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https://interviewmania.com/verbal-reasoning/mathematical-operation-and-symbol-notation/2/1 | 1,721,644,549,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00788.warc.gz | 278,180,908 | 11,448 | Mathematical operation and symbol notation
Mathematical operation and symbol notation
Direction: In the following information question, the symbols @, ©, %, S and δ are used with the following meaning as illustrated below:
'P % Q' means 'P is greater than Q'
'P δ Q' means 'P is neither greater than smallest than Q'
'P @ Q' means 'P is smallest than Q'.
'p © Q' means 'P is either smallest than or equal to Q'
'P S Q' means 'P is either greater than or equal to Q'
In each of the following questions assuming the given statements to be true, find which of the two conclusions I and II given below them is are definitely true
1. Statement :
W © D, D S B, B @ H
Conclusion :
I. H % D II. W @ B
1. W ≤ D ....(i) D ≥ B .....(ii) B > H ... (iii)
H and D can't be compared from (ii) and (iii)
Hence; I does not follows
Nor can W and B he be compared From (i) and (ii).
Hence II does not follows
Correct Option: D
W ≤ D ....(i) D ≥ B .....(ii) B > H ... (iii)
H and D can't be compared from (ii) and (iii)
Hence; I does not follows
Nor can W and B he be compared From (i) and (ii).
Hence II does not follows
1. Statements:
V % B, B S D , D © E
Conclusions:
I. E δ B II. D @ V
1. V > B ...(i) B ≥ D ...(ii) D &Le; E ...(iii)
From (ii) and (iii) E and B can't be compared.
Hence I does not follows. From (i) and (iv) V > D or D < V
Hence II follows
Correct Option: A
V > B ...(i) B ≥ D ...(ii) D ≤ E ...(iii)
From (ii) and (iii) E and B can't be compared.
Hence I does not follows. From (i) and (iv) V > D or D < V
Hence II follows
1. Statements: H S N, N % R, R @ J
Conclusions:I. R @ H II. J % H
1. H ≥ N ....(i) N > R ....(ii) R < J ...(iii)
From (i) and (ii) R > H ...(iv)
Hence R < H and I follows
But J and H can't compared From (iii) and (iv)
Hence II does not follow.
Correct Option: A
H ≥ N ....(i) N > R ....(ii) R < J ...(iii)
From (i) and (ii) R > H ...(iv)
Hence R < H and I follows
But J and H can't compared From (iii) and (iv)
Hence II does not follow.
1. Statement :
F δ T, T S M, M © R
Conclusion :
I. R S F II. M © F
1. F = T ....(i) T ≥ M ....(ii) M ≤ R .....(iii)
From (i) and (ii) F ≥ M ....(iv)
Hence M ≤ F and II follows
But R and F can't be compared From (iii) and (iv)
Hence I does not follow
Correct Option: B
F = T ....(i) T ≥ M ....(ii) M ≤ R .....(iii)
From (i) and (ii) F ≥ M ....(iv)
Hence M ≤ F and II follows
But R and F can't be compared From (iii) and (iv)
Hence I does not follow
1. Statement :
N S T, T δ H, N @ W
Conclusion :
I. W % T II. H © N
1. N ≥ T ... (i) T = H ...(ii) N < w .....(iii)
Combining these we get W > N ≥ T = H
Hence W > T and I follows.
Also H ≤ N and II follows
Correct Option: E
N ≥ T ... (i) T = H ...(ii) N < w .....(iii)
Combining these we get W > N ≥ T = H
Hence W > T and I follows.
Also H ≤ N and II follows | 958 | 2,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-30 | latest | en | 0.920389 |
https://www.theonda.org/maths-fractions-worksheets-for-grade-equivalent-models-fractions_equivalent_visual_models_simplified_both_00/fractions-maths-worksheets-free-math-four-digit-div/ | 1,596,570,303,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735882.86/warc/CC-MAIN-20200804191142-20200804221142-00451.warc.gz | 805,847,879 | 8,867 | # Fractions Maths Worksheets Free Math Four Digit Div
Iefan February 21, 2020 Worksheet
There are many types of writing worksheets. There is the cursive writing worksheets and the kindergarten worksheets. The latter is more on letter writing and number writing. This is typically given to kids of aged four to seven to first teach them how to write. Through these worksheets, they learn muscle control in their fingers and wrist by repeatedly following the strokes of writing each letter.
Before creating the worksheet for children, it is important to understand why the worksheet is being made. Is there a message to be conveyed? Can students record information that can be understood later? Is it being created to just teach a basic concept to little children? A well designed worksheet will make its objective clear. The different aspects that should influence the design of the worksheet are the age, ability and motivation of the students.
Teachers and parents basically are the primary users of worksheets. It is an effective tool in helping children learn how to write. These writing worksheets have traceable patterns of the different strokes of writing letters. By tracing these patterns, kids slowly learn how a letter is structured.
There are also worksheets that teach how to read. It includes the basic sounds each letter produce. Kids try to read the words displayed before them. In the First Alphabet worksheet, kids learn how to write the alphabet. And in the First Animals worksheet, kids try to recognize the animals in the picture and learn the names of these animals.
### Adding Fractions With Common Denominators Worksheet
Mar 06, 2020
#### Converting Fractions To Decimals Worksheet
Jun 24, 2020
##### Long Division Worksheets With Decimals
Jul 30, 2020
###### Multiplying A Fraction By A Whole Number Worksheet
Mar 07, 2020
While worksheets for homeschool can assist in home schooling, they cannot take the place of a proper homeschool curriculum. One disadvantage they have is that they often focus on one subject area only, without integrating the whole curriculum. They can also be simplistic and give the impression that the student understands more than he actually does.
There are also teachers who use these kinds of techniques to teach in a more animated manner. The idea is to keep children interested because without their attention, it is difficult to make them absorb what you are trying to teach.
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Mar 06, 2020 | 637 | 2,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-34 | latest | en | 0.943139 |
http://www.madsci.org/posts/archives/2001-11/1005230854.Ph.r.html | 1,542,247,980,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742338.13/warc/CC-MAIN-20181115013218-20181115035218-00118.warc.gz | 456,033,967 | 2,737 | ### Re: Why are electron orbitals so huge rel. to the size of a nucleus?
Date: Wed Nov 7 14:58:16 2001
Posted By: Michael Wohlgenannt, Grad student, Department of Theoretical Physics , University of Munich, Germany
Area of science: Physics
ID: 1003973239.Ph
Message:
```
Hi Matt,
thank you for your question. You are asking for the "size" of bound states
considering various interactions. You are right, the size of bound states
such as protons (bound state of quarks), nucleus (bound state of protons
and neutrons) and atoms (bound state of nuclei and electrons) depends above
all on the strength of the force involved. The strong nuclear force binds
the quarks, they are bound in protons and neutrons (and all other hadrons).
This force is the strongest of them all. Some residual force of the strong
force between the quarks keeps protons and neutrons together. This force is
much weaker than the strong force and can be compared to the van der Waals
force concerning electromagnetic interaction. But it is still rather
strong. The electromagnetic interaction is responsible for the electron
orbitals. Compared to the strong force between quarks and hadrons this
force is very weak, therefore the orbits are comparatively large.
Let us stick to the electromagnetic interaction for a second. The socalled
Bohr radius is a measure for the size of electron orbits. In fact, it is
the radius of the ground state in a hydrogen atom. It is given by the
formula
h^2
r = -----------------,
4(pi)^2 m e^2
where h is Planck's constant, m is the mass of the electron, e is the
electron charge. We see (in case of electromagnetism) that increasing the
mass of the electron m decreases the size of the orbit (Bohr radius) r.
Further is the magnitude of the electron charge e a measure for the
strength of the electromagnetic interaction. So increasing the strength of
the electromagnetic force (i.e., increasing the charge) again decreases the
greetings, Michael.
```
Current Queue | Current Queue for Physics | Physics archives | 471 | 2,024 | {"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} | 2.84375 | 3 | CC-MAIN-2018-47 | latest | en | 0.891167 |
https://slideplayer.com/slide/4847490/ | 1,582,827,156,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146744.74/warc/CC-MAIN-20200227160355-20200227190355-00515.warc.gz | 529,117,583 | 19,182 | # PDAs => CFGs Sipser 2.2 (pages 119-122). Last time…
## Presentation on theme: "PDAs => CFGs Sipser 2.2 (pages 119-122). Last time…"— Presentation transcript:
PDAs => CFGs Sipser 2.2 (pages 119-122)
Last time…
CS 311 Mount Holyoke College 3 Recognizing context-free languages Lemma 2.21: If a language is context-free, then some pushdown automaton recognizes it. Proof:
Today…
CS 311 Mount Holyoke College 5 A full characterization Theorem 2.20: A language is context-free if and only if some pushdown automaton recognizes it. => by Lemma 2.21 Now we go backwards!
CS 311 Mount Holyoke College 6 The proof Let P = (Q, Σ, Γ, δ, q 0, F) be a pushdown automaton. Assume WLOG (Without Loss Of Generality) – P has exactly one accept state q accept – P empties its stack before accepting –Each transition does either a push or a pop (but not both)
CS 311 Mount Holyoke College 7 We build a grammar G… Given P = (Q, Σ, Γ, δ, q 0, F) Construct G = (V, Σ, R, S), where – V = {A pq | p,q ∈ Q} Idea: design the rules so that A pq generates all strings that take P from p with empty stack to q with empty stack – S = A q 0, q accept
Designing the rules
CS 311 Mount Holyoke College 9 P 's operation on strings of A pq Since P starts and ends with an empty stack: –The first move from p must be a push –The last move to q must be a pop Along the way, either: 1.The stack never becomes empty 2.There is some intermediate state where the stack is empty
CS 311 Mount Holyoke College 10 Case 1: The stack never empties between p and q On the first move from p –Let r be the state moving to –Let a be the input symbol read –Let t be the stack symbol pushed On the last move to q –Let s be the state moving from –Let b be the input symbol read –It must be the case that t is the stack symbol popped
CS 311 Mount Holyoke College 11 Capturing this behavior Model with the rule A pq → aA rs b
CS 311 Mount Holyoke College 12 Case 2: the stack empties along the way from p to q Let r be the state where the stack is empty Model with the rule A pq → A pr A rq
CS 311 Mount Holyoke College 13 Formally phrasing the rules If (r, t) ∈ δ(p, a, ε) and (q, ε) ∈ δ(s, b, t) then add the rule A pq → aA rs b For each p, q, r, ∈ Q, add the rule A pq → A pr A rq For each p ∈ Q, add the rule A pp → ε
CS 311 Mount Holyoke College 14 Proving we were right A pq generates x if and only if x can bring P from p with empty stack to q with empty stack => (Claim 2.30) If A pq generates x, then x can bring P from p with empty stack to q with empty stack Proof –By induction on the number of steps in the derivation of x from A pq
CS 311 Mount Holyoke College 15 And now the other way! Claim 2.31: If x can bring P from p with empty stack to q with empty stack, then A pq generates x Proof –By induction on the number of steps in the computation of P | 804 | 2,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-10 | latest | en | 0.828378 |
https://www.coursehero.com/file/p24s9p1/The-labour-market-equilibrium-is-now-achieved-at-a-new-level-of-nominal-wage-W1/ | 1,632,422,107,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057427.71/warc/CC-MAIN-20210923165408-20210923195408-00380.warc.gz | 769,561,219 | 136,505 | # The labour market equilibrium is now achieved at a
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The labour market equilibrium is now achieved at a new level of nominal wage W1 and employment (N1). This leads to a decline of output level due to the decrease in labour productivity, from Y0 to Y1. The vertical aggregate supply curve shifts to the left. The increase in the nominal wage ( W 0 ¿ W 1 ¿ outweighs the increase in the price ( P o ¿ P 1 ¿ . Therefore, the net effect on the real wage will be increasing. In short, increasing marginal income tax rate affects the output level. FIGURE 5.1 A There are sometimes special case of full employment happening in real world. In this context, we would still be focusing on the marginal tax rate as discussed from the diagram previously. When employment has reached its full capacity, a cut in marginal tax rate shall impose effect on labour productivity instead of labour supply curve. A cut in the marginal tax rate will increase the disposable income available to every household. Suppose an increase in income will increase the labour supply, but at full employment where no excess or extra
labour will be demanded, labour productivity will be increasing. Labour who gain more income will be willing to work more efficiently thus increasing the whole labour productivity. The figure above illustrates the effect of increasing in labour productivity. Initially, equilibrium in labor market achieve when ND 0 = NS 0 at point E 0 with real wage of w 0 p and price level at P 0. Employment will be at N 0 and produce output at Y 0 . An equilibrium is achieved with price = P 0 and output = Y 0. When there is increase in labour productivity, production function shift from Y 0 = f ( ´ K ,N ¿ to Y 1 = f ( ´ K ,N ¿ . Labor demand increase shift ND curve from ND 0 to ND 1 (because of W P =MPN). New labor equilibrium achieves at NS 0 =ND 1 at point E 1 . Real wage increases from w 0 p to w 1 p as the willingness for the worker to work increase. Employment rate increase from N 0 to N 1 . Aggregate supply increase shift AS curve from AS 0 to AS 1 . The price drop from P 0 to P 1 . Output increase from Y 0 to Y 1. In short, increasing marginal income tax rate affects the output level. FIGURE 5.1 B This is the same source of financing which to increase the marginal income tax rate, but having the nominal wage instead of real wage in the labour market equilibrium. The figure above illustrates the effect of increasing in labour productivity. Initially, equilibrium in labor market achieve when ND 0 = NS 0 at point E 0 with nominal wage of W 0 and price level at P 0. Employment will be at N 0 and produce output at Y 0 . An equilibrium is achieved with price = P 0 and output = Y 0.
When there is increase in labour productivity, production function shift from Y 0 = f ( ´ K ,N ¿ to Y 1 = f ( ´ K ,N ¿ . Labor demand increase shift ND curve from Nd0 (P0 x MPN) to Nd1 (P1 x MPN) (because of W=MPN×P). New labor equilibrium achieves at NS 0 =ND 1 (P1 x MPN) at point E 1 . Nominal wage increases from W 0 to W 1 as the willingness for the worker to work increase. Employment rate increase from N 0 to N 1 . | 789 | 3,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.895271 |
https://sciencedocbox.com/Physics/98661509-Investigating-the-significance-of-a-correlation-coefficient-using-jackknife-estimates.html | 1,582,387,789,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145708.59/warc/CC-MAIN-20200222150029-20200222180029-00058.warc.gz | 543,791,424 | 28,804 | # Investigating the Significance of a Correlation Coefficient using Jackknife Estimates
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1 Iteratioal Joural of Scieces: Basic ad Applied Research (IJSBAR) ISSN (Prit & Olie) Ivestigatig the Sigificace of a Correlatio Coefficiet usig Jackkife Estimates Athoy Akpata a, Idika Okorie b a,b Departmet of Statistics, Abia State Uiversity Uturu,Nigeria a b Abstract Ofte i Applied statistics, populatio parameters are ot kow ad could be iferred usig the available sample data ad this is the uderpiig of statistical iferece. Resamplig techique such as jackkife offers effective estimates of parameters ad its asymptotic distributio. I this paper, we preset the jackkife estimate of the parameters of a simple liear regressio model with particular iterest o the correlatio coefficiet. This procedure provides a effective alterative test statistic for testig the ull hypothesis of o associatio betwee the explaatory variables ad a respose variable. Keywords: Jackkife; simple liear regressio; correlatio coefficiet; ols estimates; bias. 1. Itroductio After estimatio of parameters i applied statistics it is always crucial to assess the accuracy of the estimator by its stadard error ad costructio of cofidece itervals for the parameter [1]. Queouille i 1956 developed a cross validatio procedure kow as jackkife (leave-oe-out procedure) for estimatig the bias of a estimator [2]. Two years later this method was further exteded by Joh Tukey to estimate the variace of a estimator ad the ame Jackkife was coied for this cross validatio method [3] * Correspodig author. address: 441
4 Iteratioal Joural of Scieces: Basic ad Applied Research (IJSBAR)(2015) Volume 22, No 2, pp ,2,3,, ad evaluatig θ ols (J) ad ρ x,y (J) the least squares estimates based o the remaiig observatios [10]. The estimates of θ J ad ρ J, bias ad variace usig the pseudo values θ Ji ad ρ x,y(ji) are θ J = θ Ji (5) With bias bias = θ ols θ Ji (6) Or more succictly bias = θ ols θ J (7) Ad the variace var θ J = θ Ji θ J 2 ( 1) (8) Also, ρ x,y(j) = ρ x,y(ji) (9) With bias bias = ρ x,y ρ x,y(ji) (10) Or bias = ρ x,y ρ x,y(j) (11) Ad variace var ρ x,y(j) = ρ x,y(ji) ρ x,y(j) 2 ( 1) (12) 2.1 Algorithm for Jackkifig Simple Liear Regressio Model Steps: 444
6 Iteratioal Joural of Scieces: Basic ad Applied Research (IJSBAR)(2015) Volume 22, No 2, pp Table 2: ols Estimates S/N θ (0)Ji θ (1)Ji ρ x,y(ji) θ (0)J θ (1)J ρ x,y(j) SE θ (0)J SE θ (1)J SE ρ x,y(j) Table 3: Compariso betwee ols ad Jackkife ols Estimates Estimates ols Jackkife Bias θ (0) SE θ (0) θ (1) SE θ (1) ρ x,y SE ρ x,y Testig the sigificace of the correlatio coefficiet We shall proceed to test the sigificace of the correlatio coefficiet at 5% level of sigificace as follows: H 0 : ρ x,y = 0 H 1 : ρ x,y 0 446
### Properties and Hypothesis Testing
Chapter 3 Properties ad Hypothesis Testig 3.1 Types of data The regressio techiques developed i previous chapters ca be applied to three differet kids of data. 1. Cross-sectioal data. 2. Time series data.
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Chapter 10 Variace Estimatio 10.1 Itroductio Variace estimatio is a importat practical problem i survey samplig. Variace estimates are used i two purposes. Oe is the aalytic purpose such as costructig
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### Topic 9: Sampling Distributions of Estimators
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### This chapter focuses on two experimental designs that are crucial to comparative studies: (1) independent samples and (2) matched pair samples.
Chapter 9 & : Comparig Two Treatmets: This chapter focuses o two eperimetal desigs that are crucial to comparative studies: () idepedet samples ad () matched pair samples Idepedet Radom amples from Two | 8,438 | 30,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-10 | latest | en | 0.647266 |
https://www.storemaven.com/glossary/app-store-a-b-testing-statistics/ | 1,721,389,579,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00130.warc.gz | 848,798,867 | 26,023 | # A/B Testing Statistics
A/B testing statistics refers to the statistical model used in the conduction of an A/B test.
## What are A/B Testing Statistics
A/B testing statistics refers to the statistical model used in the conduction of an A/B test (controlled experiment) which in the App stores is the comparison in performance between two variations of an App store page. An A/B test will be used to approve or disprove a hypothesis by testing just a sample of the entire population in the live store, before using the observations collected to predict with a reasonable level of accuracy how the entire population in the live App stores will behave.
Every statistical model has a number of pre-requisites (test parameters) that need to be met in order to conduct a reliable test that shows which iteration performs better.
There are three A/B testing statistical methods that can be used in different ways. The first is the ‘frequentist’ approach which ignores any previous findings or knowledge from similar tests, using only data from the current experiment. ‘Sequential’ testing doesn’t limit sample size so the data is evaluated as it’s collected and testing can be stopped once enough data is collected. The third is the ‘Bayesian’ method of testing, used by third-party A/B testing platforms like Storemaven. This method works on the statistical theory that it’s possible to calculate the conversion rate probability of a variation based on having a model that constantly evolves and is taking into consideration the incoming data throughout the test. This results in a more accurate model that allows you to reach a conclusion with a high degree of certainty.
## Why A/B Testing Statistics are Important
Statistics is vital to the process of planning, running and evaluating A/B tests.
Simply put: failing to use the right statistical model when A/B testing would be a waste of time and money. The effective implementation of A/B testing statistics should translate to an increase in installs for the tested audience once the better performing page is applied to the entire population in the live App stores.
## A/B Testing Statistics and ASO
ASO teams need to be aware of any statistical model being used in A/B testing to ensure that it’s one that can be trusted to best serve the purpose of the test. It must produce results that can be implemented with confidence in the real App Store or Google Play Store and provide tangible benefits. Using the wrong statistical method that isn’t suited to the metrics being measured could result in having to endure the frustrating process of running tests, implementing results but without gaining any usable results initially hoped for.
## Get the ultimate ASO conversion rate optimization eBook
Everything you need to know about icons, screenshots, videos, app reviews & ratings, localization, and seasonality.
### Related Terms
• Mobile Growth
Mobile growth is the science of growing and gaining ground in reaching and retaining users in the App Store or Google Play.
• App Store Web Referral
App Store web referral is a traffic source which includes all visitors that arrive at an app listing from an external source (mobile web), most commonly the iOS Safari browser. | 620 | 3,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-30 | latest | en | 0.918898 |
https://www.coursehero.com/file/6685675/Wek1PPT2304/ | 1,516,655,084,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891539.71/warc/CC-MAIN-20180122193259-20180122213259-00469.warc.gz | 901,732,713 | 24,396 | Wek1PPT2304
# Wek1PPT2304 - 1 ADM2304 M N& R January 2012 Statistics...
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Unformatted text preview: 1 ADM2304: M, N & R January 2012 Statistics for Management II DMS 1130 Section R: Mondays-19:00-22:00 Section N: Tue 16:00-17:30, Thu-14:30-16:00 Section M: Wed 13:00-14:30, Fri 11:30-13:00 Text Book: Business Statistics, Canadian Ed. Sharpe et al Prof.: Dr. Suren Phansalker Office : DMS 5142 Office Hours: Tue 14:00-15:30 2 Lecture#1 Sample Mean Distributions: ‘t’ & ‘Z’ Distributions Dr. Suren Phansalker • Central Limit Theorem (CLT): As seen before, P-S. Laplace proved the main assertion of the CLT. However, in its modern form, it does have different forms. The following three major cases bring out the variations. • Case I of CLT: If a large sample with size, n ≥ 30, is drawn from any much larger population ‘X’, of unknown distribution, then: But, and where μ , σ , and σ 2 are the population parameters. Then simply written: It sometimes is written as: ( 29 ) ( ), ( ~ 2 X X E N X σ μ = ) ( X E n X 2 2 ) ( σ σ = n N X 2 , ~ σ μ n N X σ μ , ~ 3 • Case II of CLT: If the Population RV, ‘X’ is itself Normally Distributed, then if ‘ σ ’ is known, then for any sample size, (even ‘n’ < 30): or • Case III of CLT: If the Population RV, ‘X’ is itself Normally Distributed, then if ‘ σ ’ is unknown and must be estimated by ‘s’, the sample standard deviation, then: or • Special Condition for Case III of CLT: If ‘n’, the sample size is fairly large (n ≥ 30 or n ≥ 120), then: or In other words, the ‘t’ Distribution and ‘Z’ Distribution become almost equal. n N X 2 , ~ σ μ - n s t X n 2 1 , ~ μ n N X σ μ , ~ - n s t X n , ~ 1 μ ≈ - n s N n s t X n 2 2 1 , , ~ μ μ ≈ - n s N n s t X n , , ~ 1 μ μ 4 • Some Examples of the CLT: • Example 1 for Case I: A large sample of size 100 is taken from a population of marks on a...
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Ask a homework question - tutors are online | 745 | 2,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-05 | latest | en | 0.79355 |
https://www.reddit.com/r/MachineLearning/comments/dfevc8/discussion_sota_of_esbased_rl_algorithms/ | 1,571,826,003,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987833089.90/warc/CC-MAIN-20191023094558-20191023122058-00015.warc.gz | 1,046,636,580 | 29,039 | ×
[–] 3 points4 points (0 children)
I wrote the only known python implementation of a cooperative coevolutionary algorithm called "CoSyne" which claims to be state of the art for continuous space RL (and claims to beat ES and other less fancy neuroevolution methods)
Seemed good with my single threaded implementation and would be super neat if someone GPU parallelized it. I wouldn't be surprised if these methods (with novalty search) are objectively better than other backprop based RL techniques.
https://github.com/Hellisotherpeople/Python-Cooperative-Synapse-NeuroEvolution
[–] 4 points5 points (4 children)
So, has anyone at GECCO solved ALE yet?
[–] 1 point2 points (0 children)
After some quick googling:
https://arxiv.org/pdf/1806.05695.pdf
[–][S] 0 points1 point (2 children)
did someone create a benchmark out of it where we can just compare algorithm performance? I.e. with fixed model-architecture for comparison. We could introduce it at a gecco workshop for real-life problems. Otherwise it is probably uninteresting for many in the community, as there is a strive to have comparable benchmarks for algorithms.
[–] 0 points1 point (1 child)
ALE is a benchmark. I'm not sure what more you want, or what you mean by 'fixed model-architecture'. And if it is 'uninteresting' that is the only reason, then no one is going to be impressed by people from GECCO coming by and scoffing and claiming how inferior and inefficient the RL techniques are and how they could do so much better if ALE weren't so boring; well, 'Hic Rhodus, hic saltus'.
[–][S] 0 points1 point (0 children)
ALE is AFAIR only the ATARI-Games as an RL-Task. I.e. it is the user who has to provide the policy-model. Therefore, it is not a proper benchmark for the /algorithms/ since the chosen /model architecture/ changes the task(and it is not about comparing algorithms, but models).
I know as a core DL researcher that this distinction between algorithm and model is quite foreign for many DL people, who seem to not being able to distinguish this. But for many people in the ES community this makes it an improper benchmark for what they are interested in: objectively benchmarking how well an algorithm (read: their own algorithm) works on a wide variety of objective functions. For this, the benchmark must be comparable and to be interesting as a real-world problem, the chosen instance of the problem must be acceptable in practice. Since the ML/DL/RL community is so bad at coming up with a consensus on what model actually works for certain tasks, this is just not going to happen.
Here is for example what the community would see as proper benchmark: https://coco.gforge.inria.fr/
But hey, if it is so important to you: name tasks and model-architectures, maybe even github code and i set up the benchmark. I have soon a week downtime to do it.
[–] 1 point2 points (2 children)
Guided-ES from Google is also super interesting. Uses surrogate gradients to speed convergence. Have used it, can confirm it works as advertised for many types of problems.
[–][S] 3 points4 points (1 child)
This is actually a negative example regarding what i am talking about. The Vanilla-ES they use is one that would not be a viable option in actual ES literature. E.g. the last reference above introduces a standard implementation (eq (3), (4), (5) and (6)). The google version uses a fixed variance, which is known to introduce all kinds of problems. It also does not use the natural gradient for updating the mean. And most large-scale ES have a mechanism to learn at least a rank-1 update.
But yes, in general it is sound to introduce gradient information, even though i am not a friend of how they do it and also not a fan of the experiments.
//Edit just to show how bad the normal gradient is: when you make the sampling distance shorter, your steps in the mean increase in length. this does not make any sense. the natural gradient always leads to steps in the mean that are as long as the std-dev up to a constant.
[–] 0 points1 point (0 children)
Interesting, thanks for the reply. I’ve been using it for some things, blissfully unaware of its theoretical limitations which you point out.
[–] 1 point2 points (4 children)
A bit of a side question, but do you know of any python implementations of any of those algorithms that don't require MPI?
I only read the last reference (https://dl.acm.org/citation.cfm?id=3321724) and don't see any mention of source code or libraries.
[–][S] 2 points3 points (0 children)
standard CMA-ES:
https://github.com/CMA-ES/pycma
The implementation of the last paper is available only in C++
[–] 1 point2 points (1 child)
I wrote a kind of shitty version of CMA-ES here which is basically a translation of the Matlab code on wikipedia
[–] 0 points1 point (0 children)
/r/HelperBot_ Downvote to remove. Counter: 283433. Found a bug?
[–][S] 1 point2 points (0 children)
I have implemented a version of the code for python. Might include bugs, but works on a simple benchmark:
https://github.com/Ulfgard/fCMApy
[–] 1 point2 points (2 children)
Thank you for those references, I'll be reading.
For anyone catching up, I assume "the OpenAI paper" is Salimans et al., 2017. IMO well worth reading if you are coming from the RL/DL side of things, and in particular I would prefer (re-)reading it over later, similarly hyped papers.
There is also a recent overview article in Nature, Designing NNs through neuroevolution (Stanley et al., 2019) which I found very interesting. (I have not digged very deep into this field yet.)
Concerning CMA-ES, yes it deserves to be better known. But, last thing I know, it stops being useful above ~10'000 parameters. Has this changed?
[–] 1 point2 points (1 child)
After some skimming, apparently there is a lot of research going on to change this.
But, what does "large-scale" mean? Quotes from your references: "[...] we introduce a new large-scale testbed with dimension up to 640" (Varelas); "for d=1'000-16'000" (Krause). I can't seem to get the number from Salimans, but Atari has been played with anything between 33‘000 and 4 million parameters.
You say, "The OpenAI paper [...] algorithm would not be even considered a valid baseline anymore." But how sure are we about that? Do those new alternatives really scale up all the way to the DL Atari setup? Has anyone tried?
(PS: not a researcher myself, just curious, trying to stay up-to-date. And doing some small experiments on the side.)
[–][S] 2 points3 points (0 children)
There are implementation of large-scale variants with cost of a single iteration in O(n), e.g. Krause et.al is but essentially all large-scale variants are. Most papers, however evaluate on smaller scale since convergence-experiments require O(n) iterations and quickly become prohibitively expensive. This is because the convergence rate of all ES is O(1/n) in the number of parameters. i.e. an algorithm that requires k iterations to solve the simple quadratic function in lets say 100 dim to some precision epsilon, will require 10000*k iterations for a million parameters. But there are papers which do that, e.g. the LM-CMAES was evaluated in 200k parameters.
regarding high dimensionality effects there is not much difference between n=1000 and n=100000, the statistically interesting phase transitions in behaviour of the algorithms happen way beforehand, so if an algorithm is evaluated on 16k parameters, its parameters are probably tuned with the correct limit behaviour and it should work on arbitrary dimensions.
And it is not that so much missing to make the OpenAI algorithm a valid baseline:
1. some normalization of the function values to be invariant to at least a subset of monotonous transformations, so that just scaling the fucntion-values will destroy the convergence behaviour of your algorithms. This alone allows a good ES to use a good set of standard parameters that work on almost all functions. Why go for less?
2. Using the natural gradient. This is literally just multiplying with sigma^2 in the openAI paper so that sample-distance and step-length are on the same scale
3. any way to adapt sigma.
[–] 0 points1 point (2 children)
I’ll follow up with another silly question.
Why ES, over say genetic algorithms (GA) or probabilistic versions, i.e estimation distribution algorithms?
[–][S] 1 point2 points (1 child)
GA are discrete. EDA do not work. EDA with Gaussian Distributions will lose variance at each and every step due to the selection operator that removes samples from the set. Known since the 90s. Also ES beat the BBOB, BBComp and other benchmarks reliably.
But this is a little besides the point since EDAs are not used in RL.
[–] 0 points1 point (0 children)
GA’s are discrete?
What do you mean by this out of curiosity? Considering GA’s work well on both real, and discrete variables, or mixed.
Simulated binary crossover (SBX) goes the other way i believe as well.
[–] 0 points1 point (1 child)
This is a weird post. Bashing the Salimans paper as "not a valid baseline" is short sighted: up until that point, no one had been able to scale ES variants to that number of parameters (~4m) with any hope of convergence in a reasonable time frame. That it worked on a hard baseline experiment set, Atari, is important for growing interest in the field. It's clearly of value in and of itself as Kenneth O. Stanley's lab used a variation of the scale out genetic algorithms in the same space:
Such, Felipe Petroski, et al. "Deep neuroevolution: Genetic algorithms are a competitive alternative for training deep neural networks for reinforcement learning." arXiv preprint arXiv:1712.06567 (2017).
On one hand, I agree that the EA and DL/RL groups don't often talk with each other, but suggesting a categorical disdain for researchers for not knowing the other fields goes both ways: ES papers have been rejected from ICML and CVPR for much the same reason.
[–][S] 0 points1 point (0 children)
The first large-scale variant of CMA-ES was
Ros, R., & Hansen, N. (2008, September). A simple modification in CMA-ES achieving linear time and space complexity. In International Conference on Parallel Problem Solving from Nature (pp. 296-305). Springer, Berlin, Heidelberg.
So the achievement of the paper was trying ES on 4m parameters. but the algorithm they used is shabby. And this is my whole point. As i said: "For the first paper reintroducing this, it is okay"
[–] -2 points-1 points (2 children)
this might seem like a silly question.
what does es mean?
oh it means expert system.
[–] 3 points4 points (1 child)
Evolutionary strategies
[–][S] 0 points1 point (0 children)
correct | 2,560 | 10,705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-43 | latest | en | 0.934888 |
https://en.wikipedia.org/wiki/Talk:Gompertz%E2%80%93Makeham_law_of_mortality | 1,505,959,019,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687592.20/warc/CC-MAIN-20170921011035-20170921031035-00701.warc.gz | 659,127,312 | 11,145 | # Talk:Gompertz–Makeham law of mortality
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## Removal of category
This article was tagged with the category academic geography as the law is used in population geography and demography. AlexD 11:56, 10 August 2006 (UTC)
## Categories
Sure, and Demography remains as a category.
For example, Meteorology is a topic within Academic Geography - which means you don't have to include "clouds", "lightning", "floods", "typhoons", etc. Anyone can look at Academic Geography, then look at Meteorology, then dig into Meteorology-specific topics.
Same here. Start at Academic Geography, then Demography, then to subsections like birth rates, life tables, HIV/Aids, etc. Otherwise things could really get out of hand. Demography is included, and that covers all things demographic.
