aix-Marseille I ] , Emmanuel Russ [ Univ. Aix-Marseille III ] , Fabien Seyfert.
Given a planar domain D , the problem is to recover an analytic function from its values on a subset of the boundary of D . It is convenient to normalize D and apply in each particular case a conformal transformation to meet a “normalized” domain. In the simply connected case, which is that of the half-plane, we fix D to be the unit disk, so that its boundary is the unit circle T . We denote by Hp the Hardy space of exponent p which is the closure of polynomials in the Lp -norm on the circle if 1p< and the space of bounded holomorphic functions in D if p = . Functions in Hp have well-defined boundary values in Lp(T) , which make it possible to speak of (traces of) analytic functions on the boundary.
A standard extremal problem on the disk is [68] :
(P0 ) Let 1p and fLp(T) ; find a function gHp such that g-f is of minimal norm in Lp(T) .
When seeking an analytic function in D which approximately matches some measured values f on a sub-arc K of T , the following generalization of (P0 ) naturally arises:
(P ) Let 1p , K a sub-arc of T , fLp(K) , and M>0 ; find a function gHp such that and g-f is of minimal norm in Lp(K) under this constraint.
Here is a reference behavior capsulizing the expected behavior of the model off K , while M is the admissible error with respect to this expectation. The value of p reflects the type of stability which is sought and how much one wants to smoothen the data.
To fix terminology we generically refer to (P ) as a bounded extremal problem . The solution to this convex infinite-dimensional optimization problem can be obtained upon iteratively solving spectral equations for appropriate Hankel and Toeplitz operators, that involve a Lagrange parameter, and whose right hand-side is given by the solution to (P0 ) for some weighted concatenation of f and . Constructive aspects are described in [43] , [45] , [87] , for p = 2 , p = , and 1<p< , while the situation p = 1 is essentially open.
Various modifications of (P) have been studied in order to meet specific needs. For instance when dealing with loss-less transfer functions (see section 4.3 ), one may want to express the constraint on in a pointwise manner: |g-|M a.e. on , see [29] , [47] for p = 2 and = 0 .
The above-mentioned problems can be stated on an annular geometry rather than a disk. For p = 2 the solution proceeds much along the same lines [23] . When K is the outer boundary, (P ) regularizes a classical inverse problem occurring in nondestructive control, namely to recover a harmonic function on the inner boundary from overdetermined Dirichlet-Neumann data on the outer boundary (see sections 4.2 and 6.3 ). Interestingly perhaps, it becomes a tool to approach Bernoulli type problems for the Laplacian, where overdetermined observations are made on the outer boundary and we seek the inner boundary knowing it is a level curve of the flux (see section 6.3.3 ). Here, the Lagrange parameter indicates which deformation should be applied on the inner contour in order to improve the fit to the data.
Continuing effort is currently payed by the team to carry over bounded extremal problems and their solution to more general settings.
Such generalizations are twofold: on the one hand Apics considers 2-D diffusion equations with variable conductivity, on the other hand it investigates the ordinary Laplacian in . The targeted applications are the determination of free boundaries in plasma control and source detection in electro/magneto-encephalography (EEG/MEG, see section 6.3.2 ).
An isotropic diffusion equation in dimension 2 can be recast as a so-called real Beltrami equation [73] . This way analytic functions get replaced by “generalized” ones in problems (P0 ) and (P ). Hardy spaces of solutions, which are more general than Sobolev ones and allow one to handle Lp boundary conditions, have been introduced when 1<p< [46] . The expansions of solutions needed to constructively handle such problems have been preliminary studied in [64] , [65] . The goal is to solve the analog of (P ) in this context to approach Bernoulli-type problems (see section 6.3.1 ).
At present, bounded extremal problems for the n -D Laplacian are considered on half-spaces or balls. Following [88] , Hardy spaces are defined as gradients of harmonic functions satisfying Lp growth conditions on inner hyperplanes or spheres. From the constructive viewpoint, when p = 2 , spherical harmonics offer a reasonable substitute to Fourier expansions [13] . Only very recently were we able to define operators of Hankel type whose singular values connect to the solution of (P0 ) in BMO norms. The Lp problem also makes contact with some nonlinear PDE's, namely to the p -Laplacian. The goal is here to solve the analog of (P ) on spherical shells to approach inverse diffusion problems across a conductor layer.
#### Meromorphic and rational approximation
Keywords : meromorphic approximation, rational approximation, critical point theory, orthogonal polynomials.
Participants : Laurent Baratchart, José Grimm, Martine Olivi, Edward Saff, Herbert Stahl [ TFH Berlin ] .
Let as before D designate the unit disk, T the unit circle. We further put RN for the set of rational functions with at most N poles in D , which allows us to define the meromorphic functions in Lp(T) as the traces of functions in Hp + RN .
A natural generalization of problem (P0 ) is
(PN ) Let 1p , N0 an integer, and fLp(T) ; find a function gNHp + RN such that gN-f is of minimal norm in Lp(T) .
Problem (PN ) aims, on the one hand, at solving inverse potential problems from overdetermined Dirichlet-Neumann data, namely to recover approximate solutions of the inhomogeneous Laplace equation u = , with some (unknown) distribution, which will be discretized by the process as a linear combination of N Dirac masses. On the other hand, it is used to perform the second step of the identification scheme described in section 3.1 , namely rational approximation with a prescribed number of poles to a function analytic in the right half-plane, when one maps the latter conformally to the complement of D and solve (PN ) for the transformed function on T .
Only for p = and continuous f is it known how to solve (PN ) in closed form. The unique solution is given by the AAK theory, that allows one to express gN in terms of the singular vectors of the Hankel operator with symbol f . The continuity of gN as a function of f only holds for stronger norms than uniform, [85] .
The case p = 2 is of special importance. In particular when , the Hardy space of exponent 2 of the complement of D in the complex plane (by definition, h(z) belongs to if, and only if h(1/z) belongs to Hp ), then (PN ) reduces to rational approximation. Moreover, it turns out that the associated solution gNRN has no pole outside D , hence it is a stable rational approximant to f . However, in contrast with the situation when p = , this approximant may not be unique.
The former Miaou project (predecessor of Apics) has designed an adapted steepest-descent algorithm for the case p = 2 whose convergence to a local minimum is guaranteed; it seems today the only procedure meeting this property. Roughly speaking, it is a gradient algorithm that proceeds recursively with respect to the order N of the approximant, in a compact region of the parameter space [40] . Although it has proved rather effective in all applications carried out so far (see sections 4.2 , 4.3 ), it is not known whether the absolute minimum can always be obtained by choosing initial conditions corresponding to critical points of lower degree (as done by the Endymion software section 5.5 and RARL2 software, section 5.2 ).
In order to establish convergence results of the algorithm to the global minimum, Apics has undergone a long-haul study of the number and nature of critical points, in which tools from differential topology and operator theory team up with classical approximation theory. The main discovery is that the nature of the critical points (e.g. local minima , saddles...) depends on the decrease of the interpolation error to f as N increases [48] . Based on this, sufficient conditions have been developed for a local minimum to be unique. This technique requires strong error estimates that are often difficult to obtain, and most of the time only hold for N large. Examples where uniqueness or asymptotic uniqueness has been proved this way include transfer functions of relaxation systems (i.e., Markov functions) [49] , the exponential function, and meromorphic functions [8] . The case where f is the Cauchy integral on a hyperbolic geodesic arc of a Dini-continuous function which does not vanish “too much” has been recently answered in the positive, see section 6.7 . An analog to AAK theory has been carried out for 2p< [9] . Although not computationally as powerful, it has better continuity properties and stresses a continuous link between rational approximation in H2 and meromorphic approximation in the uniform norm, allowing one to use, in either context, techniques available from the other(When 1p<2 , problem (PN ) is still fairly open.).
A common feature to all these problems is that critical point equations express non-Hermitian orthogonality relations for the denominator of the approximant. This is used in an essential manner to assess the behavior of the poles of the approximants to functions with branched singularities which is of particular interest for inverse source problems (cf. sections 6.3.2 , 6.7 ).
In higher dimensions, the analog of problem (PN ) is the approximation of a vector field with gradients of potentials generated by N point masses instead of meromorphic functions. The issue is by no means understood at present, and is a major endeavor of future research problems.
Certain constrained rational approximation problems, of special interest in identification and design of passive systems, arise when putting additional requirements on the approximant, for instance that it should be smaller than 1 in modulus. Such questions have become over years an increasingly significant part of the team's activity (see sections 4.3 , 6.6 , 6.7 , and 6.10 ). When translated over to the circle, a prototypical formulation consists in approximating the modulus of a given function by the modulus of a rational function of degree n . When p = 2 this problem can be reduced to a series of standard rational approximation problems, but usually one needs to solve it for p = . The case where |f| is a piecewise constant function with values 0 and 1 can also be approached via classical Zolotarev problems [86] , that can be solved more or less explicitly when the pass-band consists of a single arc. A constructive solution in the case where |f| is a piecewise constant function with values 0 and 1 on several arcs (multiband filters) is one recent achievement of the team. Though the modulus of the response is the first concern in filter design, the variation of the phase must nevertheless remain under control to avoid unacceptable distortion of the signal. This is an important issue, currently under investigation within the team under contract with the CNES, see section 6.10 .
From the point of view of design, rational approximants are indeed useful only if they can be translated into physical parameter values for the device to be built. This is where system theory enters the scene, as the correspondence between the frequency response (i.e., the transfer-function) and the linear differential equations that generate this response (i.e., the state-space representation), which is the object of the so-called realization process. Since filters have to be considered as dual modes cavities, the realization issue must indeed be tackled in a 2×2 matrix-valued context that adds to the complexity. A fair share of the team's research in this direction is concerned with finding realizations meeting certain constraints (imposed by the technology in use) for a transfer-function that was obtained with the above-described techniques (see section 6.8 ).
#### Behavior of poles of meromorphic approximants and inverse problems for the Laplacian
Keywords : singularity detection, free boundary inverse problems, meromorphic and rational approximation, orthogonal polynomials, discretization of potentials.
Participants : Laurent Baratchart, Edward Saff, Herbert Stahl [ TFH Berlin ] , Maxim Yattselev.
We refer here to the behavior of the poles of best meromorphic approximants, in the Lp -sense on a closed curve, to functions f defined as Cauchy integrals of complex measures whose support lies inside the curve. If one normalizes the contour to be the unit circle T , we are back to the framework of section 3.1.2 and to problem (PN ); the invariance of the problem under conformal mapping was established in [6] . The research so far has focused on functions whose singular set inside the contour is zero or one-dimensional.
Generally speaking, the behavior of poles is particularly important in meromorphic approximation to obtain error rates as the degree goes large and also to tackle constructive issues like uniqueness. However, the original motivation of Apics is to consider this issue in connection with the approximation of the solution to a Dirichlet-Neumann problem, so as to extract information on the singularities. The general theme is thus how do the singularities of the approximant reflect those of the approximated function? The approach to inverse problem for the 2-D Laplacian that we outline here is attractive when the singularities are zero- or one-dimensional (see section 4.2 ). It can be used as a computationally cheap preliminary step to obtain the initial guess of a more precise but heavier numerical optimization.
For sufficiently smooth cracks, or pointwise sources recovery, the approach in question is in fact equivalent to the meromorphic approximation of a function with branch points, and we were able to prove [4] , [6] that the poles of the approximants accumulate in a neighborhood of the geodesic hyperbolic arc that links the endpoints of the crack, or the sources [44] . Moreover the asymptotic density of the poles turns out to be the equilibrium distribution on the geodesic arc of the Green potential and it charges the end points, that are thus well localized if one is able to compute sufficiently many zeros (this is where the method could fail). The case of more general cracks, as well as situations with three or more sources, requires the analysis of the situation where the number of branch points is finite but arbitrary, see section 6.7 ). This are outstanding open questions for applications to inverse problems (see section 6.3 ), as also the problem of a general singularity, that may be two dimensional.
Results of this type open new perspectives in non-destructive control, in that they link issues of current interest in approximation theory (the behavior of zeroes of non-Hermitian orthogonal polynomials) to some classical inverse problems for which a dual approach is thereby proposed: to approximate the boundary conditions by true solutions of the equations, rather than the equation itself (by discretization).
Let us point out that the problem of approximating, by a rational or meromorphic function, in the Lp sense on the boundary of a domain, the Cauchy transform of a real measure, localized inside the domain, can be viewed as an optimal discretization problem for a logarithmic potential according to a criterion involving a Sobolev norm. This formulation can be generalized to higher dimensions, even if the computational power of complex analysis is then no longer available, and this makes for a long-term research project with a wide range of applications. It is interesting to mention that the case of sources in dimension three in a spherical or ellipsoidal geometry, can be attacked with the above 2-D techniques as applied to planar sections (see section 6.3 ).
#### Matrix-valued rational approximation
Keywords : rational approximation, inner matrix, reproducing kernel space, realization theory.
Participants : Laurent Baratchart, Martine Olivi, José Grimm, Jean-Paul Marmorat, Bernard Hanzon, Ralf Peeters [ Univ. Maastricht ] .
Matrix-valued approximation is necessary for handling systems with several inputs and outputs, and it generates substantial additional difficulties with respect to scalar approximation, theoretically as well as algorithmically. In the matrix case, the McMillan degree (i.e., the degree of a minimal realization in the System-Theoretic sense) generalizes the degree.
The problem we want to consider reads: Let and n an integer; find a rational matrix of size m×l without poles in the unit disk and of McMillan degree at most n which is nearest possible to in (H2)m×l . Here the L2 norm of a matrix is the square root of the sum of the squares of the norms of its entries.
The approximation algorithm designed in the scalar case generalizes to the matrix-valued situation [67] . The first difficulty consists here in the parametrization of transfer matrices of given McMillan degree n , and the inner matrices (i.e., matrix-valued functions that are analytic in the unit disk and unitary on the circle) of degree n enter the picture in an essential manner: they play the role of the denominator in a fractional representation of transfer matrices (using the so-called Douglas-Shapiro-Shields factorization).
The set of inner matrices of given degree has the structure of a smooth manifold that allows one to use differential tools as in the scalar case. In practice, one has to produce an atlas of charts (parametrization valid in a neighborhood of a point), and we must handle changes of charts in the course of the algorithm. Such parametrization can be obtained from interpolation theory and Schur type algorithms, the parameters being interpolation vectors or matrices [33] , [10] , [11] . Some of these parametrizations have a particular interest for computation of realizations [10] , [11] , involved in the estimation of physical quantities for the synthesis of resonant filters. Two rational approximation codes (see sections 5.2 and 5.5 ) have been developed in the team.
Problems relative to multiple local minima naturally arise in the matrix-valued case as well, but deriving criteria that guarantee uniqueness is even more difficult than in the scalar case. The already investigated case of rational functions of the sought degree (the consistency problem) was solved using rather heavy machinery [7] , and that of matrix-valued Markov functions, that are the first example beyond rational function has made progress only recently [39] .
Let us stress that the algorithms mentioned above are first to handle rational approximation in the matrix case in a way that converges to local minima, while meeting stability constraints on the approximant.
Logo Inria | 4,973 | 23,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-14 | latest | en | 0.916405 |
https://ici2016.org/how-do-you-distribute-data-in-python/ | 1,660,118,468,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571150.88/warc/CC-MAIN-20220810070501-20220810100501-00478.warc.gz | 299,431,843 | 15,030 | # How do you distribute data in Python?
## How do you distribute data in Python?
How to fit data to a distribution in Python
1. data = np. random. normal(0, 0.5, 1000)
2. mean, var = scipy. stats. distributions. norm. fit(data)
3. x = np. linspace(-5,5,100)
4. fitted_data = scipy. stats. distributions. norm. pdf(x, mean, var)
5. hist(data, density=True)
6. plot(x,fitted_data,’r-‘) Plotting data and fitted_data.
### What are the different Python distributions?
Python Distributions Anaconda from Continuum Analytics. ChinesePython Project: Translation of Python’s keywords, internal types and classes into Chinese. Eventually allows a programmer to write Python programs in Chinese. Enthought’s EDM.
#### How do you find the probability distribution in Python?
How to calculate the probability of a random variable in a normal distribution in Python
1. x = 1.0.
2. pdf_probability = scipy. stats. norm. pdf(x, loc=0, scale=1)
3. print(pdf_probability)
4. y = 0.5.
5. cdf_probability = scipy. stats. norm. cdf(x, loc=0, scale=1) – scipy. stats. norm. cdf(y, loc=0, scale=1)
6. print(cdf_probability)
What is normal distribution Python?
The normal distribution is a form presenting data by arranging the probability distribution of each value in the data. Most values remain around the mean value making the arrangement symmetric.
What is the best Python distribution?
5 Python distributions for mastering machine learning
• Anaconda Python.
• ActivePython.
• CPython.
• Enthought Canopy.
• WinPython.
## What is distribution plot in Python?
Seaborn is a Python data visualization library based on Matplotlib. It provides a high-level interface for drawing attractive and informative statistical graphics. This article deals with the distribution plots in seaborn which is used for examining univariate and bivariate distributions.
### What is the most popular python distribution?
Anaconda Python
Anaconda Python Anaconda has come to prominence as a major Python distribution, not just for data science and machine learning but for general purpose Python development as well. Anaconda is backed by a commercial provider of the same name (formerly Continuum Analytics) that offers support plans for enterprises.
#### Is anaconda better than python?
Anaconda is the best tool in processing a large amount of data for the required purpose. Python is versatile in creating the applications needed for the data science industry.
Where is normal distribution used?
It is the most important probability distribution in statistics because it fits many natural phenomena. For example, heights, blood pressure, measurement error, and IQ scores follow the normal distribution.
What is the meaning of Python distribution?
A distribution of Python is a bundle that contains an implementation of Python along with a bunch of libraries or tools. In theory, a distribution of Python could use any implementation, although all the ones I know of use CPython.
## How do I create a Python package?
To create a package in Python, we need to follow these three simple steps: First, we create a directory and give it a package name, preferably related to its operation. Then we put the classes and the required functions in it. Finally we create an __init__.py file inside the directory, to let Python know that the directory is a package.
### What are the best packages in Python?
but you can take your GUIs to the next level using an external Python module.
• Databases.
• Web Development.
• Image and Video Manipulation.
• Data Science and Maths.
• Game Development.
• Sound.
• Microsoft Windows.
• Mac OS.
• USB and Serial Ports.
• #### What is a Python package?
A package is a collection of Python modules, i.e., a package is a directory of Python modules containing an additional __init__.py file. The __init__.py distinguishes a package from a directory that just happens to contain a bunch of Python scripts. Packages can be nested to any depth,… | 855 | 3,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-33 | latest | en | 0.775059 |
https://math.stackexchange.com/questions/2019121/consecutive-galois-extensions | 1,560,955,293,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999000.76/warc/CC-MAIN-20190619143832-20190619165832-00345.warc.gz | 527,371,485 | 35,807 | # Consecutive Galois extensions
$Proposition$: An extension $K/F$ is normal if and only if $H \triangleleft Gal(E/F)$
Now let $E/F$ be a finite galois extension and let $F=K_0 \leqslant K_1 \leqslant K_2 \leqslant .....\leqslant K_n=E$ consecutive extensions.
Let $H_i$ (for $i=0,1,2,,,n$) subgroup of $Gal(E/F)$ which corresponds to $K_i$ from the Galois correspondence (i.e $K_i \longleftarrow \longrightarrow H_i$)
Prove that $\forall i \in \{1,2,,,n\}$, we have that $K_i / K_{i-1}$ is a galois extension if and only if $H_i \triangleleft H_{i-1}$
I was going to proceed by induction to $n$.
For $n=1$ its valid because of the proposition .
Thus i assumed that the statement is valid $\forall 1<i<n$
But i find a difficulty to proceed because i cannot find the inductive step.
Can someone help me to solve this and complete the proof?
It has nothing to do with induction. Assume $F\subseteq L\subseteq K\subseteq E$ with $E/F$ Galois. Let $H$ be the subgroup fixing $L$ and $N$ the subgroup fixing $K$. The $K/L$ is Galois iff $N \triangleleft H$. Assume first that $K/L$ is Galois let $\tau \in N$ and $\sigma \in H$. Then $\sigma$ maps $K$ into $K$. So if $x\in K$ then $\sigma(x)\in K$ and thus $\sigma^{-1}\tau\sigma(x)=x$, since $\tau$ fixes $K$ and thus $\sigma^{-1}\tau\sigma\in N$.
Assume conversely that $N\triangleleft H$ then with the same notation as above, $\sigma^{-1}\tau\sigma\in N$ and so $\sigma^{-1}\tau\sigma(x)=x$. So $\tau\sigma(x)=\sigma(x)$ and so $\sigma(x)$ is fixed by $H$ and is thus in $K$.
• @capo Because if an automorphism fixes $K$, then it fixes $L$. Fixes means pointwise fixes. – Rene Schipperus Nov 18 '16 at 13:40
• Every auto of $K$ extends to an auto of $E$. Thus autos of $E$ and $K$ are the same on $K$. – Rene Schipperus Nov 18 '16 at 15:33 | 589 | 1,797 | {"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} | 3.6875 | 4 | CC-MAIN-2019-26 | latest | en | 0.705096 |
https://astrowizici.st/teaching/phs5000/5/ | 1,709,061,914,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474686.54/warc/CC-MAIN-20240227184934-20240227214934-00587.warc.gz | 96,173,723 | 14,281 | # Markov chain Monte Carlo sampling
Data analysis and machine learning
In the next hour50 minutes, technically. you will:
• learn (or be reminded of) what Markov chain Monte Carlo (MCMC) is;
• learn (or be reminded of) how some variations of MCMC exist, including:
• the Metropolis method;
• affine-invariant methods like the emcee implementation;
• Hamiltonian methods.