Thanks
## equation
Am I wrong in thinking that the equation for this is something like p(s) = α + (1-α)e(s- 100)/β, where s=age of death, α=accident rate & β=death rate, and this should be on the page somewhere? Bueller 007 (talk) 14:31, 9 January 2008 (UTC)
• Can't someone explain this in laymans terms? Even the equations have no explanation of the variables involved. --71.245.164.83 (talk) 01:43, 22 November 2010 (UTC)
## New Findings
A recent news report appears to set at naught some assertions in this article: "Leonid A. Gavrilov and Natalia S. Gavrilova of NORC at the University of Chicago, formerly known as the National Opinion Research Center, said the findings contradict a long-held belief that the mortality rate of Americans flattens out after age 80." Lacking familiarity with this field, I leave any needed edit to better minds than mine. — Preceding unsigned comment added by Orthotox (talkcontribs) 21:56, 19 February 2012 (UTC) | 619 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | latest | en | 0.957308 |
https://www.education.com/lesson-plans/second-grade/data-and-graphing/ | 1,627,115,098,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150134.86/warc/CC-MAIN-20210724063259-20210724093259-00116.warc.gz | 747,142,878 | 28,323 | 487 filtered results
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Bar Graphs: Interpreting Data
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Bar Graphs: Interpreting Data
In this lesson, students will make bar graphs and interpret data using real-life data from other students. They will get practice writing and answering survey questions.
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Let's Collect Data!
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Let's Collect Data!
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Making Bar Graphs: Graph It!
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Making Bar Graphs: Graph It!
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Math + Love = Valentine's Day Bar Graph
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Math + Love = Valentine's Day Bar Graph
What's more romantic than a bar graph on Valentine's Day? In this Valentine's Day lesson plan, students will use information from a data set to create their own tally charts and bar graphs and then analyze these graphs to answer questions.
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Collect, Organize, Display, Analyze
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Collect, Organize, Display, Analyze
In this lesson, your students will collect and organize data to make a beautiful bar graph! They will also practice analyzing the graphs with the information they have.
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How Many More?
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How Many More?
Subtraction is a lot more than just “take away.” In this lesson, students develop an understanding of subtraction as comparison by creating and interpreting data from bar graphs.
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How Big is the Playground?
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How Big is the Playground?
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Line Plots: Representing the Length of Classroom Items
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Line Plots: Representing the Length of Classroom Items
In this lesson, your students will measure the lengths of items and then make a line plot to show the measurement data. They will get hands-on by measuring and surveying the class.
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Understanding Line Plots
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Understanding Line Plots
This lesson provides students with the opportunity to critically think about data. Use this lesson alongside Line Plots: Representing the Length of Classroom Items or as a stand alone lesson.
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Sub plans just got a whole lot easier! In this second grade, week-long sub packet, subs can supercharge learning with "All About Me" themed lessons and activities. The “All About Me” theme is the perfect opportunity to reinforce foundational skills like understanding character traits, taking and interpreting meaningful data, and writing from the heart.
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Anansi and the Turtle and Me!
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Anansi and the Turtle and Me!
Anansi invites his friend Turtle to dinner at his house ... but does he treat him like a friend? In this reading comprehension lesson, your class practices questioning, retelling, and identifying the moral in "Anansi and the Turtle."
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Composing and Decomposing
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Composing and Decomposing
Use this lesson to introduce composing and decomposing as part of your students’ addition and subtraction toolkits.
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Stop and Jot!
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Stop and Jot!
Use this interactive activity to help students learn blends and digraphs and the active reading strategy stop and jot! This lesson can be used as a stand alone activity or a support lesson.
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Producers and Consumers!
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Producers and Consumers!
Use this lesson to have your students learn about producers and consumers by pretending to be one.
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Summaries and Predictions
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Summaries and Predictions
Teach your second graders how to write a prediction and summarize book chapters in this exciting lesson!
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101 and Out!
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101 and Out!
Get the dice out and help students get targeted practice in adding 1 and 10! In this fun place value game, students keep adding to try to get to a maximum score of 100. But be careful. If you go over 100 you’re out!
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Problem Solving and Cooperation
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Problem Solving and Cooperation
Group work can build cooperation and problem-solving skills inside and outside of the classroom! In this lesson, students will discuss cooperation and practice using cooperative techniques in human knot and tower-building activities.
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Compare and Contrast Clocks
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Compare and Contrast Clocks
Help students develop compare and contrast skills while they examine digital and analog clocks. This lesson can be used independently or in conjunction with the lesson Time to Tell Time: Showing and Writing Time.
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Cooperative Commerce and Geography
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Cooperative Commerce and Geography
Show students what buying and selling looked like before credit cards! This lesson explores commerce with the book *Ox-Man Cart* and a United States regional food production map. Students will answer questions and draw conclusions.
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How and Why Stories
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How and Why Stories
In this lesson, students have a great time using their imaginations to collaborate and create their own story following the "how and why" style of common folktales.
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This lesson helps students learn about asking and answering questions about a text. It also exposes them to valuable lessons about trying to figure out their dreams and not giving up along the way.
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Appreciating Diversity and Differences
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Appreciating Diversity and Differences
In this lesson, students will hear the story *Lovely* and work in small groups to create a book or a symbol of appreciation for others' differences. Students will consider how appreciating differences supports community and relationships. | 1,270 | 6,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-31 | longest | en | 0.884818 |
https://www.thestudentroom.co.uk/showthread.php?t=106984 | 1,534,645,265,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221214538.44/warc/CC-MAIN-20180819012213-20180819032213-00572.warc.gz | 1,035,034,137 | 40,178 | You are Here: Home >< Maths
# p5 june 2004 help ! watch
1. Part B
can someone explain plz ?
Attached Files
2. Doc1.doc (53.0 KB, 110 views)
3. Sub the equation of the normal back into xy = c²:
=> t²x² + (c/t)x - (ct³)x = c²
t³ + (c-ct4)x - tc² = 0
Now use the quadratic equation to solve for x:
x = {(ct4-c) ± √[(c-ct4)² + 4c²t4] } / 2t³
If you simplify the bit in the root, you get: √[c²(t8+2t4+1)]
= √[c²(t4+1)²]
= ct4 + c
So, x = [ct4-c + (ct4+c)]/2t³ or x = [ct4-c - (ct4+c)]/2t³
After simplification,
x = ct or x = -c/t³
x = ct is for the point P, therefore x = -c/t³ is for the point Q.
When x = -c/t³ , (-c/t³)y = c²
y = -ct³
Therefore, Q( -c/t³ , -ct³ )
4. thanks (bloody hell thats long ) do u know wat u need to get A grade on that paper ???? out of 75 ?????
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Top tips from students who have already aced their exams | 526 | 1,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-34 | latest | en | 0.838623 |
https://www.physicsforums.com/threads/trig-derivatives.209054/ | 1,508,431,981,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823350.23/warc/CC-MAIN-20171019160040-20171019180040-00128.warc.gz | 1,001,151,276 | 15,140 | # Trig derivatives
1. Jan 15, 2008
### Wonderballs
1. The problem statement, all variables and given/known data
$$\int$$e$$^{2x}$$sin(3x)dx
2. Relevant equations
integration by parts formula
3. The attempt at a solution
= 9/10e$$^{2x}$$(sin(3x)-1/3cos(3x))
I don't think I am taking the derivative of sin(3x) correctly
thanks for taking a look :)
2. Jan 15, 2008
### rock.freak667
$$\int e^{2x}sin(3x)dx$$
u=e^2x so that du=2e^2x dx
dv=sin(3x)dx ; v=-1/3 cos(3x)
but either way you would have to do integration by parts twice so I am thinking that is not the answer(BUT I can be wrong as I did not really work it out)
3. Jan 15, 2008
### silver-rose
Hint :
What you do is you integrate it by parts two times.
A multiple of the original integral will appear in the answer.
4. Jan 15, 2008
### Wonderballs
I tried it with u = e^2x ; dv = sin3xdx
I think i got the multiple and im stuck here:
= -1/3e$$^{2x}cos(3x) + 4/9$$$$\int$$sin(3x)e$$^{2x}$$
5. Jan 15, 2008
### rock.freak667
you are supposed to get
$$\int e^{2x}sin(3x)dx=\frac{-e^{2x}}{3}cos(3x)+ \frac{2}{3}\int e^{2x}cos(3x)dx$$
then do integration by parts again
6. Jan 15, 2008
### Wonderballs
I got it, thanks guys
1/13 e^2x (2sin3x - 3cos3x) = integral | 464 | 1,247 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-43 | longest | en | 0.802323 |
https://tex.stackexchange.com/users/235725/ver%C3%B3nica-rmz | 1,624,117,393,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487648373.45/warc/CC-MAIN-20210619142022-20210619172022-00117.warc.gz | 497,331,921 | 24,823 | 299
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$\color{black}{\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ 2\left( x\right )}{x^2} \mathrm{d}x}$
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Posts 1 | 561 | 1,400 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-25 | latest | en | 0.406332 |
http://mathhelpforum.com/trigonometry/40034-simplifying-trig-expression.html | 1,481,124,629,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542213.61/warc/CC-MAIN-20161202170902-00199-ip-10-31-129-80.ec2.internal.warc.gz | 172,243,760 | 10,377 | # Thread: Simplifying a trig expression
1. ## Simplifying a trig expression
[(1 / (cos^2 (x)) - x]*sin(x)
Can someone help me simplify this? Thanks,
Kim
2. Probably not. Why do you think it possible?
3. Originally Posted by Kim Nu
[(1 / (cos^2 (x)) - x]*sin(x)
Can someone help me simplify this? Thanks,
Kim
You've done a good job trying to line the brackets up, but they are unfortunately not correct. The "(" to the left of the "1" is never closed.
4. Originally Posted by Kim Nu
[(1 / (cos^2 (x)) - x]*sin(x)
Can someone help me simplify this? Thanks,
Kim
I cannot see any simplification other than
$\bigg[\frac{1}{\cos^2(x)}-x\bigg]\cdot\sin(x)=\frac{\sin(x)}{\cos^2(x)}-x\cdot\sin(x)=\tan(x)\cdot\sec(x)-x\cdot\sin(x)$ | 232 | 735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2016-50 | longest | en | 0.892841 |
https://codereview.stackexchange.com/questions/234152/sum-of-primes-under-a-certain-number | 1,713,741,092,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818067.32/warc/CC-MAIN-20240421225303-20240422015303-00170.warc.gz | 152,764,297 | 46,199 | # Sum of primes under a certain number
I have tried two different methods to find the sum of primes under a certain number. However both of these methods end up taking a very long time for large numbers. How can I optimise either of these two methods?
Method 1 (Sieve of Eratosthenes):
import time
import math
primes_in_range = [2]
n = int(input("Enter the range of your sieve"))
start = time.time()
for x in range(3, n+1, 2):
primes_in_range.append(x)
d = 3
while d <= math.sqrt(n):
if d in primes_in_range:
for j in range(2 * d, n + 1, d):
if j in primes_in_range:
primes_in_range.remove(j)
d += 2
print(sum(primes_in_range))
end = time.time()
print(end - start)
Method 2:
import time
import math
def is_prime(n):
if n>1:
for i in range (2, int(math.sqrt(n)) + 1):
if (n % i) == 0:
return False
else:
return True
else:
return False
n = int(input("Enter the range of your sieve"))
start = time.time()
total = 2
i = 3
while i < n:
if is_prime(i):
total += i
i += 2
print(total)
end = time.time()
print(end -start)
Method 1 (Sieve of Eratosthenes) is definitely the faster method to use.
Starting point on my computer, summing primes less than 100,000 takes 39.916 seconds.
## List to Set
The problem is basically this line:
primes_in_range = [2]
This creates a list, to which you add all odd numbers above 1, and then repeatedly search for existence of a number within (d in primes_in_range), then for all multiples of that number, search for the multiple (j in primes_in_range), and then search for it a second time in the process of removing it (primes_in_range.remove(j)). Searching a list is an $$\O(n)\$$ operation. When you do it approximately once for every number in the list, your algorithm devolves to $$\O(n^2)\$$.
On top of this, you’re computingmath.sqrt(n) many, many times - a relatively expensive operation - yet the value does not change. You only need to compute it once.
The second item you can solve by computing the $$\\sqrt n\$$ once, and store it in a local variable.
For the first issue, there are a few possibilities. One, you could use a set, instead of a list. A set has approximately $$\O(1)\$$ lookup time, which is much faster than $$\O(n)\$$.
primes_in_range = { 2 }
Since sets are not ordered, you don’t "append()" items to it, you "add" them instead.
Time has now dropped to 0.020511 seconds
Since the time has dropped to under a second, let's increase to primes less than 10,000,000 for better time measurements.
Time for summing primes less than 10,000,000: 3.519 seconds
## Search to Remove
When removing, you are checking for the existence of the item in primes_in_range before removing it, since removing it when it is not present will raise an exception. The set class has a discard method, which will only remove an item if present. No need to check for it to be present ahead of time.
Time for summing primes less than 10,000,000: 3.453 seconds
## Removing Multiples
When you find a prime, d, you cross off multiples of that prime, starting at 2*d, and going 3*d, 4*d, 5*d, and so on. But when you determined 3 was a prime number, you crossed off all multiples of 3, so crossing off 3*d is a waste of time, for d > 3. Similarly, crossing off multiples of the prime 5 means crossing off 5*d for d > 5 is also a waste of time. The first multiple of d that hasn't been crossed off will be d*d, so you can remove starting at that point:
for j in range(d * d, n + 1, d):
Time for summing primes less than 10,000,000: 3.305 seconds
Similarly, since you are only considering odd prime candidates in your sieve, you don't need to cross off any even multiples. Starting at d*d, the next multiple you need to remove would be (d+2)*d, not (d+1)*d. Your step size can be 2*d.
for j in range(d * d, n + 1, 2 * d):
Time for summing primes less than 10,000,000: 2.105 seconds
## Loop Like a Native
Python has its own looping, which includes incrementing loop indexes. This native Python looping is usually faster than writing loops with their own index manipulation:
primes_in_range = {2}
primes_in_range.update(x for x in range(3, n+1, 2))
for d in range(3, math.isqrt(n) + 1, 2):
if d in primes_in_range:
primes_in_range.difference_update(j for j in range(d * d, n + 1, 2*d))
Here, constructing the set is done with a generator expression in primes_in_range.update(...), and removing prime multiples is done with a generator expression in primes_in_range.difference_update(...) which adds and removes entire swathes of items in a single operation:
Time for summing primes less than 10,000,000: 2.049 seconds
## Direct Indexing
However, a set is still an expensive memory structure to maintain. It requires hashing the values which are being added, creating bins to store the values in, rebinning as the set size changes, and so on. An array of flags is much more efficient, time-wise:
primes_in_range = [False] * n
primes_in_range[2] = True
for x in range(3, n, 2):
primes_in_range[x] = True
for d in range(3, math.isqrt(n) + 1, 2):
if primes_in_range[d]:
for j in range(d * d, n + 1, 2 * d):
primes_in_range[j] = False
total = sum(idx for idx, flag in enumerate(primes_in_range) if flag)
The first line, above, allocates an array of n items, each containing the flag value False. Then, indexing into the array is an $$\O(1)\$$ operation. Set the flag to True for all odd numbers above 3, and then proceed to set the appropriate flags to False based on the Sieve of Eratosthenes algorithm.
Time for summing primes less than 10,000,000: 1.398 seconds
## Bit Array
That last approach is a little inefficient when it comes to memory. It requires a list of n 28-byte items to store one True/False flag. These flags could be packed 8 to a byte, resulting in a 99.5% reduction in required memory! The bitarray package provides such an array of flags.
from math import isqrt
from bitarray import bitarray
def sum_of_primes_below(n):
primes_in_range = bitarray(n)
primes_in_range.setall(False)
primes_in_range[2] = True # 2 is a prime
primes_in_range[3::2] = True # Odd numbers starting at 3 are prime candidates
for d in range(3, isqrt(n) + 1, 2):
if primes_in_range[d]:
primes_in_range[d*d::2*d] = False # Reset multiples of prime candidate
return sum(idx for idx, flag in enumerate(primes_in_range) if flag)
Time for summing primes less than 10,000,000: 0.455 seconds
• He’s also only removing the odd multiples of x because the even ones are not there anyway. Saves a factor 2 again. Dec 17, 2019 at 13:25
• Nice work - that's something like 5 orders of magnitude faster than the original. Dec 18, 2019 at 9:14
If you're making a text interface, remember to put a prompt character with spaces like > so I know where to type. Even better would be to get an argument from sys.argv but I digress. Also, you should specify that specifically an integer greater than 1 is required, and raise an Exception otherwise.
After reading the pseudocode on Wikipedia your implementation is perfect. However your code is hard to read: include some comments. primes_in_range - in range of what? d - what is "d"? Since we have the ability to write variable names we should take advantage of it.
As for speed optimizations: in your first program you are storing the primes into a list. If you read the time complexity for Python collections, set is more efficient for deleting items - O(1) instead of O(n). After using a set instead of a list, I managed to significantly reduce the time it took for your code to run by more than 1000X.
In fact, after using a set instead of a list I was able to calculate primes up to 1,000,000 in 0.30 seconds, whereas using the list did not finish even after 20 minutes on my Core i7.
Your second solution has nothing horribly wrong with it, it's simply just a brute force algorithm. That's why it takes forever. Consider trying the Sieve of Atkin algorithm.
The problem with your sieve implementation is that the operations are not primitive operations but are rather time consuming.
You create an array of integers. Let's say n = 1 billion. You add 500 million integers. Each operation of adding an integer checks if there is enough space, increases the array size if needed, and adds a number. That’s slow but reasonably harmless.
Removing the non-primes is the killer. The harmless looking if j in primes_in_range visits all elements of the array until it finds the matching one. That can be tens of millions array elements checked for each multiple you are removing. And the remove(j) will find j again, remove it, and then fill the gap in the array by moving all the following items one position forward.
What you do instead is create an array of Boolean values, and set these Boolean to true or false, which will be a primitive and therefore fast operation.
An implementation in C or C++ would be expected to handle a few hundred million integers per second.
Measure the time; as long as it grows quadratically with n, you are still doing many non-primitive operations.
For your second algorithm which I suppose is slow, but not fatally slow like the first one, you can actually perform some brute force optimisations:
You are trying out whether n is divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10, etc. Now if you think about it, n cannot fail the test of divisibility by 4, 6, 8 or 10, because it would have been divisible by 2 already. n also cannot fail the division test for 9, 15, 21, 27 etc. because it would have been divisible by 3 already.
If you check first whether n is divisible by 2, 3 and 5, then you only need to check divisibility by other numbers that are not divisible by 2, 3 or 5. Only 8 out of every 30 consecutive numbers are not divisible by 2, 3, or 5, and that is the numbers 30k+1, 30k+7, 30k+11, 30k+13, 30k+17, 30k+19, 30k+23 and 30k+29. If you only check divisibility by these numbers, then only 8 out of 30 tests are needed. That will be about 4 times faster.
You can do even better by only checking divisibility by primes. Store all the small primes in an array, and only check divisibility by primes. If n > p^2 for the largest stored prime p, you have to find the next prime first.
And please make sure that your algorithm gives the correct result for n = 1, n = 0, n < 0. | 2,681 | 10,264 | {"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": 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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-18 | latest | en | 0.768548 |
https://whatsupwhimsy.com/quadratic-examples-solutions-40 | 1,675,074,316,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499816.79/warc/CC-MAIN-20230130101912-20230130131912-00583.warc.gz | 606,900,235 | 6,964 | The formula for a quadratic equation is used to find the roots of the equation. Since quadratics have a degree equal to two, therefore there will be two solutions for the
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For a quadratic equation a x 2 + b x + c = 0, the values of x that are the solutions of the equation are given by: x = − b ± b 2 − 4 a c 2 a. For the quadratic formula to work, we must always put
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In general, if α is a root of the quadratic equation ax 2 + bx + c = 0, a ≠ 0; then, aα 2 + bα + c = 0. We can also say that x = α is a solution of the
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How To Solve Quadratic Equations Using The
The quadratic formula is used to solve quadratic equations. Consider a quadratic equation in standard form: ax2 + bx + c = 0 a x 2 + b x + c = 0. You may also see the standard form called a general quadratic equation, or the general form. So
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Looking for a little help with your math homework? Check out our online calculation assistance tool! | 537 | 2,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-06 | latest | en | 0.919629 |
http://www.mathworks.com/matlabcentral/fileexchange/38318-gstd | 1,432,584,142,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928586.49/warc/CC-MAIN-20150521113208-00108-ip-10-180-206-219.ec2.internal.warc.gz | 573,734,437 | 8,963 | Code covered by the BSD License
### Highlights from gstd
5.0
5.0 | 1 rating Rate this file 7 Downloads (last 30 days) File Size: 2.74 KB File ID: #38318 Version: 1.6
# gstd
### Antonio Trujillo-Ortiz (view profile)
24 Sep 2012 (Updated )
Standard deviation of a grouped sample.
File Information
Description
In some scientific works, once the data have been gathered from a population of interest, it is often difficult to get a sense of what the data indicate when they are presented in an unorganized fashion.
Assembling the raw data into a meaningful form, such as a frequency distribution, makes the data easier to understand and interpret. It is in the context of frequency distributions that the importance of conveying in a succinct way numerical information contained in the data is encountered.
So, grouped data is data that has been organized into groups known as classes. The raw dataset can be organized by constructing a table showing the frequency distribution of the variable (whose values are given in the raw dataset). Such a frequency table is often referred to as grouped data.
Here, we developed a m-code to calculate the standard deviation of a grouped data.
One can input the returns or modified vectors n and xout containing the frequency counts and the bin locations of the hist m-function, in a column form matrix.
Normalizes Y by (N-1), where N is the sample size. This is an unbiased estimator of the standard deviation of the population from which X is drawn. Y = STD(X,1) normalizes by N and produces the square root of the second moment of the sample about its mean. STD(X,0) is the same as STD(X).
Variance calculation uses the formula,
S = SQRT(Sum(F*(MC - M)^2)/D)
where:
F = class frequency
MC = mark class
M = grouped mean
D = N - 1 or N, whether normalizes by n-1 or by n
N = sample size [=sum(F)]
--In orden to run it you must first download the m-file gvar at:
http://www.mathworks.com/matlabcentral/fileexchange/38281-gvar
Syntax: function y = gstd(x)
Inputs:
x - data matrix (Size of matrix must be n-by-2; absolut frequency=column 1, class mark=column 2)
n - [normalized by n-1] = 0 (default), [normalized by n] = 1
Outputs:
y - standard deviation of the values in x
Required Products Statistics and Machine Learning Toolbox
MATLAB release MATLAB 7.10 (R2010a)
04 Sep 2014 Wajahat
### Wajahat (view profile)
Excellent. A very quick and tool to calculate the Std.Dev. of grouped data or histograms.
Its strange that Matlab Statistics Toolbox does not include a function for this. Or at least I could not find one.
26 Sep 2012 Antonio Trujillo-Ortiz
### Antonio Trujillo-Ortiz (view profile)
Bug was removed.
Comment only
24 Sep 2012 1.1
It was added an appropriate format to cite this file.
26 Sep 2012 1.4
Text was improved.
26 Sep 2012 1.2
Text was improved.
26 Sep 2012 1.6
Text was improved. | 719 | 2,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | longest | en | 0.898085 |
https://www.physicsforums.com/tags/prime-numbers/ | 1,716,974,702,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00863.warc.gz | 807,800,923 | 26,941 | What is Prime numbers: Definition and 155 Discussions
A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. A natural number greater than 1 that is not prime is called a composite number. For example, 5 is prime because the only ways of writing it as a product, 1 × 5 or 5 × 1, involve 5 itself.
However, 4 is composite because it is a product (2 × 2) in which both numbers are smaller than 4. Primes are central in number theory because of the fundamental theorem of arithmetic: every natural number greater than 1 is either a prime itself or can be factorized as a product of primes that is unique up to their order.
The property of being prime is called primality. A simple but slow method of checking the primality of a given number
n
{\displaystyle n}
, called trial division, tests whether
n
{\displaystyle n}
is a multiple of any integer between 2 and
n
{\displaystyle {\sqrt {n}}}
. Faster algorithms include the Miller–Rabin primality test, which is fast but has a small chance of error, and the AKS primality test, which always produces the correct answer in polynomial time but is too slow to be practical. Particularly fast methods are available for numbers of special forms, such as Mersenne numbers. As of December 2018 the largest known prime number is a Mersenne prime with 24,862,048 decimal digits.There are infinitely many primes, as demonstrated by Euclid around 300 BC. No known simple formula separates prime numbers from composite numbers. However, the distribution of primes within the natural numbers in the large can be statistically modelled. The first result in that direction is the prime number theorem, proven at the end of the 19th century, which says that the probability of a randomly chosen number being prime is inversely proportional to its number of digits, that is, to its logarithm.
Several historical questions regarding prime numbers are still unsolved. These include Goldbach's conjecture, that every even integer greater than 2 can be expressed as the sum of two primes, and the twin prime conjecture, that there are infinitely many pairs of primes having just one even number between them. Such questions spurred the development of various branches of number theory, focusing on analytic or algebraic aspects of numbers. Primes are used in several routines in information technology, such as public-key cryptography, which relies on the difficulty of factoring large numbers into their prime factors. In abstract algebra, objects that behave in a generalized way like prime numbers include prime elements and prime ideals.
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I announce a playful competition :smile: Who can find the largest prime number with the programmed code? I found the number 2249999999999999981 with the Python code. I first tabulated the truth value of the prime numerosity of numbers smaller than 1.5 billion using Erasthonene's sieve, and then...
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How should I write an account of prime numbers in arithmetic progressions? Assuming this account should be in the form of an essay of at least ## 500 ## words. Should I apply the formula ## a_{n}=3+4n ## for ## 0\leq n\leq 2 ##? Can anyone please provide any idea(s)?
5. I Union of Prime Numbers & Non-Powers of Integers: Usage & Contexts
Is there a name for the union of {prime numbers} and {integers that are not powers of integers}? For example, we would include 2, 3, 5, 7, 11... And also 6, 10, 12... But we exclude 2^n, 3^n, ... and 6^n , 10^n , etc. What are some interesting contexts where this set crops up?
6. Determine (with proof) the set of all prime numbers
Proof: Let ## p ## be the prime divisor of two successive integers ## n^{2}+3 ## and ## (n+1)^{2}+3 ##. Then ## p\mid [(n+1)^{2}+3-(n^{2}+3)]\implies p\mid (2n+1) ##. Observe that ## p\mid (n^{2}+3) ## and ## p\mid (2n+1) ##. Now we see that ## p\mid [(n^{2}+3)-3(2n+1)]\implies p\mid...
7. A I think I discovered a pattern for prime numbers
I wrote a program that implements the pattern and finds the primes automatically. It worked up to 70 million then it crashed because program holds data in RAM so it can be fixed. It found all the primes up to 70 million and found no exception. I won't explain the pattern because its so...
8. If ## p ## and ## p^{2}+8 ## are both prime numbers, prove that....
Proof: Suppose ## p ## and ## p^{2}+8 ## are both prime numbers. Since ## p^{2}+8 ## is prime, it follows that ## p ## is odd, so ## p\neq 2 ##. Let ## p>3 ##. Then ## p^{2}\equiv 1 \mod 3 ##, so ## p^{2}+8\equiv 0 \mod 3 ##. Note that ## p^{2}+8 ## can only be prime for ## p=3 ##. Thus ##...
9. I Is Riemann's Zeta at 2 Related to Pi through Prime Numbers?
I just saw that one of the ways of calculating Pi uses the set of prime numbers. This must sound crazy even to people who understand it, is it possible that this can be explained in terms that I, a mere mortal can understand or it is out of reach for non mathematicians?
10. Find all prime numbers that divide 50
Proof: Note that all primes less than 50 will divide 50!, because each prime is a term of 50!. Applying the Fundamental Theorem of Arithmetic produces: Each term k of 50! that is non-prime has a unique prime factorization. Since 48, 49 and 50 are not primes, it follows that all primes...
11. What is the fastest algorithm for finding large prime numbers?
Dear PF Forum, Can someone help me with the algorithm for finding a very large prime number? In RSA Encryption (1024 bit? 2048?, I forget, should look it up at wiki for that), Private Keys is a - two prime number packet. Now, what I wonder is, what algorithm that the computer use to find those...
12. Cryptography problem involving prime numbers
Here is my attempt When we raise both sides to the power (p-1)/2, we get x^(p-1)= -1^[(p-1)/2](modp) Looking at p=3(mod4), the possible values of p are {3, 7, 11, 19, 23, 31...}. Putting these values of p into (p-1)/2 we get odd integers. {1, 3, 5, 9, 11, 15...}. So we have x^(p-1) =...
13. Consecutive integers and relatively prime numbers
Summary:: Interested in the history of the proof. Consecutive integer numbers are always relatively prime to each other. Does anyone know when this was proved? Was this known since Euclid's time or was this proved in modern times?
14. B Aren't There Any Formulas for Prime Numbers?
Hello all, We know that following formulas failed to produced all prime numbers for any given whole number ##n##: ##f(n) = n^2 - n - 41##, failed for ##n = 41~(f = 1681)## ##g(n) = 2^(2^n) + 1##, failed for ##n = 5~(g = 4,294,967,297)## ##m(n) = 2^n - 1##, failed for ##n = 67~(m =...
15. B Do Prime Numbers Follow a Pattern?
Hello everyone! I was going through a simple high school level mathematics book and got to the following question: n2 - n + 41 is a prime for all positive integers n. You're supposed to find a counter-example and prove the statement false. You could of course sit and enter different...
16. Prime numbers and divisibility by 12
Homework Statement Prove that if ##p## is a prime number and if ##p>5## then ##p^2-37## is divisible by ##12## Homework EquationsThe Attempt at a Solution So I think that the number ##p^2-37## should be expressed in a way that we can clearly see that it is divisible by 3 and by 2 twice...
17. I |Li(x) - pi(x)| goes to 0 under RH?
Extremely quick question: According to http://mathworld.wolfram.com/PrimeNumberTheorem.html, the Riemann Hypothesis is equivalent to |Li(x)-π(x)|≤ c(√x)*ln(x) for some constant c. Am I correct that then c goes to 0 as x goes to infinity? Does any expression exist (yet) for c? Thanks.
18. A Is it possible to locate prime numbers through addition only
I was reading an old thread about multiplying successive prime numbers adding 1 to obtain another prime number. I have worked with prime numbers for several years now and have developed what I best call a bi-linear advancement. It is an open-ended sieve of Eratosthenes. After many, many hours...
19. B Simple Question About Term(s) re: Fermat
Since Fermat, the French magistrate & noted mathematician, expounded : all odd integers ,(2n+1) where n≥0, are representable by the difference of TWO squares...
20. Analysis question -- Aren't all prime numbers not a product of primes?
Homework Statement I don't understand the lemma. Homework EquationsThe Attempt at a Solution Isn't all prime number not a product of primes? The lemma doesn't make sense to me... Moreover, if m=2, m-1 is smaller than 2, the inequality also doesn't make sense. Please help me
21. MHB Determine all prime numbers p such that the total number of positive divisors of A=p^2+1007 (including 1 and A) is less than 7 .
Determine all prime numbers $p$ such that the total number of positive divisors of $A = p^2 + 1007$ (including $1$ and $A$) is less than $7$.
22. MHB Set of 2015 Consecutive Positive Ints with 15 Primes
Is there a set of $2015$ consecutive positive integers containing exactly $15$ prime numbers?
23. How many ways one can put prime numbers to form 3 digit NIP?
Homework Statement as listed above the question is how many and which three digit NIP can be formed whit the use of prime numbers[/B]Homework Equations nothing currently trying to understand[/B]The Attempt at a Solution well i have found at least 168 primer numbers below 1000 i mean in the...
24. B Factoring a number and prime numbers?
A number can be factored into a product of its component factors A number can be factored into a product of its prime . But, What exactly is a prime number ? Prime numbers are numbers greater than 1 that are evenly divisible only by themselves and 1 Is it a number that can only be evenly...
25. B Are Prime Pairs Ending in 9 or 1 Derived from 20x2-1 Centred 10-Gonal Primes?
45x2+15x +/-1. ... 59;61 ,209;211 ,449;451 ,779;781 ... 45x2- 15x +/-1 ... 29;31 ,149;151 ,359;361 ,659;661 ... Derived from 20x2-1 can only have factors ending in the digit _1, or _9 .
26. How to Identify Prime Numbers Using Eratosthenes in C#?
Mod note: added code tags using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace ConsoleApplication1 { class Program { static void Main(string[] args) { List<int> number = new List<int>(1000); // int list for 1000 numbers...
27. Checking a proof of a basic property of prime numbers
Homework Statement Prove: If p is prime and m, n are positive integers such that p divides mn, then either p divides n or p divides m. Is anyone willing to look through this proof and give me comments on the following: a) my reasoning within the strategy I chose (validity, any constraints or...
28. B Connections with Prime numbers and Quantum Physics?
Hello I'm hard at work trying to find a pattern for the prime numbers and this keeps cropping up. To be honest though, to me it comes across like pseudo science. I mean I never really hear people talk about it. This seems an obvious thing to look into but I don't know anyone who does. Prime...
29. What properties do prime numbers exhibit?
Mod note: moved from a homework section What properties do prime numbers exhibit which can be used in proofs to define them? Like rational numbers have a unique property that they can be expressed as a quotient of a/b. Even numbers have a unique property of divisibility by 2 and thus they can be...