• learn (or be reminded of) how to tune MCMC samplers;
• learn (or be reminded of) how to diagnose outputs from MCMC samplers;
• implement your own Metropolis MCMC sampler; and
• implement your own Hamiltonian Monte Carlo sampler.
## What is MCMC and why would we use it?
Markov chain Monte Carlo (MCMC) methods are for approximating integrals.
MCMC methods are used to draw samples from a continuous random variable, where the probability density for each sample will be proportional to some known function. In most cases that you (the physicist) will use MCMC, the continuous random variable you are drawing from is the posterior probability, and the integral that you are approximating is the posterior probability density function (pdf; the area under the curve of posterior probability).
The Markov chain part of the name is a stochastic process that describes a sequence of events, where each event depends only on the outcome of the immediately previous event. You can think of it as walking, where after each step you roll a four-sided die to decide where to move next: north, east, south, or west. Because each roll of the dice is independent, your current position only depends on your last position, and the roll of the dice.
The Monte Carlo part of the name describes randomness. Let's say that we roll the die as before, but before taking a step we look in that direction and then decide if we actually want to step there or not. For theatrical purposes we will say that there are dropbears scattered around you, and dropbears are known to chase humans who have walking patterns that don't follow a Markov chain Monte Carlo method.Dropbears are surprisingly passionate about detailed balance. After each roll of the die you look in that direction. If there's no dropbear in that direction then you take a step forward in that direction. If there is a dropbear in that direction then you don't want them to think that you're afraid of them, so you roll another six-faced die: if you roll evens, you step ahead. If you roll odds, you stay where you are.
That's (almost) a Markov chain Monte Carlo algorithm. Except for the dropbearsWell, unless you're in Australia.; instead we evaluate the posterior probability. If the posterior probability is higher at the proposed step, we move there. But if the probability at the proposed step is lower than the current probability we randomly decide whether to move there or not. If you think about this procedure a little bit you will realise that it seems so remarkably simple that it could not possibly approximate the posterior probability distribution. Yet it does! There is no great explanation I have come across that so simply explains why this works, but here goes: the Markov chain produces a unique stationary distribution, and the density of the samples will be proportional to the probability of those samples.
There are many variations of MCMC available, and many are tailored for particular problems (e.g., multi-modal problems). All of them share two properties:
1. how to decide on the new proposed step in parameter space, and
2. how to decide on whether to accept or reject that proposal.
You can't do this ad hoc. There are Rules™. If your answers to (1) and (2) above don't follow the Rules™ then you lose all guarantees for your sampler to accurately approximate the integral. The first Rule™ is that of detailed balance: it must be equally likely to move from point $$A$$ in parameter space to point $$B$$, as it is to move from point $$B$$ to point $$A$$. In other words, the probability of movement must be just as reversible. Your method for proposing a new step in parameter space must follow this Rule™, or it must correct for detailed balance in the accept/reject step. You can't easily do the correction causally, either: you must be able to prove that the correction is (well,..) correct from a mathematical perspective. You must also be able to show that the proposal distribution $$q(\theta'|\theta_k)$$ is properly normalised: the integration over $$\theta'$$ should equal unity. Lastly, you must make the proposals in the same coordinates which your priors are specified. Otherwise you must add a Jacobian evaluation to the log probability in order to account for the change of variables.
## The Metropolis algorithm
The Metropolis algorithmNot the Metropolis-Hasting algorithm: that has another multiplicative ratio. is the simplest MCMC algorithm available. Let us first choose some arbitrary point $$\theta_k$$ to be the initial value. We must also chose some proposal density $$q(\theta'|\theta_k)$$ that will propose a new position $$\theta'$$, given the current position $$\theta_k$$. The proposal distribution we adopt will be symmetric such that $$q(\theta'|\theta_k) = q(\theta_k | \theta')$$ and it will obey detailed balance. For simplicity we will choose $$q(\theta'|\theta_k)$$ to be a Gaussian distribution centred at $$\theta_k$$ so that points near $$\theta_k$$ are more likely to be proposed next.
With the current position $$\theta_k$$ and the proposal distribution $$q(\theta'|\theta_k)$$ the algorithm is as follows:
1. Propose a position $$\theta'$$ by drawing from the proposal distribution $$q(\theta'|\theta_k)$$.
2. Compare the proposed position with the current position by calculating the acceptance ratio $$\alpha = f(\theta')/f(\theta_k)$$.
3. Draw a random number $$u \sim \mathcal{U}\left(0, 1\right)$$:
• If $$u \leq \alpha$$ then we accept the proposed position and set $$\theta_{k+1} = \theta'$$.
• If $$u > \alpha$$ then we reject the proposal and keep the existing one by setting $$\theta_{k+1} = \theta_{k}$$.
Consider what happens when we move from an unprobable position to a highly probable position, where the acceptance ratio $$\alpha >> 1$$ exceeds the maximum bound for $$u$$. If the proposed position $$\theta'$$ has a higher posterior probability than at the current position it will be accepted. But if the proposed position $$\theta'$$ has a lower posterior probability than the current position, then there is still a chance it will be accepted.
Let's implement this in code. First we will generate some faux data where we know the true value, which is essential for model checking! We will adopt the following model $y \sim \mathcal{N}\left(\mu, \sigma\right)$ where the data $$y$$ are drawn from a Gaussian distribution centered on $$\mu$$ with standard deviation $$\sigma$$, but the data $$y$$ also have some uncertainty of $$\sigma_y$$. The unknown model parameters are those of the Gaussian: $$\theta = \{\mu, \sigma\}$$. import numpy as np np.random.seed(0) # for reproducibility # Bring some truth into this virtual world. mu_t, sigma_t = truths = [0.873, 5.323] # Generate random data. N = 100 # number of data points. y = np.random.normal(mu_t, sigma_t, size=N) y_err = 0.2 * np.random.rand(N) y += y_err * np.random.randn(N)
Now let's define our log prior, log likelihood, and log probability functions: def ln_prior(theta): return 0 def ln_likelihood(theta, y, y_err): mu, sigma = theta ivar = 1.0/(sigma**2 + y_err**2) # Omitting constant terms: return -0.5 * np.sum((mu - y)**2 * ivar) + np.sum(0.5 * np.log(ivar)) def ln_probability(theta, y, y_err): return ln_prior(theta) + ln_likelihood(theta, y, y_err)
And we're ready to implement the Metropolis MCMC algorithm: # This should be part of stdlib. from tqdm import tqdm # We will run MCMC for 10,000 iterations n_iterations = 10000 n_dim = len(truths) args = (y, y_err) # Initialise the positions. positions = np.empty((n_iterations, n_dim)) positions[0] = np.random.uniform(0, 1, size=2) # Define the proposal distribution. def q(theta_k): """ Draw from the proposal distribution. """ return np.random.multivariate_normal(theta_k, np.eye(2)) # Sample! for k, theta_k in tqdm(enumerate(positions[:-1]), total=n_iterations - 1): # Propose a new position. theta_ = q(theta_k) # Calculate the acceptance ratio. log_acceptance_ratio = ln_probability(theta_, *args) \ - ln_probability(theta_k, *args) acceptance_ratio = np.exp(log_acceptance_ratio) # Draw a random number. u = np.random.uniform(0, 1) if u <= acceptance_ratio: # Accept the proposal. positions[k + 1] = theta_ else: # Reject the proposal. positions[k + 1] = theta_k
The astute among you will see that we made some decisions without explicitly talking about them. One of those is the number of iterations: 10,000. Because we have a Markov chain, the position of each step depends on the previous step. But in the end what we want are independent samples of the posterior density function. How can we have an independent sample if each position depends directly on the last position? And how can the sampling be independent if we decided on the initialisation point? We will discuss this in more detail below, but the short answer is that you need to run MCMC for a long time.But not too long! Don't waste energy. If we ignore the Markov property and just focus on initialisation, then you can see that you would have to wait a while for the walker to move from the initialisation point and start exploring the posterior density. We refer to this as burn-in. The burn-in phase is not representative of the pdf: it must be discarded.
But how long is the burn-in? This relates to the Markov property (and many other things). Each sample is dependent on the previous sample. But the 100th sample depends very little (or at all) on the 1st sample. That means the chain has some auto-correlation length where nearby samples are correlated with each other. If you have a long correlation length then it means you don't have many effective independent samples of the posterior pdf because your samples are correlated with each other. If you have short correlation lengths then you will have many effective independent samples (for the same length chain). Samplers that produce shorter auto-correlation lengths for the same number of objective function evaluations are more efficient. We discuss this in more detail below.
The other decision we made without discussing it relates to the proposal distribution $$q(\theta'|\theta_k)$$. We explicitly stated that we would adopt a Gaussian distribution centered on the point $$\theta_k$$, but we made no mention of the covariance matrix for that proposal distribution. In the code we have adopted an identity matrix $\mathbf{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which effectively defines the scale of the proposal step in each direction. We get to choose this step size, and it can be different for each parameter!Doesn't this violate detailed balance? Why or why not? For illustrative purposes here we have just set it as 1, but it doesn't have to be that way.
Let's plot the value of the model parameters at each iteration. import matplotlib.pyplot as plt fig, axes = plt.subplots(2, 1, figsize=(8, 4)) labels = (r"$\mu$", r"$\sigma$") for i, (ax, label) in enumerate(zip(axes, labels)): ax.plot(positions[:, i], c="k", lw=1) ax.axhline(truths[i], c="tab:blue", lw=2) ax.set_xlabel(r"Step") ax.set_ylabel(label) # So we can see the initial behaviour: ax.set_xlim(-100, n_iterations) fig.tight_layout()
You can see that the walker quickly moves from the initial position and settles around the true values shown in blue. That looks sensible. For now we will not investigate any diagnostics, and simply say that is good enough because we don't intend to use these samples for any proper (scientific) inference.
### Affine-invariant methods
There are many MCMC samplers but there is barely time in this class to introduce three. The second is the affine-invariant MCMC sampler. This method is popular because its proposal distribution makes it able to sample highly correlated parameters just as efficiently as if it were sampling a unit normal distribution, and it is highly amenable to parallelisation: allowing to efficiently spread the inference load across multiple CPU cores. The method has been popularised by the excellent emcee Python package.
Affine-invariant methods use multiple walkers (chains) to propose a new position. At each step the walkers will draw a line between another randomly chosen walker and a new proposal is generated somewhere along that line. This reduces the number of hyperparameters (e.g., like those along the diagonal of the proposal distribution) to just 1 or 2. This implementation is what makes sampling correlated parameters so efficient, because the proposals can be drawn along some line between walkers. The proposal itself does not follow detailed balance since there is a preferred direction for proposals to be made. However, the correction to this proposal is done in the accept/reject step. The growth of the walkers is controlled by $$\alpha$$, a stretch parameter that controls how quickly the walkers will expand into the typical set. It's called a stretch parameter because it's easy for parameters to stretch out, but harder for them to collapse back in. That means it's a good idea to initialise your walkers in a very small ball in parameter space (e.g. at the optimised point) and watch them stretch out in parameter space.
Affine invariant samplers are excellent if you have relatively low numbers of model parameters, and you don't have gradient information for your log probability function. Affine invariant samplers are not suitable for high dimensional problems, and can fail quietly.