30. I Question about the gaps between prime numbers
Is there any prime number pn, such that it has a relationship with the next prime number pn+1 p_{n+1} > p_{n}^2 If not, is there any proof saying a prime like this does not exist? I have the exact same question about this relation: p_{n+1} > 2p_{n}
31. A Equation with three consecutive prime numbers
Solve the equation np_n+(n+1)p_{n+1}+(n+2)p_{n+2}=p^2_{n+2} where n\in \mathbb N^* and p_n , p_{n+1} , p_{n+2} are three consecutive prime numbers. ------------------------------------- A solution is n=2,p_2=3,p_3=5,p_4=7. May be other solutions?
32. A Is this product always greater than these sums?
I've been working on a problem for a couple of days now and I wanted to see if anyone here had an idea whether this was already proven or where I could find some guidance. I feel this problem is connected to the multinomial theorem but the multinomial theorem is not really what I need . Perhaps...
33. MHB Fascinating Discovery about Prime Numbers
I found this article fascinating. Imagine overlooking something so simple for so long!
34. Print Highly Prime Numbers in an Input Interval | C Program
Homework Statement Write a program that will print all highly prime numbers from the input interval <a,b>. Prime number is highly prime if deletion of every digit from right is a prime. Example: 239 is highly prime because 239,23,2 are primes. 2. The attempt at a solution Could someone point...
35. MHB Beginner Exercises on Prime numbers and Equality / Inequality proofs
Does anyone know a good resource for exercises on these topics?
36. MHB Prime numbers proof by contradiction
For prime numbers, $a$, $b$, $c$, $a^2 + b^2 \ne c^2$. Prove this by contradiction. So, I get that $a^2 = c^2 - b^2 = (c - b)(c +b)$ And I get that prime numbers are the product of 2 numbers that are either greater than one, or less than the prime numbers. But I'm unsure how to go from here.
37. Comp Sci C++ Sum of prime numbers in matrix
Homework Statement My Program is not showing the sum value or not returning it. A blank space is coming.Why that is so? Homework Equations Showing the attempt below in form of code. The Attempt at a Solution #include<iostream.h> #include<conio.h> Prime_Sum(int arr[30][30],int m, int n); void...
38. Is This Simple Algorithm the Key to Finding the Next Largest Prime Number?
I have a simple algorithm that appears to generate many primes (or semi-primes with relatively large factors). By 'relatively large', I mean large in relation to inputs. I have tested this algorithm for small values, and of the forty (six-digit) numbers produced, 22 are prime, 16 are...
39. Prime Numbers Between Two Quadratics: A Useful Result?
Would it be a useful result to know there is at least one prime between 16x^2+4x-1 and 16x^2+8x-5 for any odd natural number x?
40. A Is there a formula for generating prime numbers and proving their primality?
I have figured out a formula that generates prime numbers along with the proof that all such generated numbers are primes. The way it works is that you have to input consecutive prime numbers staring from 2 and ending at some Pn. And no it's not primorial minus or plus 1. Is this of any value...
41. Convergence of a sum over primes
I am trying to understand a condition for a nonincreasing sequence to converge when summed over its prime indices. The claim is that, given a_n a nonincreasing sequence of positive numbers, then \sum_{p}a_p converges if and only if \sum_{n=2}^{\infty}\frac{a_n}{\log(n)} converges. I have tried...
42. Are there prime numbers n for which S=/0?
We have the set:S={1<a<n:gcd(a,n)=1,a^(n-1)=/1(modn)} Are there prime numbers n for which S=/0?After this, are there any composite numbers n for which S=0? (with =/ i mean the 'not equal' and '0' is the empty set) for the first one i know that there are no n prime numbers suh that S to be not...
43. MHB Proving Prime Numbers in Quadratic Imaginary Fields
Hi, I need your help with the next two problems: 1) If p is a prime number such that p\equiv{3}\;mod\;4, prove that \sqrt{-p} is prime in \mathbb{Z}[\sqrt[ ]{-p}] and in \mathbb{Z}[\displaystyle\frac{1+\sqrt[ ]{-p}}{2}] too. 2) 2) We have d > 1 a square-free integer. Consider the quadratic...
44. Are There Infinitely Many Prime Numbers Written as ak+b?
Homework Statement Prove that there are infinitely many prime numbers written ##ak+b##, with ##a,b,k## integers greater than 1 Homework EquationsThe Attempt at a Solution Please could you tell me if you agree with that proof ? By contradiction: Assume that there is an integer ##k## such that...
45. Prime Numbers as Ortho-normal basis for all numbers
Hi, Can we treat prime numbers as an Ortho-normal basis of "Infinite" dimensions to represent every possible number. Treating numbers as vectors. Thanks.
46. R-Simplex's the number 5 and prime numbers.
So if you do a search for R-Simplexs you should find that. RSimplex(n,d)=Pochhammer(n,d)/d! Well so to does RSimplex(n,d)=If(n<d, Pochhammer(d+1,n-1)/n!, Pochhammer(n,d)/d!) Or something like that my maths package is down so I'm not sure quite how it works. Anyway the relationship between...
47. Do Two Primes Divide All Binomial Coefficients for Any n?
Homework Statement Is it true that for each ##n\geq 2## there are two primes ##p, q \neq 1## that divide every ##\binom{n}{k}## for ##1\leq k\leq n-1##?Examples: For ##n=6: \binom{6}{1}=6; \binom{6}{2}=15; \binom{6}{3}=20; \binom{6}{4}=15; \binom{6}{5}=6.## So we can have ##p=2## and...
48. The Even Primes: Exploring the Fascinating World of Prime Numbers
Just want to know if there are applications in the derivation of prime numbers. My instructor and the textbook that we are using seems to be obsessed with it, there is at least one problem about deriving prime numbers in each chapter. And also different versions like palindromic prime, emirp...
49. What Is the Product of Primes for the Integer 23?
Homework Statement My textbook says any integer greater than 1 is a product of primes. Wouldn't that mean that there are no prime numbers? What is the product of primes that create the integer 23? Homework Equations The Attempt at a Solution
50. Exploring the Prime Number Theorem: A Comprehensive Guide
Can I get the link to the prime number theorem? | 4,769 | 18,495 | {"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} | 3.78125 | 4 | CC-MAIN-2024-22 | latest | en | 0.957642 |
http://kineme.net/Discussion/DevelopingCompositions/GLLineStructurescale3D | 1,611,147,244,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703520883.15/warc/CC-MAIN-20210120120242-20210120150242-00541.warc.gz | 53,161,292 | 8,219 | # GL Line Structure -- scale in 3D?
Using the Kineme GL Line Structure inside an iterator, inside a trackball -- Is there a way to have a line drawing and scaling in 3D space using the Kineme GL Line Structure? Or is there another tool for this job?
## Comment viewing options
### Re: GL Line Structure -- scale in 3D?
hi,
try Kineme GL Scale. (in SL the 3dtransform has scale option as well)
d.
### Re: GL Line Structure -- scale in 3D?
I haven't used either patch, but I know a little about OpenGL and very little about math.
In the mathematical sense, as well as to OpenGL, a line only extends in one dimension. IIRC, in an OpenGL line structure, you can set the stroke width, but that's a uniform setting. It doesn't scale by perspective.
What you need would be an extruded cylinder (or extruded + shape) of polygons.
Again, that's how I understand it, but I may be too out of touch.
### Re: GL Line Structure -- scale in 3D?
Thanks for the replies. I am using SL. 3D transformation and GL Scale do not do the deed. Makes sense that the line is 1D. I guess there are no other line tools with width responsive to 3D? I'll look into extruding a cylinder or polygon -- done that before with the 3D parametric solid tool. Maybe could use the built in cylinder, too. Good tip, thanks.
### Re: GL Line Structure -- scale in 3D?
Try using the 'Kineme GL Depth Sort Sprite' together with Iterator and Replicate in Space:
See the composition in action here: http://www.youtube.com/watch?v=XWBeowqLO70
PreviewAttachmentSize
FreeMotion.qtz8.3 KB
### Re: GL Line Structure -- scale in 3D?
This is an example of using the GL scale to change apparent z scaling on group of lines. Amp up the z scale and see what I mean.
This is the basic macro I started my "gt lines app" stuff with. The javascript queue can also be changed to add in a manual z val, instead of writing it at 0, if a decent way for controlling the z was established (maybe 2 finger scroll or something... might be good for multitouch actually!).
(FWIW, the composition above starts glowing and then plummets to sub 5fps... I think something may be awry.)
PreviewAttachmentSize
line zigzag_gt.qtz34.1 KB
### Re: GL Line Structure -- scale in 3D?
yes you're right, the FreeMotion.qtz composition will drop in fps if the midi control data isn't input. I should have mentioned that a signal pulse (from a MIDI control source in this case) feeding into the Integrator patches will reset them, otherwise the number of copies will keep increasing hence the drop in fps.
### Re: GL Line Structure -- scale in 3D?
Thanks so much! I'll have a look later today.
### Re: GL Line Structure -- scale in 3D?
Both comps are interesting to see. Thanks again.
GT -- That's a useful lesson in there. Actually, I got a little giddy when I saw that one working. F-in' awesome! Very generous. Puts me on the right path and more. | 713 | 2,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.897785 |
https://blog.madebymunsters.com/epub/introduction-to-plane-algebraic-curves | 1,610,932,134,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514046.20/warc/CC-MAIN-20210117235743-20210118025743-00353.warc.gz | 255,341,878 | 9,371 | # Introduction to Plane Algebraic Curves by Ernst Kunz,Richard G. Belshoff
By Ernst Kunz,Richard G. Belshoff
This paintings treats an advent to commutative ring conception and algebraic airplane curves, requiring of the scholar just a uncomplicated wisdom of algebra, with all the algebraic evidence amassed into a number of appendices that may be simply mentioned, as wanted. Kunz's confirmed belief of training themes in commutative algebra including their purposes to algebraic geometry makes this e-book considerably various from others on airplane algebraic curves. The exposition specializes in the in basic terms algebraic points of aircraft curve thought, leaving the topological and analytical viewpoints within the historical past, with merely informal references to those topics and recommendations for additional interpreting. most crucial to this article: Emphasizes and makes use of the speculation of filtered algebras, their graduated earrings and Rees algebras, to infer simple evidence concerning the intersection conception of aircraft curves provides residue idea within the affine airplane and its functions to intersection concept equipment of facts for the Riemann–Roch theorem comply with the presentation of curve conception, formulated within the language of filtrations and linked graded earrings Examples, workouts, figures and proposals for additional learn around out this quite self-contained textbook
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Extra resources for Introduction to Plane Algebraic Curves
Sample text | 657 | 3,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-04 | latest | en | 0.9187 |
https://web2.0calc.com/questions/help-please_78983 | 1,627,748,839,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00577.warc.gz | 606,303,699 | 5,410 | +0
0
52
1
Write the following polar equation in rectangular form.
r = sec Θ
May 28, 2021
#1
+121004
+1
sec θ = r / x
So
r = r / x divide out r
1 = 1/ x
x = 1
May 28, 2021 | 94 | 215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-31 | longest | en | 0.42796 |
https://study.com/academy/practice/quiz-worksheet-reducing-simplifying-improper-fractions.html | 1,571,463,110,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986688826.38/warc/CC-MAIN-20191019040458-20191019063958-00332.warc.gz | 718,679,551 | 27,873 | # How to Reduce or Simplify Improper Fractions Video
Instructions:
question 1 of 3
### Which fraction is equivalent to 15/12 ?
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### 2. Which 4 sets of fractions are equivalent?
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Improper fractions can look a little tricky at first, but if you can simplify them it can be easier. In this quiz, you'll determine equivalent reduced fractions and reduce others.
## Quiz & Worksheet Goals
This improper fraction quiz tests your ability to:
• Recognize an equivalent to a given fraction
• Determine which of given options is a fraction greater than 1
• Identify which group of fractions is equivalent
• Know which of given fractions cannot be reduced further
• Reduce a fraction to its lowest terms
## Skills Practiced
• Reading comprehension - draw from the most relevant information from the lesson on reducing and simplifying fractions
• Critical thinking - use this information to consider if a group of fractions is or is not equivalent
• Problem solving - take what you've learned and reduce given improper fractions | 303 | 1,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-43 | longest | en | 0.909013 |
http://stackoverflow.com/questions/6972515/python-conditional-statements/6972533 | 1,464,365,770,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049276964.14/warc/CC-MAIN-20160524002116-00003-ip-10-185-217-139.ec2.internal.warc.gz | 279,682,846 | 22,353 | # Python conditional statements
``````def Fitness(a, b, c):
if ((a&b&c) >= 4) & ((a+b+c) >= 13):
return('Gold')
if ((a&b&c) >= 3) & ((a+b+c) >= 10):
return('Silver')
if ((a&b&c) >= 2) & ((a+b+c) >= 07):
return('Pass')
else:
return('Fail')
``````
now the Problem is when `Fitness(2,2,5)` is given, the control jumps to default ie. 'Fail'. Where is the actual output is 'Pass'. ?
-
Could you explain the logic you're trying to implement? – Tim Pietzcker Aug 7 '11 at 11:41
Note that
``````a&b&c >= 2
``````
is different from
``````a>=2 and b>=2 and c>=2.
``````
I think you mean the second one, i.e. all values are larger than two. (The first one does a binary and with all your values and compares this to the value two.)
-
+1 I think this is probably what the OP intends. – Chris Lutz Aug 7 '11 at 11:40
Ah! +1 for best use of crystal ball today! – Tim Pietzcker Aug 7 '11 at 11:42
You forgot the second `&`. – quantum Sep 22 '12 at 1:14
Use `and` instead of `&` (binary and). And don't write `07` - numbers starting with `0` may be interpreted as octal depending on your version of Python.
Together with Howard's inspiration, I'd suggest this:
``````def Fitness(a, b, c):
if all(x>=4 for x in (a,b,c)) and (a+b+c) >= 13:
return('Gold')
if all(x>=3 for x in (a,b,c)) and (a+b+c) >= 10:
return('Silver')
if all(x>=2 for x in (a,b,c)) and (a+b+c) >= 7:
return('Pass')
return('Fail')
``````
Also, it's sad that you're not awarding a bronze medal...
-
I'm not sure this should be causing the problem. `True` is 1 and `False` is 0, so `True & True` should still give `True`, `True & False` should still give `False`, etc. – Chris Lutz Aug 7 '11 at 11:37
hmm , it doesn't worked. same as before – Dickson Xavier Aug 7 '11 at 11:39
@Chris: Right. I've just checked that after many subtle corrections to his code to get it to run at all. Clarification by the OP needed. – Tim Pietzcker Aug 7 '11 at 11:40
hmm...(a&b&c)>= no, causes problems, if a=b it goes to default condition. i m not sure!! – Dickson Xavier Aug 7 '11 at 11:46
used (a>=no) & (b>=no) & .... instead, i like to know why the latter doesn't work – Dickson Xavier Aug 7 '11 at 11:47
Where is the actual output is 'Pass'.
No it isn't. 2 (0b010) & 2 & 5 (0b101) is 0, so the expressions will all fail even if you change the `&` separating the two terms to `and`. Perhaps you meant to use a completely different expression?
-
By "where the actual output is", he seems to mean "where the actual output should be". – Karl Knechtel Aug 7 '11 at 14:33
I'd post this as a comment to Tim Pietzcker's answer, since it's just an improvement on it, but the comment field would mess up the formatting, so I'm making it a separate answer.
``````def fitness(*args):
arg_sum = sum(args)
smallest_arg = min(args)
if smallest_arg >= 4 and arg_sum >= 13:
return 'Gold'
if smallest_arg >= 3 and arg_sum >= 10:
return 'Silver'
if smallest_arg >= 2 and arg_sum >= 7:
return 'Pass'
return 'Fail'
``````
First the important stuff:
1. Enclosing the return value in parenthesis like in `return('Gold')` is not wrong per se, but will confuse the reader; `return` is a statement, not a function. It works nontheless because the parenthesis are in effect silently ignored. If what you wanted was in fact to return a tuple with a single string element, do `return ('Gold',)`.
2. DRY: Compute the sum and smallest element of the arguments once, and use the computed values in the `if` blocks, as shown above.
3. Since the arguments are all treated equally and together, gather them in an `*args` tuple. This will also generalize the function to work with any number of arguments. If that should not be allowed, add this to the beginning of the function:
``````if len(args) != 3:
raise TypeError("Need exactly 3 arguments, got {0}".format(len(args)))
``````
Finally I'll just mention that functions should have `lower_case_names` according to the python style guide. Following the style guide is of course not mandatory, just recommended. :)
-
``````def fitness(*args):
arg_sum = sum(args)
smallest_arg = min(args)
if smallest_arg >= 4 and arg_sum >= 13:
return 'Gold'
if smallest_arg >= 3 and arg_sum >= 10:
return 'Silver'
if smallest_arg >= 2 and arg_sum >= 7:
return 'Pass'
return 'Fail'
``````
- | 1,271 | 4,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2016-22 | latest | en | 0.843295 |
https://expertinstudy.com/t/UxkEvHb4hNQEDG | 1,685,973,268,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652116.60/warc/CC-MAIN-20230605121635-20230605151635-00107.warc.gz | 281,847,497 | 5,163 | # Mr B aged 63 years, has earned rupees 75,00,000 out of his business. His ex- wife gifted him cash in account worth rupees 6 lakh. He spent a total of rupees 15 lakh during a family trip. He won a lottery of 19 lakh rupees. Out of happiness he gifted his wife cash of rupees 450000. He bought a life insurance policy and paid a premium of 50000 annually. He paid tuition fee for his daughter for rupees 50000. He invested rupees 80000 in PPF. One of his existing life insurance policy got matured and he received a total of rupees 32 lakh. Seeing his health conditions the doctor adviced him a medical test of rupees 6000 following he bought a health insurance for rupees 27000. Calculate his taxable income and tax liability.
Mr B's annual income = 75,00,000 .
Family trip = 15 lakh.
lottery = 19 lakh rupees.
he gifted his wife cash = 450000.
life insurance policy = 50000
tuition fee for his daughter= 50000.
Investment in PPF= 80000 .
existing life insurance policy got matured =32 lakh.
medical test = 6000
health insurance = 27000.
To find:
His taxable income and tax liability
Solution:
Taxable income = Revenue gift cash-Family trip Lottery money-gift-Life insurance premium -tution fee for daughter - PPF investment Life insurance matured - Health insurance
= 7500000 600000 - 1500000 1900000-4500000-50000-50000-80000 3200000-6000-27000
=11137000
Tax liability = 30% of 11137000 = Rs. 33,41,100
Health cess = 4% of 33000 = 1320
Educational cess = 2000
Total tax liability = Rs. 33,44,420
Hence, taxable income is Rs. 1,11,37,000 and tax liability is Rs. 33,44,420.
1 | 456 | 1,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-23 | latest | en | 0.966321 |
https://www.sawaal.com/time-and-work-questions-and-answers/two-boys-and-a-girl-can-do-a-work-in-5-days-while-a-boy-and-2-girls-can-do-it-in-6-days-if-the-boy-i_11169 | 1,537,291,353,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155634.45/warc/CC-MAIN-20180918170042-20180918190042-00152.warc.gz | 855,298,487 | 16,430 | 5
Q:
# Two boys and a girl can do a work in 5 days, while a boy and 2 girls can do it in 6 days. If the boy is paid at the rate of 28$a week, what should be the wages of the girl a week ? A) 24$ B) 22 $C) 16$ D) 14 $Answer: C) 16$
Explanation:
Let the 1 day work of a boy=b and a girl=g, then
2b + g = 1/5 ---(i) and
b + 2g = 1/6 ---(ii)
On solving (i) & (ii), b=7/90, g=2/45
As payment of work will be in proportion to capacity of work and a boy is paid $28/week, so a girl will be paid $28×245790$ = 16$.
Q:
Raghu can do a job in 12 days alone and Sam can do the same job in 15 days alone. A third person Aru whose efficiency is two-third of efficiency of both Ram and Shyam together, can do the same job in how many days alone?
A) 10 B) 12 C) 13 D) 15
Explanation:
One day work of Raghu and Sam together = 1/12 + 1/15 = 9/60 = 3/20
Aru efficiency = 2/3 of (Raghu + Sam)
Number of days required for Aru to do thw work alone = 3/2 x 20/3 = 10 days.
3 332
Q:
12 men complete a work in 14 days. 5 days after they had started working, 3 men join them. How many more days will all of them take to complete the remaining work?
A) 5 B) 4 C) 3 D) 2
Explanation:
Let p be the required number of days.
From the given data,
12 x 14 = 12 x 5 + (12+3) x p
12 x 14 = 12[5 + 3p]
14 = 5 + 3p
3p = 9
p = 3 days.
Hence, more 3 days all of them take to complete the remaining work.
0 497
Q:
Lasya alone can do a work in 16 days. Srimukhi’s efficiency is 20 % lesser than that of Laya. If Rashmi and Srimukhi together can do the same work in 12 days, then find the efficiency ratio of Rashmi to that of Lasya?
A) 19 : 7 B) 30 : 19 C) 8 : 15 D) 31 : 17
Explanation:
Given Lasya can do a work in 16 days.
Now, time taken by Srimukhi alone to complete the work = 16 x 100/80
Time taken by Rashmi = n days
=> (12 x 20)/(20 - 12) = (12 x 20)/n
=> n= 30 days.
Required ratio of efficiencies of Rashmi and Lasya = 1/30 :: 1/16 = 8 : 15.
5 692
Q:
5 boys and 3 girls can together cultivate a 23 acre field in 4 days and 3 boys and 2 girls together can cultivate a 7 acre field in 2 days. How many girls will be needed together with 7 boys, if they cultivate 45 acres of field in 6 days?
A) 4 B) 3 C) 2 D) 1
Explanation:
Let workdone 1 boy in 1 day be b
and that of 1 girl be g
From the given data,
4(5b + 3g) = 23
20b + 12g = 23 .......(a)
2(3b + 2g) = 7
6b + 4g = 7 ........(b)
Solving (a) & (b), we get
b = 1, g = 1/4
Let number og girls required be 'p'
6(7 x 1 + p x 1/4) = 45
=> p = 2.
Hence, number of girls required = 2
5 607
Q:
70000 a year is how much an hour?
A) 80 B) 8 C) 0.8 D) 0.08
Explanation:
Given for year = 70000
=> 365 days = 70000
=> 365 x 24 hours = 70000
=> 1 hour = ?
70000/365x24 = 7.990 = 8
2 646
Q:
A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained?
A) 1/5 B) 1/6 C) 1/7 D) 1/8
Explanation:
Total work is given by L.C.M of 72, 48, 36
Total work = 144 units
Efficieny of A = 144/72 = 2 units/day
Efficieny of B = 144/48 = 3 units/day
Efficieny of C = 144/36 = 4 units/day
According to the given data,
2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100
3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54
p = 198/16.5
p = 12 days.
Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day
(C+D) in p days = (4 + 6) x 12 = 120 unit
Remained part of work = (144-120)/144 = 1/6.
6 1820
Q:
10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?
A) 215 days B) 225 days C) 235 days D) 240 days
Explanation:
Given that
(10M + 15W) x 6 days = 1M x 100 days
=> 60M + 90W = 100M
=> 40M = 90W
=> 4M = 9W.
From the given data,
1M can do the work in 100 days
=> 4M can do the same work in 100/4= 25 days.
=> 9W can do the same work in 25 days.
=> 1W can do the same work in 25 x 9 = 225 days.
Hence, 1 woman can do the same work in 225 days.
9 2320
Q:
A,B,C can complete a work in 15,20 and 30 respectively.They all work together for two days then A leave the work,B and C work some days and B leaves 2 days before completion of that work.how many days required to complete the whole work?
Given A,B,C can complete a work in 15,20 and 30 respectively.
The total work is given by the LCM of 15, 20, 30 i.e, 60.
A's 1 day work = 60/15 = 4 units
B's 1 day work = 60/20 = 3 units
C's 1 day work = 60/30 = 2 units
(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units
Let B + C worked for x days = (3 + 2) x = 5x units
C worked for 2 days = 2 x 2 = 4 units
Then, 18 + 5x + 4 = 60
22 + 5x = 60
5x = 38
x = 7.6
Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days. | 1,911 | 5,026 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2018-39 | longest | en | 0.892121 |
http://mathhelpforum.com/calculus/143028-eigenfunction-expansion-print.html | 1,527,330,512,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867416.82/warc/CC-MAIN-20180526092847-20180526112847-00602.warc.gz | 186,462,916 | 3,156 | # Eigenfunction Expansion
• May 4th 2010, 11:17 AM
Aryth
Eigenfunction Expansion
Use the orthogonality relation to find expressions for the coefficients $\displaystyle c_n$ in the eigenfunction expansion of a function f(x):
$\displaystyle f(x) \approx \sum_{n=1}^{\infty} c_n\phi_n$
• May 4th 2010, 11:19 AM
Bruno J.
Hint : what is $\displaystyle <f, \phi_n>$?
• May 4th 2010, 11:40 AM
Aryth
Isn't it: $\displaystyle \int_a^b w(x)f(x)\phi_n(x)~dx$?
Sorry if I don't understand how it helps... I do know that $\displaystyle c_n$ is a quotient of inner products... I'm afraid I don't know how to show it.
• May 4th 2010, 11:48 AM
Bruno J.
Quote:
Originally Posted by Aryth
Isn't it: $\displaystyle \int_a^b w(x)f(x)\phi_n(x)~dx$?
Sorry if I don't understand how it helps... I do know that $\displaystyle c_n$ is a quotient of inner products... I'm afraid I don't know how to show it.
Well that's the left side of the equation, indeed. But what if you replace $\displaystyle f$ by its expansion in the expression $\displaystyle <f, \phi_n>$, and then use the linearity of the inner product?
• May 4th 2010, 11:56 AM
Aryth
Ok... So... We get:
$\displaystyle <f,\phi_n(x)> = \sum_{n=1}^{\infty}c_n \int_a^b w(x)\phi_n(x)^2~dx = \sum_{n=1}^{\infty}c_n <\phi_n,\phi_n>$
Which means that:
$\displaystyle \frac{<f,\phi_n(x)>}{<\phi_n,\phi_n>} = \sum_{n=1}^{\infty} c_n$
Is that along the right lines?
• May 4th 2010, 12:04 PM
Bruno J.
Be careful! You're using $\displaystyle n$ both as a fixed integer and as a summation index! Here's what your corrected post looks like :
Quote:
Originally Posted by Aryth
Ok... So... We get:
$\displaystyle <f,\phi_n(x)> = \sum_{j=1}^{\infty}c_j <\phi_j,\phi_n> = c_n <\phi_n, \phi_n>$
because all the terms cancel (by orthogonality), except $\displaystyle <\phi_n, \phi_n>$.
• May 4th 2010, 12:05 PM
Aryth
Ah, I see. I appreciate the help. Thanks a lot.
• May 4th 2010, 12:06 PM
Bruno J.
Quote:
Originally Posted by Aryth
Ah, I see. I appreciate the help. Thanks a lot.
Most welcome! Good luck. (Wink) | 704 | 2,044 | {"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} | 3.875 | 4 | CC-MAIN-2018-22 | latest | en | 0.862874 |
http://numerologysecrets.net/pythagorean-numerology/ | 1,544,653,783,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824119.26/warc/CC-MAIN-20181212203335-20181212224835-00158.warc.gz | 204,028,158 | 14,490 | Pythagorean numerology is based in Ancient Greek teachings that were introduced during the time of Pythagoras. This is a well-known Greek mystic, mathematician, singer and physicist who had a fascination with numbers from a very young age. He is the father of the Pythagoras theorem among many other pivotal mathematical discoveries.
Pythagorean numerology takes into account numbers as well as the alphabet. Each letter in the alphabet is assigned a different number in increments of 9. This then becomes a standard with which numerologists are able to find out various aspects of your personality as well as what your future holds for you.
## Pythagoras and Numbers
Pythagoras was of the idea that the world could be understood by looking at numbers and then using that information in a precise, mathematical ways. He was a master diviner who was sought after by kings, princes, government heads and influential businessmen.
Pythagoras was often heard saying that ‘all things are numbers’ as well as that the first four digits are viewed as the essence of the monad, or divinity (that is, 1, 2, 3 and 4). He also was of the idea that these four numbers contained the answers of the universe, and that the manipulation of these numbers can produce all other numbers.
Pythagorean numerology uses the triangle figure to come up with equations that tell a story. For example, a triangle could be built having rows of fours, threes, or twos. These rows would be assigned numbers that could be added up to come up with a singular prime number. The numbers on the edges of the triangle usually represent central issues of whatever dilemma you’re facing, and the numbers in the middle point to solutions to your problem. The numbers near the top of the triangle represent influences that you cannot control, and the number in the middle of the triangle represents the solution you can look forward to once you’ve solved the problem.
In order to get the correct value of your name in Pythagorean numerology, you must take into account your first and last name. In addition, you should write down your name as it appears on your certificate, following the same exact order. Once you’ve done that, assign each letter its corresponding Pythagorean number, and then add them up in order to come up with your total figure. If you come up with a two or three-digit number, you should add these digits so you can end up with a single number.
## Conclusion
The final number is known as your life path, and it is considered as something that can tell you a lot about your natural abilities as well as what to do in order to get ahead in life. It also gives you insight into the three stages of your life and what you must do in each to get a certain outcome. Life paths range from 1 to 9, and they are distinct from each other. For the most part, people have only one life path, and it resonates deeply with who they are at their core. Pythagorean numerology is thus one of the more accurate types out there, giving those who seek it an accurate representation of their personality and needs. | 644 | 3,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-51 | longest | en | 0.974661 |
http://www.bruinsnation.com/2011/1/18/1941522/plus-minus-stats-from-the-games-against-oregon-and-osu | 1,408,552,482,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500811391.43/warc/CC-MAIN-20140820021331-00297-ip-10-180-136-8.ec2.internal.warc.gz | 281,163,138 | 24,056 | ## +/- (Plus/Minus) Stats from the games against Oregon and OSU
I was inspired by LVBruin's fanpost to determine some +/- calculations for certain lineups. I noticed that our own official athletics site provides detailed play-by-plays of each game, which I hypothesize are generated in an automated manner. I programmed a python script to parse these play-by-plays and calculate the +/- for every possible combination of our players out on the floor. Fortunately, all the play-by-play data is provided in a basic HTML page, which has been fairly painless to process. Sadly, there are a few errors in the data. I once found that the play-by-play indicated that Malcolm Lee was subbed out and he subsequently made an offensive rebound.
LVBruins said that some of his data was lost in spreadsheets, so I thought I might remedy that by writing an automated script. This is my first try at attempting a program like this, so it's possible it may be inaccurate. I did not test it thoroughly but I did note that the minutes played by all possible rotations seen during the game add up to 40. Also, it does not handle overtime situations but I can add that functionality later if it happens.
Find the stats I collected after the jump.
Below are +/- stats for every 5-man line up that played at least 2 minutes together on the floor.
It is listed as the names of the players in the rotation, the +/- per minute, the actual +/-, and the minutes played.
@ Oregon State:
Honeycutt, Jones, Lee, Nelson, Smith: 0.26, 2, 7.77 minutes
Honeycutt, Jones, Lane, Lee, Nelson: 1.4, 1, 7.33 minutes
Anderson, Honeycutt, Lane, Lee, Smith: 0, 0, 3.05 minutes
Anderson, Honeycutt, Lee, Nelson, Smith: 0.0, 0, 2.93 minutes
Jones, Lamb, Lane, Lee, Smith: 1.34, 3, 2.23 minutes
Anderson, Honeycutt, Jones, Lamb, Nelson: 0.0, 0, 2.22 minutes
@ Oregon:
Anderson, Honeycutt, Jones, Lee, Smith: 1.32, 13, 9.87 minutes
Anderson, Honeycutt, Jones, Lee, Nelson: -0.31, -2, 6.42 minutes
Anderson, Honeycutt, Lamb, Lee, Smith: 1.81, 5, 4.23 minutes
Anderson, Honeycutt, Lane, Lee, Smith: 0.0, 0, 3.7 minutes
Anderson, Honeycutt, Jones, Lane, Lee: -2.39, -8, 3.35 minutes
Honeycutt, Jones, Lee, Nelson, Smith: -1.69, -4, 2.37 minutes
Honeycutt, Jones, Lane, Lee, Nelson: -1.85, -4, 2.17 minutes
Naturally the +/- in the Oregon game is higher for lineups without Reeves because he was scoreless that night. The stats look pretty awful for the game against OSU though, thanks to our terrible second half.
Let me know if you have any suggestions or ideas on how I can improve the data collection process. Should I present it visually like LVBruin did? Maybe you guys would like to see an aggregate +/- for every combination over the course of the season so far? That'll take a bit more time on my end to code up, but right now, I can provide +/- for the lineups seen in individual games. I could also redo the program in Java to I can make a standalone application for other users to download and use, which would take some time but would likely be good practice for my career.
<em>This is a FanPost and does not necessarily reflect the views of BruinsNation's (BN) editors. It does reflect the views of this particular fan though, which is as important as the views of BN's editors.</em>
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We have our own Community Guidelines at Bruins Nation. You should read them. | 988 | 3,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2014-35 | latest | en | 0.932018 |
https://math.stackexchange.com/questions/443758/write-an-expression-in-powers-of-x1-and-y-1-for-x2xyy2 | 1,718,671,221,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00089.warc.gz | 339,559,070 | 36,745 | # Write an expression in powers of $(x+1)$ and $(y-1)$ for $x^2+xy+y^2$
Write an expression in powers of $(x+1)$ and $(y-1)$ for $x^2+xy+y^2$
I calculated
$f_x=2x+y$
$f_{xx}=2$
$f_y=x+2y$
$f_{yy}=2$
And then what I need to do?
What is the formula to solve the question ?