### Hamiltonian Monte Carlo
In the optimisation class we emphasisedPerhaps too much. how important it is to know the gradient (or curvature) of the objective function to efficiently move through parameter space. MCMC is inherently more expensive than optimisation, yet so far we have used no gradient information! That seems like an oversight. You might think that we are ignoring gradient information because our proposal distribution $$q(\theta'|\theta_k)$$ must obey detailed balance, since the proposals have to be equally likely in different directions and including gradient information would present a preferential direction. But we also saw that the affine-invariant proposal distribution disobeys detailed balance!And corrects for it in the accept/reject step. Can we use curvature information to make a more efficient sampler?
The answer is an overwhelming yes, but the sampler must still obey detailed balance. The way we will do this is by introducing new variables that will help guide the proposal distribution along the curvature of the objective function, but make sure that the target distribution (the posterior) remains independent of the variables we have introduced. Another way of thinking of this is that the variables we introduce will be marginalised out. It's not strictly a marginalisation, rather a factorisation, but it may help you imagine the system.
$$\renewcommand{\vec}[1]{\boldsymbol{#1}} \newcommand{\transpose}{^{\scriptscriptstyle \top}}$$
You will recall that a Hamiltonian uniquely defines the evolution of a system with time. Specifically, Hamiltonian dynamics describes how potential energy is converted to kinetic energy (and vice versa). At time $$t$$, a particle in some system has a location $$\vec{x}$$ and momentum $$\vec{p}$$, and the total energy of the system is given by the Hamiltonian $\mathcal{H}\left(\vec{x}, \vec{p}\right) = U(\vec{x}) + K(\vec{p})$ which remains constant with time. Here we follow the usual nomenclature where $$U(\vec{x})$$ describes the potential energy and $$K(\vec{p})$$ is the kinetic energy. The location $$\vec{x}$$ is our location in the parameter space, which we previously called $$\vec{\theta}$$, $$\vec{\theta} \equiv \vec{x}$$. We will introduce the momentum vector $$\vec{p}$$ as auxilary (nuisance) variables, which we have yet to define. The momentum vector $$\vec{p}$$ has the same dimensionality as $$\vec{x}$$, so we are immediately duplicating the number of parameters in the problem!From already doing this in earlier lectures, I hope you are coming to appreciate that this is not always a bad thing. It can be a Very Good Thing™!
The time evolution of the system is described by the set of differential equations $\frac{d\vec{x}}{dt} = +\frac{\partial\mathcal{H}}{\partial\vec{p}} \quad \textrm{and} \quad \frac{d\vec{p}}{dt} = -\frac{\partial\mathcal{H}}{\partial\vec{x}} \quad .$ We want to develop a Hamiltonian function $$\mathcal{H}(\vec{x},\vec{p})$$ that will let us efficiently explore the target distribution $$p(\vec{x})$$.
Imagine if a particle were moving through parameter space and its movement followed a Hamiltonian. If the target density were a two dimensional Gaussian with an identity matrix as the covariance matrix, then a particle moving through this density (with constant energy) would be akin to a particle moving in a circular motion around the center of the Gaussian. In other words, it would be like the particle is orbiting around the Gaussian. But we don't want the particle to just explore one slice (or one line of constant energy) in the parameter space: we want the particle to explore all of parameter space. For this reason, we need to consider particles to have a distribution of energies where the particles with that distribution of energies would efficiently sample the target density.
For reasons outside the scope of this class, we will use the canonical distribution. With an energy function $$E(\vec{\theta})$$ with some set of parameters $$\vec{\theta}$$, the canonical distribution is $p(\vec{\theta}) = \frac{1}{Z}\exp\left[-E\left(\vec{\theta}\right)\right]$ where $$Z$$ is a normalising constantOtherwise known as the partition function. that we don't care about: the normalisation constant just scales the pdf so that the integral of the pdf is one. The canonical distribution for the Hamiltonian is then $\begin{eqnarray} p(\vec{x},\vec{p}) &\propto& \exp\left[-\mathcal{H}(\vec{x},\vec{p})\right] \\ p(\vec{x},\vec{p}) &\propto& \exp\left[-U(\vec{x}) - K(\vec{p})\right] \\ p(\vec{x},\vec{p}) &\propto& \exp\left[-U\left(\vec{x}\right)\right]\exp\left[-K\left(\vec{p}\right)\right] \end{eqnarray}$ or in other words our Hamiltonian is separable: $p(\vec{x},\vec{p}) \propto p(\vec{x})p(\vec{p})$ That means that the position parameters $$\vec{x}$$ and the momentum parameters we've introduced $$\vec{p}$$ are independent. That means we can choose anything we want for the distribution of $$\vec{p}$$ without affecting the position parameters $$\vec{x}$$!If you're not shocked by this you should probably start reading again from the top. A common choice is a zero-mean Gaussian distribution with unit variance, $p(\vec{p}) \propto \frac{\vec{p}\transpose\vec{p}}{2}$ because it is equivalent to solving a Hamiltonian for a simple harmonic oscillator and finding a kinetic energy: $K(\vec{p}) = \frac{\vec{p}\transpose\vec{p}}{2} \quad .$ Now that we have a function for the kinetic energy, we still need the potential energy function $$U(\vec{x})$$. When combined with the canonical function, we need the potential energy function that gives an unscaled version of the target distribution $$p(\vec{x})$$. Let us define $U(\vec{x}) = -\log{p\left(\vec{x}\right)}$ such that $$p(\vec{x}) \propto \exp\left[-U(\vec{x})\right] = \exp\left[\log{p(\vec{x})}\right] = p(\vec{x})$$.
We have defined our Hamiltonian. But how will we use it in practice?And is it actually useful, or is everything we have done so far just mathematical trickery? Starting from the current position $$\vec{x}_k$$, the most simplistic Hamiltonian Monte Carlo algorithm proceeds as follows:
1. Draw from the momentum distribution $$p(\vec{p})$$ to set the momentum $$p(\vec{p}_k)$$ for the current position $$\vec{x}_k$$: $$\vec{p}_k \sim \mathcal{N}\left(\vec{0}, \vec{I}\right)$$.
2. Given the current position $$\vec{x}_k$$ and momentum $$\vec{p}_k$$, integrate the motion of the fictitious particle for $$L$$ steps with a step-size of $$\delta$$ (which we have yet to define), requiring that the total energy of the Hamiltonian remains constant. The position and momentum of the particle at the end of the integration are $$\vec{x}'$$ and $$\vec{p}'$$, respectively.
3. If we always integrated forward in time then it would not maintain detailed balance. To correct for this we flip the momentum vector of the particle at the end of integration so that it becomes $$-\vec{p}'$$. This alternates the direction of the particle at each step, equivalent to integrating forward in time and then backward et cetera.
4. Calculate an acceptance ratio $$\alpha$$ for the proposed position and momentum $$(\vec{x}',-\vec{p}')$$compared to the initial position and momentum vector $$(\vec{x}_k,\vec{p}_k)$$ Just like Metropolis! $\alpha = \exp\left[-U(\vec{x}') + U(\vec{x}_k) - K(-\vec{p}') + K(\vec{p}_k)\right]$
5. Draw a random number $$u \sim \mathcal{U}\left(0, 1\right)$$
• If $$\alpha >= u$$, accept the proposed state and set $$\vec{x}_{k+1} = \vec{x}'$$.
• If $$\alpha < u$$, reject the proposed state and set $$\vec{x}_{k+1} = \vec{x}_k$$
If our integrator was well behaved then we would almost always end up accepting the proposed distribution. That's because we are drawing from a symmetric momentum distribution that helps guide the particle along some constant energy, and is independent of the position vector. The main reason why we wouldn't accept a proposal would be because we are using a numerical integrator (not an exact integration), and the final position of the particle will not always exactly maintain constant energy.
Since constant energy is so important to us, when integrating the motion of the particle you should use a symplectic integrator. Specifically, the leapfrog integration scheme is well-suited to this problem because the position and momentum vectors are independent of each other.
## Tuning
Create a new file and run the example code for the Metropolis MCMC sampler above. First make sure that you get the same results as shown in the figure. Once you're satisfied with that, copy the file and change the value of sigma_t to 0.1. What happens? Why?
Here's a hint. Plot your acceptance ratios with time by using the following code: acceptance_ratios = np.empty(n_iterations) for k, theta_k in tqdm(enumerate(positions[:-1]), total=n_iterations - 1): # Propose a new position. theta_ = q(theta_k) # Calculate the acceptance ratio. log_acceptance_ratio = ln_probability(theta_, *args) \ - ln_probability(theta_k, *args) acceptance_ratio = np.exp(log_acceptance_ratio) # Store the acceptance ratio. acceptance_ratios[k] = acceptance_ratio # Draw a random number. u = np.random.uniform(0, 1) if u <= acceptance_ratio: # Accept the proposal. positions[k + 1] = theta_ else: # Reject the proposal. positions[k + 1] = theta_k # Plot the acceptance ratios. fig, ax = plt.subplots() ax.plot(acceptance_ratios) What do they look like? Naively, what do you expect the acceptance ratios to look like?
The reason why things have gone to shit is because the step size for our proposal distribution $$q(\theta'|\theta_k)$$ is too large. Recall we adopted the identity matrix for the covariance matrix, which means that each time we are taking a step size of order 1, but the scale of the parameters is about one tenth that. That means even though we moving through parameter space, each time we are taking a very big step, much bigger than what we need to. If we are at a point in parameter space $$\vec{\theta}_k$$ where the posterior has lots of support (high probability) then when we propose a new point we will almost always move far away from the true value. How do we tune the step size so that it takes the appropriate step size?
During burn-in most Metropolis samplers will vary the scale length of the proposal distribution until they reach mean acceptance ratios of between 0.25 to 0.50. You can have a single scale length, or different scales for each dimension. But you can only do this during burn in! There are otherProbably better. ways to tune the sampler that focus on the autocorrelation length of the samples. That's probably a better idea because you only really care about the autocorrelation length (and not the acceptance ratios), but it can be harder to do in practice.
For Hamiltonian Monte Carlo samplers the acceptance fraction will always be near one. But there are still tuning knobs to play with, including the covariance matrix of the proposal distribution, the number of steps to integrate the motion of a particle for, and the step size of integration. In this class (and the related assignment) I recommend you try to vary these parameters to and diagnose the impacts it has on the posterior.
## Diagnosing outputs
MCMC is just a method. It will (almost always) generate numbers. But just because it generates numbers doesn't mean you should believe those numbers! Just like how optimisation can fail quietly in subtle ways, MCMC can fail spectacularly in very quiet ways. But MCMC is worse: with optimisation you can always cross-check (against different initialisations, grid evaluations) for whether things make sense, and with convex problems you have mathematical guarantees that optimisation worked correctly. With MCMC there are no such guarantees.
Not only are there no guarantees, but even if you wanted to prove that an MCMC chain has converged to the typical set, you can't. It is impossible. You should always be skeptical of an MCMC chain, and you should inspect as many things as possible to determine whether it seems consistent with being converged.