Let $u=x+1$, $v=y-1$, then $$x^2+xy+y^2=(u-1)^2+(u-1)(v+1)+(v+1)^2=u^2+v^2+uv-u+v+1=\;\cdots$$
• This is simpler than my suggestion! Commented Jul 14, 2013 at 23:52
• So easy? Thank you :)
– 1190
Commented Jul 15, 2013 at 10:50
Since $x^2+xy+y^2$ is a polynomial of degree $2$, the answer is the Taylor expansion of $x^2+xy+y^2$ about the point $(-1,1)$. Let $f(x,y)=x^2+xy+y^2$.
Note that $f_x(x,y)=2x+y$, $f_y(x,y)=2y+x$, $f_{xx}(x,y)=f_{yy}(x,y)=2$ and $f_{xy}(x,y)=1$. All higher order partial derivatives are identically $0$. So the Taylor expansion is $$f(-1,1)+f_x(-1,1)(x+1)+f_y(-1,1)(y-1)+\frac{1}{2!}\left(f_{xx}(-1,1)(x+1)^2 +2f_{xy}(-1,1)(x+1)(y-1)+f_{yy}(-1,1)(y-1)^2\right).$$
It looks messy, isn't too bad. We have $f(-1,1)=1$, $f_x(-1,1)=-1$, and $f_y(-1,1)=1$. The second-order partial derivatives are constant so are already evaluated at $(-1,1)$.
Remark: Note that the approach by Norbert is quite a bit quicker. It has the further great advantage of being less likely to produce an error.
The only advantage of the procedure above is that it lets you practice a bit with the second-order Taylor polynomial.
• You explained in detail. So good:)
– 1190
Commented Jul 15, 2013 at 10:51 | 574 | 1,489 | {"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} | 4.3125 | 4 | CC-MAIN-2024-26 | latest | en | 0.802493 |
https://www.jiskha.com/questions/1731283/how-is-orwells-animal-farm-an-allegory-answer-a-orwells-animal-farm-is-an-allegory | 1,632,551,052,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057598.98/warc/CC-MAIN-20210925052020-20210925082020-00120.warc.gz | 835,300,932 | 5,448 | # english
How is Orwell's Animal Farm an allegory?
Answer:a. Orwell’s Animal Farm is an allegory because there are two political overtones within this story. On the surface you have a story that depicts a tale of a society of animals that shape into a dictatorial rule. The second overtone deals with animals that stand for a representation of Orwell’s way of constructing how he felt about the totalitarianism in Soviet Russia. For example, after serious disregard by their owner, the animals revolt and throw out Mr. Jones and his wife from the farm. This shows how it is an allegory because creatures are yearning for freedom but in the end become depraved by accepting the very power that had initially persecuted them. This shows in the text when “Old Major, the prize white boar “states, “Man is the only creature that consumes with producing.” He inspires the animals to “work nigh and day, body and soul, for over-throw of the human race.”
1. 👍
2. 👎
3. 👁
1. It's an allegory in that the animals on the farm represent human activity. If it was not an allegory, the characters would be humans, not animals. Did you read the definition of an allegory that I posted to you? Did you understand it?
1. 👍
2. 👎
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$x=16$
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http://forum.allaboutcircuits.com/threads/9-0-counter-running-lights-that-goes-faster-as-the-segment-ticks-at-5-until-0.121605/ | 1,484,884,473,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280774.51/warc/CC-MAIN-20170116095120-00438-ip-10-171-10-70.ec2.internal.warc.gz | 106,735,165 | 14,411 | # 9-0 counter. Running lights that goes faster as the segment ticks at 5 until 0
Discussion in 'Homework Help' started by jzz1g, Mar 5, 2016.
1. ### jzz1g Thread Starter New Member
Mar 5, 2016
3
0
Hello!
I already have done the 9-0 counter using JK F/F, decoder, seven segment display and 555.
At the running lights part, i'm planning to do it using JK or D type F/F.
Currently I have a seperate 555 timer for the 9-0 counter and another seperate 555 timer for the running lights. but my only problem is how will the timer react when the segment ticks at 5 so it will go faster.
2. ### WBahn Moderator
Mar 31, 2012
18,087
4,917
Are we supposed to just somehow know what this "running lights part" is?
Why should anything go faster when the "segment ticks at 5"?
We are not mind readers. You need to describe your project/problem in enough detail so that we have a clue what you are talking about.
3. ### jzz1g Thread Starter New Member
Mar 5, 2016
3
0
I'm sorry for not giving enough details.
It's a simulation of a person crossing a road.
The person is represent by an LED (which will be represented by running lights)
The 9-0 represents the time that the person has to cross the road.
During the 9-6 part of the counter, the LED should run as fast as how the segment is ticking from 9-6.
Once the counter ticks at 5 until 0, the LED should run faster than how it was running from 9-6 without affecting the speed of the 9-0 countdown segment.
I'm using two seperate 555 timers for this circuit. One for the 9-0 countdown and one for the running lights (the person crossing the road).
I'm currently having a problem on how to make it go faster when the the counter reaches 5.
We are only allowed to use JK F/F, D F/F, common gates, resistors, capacitors, decoder, seven segment and 555 timer.
4. ### WBahn Moderator
Mar 31, 2012
18,087
4,917
There are a number of ways. One would be to use the output of the counter to increase the frequency of the second 555 timer.
Are you allowed to use diodes?
Another would be for the running counter to count by a greater increment once the timer counter reaches 5.
5. ### jzz1g Thread Starter New Member
Mar 5, 2016
3
0
Hello again.
"One would be to use the output of the counter to increase the frequency of the second 555 timer."
I don't get the part of the "output of the counter"
In the counter I used JK F/F, bcd to seven segment decoder and 555 timer to do it. Which output should I use there?
Also I just asked my professor and we are allowed to use diodes.
As for this "running counter to count by a greater increment once the timer counter reaches 5."
The counter for the running lights right?
6. ### WBahn Moderator
Mar 31, 2012
18,087
4,917
The first counter has a four bit output whose values are 9 down to 0, correct? The speed that you want the second timer to run at depends on the output of the first counter, correct?
So can you design a circuit that takes the output of the first counter and produces a HI pulse if the second timer is supposed to run fast and a LO if it is supposed to run slow? | 796 | 3,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-04 | latest | en | 0.934046 |
http://www.advogato.org/person/cananian/diary.html?start=89 | 1,492,941,887,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118519.29/warc/CC-MAIN-20170423031158-00366-ip-10-145-167-34.ec2.internal.warc.gz | 435,353,947 | 21,510 | # Older blog entries for cananian (starting at number 89)
2012 Mystery Hunt
This year I was a member of Codex, the writing team for the 2012 Mystery Hunt. I'm going to describe some of the puzzles I wrote for "The Producers" hunt, in release order. BEWARE SPOILERS!
1. One of the early theme proposals for our hunt was "Alice in Wonderland." Casting about for novel meta ideas, I hit upon the idea of a round with purely numeric answers, 1 through 29,394, which would resolve to words via "looking glass numbers"—that is, numbering all the words in "Through the Looking Glass". It occurred to me that you could make your numbering system self-descriptive if you used certain words; for example, if you wanted to make clear that hyphenated words should be counted as one (instead of two), you could include "great" and "half" on either side of "arm-chair". The numbering of "great" (164) and "half" (166) would make it clear that "arm-chair" should be treated as a single number (165).
This didn't survive as a meta, but it eventually became a puzzle, called 1207 1370 (which translates to "Looking-Glass Words" using its enumeration system). It also served to ensure that teams had a good wordlist by the time they got to the Charles Dodgson meta...
2. Blinkenlights. A recursive-structured puzzle inspired by (but not reaching the greatness of) Derek Kisman's Maze from Setec's '05 Hunt. If anyone is mourning the lack of Jonathan Coulton-related puzzles from this year's hunt, blame me: I stole the answer PROTECTORS which Andrew Lin had earmarked for a JoCo puzzle. ("Did I say overlords?")
3. Caterpillars. I like giving physical objects to teams. This was another failed meta—you would have assembled the pieces out of words, then would have to assemble the jigsaw from the word-pieces. The location of the caterpillars' heads in the final assembly would spell out the final meta answer using an overlay. But the puzzle is more fun with tangible pieces, I think.
4. B.J. Blazkowicz in ‘Wintertime for Hitler’. I was writing the meta for this round and trying to find non-dictionary words. I needed "CAR..." as a prefix to make the chess game work, which suggested CARMACK as an answer, and the puzzle just wrote itself from there. Scott Handelman contributed the title. This puzzle was going to be distributed on 3.5" disks (remember how I said I like giving teams physical objects?), but the last 3.5" floppy disk puzzle was Blue Steel in '06. (Redundant Obsolescence doesn't count, since the 5 1/4" disk was redundant.) The past six years have not been kind to the 3.5" floppy; ultimately we decided we didn't want to deny teams the pleasure of playing the game because they couldn't locate a floppy drive. It's more important that puzzles be fun than hard!
5. Charles Lutwidge Dodgson meta. I began writing this puzzle immediately after the 2011 hunt, dissatisfied with the mechanism and final clue phrase of that year's Racking Your Brains. I thought I could write a better puzzle using Scrabble Solitaire as a mechanism.
Slightly later it became part of the "Alice in Wonderland" theme proposal, with Jabberwocky words. Then I spent a couple of months away from the hunt, getting married.
Upon returning we badly needed critic metas so I dusted off the puzzle, adding an Alice chess frontend yielding the tile string in order to make it a shell meta. The puzzle can still be solved as pure Scrabble Solitaire (ie, without the given "scores after each play") but it's easier for humans to solve with the frequent checkpoints given. For what it's worth, I constructed the chess game with a reasonably-deep alpha-beta search, so all the moves "make sense" as much as is possible given the constraints of the puzzle. And it ends in a clean checkmate, obviously... I have no idea how BENOISY snuck in there.
6. Ben Bitdiddle meta. The idea of making an electronic circuit which was impossible to assemble incorrectly had been in my "Mystery Hunt ideas" folder for years. A coworker at OLPC mentioned the odd power-pin configuration of the PIC chips one day, which gave me the "flip" mechanism. Brainstorming with Andrew Lin brought it the rest of the way.
I promise never to abuse an optoisolator in this way again.
(Of course it turned out when constructing this puzzle that Ben Bitdiddle really needed to use the show answers CARPAL and THESOUTH because of their length in morse code, so I ended up having to rewrite parts of Dodgson to make Bitdiddle work. In the rewrite CARPAL became CARMACK... and B.J. Blazkowitcz was born.)
7. JFK SHAGS A SAD SLIM LASS. One of my earliest puzzle submissions was, "A puzzle contained only in its title." Again, the fabulous Codex editor team turned this into a real puzzle.
Some puzzles I enjoyed editing:
1. Revisiting History — I commissioned a Doctor Who-themed puzzle for the answer TORCHWOOD (see the final clue phrase for the reason why) and contributed the "location of the word 'who'" mechanism.
2. Gibberish and More Gibberish. I liked the idea for this puzzle enough that I shoehorned a suitable answer into the Charles Dodgson meta... and then had to do some heavy lifting to get the puzzle finished and into the hunt.
3. Sounds Good to Me. It was immediately obvious this was a brilliant idea from Seth Schoen. But the twin barriers of toki pona and hiragana threatened to make it unsolvable. I'd like to think I played a role in making this an accessible and solvable puzzle.
4. Itinerant People of America. Same deal. Squiggles had bequeathed the world the facial expression described as, "That's my brain leaving out the back door while my face distracts you." My contribution here was solely instilling the fear of God into the authors. Scott Handleman describes how he and Emily Morgan took that advice and constructed a kick-ass puzzle.
And that's it for my puzzles! I also did a heck of a lot of other stuff for the hunt; I hope y'all enjoyed it. (My own favorite part was the wrap-up, since all my responsibilities had been discharged by then. I could just watch Patrick rock my hat and accordion, play along on ukulele, and sing tenor with Francis at the end.)
Syndicated 2012-01-18 23:09:41 (Updated 2012-01-18 23:11:16) from Dr. C. Scott Ananian
A collection of Nell demos
Here are some banged-together demos of various pieces of One Laptop per Child's Project Nell. The ultimate goal is a Nell demo for CES in January 2012, but these bits should be considered as tech demos, benchmarks, and proofs of concept, not actual pieces of that demo (yet).
Most of these demos require WebGL support. Visit get.webgl.org for information about enabling WebGL in your browser; there is WebGL support in Chrome, Firefox, Safari, and Opera—although it often requires enabling experimental features in the browser preferences.
• Tiles. Performance benchmark for a tile-based home screen. "Apps" are "locations" on your world map, which you can customize as you like. (Here's an interesting blog entry discussing world-creation for kids.) Day/night would ultimately reflect current time, although they've been greatly sped up in this demo. Lots of rough edges and missing UI, but all the textured triangles are present, so it should be an accurate benchmark.
(Drag with left mouse button to rotate, middle mouse button to zoom, right mouse button to pan.)
• Nell at home. Basic idea (including transition) for activities which include dialog with Nell or story-telling.
Standalone model viewers: Castle (from blendswap), "Nell" (Sintel, from blendswap), Alternate (lightweight) Nell model, Alternate (heavyweight) house model (from blendswap).
In model viewers: drag with left mouse button to rotate, middle mouse button to zoom, right mouse button to pan.
• Music maker. Uses WebGL and the Web Audio APIs to let you draw and perform music.
Inspired by André Michelle's ToneMatrix and Karplus-Strong Guitar (see also wiki and this 2008 Linux Audio Conference paper), as well as DinahMoe's ToneCraft and the Tenori-on.
• Quake on XO-1.75 (video). Of course we need to actually run WebGL with good performance on XO hardware. Jon Nettleton has been working hard on our GL drivers, enabling the GPU on the XO-1.75 hardware for the first time. This Quake demo shows his progress—don't worry, Quake is not actually part of the Nell demo! (We have a GPU in the XO-1.5 as well, which hasn't yet been utilized.)
• Codify—not one of our demos (it's a commercial iPad app) but it demonstrates the direction we'd like to push Pippy.
Coming soon: TurtleArt and Implode for the web. We've started converting them to GTK3 in preparation for hoisting them bodily onto the interwebs. Here's the source code repository for the TurtleArt port if you'd like to watch or participate in this hackage. (See repl.it for one of the more unusual ways to get Python running in the web context.) The rest of the demo source code is on github (or just "View Source" in your browser).
Syndicated 2011-11-01 22:28:14 from Dr. C. Scott Ananian
Introducing Nell
Between now and January CES, Chris Ball and I will be building Nell for the OLPC XO-3 tablet. Nell is a name, not an acronym, but if you want to pronounce it as "Narrative Environment for Learning Learning," I won't stop you.
Nell's development will be demo-oriented—we're going to try to write the most interesting bits first and learn as we go—so don't be upset if you don't see support right away for legacy Sugar activities ("Sweet Nell"), robust sharing support, mesh networking, or whatever your favorite existing feature is. They'll come, but the new crazy stuff is what we need to evaluate first.
Here are four of the big ideas behind Nell, along with pointers to some of our sources of inspiration.
Narrative. I probably don't need to restate that Neil Stephenson's "The Diamond Age" has been hugely influential, and we also owe a large debt to interactive fiction and the Boston IF group in particular. (Check out the talks from our "Narrative Interfaces day" at OLPC.) Wide Ruled (conference paper) and Mark Riedl at Georgia Tech have demonstrated interesting approaches to story representation. I'm also looking forward to the results of the Experimental Game Play group's September Story Game competition.
Emotion. The Radiolab podcast "Talking To Machines" crystallized my thinking about emotionally-attractive environments. The discussion with Caleb Chung, the creator of Furby, is particularly apropos. Caleb's goal is to make things which kids want to "play with for a long time," and he contributes his three rules for creating things which "feel alive": it must (1) feel and show emotions, (2) be aware of itself and its environment, and (3) have behaviors which change over time. Furby's pursuit of these goals include expressive eyes and ears, crying when held upside down, reacting to loud noises, and gradually switching from Furbish to English for its utterances. A living thing emits a constant stream of little surprises. Expect to see Nell put the XO-3's microphone and accelerometer to good use.
Talking and Listening. The "Talking To Machines" podcast also discusses ELIZA and Cleverbot, which dovetails with my interest in the popular Speak activity for Sugar and related toys like Talking Tomcat for mobile phones. The key insight here is that a little bit of "cheap trick" AI can go a long way toward making a personable and engaging system. We want Nell to feel like a friend. Recent work by the Common Sense Computing Initiative at MIT's Media Lab shows how we can reset this on a sounder basis and use mostly-unstructured input to allow the system to grow and learn (creating "behaviors changing over time"). In particular, I'll cite ConceptNet for its database and practical NLP tools, and inspiration from "Empathy Buddy," "StoryFighter," and the other projects described in their Beating Common Sense paper. It's also worth noting that open source speech tools are good and getting better (the VoxForge site points to most of them); also interesting is this technique for matching a synthesized voice to that of the user.
Collecting, nurturing, and rewarding. Collector games such as Pocket Frogs and Flower Garden are sticky activities which encourage kids to come back to the device and continue working toward a goal over a long period of time. Memrise is educational software illustrating this technique: its users tend a garden of flowers by mastering a set of flash cards. Nell will incorporate the sticky aspects of such games, possibly also integrating the Mozilla Open Badges infrastructure into an achievement/reward system.
I hope this has given you a general sense of the direction of our Nell project. In future blog posts I'll drill down into implementation details, demonstration storyboards, and other more concrete facets of Nell.
Syndicated 2011-10-01 05:18:05 (Updated 2011-10-01 05:45:36) from Dr. C. Scott Ananian
15 Jun 2011 (updated 18 Jun 2011 at 06:04 UTC) »
Narrative Interfaces
One Laptop per Child creates student-centric learning experiences. Our current software stack, however, is somewhat "shallow". When you turn on the XO, all the content is immediately available but there is no path or guidance provided. Nothing suggests what you should try first, or indicates an order to progress through the activities provided. Everything is available, but there's no built-in journey. No plot. How can we improve this?
This Friday (June 17) at 2pm Eastern we're inviting some folks over to OLPC's new offices at the American Twine building to discuss Narrative Interfaces, as part of the proposed XO-3 software stack. Nick Montfort will give a short talk on Curveship, his model-based interactive fiction system, and Chris Ball will present some related recent hacking. Angela Chang will present her Tinkerbooks early-literacy platform, which allows kids to interactively change the written story on the page. And I'll discuss Neal Stephenson's novel The Diamond Age (a recap of a short talk I gave at EduJAM in Uruguay), and give concrete suggestions for how Diamond Age's Primer might influence the software architecture for the XO-3. (I might even reveal how to make software testing semantically indistinguishable from writing a game!) Chris Ball and I have also been collecting best-of-breed "comic books that teach you something" as examples of educational narrative; we'll pass those around and post a reading list after the event.
The real point of this meeting isn't the talks, per se, but the discussions to follow. We're trying to gather folks who know a lot more about this stuff than we do, in order to learn from them and be inspired. We don't have a lot of space, unfortunately, so I'm going to have to ask for RSVPs from those who wish to attend. If you're in the Boston area and feel like you have something to contribute (and especially if you have created/could create Creative Commons-licensed content for education), drop me a line at cscott at laptop dot org. Describing what you can contribute to the discussion will help break ties if space is inadequate.
We will also live-stream the meeting at ustream.tv/channel/cscottnet. Afterwards we'll post higher-quality video and a list of cited works. Thanks in advance to everyone who will participate, online and off!
UPDATE: video now up; see this writeup on Chris Ball's blog.
Syndicated 2011-06-15 21:27:50 (Updated 2011-06-18 05:12:14) from Dr. C. Scott Ananian
24 May 2011 (updated 29 May 2011 at 23:05 UTC) »
Small systems... and distributed ones
Today I stumbled across some very interesting projects by Nickolay Platonov which I'd like to discuss in an OLPC context.
I've been hacking away at TurtleScript fueled partly by a drive for minimalism: a small system is a learnable system. To that end, the language is based on Douglas Crockford's "Simplified JavaScript" (as recognized by this top-down operator precedence parser) which tries hard to include only JavaScript's "Good Parts". For example, the initial code for the TurtleScript system uses prototype inheritance (via `Object.create`) rather than classical class-style inheritance. In fact, the `new` operator isn't even included in the Simplified JavaScript/TurtleScript language.
In a discussion of TurtleScript Alan Kay mentioned, "It is very tricky to retain/maintain readability (so the first Smalltalk was also an extensible language not just semantically but syntactically)." And today I stumbled across the Joose library, which gives a very readable syntax for traditional class-based inheritance. It backs this up with a robust meta-object protocol, introspection, and lots of nice features borrowed from Perl 6, CLOS, and Smalltalk. The syntax ought to work fine with the limited tile set of TurtleScript, although I might have to add a tile for the `new` operator.
However, adding Joose raises some questions. Is the increase in readability worth the addition of such a large library to the system? What impact will this have on understanding problems and debugging? Is a return to class-based inheritance a positive change? (There have been arguments that the make-a-clone-and-change-it practice of prototype inheritance is easier to understand for new learners.) Can a larger overall system actually be easier to understand?
And once we're looking at libraries... Nickolay Platonov is now working on Syncler, based on the Bayou system. Unobstructive replication mechanisms would certainly make it easier to construct the sorts of collaborative applications we've wanted for Sugar. I have two concerns with Syncler's current state. First, the use of explicit continuation-passing style greatly impairs readability. The JavaScript `yield` keyword helps a lot when writing asynchronous code. (It's not supported by all JavaScript engines, but `yield` wouldn't be hard to add to TurtleScript.) Second, Syncler's event model uses explicit callbacks. I've been greatly impressed with the Flapjax event model (and its strongly-typed functional cousin). Both of these changes ought to make asynchronous code much more readable—and isn't that an important part of grokking a system?
Syndicated 2011-05-24 16:22:51 (Updated 2011-05-29 22:48:04) from Dr. C. Scott Ananian
Turtles All The Way Down
Inspired by Bert Freudenberg, Ian Piumarta, and Walter Bender, I started hacking on "Turtles All The Way Down" (aka TurtleScript) on the plane back from Uruguay. Now there's a nice rendering demo to show what a tile-based editor for JavaScript might look like, as well as a bytecode compiler and interpreter for the language. The bytecode instruction set is still too large; encouraged by Craig Chambers' work on SELF I think I ought to be able to replace all the unary and binary operators, conditional jumps, and slot selectors by a single `mapof` operator. I can put a better object model on the interpreter, too; I've written some notes on the matter.
The question is: does this really have educational value? "Turtles all the way down" is a great slogan, and a fine way to teach a graduate-level class on compiler technology, but I feel that the higher-level UI for tile-based program editing is the really useful thing for tablet computing. I'm a compiler geek and love the grungy underbelly of this stuff, but I keep reminding myself I should really be spending more time building a beautiful fluffy surface.
Syndicated 2011-05-20 06:06:15 from Dr. C. Scott Ananian
Next Steps for New Technologies
I've reached the end of the month. I've accomplished my Android and NativeClient-related goals, but didn't get the time to do as much mesh and python investigation as I'd wanted. Here are some ideas for next month's work. (Next week I'll be in Uruguay for EduJAM.)
### GObject Introspection (Android or NaCl)
1. Start by porting libffi. An android port would be straightforward, but since libffi involves code generation (ARM, x86), this is going to require a bit of assembly magic and the new "JIT"/"shared library" support in the NaCl plugin.
2. Then port gobject-introspection. GObject-Introspection relies on libffi for its guts, but the hard part of this port will be refactoring g-i's build process, which is not cross-compilation friendly. Might need to rewrite some tools. If targeting NaCl, you might consider finishing the code allowing execution of unsandboxed NaCl binaries.
3. Turn gobject-introspection on its head: generate GIR and a C binding for the platform "native" interface. For NaCl, this would be a GObject-using C-level binding of the browser-native DOM; for Android, this would be a GIR binding of the native Android APIs. These bindings should be mostly automatically generated, since they will need to continue tracking successive native platform releases/HTML5 features.
4. Demos! Change browser DOM from Python, write native Android apps in Python. Add a gobject-introspection binding to cforth, then do the same from forth. (Forth might be a simpler place to start than Python. Or not.)
### GTK (Android or NaCl)
1. Build on the cairo/pango port to proceed to a full GTK backend for Android/NaCl. These backends ought to be upstreamable. The NaCl port should be based on the broadway work: the cairo canvas would be drawn to more directly, but a lot of the mechanism which captures JavaScript events and translates them into the GTK event loop could probably be reused.
2. Demo: "Hello GTK world" in Android/NaCl.
### Sugar partitioning.
Bring Sugar closer to being a true multi-language multi-library platform.
1. Refactor sugar modules (for example, sugar toolbar widget) as standalone C libraries. Basic idea is to embed Python and export a C API, while preserving as much of the code as possible. Python libraries now invoke this library via g-i-r instead of directly. The python embedding tool is probably a useful standalone product.
2. Rewrite "Hello, Sugar" activity in C (or vala), using `#include` for `import` and GObject inheritance instead of python inheritance. Use this as a guide to pull apart sugar into modules (as above) to make this code actually work as written.
### Miscellanous topics
1. ChromeOS w/ touch support.
Find an appropriate machine, do an installation, what are the roadblocks/rough spots? Can we install on XO-1.75 as a testbed?
2. TurtleArt as JavaScript viewer/editor.
Revisit TurtleScript work, but skip over the time-consuming "construct an editor" step by reusing the (excellent) TurtleArt code.
3. Mesh: android olsrd frontend, build testbed, research 802.11 DCF issues.
### Summary
There are four rough topics here; I might try to continue the breadth-first search by spending a week on each. It might be more satisfying to downselect two of these issues and spend two weeks on each.
Syndicated 2011-04-29 15:47:57 from C. Scott Ananian
Pango/Android -vs- Pango/NaCl
At the end of my Sugar/Android week, I had a simple Pango-on-Cairo demo running. This was built on a stack of ported libraries, including gettext, pixman, freetype, libxml2, fontconfig, and glib, as well as cairo and pango. You can run the demo yourself by sideloading pango-demo.apk onto your Android device (tested on a Motorola Xoom), and you can browse the source code to see what it entailed (here's the scariest part). (I was inspired by Akita Noek's android-cairo project, but I ended up reworking the build scheme and redoing most of the ports.)
It made sense to start my Sugar/NaCl investigation by porting the same demo application to Native Client. The same stack of ported libraries was involved, although it was easy to include more functionality in the Native Client ports, including threading and PNG/PS/PDF support in cairo. The source code is a fork from the upstream naclports project, and the process was generally much cleaner. (But see my previous post for some caveats regarding naclports.) If you're using Chrome 10 or 11, you can run the demo in your browser (follow the instructions on that page). The Wesnoth team has a parallel project which ported some of these libraries as well, but not in an upstreamable manner.
The demo app uses cairo to draw the background, an animated X, and some basic text in the center; it uses Pango's advanced international text support to draw properly-shaped Persian text in a circle around it. The center text is the "proper" bilingual Greek/Japanese written form of "pango"; the text around the edges is the Persian name of the internationalization library, "harfbuzz". Note that the Persian text is written right-to-left—and that I didn't put a full CJK font in the NaCl app, so the Japanese "go" character is missing. The Android port rebuilds the font cache at each startup, so it loads rather slowly; the NaCl port contains a prebuilt font cache so it starts more quickly.
Both ports took about two weeks. I blew my original schedule, partly due to the Patriot's day holiday, and partly because I'd given Android about a week's head start by tinkering on it before my original schedule post. The framerate of the demo is much better on NaCl (so fast that the edges of the animated X look choppy in the screenshot), but the hardware isn't easily comparable, so the comparison doesn't really tell us much. The porting effort was certainly more pleasant on NaCl, since newlib is a much more complete libc than Android's "Bionic"—but having gdb available made debugging on Android easier. (There is an unintegrated NaCl branch that integrates NaCl gdb in the browser, though!)
Much of the GNOME/POSIX library stack assumes access to a filesystem tree and does file-based configuration. In our demo application, fontconfig was the most culpable party: it wanted to load a configuration file describing font locations and naming, then to load the fonts themselves from the file system, and finally to write a cache file describing what it found back to the file system. Most ported software is going to want similar access—even if you store the user's own documents in a Journal, software still expects to find configuration, caches, and other data in a filesystem.
Android provides the POSIX filesystem APIs, but the filesystem an app can touch is segmented and sandboxed. As discussed previously, Android's Opaque Binary Blob feature may allow you to create a app-specific filesystem, but this doesn't let you share (for example) fonts and font configuration between activities. NaCl might eventually provide a similar unshared mechanism based on the HTML5 AppCache.
The preferred solution is more limited, but more flexible: no built-in filesystem APIs are used (or in NaCl's case, provided!) at all. Instead, you provide your own implementation of the POSIX file APIs (either via the --wrap linker indirection or through an appropriate backend to newlib/glibc/glib). For the NaCl demo app, I wrote a rather-elaborate in-memory filesystem --- only to find that an even-more-elaborate one already existed in naclports. But the longer-term solution uses message-passing (SRPC in NaCl, intents in Android) to implement these POSIX APIs. In Native Client, the implementation would be in browser-side JavaScript, which would then allow you to share parts of the filesystem tree between activities and/or map it into (cached) web-addressed resources. In either case, your application still sees the bog-standard POSIX API it expects.
More problematic are the networking APIs. Here Android provides a pretty standard socket library, while Native Client provides nothing at all. Using a browser-based implementation, as for the file APIs, will work fine for HTTP, WebSockets and even P2P via the HTML5 P2P APIs. But it's not clear that (for example) glib's elaborate asynchronous DNS name resolver implementation can (or should!) be implemented in a NaCl port.
In the end, the porting effort and abstraction shifts needed for Native Client and Android are roughly comparable. I expect Native Client will hold a strong edge in allowing close integration with web standards and web technologies. Android will probably continue to hold an edge in third-party application support and platform maturity.
Syndicated 2011-04-29 15:08:56 from C. Scott Ananian
Sugar-on-Native Client investigation
This post will describe the state of Native Client in general, based on week 2 of my original four week plan. In the next post, I'll link to my work so far, and compare the Native Client and the Android efforts. Recapping, the end goal of these explorations is a platform for the next generation of the Sugar learning environment.
To begin, the Native Client (NaCl) plugin is fairly mature in a number of areas. Version 0.2 of the NaCl SDK was recently released (a version number which substantiates the "fairly" in my previous sentence), and the NativeClient plugin is currently shipping in Chrome (versions 10 and 11), although you have to manually turn on a preference in the `about:flags` dialog to enable it. The NaCl toolchain is much more standard than the Android NDK toolchain I discussed previously, and the robust naclports tree shows that the patches required for NaCl ports of common packages tend not to be too evil. The Tcl interpreter and Qt tookit port demos show that fairly complex pieces of code can be deployed today on NaCl.
On the other hand, there are three main difficulties:
1. The default NaCl toolchain uses newlib as its standard C library. This is consistent with Google's preference for BSD-licensed code in SDKs they provide to the public (see the discussion of Bionic in the Android SDK). However, there also exists a branch of the SDK which uses glibc. The glibc branch supports several additional features, like shared library support. However, it is unclear whether this will ever be a "supported" part of the SDK. If glibc does become supported, it is unlikely ever to be the only supported libc; the BSD-licensed newlib will need to remain available as an option. (Yes, the LGPL license of glibc shouldn't inspire such paranoia, but Google has elected not to undertake the education of all prospective third-party developers.)
2. The naclports project, although fairly robust, is driven between the Scylla and Charybdis of compatibility. The goal is that all the code in naclports be buildable at all times on all three major platforms: Windows, Mac, and Linux. Further, it should support both x86_32 and x86_64 backends, and ideally ARM and pNaCl as well. It's auto-built against the latest SDK sources, but should also work on the latest released SDK. And with the addition of the glibc/newlib split discussed above, the possible build targets are multiplied further. Needless to say, keeping the tree building against such a large number of variants is not an easy task, and naclports is usually broken in some way. In practice, most developers seem to pay attention to some subset (say, x86_32/newlib/Linux host), but it's hard to push patches upstream without worrying about breaking some obscure target. It might be best to base future work on a proper package technology, like (say) dpkg-cross.
3. In general, a lot of interesting NaCl development has occurred on branches that are not easily integrated. I've already mentioned glibc support, which is a toolchain branch; shared library support is on another branch that requires a new chromium plugin as well. At various times different means have been implemented to run NaCl binaries "natively" outside the sandbox (for example, in order to test some feature at build time, or auto-generate some piece of code via introspection). These efforts live on abandoned branches, while the "official" means to do this is incomplete. Similarly, a lot of interesting NaCl work used the now-abandoned legacy "NPAPI" plugin interface to interact with the browser. It was followed by the "Pepper" plugin interface, which was itself abandoned. Current work uses the Pepper2 browser plugin APIs, which (unfortunately) have not yet been implemented in non-Chrome browsers and continue to flux about. Many interesting browser interactions exist only in deprecated Pepper APIs, not having yet been built into Pepper2. ARM and pNaCl work also appears to be on unintegrated branches. There are a number of different gdb support strategies.
None of these difficulties is insurmountable—and in fact, some are side-effects of the desirable active development and productization of Native Client. To date I've done my work on the (more compatible) SDK v0.1 and the (more upstreamable) newlib library. So far newlib has not been a huge obstacle, and this basis allows my patches and ports to be more broadly useful. This might change in the future—certainly at some point we need to move to ARM and/or pNaCl for XO-3, which will probably require building chrome and the NaCl toolchain from scratch. At that point, it may be worth further exploring the non-mainstream branches.
Syndicated 2011-04-29 14:24:36 (Updated 2011-04-29 16:43:48) from C. Scott Ananian
How does the iPad "use the iPhone's GPS"?
A few months ago, a number of stories came out covering the iPad's remarkable-seeming ability to share the GPS of a tethered iPhone. Apple's latest location database FAQ confirms the suspicions I voiced at the time: there's no actual GPS sharing involved. Instead, Apple is using the simultaneous GPS and Wifi radios on your iPhone to "crowd-source" what I'll call a "skyhook" database (after the first company to publicly use the technique). This correlates Wifi base station identifiers with their GPS locations in real time -- including (most likely) the real time location of the "base station" created by the iPhone when it is in tethering mode. All nearby Apple devices use this database to compute their location (based on all visible wifi base stations). Since the nearby device sees the iPhone's "base station" and the iPhone is busily updating the position of that "base station" in real time (along with all the other base stations the iPhone can see), the iPad (lacking a GPS of its own) gains the apparent magical ability to compute a very accurate position for itself.
The real interesting part of this story involves user consent and privacy—do iPhone users generally know that their devices are registering their location in Apple's database in real time whenever tethering is turned on? Any device which can query Apple's location database for the MAC address of your iPhone can track the position of your iPhone whenever you are tethering. That's basically what the magical ability of the iPad/iPhone pair tells us. Did you know that?
Syndicated 2011-04-27 18:30:23 (Updated 2011-04-27 18:30:50) from C. Scott Ananian
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If you're a C programmer with some spare time, take a look at the mod_virgule project page and help us with one of the tasks on the ToDo list! | 7,806 | 35,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.964418 |
https://www.reference.com/science/convert-square-feet-feet-abaf5d068e99ad29 | 1,485,209,305,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560283008.19/warc/CC-MAIN-20170116095123-00217-ip-10-171-10-70.ec2.internal.warc.gz | 969,403,967 | 20,490 | Q:
How do you convert square feet into feet?
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Square feet cannot be converted into feet because the two units measure different attributes of objects. Square feet are units of measurement for area, the space covered by an object, while feet are units of measurement for length and other measurements of the distance between two points.
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A square foot is defined as the area of a square that measures one foot on each of its four sides. The formula for calculating the area of a rectangle is expressed as length times width; the designations are interchangeable for the purpose of this calculation because both refer to the distance between two points. In squares, which are a subset of rectangles, all four sides measure the same and need not be designated as length or width. The formula for area of a square is a², or the length of any side (represented by the variable a) multiplied by itself, or "squared."
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PEOPLE SEARCH FOR | 368 | 1,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-04 | longest | en | 0.921982 |
https://www.mail-archive.com/perl6-language@perl.org/msg32008.html | 1,542,048,429,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741016.16/warc/CC-MAIN-20181112172845-20181112194845-00192.warc.gz | 946,808,645 | 3,477 | # Re: Ordering in \bbold{C}
```Maybe it's just me, but I don't see the value of having some
*arbitrary* predefined order for complex numbers. If people really
want to order their complex numbers, let them do it themselves in
whatever way they want.```
```
Leon
On Mon, Mar 29, 2010 at 6:10 AM, Darren Duncan <dar...@darrenduncan.net> wrote:
> I was actually thinking, during the previous thread involving Complex
> numbers ...
>
> It may not have any practical use, but if one wanted to define an ordering
> for complex numbers that was deterministic and relatively unbiased, a way to
> do this would be based on what I'll call for now the "spiral distance".
>
> Conceptually, you take an infinite length spiral line that starts at and is
> centered on the origin, where for each turn the current spot on the spiral
> increases an infinitesimal radius from the origin, or a distance approaching
> zero, in the calculus sense. Complex numbers closer to the origin on the
> spiral will be ordered earlier than those further from the spiral.
>
> Actually calculating this is a simple comparison of the radius and angle
> components of the two complex numbers in the polar coordinate system. If
> the radius value is different, then the one with the smaller radius is
> ordered before the one with the larger; if the two radius values are the
> same, then the one with the smaller angle is ordered first; if both are the
> same, then the two complex numbers are equal.
>
> The math is just as simple as a naive comparison that just compares the real
> component and then imaginary component in a cartesian coordinate system, but
> the result is much more reasonable I think.
>
> This whole principle of "distance from origin" method of ordering does also,
> I suspect, scale to any number of dimensions; the one-dimensional version is
> simply comparing first the absolute value of the two numbers, and then
> saying that either the positive or negative version orders first.
>
> -- Darren Duncan
>
``` | 456 | 2,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-47 | latest | en | 0.920364 |
https://www.cfd-online.com/Forums/openfoam/78236-conditional-operation-geometric-field.html | 1,503,110,925,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105291.88/warc/CC-MAIN-20170819012514-20170819032514-00426.warc.gz | 870,084,749 | 16,206 | # Conditional Operation on a Geometric Field
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July 15, 2010, 16:52 Conditional Operation on a Geometric Field #1 New Member Sachin Shanbhag Join Date: Nov 2009 Posts: 3 Rep Power: 9 Sponsored Links This may be a relatively simple question. I am simplifying the actual problem, for clarity. Say variable X is of type volScalarField, and takes values between 0 and 1.0. I want to define a dependent variable Y (type volScalarField) whose definition depends on the local value of X. Something like if(X>0.5) { Y = exp(-X); } else { Y = 2*X; } I tried to do this literally, and it said something like error: no match for 'operator>' in 'Foam:perator*(const Foam::GeometricField&, const Foam::scalar&) [...]
July 15, 2010, 17:55 #2 Senior Member Marco A. Turcios Join Date: Mar 2009 Location: Vancouver, BC, Canada Posts: 734 Rep Power: 21 Because this operation has to be done cell by cell, you need to use something like: forAll (X, celli) { if(X[celli]>0.5) { Y[celli] = exp(-X[celli]);} else { Y[celli] = 2*X[celli];} } If the field has dimensions, then you're going to have to make sure you deal with them properly, but other than that I think that's how it goes down.
July 16, 2010, 12:02 #3 New Member Sachin Shanbhag Join Date: Nov 2009 Posts: 3 Rep Power: 9 Thanks mturcios.
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https://chemistry.stackexchange.com/questions/74964/solving-a-kinetic-rate-law | 1,563,568,838,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526359.16/warc/CC-MAIN-20190719202605-20190719224605-00280.warc.gz | 363,863,738 | 35,545 | # Solving a kinetic rate law
Here is a problem I just don't see how to get around... The following equation describes the oxidation by $\ce{O_2}$ of a phenolic compound, catalysed by an enzyme $\ce{E}$.
$\ce{4R–OH + O_2 ->[r] 4R–O. +H_2O}$
$r=k\frac{\ce{[R-OH]}}{\ce{K_M +[R-OH]}}[\ce{E}][\ce{O_2}]$
Rate $r$ is the speed at which $[\ce{R–OH}]$ is consumed. It tells us that our reaction follows a Michaelis-Menten kinetic law with respect to our substrate $[\ce{R–OH}]$ and first order kinetics with respect to oxygen.
First, we are asked to find a way by which the rate equation could be simplified by approximation.
We are told that initially, $[\ce{R–OH}] = 10 mol/m^3$ and that we want a $95\%$ conversion of that reactant. We also know that $K_M =0.04 mol/m^3$. Therefore $[\ce{R–OH}]$ can vary between being $250$ times and $12.5$ times greater than $K_M$, so we could approximate the problem by getting rid of $K_M$ which would leave us with the following rate: $r=k[\ce{E}][\ce{O_2}]$. Since $[\ce{E}]$ is constant, we could write $r=k'[\ce{O_2}]$. (This sort of trick has been used before in my chemical engineering class.)
The next question is the one I am struggling with.
We are asked to use our new kinetic rate law to write down the material balance equation expressing how $[\ce{R–OH}]$ would changes with time in a batch reactor (perfectly stirred, no input or output), then solve for $[\ce{R–OH}]$.
So I would be inclined to write $\frac{d[\ce{ROH}]}{dt}=-r=-k'[\ce{O_2}]$ but then I don't see how this can be solved since we have two different variables.
From the stoichiometry we can see $\frac{d[\ce{ROH}]}{dt}=4\frac{d[\ce{O_2}]}{dt}$ therefore $\frac{d[\ce{O_2}]}{dt} = -\frac{1}{4}k'[\ce{O_2}]$ and this can be integrated to give us $[\ce{O_2}]=[\ce{O_2}]_0e^{-k't/4}$. But how can the rate law be solved for $[\ce{R–OH}]$ ?
Any insight into this problem would be hugely appreciated.
• Hello, and Welcome to StackExchange. Looks like a pretty good question for a first post. You can write chemical equations using \ce. For example, $$\ce{NH4OH <=>[k_1][k_2] NH4+ + OH-}$$ would give: $$\ce{NH4OH <=>[k_1][k_2] NH4+ + OH-}$$ Good luck! – Pritt Balagopal May 22 '17 at 14:55
• In your simplification of the rxn. equation, you are assuming that the change in [R-OH] negligibly impacts the rate, so your mass balance would reflect that with the new rate law. Have you tried writing an expression for the concentration of $[\ce{O_2}]$ in terms of R-OH, knowing that overall mass is conserved? – J. Ari May 22 '17 at 15:38
• Is there a reason why you cannot relate the concentration of oxygen with the concentration of the alcohol? – CoffeeIsLife May 23 '17 at 6:38 | 858 | 2,695 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-30 | latest | en | 0.923186 |
https://search.yahoo.com/mobile/s?p=geographic+coordinate+system+wikipedia+in+english+translation+dictionary&ei=UTF-8&fr2=p%3As%2Cv%3Aw%2Cm%3Ars-bottom%2Cct%3Agossip&_tsrc=apple&b=29&pz=7&bct=0&xargs=0 | 1,685,496,557,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00194.warc.gz | 588,260,484 | 42,839 | # Yahoo Web Search
1. ## About 415,000 search results
1. ### en.wikipedia.org › wiki › Geographic_coordinate_systemGeographic coordinate system - Wikipedia
The geographic coordinate system (GCS) is a spherical or geodetic coordinates system for measuring and communicating positions directly on the Earth as latitude and longitude. It is the simplest, oldest and most widely used of the various spatial reference systems that are in use, and forms the basis for most others.
2. ### simple.wikipedia.org › wiki › Geographic_coordinateGeographic coordinate system - Simple English Wikipedia, the ...
A geographical coordinate system is a coordinate system. This means that every place can be specified by a set of three numbers, called coordinates. A full circle can be divided into 360 degrees (or 360°); this was first done by the Babylonians; Ancient Greeks, like Ptolemy later extended the theory. Today, degrees are divided further.
3. ### static.hlt.bme.hu › semantics › externalGeographic coordinate system - Wikipedia - BME
A geographic coordinate system is a coordinate system that enables every location on Earth to be specified by a set of numbers, letters or symbols. [n 1] The coordinates are often chosen such that one of the numbers represents a vertical position and two or three of the numbers represent a horizontal position; alternatively, a geographic positio...
4. ### www.esri.com › coordinate-systems-differenceCoordinate Systems: What's the Difference? - Esri
Mar 3, 2020 · A geographic coordinate system (GCS) is a reference framework that defines the locations of features on a model of the earth. It’s shaped like a globe—spherical. Its units are angular, usually degrees. A projected coordinate system (PCS) is flat.
5. ### wiki.gis.com › State_Plane_Coordinate_SystemState Plane Coordinate System - GIS Wiki | The GIS Encyclopedia
Sep 23, 2018 · The State Plane Coordinate System (SPS or SPCS) is a set of 126 geographic zones or coordinate systems designed for specific regions of the United States. Each state contains one or more state plane zones, the boundaries of which usually follow county lines.
6. ### www.esri.com › arcgis-blog › productsAbout geographic transformations and how to choose the ... - Esri
May 6, 2009 · Unproject the data to geodetic latitude and longitude using the same GCS. Transform the data to geodetic latitude and longitude using the new GCS. Project the data to the new PCS using the new GCS. You can see that when you select the PCS you want to use, you also need to select the geographic transformation (step 3 above) because there are ...
7. ### www.wikidata.org › wiki › Q22664geographic coordinate system - Wikidata
May 20, 2023 · Also known as. English. geographic coordinate system. system to specify locations on Earth. Geo Position. Geo-coordinates. Geographic References. Geo coordinates. Geo-co-ordinates.
8. ### desktop.arcgis.com › en › arcmapGeographic transformation methods—ArcMap | Documentation - Esri
A geographic transformation always converts geographic (latitude–longitude) coordinates. Some methods convert the geographic coordinates to geocentric (X,Y,Z) coordinates, transform the X,Y,Z coordinates, and convert the new values back to geographic coordinates. These include the Geocentric Translation, Molodensky, and Coordinate Frame methods.
9. ### www.ibm.com › docs › enGeographic coordinate systems - IBM
A geographic coordinate system The lines that run east and west each have a constant latitude value and are called parallels. They are equidistant and parallel to one another, and form concentric circles around the earth. The equatoris the largest circle and divides the earth in half. It is equal in distance
10. ### www.scribd.com › document › 275325549Geographic Coordinate System | PDF | Latitude | Longitude
A geographic coordinate system is a coordinate sys- the motion, while France and Brazil abstained. France tem that enables every location on the Earth to be spec- adopted Greenwich Mean Time in place of local deteried by a set of numbers or letters. [n 1] The coordinates minations by the Paris Observatory in 1911.
11. ### developers.arcgis.com › documentation › spatialSpatial references | Documentation | ArcGIS Developers
A geographic coordinate system (GCS) uses a three-dimensional ellipsoidal surface to define locations on the Earth. These coordinates are based on angles from the center of the Earth to the surface. Typically GCSs use latitude and longitude specified in degrees. There are three parts to a geographic coordinate system:
12. ### pro.arcgis.com › en › pro-appGeographic datum transformations—ArcGIS Pro | Documentation
Other versions | Help archive A geographic datum transformation is a calculation used to convert between two geographic coordinate systems to ensure that data is properly aligned within a map. Geographic coordinate systems describe how locations on the earth are placed on a hypothetical reference spheroid.
13. ### www.esri.com › arcgis-blog › productsGeographic vs Projected Coordinate Systems - Esri
A geographic coordinate system (GCS) is used to define locations on a model of the surface of the earth. The GCS uses a network of imaginary lines (longitude and latitude) to define locations. This network is called a graticule. So why isn’t knowing the latitude and longitude of a location good enough to know where it is? | 1,138 | 5,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.785307 |
https://admin.clutchprep.com/chemistry/practice-problems/43931/what-is-the-boiling-point-of-a-solution-of-0-150-mole-al-no3-3-in-0-500-kg-water | 1,597,407,728,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739211.34/warc/CC-MAIN-20200814100602-20200814130602-00576.warc.gz | 199,234,270 | 22,653 | # Problem: What is the boiling point of a solution of 0.150 mole Al(NO3)3 in 0.500 kg water? Kb of water is 0.512 °C/m and the boiling point of water is 100.00°C A) 100.15 °C B) 100.61 °C C) 99.85 °C D) 101.50°C E) 99.39°C
🤓 Based on our data, we think this question is relevant for Professor Senevirathne's class at SJU.
###### Problem Details
What is the boiling point of a solution of 0.150 mole Al(NO3)3 in 0.500 kg water? Kb of water is 0.512 °C/m and the boiling point of water is 100.00°C
A) 100.15 °C
B) 100.61 °C
C) 99.85 °C
D) 101.50°C
E) 99.39°C
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Boiling Point Elevation concept. If you need more Boiling Point Elevation practice, you can also practice Boiling Point Elevation practice problems.
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Based on our data, we think this problem is relevant for Professor Senevirathne's class at SJU. | 386 | 1,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-34 | latest | en | 0.900168 |
https://www.futureschool.com/international-baccalaureate-curriculum/mathematics-vectors-topic-5/ | 1,550,477,377,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484772.43/warc/CC-MAIN-20190218074121-20190218100121-00311.warc.gz | 839,875,651 | 8,989 | ### Topic 5 – Vectors Mathematics
# TOPIC TITLE
1 Study Plan Study plan – Topic 5 – Vectors
Objective: On completion of the course formative assessment a tailored study plan is created identifying the lessons requiring revision.
2 Vectors Vectors
Objective: On completion of the lesson the student will be able to represent a vector in matrix and diagrammatic form, as well as add two vectors using matrices and/or a diagram.
3 Simultaneous equations Number of solutions (Stage 2)
Objective: On completion of the lesson of the lesson the student will identify simultaneous equations that are consistent, inconsistent or the same.
4 Vectors 2 vector addition in 2 and 3D (stage 2)
Objective: On completion of the lesson the student will understand and use component forms for vector resolution.
5 Linear systems Optimal solutions (Stage 2) – Vectors
Objective: On completion of the lesson the student will understand the process of linear programming to find optimal solutions.
6 Linear systems Linear systems with matrices (Stage 2)
Objective: On completion of the lesson the student will process matrices formed from linear systems of equations.
7 Linear systems Row-echelon form (Stage 2)
Objective: On completion of the lesson the student will process matrices formed from linear systems of equations using the row-echelon form.
8 Linear systems Gauss Jordan elimination method (Stage 2)
Objective: On completion of the lesson the student will process matrices formed from linear systems of equations using the Gauss Jordan elimination method.
9 Exam Exam – Topic 5 – Vectors
Objective: Exam | 325 | 1,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-09 | longest | en | 0.872165 |
http://oeis.org/A030133 | 1,558,300,174,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255165.2/warc/CC-MAIN-20190519201521-20190519223521-00285.warc.gz | 148,662,799 | 3,995 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A030133 a(n+1) = sum of digits of (a(n) + a(n-1)). 6
2, 1, 3, 4, 7, 2, 9, 2, 2, 4, 6, 1, 7, 8, 6, 5, 2, 7, 9, 7, 7, 5, 3, 8, 2, 1, 3, 4, 7, 2, 9, 2, 2, 4, 6, 1, 7, 8, 6, 5, 2, 7, 9, 7, 7, 5, 3, 8, 2, 1, 3, 4, 7, 2, 9, 2, 2, 4, 6, 1, 7, 8, 6, 5, 2, 7, 9, 7, 7, 5, 3, 8, 2, 1, 3, 4, 7, 2, 9, 2, 2, 4, 6, 1, 7, 8, 6, 5, 2, 7, 9, 7, 7, 5, 3, 8, 2, 1, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS a(n) = A010888(A000032(n)). - Reinhard Zumkeller, Aug 20 2011 Similar to the digital roots of several Fibonacci sequences, this digital root sequence for Lucas numbers (A000032) has period 24 with digits summing to 117. Decimal expansion of 23719213606865169775282 / 111111111111111111111111 = 0.[213472922461786527977538] (periodic). - Daniel Forgues, Feb 27 2017 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 0..10000 Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1). FORMULA a(n+24) = a(n); a(A017593(n)) = 9. - Reinhard Zumkeller, Jul 04 2007 MATHEMATICA Transpose[NestList[{Last[#], Total[IntegerDigits[Total[#]]]}&, {2, 1}, 100]] [[1]] (* Harvey P. Dale, Jul 25 2011 *) PROG (Haskell) a030133 n = a030133_list !! n a030133_list = 2 : 1 : map a007953 (zipWith (+) a030133_list \$ tail a030133_list) -- Reinhard Zumkeller, Aug 20 2011 (PARI) V=[2, 1]; for(n=1, 100, V=concat(V, sumdigits(V[n]+V[n+1]))); V \\ Derek Orr, Feb 27 2017 CROSSREFS Cf. A030132, A007953, A049341. Sequence in context: A325271 A060214 A259773 * A139374 A111958 A192439 Adjacent sequences: A030130 A030131 A030132 * A030134 A030135 A030136 KEYWORD nonn,base,nice,easy AUTHOR STATUS approved
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Last modified May 19 17:08 EDT 2019. Contains 323395 sequences. (Running on oeis4.) | 943 | 2,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.495367 |
https://www.scribd.com/document/340319939/Cost-Analysis-Tool | 1,539,950,706,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512395.23/warc/CC-MAIN-20181019103957-20181019125457-00173.warc.gz | 1,058,872,394 | 39,975 | # EXAMPLE
Asset Type LIFE CYCLE COST ANALYSIS TOOL
Useful Life Current Maintenance Practices
First Cost BASED ON ______________________
Projected
Brake Total
Inflate Rate *1 ASSET LIFE PM & Inspection Engine RR Trans. RR program Tire program Road Calls Ownership
# per Year Year Year Year # per
PM & Insp. year Cost (x) Cost (x) Cost (x) Cost (x) Cost year Cost
Engine R/R Year 1 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Trans R/R Year 2 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Brake program Year 3 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Tire program Year 4 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Miles/Year Year 5 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Miles/PM Year 6 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Engine Miles Year 7 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Trans Miles Year 8 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Brake Miles Year 9 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Tire Miles Year 10 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Road Calls/per call Year 11 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Multipliers based on inflation Year 12 \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Year 1 1.00
Year 2 1 TOTALS \$ - \$ - \$ - \$ - \$ - \$ - \$ -
Year 3 1
Year 4 1.0000
Year 5 1.0000 *1 The rate used as the inflation factor is agency specific, consult your finance officer for the applicable rate in your area
Year 6 1.0000
Year 7 1.0000
Year 8 1.0000
Year 9 1.0000
Year 10 1.0000
Year 11 1.0000
Year 12 1.0000
0000 *1 Inclusion of and inflation rate is optional in calculating costs Year 6 1. \$ .00 Year 2 1 TOTALS \$ . \$ - Brake Miles Year 9 \$ . \$ . \$ . \$ . \$ . \$ . \$ - Miles/PM Year 6 \$ . \$ . \$ . \$ - Tire Program Year 4 \$ .0000 Year 9 1. \$ . \$ . \$ . \$ . \$ . \$ . \$ . \$ . \$ .0000 Year 5 1. \$ . \$ . \$ . \$ . \$ - Road Calls/per call Year 11 \$ . \$ . \$ . \$ .0000 Year 10 1. \$ . \$ . \$ - Year 3 1 Year 4 1. \$ . \$ . \$ . \$ - Brake Program Year 3 \$ .0000 Year 7 1.0000 Year 12 1. \$ . \$ . \$ . \$ .0000 Year 8 1. \$ . \$ .0000 . \$ - Multipliers based on inflation Year 12 \$ . \$ . \$ . \$ . \$ . \$ - Trans Miles Year 8 \$ . \$ - Tire Miles Year 10 \$ . \$ . \$ . \$ - Miles/Year Year 5 \$ . RR program Tire program Road Calls # # per Year Year Year Year per PM & Insp. \$ - Engine Miles Year 7 \$ . \$ . EXAMPLE Asset Type LIFE CYCLE COST ANALYSIS TOOL Useful Life Alternate Mainteance Practices First Cost BASED ON ______________________________ Brake Inflate Rate *1 ASSET LIFE PM & Inspection Engine RR Trans. year Cost (x) Cost (x) Cost (x) Cost (x) Cost year Cost Engine RR Year 1 \$ . \$ . \$ . \$ . \$ . \$ . \$ . \$ - Trans RR Year 2 \$ . \$ . \$ . \$ .0000 Year 11 1. \$ . \$ - Year 1 1.
Projected Total Ownership \$ - \$ - \$ - \$ - \$ - \$ - \$ - \$ - \$ - \$ - \$ - \$ - \$ - .
816 Year 1 \$ . \$ - Total \$ . \$ - 9 \$ . \$109.252 Practices. \$ - 11 \$ . \$ - \$1 4 \$ . \$253. \$ - \$1 10 \$ . \$253. \$ - 8 \$ . \$ - 3 \$ .252 Projected Cost Alternate Maintenance Practices. Life Cycle Cost Analysis Chart Projected Projected Cost Cost- \$1 Current Alternate Maintenan Maintenanc ce e Practices. \$ - \$1 7 \$ . \$ - \$0 12 \$ . \$ - 6 \$ . \$1 \$109. \$ - 5 \$ . \$ - \$0 \$0 \$0 \$- 1 2 3 4 5 6 7 8 9 10 11 12 Projected Cost-Current Maintenance Practices. \$ - \$1 2 \$ .816 . | 1,091 | 3,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-43 | latest | en | 0.188363 |
https://www.coursehero.com/file/6718051/Pre-Calc-Exam-Notes-76/ | 1,481,210,990,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541529.0/warc/CC-MAIN-20161202170901-00486-ip-10-31-129-80.ec2.internal.warc.gz | 917,266,503 | 72,539 | Pre-Calc Exam Notes 76
# Pre-Calc Exam Notes 76 - . 5. Use 15 = 45 30 to nd the...
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76 Chapter 3 Identities §3.2 Solution: Multiply the top and bottom of r 1 2 s by sin θ 1 sin θ 2 to get: r 1 2 s = n 1 cos θ 1 n 2 cos θ 2 n 1 cos θ 1 + n 2 cos θ 2 · sin θ 1 sin θ 2 sin θ 1 sin θ 2 = ( n 1 sin θ 1 ) sin θ 2 cos θ 1 ( n 2 sin θ 2 ) cos θ 2 sin θ 1 ( n 1 sin θ 1 ) sin θ 2 cos θ 1 + ( n 2 sin θ 2 ) cos θ 2 sin θ 1 = ( n 1 sin θ 1 ) sin θ 2 cos θ 1 ( n 1 sin θ 1 ) cos θ 2 sin θ 1 ( n 1 sin θ 1 ) sin θ 2 cos θ 1 + ( n 1 sin θ 1 ) cos θ 2 sin θ 1 (by Snell’s law) = sin θ 2 cos θ 1 cos θ 2 sin θ 1 sin θ 2 cos θ 1 + cos θ 2 sin θ 1 = sin ( θ 2 θ 1 ) sin ( θ 2 + θ 1 ) The last two examples demonstrate an important aspect of how identities are used in practice: recognizing terms which are part of known identities, so that they can be factored out. This is a common technique. Exercises 1. Verify the addition formulas (3.12) and (3.13) for A = B = 0 . For Exercises 2 and 3, ±nd the exact values of sin ( A + B ), cos ( A + B ), and tan ( A + B ). 2. sin A = 8 17 , cos A = 15 17 , sin B = 24 25 , cos B = 7 25 3. sin A = 40 41 , cos A = 9 41 , sin B = 20 29 , cos B = 21 29 4. Use 75 = 45 + 30 to ±nd the exact value of sin 75
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Unformatted text preview: . 5. Use 15 = 45 30 to nd the exact value of tan 15 . 6. Prove the identity sin + cos = r 2 sin ( + 45 ) . Explain why this shows that r 2 sin + cos r 2 for all angles . For which between 0 and 360 would sin + cos be the largest? For Exercises 7-14, prove the given identity. 7. cos ( A + B + C ) = cos A cos B cos C cos A sin B sin C sin A cos B sin C sin A sin B cos C 8. tan ( A + B + C ) = tan A + tan B + tan C tan A tan B tan C 1 tan B tan C tan A tan C tan A tan B 9. cot ( A + B ) = cot A cot B 1 cot A + cot B 10. cot ( A B ) = cot A cot B + 1 cot B cot A...
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Ask a homework question - tutors are online | 772 | 2,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2016-50 | longest | en | 0.534602 |
https://vivaelsoftwarelibre.com/product/course-generalized-linear-models-in-r/ | 1,571,308,700,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986673538.21/warc/CC-MAIN-20191017095726-20191017123226-00307.warc.gz | 763,654,681 | 41,346 | # Course Generalized Linear Models in R
The Generalized Linear Model course expands your regression toolbox to include logistic and Poisson regression.
DescriptionChaptersExercisesInstructor
This online course about Generalized Linear Models in R covers a key part of what a future data analyst would require.
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This chapter teaches you about interpreting GLM coefficients and plotting GLMs using ggplot2.
Chapter 4: Multiple regression with GLMs
In this chapter, you will learn how to do multiple regression with GLMs in R.
Chapter 5:
Chapter 6:
Chapter 7:
Chapter 8:
Chapter 9:
Chapter 10:
Chapter 11:
Chapter 12:
Richard Erickson
Quantitative Ecologist
Richard helps people to experience and understand their increasingly numerical world. For his day job he develops new quantitative methods for monitoring and controlling invasive species as well as helping other scientists analyze and understand their data. He has worked on diverse datasets ranging from continent wide species distributions to pesticides in playa wetlands. After hours, he teaches SCUBA Diving as a NAUI Instructor. He has been a “UserR” since 2007.