The first thing you can check are trace plots. This is like what we plotted earlier: showing the parameter value at each step of the MCMC chain. If you have multiple chains then show them in different colours or on different figures, unless you're using emcee. Ideally you should see your chains moving from their initialisation value and settling in to some part of parameter space. There shouldn't be any walker that is trailing off far away from all others.
If your chain looks to be settling into the typical set, then it's worth taking the second half of your samples (e.g., discarding the first half as burn-in) and calculating the autocorrelation length of those samples. The number of effective samples you have will be the number of MCMC samples divided by the autocorrelation length. However, you should continue to be skeptical: in order to measure the autocorrelation length you need at least a few autocorrelation lengths of samples! So if you have 1,000 samples and the true autocorrelation length is 5,000 then your estimate of the autocorrelation will be wrong and you could incorrectly think that your chain has converged. In practice the autocorrelation length hasn't been estimated correctly.
Other things you can do with the trace include evaluating the variance of the last half of your samples compared to the first half. Or even split up your chain into quarters and evaluate the variance in each quarter worth of samples. You should see the variance in the first quarter worth of samples be larger than all others. That's why it's the burn-in: the walker is still settling in to the typical set and you should discard those samples.
If you started your walker from a very good position (and perhaps you're using emcee with a ball of walkers around one spot) then you may see the variance of the first quarter of data be smaller than the last quarter because the walkers are still stretching out in parameter space. That doesn't mean that your first quarter of samples are worth keeping! You still have to throw them away! Always discard the burn-in.
People like making corner plots to show the posterior. The great thing about MCMC is that if you have many nuisance parameters that you don't care about then you can just exclude them from your corner plot and by doing so you are effectively showing a marginalised posterior distribution. Corner plots can be instructive to understand model pathologies, the effects of priors, et cetera.
Other than the autocorrelation length, the other metric that is worth calculating is the so-called Geman-Rubin convergence diagnostic, or $$\hat{r}$$. The idea goes that if you have multiple MCMC chains, and those chains are converged, then the variance between the chains and within the chains should be similar. More formally consider if we had $$M$$ chains each of length $$N$$. The sample posterior mean for the $$m$$-th chain is $$\hat{\theta}_m$$ and $$\hat{\sigma}_m$$ is the variance of the $$m$$-th chain. The overall posterior mean (across all chains) is $\hat{\theta} = \frac{1}{M}\sum_{m=1}^{M} \hat{\theta}_m \quad .$ The between-chain variance $$B$$ and within-chain variance $$W$$ are defined as $B = \frac{N}{M-1}\sum_{m=1}^{M}\left(\hat{\theta}_m-\hat{\theta}\right)^2$ and $W = \frac{1}{M} \sum_{m=1}^{M} \hat{\sigma}_m^2 \quad .$ The pooled variance $$\hat{V}$$ is an unbiased estimator of the marginal posterior variance of $$\theta$$ $\hat{V} = \frac{N-1}{N}W + \frac{M + 1}{MN}B$ and the potential scale reduction factor (better known as $$\hat{r}$$, or the Gelman-Rubin diagnostic) is given by $\hat{r} = \sqrt{\frac{\hat{d} + 3}{\hat{d} + 1}\frac{\hat{V}}{W}}$ where $$\hat{d}$$ is the degrees of freedom. If the $$M$$ chains have converged to the target posterior distribution then $$\hat{r} \approx 1$$.
Remember: you should do all of these diagnostic tests, but you can never prove that your chains are converged. If you're skeptical, try re-initialisating your MCMC from different locations, or running a longer chain, or using a different sampler, or different model parameterisation. In general most samplers will always fail quietly. Hamiltonian Monte Carlo methods are the general exception: when they fail, they do so in such a spectacular way that it is usually obvious that they have failed. In practice I use Hamiltonian Monte Carlo with at least two MCMC chains. Each chain is independent of each other. So if the two chains sample very different posteriors, then I know something is definitely wrong!
## Why shouldn't I use MCMC?
People often use MCMC when they shouldn't. Usually this is under some false belief that MCMC will give them all the answers to their data analysis. There is a great pedagocial paper on this, which all practicioners of MCMC should read. Here are a handful of common reasons why you should not use MCMC:
1. To sample all of the parameter space. MCMC will not sample your entire parameter space! It will only try to move towards the typical set, but if your problem is multi-modal (i.e., there are islands of probability support that are separated by regions of extremely low probability) then there is every likelihood that your MCMC sampler will sample one of those probability islands and never move to another probability island. The same is true for the rest of your parameter space: just because you can draw samples from your MCMC method doesn't mean the sampler has traversed all of parameter space!
2. To perform better optimisation. You should never use MCMC for practical optimisation. MCMC is probably (or provably?) the slowest method to use to optimise some function. If you think your optimisation is bad, MCMC will not fix that! You should instead consider a search strategy (not an optimisation strategy) that reasonably trials some large volume of your parameter space. Gridding is a common example.
3. To model some intermediate step only. There are examples in the literature where some expensive MCMC will be performed on an intermediate step in the data processing. For example, some intermediate normalisation procedure or something of that nature. Then they will take the mean of the posterior samples in parameters and discard everything else! Why did you perform MCMC if you are not going to preserve uncertainty in later steps? Your later steps are conditioned on the single point estimate you took from your MCMC samples. The fact that you did MCMC just means you burnt a few forests unnecessarily.
In short: you should only be doing MCMC if you have a really. good. reason. If you're a MCMC practicioner, ask yourself what did I do with the posterior samples of the last MCMC I performed? Did percentiles from that sample appear in some published literature? Good. If not, what decision did that sampling inform? What if you could have derived reasonable estimates of uncertainties for 1/1,000,000th of the cost? Would that have been sufficient? | 7,601 | 32,734 | {"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": 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-2024-10 | latest | en | 0.908552 |
https://discourse.julialang.org/t/permuting-the-rows-of-a-matrix-based-on-given-a-given-permutation/41444 | 1,669,603,673,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00020.warc.gz | 239,873,947 | 7,366 | # Permuting the rows of a matrix based on given a given permutation
Suppose I am given an mXn matrix A, and I have a permutation of {1,…,m}, for example by
``````p = sortperm(A[:,1]);
``````
I now wish to rearrange the wors of `A` based on `p`. How do I do that ? The `permute!()` function only applies to vectors, as shown here. I can always loop over the columns, as in
``````for j in 1:size(A,2)
permute!(A[:,j], p);
end
``````
but is there a better way ?
1 Like
Note that this permutes a copy of the column, unless you switch to using views (e.g. wrapping the code block or the line in a `@views` macro call).
For an alternative approach, you could write something similar to the `Base.permutecols!!` function (which permutes columns, not rows). (Since Julia’s `Array` is column-major, however, permuting rows with this algorithm will probably not be as efficient as permuting columns.) Or your could rearrange your code to swap dimensions on your array, and then call the `Base.permutecols!!` function directly. (It has a double `!!` because it also overwrites the permutation array `p`.)
3 Likes
Is this a (new) convention? Is it meant to signify that more than one passed argument is being mutated, or is its meaning more specific?
A couple of Base functions use it to indicate that multiple arguments are mutated, but I don’t think it’s a very widespread convention in Julia.
2 Likes
You can use `sortslices`, it sorts by rows, columns, or any row/column specified by the user.
``````julia> A = rand(0:9,5,5)
5×5 Array{Int64,2}:
4 6 4 6 3
2 7 5 5 2
7 7 0 4 3
4 5 2 5 6
6 5 8 1 0
julia> sortslices(A, dims=1)
5×5 Array{Int64,2}:
2 7 5 5 2
4 5 2 5 6
4 6 4 6 3
6 5 8 1 0
7 7 0 4 3
``````
1 Like
Thank you. Although I was looking for an in-place algorithm, this function is a very useful to learn. From the examples that you provided, it seems that the sorting is only according to the first column or row. But what if I try to rearrange the rows in in increasing order of the second column ? You said that it can be done, could you give me the code to do it ? Julia’s documentation on sortclices does not provide a solution.
Thank you, however the function permutecolums itself did not work.
``````julia> permutecols!!(A,p)
ERROR: UndefVarError: permutecols!! not defined
Stacktrace:
[1] top-level scope at none:0
``````
In fact, it seems that Base does not have any such function defined.
``````julia> Base.permute
permute! permute!! permutedims permutedims!
``````
In the hyperlink that you provided, the function was defined in the Combinatorics.jl package. importing the package also did not work. I tried copying the function definition and explicitly defining `permutecolumns!!`. But this gives the error
``````julia> permutecols!!(A,p)
ERROR: UndefVarError: require_one_based_indexing not defined
Stacktrace:
[1] permutecols!!(::Array{Float64,2}, ::Array{Int64,1}) at .\REPL[22]:2
[2] top-level scope at none:0
``````
Yes, as I said you can sort by any row/column you want which can be very useful. Just override the `lt` function to specify how the new comparison is done. Searching for documentation inside the REPL is so easy, just type `?sortslices` and you get good examples.
``````julia> A = rand(0:9,5,5)
5×5 Array{Int64,2}:
6 5 5 0 0
3 4 5 9 5
0 2 5 0 5
9 9 1 7 2
3 1 2 4 7
julia> sortslices(A, dims=1, lt=(x,y)->isless(x[2],y[2]))
5×5 Array{Int64,2}:
3 1 2 4 7
0 2 5 0 5
3 4 5 9 5
6 5 5 0 0
9 9 1 7 2
``````
1 Like
Thanks. I did search for sortslices. But there is no way anyone can possibly understand from the documentation alone that `lt=(x,y)->isless(x[2],y[2]))` is supposed to mean ``by second column".
It was added in Julia 1.2.
1 Like
You can also write `sortslices(A, dims=1, by=row->row[2])`, although I agree it’s not super-obvious what’s going to happen.
2 Likes
I see
I think this code doesn’t work b/c you haven’t specified `j` | 1,312 | 3,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-49 | latest | en | 0.879115 |
https://forums.wolfram.com/mathgroup/archive/2011/Sep/msg00027.html | 1,726,778,228,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00404.warc.gz | 235,385,485 | 7,422 | Mathematica collect function,
• To: mathgroup at smc.vnet.net
• Subject: [mg121189] Mathematica collect function,
• From: HOANG <hoangnhuy83 at yahoo.com>
• Date: Sat, 3 Sep 2011 08:05:24 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
```Hi all,
I would like to use the collect function for this:
(1 + 2 Z + Z Z.U)^2 + (1 + 2 Z + Z Z.U) (Z (1 + Z) + Z.U (Z + Z Z.U)) + (1 + Z + Z.U + (Z.U)^2) (Z (1 + Z) + Z.U (Z + Z Z.U)) + (Z (1 + Z) + Z.U (Z + Z Z.U)) (Z (1 + Z.U) + Z.U (Z + (Z.U)^2))
This is what I tried in Mathematica:
Collect[(1 + 2*Z + Z* Z*U)^2 + (1 + 2*Z + Z* Z*U)* (Z* (1 + Z) + Z*U* (Z + Z* Z*U)) + (1 + Z + Z*U + (Z*U)^2)* (Z *(1 + Z) + Z*U (Z + Z *Z*U)) + (Z* (1 + Z) +
Z*U* (Z + Z* Z*U)) *(Z* (1 + Z*U) + Z*U *(Z + (Z*U)^2), Z]
But it does not work
However, it works when I tried this one:
Collect[(1 + Z)^2 + (1 + Z)*(Z + Z*Z*U) + (1 + Z*U)* (Z + Z*Z*U) + (Z +Z*Z*U)*(Z +(Z*U)^2) + Z *((1 + Z)^2 + (1 + Z*U)* (Z + Z* Z*U)) + Z*U *((1 + Z)* (Z + Z*Z*U) + (Z + Z* Z*U)* (Z + (Z*U)^2)), Z]
Someone knows why? thanks for your answer
```
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• Next by thread: Re: Mathematica collect function, | 575 | 1,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.691825 |
http://docplayer.net/38837286-Solutions-to-problem-set-6.html | 1,545,226,126,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376832330.93/warc/CC-MAIN-20181219130756-20181219152756-00543.warc.gz | 76,577,481 | 25,238 | # SOLUTIONS TO PROBLEM SET 6
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1 SOLUTIONS TO PROBLEM SET SPRING 16 Note the difference of conventions: these solutions adopt that the characteristic polynomial of a matrix A is det A xi while the lectures adopt the convention that it is det ti A. The difference between the two is the sign 1 n. As far as the answers are concerned, it only affects problem 1. 1 What is the constant term of the characteristic polynomial of a square matrix? Why? The constant term of a polynomial P x is its value when x. By definition of the characteristic polynomial, its value when x is the determinant of the matrix. Answer: The determinant of the matrix. Compute the eigenvalues of the matrix Compute the characteristic polynomial x 1 det 1 x 1 x x 1 x det 1 x 1 1 det x 1 1 x 1 x 1 x x x 1 x x det 1 det 1 1 det 1 x x 1 x x xx 1 + x x 1 x 4 3x + 1. The eigenvalues are the roots of this polynomial. Let y x. Then y is a root of the quadratic polynomial y 3y + 1, and y. Solving the quadratic polynomial, we get y 3± 5. Both roots are positive, so given that x ± y we get four possibilities for x. 3+ Answer: 5 3, 5 3+, 5 3, 5. Or alternatively: ±1± 5 note that etc.. 1
2 18.6 SPRING 16 3 The following matrices have only one eigenvalue: 1. What are the dimensions of the eigenspaces in each case? 1 1, 1 1, 1, For a matrix A, the eigenspace with eigenvalue λ is the kernel of the matrix A λi. Here we have λ 1, so we subtract I from each of the matrices above:, 1, 1, 1 1 and find the dimensions of the kernels. The ranks of these matrices are,,, 1 respectively, so by the rank-nullity theorem the dimensions of the kernels are 3, 1, 1,. Answer: 3, 1, 1,. 4 Is the matrix diagonalizable? 1 Short solution: Yes, because it is symmetric. Symmetric matrices are always diagonalizable. Long solution: An n n matrix is diagonalizable if and only if the dimensions of its eigenspaces add up to n. Let us find the eigenvalues and then find the dimension of the eigenspace for each eigenvalue. To find the eigenvalues, let us write the characteristic polynomial. 1 x 1 det 1 1 x 1 1 x det 1 det 1 x 1 x1 x 1 1 x 1 x1 x 1 xx x 1. The roots of the quadratic polynomial x x 1 are ± 8 1 ±, therefore the eigenvalues of the matrix are 1, 1+, 1. We have three distinct eigenvalues, and each of them has an eigenspace of dimension at least 1, so the sum of the dimensions is at least 3. On the other hand, in a 3-dimensional space this sum is at most 3, so it equals 3, so the matrix is diagonalizable. Note that by the same argument an n n matrix with n distinct eigenvalues is always diagonalizable. Answer: Yes. 5 If n is odd, then every n n matrix has at least one eigenvector in R n. Why? The degree of the characteristic polynomial of an n n matrix equals n. polynomial of degree n always has a real root. If n is odd, a
3 SOLUTIONS TO PROBLEM SET Suppose n, and consider the n n matrix A α i,j whose entries are given by { 1 if j i + 1; α i,j otherwise. a Write a formula for the entries of the matrix A k for k n. b For 1 k n, compute the eigenvalues and eigenspaces of A k. Let e i be the ith vector of the standard basis. Then Ae 1, Ae e 1, Ae 3 e etc. Therefore, A e i for i 1, and A e i e i for i 3. Similarly, A k e i for i 1,..., k, and A k e i e i k for k < i n. Using this, we can write a formula for the entries of A k : we have A k i,j 1 if j i + k and otherwise. Note that this also covers k, when A I. For the eigenvectors of the matrix A k, note that the vectors e 1,..., e k are eigenvectors with eigenvalue, so any vector of the form k i1 a i e i is an eigenvector with eigenvalue. Suppose there is some other eigenvector v n b j e j, and let m > k be the highest index with b m we j1 can tell that m > k from the assumption that v is not a linear combination of e 1,..., e k. Then A k v m b j A k e j m b j e j k, in particular, the component along e m equals. Therefore, v j1 jk+1 can only be an eigenvector if its eigenvalue is. On the other hand, b m e m k, so eigenvalue is not a possibility either. We conclude that there are no eigenvectors other than the linear combinations of e 1,..., e k. Answer: a A k i,j 1 if j i + k and otherwise; b the only eigenvalue of A k is, and the space of eigenvectors is the span of e 1,..., e k. 7 Suppose x R n a unit vector. Recall from Exam III the Householder matrix H I x x T and the hyperplane N : { v R n v x } which is the orthogonal complement to x. a If you werent able to show that for any w R n, one has π N w π N H w on Exam III, please write up a proof here in your own words! b Prove that for any w R n, one also has w π N w π N H w H w. Explain what H does geometrically; draw a picture for n and n 3. c Purely from geometry, compute the eigenvalues and eigenspaces of H. You dont have to compute any determinants for this. Is H diagonalizable? The n n matrix x x T is the matrix of the projection of the space R n onto the line spanned by the vector x. Note: this is not to be confused with x T x, which is a number equal to x x. Denote this projection by π x. The line spanned by x and the hyperplane N are orthogonal complements, so for any w R n we have w π N w + π x w. The vector π N w w π x w is the projection of w onto N; the vector H w w π x w is the reflection of w with respect to N. In other words, H w π N w π x w. Now it is clear that the vectors w and H w, its reflection with respect to N, have the same projection onto N part a. Moreover, w π N w π x w π x H w π N H w H w part b. Finally, as a reflection, H has eigenvalues 1 and 1. The eigenspace for eigenvalue 1 is the hyperplane N, and the eigenspace for eigenvalue 1 is the line spanned by x. Since their dimensions add up to n, the matrix H is diagonalizable part c.
4 SPRING 16 8 For every permutation σ Σ 3, compute the eigenvalues and eigenspaces of the 3 3 matrix P σ. For σ id, we have P σ I, so the only eigenvalue is 1 and the corresponding eigenspace is R 3. The two matrices corresponding to the two cycles of length 3 are rotations around the line x y z, so the only eigenvalue is 1, and the corresponding eigenspace is the line x y z. The three transpositions correspond to reflections, so the eigenvalues are 1 with multiplicity 1 and 1 with multiplicity. Answer: let the coordinates on R 3 be x, y, z, in this order; permutations given in cycle notation. σ id, P σ 1 1. Eigenvalue 1: eigenspace R σ 1, P σ 1. Eigenvalue 1: eigenspace x y a plane, eigenvalue 1: 1 eigenspace x + y, z a line. 1 σ 13, P σ 1. Eigenvalue 1: eigenspace x z a plane, eigenvalue 1: 1 eigenspace x + z, y a line. 1 σ 3, P σ 1. Eigenvalue 1: eigenspace y z a plane, eigenvalue 1: 1 eigenspace y + z, x a line. 1 σ 13, P σ 1. Eigenvalue 1: eigenspace x y z a line. 1 1 σ 13, P σ 1. Eigenvalue 1: eigenspace x y z a line. 1 9 Does the matrix have any real eigenvalues? Compute the characteristic polynomial: 1 x 1 det x det 1 x det 1 x det 1 x 1 x 1 1 det x x 1 1 x 1 x1 x1 x x + 1 x + 1.
5 SOLUTIONS TO PROBLEM SET 6 5 Let t 1 x. Then the polynomial takes form ttt t + t + 1 t 4 + 3t + 1. This polynomial has no real roots since both t and t 4 are always non-negative, so neither does the characteristic polynomial of the matrix. Therefore, this matrix has no real eigenvalues. Answer: No. 1 What is the characteristic polynomial of the matrix ? c det From the previous problem sets we know that the rank of this matrix is we can also easily find this out by row reduction. Therefore, by the rank-nullity theorem, the kernel of this matrix has dimension. By definition, the kernel of a matrix is the eigenspace for eigenvalue, so we know that the geometric multiplicity of is. Hence the algebraic multiplicity of is at least, i.e. the polynomial is divisible by x. Therefore, it has the form ax 4 + bx 3 + cx. Let us look at the three coefficients a, b, c. The characteristic polynomial consists of all possible terms of the following form: choose n entries in the matrix A so that the combination has exactly one entry from each row and each column; for each entry on the diagonal make an additional choice, taking either a ii or x; multiply all together and add sign. Among these terms, the ones which have x in degree k are of the following form: choose k entries from the diagonal that s where your x s come from; from the matrix A, cross out the columns and rows that contain these entries; take the determinant of the n k n k matrix you got; multiply by x k. In this manner, we see that the degree n term is always 1 n x n, and the degree n 1 term is 1 n 1 a ii x n 1 the value a ii is also called the trace of the matrix. Thus we have a 1, b , and Answer: x 4 1x 3 + 9x
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### 3 Orthogonal Vectors and Matrices
3 Orthogonal Vectors and Matrices The linear algebra portion of this course focuses on three matrix factorizations: QR factorization, singular valued decomposition (SVD), and LU factorization The first | 9,014 | 34,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-51 | latest | en | 0.804067 |
https://www.answers.com/Q/Square_centimeter_to_square_feet | 1,603,540,308,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882581.13/warc/CC-MAIN-20201024110118-20201024140118-00125.warc.gz | 636,219,736 | 40,194 | Area
Math and Arithmetic
# Square centimeter to square feet?
234
###### 2010-03-23 10:45:18
Divide square centimeters by 929 to get square feet.
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## Related Questions
There are 5.472 square feet in 5,084 square centimeters. 5,084 square centimeters x 0.00107639104 square feet/1 square centimeter = 5.472 square feet 1 square centimeter = 0.00107639104 square feet
1 square centimeter is 1/929 (0.0010764) of a square foot.
If you are asking how many square feet is in a centimeter, it cannot be converted. A centimeter is a measure of length and square feet is a measure of area.