Collaborators
#R #Python #MachineLearning #BigData #DataAnalysis | 508 | 2,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.902532 |
https://www.findstat.org/MapsDatabase/Mp00090.txt/ | 1,696,021,212,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-40/segments/1695233510528.86/warc/CC-MAIN-20230929190403-20230929220403-00197.warc.gz | 843,864,323 | 7,090 | ************************************************************************ * www.FindStat.org - The Combinatorial Statistic Finder * * * * Copyright (C) 2013 The FindStatCrew * * * * This information is distributed in the hope that it will be * * useful, but WITHOUT ANY WARRANTY; without even the implied * * warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. * ************************************************************************ ------------------------------------------------------------------------ Map identifier: Mp00090 ------------------------------------------------------------------------ Map name: cycle-as-one-line notation ------------------------------------------------------------------------ Domain: Permutations ------------------------------------------------------------------------ Codomain: Permutations ------------------------------------------------------------------------ Description: Return the permutation obtained by concatenating the cycles of a permutation, each written with minimal element first, sorted by minimal element. ------------------------------------------------------------------------ References: [1] Can, M. B., Cherniavsky, Y. Omitting parentheses from the cyclic notation [[MathSciNet:3416856]] ------------------------------------------------------------------------ Code: def mapping(pi): return Permutation(sum([list(t) for t in pi.cycle_tuples()],[])) ------------------------------------------------------------------------ Map images: [1] => [1] [1,2] => [1,2] [2,1] => [1,2] [1,2,3] => [1,2,3] [1,3,2] => [1,2,3] [2,1,3] => [1,2,3] [2,3,1] => [1,2,3] [3,1,2] => [1,3,2] [3,2,1] => [1,3,2] [1,2,3,4] => [1,2,3,4] [1,2,4,3] => [1,2,3,4] [1,3,2,4] => [1,2,3,4] [1,3,4,2] => [1,2,3,4] [1,4,2,3] => [1,2,4,3] [1,4,3,2] => [1,2,4,3] [2,1,3,4] => [1,2,3,4] [2,1,4,3] => [1,2,3,4] [2,3,1,4] => [1,2,3,4] [2,3,4,1] => [1,2,3,4] [2,4,1,3] => [1,2,4,3] [2,4,3,1] => [1,2,4,3] [3,1,2,4] => [1,3,2,4] [3,1,4,2] => [1,3,4,2] [3,2,1,4] => [1,3,2,4] [3,2,4,1] => [1,3,4,2] [3,4,1,2] => [1,3,2,4] [3,4,2,1] => [1,3,2,4] [4,1,2,3] => [1,4,3,2] [4,1,3,2] => [1,4,2,3] [4,2,1,3] => [1,4,3,2] [4,2,3,1] => [1,4,2,3] [4,3,1,2] => [1,4,2,3] [4,3,2,1] => [1,4,2,3] [1,2,3,4,5] => [1,2,3,4,5] [1,2,3,5,4] => [1,2,3,4,5] [1,2,4,3,5] => [1,2,3,4,5] [1,2,4,5,3] => [1,2,3,4,5] [1,2,5,3,4] => [1,2,3,5,4] [1,2,5,4,3] => [1,2,3,5,4] [1,3,2,4,5] => [1,2,3,4,5] [1,3,2,5,4] => [1,2,3,4,5] [1,3,4,2,5] => [1,2,3,4,5] [1,3,4,5,2] => [1,2,3,4,5] [1,3,5,2,4] => [1,2,3,5,4] [1,3,5,4,2] => [1,2,3,5,4] [1,4,2,3,5] => [1,2,4,3,5] [1,4,2,5,3] => [1,2,4,5,3] [1,4,3,2,5] => [1,2,4,3,5] [1,4,3,5,2] => [1,2,4,5,3] [1,4,5,2,3] => [1,2,4,3,5] [1,4,5,3,2] => [1,2,4,3,5] [1,5,2,3,4] => [1,2,5,4,3] [1,5,2,4,3] => [1,2,5,3,4] [1,5,3,2,4] => [1,2,5,4,3] [1,5,3,4,2] => [1,2,5,3,4] [1,5,4,2,3] => [1,2,5,3,4] [1,5,4,3,2] => [1,2,5,3,4] [2,1,3,4,5] => [1,2,3,4,5] [2,1,3,5,4] => [1,2,3,4,5] [2,1,4,3,5] => [1,2,3,4,5] [2,1,4,5,3] => [1,2,3,4,5] [2,1,5,3,4] => [1,2,3,5,4] [2,1,5,4,3] => [1,2,3,5,4] [2,3,1,4,5] => [1,2,3,4,5] [2,3,1,5,4] => [1,2,3,4,5] [2,3,4,1,5] => [1,2,3,4,5] [2,3,4,5,1] => [1,2,3,4,5] [2,3,5,1,4] => [1,2,3,5,4] [2,3,5,4,1] => [1,2,3,5,4] [2,4,1,3,5] => [1,2,4,3,5] [2,4,1,5,3] => [1,2,4,5,3] [2,4,3,1,5] => [1,2,4,3,5] [2,4,3,5,1] => [1,2,4,5,3] [2,4,5,1,3] => [1,2,4,3,5] [2,4,5,3,1] => [1,2,4,3,5] [2,5,1,3,4] => [1,2,5,4,3] [2,5,1,4,3] => [1,2,5,3,4] [2,5,3,1,4] => [1,2,5,4,3] [2,5,3,4,1] => [1,2,5,3,4] [2,5,4,1,3] => [1,2,5,3,4] [2,5,4,3,1] => [1,2,5,3,4] [3,1,2,4,5] => [1,3,2,4,5] [3,1,2,5,4] => [1,3,2,4,5] [3,1,4,2,5] => [1,3,4,2,5] [3,1,4,5,2] => [1,3,4,5,2] [3,1,5,2,4] => [1,3,5,4,2] [3,1,5,4,2] => [1,3,5,2,4] [3,2,1,4,5] => [1,3,2,4,5] [3,2,1,5,4] => [1,3,2,4,5] [3,2,4,1,5] => [1,3,4,2,5] [3,2,4,5,1] => [1,3,4,5,2] [3,2,5,1,4] => [1,3,5,4,2] [3,2,5,4,1] => [1,3,5,2,4] [3,4,1,2,5] => [1,3,2,4,5] [3,4,1,5,2] => [1,3,2,4,5] [3,4,2,1,5] => [1,3,2,4,5] [3,4,2,5,1] => [1,3,2,4,5] [3,4,5,1,2] => [1,3,5,2,4] [3,4,5,2,1] => [1,3,5,2,4] [3,5,1,2,4] => [1,3,2,5,4] [3,5,1,4,2] => [1,3,2,5,4] [3,5,2,1,4] => [1,3,2,5,4] [3,5,2,4,1] => [1,3,2,5,4] [3,5,4,1,2] => [1,3,4,2,5] [3,5,4,2,1] => [1,3,4,2,5] [4,1,2,3,5] => [1,4,3,2,5] [4,1,2,5,3] => [1,4,5,3,2] [4,1,3,2,5] => [1,4,2,3,5] [4,1,3,5,2] => [1,4,5,2,3] [4,1,5,2,3] => [1,4,2,3,5] [4,1,5,3,2] => [1,4,3,5,2] [4,2,1,3,5] => [1,4,3,2,5] [4,2,1,5,3] => [1,4,5,3,2] [4,2,3,1,5] => [1,4,2,3,5] [4,2,3,5,1] => [1,4,5,2,3] [4,2,5,1,3] => [1,4,2,3,5] [4,2,5,3,1] => [1,4,3,5,2] [4,3,1,2,5] => [1,4,2,3,5] [4,3,1,5,2] => [1,4,5,2,3] [4,3,2,1,5] => [1,4,2,3,5] [4,3,2,5,1] => [1,4,5,2,3] [4,3,5,1,2] => [1,4,2,3,5] [4,3,5,2,1] => [1,4,2,3,5] [4,5,1,2,3] => [1,4,2,5,3] [4,5,1,3,2] => [1,4,3,2,5] [4,5,2,1,3] => [1,4,2,5,3] [4,5,2,3,1] => [1,4,3,2,5] [4,5,3,1,2] => [1,4,2,5,3] [4,5,3,2,1] => [1,4,2,5,3] [5,1,2,3,4] => [1,5,4,3,2] [5,1,2,4,3] => [1,5,3,2,4] [5,1,3,2,4] => [1,5,4,2,3] [5,1,3,4,2] => [1,5,2,3,4] [5,1,4,2,3] => [1,5,3,4,2] [5,1,4,3,2] => [1,5,2,3,4] [5,2,1,3,4] => [1,5,4,3,2] [5,2,1,4,3] => [1,5,3,2,4] [5,2,3,1,4] => [1,5,4,2,3] [5,2,3,4,1] => [1,5,2,3,4] [5,2,4,1,3] => [1,5,3,4,2] [5,2,4,3,1] => [1,5,2,3,4] [5,3,1,2,4] => [1,5,4,2,3] [5,3,1,4,2] => [1,5,2,3,4] [5,3,2,1,4] => [1,5,4,2,3] [5,3,2,4,1] => [1,5,2,3,4] [5,3,4,1,2] => [1,5,2,3,4] [5,3,4,2,1] => [1,5,2,3,4] [5,4,1,2,3] => [1,5,3,2,4] [5,4,1,3,2] => [1,5,2,4,3] [5,4,2,1,3] => [1,5,3,2,4] [5,4,2,3,1] => [1,5,2,4,3] [5,4,3,1,2] => [1,5,2,4,3] [5,4,3,2,1] => [1,5,2,4,3] [1,2,3,4,5,6] => [1,2,3,4,5,6] [1,2,3,4,6,5] => [1,2,3,4,5,6] [1,2,3,5,4,6] => [1,2,3,4,5,6] [1,2,3,5,6,4] => [1,2,3,4,5,6] [1,2,3,6,4,5] => [1,2,3,4,6,5] [1,2,3,6,5,4] => [1,2,3,4,6,5] [1,2,4,3,5,6] => [1,2,3,4,5,6] [1,2,4,3,6,5] => [1,2,3,4,5,6] [1,2,4,5,3,6] => [1,2,3,4,5,6] [1,2,4,5,6,3] => [1,2,3,4,5,6] [1,2,4,6,3,5] => [1,2,3,4,6,5] [1,2,4,6,5,3] => [1,2,3,4,6,5] [1,2,5,3,4,6] => [1,2,3,5,4,6] [1,2,5,3,6,4] => [1,2,3,5,6,4] [1,2,5,4,3,6] => [1,2,3,5,4,6] [1,2,5,4,6,3] => [1,2,3,5,6,4] [1,2,5,6,3,4] => [1,2,3,5,4,6] [1,2,5,6,4,3] => [1,2,3,5,4,6] [1,2,6,3,4,5] => [1,2,3,6,5,4] [1,2,6,3,5,4] => [1,2,3,6,4,5] [1,2,6,4,3,5] => [1,2,3,6,5,4] [1,2,6,4,5,3] => [1,2,3,6,4,5] [1,2,6,5,3,4] => [1,2,3,6,4,5] [1,2,6,5,4,3] => [1,2,3,6,4,5] [1,3,2,4,5,6] => [1,2,3,4,5,6] [1,3,2,4,6,5] => [1,2,3,4,5,6] [1,3,2,5,4,6] => [1,2,3,4,5,6] [1,3,2,5,6,4] => [1,2,3,4,5,6] [1,3,2,6,4,5] => [1,2,3,4,6,5] [1,3,2,6,5,4] => [1,2,3,4,6,5] [1,3,4,2,5,6] => [1,2,3,4,5,6] [1,3,4,2,6,5] => [1,2,3,4,5,6] [1,3,4,5,2,6] => [1,2,3,4,5,6] [1,3,4,5,6,2] => [1,2,3,4,5,6] [1,3,4,6,2,5] => [1,2,3,4,6,5] [1,3,4,6,5,2] => [1,2,3,4,6,5] [1,3,5,2,4,6] => [1,2,3,5,4,6] [1,3,5,2,6,4] => [1,2,3,5,6,4] [1,3,5,4,2,6] => [1,2,3,5,4,6] [1,3,5,4,6,2] => [1,2,3,5,6,4] [1,3,5,6,2,4] => [1,2,3,5,4,6] [1,3,5,6,4,2] => [1,2,3,5,4,6] [1,3,6,2,4,5] => [1,2,3,6,5,4] [1,3,6,2,5,4] => [1,2,3,6,4,5] [1,3,6,4,2,5] => [1,2,3,6,5,4] [1,3,6,4,5,2] => [1,2,3,6,4,5] [1,3,6,5,2,4] => [1,2,3,6,4,5] [1,3,6,5,4,2] => [1,2,3,6,4,5] [1,4,2,3,5,6] => [1,2,4,3,5,6] [1,4,2,3,6,5] => [1,2,4,3,5,6] [1,4,2,5,3,6] => [1,2,4,5,3,6] [1,4,2,5,6,3] => [1,2,4,5,6,3] [1,4,2,6,3,5] => [1,2,4,6,5,3] [1,4,2,6,5,3] => [1,2,4,6,3,5] [1,4,3,2,5,6] => [1,2,4,3,5,6] [1,4,3,2,6,5] => [1,2,4,3,5,6] [1,4,3,5,2,6] => [1,2,4,5,3,6] [1,4,3,5,6,2] => [1,2,4,5,6,3] [1,4,3,6,2,5] => [1,2,4,6,5,3] [1,4,3,6,5,2] => [1,2,4,6,3,5] [1,4,5,2,3,6] => [1,2,4,3,5,6] [1,4,5,2,6,3] => [1,2,4,3,5,6] [1,4,5,3,2,6] => [1,2,4,3,5,6] [1,4,5,3,6,2] => [1,2,4,3,5,6] [1,4,5,6,2,3] => [1,2,4,6,3,5] [1,4,5,6,3,2] => [1,2,4,6,3,5] [1,4,6,2,3,5] => [1,2,4,3,6,5] [1,4,6,2,5,3] => [1,2,4,3,6,5] [1,4,6,3,2,5] => [1,2,4,3,6,5] [1,4,6,3,5,2] => [1,2,4,3,6,5] [1,4,6,5,2,3] => [1,2,4,5,3,6] [1,4,6,5,3,2] => [1,2,4,5,3,6] [1,5,2,3,4,6] => [1,2,5,4,3,6] [1,5,2,3,6,4] => [1,2,5,6,4,3] [1,5,2,4,3,6] => [1,2,5,3,4,6] [1,5,2,4,6,3] => [1,2,5,6,3,4] [1,5,2,6,3,4] => [1,2,5,3,4,6] [1,5,2,6,4,3] => [1,2,5,4,6,3] [1,5,3,2,4,6] => [1,2,5,4,3,6] [1,5,3,2,6,4] => [1,2,5,6,4,3] [1,5,3,4,2,6] => [1,2,5,3,4,6] [1,5,3,4,6,2] => [1,2,5,6,3,4] [1,5,3,6,2,4] => [1,2,5,3,4,6] [1,5,3,6,4,2] => [1,2,5,4,6,3] [1,5,4,2,3,6] => [1,2,5,3,4,6] [1,5,4,2,6,3] => [1,2,5,6,3,4] [1,5,4,3,2,6] => [1,2,5,3,4,6] [1,5,4,3,6,2] => [1,2,5,6,3,4] [1,5,4,6,2,3] => [1,2,5,3,4,6] [1,5,4,6,3,2] => [1,2,5,3,4,6] [1,5,6,2,3,4] => [1,2,5,3,6,4] [1,5,6,2,4,3] => [1,2,5,4,3,6] [1,5,6,3,2,4] => [1,2,5,3,6,4] [1,5,6,3,4,2] => [1,2,5,4,3,6] [1,5,6,4,2,3] => [1,2,5,3,6,4] [1,5,6,4,3,2] => [1,2,5,3,6,4] [1,6,2,3,4,5] => [1,2,6,5,4,3] [1,6,2,3,5,4] => [1,2,6,4,3,5] [1,6,2,4,3,5] => [1,2,6,5,3,4] [1,6,2,4,5,3] => [1,2,6,3,4,5] [1,6,2,5,3,4] => [1,2,6,4,5,3] [1,6,2,5,4,3] => [1,2,6,3,4,5] [1,6,3,2,4,5] => [1,2,6,5,4,3] [1,6,3,2,5,4] => [1,2,6,4,3,5] [1,6,3,4,2,5] => [1,2,6,5,3,4] [1,6,3,4,5,2] => [1,2,6,3,4,5] [1,6,3,5,2,4] => [1,2,6,4,5,3] [1,6,3,5,4,2] => [1,2,6,3,4,5] [1,6,4,2,3,5] => [1,2,6,5,3,4] [1,6,4,2,5,3] => [1,2,6,3,4,5] [1,6,4,3,2,5] => [1,2,6,5,3,4] [1,6,4,3,5,2] => [1,2,6,3,4,5] [1,6,4,5,2,3] => [1,2,6,3,4,5] [1,6,4,5,3,2] => [1,2,6,3,4,5] [1,6,5,2,3,4] => [1,2,6,4,3,5] [1,6,5,2,4,3] => [1,2,6,3,5,4] [1,6,5,3,2,4] => [1,2,6,4,3,5] [1,6,5,3,4,2] => [1,2,6,3,5,4] [1,6,5,4,2,3] => [1,2,6,3,5,4] [1,6,5,4,3,2] => [1,2,6,3,5,4] [2,1,3,4,5,6] => [1,2,3,4,5,6] [2,1,3,4,6,5] => [1,2,3,4,5,6] [2,1,3,5,4,6] => [1,2,3,4,5,6] [2,1,3,5,6,4] => [1,2,3,4,5,6] [2,1,3,6,4,5] => [1,2,3,4,6,5] [2,1,3,6,5,4] => [1,2,3,4,6,5] [2,1,4,3,5,6] => [1,2,3,4,5,6] [2,1,4,3,6,5] => [1,2,3,4,5,6] [2,1,4,5,3,6] => [1,2,3,4,5,6] [2,1,4,5,6,3] => [1,2,3,4,5,6] [2,1,4,6,3,5] => [1,2,3,4,6,5] [2,1,4,6,5,3] => [1,2,3,4,6,5] [2,1,5,3,4,6] => [1,2,3,5,4,6] [2,1,5,3,6,4] => [1,2,3,5,6,4] [2,1,5,4,3,6] => [1,2,3,5,4,6] [2,1,5,4,6,3] => [1,2,3,5,6,4] [2,1,5,6,3,4] => [1,2,3,5,4,6] [2,1,5,6,4,3] => [1,2,3,5,4,6] [2,1,6,3,4,5] => [1,2,3,6,5,4] [2,1,6,3,5,4] => [1,2,3,6,4,5] [2,1,6,4,3,5] => [1,2,3,6,5,4] [2,1,6,4,5,3] => [1,2,3,6,4,5] [2,1,6,5,3,4] => [1,2,3,6,4,5] [2,1,6,5,4,3] => [1,2,3,6,4,5] [2,3,1,4,5,6] => [1,2,3,4,5,6] [2,3,1,4,6,5] => [1,2,3,4,5,6] [2,3,1,5,4,6] => [1,2,3,4,5,6] [2,3,1,5,6,4] => [1,2,3,4,5,6] [2,3,1,6,4,5] => [1,2,3,4,6,5] [2,3,1,6,5,4] => [1,2,3,4,6,5] [2,3,4,1,5,6] => [1,2,3,4,5,6] [2,3,4,1,6,5] => [1,2,3,4,5,6] [2,3,4,5,1,6] => [1,2,3,4,5,6] [2,3,4,5,6,1] => [1,2,3,4,5,6] [2,3,4,6,1,5] => [1,2,3,4,6,5] [2,3,4,6,5,1] => [1,2,3,4,6,5] [2,3,5,1,4,6] => [1,2,3,5,4,6] [2,3,5,1,6,4] => [1,2,3,5,6,4] [2,3,5,4,1,6] => [1,2,3,5,4,6] [2,3,5,4,6,1] => [1,2,3,5,6,4] [2,3,5,6,1,4] => [1,2,3,5,4,6] [2,3,5,6,4,1] => [1,2,3,5,4,6] [2,3,6,1,4,5] => [1,2,3,6,5,4] [2,3,6,1,5,4] => [1,2,3,6,4,5] [2,3,6,4,1,5] => [1,2,3,6,5,4] [2,3,6,4,5,1] => [1,2,3,6,4,5] [2,3,6,5,1,4] => [1,2,3,6,4,5] [2,3,6,5,4,1] => [1,2,3,6,4,5] [2,4,1,3,5,6] => [1,2,4,3,5,6] [2,4,1,3,6,5] => [1,2,4,3,5,6] [2,4,1,5,3,6] => [1,2,4,5,3,6] [2,4,1,5,6,3] => [1,2,4,5,6,3] [2,4,1,6,3,5] => [1,2,4,6,5,3] [2,4,1,6,5,3] => [1,2,4,6,3,5] [2,4,3,1,5,6] => [1,2,4,3,5,6] [2,4,3,1,6,5] => [1,2,4,3,5,6] [2,4,3,5,1,6] => [1,2,4,5,3,6] [2,4,3,5,6,1] => [1,2,4,5,6,3] [2,4,3,6,1,5] => [1,2,4,6,5,3] [2,4,3,6,5,1] => [1,2,4,6,3,5] [2,4,5,1,3,6] => [1,2,4,3,5,6] [2,4,5,1,6,3] => [1,2,4,3,5,6] [2,4,5,3,1,6] => [1,2,4,3,5,6] [2,4,5,3,6,1] => [1,2,4,3,5,6] [2,4,5,6,1,3] => [1,2,4,6,3,5] [2,4,5,6,3,1] => [1,2,4,6,3,5] [2,4,6,1,3,5] => [1,2,4,3,6,5] [2,4,6,1,5,3] => [1,2,4,3,6,5] [2,4,6,3,1,5] => [1,2,4,3,6,5] [2,4,6,3,5,1] => [1,2,4,3,6,5] [2,4,6,5,1,3] => [1,2,4,5,3,6] [2,4,6,5,3,1] => [1,2,4,5,3,6] [2,5,1,3,4,6] => [1,2,5,4,3,6] [2,5,1,3,6,4] => [1,2,5,6,4,3] [2,5,1,4,3,6] => [1,2,5,3,4,6] [2,5,1,4,6,3] => [1,2,5,6,3,4] [2,5,1,6,3,4] => [1,2,5,3,4,6] [2,5,1,6,4,3] => [1,2,5,4,6,3] [2,5,3,1,4,6] => [1,2,5,4,3,6] [2,5,3,1,6,4] => [1,2,5,6,4,3] [2,5,3,4,1,6] => [1,2,5,3,4,6] [2,5,3,4,6,1] => [1,2,5,6,3,4] [2,5,3,6,1,4] => [1,2,5,3,4,6] [2,5,3,6,4,1] => [1,2,5,4,6,3] [2,5,4,1,3,6] => [1,2,5,3,4,6] [2,5,4,1,6,3] => [1,2,5,6,3,4] [2,5,4,3,1,6] => [1,2,5,3,4,6] [2,5,4,3,6,1] => [1,2,5,6,3,4] [2,5,4,6,1,3] => [1,2,5,3,4,6] [2,5,4,6,3,1] => [1,2,5,3,4,6] [2,5,6,1,3,4] => [1,2,5,3,6,4] [2,5,6,1,4,3] => [1,2,5,4,3,6] [2,5,6,3,1,4] => [1,2,5,3,6,4] [2,5,6,3,4,1] => [1,2,5,4,3,6] [2,5,6,4,1,3] => [1,2,5,3,6,4] [2,5,6,4,3,1] => [1,2,5,3,6,4] [2,6,1,3,4,5] => [1,2,6,5,4,3] [2,6,1,3,5,4] => [1,2,6,4,3,5] [2,6,1,4,3,5] => [1,2,6,5,3,4] [2,6,1,4,5,3] => [1,2,6,3,4,5] [2,6,1,5,3,4] => [1,2,6,4,5,3] [2,6,1,5,4,3] => [1,2,6,3,4,5] [2,6,3,1,4,5] => [1,2,6,5,4,3] [2,6,3,1,5,4] => [1,2,6,4,3,5] [2,6,3,4,1,5] => [1,2,6,5,3,4] [2,6,3,4,5,1] => [1,2,6,3,4,5] [2,6,3,5,1,4] => [1,2,6,4,5,3] [2,6,3,5,4,1] => [1,2,6,3,4,5] [2,6,4,1,3,5] => [1,2,6,5,3,4] [2,6,4,1,5,3] => [1,2,6,3,4,5] [2,6,4,3,1,5] => [1,2,6,5,3,4] [2,6,4,3,5,1] => [1,2,6,3,4,5] [2,6,4,5,1,3] => [1,2,6,3,4,5] [2,6,4,5,3,1] => [1,2,6,3,4,5] [2,6,5,1,3,4] => [1,2,6,4,3,5] [2,6,5,1,4,3] => [1,2,6,3,5,4] [2,6,5,3,1,4] => [1,2,6,4,3,5] [2,6,5,3,4,1] => [1,2,6,3,5,4] [2,6,5,4,1,3] => [1,2,6,3,5,4] [2,6,5,4,3,1] => [1,2,6,3,5,4] [3,1,2,4,5,6] => [1,3,2,4,5,6] [3,1,2,4,6,5] => [1,3,2,4,5,6] [3,1,2,5,4,6] => [1,3,2,4,5,6] [3,1,2,5,6,4] => [1,3,2,4,5,6] [3,1,2,6,4,5] => [1,3,2,4,6,5] [3,1,2,6,5,4] => [1,3,2,4,6,5] [3,1,4,2,5,6] => [1,3,4,2,5,6] [3,1,4,2,6,5] => [1,3,4,2,5,6] [3,1,4,5,2,6] => [1,3,4,5,2,6] [3,1,4,5,6,2] => [1,3,4,5,6,2] [3,1,4,6,2,5] => [1,3,4,6,5,2] [3,1,4,6,5,2] => [1,3,4,6,2,5] [3,1,5,2,4,6] => [1,3,5,4,2,6] [3,1,5,2,6,4] => [1,3,5,6,4,2] [3,1,5,4,2,6] => [1,3,5,2,4,6] [3,1,5,4,6,2] => [1,3,5,6,2,4] [3,1,5,6,2,4] => [1,3,5,2,4,6] [3,1,5,6,4,2] => [1,3,5,4,6,2] [3,1,6,2,4,5] => [1,3,6,5,4,2] [3,1,6,2,5,4] => [1,3,6,4,2,5] [3,1,6,4,2,5] => [1,3,6,5,2,4] [3,1,6,4,5,2] => [1,3,6,2,4,5] [3,1,6,5,2,4] => [1,3,6,4,5,2] [3,1,6,5,4,2] => [1,3,6,2,4,5] [3,2,1,4,5,6] => [1,3,2,4,5,6] [3,2,1,4,6,5] => [1,3,2,4,5,6] [3,2,1,5,4,6] => [1,3,2,4,5,6] [3,2,1,5,6,4] => [1,3,2,4,5,6] [3,2,1,6,4,5] => [1,3,2,4,6,5] [3,2,1,6,5,4] => [1,3,2,4,6,5] [3,2,4,1,5,6] => [1,3,4,2,5,6] [3,2,4,1,6,5] => [1,3,4,2,5,6] [3,2,4,5,1,6] => [1,3,4,5,2,6] [3,2,4,5,6,1] => [1,3,4,5,6,2] [3,2,4,6,1,5] => [1,3,4,6,5,2] [3,2,4,6,5,1] => [1,3,4,6,2,5] [3,2,5,1,4,6] => [1,3,5,4,2,6] [3,2,5,1,6,4] => [1,3,5,6,4,2] [3,2,5,4,1,6] => [1,3,5,2,4,6] [3,2,5,4,6,1] => [1,3,5,6,2,4] [3,2,5,6,1,4] => [1,3,5,2,4,6] [3,2,5,6,4,1] => [1,3,5,4,6,2] [3,2,6,1,4,5] => [1,3,6,5,4,2] [3,2,6,1,5,4] => [1,3,6,4,2,5] [3,2,6,4,1,5] => [1,3,6,5,2,4] [3,2,6,4,5,1] => [1,3,6,2,4,5] [3,2,6,5,1,4] => [1,3,6,4,5,2] [3,2,6,5,4,1] => [1,3,6,2,4,5] [3,4,1,2,5,6] => [1,3,2,4,5,6] [3,4,1,2,6,5] => [1,3,2,4,5,6] [3,4,1,5,2,6] => [1,3,2,4,5,6] [3,4,1,5,6,2] => [1,3,2,4,5,6] [3,4,1,6,2,5] => [1,3,2,4,6,5] [3,4,1,6,5,2] => [1,3,2,4,6,5] [3,4,2,1,5,6] => [1,3,2,4,5,6] [3,4,2,1,6,5] => [1,3,2,4,5,6] [3,4,2,5,1,6] => [1,3,2,4,5,6] [3,4,2,5,6,1] => [1,3,2,4,5,6] [3,4,2,6,1,5] => [1,3,2,4,6,5] [3,4,2,6,5,1] => [1,3,2,4,6,5] [3,4,5,1,2,6] => [1,3,5,2,4,6] [3,4,5,1,6,2] => [1,3,5,6,2,4] [3,4,5,2,1,6] => [1,3,5,2,4,6] [3,4,5,2,6,1] => [1,3,5,6,2,4] [3,4,5,6,1,2] => [1,3,5,2,4,6] [3,4,5,6,2,1] => [1,3,5,2,4,6] [3,4,6,1,2,5] => [1,3,6,5,2,4] [3,4,6,1,5,2] => [1,3,6,2,4,5] [3,4,6,2,1,5] => [1,3,6,5,2,4] [3,4,6,2,5,1] => [1,3,6,2,4,5] [3,4,6,5,1,2] => [1,3,6,2,4,5] [3,4,6,5,2,1] => [1,3,6,2,4,5] [3,5,1,2,4,6] => [1,3,2,5,4,6] [3,5,1,2,6,4] => [1,3,2,5,6,4] [3,5,1,4,2,6] => [1,3,2,5,4,6] [3,5,1,4,6,2] => [1,3,2,5,6,4] [3,5,1,6,2,4] => [1,3,2,5,4,6] [3,5,1,6,4,2] => [1,3,2,5,4,6] [3,5,2,1,4,6] => [1,3,2,5,4,6] [3,5,2,1,6,4] => [1,3,2,5,6,4] [3,5,2,4,1,6] => [1,3,2,5,4,6] [3,5,2,4,6,1] => [1,3,2,5,6,4] [3,5,2,6,1,4] => [1,3,2,5,4,6] [3,5,2,6,4,1] => [1,3,2,5,4,6] [3,5,4,1,2,6] => [1,3,4,2,5,6] [3,5,4,1,6,2] => [1,3,4,2,5,6] [3,5,4,2,1,6] => [1,3,4,2,5,6] [3,5,4,2,6,1] => [1,3,4,2,5,6] [3,5,4,6,1,2] => [1,3,4,6,2,5] [3,5,4,6,2,1] => [1,3,4,6,2,5] [3,5,6,1,2,4] => [1,3,6,4,2,5] [3,5,6,1,4,2] => [1,3,6,2,5,4] [3,5,6,2,1,4] => [1,3,6,4,2,5] [3,5,6,2,4,1] => [1,3,6,2,5,4] [3,5,6,4,1,2] => [1,3,6,2,5,4] [3,5,6,4,2,1] => [1,3,6,2,5,4] [3,6,1,2,4,5] => [1,3,2,6,5,4] [3,6,1,2,5,4] => [1,3,2,6,4,5] [3,6,1,4,2,5] => [1,3,2,6,5,4] [3,6,1,4,5,2] => [1,3,2,6,4,5] [3,6,1,5,2,4] => [1,3,2,6,4,5] [3,6,1,5,4,2] => [1,3,2,6,4,5] [3,6,2,1,4,5] => [1,3,2,6,5,4] [3,6,2,1,5,4] => [1,3,2,6,4,5] [3,6,2,4,1,5] => [1,3,2,6,5,4] [3,6,2,4,5,1] => [1,3,2,6,4,5] [3,6,2,5,1,4] => [1,3,2,6,4,5] [3,6,2,5,4,1] => [1,3,2,6,4,5] [3,6,4,1,2,5] => [1,3,4,2,6,5] [3,6,4,1,5,2] => [1,3,4,2,6,5] [3,6,4,2,1,5] => [1,3,4,2,6,5] [3,6,4,2,5,1] => [1,3,4,2,6,5] [3,6,4,5,1,2] => [1,3,4,5,2,6] [3,6,4,5,2,1] => [1,3,4,5,2,6] [3,6,5,1,2,4] => [1,3,5,2,6,4] [3,6,5,1,4,2] => [1,3,5,4,2,6] [3,6,5,2,1,4] => [1,3,5,2,6,4] [3,6,5,2,4,1] => [1,3,5,4,2,6] [3,6,5,4,1,2] => [1,3,5,2,6,4] [3,6,5,4,2,1] => [1,3,5,2,6,4] [4,1,2,3,5,6] => [1,4,3,2,5,6] [4,1,2,3,6,5] => [1,4,3,2,5,6] [4,1,2,5,3,6] => [1,4,5,3,2,6] [4,1,2,5,6,3] => [1,4,5,6,3,2] [4,1,2,6,3,5] => [1,4,6,5,3,2] [4,1,2,6,5,3] => [1,4,6,3,2,5] [4,1,3,2,5,6] => [1,4,2,3,5,6] [4,1,3,2,6,5] => [1,4,2,3,5,6] [4,1,3,5,2,6] => [1,4,5,2,3,6] [4,1,3,5,6,2] => [1,4,5,6,2,3] [4,1,3,6,2,5] => [1,4,6,5,2,3] [4,1,3,6,5,2] => [1,4,6,2,3,5] [4,1,5,2,3,6] => [1,4,2,3,5,6] [4,1,5,2,6,3] => [1,4,2,3,5,6] [4,1,5,3,2,6] => [1,4,3,5,2,6] [4,1,5,3,6,2] => [1,4,3,5,6,2] [4,1,5,6,2,3] => [1,4,6,3,5,2] [4,1,5,6,3,2] => [1,4,6,2,3,5] [4,1,6,2,3,5] => [1,4,2,3,6,5] [4,1,6,2,5,3] => [1,4,2,3,6,5] [4,1,6,3,2,5] => [1,4,3,6,5,2] [4,1,6,3,5,2] => [1,4,3,6,2,5] [4,1,6,5,2,3] => [1,4,5,2,3,6] [4,1,6,5,3,2] => [1,4,5,3,6,2] [4,2,1,3,5,6] => [1,4,3,2,5,6] [4,2,1,3,6,5] => [1,4,3,2,5,6] [4,2,1,5,3,6] => [1,4,5,3,2,6] [4,2,1,5,6,3] => [1,4,5,6,3,2] [4,2,1,6,3,5] => [1,4,6,5,3,2] [4,2,1,6,5,3] => [1,4,6,3,2,5] [4,2,3,1,5,6] => [1,4,2,3,5,6] [4,2,3,1,6,5] => [1,4,2,3,5,6] [4,2,3,5,1,6] => [1,4,5,2,3,6] [4,2,3,5,6,1] => [1,4,5,6,2,3] [4,2,3,6,1,5] => [1,4,6,5,2,3] [4,2,3,6,5,1] => [1,4,6,2,3,5] [4,2,5,1,3,6] => [1,4,2,3,5,6] [4,2,5,1,6,3] => [1,4,2,3,5,6] [4,2,5,3,1,6] => [1,4,3,5,2,6] [4,2,5,3,6,1] => [1,4,3,5,6,2] [4,2,5,6,1,3] => [1,4,6,3,5,2] [4,2,5,6,3,1] => [1,4,6,2,3,5] [4,2,6,1,3,5] => [1,4,2,3,6,5] [4,2,6,1,5,3] => [1,4,2,3,6,5] [4,2,6,3,1,5] => [1,4,3,6,5,2] [4,2,6,3,5,1] => [1,4,3,6,2,5] [4,2,6,5,1,3] => [1,4,5,2,3,6] [4,2,6,5,3,1] => [1,4,5,3,6,2] [4,3,1,2,5,6] => [1,4,2,3,5,6] [4,3,1,2,6,5] => [1,4,2,3,5,6] [4,3,1,5,2,6] => [1,4,5,2,3,6] [4,3,1,5,6,2] => [1,4,5,6,2,3] [4,3,1,6,2,5] => [1,4,6,5,2,3] [4,3,1,6,5,2] => [1,4,6,2,3,5] [4,3,2,1,5,6] => [1,4,2,3,5,6] [4,3,2,1,6,5] => [1,4,2,3,5,6] [4,3,2,5,1,6] => [1,4,5,2,3,6] [4,3,2,5,6,1] => [1,4,5,6,2,3] [4,3,2,6,1,5] => [1,4,6,5,2,3] [4,3,2,6,5,1] => [1,4,6,2,3,5] [4,3,5,1,2,6] => [1,4,2,3,5,6] [4,3,5,1,6,2] => [1,4,2,3,5,6] [4,3,5,2,1,6] => [1,4,2,3,5,6] [4,3,5,2,6,1] => [1,4,2,3,5,6] [4,3,5,6,1,2] => [1,4,6,2,3,5] [4,3,5,6,2,1] => [1,4,6,2,3,5] [4,3,6,1,2,5] => [1,4,2,3,6,5] [4,3,6,1,5,2] => [1,4,2,3,6,5] [4,3,6,2,1,5] => [1,4,2,3,6,5] [4,3,6,2,5,1] => [1,4,2,3,6,5] [4,3,6,5,1,2] => [1,4,5,2,3,6] [4,3,6,5,2,1] => [1,4,5,2,3,6] [4,5,1,2,3,6] => [1,4,2,5,3,6] [4,5,1,2,6,3] => [1,4,2,5,6,3] [4,5,1,3,2,6] => [1,4,3,2,5,6] [4,5,1,3,6,2] => [1,4,3,2,5,6] [4,5,1,6,2,3] => [1,4,6,3,2,5] [4,5,1,6,3,2] => [1,4,6,2,5,3] [4,5,2,1,3,6] => [1,4,2,5,3,6] [4,5,2,1,6,3] => [1,4,2,5,6,3] [4,5,2,3,1,6] => [1,4,3,2,5,6] [4,5,2,3,6,1] => [1,4,3,2,5,6] [4,5,2,6,1,3] => [1,4,6,3,2,5] [4,5,2,6,3,1] => [1,4,6,2,5,3] [4,5,3,1,2,6] => [1,4,2,5,3,6] [4,5,3,1,6,2] => [1,4,2,5,6,3] [4,5,3,2,1,6] => [1,4,2,5,3,6] [4,5,3,2,6,1] => [1,4,2,5,6,3] [4,5,3,6,1,2] => [1,4,6,2,5,3] [4,5,3,6,2,1] => [1,4,6,2,5,3] [4,5,6,1,2,3] => [1,4,2,5,3,6] [4,5,6,1,3,2] => [1,4,2,5,3,6] [4,5,6,2,1,3] => [1,4,2,5,3,6] [4,5,6,2,3,1] => [1,4,2,5,3,6] [4,5,6,3,1,2] => [1,4,3,6,2,5] [4,5,6,3,2,1] => [1,4,3,6,2,5] [4,6,1,2,3,5] => [1,4,2,6,5,3] [4,6,1,2,5,3] => [1,4,2,6,3,5] [4,6,1,3,2,5] => [1,4,3,2,6,5] [4,6,1,3,5,2] => [1,4,3,2,6,5] [4,6,1,5,2,3] => [1,4,5,2,6,3] [4,6,1,5,3,2] => [1,4,5,3,2,6] [4,6,2,1,3,5] => [1,4,2,6,5,3] [4,6,2,1,5,3] => [1,4,2,6,3,5] [4,6,2,3,1,5] => [1,4,3,2,6,5] [4,6,2,3,5,1] => [1,4,3,2,6,5] [4,6,2,5,1,3] => [1,4,5,2,6,3] [4,6,2,5,3,1] => [1,4,5,3,2,6] [4,6,3,1,2,5] => [1,4,2,6,5,3] [4,6,3,1,5,2] => [1,4,2,6,3,5] [4,6,3,2,1,5] => [1,4,2,6,5,3] [4,6,3,2,5,1] => [1,4,2,6,3,5] [4,6,3,5,1,2] => [1,4,5,2,6,3] [4,6,3,5,2,1] => [1,4,5,2,6,3] [4,6,5,1,2,3] => [1,4,2,6,3,5] [4,6,5,1,3,2] => [1,4,2,6,3,5] [4,6,5,2,1,3] => [1,4,2,6,3,5] [4,6,5,2,3,1] => [1,4,2,6,3,5] [4,6,5,3,1,2] => [1,4,3,5,2,6] [4,6,5,3,2,1] => [1,4,3,5,2,6] [5,1,2,3,4,6] => [1,5,4,3,2,6] [5,1,2,3,6,4] => [1,5,6,4,3,2] [5,1,2,4,3,6] => [1,5,3,2,4,6] [5,1,2,4,6,3] => [1,5,6,3,2,4] [5,1,2,6,3,4] => [1,5,3,2,4,6] [5,1,2,6,4,3] => [1,5,4,6,3,2] [5,1,3,2,4,6] => [1,5,4,2,3,6] [5,1,3,2,6,4] => [1,5,6,4,2,3] [5,1,3,4,2,6] => [1,5,2,3,4,6] [5,1,3,4,6,2] => [1,5,6,2,3,4] [5,1,3,6,2,4] => [1,5,2,3,4,6] [5,1,3,6,4,2] => [1,5,4,6,2,3] [5,1,4,2,3,6] => [1,5,3,4,2,6] [5,1,4,2,6,3] => [1,5,6,3,4,2] [5,1,4,3,2,6] => [1,5,2,3,4,6] [5,1,4,3,6,2] => [1,5,6,2,3,4] [5,1,4,6,2,3] => [1,5,2,3,4,6] [5,1,4,6,3,2] => [1,5,3,4,6,2] [5,1,6,2,3,4] => [1,5,3,6,4,2] [5,1,6,2,4,3] => [1,5,4,2,3,6] [5,1,6,3,2,4] => [1,5,2,3,6,4] [5,1,6,3,4,2] => [1,5,4,3,6,2] [5,1,6,4,2,3] => [1,5,2,3,6,4] [5,1,6,4,3,2] => [1,5,3,6,2,4] [5,2,1,3,4,6] => [1,5,4,3,2,6] [5,2,1,3,6,4] => [1,5,6,4,3,2] [5,2,1,4,3,6] => [1,5,3,2,4,6] [5,2,1,4,6,3] => [1,5,6,3,2,4] [5,2,1,6,3,4] => [1,5,3,2,4,6] [5,2,1,6,4,3] => [1,5,4,6,3,2] [5,2,3,1,4,6] => [1,5,4,2,3,6] [5,2,3,1,6,4] => [1,5,6,4,2,3] [5,2,3,4,1,6] => [1,5,2,3,4,6] [5,2,3,4,6,1] => [1,5,6,2,3,4] [5,2,3,6,1,4] => [1,5,2,3,4,6] [5,2,3,6,4,1] => [1,5,4,6,2,3] [5,2,4,1,3,6] => [1,5,3,4,2,6] [5,2,4,1,6,3] => [1,5,6,3,4,2] [5,2,4,3,1,6] => [1,5,2,3,4,6] [5,2,4,3,6,1] => [1,5,6,2,3,4] [5,2,4,6,1,3] => [1,5,2,3,4,6] [5,2,4,6,3,1] => [1,5,3,4,6,2] [5,2,6,1,3,4] => [1,5,3,6,4,2] [5,2,6,1,4,3] => [1,5,4,2,3,6] [5,2,6,3,1,4] => [1,5,2,3,6,4] [5,2,6,3,4,1] => [1,5,4,3,6,2] [5,2,6,4,1,3] => [1,5,2,3,6,4] [5,2,6,4,3,1] => [1,5,3,6,2,4] [5,3,1,2,4,6] => [1,5,4,2,3,6] [5,3,1,2,6,4] => [1,5,6,4,2,3] [5,3,1,4,2,6] => [1,5,2,3,4,6] [5,3,1,4,6,2] => [1,5,6,2,3,4] [5,3,1,6,2,4] => [1,5,2,3,4,6] [5,3,1,6,4,2] => [1,5,4,6,2,3] [5,3,2,1,4,6] => [1,5,4,2,3,6] [5,3,2,1,6,4] => [1,5,6,4,2,3] [5,3,2,4,1,6] => [1,5,2,3,4,6] [5,3,2,4,6,1] => [1,5,6,2,3,4] [5,3,2,6,1,4] => [1,5,2,3,4,6] [5,3,2,6,4,1] => [1,5,4,6,2,3] [5,3,4,1,2,6] => [1,5,2,3,4,6] [5,3,4,1,6,2] => [1,5,6,2,3,4] [5,3,4,2,1,6] => [1,5,2,3,4,6] [5,3,4,2,6,1] => [1,5,6,2,3,4] [5,3,4,6,1,2] => [1,5,2,3,4,6] [5,3,4,6,2,1] => [1,5,2,3,4,6] [5,3,6,1,2,4] => [1,5,2,3,6,4] [5,3,6,1,4,2] => [1,5,4,2,3,6] [5,3,6,2,1,4] => [1,5,2,3,6,4] [5,3,6,2,4,1] => [1,5,4,2,3,6] [5,3,6,4,1,2] => [1,5,2,3,6,4] [5,3,6,4,2,1] => [1,5,2,3,6,4] [5,4,1,2,3,6] => [1,5,3,2,4,6] [5,4,1,2,6,3] => [1,5,6,3,2,4] [5,4,1,3,2,6] => [1,5,2,4,3,6] [5,4,1,3,6,2] => [1,5,6,2,4,3] [5,4,1,6,2,3] => [1,5,2,4,6,3] [5,4,1,6,3,2] => [1,5,3,2,4,6] [5,4,2,1,3,6] => [1,5,3,2,4,6] [5,4,2,1,6,3] => [1,5,6,3,2,4] [5,4,2,3,1,6] => [1,5,2,4,3,6] [5,4,2,3,6,1] => [1,5,6,2,4,3] [5,4,2,6,1,3] => [1,5,2,4,6,3] [5,4,2,6,3,1] => [1,5,3,2,4,6] [5,4,3,1,2,6] => [1,5,2,4,3,6] [5,4,3,1,6,2] => [1,5,6,2,4,3] [5,4,3,2,1,6] => [1,5,2,4,3,6] [5,4,3,2,6,1] => [1,5,6,2,4,3] [5,4,3,6,1,2] => [1,5,2,4,6,3] [5,4,3,6,2,1] => [1,5,2,4,6,3] [5,4,6,1,2,3] => [1,5,2,4,3,6] [5,4,6,1,3,2] => [1,5,3,6,2,4] [5,4,6,2,1,3] => [1,5,2,4,3,6] [5,4,6,2,3,1] => [1,5,3,6,2,4] [5,4,6,3,1,2] => [1,5,2,4,3,6] [5,4,6,3,2,1] => [1,5,2,4,3,6] [5,6,1,2,3,4] => [1,5,3,2,6,4] [5,6,1,2,4,3] => [1,5,4,2,6,3] [5,6,1,3,2,4] => [1,5,2,6,4,3] [5,6,1,3,4,2] => [1,5,4,3,2,6] [5,6,1,4,2,3] => [1,5,2,6,3,4] [5,6,1,4,3,2] => [1,5,3,2,6,4] [5,6,2,1,3,4] => [1,5,3,2,6,4] [5,6,2,1,4,3] => [1,5,4,2,6,3] [5,6,2,3,1,4] => [1,5,2,6,4,3] [5,6,2,3,4,1] => [1,5,4,3,2,6] [5,6,2,4,1,3] => [1,5,2,6,3,4] [5,6,2,4,3,1] => [1,5,3,2,6,4] [5,6,3,1,2,4] => [1,5,2,6,4,3] [5,6,3,1,4,2] => [1,5,4,2,6,3] [5,6,3,2,1,4] => [1,5,2,6,4,3] [5,6,3,2,4,1] => [1,5,4,2,6,3] [5,6,3,4,1,2] => [1,5,2,6,3,4] [5,6,3,4,2,1] => [1,5,2,6,3,4] [5,6,4,1,2,3] => [1,5,2,6,3,4] [5,6,4,1,3,2] => [1,5,3,4,2,6] [5,6,4,2,1,3] => [1,5,2,6,3,4] [5,6,4,2,3,1] => [1,5,3,4,2,6] [5,6,4,3,1,2] => [1,5,2,6,3,4] [5,6,4,3,2,1] => [1,5,2,6,3,4] [6,1,2,3,4,5] => [1,6,5,4,3,2] [6,1,2,3,5,4] => [1,6,4,3,2,5] [6,1,2,4,3,5] => [1,6,5,3,2,4] [6,1,2,4,5,3] => [1,6,3,2,4,5] [6,1,2,5,3,4] => [1,6,4,5,3,2] [6,1,2,5,4,3] => [1,6,3,2,4,5] [6,1,3,2,4,5] => [1,6,5,4,2,3] [6,1,3,2,5,4] => [1,6,4,2,3,5] [6,1,3,4,2,5] => [1,6,5,2,3,4] [6,1,3,4,5,2] => [1,6,2,3,4,5] [6,1,3,5,2,4] => [1,6,4,5,2,3] [6,1,3,5,4,2] => [1,6,2,3,4,5] [6,1,4,2,3,5] => [1,6,5,3,4,2] [6,1,4,2,5,3] => [1,6,3,4,2,5] [6,1,4,3,2,5] => [1,6,5,2,3,4] [6,1,4,3,5,2] => [1,6,2,3,4,5] [6,1,4,5,2,3] => [1,6,3,4,5,2] [6,1,4,5,3,2] => [1,6,2,3,4,5] [6,1,5,2,3,4] => [1,6,4,2,3,5] [6,1,5,2,4,3] => [1,6,3,5,4,2] [6,1,5,3,2,4] => [1,6,4,3,5,2] [6,1,5,3,4,2] => [1,6,2,3,5,4] [6,1,5,4,2,3] => [1,6,3,5,2,4] [6,1,5,4,3,2] => [1,6,2,3,5,4] [6,2,1,3,4,5] => [1,6,5,4,3,2] [6,2,1,3,5,4] => [1,6,4,3,2,5] [6,2,1,4,3,5] => [1,6,5,3,2,4] [6,2,1,4,5,3] => [1,6,3,2,4,5] [6,2,1,5,3,4] => [1,6,4,5,3,2] [6,2,1,5,4,3] => [1,6,3,2,4,5] [6,2,3,1,4,5] => [1,6,5,4,2,3] [6,2,3,1,5,4] => [1,6,4,2,3,5] [6,2,3,4,1,5] => [1,6,5,2,3,4] [6,2,3,4,5,1] => [1,6,2,3,4,5] [6,2,3,5,1,4] => [1,6,4,5,2,3] [6,2,3,5,4,1] => [1,6,2,3,4,5] [6,2,4,1,3,5] => [1,6,5,3,4,2] [6,2,4,1,5,3] => [1,6,3,4,2,5] [6,2,4,3,1,5] => [1,6,5,2,3,4] [6,2,4,3,5,1] => [1,6,2,3,4,5] [6,2,4,5,1,3] => [1,6,3,4,5,2] [6,2,4,5,3,1] => [1,6,2,3,4,5] [6,2,5,1,3,4] => [1,6,4,2,3,5] [6,2,5,1,4,3] => [1,6,3,5,4,2] [6,2,5,3,1,4] => [1,6,4,3,5,2] [6,2,5,3,4,1] => [1,6,2,3,5,4] [6,2,5,4,1,3] => [1,6,3,5,2,4] [6,2,5,4,3,1] => [1,6,2,3,5,4] [6,3,1,2,4,5] => [1,6,5,4,2,3] [6,3,1,2,5,4] => [1,6,4,2,3,5] [6,3,1,4,2,5] => [1,6,5,2,3,4] [6,3,1,4,5,2] => [1,6,2,3,4,5] [6,3,1,5,2,4] => [1,6,4,5,2,3] [6,3,1,5,4,2] => [1,6,2,3,4,5] [6,3,2,1,4,5] => [1,6,5,4,2,3] [6,3,2,1,5,4] => [1,6,4,2,3,5] [6,3,2,4,1,5] => [1,6,5,2,3,4] [6,3,2,4,5,1] => [1,6,2,3,4,5] [6,3,2,5,1,4] => [1,6,4,5,2,3] [6,3,2,5,4,1] => [1,6,2,3,4,5] [6,3,4,1,2,5] => [1,6,5,2,3,4] [6,3,4,1,5,2] => [1,6,2,3,4,5] [6,3,4,2,1,5] => [1,6,5,2,3,4] [6,3,4,2,5,1] => [1,6,2,3,4,5] [6,3,4,5,1,2] => [1,6,2,3,4,5] [6,3,4,5,2,1] => [1,6,2,3,4,5] [6,3,5,1,2,4] => [1,6,4,2,3,5] [6,3,5,1,4,2] => [1,6,2,3,5,4] [6,3,5,2,1,4] => [1,6,4,2,3,5] [6,3,5,2,4,1] => [1,6,2,3,5,4] [6,3,5,4,1,2] => [1,6,2,3,5,4] [6,3,5,4,2,1] => [1,6,2,3,5,4] [6,4,1,2,3,5] => [1,6,5,3,2,4] [6,4,1,2,5,3] => [1,6,3,2,4,5] [6,4,1,3,2,5] => [1,6,5,2,4,3] [6,4,1,3,5,2] => [1,6,2,4,3,5] [6,4,1,5,2,3] => [1,6,3,2,4,5] [6,4,1,5,3,2] => [1,6,2,4,5,3] [6,4,2,1,3,5] => [1,6,5,3,2,4] [6,4,2,1,5,3] => [1,6,3,2,4,5] [6,4,2,3,1,5] => [1,6,5,2,4,3] [6,4,2,3,5,1] => [1,6,2,4,3,5] [6,4,2,5,1,3] => [1,6,3,2,4,5] [6,4,2,5,3,1] => [1,6,2,4,5,3] [6,4,3,1,2,5] => [1,6,5,2,4,3] [6,4,3,1,5,2] => [1,6,2,4,3,5] [6,4,3,2,1,5] => [1,6,5,2,4,3] [6,4,3,2,5,1] => [1,6,2,4,3,5] [6,4,3,5,1,2] => [1,6,2,4,5,3] [6,4,3,5,2,1] => [1,6,2,4,5,3] [6,4,5,1,2,3] => [1,6,3,5,2,4] [6,4,5,1,3,2] => [1,6,2,4,3,5] [6,4,5,2,1,3] => [1,6,3,5,2,4] [6,4,5,2,3,1] => [1,6,2,4,3,5] [6,4,5,3,1,2] => [1,6,2,4,3,5] [6,4,5,3,2,1] => [1,6,2,4,3,5] [6,5,1,2,3,4] => [1,6,4,2,5,3] [6,5,1,2,4,3] => [1,6,3,2,5,4] [6,5,1,3,2,4] => [1,6,4,3,2,5] [6,5,1,3,4,2] => [1,6,2,5,4,3] [6,5,1,4,2,3] => [1,6,3,2,5,4] [6,5,1,4,3,2] => [1,6,2,5,3,4] [6,5,2,1,3,4] => [1,6,4,2,5,3] [6,5,2,1,4,3] => [1,6,3,2,5,4] [6,5,2,3,1,4] => [1,6,4,3,2,5] [6,5,2,3,4,1] => [1,6,2,5,4,3] [6,5,2,4,1,3] => [1,6,3,2,5,4] [6,5,2,4,3,1] => [1,6,2,5,3,4] [6,5,3,1,2,4] => [1,6,4,2,5,3] [6,5,3,1,4,2] => [1,6,2,5,4,3] [6,5,3,2,1,4] => [1,6,4,2,5,3] [6,5,3,2,4,1] => [1,6,2,5,4,3] [6,5,3,4,1,2] => [1,6,2,5,3,4] [6,5,3,4,2,1] => [1,6,2,5,3,4] [6,5,4,1,2,3] => [1,6,3,4,2,5] [6,5,4,1,3,2] => [1,6,2,5,3,4] [6,5,4,2,1,3] => [1,6,3,4,2,5] [6,5,4,2,3,1] => [1,6,2,5,3,4] [6,5,4,3,1,2] => [1,6,2,5,3,4] [6,5,4,3,2,1] => [1,6,2,5,3,4] [1,2,3,4,5,6,7] => [1,2,3,4,5,6,7] [1,2,3,4,5,7,6] => [1,2,3,4,5,6,7] [1,2,3,4,6,5,7] => [1,2,3,4,5,6,7] [1,2,3,4,6,7,5] => [1,2,3,4,5,6,7] [1,2,3,4,7,5,6] => [1,2,3,4,5,7,6] [1,2,3,4,7,6,5] => [1,2,3,4,5,7,6] [1,2,3,5,4,6,7] => [1,2,3,4,5,6,7] [1,2,3,5,4,7,6] => [1,2,3,4,5,6,7] [1,2,3,5,6,4,7] => [1,2,3,4,5,6,7] [1,2,3,5,6,7,4] => [1,2,3,4,5,6,7] [1,2,3,5,7,4,6] => [1,2,3,4,5,7,6] [1,2,3,5,7,6,4] => [1,2,3,4,5,7,6] [1,2,3,6,4,5,7] => [1,2,3,4,6,5,7] [1,2,3,6,4,7,5] => [1,2,3,4,6,7,5] [1,2,3,6,5,4,7] => [1,2,3,4,6,5,7] [1,2,3,6,5,7,4] => [1,2,3,4,6,7,5] [1,2,3,6,7,4,5] => [1,2,3,4,6,5,7] [1,2,3,6,7,5,4] => [1,2,3,4,6,5,7] [1,2,3,7,4,5,6] => [1,2,3,4,7,6,5] [1,2,3,7,4,6,5] => [1,2,3,4,7,5,6] [1,2,3,7,5,4,6] => [1,2,3,4,7,6,5] [1,2,3,7,5,6,4] => [1,2,3,4,7,5,6] [1,2,3,7,6,4,5] => [1,2,3,4,7,5,6] [1,2,3,7,6,5,4] => [1,2,3,4,7,5,6] [1,2,4,3,5,6,7] => [1,2,3,4,5,6,7] [1,2,4,3,5,7,6] => [1,2,3,4,5,6,7] [1,2,4,3,6,5,7] => [1,2,3,4,5,6,7] [1,2,4,3,6,7,5] => [1,2,3,4,5,6,7] [1,2,4,3,7,5,6] => [1,2,3,4,5,7,6] [1,2,4,3,7,6,5] => [1,2,3,4,5,7,6] [1,2,4,5,3,6,7] => [1,2,3,4,5,6,7] [1,2,4,5,3,7,6] => [1,2,3,4,5,6,7] [1,2,4,5,6,3,7] => [1,2,3,4,5,6,7] [1,2,4,5,6,7,3] => [1,2,3,4,5,6,7] [1,2,4,5,7,3,6] => [1,2,3,4,5,7,6] [1,2,4,5,7,6,3] => [1,2,3,4,5,7,6] [1,2,4,6,3,5,7] => [1,2,3,4,6,5,7] [1,2,4,6,3,7,5] => [1,2,3,4,6,7,5] [1,2,4,6,5,3,7] => [1,2,3,4,6,5,7] [1,2,4,6,5,7,3] => [1,2,3,4,6,7,5] [1,2,4,6,7,3,5] => [1,2,3,4,6,5,7] [1,2,4,6,7,5,3] => [1,2,3,4,6,5,7] [1,2,4,7,3,5,6] => [1,2,3,4,7,6,5] [1,2,4,7,3,6,5] => [1,2,3,4,7,5,6] [1,2,4,7,5,3,6] => [1,2,3,4,7,6,5] [1,2,4,7,5,6,3] => [1,2,3,4,7,5,6] [1,2,4,7,6,3,5] => [1,2,3,4,7,5,6] [1,2,4,7,6,5,3] => [1,2,3,4,7,5,6] [1,2,5,3,4,6,7] => [1,2,3,5,4,6,7] [1,2,5,3,4,7,6] => [1,2,3,5,4,6,7] [1,2,5,3,6,4,7] => [1,2,3,5,6,4,7] [1,2,5,3,6,7,4] => [1,2,3,5,6,7,4] [1,2,5,3,7,4,6] => [1,2,3,5,7,6,4] [1,2,5,3,7,6,4] => [1,2,3,5,7,4,6] [1,2,5,4,3,6,7] => [1,2,3,5,4,6,7] [1,2,5,4,3,7,6] => [1,2,3,5,4,6,7] [1,2,5,4,6,3,7] => [1,2,3,5,6,4,7] [1,2,5,4,6,7,3] => [1,2,3,5,6,7,4] [1,2,5,4,7,3,6] => [1,2,3,5,7,6,4] [1,2,5,4,7,6,3] => [1,2,3,5,7,4,6] [1,2,5,6,3,4,7] => [1,2,3,5,4,6,7] [1,2,5,6,3,7,4] => [1,2,3,5,4,6,7] [1,2,5,6,4,3,7] => [1,2,3,5,4,6,7] [1,2,5,6,4,7,3] => [1,2,3,5,4,6,7] [1,2,5,6,7,3,4] => [1,2,3,5,7,4,6] [1,2,5,6,7,4,3] => [1,2,3,5,7,4,6] [1,2,5,7,3,4,6] => [1,2,3,5,4,7,6] [1,2,5,7,3,6,4] => [1,2,3,5,4,7,6] [1,2,5,7,4,3,6] => [1,2,3,5,4,7,6] [1,2,5,7,4,6,3] => [1,2,3,5,4,7,6] [1,2,5,7,6,3,4] => [1,2,3,5,6,4,7] [1,2,5,7,6,4,3] => [1,2,3,5,6,4,7] [1,2,6,3,4,5,7] => [1,2,3,6,5,4,7] [1,2,6,3,4,7,5] => [1,2,3,6,7,5,4] [1,2,6,3,5,4,7] => [1,2,3,6,4,5,7] [1,2,6,3,5,7,4] => [1,2,3,6,7,4,5] [1,2,6,3,7,4,5] => [1,2,3,6,4,5,7] [1,2,6,3,7,5,4] => [1,2,3,6,5,7,4] [1,2,6,4,3,5,7] => [1,2,3,6,5,4,7] [1,2,6,4,3,7,5] => [1,2,3,6,7,5,4] [1,2,6,4,5,3,7] => [1,2,3,6,4,5,7] [1,2,6,4,5,7,3] => [1,2,3,6,7,4,5] [1,2,6,4,7,3,5] => [1,2,3,6,4,5,7] [1,2,6,4,7,5,3] => [1,2,3,6,5,7,4] [1,2,6,5,3,4,7] => [1,2,3,6,4,5,7] [1,2,6,5,3,7,4] => [1,2,3,6,7,4,5] [1,2,6,5,4,3,7] => [1,2,3,6,4,5,7] [1,2,6,5,4,7,3] => [1,2,3,6,7,4,5] [1,2,6,5,7,3,4] => [1,2,3,6,4,5,7] [1,2,6,5,7,4,3] => [1,2,3,6,4,5,7] [1,2,6,7,3,4,5] => [1,2,3,6,4,7,5] [1,2,6,7,3,5,4] => [1,2,3,6,5,4,7] [1,2,6,7,4,3,5] => [1,2,3,6,4,7,5] [1,2,6,7,4,5,3] => [1,2,3,6,5,4,7] [1,2,6,7,5,3,4] => [1,2,3,6,4,7,5] [1,2,6,7,5,4,3] => [1,2,3,6,4,7,5] [1,2,7,3,4,5,6] => [1,2,3,7,6,5,4] [1,2,7,3,4,6,5] => [1,2,3,7,5,4,6] [1,2,7,3,5,4,6] => [1,2,3,7,6,4,5] [1,2,7,3,5,6,4] => [1,2,3,7,4,5,6] [1,2,7,3,6,4,5] => [1,2,3,7,5,6,4] [1,2,7,3,6,5,4] => [1,2,3,7,4,5,6] [1,2,7,4,3,5,6] => [1,2,3,7,6,5,4] [1,2,7,4,3,6,5] => [1,2,3,7,5,4,6] [1,2,7,4,5,3,6] => [1,2,3,7,6,4,5] [1,2,7,4,5,6,3] => [1,2,3,7,4,5,6] [1,2,7,4,6,3,5] => [1,2,3,7,5,6,4] [1,2,7,4,6,5,3] => [1,2,3,7,4,5,6] [1,2,7,5,3,4,6] => [1,2,3,7,6,4,5] [1,2,7,5,3,6,4] => [1,2,3,7,4,5,6] [1,2,7,5,4,3,6] => [1,2,3,7,6,4,5] [1,2,7,5,4,6,3] => [1,2,3,7,4,5,6] [1,2,7,5,6,3,4] => [1,2,3,7,4,5,6] [1,2,7,5,6,4,3] => [1,2,3,7,4,5,6] [1,2,7,6,3,4,5] => [1,2,3,7,5,4,6] [1,2,7,6,3,5,4] => [1,2,3,7,4,6,5] [1,2,7,6,4,3,5] => [1,2,3,7,5,4,6] [1,2,7,6,4,5,3] => [1,2,3,7,4,6,5] [1,2,7,6,5,3,4] => [1,2,3,7,4,6,5] [1,2,7,6,5,4,3] => [1,2,3,7,4,6,5] [1,3,2,4,5,6,7] => [1,2,3,4,5,6,7] [1,3,2,4,5,7,6] => [1,2,3,4,5,6,7] [1,3,2,4,6,5,7] => [1,2,3,4,5,6,7] [1,3,2,4,6,7,5] => [1,2,3,4,5,6,7] [1,3,2,4,7,5,6] => [1,2,3,4,5,7,6] [1,3,2,4,7,6,5] => [1,2,3,4,5,7,6] [1,3,2,5,4,6,7] => [1,2,3,4,5,6,7] [1,3,2,5,4,7,6] => [1,2,3,4,5,6,7] [1,3,2,5,6,4,7] => [1,2,3,4,5,6,7] [1,3,2,5,6,7,4] => [1,2,3,4,5,6,7] [1,3,2,5,7,4,6] => [1,2,3,4,5,7,6] [1,3,2,5,7,6,4] => [1,2,3,4,5,7,6] [1,3,2,6,4,5,7] => [1,2,3,4,6,5,7] [1,3,2,6,4,7,5] => [1,2,3,4,6,7,5] [1,3,2,6,5,4,7] => [1,2,3,4,6,5,7] [1,3,2,6,5,7,4] => [1,2,3,4,6,7,5] [1,3,2,6,7,4,5] => [1,2,3,4,6,5,7] [1,3,2,6,7,5,4] => [1,2,3,4,6,5,7] [1,3,2,7,4,5,6] => [1,2,3,4,7,6,5] [1,3,2,7,4,6,5] => [1,2,3,4,7,5,6] [1,3,2,7,5,4,6] => [1,2,3,4,7,6,5] [1,3,2,7,5,6,4] => [1,2,3,4,7,5,6] [1,3,2,7,6,4,5] => [1,2,3,4,7,5,6] [1,3,2,7,6,5,4] => [1,2,3,4,7,5,6] [1,3,4,2,5,6,7] => [1,2,3,4,5,6,7] [1,3,4,2,5,7,6] => [1,2,3,4,5,6,7] [1,3,4,2,6,5,7] => [1,2,3,4,5,6,7] [1,3,4,2,6,7,5] => [1,2,3,4,5,6,7] [1,3,4,2,7,5,6] => [1,2,3,4,5,7,6] [1,3,4,2,7,6,5] => [1,2,3,4,5,7,6] [1,3,4,5,2,6,7] => [1,2,3,4,5,6,7] [1,3,4,5,2,7,6] => [1,2,3,4,5,6,7] [1,3,4,5,6,2,7] => [1,2,3,4,5,6,7] [1,3,4,5,6,7,2] => [1,2,3,4,5,6,7] [1,3,4,5,7,2,6] => [1,2,3,4,5,7,6] [1,3,4,5,7,6,2] => [1,2,3,4,5,7,6] [1,3,4,6,2,5,7] => [1,2,3,4,6,5,7] [1,3,4,6,2,7,5] => [1,2,3,4,6,7,5] [1,3,4,6,5,2,7] => [1,2,3,4,6,5,7] [1,3,4,6,5,7,2] => [1,2,3,4,6,7,5] [1,3,4,6,7,2,5] => [1,2,3,4,6,5,7] [1,3,4,6,7,5,2] => [1,2,3,4,6,5,7] [1,3,4,7,2,5,6] => [1,2,3,4,7,6,5] [1,3,4,7,2,6,5] => [1,2,3,4,7,5,6] [1,3,4,7,5,2,6] => [1,2,3,4,7,6,5] [1,3,4,7,5,6,2] => [1,2,3,4,7,5,6] [1,3,4,7,6,2,5] => [1,2,3,4,7,5,6] [1,3,4,7,6,5,2] => [1,2,3,4,7,5,6] [1,3,5,2,4,6,7] => [1,2,3,5,4,6,7] [1,3,5,2,4,7,6] => [1,2,3,5,4,6,7] [1,3,5,2,6,4,7] => [1,2,3,5,6,4,7] [1,3,5,2,6,7,4] => [1,2,3,5,6,7,4] [1,3,5,2,7,4,6] => [1,2,3,5,7,6,4] [1,3,5,2,7,6,4] => [1,2,3,5,7,4,6] [1,3,5,4,2,6,7] => [1,2,3,5,4,6,7] [1,3,5,4,2,7,6] => [1,2,3,5,4,6,7] [1,3,5,4,6,2,7] => [1,2,3,5,6,4,7] [1,3,5,4,6,7,2] => [1,2,3,5,6,7,4] [1,3,5,4,7,2,6] => [1,2,3,5,7,6,4] [1,3,5,4,7,6,2] => [1,2,3,5,7,4,6] [1,3,5,6,2,4,7] => [1,2,3,5,4,6,7] [1,3,5,6,2,7,4] => [1,2,3,5,4,6,7] [1,3,5,6,4,2,7] => [1,2,3,5,4,6,7] [1,3,5,6,4,7,2] => [1,2,3,5,4,6,7] [1,3,5,6,7,2,4] => [1,2,3,5,7,4,6] [1,3,5,6,7,4,2] => [1,2,3,5,7,4,6] [1,3,5,7,2,4,6] => [1,2,3,5,4,7,6] [1,3,5,7,2,6,4] => [1,2,3,5,4,7,6] [1,3,5,7,4,2,6] => [1,2,3,5,4,7,6] [1,3,5,7,4,6,2] => [1,2,3,5,4,7,6] [1,3,5,7,6,2,4] => [1,2,3,5,6,4,7] [1,3,5,7,6,4,2] => [1,2,3,5,6,4,7] [1,3,6,2,4,5,7] => [1,2,3,6,5,4,7] [1,3,6,2,4,7,5] => [1,2,3,6,7,5,4] [1,3,6,2,5,4,7] => [1,2,3,6,4,5,7] [1,3,6,2,5,7,4] => [1,2,3,6,7,4,5] [1,3,6,2,7,4,5] => [1,2,3,6,4,5,7] [1,3,6,2,7,5,4] => [1,2,3,6,5,7,4] [1,3,6,4,2,5,7] => [1,2,3,6,5,4,7] [1,3,6,4,2,7,5] => [1,2,3,6,7,5,4] [1,3,6,4,5,2,7] => [1,2,3,6,4,5,7] [1,3,6,4,5,7,2] => [1,2,3,6,7,4,5] [1,3,6,4,7,2,5] => [1,2,3,6,4,5,7] [1,3,6,4,7,5,2] => [1,2,3,6,5,7,4] [1,3,6,5,2,4,7] => [1,2,3,6,4,5,7] [1,3,6,5,2,7,4] => [1,2,3,6,7,4,5] [1,3,6,5,4,2,7] => [1,2,3,6,4,5,7] [1,3,6,5,4,7,2] => [1,2,3,6,7,4,5] [1,3,6,5,7,2,4] => [1,2,3,6,4,5,7] [1,3,6,5,7,4,2] => [1,2,3,6,4,5,7] [1,3,6,7,2,4,5] => [1,2,3,6,4,7,5] [1,3,6,7,2,5,4] => [1,2,3,6,5,4,7] [1,3,6,7,4,2,5] => [1,2,3,6,4,7,5] [1,3,6,7,4,5,2] => [1,2,3,6,5,4,7] [1,3,6,7,5,2,4] => [1,2,3,6,4,7,5] [1,3,6,7,5,4,2] => [1,2,3,6,4,7,5] [1,3,7,2,4,5,6] => [1,2,3,7,6,5,4] [1,3,7,2,4,6,5] => [1,2,3,7,5,4,6] [1,3,7,2,5,4,6] => [1,2,3,7,6,4,5] [1,3,7,2,5,6,4] => [1,2,3,7,4,5,6] [1,3,7,2,6,4,5] => [1,2,3,7,5,6,4] [1,3,7,2,6,5,4] => [1,2,3,7,4,5,6] [1,3,7,4,2,5,6] => [1,2,3,7,6,5,4] [1,3,7,4,2,6,5] => [1,2,3,7,5,4,6] [1,3,7,4,5,2,6] => [1,2,3,7,6,4,5] [1,3,7,4,5,6,2] => [1,2,3,7,4,5,6] [1,3,7,4,6,2,5] => [1,2,3,7,5,6,4] [1,3,7,4,6,5,2] => [1,2,3,7,4,5,6] [1,3,7,5,2,4,6] => [1,2,3,7,6,4,5] [1,3,7,5,2,6,4] => [1,2,3,7,4,5,6] [1,3,7,5,4,2,6] => [1,2,3,7,6,4,5] [1,3,7,5,4,6,2] => [1,2,3,7,4,5,6] [1,3,7,5,6,2,4] => [1,2,3,7,4,5,6] [1,3,7,5,6,4,2] => [1,2,3,7,4,5,6] [1,3,7,6,2,4,5] => [1,2,3,7,5,4,6] [1,3,7,6,2,5,4] => [1,2,3,7,4,6,5] [1,3,7,6,4,2,5] => [1,2,3,7,5,4,6] [1,3,7,6,4,5,2] => [1,2,3,7,4,6,5] [1,3,7,6,5,2,4] => [1,2,3,7,4,6,5] [1,3,7,6,5,4,2] => [1,2,3,7,4,6,5] [1,4,2,3,5,6,7] => [1,2,4,3,5,6,7] [1,4,2,3,5,7,6] => [1,2,4,3,5,6,7] [1,4,2,3,6,5,7] => [1,2,4,3,5,6,7] [1,4,2,3,6,7,5] => [1,2,4,3,5,6,7] [1,4,2,3,7,5,6] => [1,2,4,3,5,7,6] [1,4,2,3,7,6,5] => [1,2,4,3,5,7,6] [1,4,2,5,3,6,7] => [1,2,4,5,3,6,7] [1,4,2,5,3,7,6] => [1,2,4,5,3,6,7] [1,4,2,5,6,3,7] => [1,2,4,5,6,3,7] [1,4,2,5,6,7,3] => [1,2,4,5,6,7,3] [1,4,2,5,7,3,6] => [1,2,4,5,7,6,3] [1,4,2,5,7,6,3] => [1,2,4,5,7,3,6] [1,4,2,6,3,5,7] => [1,2,4,6,5,3,7] [1,4,2,6,3,7,5] => [1,2,4,6,7,5,3] [1,4,2,6,5,3,7] => [1,2,4,6,3,5,7] [1,4,2,6,5,7,3] => [1,2,4,6,7,3,5] [1,4,2,6,7,3,5] => [1,2,4,6,3,5,7] [1,4,2,6,7,5,3] => [1,2,4,6,5,7,3] [1,4,2,7,3,5,6] => [1,2,4,7,6,5,3] [1,4,2,7,3,6,5] => [1,2,4,7,5,3,6] [1,4,2,7,5,3,6] => [1,2,4,7,6,3,5] [1,4,2,7,5,6,3] => [1,2,4,7,3,5,6] [1,4,2,7,6,3,5] => [1,2,4,7,5,6,3] [1,4,2,7,6,5,3] => [1,2,4,7,3,5,6] [1,4,3,2,5,6,7] => [1,2,4,3,5,6,7] [1,4,3,2,5,7,6] => [1,2,4,3,5,6,7] [1,4,3,2,6,5,7] => [1,2,4,3,5,6,7] [1,4,3,2,6,7,5] => [1,2,4,3,5,6,7] [1,4,3,2,7,5,6] => [1,2,4,3,5,7,6] [1,4,3,2,7,6,5] => [1,2,4,3,5,7,6] [1,4,3,5,2,6,7] => [1,2,4,5,3,6,7] [1,4,3,5,2,7,6] => [1,2,4,5,3,6,7] [1,4,3,5,6,2,7] => [1,2,4,5,6,3,7] [1,4,3,5,6,7,2] => [1,2,4,5,6,7,3] [1,4,3,5,7,2,6] => [1,2,4,5,7,6,3] [1,4,3,5,7,6,2] => [1,2,4,5,7,3,6] [1,4,3,6,2,5,7] => [1,2,4,6,5,3,7] [1,4,3,6,2,7,5] => [1,2,4,6,7,5,3] [1,4,3,6,5,2,7] => [1,2,4,6,3,5,7] [1,4,3,6,5,7,2] => [1,2,4,6,7,3,5] [1,4,3,6,7,2,5] => [1,2,4,6,3,5,7] [1,4,3,6,7,5,2] => [1,2,4,6,5,7,3] [1,4,3,7,2,5,6] => [1,2,4,7,6,5,3] [1,4,3,7,2,6,5] => [1,2,4,7,5,3,6] [1,4,3,7,5,2,6] => [1,2,4,7,6,3,5] [1,4,3,7,5,6,2] => [1,2,4,7,3,5,6] [1,4,3,7,6,2,5] => [1,2,4,7,5,6,3] [1,4,3,7,6,5,2] => [1,2,4,7,3,5,6] [1,4,5,2,3,6,7] => [1,2,4,3,5,6,7] [1,4,5,2,3,7,6] => [1,2,4,3,5,6,7] [1,4,5,2,6,3,7] => [1,2,4,3,5,6,7] [1,4,5,2,6,7,3] => [1,2,4,3,5,6,7] [1,4,5,2,7,3,6] => [1,2,4,3,5,7,6] [1,4,5,2,7,6,3] => [1,2,4,3,5,7,6] [1,4,5,3,2,6,7] => [1,2,4,3,5,6,7] [1,4,5,3,2,7,6] => [1,2,4,3,5,6,7] [1,4,5,3,6,2,7] => [1,2,4,3,5,6,7] [1,4,5,3,6,7,2] => [1,2,4,3,5,6,7] [1,4,5,3,7,2,6] => [1,2,4,3,5,7,6] [1,4,5,3,7,6,2] => [1,2,4,3,5,7,6] [1,4,5,6,2,3,7] => [1,2,4,6,3,5,7] [1,4,5,6,2,7,3] => [1,2,4,6,7,3,5] [1,4,5,6,3,2,7] => [1,2,4,6,3,5,7] [1,4,5,6,3,7,2] => [1,2,4,6,7,3,5] [1,4,5,6,7,2,3] => [1,2,4,6,3,5,7] [1,4,5,6,7,3,2] => [1,2,4,6,3,5,7] [1,4,5,7,2,3,6] => [1,2,4,7,6,3,5] [1,4,5,7,2,6,3] => [1,2,4,7,3,5,6] [1,4,5,7,3,2,6] => [1,2,4,7,6,3,5] [1,4,5,7,3,6,2] => [1,2,4,7,3,5,6] [1,4,5,7,6,2,3] => [1,2,4,7,3,5,6] [1,4,5,7,6,3,2] => [1,2,4,7,3,5,6] [1,4,6,2,3,5,7] => [1,2,4,3,6,5,7] [1,4,6,2,3,7,5] => [1,2,4,3,6,7,5] [1,4,6,2,5,3,7] => [1,2,4,3,6,5,7] [1,4,6,2,5,7,3] => [1,2,4,3,6,7,5] [1,4,6,2,7,3,5] => [1,2,4,3,6,5,7] [1,4,6,2,7,5,3] => [1,2,4,3,6,5,7] [1,4,6,3,2,5,7] => [1,2,4,3,6,5,7] [1,4,6,3,2,7,5] => [1,2,4,3,6,7,5] [1,4,6,3,5,2,7] => [1,2,4,3,6,5,7] [1,4,6,3,5,7,2] => [1,2,4,3,6,7,5] [1,4,6,3,7,2,5] => [1,2,4,3,6,5,7] [1,4,6,3,7,5,2] => [1,2,4,3,6,5,7] [1,4,6,5,2,3,7] => [1,2,4,5,3,6,7] [1,4,6,5,2,7,3] => [1,2,4,5,3,6,7] [1,4,6,5,3,2,7] => [1,2,4,5,3,6,7] ----------------------------------------------------------------------------- Created: Jan 29, 2020 at 13:25 by FindStatCrew ----------------------------------------------------------------------------- Last Updated: Jun 16, 2021 at 12:36 by Martin Rubey | 33,154 | 39,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.5739 |
https://www.tutorialspoint.com/p-observe-the-following-pattern-br-1-2-frac-1-6-br-1-2-plus2-2-frac-1-6-br-1-2-plus2-2-plus3-2-frac-1-6-br-1-2-plus2-2-plus3-2-plus4-2-frac-1-6-br-and-find-the-values-of-each-of-the-following-br-i-1-2-plus-2-2-plus-3-2-plus-4-2-plus-plus-10-2-br-ii-5-2-plus-6-2-plus-7-2-plus-8-2-plus-9-2-plus-10-2-plus-11-2-plus-12-2-p | 1,685,314,890,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00634.warc.gz | 1,151,726,582 | 11,252 | # Observe the following pattern$1^{2}=\frac{1}{6}[1 \times(1+1) \times(2 \times 1)+1)] $$1^{2}+2^{2}=\frac{1}{6}[2 \times(2+1) \times(2 \times 2)+1)]$$ 1^{2}+2^{2}+3^{2}=\frac{1}{6}[3 \times(3+1) \times(2 \times 3)+1)]$$1^{2}+2^{2}+3^{2}+4^{2}=\frac{1}{6}[4 \times(4+1) \times(2 \times 4)+1)]$and find the values of each of the following:(i) $1^2 + 2^2 + 3^2 + 4^2 +…………… + 10^2$(ii)$5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2$
To do:
We have to find the values of the given series.