The length centimeter has nothing to do with an area of square feet. Only square centimeters or square feet could be a solution, if you mean a flat area.
1 200 square centimeter = 1.291 669 25 square foot
1 square foot = 929.0304 square centimetres.1 square foot = 929 square centimetres
You cannot change square centimeter to centimeter. The square centimeter is a measure of area and the centimeter is a measure of distance.
One square centimeter is 100 square millimeters.
125 cm times 107 cm = 13375 square centimeter. 13375 square centimeter are 14.39673 square feet. 1 square foot = 9.290 304×10−2 m²
There are 100 square millimeters in one square centimeter.
10 percent of 3 square centimeter is 0.3 square centimeter.
A square meter is quite a bit larger then a square centimeter. For example the area of a square with sides 1 meter long (about three feet) is a square meter. A square with sides 1 centimeter long (a little less then a half inch) is a square centimeter. You could fit 10,000 square centimeters in 1 square meter. They are similar in that they are both measures of area in the metric system.
A square with sides that are one centimeter long
different between centimeter square centimeter is that cm is a unit of length and cm square is used for measuring area
A square centimeter is a unit of area; a centimeter is a unit of length or distance.
1 square centimeter equals 1 centimeter squared
There are 0.0001 square meters in a square centimeter, since a square meter is made up of 10,000 square centimeters.
Answer: zero 1 centimeter square[d] - 1 square centimeter= 0 Notes:: a centimeter squared is 1 cm * 1 cm = 1 cm^2. This area is equal to 1 square centimeter (i.e.: an area that is 1 cm x 1 cm). a square centimeter is an area 1 cm * 1 cm
A square centimeter measures area, and a centimeter measures length, so they are not exactly comparable in size.
centimeter - unit of lengthsquare centimeter - unit of areacubic centimeter - unit of volume - equal to milliliter
###### Math and ArithmeticAreaUnits of MeasureJobs & EducationPercentages, Fractions, and Decimal ValuesSchool SubjectsScience
Copyright ยฉ 2020 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. | 785 | 2,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-45 | latest | en | 0.851318 |
http://photonics101.com/light-matter-interactions/the-dispersion-relation-of-a-magnetized-plasma | 1,603,633,115,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889173.38/warc/CC-MAIN-20201025125131-20201025155131-00150.warc.gz | 83,192,064 | 9,161 | ## The Dispersion Relation of a Magnetized Plasma
It is known that a free electron gas prone to a periodic electric field obeys oscillations. One might ask what happens if an additional external magnetic field is acting on the plasma. In this problem we will discover how the magnetic field changes the dispersion relation. This will later allow us to understand how the magnetic field in the universe is measured using the "Faraday Rotation".
## Problem Statement
We want to assume an electron in a constant magnetic field parallel to the $$z$$ axis, $$\mathbf{B}=B_{0}\mathbf{e}_{z}$$. The free (non-interacting) electrons shall be driven by a left/right circularly polarized plane wave propagating in $$z$$-direction, see figure on the right:
Find the equations of motion for the electron. Use the assumptions that the range of the electron movement is much smaller than the wavelength and that its speed is much slower than the speed of light. Solve the equations of motion to find the electron displacement. Then, calculate the dispersion relation of the system assuming an electron density $$n_{e}$$.
This problem is very helpful to understand "The Faraday Rotation - How an Electron Gas and a Magnetic Field Rotate a Plane Wave" and a generalization of the Drude model which describes the electrodynamic properties of metals.
## Hints
Which force is responsible for the movement of a charged body in an electromagnetic field?
How is the electron displacement related to the polarization?
What is the connection between polarization and permittivity?
## Show Solution
Before we go into details lets imagine what might be happening. We can already guess that a single electron will follow the movement of the electric field of the wave. That means, it will make a circular motion. Such a motion however implies a magnetic moment $$\boldsymbol{\mu}$$ that will interact with the magnetic field $$\mathbf{}B$$. $$\boldsymbol{\mu}$$ and $$\mathbf{}B$$ being parallel is an energetically favorable situation (left picture below) whereas the contrary holds for the antiparallel case (right picture). Now we have to figure out what this means in mathematical terms and how it is reflected in the dispersion relation.
We can divide our efforts into three steps. It is very useful to make the calculations in a complex notation. To do so, we first have to know howa circular polarized monochromatic wave is expressed in this way. Next we calculate the electron displacement using the Lorentz force. At last we have to use our results to calculate the polarization and thus the relative permittivity.
## Circular Polarized Light - Complex Notation
In real space, a right (left) circularly polarized plane may be defined by$\begin{eqnarray*}\mathbf{E}^{r/l}\left(\mathbf{r},t\right) & = & E_{0}\left\{ \cos\left(kz-\omega t\right)\mathbf{e}_{x}\pm\sin\left(kz-\omega t\right)\mathbf{e}_{y}\right\} \ .\end{eqnarray*}$Remark: This choice is basically convention. The one I used here should be the same as used in optics textbooks. It is based on the interpretation that a right-handed polarization is a clockwise rotation of the electric field in a spatially fixed plane viewing in the direction of the source. However, engineers use the opposite convention as defined in the "IEEE Standard Definition of Terms for Antennas".
Using $$\exp\left(\mathrm{i}x\right)=\cos x+\mathrm{i}\sin x$$, we may write:$\begin{eqnarray*} \mathbf{E}^{r/l}\left(\mathbf{r},t\right) & = & \frac{1}{2}E_{0}\Re\left\{ \exp\left(\mathrm{i}\left(kz-\omega t\right)\right)\mathbf{e}_{x}\mp\mathrm{i}\exp\left(\mathrm{i}\left(kz-\omega t\right)\right)\mathbf{e}_{y}\right\} \ .\\ & = & \frac{1}{2}E_{0}\Re\left\{ \exp\left(\mathrm{i}\left(kz-\omega t\right)\right)\left[\mathbf{e}_{x}\mp\mathrm{i}\mathbf{e}_{y}\right]\right\} \ . \end{eqnarray*}$Hence, the part in brackets corresponds to circular polarized plane waves in the complex notation.
## The Electron Displacement
The equations of motion are given due to the Lorentz force:$\begin{eqnarray*} m_{e}\ddot{\mathbf{r}}_{l/r} & = & -e\left[\mathbf{E}\left(\mathbf{r},t\right)+\dot{\mathbf{r}}_{l/r}\times\mathbf{B}\left(\mathbf{r},t\right)\right]\\ & = & -e\left[E_{0}\exp\left(\mathrm{i}\left(kz-\omega t\right)\right)\left[\mathbf{e}_{x}\pm\mathrm{i}\mathbf{e}_{y}\right]+\dot{\mathbf{r}}_{l/r}\times B_{0}\mathbf{e}_{z}\right]\\ & \approx & -e\left[E_{0}\exp\left(-\mathrm{i}\omega t\right)\left[\mathbf{e}_{x}\pm\mathrm{i}\mathbf{e}_{y}\right]+\dot{\mathbf{r}}_{l/r}\times B_{0}\mathbf{e}_{z}\right]\ . \end{eqnarray*}$We have used in the last step that $$kz\approx0$$ because the electron motion shall be small compared to the wavelength. Furthermore, due to the assumed slow motion any relativistic effects are neglected since we just use the velocity here. Obviously the electron's motion can be studied in the x-y-plane since there is no force in $$z$$ direction. The electric field obeys a $$\exp\left(-\mathrm{i}\omega t\right)$$ dependency. Since further the magnetic field is constant, we can use an ansatz of the form$\begin{eqnarray*} \mathbf{r}_{l/r}\left(t\right) & = & \mathbf{r}_{l/r,0}\exp\left(-\mathrm{i}\omega t\right) \end{eqnarray*}$yielding$\begin{eqnarray*} -m_{e}\omega^{2}\mathbf{r}_{l/r,0} & = & -e\left[E_{0}\left[\mathbf{e}_{x}\pm\mathrm{i}\mathbf{e}_{y}\right]-\mathrm{i}\omega B_{0}\mathbf{r}_{l/r,0}\times \mathbf{e}_{z}\right] \end{eqnarray*}$where we have cancelled the time dependency in all factors. We can see that this equation only has a solution if $$\mathbf{r}_{l/r,0}=r_{l/r,0}\left(\mathbf{e}_{x}\pm\mathrm{i}\mathbf{e}_{y}\right)$$.Using $$\left(\mathbf{e}_{x}\pm\mathrm{i}\mathbf{e}_{y}\right)\times\mathbf{e}_{z}=-\mathbf{e}_{y}\pm\mathrm{i}\mathbf{e}_{x}=\pm\mathrm{i}\left(\mathbf{e}_{x}\pm\mathrm{i}\mathbf{e}_{y}\right)$$we find$\begin{eqnarray*}-m_{e}\omega^{2}r_{l/r,0} & = & -e\left[E_{0}\pm\omega B_{0}r_{l/r,0}\right]\ ,\\ r_{l/r} & = & \frac{eE_{0}}{m_{e}\omega^{2}\mp e\omega B_{0}}=\frac{eE_{0}}{m_{e}}\frac{1}{\omega\left(\omega\mp\Omega\right)} \end{eqnarray*}$with the well-known (non-relativistic) cyclotron frequency $$\Omega=eB_{0}/m_{e}$$. Note that using a monochromatic plane-wave is equivalent to a calculation in frequency space with only one frequency component. Thus, $$\omega$$ is not only a parameter but rather a variable.
Next, we will use this result to calculate the dispersion relation.
## Dispersion Relation
While analysing the movement of a free electron we implicitly implied a linear response to the driving electric field. Then, the electric displacement field is given by the relation$\begin{eqnarray*} \mathbf{D}\left(\mathbf{r},\omega\right) & = & \epsilon_{0}\mathbf{E}\left(\mathbf{r},\omega\right)+\mathbf{P}\left(\mathbf{r},\omega\right)\\ & = & \epsilon_{0}\mathbf{E}\left(\mathbf{r},\omega\right)+\epsilon_{0}\chi\left(\mathbf{r},\omega\right)\mathbf{E}\left(\mathbf{r},\omega\right)\\ & = & \epsilon_{0}\epsilon_{r}\left(\mathbf{r},\omega\right)\mathbf{E}\left(\mathbf{r},\omega\right)\ . \end{eqnarray*}$We have used the linear susceptibility $$\chi\left(\mathbf{r},\omega\right)$$ and the relative permittivy $$\epsilon_{r}\left(\mathbf{r},\omega\right)$$ to express the polarization $$\mathbf{P}\left(\mathbf{r},\omega\right)$$ in terms of $$\mathbf{E}\left(\mathbf{r},\omega\right)$$. Note that the latter equation does not make sense in a time representation. There, one has to consider the response of the medium which usually involves an integration over the past. In our case we know that the polarization is given by$\begin{eqnarray*} p_{l/r}\left(\omega\right) & = & -er_{l/r}\left(\omega\right)\\ & = & -\frac{e^{2}E_{0}}{m_{e}}\frac{1}{\omega\left(\omega\mp\Omega\right)}\ .\end{eqnarray*}$The latter equation is scalar since we keep in mind that this polarization holds with respect to left and right circular polarized light. However, a plasma containing only of one electron may be slightly underpopulated. For the overall polarization we have to incorporate an electron density $$n_{e}$$. Then we can use the relation of $$\mathbf{P}$$ and $$\mathbf{E}$$ to finally arrive at the susceptibility $$\chi_{l/r}$$. Again using scalar quantities we find$\begin{eqnarray*} P_{l/r}\left(\omega\right) & = & -\frac{n_{e}e^{2}}{m_{e}}\frac{E_{0}}{\omega\left(\omega\mp\Omega\right)}\\ & = & \epsilon_{0}\chi_{l,r}\left(\omega\right) \left|E_{l/r}\left(\omega\right)\right|\ \text{, so}\\ \chi_{l,r}\left(\omega\right) & = & \frac{P_{l/r}\left(\omega\right)}{\epsilon_{0}\chi_{l,r}\left(\omega\right)E_{0}}\\ & = & -\frac{n_{e}e^{2}}{\epsilon_{0}m_{e}}\frac{1}{\omega\left(\omega\mp\Omega\right)}\ . \end{eqnarray*}$Here, we discover the very important plasma frequency$\begin{eqnarray*} \omega_{p}^{2} & = & \frac{n_{e}e^{2}}{\epsilon_{0}m_{e}} \end{eqnarray*}$which is a characteristic quantity for free-electron oscillations.
Using the determined susceptibility we finally find the dispersion relation$\begin{eqnarray*} \epsilon_{r}^{l/r}\left(\omega\right) & = & 1+\chi_{l/r}\left(\omega\right)\\ & = & 1-\frac{\omega_{p}^{2}}{\omega\left(\omega\mp\Omega\right)}\ . \end{eqnarray*}$This dispersion relation only has two parameters: the plasma and cyclotron frequencies $$\omega_p$$ and $$\Omega$$, respectively. This makes it easy to understand its characteristics, see the figure below.
The dispersion relation of a magnetized plasma with respect to left and right polarized plane waves. Here a ratio of $$\Omega/\omega_p = 0.5$$ has been chosen meaning a relatively high magnetic field. One can see that in the limit $$\omega\rightarrow\infty$$ both solutions converge. However for low energies one can see a clear distinction. Interestingly, the left polarization is highly dispersive for low frequencies and obeys a dielectric behavior for $$\omega < \Omega$$. On the contrary, the right polarization acts like a metal. For comparison the dispersion without magnetic field is also shown. | 2,923 | 9,944 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 4.0625 | 4 | CC-MAIN-2020-45 | latest | en | 0.892284 |
http://coolconversion.com/volume/57-cubic+centimeter-to-cubic+inch | 1,527,414,981,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868239.93/warc/CC-MAIN-20180527091644-20180527111644-00507.warc.gz | 65,924,467 | 22,600 | Please get in touch with us if you:
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# 57 Cubic Centimeters to Cubic Inches Conversion
## How many cubic inches in 57 cubic centimeters?
There are 3.5 cubic inches in 57 cubic centimeters
To convert cubic centimeters to cubic inches, just multiply the value in cubic centimeters by the conversion factor 0.061. So, 57 cubic centimeters times 0.061 is equal to 3.5 cubic inches. Use our calculator to learn how to convert cubic centimeters to cubic inches. Note: these values are rounded to 2 significant figures.
### Cubic Centimeter to Cubic Inch Converter
Enter values here: Results here:
=
Detailed result here
To calculate a cubic centimeter value to the corresponding value in cubic inches, just multiply the quantity in cubic centimeter by 0.061 (the conversion factor).
Here is the formula:
Value in cubic inch = value in cubic centimeter * 0.061023744094732
Supose you want to convert 57 cubic centimeter into cubic inch. In this case you will have:
Value in cubic inch = 57 * 0.061023744094732 = 3.4783534133997 (cubic inch)(s)
Using this converter you can get answers to questions like:
1. How many cubic centimeters are in 57 cubic inches?
2. 57 cubic centimeters are equal to how many cubic inches?
3. how much are 57 cubic centimeter in cubic inches?
4. How to convert cubic centimeters to cubic inches?
5. What is the conversion factor to convert from cubic centimeters to cubic inches?
6. How to transform cubic centimeters in cubic inches?
7. What is the cubic centimeters to cubic inches conversion formula? Among others.
## Values Near 57 cubic centimeter in cubic inch
cubic centimetercubic inch
492.9901634606419
503.0511872047366
513.1122109488313
523.1732346929261
533.2342584370208
543.2952821811155
553.3563059252103
563.417329669305
573.4783534133997
583.5393771574945
593.6004009015892
603.6614246456839
613.7224483897787
623.7834721338734
633.8444958779681
643.9055196220629
653.9665433661576
Note: some values may be rounded.
### Disclaimer
While every effort is made to ensure the accuracy of the information provided on this website, we offer no warranties in relation to these informations. | 592 | 2,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-22 | latest | en | 0.648043 |
https://www.numbers-figures.com/cosine-of-21-twentyone | 1,701,780,639,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100551.17/warc/CC-MAIN-20231205105136-20231205135136-00428.warc.gz | 1,036,979,237 | 4,238 | # cosine of 21 twentyone
The cosine of 21 is:
-0.54772926022427
## What is the cosine of a number?
Cosine belongs to the trigonometric functions of mathematics and is used among other things in geometry as a number.
The word cosine comes from Latin and is derived from "complementi sinus" - the complementary sine.
The cosine is used as a number, for example, in the triangle calculation of mathematics to calculate the complementary angle of the sine. | 104 | 458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-50 | latest | en | 0.911537 |
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# LECTURE NOTES
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1 Introduction We shall consider continuous maps f of an interval I into itself and the discrete dynamical system xn 1 f xn n 0 1 1 We apply 1 1 to an initial condition x0 to generate its orbit the sequence x0 x1 x2 A convenient way to visualize the orbit is to draw the cobweb diagram bouncing between the graph of f and the line y x see the figures below The orbit is generated by repeated composition of functions We denote k fold composition by f k that is f 2 x f f x f 3 x f f f x etc A fixed point or equilibrium is a solution to x f x 1 2 A periodic orbit with minimal or prime period p is an orbit such that xk xk mod p and x0 x1 xp 1 are distinct Clearly f p xk xk 0 k p 1 Graphically fixed points are given by the intersection between the graph of f and the line y x An intersection between the graph of f p and the line y x give points with period not greater than p We typically study questions of asymptotic behavior such as convergence to a fixed point or periodic orbit Let us note at the outset that the dynamics are trivial if f is monotone and I is bounded In this case the orbit is a bounded monotone sequence and every initial condition converges to a fixed point We thus consider unimodal maps ie maps with a single hump The fundamental example is the family of logistic maps f x rx 1 x r 0 4 1 3 f takes a unique maximum value r 4 at the critical point x 1 2 thus the restriction to r 4 The orbit diagram of the logistic map has become an icon for chaos see Figure 1 1 This is a numerically generated picture that describes the asymptotic behavior of a typical initial condition as r is varied The complexity of the orbit diagram for this simple example is striking 1 1 Stability and bifurcations Linear maps f x ax admit the exact solution xn an x0 The origin is an attracting fixed point if a 1 and repelling if a 1 Observe that if a 0 the dynamics are oscillatory that is xk and xk 1 have opposite sign In particular there are two distinct forms of neutral stability if a 1 1 Figure 1 1 Orbit diagrams for the logistic and sine maps every point is fixed and if a 1 every orbit is of period 2 Linear maps are used to describe the stability of fixed points If x is a fixed point the characteristic multiplier at x is f 0 x If f 0 x 6 1 the fixed point is hyperbolic and the characteristic multiplies determines stability Exercise 1 We may expect bifurcations when f 0 x 1 In the logistic family the fixed point x r 1 1 r loses stability to a period 2 orbit in a period doubling bifurcation at r 3 This is illustrated graphically in Figures 1 2 1 4 When r 3 trajectories spiral into the fixed point The rate of approach is very slow as r approaches 2 as is seen from the density of orbits in Figure 1 3 When r 3 the fixed point has lost stability to period 2 orbit This corresponds to new fixed points of f 2 seen in the lower half of Figure 1 4 The period 2 orbit then loses stability in another period doubling bifurcation at r 1 6 Exercise 2 Trajectories when r is just below this bifurcation point are illustrated in Figure 1 5 These are the first two steps in a period doubling cascade at parameter values rn At this value an orbit with period 2n 1 loses stability to an orbit with period 2n in a period doubling bifurcation The first few terms in the 2 Figure 1 2 A stable fixed point with r 3 3 Figure 1 3 Slow approach to the fixed point as r 3 4 Figure 1 4 Birth of a stable period 2 orbit for r 3 5 Figure 1 5 Loss of stability of period 2 orbit near r r1 6 increasing sequence rn are 3 1 6 3 54409 3 5644 3 568759 The bifurcation values accumulate at r 3 566946 The orbit diagram of the logistic map include windows of chaos and chaotic bands beyond r 1 2 Universality and renormalization The most remarkable feature of the orbit diagram of the logistic is that its essential features depend only on minimal properties That is all families of the form rf1 x where f1 I I is unimodal with a nondegenerate maximum have similar orbit diagrams Roughly speaking this is what is meant by universality To illustrate this point Figure 1 1 compares the orbit diagram for the sine family f x r sin x r 0 1 and the logistic map In both orbit diagrams we see the same period doubling cascade and similar windows of order and chaos Metropolis Stein and Stein discovered that the ordering of these periodic orbits depends only on the fact that f is unimodal and continuous This is an example of qualitative or combinatorial universality A more dramatic quantitative feature is the following Let rn and r n denote the values of the period doubling bifurcations for the logistic and sine map respectively Numerical experiments reveal the amazing fact that lim r n 1 r n rn 1 rn lim 4 669201609102290 rn 1 n r n r n 1 n rn 1 4 The number is known as Feigenbaum s constant in honor of his penetrating analysis of this quantitative universality The central feature of Feigenbaum s analysis is the notion of renormalization To explain this idea we consider a family of maps f r x undergoing a cascade of period doubling bifurcation at the values rn We assume f r maps the interval I 1 1 into itself and has a critical point at 0 The characteristic multiplier of the orbit with period 2n decreases from 1 at rn to 1 at rn 1 We focus on the superstable orbit with characteristic multiplier 0 at the value Rn rn rn 1 It is easy to compute the values R0 and R1 for the logistic map Exercise 4 Let us compare the graphs of f R0 x Figure 1 6 and f 2 R1 x the lower half of Figure 1 7 The main idea is that by restricting the graph of f 2 R1 x to a smaller interval we again obtain a unimodal map More precisely the assumption that the period 2 orbit of f 2 R1 x is superstable 7 Figure 1 6 Superstable fixed point at r R0 8 implies that the interval 0 f R1 0 is positively invariant consider an orbit diagram in the upper half of Figure 1 7 Let 1 f R1 0 denote the length of this interval We restrict f 2 R1 to the interval 1 1 and rescale the x axis …
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