Solution:
We observe that,
$1^{2}=\frac{1}{6}[1 \times(1+1) \times(2 \times 1)+1)]$
$1^{2}+2^{2}=\frac{1}{6}[2 \times(2+1) \times(2 \times 2)+1)]$
$1^{2}+2^{2}+3^{2}=\frac{1}{6}[3 \times(3+1) \times(2 \times 3)+1)]$
$1^{2}+2^{2}+3^{2}+4^{2}=\frac{1}{6}[4 \times(4+1) \times(2 \times 4)+1)]$
Therefore,
(i) $1^{2}+2^{2}+3^{2}+4^{2}+\ldots .+10^{2}=\frac{1}{6}\{10 \times(10+1) \times(2 \times 10+1)]$
$=\frac{1}{6}[10 \times 11 \times 21]$
$=\frac{10 \times 11 \times 21}{6}$
$=\frac{2310}{6}$
$=385$
(ii) $5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}+11^{2}+12^{2}=[1^{2}+2^{2}+3^{2}+4^{2}+\ldots . .+12^{2}]-[1^{2}+2^{2}+3^{2}+4^{2}]$
$=\frac{1}{6}[12 \times(12+1) \times(2 \times 12+1)]-\frac{1}{6}[4 \times(4+1) \times(2 \times 4+1)]$
$=\frac{1}{6}[12 \times13 \times25]-\frac{1}{6}[4 \times5\times9]$
$=650-30$
$=620$
Tutorialspoint
Simply Easy Learning | 743 | 1,368 | {"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} | 4.4375 | 4 | CC-MAIN-2023-23 | latest | en | 0.276201 |
https://in.mathworks.com/matlabcentral/answers/439567-how-to-fit-a-line-through-this-data-smartly | 1,660,426,381,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00233.warc.gz | 299,321,895 | 36,046 | # How to fit a line through this data "smartly" ?
13 views (last 30 days)
Luca Amerio on 13 Jan 2019
Commented: Image Analyst on 13 Jan 2019
I have the x-y data shown in the figure below (for the curious, it is the logarithm of the amplitude of a Hilbert transform).
I'm trying to write an algorithm to automatically get the slope of the initial portion of the plot, i.e. before the noise overcomes the signal.
What's causing the task to be even more challenging, are the small saw-tooth jumps that you can see in the zoomed-in portion on the right.
I tried several approaches, but could not find a reliable way to automatically fit a line only of the left-hand side linear portion.
Do you have any suggestion?
PS: I know it would help, but I don't (and cannot have) the curve fitting toolbox.
Luca Amerio on 13 Jan 2019
The data should now be attached.
Yes, the data will always start with the sloping portion as this is the logarithm of the magnitude of the hilbert transform of a decay, so the initial portion will always be the one with the highest S/N ratio
Star Strider on 13 Jan 2019
Edited: Star Strider on 13 Jan 2019
One approach (using the Signal Processing Toolbox medfilt1 function):
X = D.X;
Y = D.Y;
Ts = mean(diff(X));
YF = medfilt1(Y, 2500);
gt0 = find(dYdX > -0.001, 1, 'first');
b = polyfit(X(1:gt0),Y(1:gt0),1);
Yf = polyval(b, X(1:gt0));
figure
plot(X, Y)
hold on
plot(X(1:gt0), Yf,, ':r', 'LineWidth',2)
hold off
text(10, -5, sprintf('y = %6.3f x %6.3f', b))
This code does a median filter (medfilt1) on ‘Y’, then uses the gradient function to calculate the derivative. It then uses a relatively unsophisticated threshold (the find call) to determine the end of the relatively linear section of the data, does a linear regression on that section, and plots the result.
I cannot determine how well it will work with any other data you have, so you will likely need to experiment with it.
EDIT — (13 Jan 2019 at 18:05)
EDIT #2 — (13 jan 2019 at 18:50)
Another option, using the Signal Processing Toolbox findchangepts function:
X = D.X;
Y = D.Y;
ipt = findchangepts(X, 'Statistic','std');
b = polyfit(X(1:ipt), Y(1:ipt), 1);
Yf = polyval(b, X(1:ipt));
figure
plot(X, Y)
hold on
plot(X(1:ipt), Yf, ':r', 'LineWidth',2)
hold off
text(10, -5, sprintf('y = %6.3f x %6.3f', b))
This may be more robust. It produces similar statistics and plot.
Image Analyst on 13 Jan 2019
You said in your comment to me that you're getting a new curve from the server every 10 minutes. When you run an experiment there are some parameters you want to keep constant, and others that you allow to vary. Have you thought about whether, theoretically, it is better to compute the slope between fixed limits, or whether you want to allow the right-most index of the fitting data to vary in a way that depends on the variable data itself? I suppose arguments could be made either way, depending on what the overall situation is (which we don't know but you do).
John D'Errico on 13 Jan 2019
Edited: John D'Errico on 13 Jan 2019
A serious problem is your data also starts with a strongly nonlinear portion.
You essentially need to trim it at both ends. I'd suggest the trick is to use a Savitsky-Golay style filter, combined with a simple call to median. Median will implicitly trim those ends for you, and because we use a filter of a fairly short span, that allows median to do some smoothing too. And the linear fits are short enough they will give you sufficient points inside that quasi-linear region.
The step is constant at 0.002.
dx = 0.002;
nfilt = 20;
M = [dx*(0:nfilt-1)',ones(nfilt,1)];
Mp = pinv(M);
slopeFilter = Mp(1,:);
allSlopes = conv(Y,slopeFilter,'valid');
histogram(allSlopes,1000)
linearSegmentSlope = median(allSlopes)
linearSegmentSlope =
0.15569
As you see, the histogram has a very high spike, as we would expect. A simple median will give us a good estimate now. You could change the filter length, and the result should be fairly robust to that value, because the essentially linear segment isquite long.
Luca Amerio on 13 Jan 2019
I'm really impressed by your solution. Would you a link to understand a bit better what "a Savitsky-Golay style filter, combined with a simple call to median" is and what is that M matrix you compute in your code?
At the moment it looks like black magic to me!
Image Analyst on 13 Jan 2019
I plotted the data then it looked like the squirrely stuff stopped around element 2300 or so, and started to get fairly linear after that. So I found the slopes from element 2300 to the end of the data and plotted it.
fontSize = 15;
x = s.X;
y = s.Y;
subplot(1, 2, 1);
plot(x, y, 'b-');
grid on;
title('Original Data', 'FontSize', fontSize);
xlabel('x', 'FontSize', fontSize);
ylabel('y', 'FontSize', fontSize);
startingElement = 2310;
for k = 2310 : 100 : length(x) - 1
thisX = x((startingElement-10):k);
thisY = y((startingElement-10):k);
coefficients = polyfit(thisX, thisY, 1);
slope(k) = coefficients(1);
end
slope(1) = slope(2);
subplot(1, 2, 2);
plot(x2, y2, 'b-');
grid on;
caption = sprintf('Slope from element %d to x', startingElement);
title(caption, 'FontSize', fontSize);
xlabel('x', 'FontSize', fontSize);
ylabel('Slope', 'FontSize', fontSize);
As you can see there is sort of a flat place for slopes, where they're fairly constant, but not quite. So it kind of becomes a judgment call as to what slope you want to use, or where you want to stipulate that the slopes are starting to move away from a "constant" value. There is no clear winner - it's not like the slopes are totally constant and then abruptly switch to a different value. There is more of a gradual change of slope as you progress into the noisy area. Since there is no clear winner, you can just pick wherever you want to stop. How about around x=30? Anything wrong with that value of -0.165?
Luca Amerio on 13 Jan 2019
Absolutelly not. Nothing wrong with x=30.
I know there is a certain degree of arbitrariness in this. Since this will need to be repeated on a server automatically every 10 minutes, however, I was trying to get something robust that does not neet human intervention with ginput or "looking at the data".
I've to be honest saying that I'm a bit surprised by the change in slope in your second plot from 10 to 15 and back to 20. Looking at the data I can't really tell why fitting data up to 15 is different from fitting it to 10 or 20. The data seams pretty linear in that portion...
I will investigate this
Thank you for the help anyway
R2017b
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Start Hunting! | 1,776 | 6,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-33 | latest | en | 0.879535 |
https://eyebulb.com/how-many-types-of-curve-fitting-are-there/ | 1,722,907,390,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640461318.24/warc/CC-MAIN-20240806001923-20240806031923-00651.warc.gz | 189,697,268 | 36,853 | # How many types of curve fitting are there?
## How many types of curve fitting are there?
Comparing the Curve-Fitting Effectiveness of the Different Models
Model R-squared Biased fits
Semi-Log 98.6 Yes
Reciprocal – Linear 90.4 Yes
Linear 84.0 Yes
### Which method is applicable for best fit of a curve?
In regression analysis, curve fitting is the process of specifying the model that provides the best fit to the specific curves in your dataset.
#### How do you smooth a curve in MATLAB?
Curve Fitting Toolbox™ allows you to smooth data using methods such as moving average, Savitzky-Golay filter and Lowess models or by fitting a smoothing spline. Smooth data interactively using the Curve Fitting app or at the command line using the smooth function.
How do you select data in a curve fitting tool in MATLAB?
To select data to fit, use the drop-down lists in the Curve Fitting app to select variables in your MATLAB® workspace.
1. To fit curves: Select X data and Y data. Select only Y data to plot Y against index ( x=1:length( y ) ).
2. To fit surfaces, select X data, Y data and Z data.
How do you cite a Curve Fitting Toolbox in MATLAB?
Citation in Harvard style The MathWorks, I., 2020. Curve Fitting Toolbox, Natick, Massachusetts, United State. Available at: https://www.mathworks.com/products/curvefitting.html.
## What is a 4 parameter curve fit?
Four parameter logistic (4PL) curve is a regression model often used to analyze bioassays such as ELISA. They follow a sigmoidal, or “s”, shaped curve. This type of curve is particularly useful for characterizing bioassays because bioassays are often only linear across a specific range of concentration magnitudes.
### How do you make a curve fit perfectly?
In order to make perfect fit, we must consider error estimates as well. Perfect fit means, the curve should fit the original curve without showing any errors (such as centering and scaling erros) in that perticular degree of polynomial. Perfect fit can always be a best fit but best fit can not be a perfect fit.
#### What is linear fitting in MATLAB?
These scripts should be in the directory folder where you are using Matlab. Linear Fit file %Load this into Matlab to excute function [ outStruct ] = linfit ( x, y, dy ) %LINFIT Performs a Linear Fit on data and calculates % uncertainty in fits. Fit is y = A + B*x % % Part of the Physics 111 MATLAB Fitting Toolkit – 2009 % % INPUTS: x, y,…
What does polyval do MATLAB?
Polyfit and Polyval. Polyfit is a Matlab function that computes a least squares polynomial for a given set of data. Polyfit generates the coefficients of the polynomial, which can be used to model a curve to fit the data. Polyval evaluates a polynomial for a given set of x values.
What is curve fitting in numerical analysis?
Curve fitting is the process of constructing a curve, or mathematical function, that has the best fit to a series of data points, possibly subject to constraints. Curve fitting can involve either interpolation, where an exact fit to the data is required, or smoothing, in which a “smooth” function is constructed that approximately fits the data.
## How does the “fittype” function work?
The fittype function determines input arguments by searching the fit type expression input for variable names. fittype assumes x is the independent variable, y is the dependent variable, and all other variables are coefficients of the model. x is used if no variable exists.
25/05/2020 | 776 | 3,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.846014 |
https://constrblog.wordpress.com/2015/11/16/brief-overview-of-pin-jointed-truss-and-different-analysis-methods/ | 1,501,202,294,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549436316.91/warc/CC-MAIN-20170728002503-20170728022503-00405.warc.gz | 634,295,427 | 36,118 | # Brief overview of Pin-jointed Truss and different Analysis methods
Truss belongs to a structure that contains straight members to build up one or more triangular units. Generally Pin-joints are used to tie the members of truss at the end. The joints concerning a truss are defined as nodes. External forces and reactions can only function at the nodes and produce forces in the members to be axial forces (tensile or compressive).
If all the members and nodes are positioned inside a two dimensional plane, it is called plane truss. If a truss consists of members and nodes and expands into three dimensions, it is called space truss.
Trusses are inherent parts of various structures like bridges, roof supports, transmission towers, space stations etc.
Truss are defined as:-
a) Statically establish; all the unidentified forces (support reaction and member forces) can be specified with the use of equations of static equilibrium. (provided m+r = 2j).
b) Indeterminate; equations of static equilibrium can’t solely find out unknown forces, provided m+r> 2j.
c) Unstable; not appropriate to bear load; provided m+r < 2j; Where ‘m’ denotes member’s no in a truss; ‘r’ denotes number of reaction components; And ‘j’ denotes number of joints in a truss;
Analysis methods:
Two methods are available for determining the forces in the members concerning a truss;
(i) Methods of Joints : This method focuses on the equilibrium of the all joints concerning the truss. There exists two equations of static equilibrium ?Fx & ?Fy.So one has to start with the joint containing lower than 2 unknown forces. Go through problem 3-1 (civilengineer.webinfolist.com)
(ii) This method is useful while finding out the forces in a few members. Under this method, a fictitious is passed over the members in which forced should be set and to obtain unidentified forces, select the equilibrium of the left hand side or the right hand side of the truss.Go through problem 3-2 (civilengineer.webinfolist.com)
~~~~~~~~~~~~~~~~~~~~~~ | 446 | 2,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.883598 |
https://snipplr.com/view/46621/implementacija-binarnog-stabla-pomocu--bpolja-i-cpokazivaca | 1,642,396,146,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300289.37/warc/CC-MAIN-20220117031001-20220117061001-00719.warc.gz | 580,625,435 | 8,567 | # Posted By
tomislav_matic on 01/06/11
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# implementacija binarnog stabla pomocu B.polja i C.pokazivaca
/ Published in: C++
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`#include "c_zadatak.h"//#include "b_zadatak.h"#include <iostream> using namespace std; int main(void) { btree T; cout << "Test 1, init" << endl; InitB(1, &T); cout << "Test gotov " << endl; cout << "Test 2, init, label, root (ispis: 1)" << endl; cout << LabelB(RootB(&T), &T) << endl; cout << "Test gotov " << endl; cout << "Test 3, create left, create right (ispis: 1, 1, 0)" << endl; cout << CreateLeftB(42, RootB(&T), &T) << endl; cout << CreateRightB(43, RootB(&T), &T) << endl; cout << CreateLeftB(42, RootB(&T), &T) << endl; cout << "Test gotov " << endl; cout << "Test 4, delete" << endl; DeleteB(LeftChildB(RootB(&T), &T), &T); cout << "Test gotov " << endl; cout << "Test 5, delete (ispisuje 1)" << endl; cout << CreateLeftB(7, RootB(&T), &T) << endl; cout << "Test gotov " << endl; cout << "Test 6, parent (ispisuje 1)" << endl; cout << LabelB(ParentB(RightChildB(RootB(&T), &T), &T), &T) << endl; cout << "Test gotov " << endl; cout << "Test 7, change label (ispisuje 666)" << endl; ChangeLabelB(666, RootB(&T), &T); cout << LabelB(RootB(&T), &T) << endl; cout << "Test gotov " << endl; return 0;} //Biblioteka B.polja /// struct element { int label; int used;}; struct bt { element elements[10000];}; typedef bt btree;typedef int node; node ParentB(node n, btree *T) { return n/2;} node RootB(btree *T) { return 1;} void InitB(int x, btree *T) { T->elements[1].used = 1; T->elements[1].label = x; for (int i = 2; i < 10000; i++) { T->elements[i].used = 0; }} node LeftChildB(node n, btree *T) { if (T->elements[n*2].used == 1) return n*2; else return 0;} node RightChildB(node n, btree *T) { if (T->elements[n*2 +1].used == 1) return n*2 +1; else return 0;} int LabelB(node n, btree *T) { return T->elements[n].label; } void ChangeLabelB(int x, node n, btree *T) { T->elements[n].label = x;} void DeleteB(node n, btree *T) { if (LeftChildB(n, T)) { DeleteB(LeftChildB(n, T), T); } if (RightChildB(n, T)) { DeleteB(RightChildB(n, T), T); } T->elements[n].used = 0;} bool CreateLeftB(int x, node n, btree *T) { if (LeftChildB(n, T) != 0) { return false; } else { T->elements[n*2].used = 1; T->elements[2*n].label = x; return true; }} bool CreateRightB(int x, node n, btree *T) { if (RightChildB(n, T) != 0) { return false; } else { T->elements[n*2 +1].used = 1; T->elements[2*n +1].label = x; return true; }} Biblioteka C.pokazivaca #include <cstdlib> struct element { int label; element *left,*right;}; typedef element *node;typedef element btree; node RootB(btree *T) { return T;} void InitB(int x, btree *T) { T->label = x; T->left = NULL; T->right = NULL;} node LeftChildB(node n, btree *T) { return (n->left);} node RightChildB(node n, btree *T) { return (n->right);} node rekurzivna(node cvor, node trazeni, btree *T) { if (cvor->left == trazeni || cvor->right == trazeni) { return cvor; } node tmp; tmp = rekurzivna(cvor->left, trazeni, T); if (tmp != NULL) return tmp; tmp = rekurzivna(cvor->right, trazeni, T); if (tmp != NULL) return tmp; return NULL;} node ParentB(node n, btree *T) { return rekurzivna(RootB(T), n, T);} int LabelB(node n, btree *T) { return (n->label);} void ChangeLabelB(int x, node n, btree *T) { n->label = x;} void DeleteB(node n, btree *T) { if (n->left != NULL) { DeleteB(n->left, T); } if (n->right != NULL) { DeleteB(n->right, T); } node Parent = ParentB(n, T); if (Parent->left == n) { Parent->left = NULL; } else { Parent->right = NULL; } delete n;} bool CreateLeftB(int x, node n, btree *T) { if (n->left == NULL) { n->left = new element; n->left->left = NULL; n->left->right = NULL; n->left->label = x; return 1; } return 0;} bool CreateRightB(int x, node n, btree *T) { if (n->right == NULL) { n->right = new element; n->right->left = NULL; n->right->right = NULL; n->right->label = x; return 1; } return 0;}` | 1,614 | 4,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-05 | latest | en | 0.296618 |
https://ninjasquad.org/the-6s-math-puzzle-coding-n-concepts/ | 1,679,734,175,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00135.warc.gz | 493,150,387 | 21,398 | The 6s Math Puzzle – Coding N Concepts
The 6s Math Puzzle – Coding N Concepts | ninjasquad
This is a pure mathematical puzzle which evaluate your ability to solve maths equation. This is a great problem for building number sense.
Puzzle
The challenge is to make each below equation true using common mathematical operations. You cannot introduce any new digits (so the cube root ∛ is not allowed since it has a 3).
Hint: common mathematical operations are `+` `-` `x` `/` `!` `√`
``````0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6
``````
Solution
Note: Please try on your own before looking at the solution. Also some of the equations has multiple solutions.
Let’s start with some easier ones.
``````2 + 2 + 2 = 6
6 + 6 – 6 = 6
6 × (6/6) = 6
7 – 7/7 = 6
5 + 5/5 = 6
``````
Here are a few solutions for 3.
``````3 × 3 – 3 = 6
3! + 3 – 3 = 6
3! × (3/3) = 6
√(3 × 3) + 3 = 6
``````
For the number 9, we can use a trick. Since √(9) = 3, we can take the square root of each number, so the problem is equivalent to solving 3 3 3 = 6, which we have just solved! So we can use any of those solutions, or we can find others too.
``````√9 × √9 – √9 = 6
(√9)! + √9 – √9 = 6
(√9)! × √9/√9 = 6
(√9 × √9/√9)! = 6
``````
We can do a similar trick for 4. Since √4 = 2, we can use the solution for 2 2 2 = 6. But there are other solutions too.
``````√4 + √4 + √4 = 6
(4 – 4/4)! = 6
(√4 + 4/4)! = 6
``````
Now we just have a couple more to solve and we will use 3! = 6 in many of the answers. We can solve for 10 as:
``````(√(10 – 10/10))! = 6
``````
Then we have the 1 solution:
``````(1 + 1 + 1)! = 6
``````
To solve 0, we use the fact that 0! = 1, and then we have reduced the problem to solving 1 1 1 = 6, which was previously solved.
``````(0! + 0! + 0!)! = 6
``````
We just have one more to solve, which many people consider to be the hardest. One way to solve uses nested square roots.
``````8 – √(√(8 + 8)) = 6
``````
The other method uses 3! = 6.
``````(√(8 + 8/8))! = 6
``````
And we are done! Here are the above solutions listed in numerical order. This is not a comprehensive list. you might have found another way too!
``````(0! + 0! + 0!)! = 6
(1 + 1 + 1)! = 6
2 + 2 + 2 = 6
3 × 3 – 3 = 6
3! + 3 – 3 = 6
3! × (3/3) = 6
√(3 × 3) + 3 = 6
√4 + √4 + √4 = 6
(4 – 4/4)! = 6
(√4 + 4/4)! = 6
5 + 5/5 = 6
6 + 6 – 6 = 6
6 ×(6/6) = 6
7 – 7/7 = 6
8 – √(√(8 + 8)) = 6
(√(8 + 8/8))! = 6
√9 × √9 – √9 = 6
(√9)! + √9 – √9 = 6
(√9)! × √9/√9 = 6
(√9 × √9/√9)! = 6
(√(10 – 10/10))! = 6
``````
Source: Internet
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X | 1,128 | 2,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-14 | longest | en | 0.935304 |
http://openstudy.com/updates/564217e8e4b0d434ade6ba43 | 1,519,312,230,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814124.25/warc/CC-MAIN-20180222140814-20180222160814-00523.warc.gz | 255,688,653 | 8,626 | • anonymous
Which equation can be used to solve the problem? Maura spent 1-1/2hours on science homework. This was 2/3 of the total time she spent on homework. How Much Time Did She Spend On Homework? A.h=2/3*1-1/2 B.1-1/2*h - 2/3 c.h-2/3=1-1/2 d.2/3h=1-1/2
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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Looking for something else?
Not the answer you are looking for? Search for more explanations. | 345 | 1,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-09 | latest | en | 0.371925 |
http://cboard.cprogramming.com/c-programming/143435-****-exe-has-stopped-working.html | 1,475,009,726,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661213.28/warc/CC-MAIN-20160924173741-00128-ip-10-143-35-109.ec2.internal.warc.gz | 42,753,360 | 13,501 | # Thread: ****.exe has stopped working
1. ## ****.exe has stopped working
I wrote a code below to find out the prime factors using recursion. it runs fine but terminates with a message: primfactor.exe has stopped working. check for online solution.. any idea why so?
Code:
```/*prime factor by recursion*/
#include<stdio.h>
int rec (int,int); /* this semi colon is must*/
int main(void)
{
int n;
printf("Enter any positive integer for prime factorization ? ");
scanf("%d",&n);
rec(n,2);
return (0);
}
int rec(int n, int l)
{
while (1)
{
if (n%l==0)
{
printf("%d ",l);
n=n/l;
}
else
{
if (n==l)
break;
rec(n,l+1);
}
}
return (0);
}```
2. You use "while(1)" OR recursion - not both together at the same time.
while(1) calling another while(1) calling another while(1) calling another while(1) is a special brand of infinity.
3. Salem, I feel while(1) is must in this case since we don't know the number of loops required. I put the if condition at right place and got the code running fine. here is the edited version:
Code:
```/*prime factor by recursion*/
#include<stdio.h>
int rec (int,int); /* this semi colon is must*/
int main(void)
{
int n;
printf("Enter any positive integer for prime factorization ? ");
scanf("%d",&n);
rec(n,2);
return (0);
}
int rec(int n, int l)
{
while (1)
{
if (n%l==0)
{
printf("%d ",l);
n=n/l;
}
else
break;
}
if (n>l)
rec(n,l+1);
if (n==l || n<l)
return (0);
}```
4. You are thinking iteratively, not recursively.
The number of times you recursively call the function doesn't need to be known. What needs to be known is:
1) that you have in fact a stopping point for the recursion. It's essential that the recursion not go horribly deep with nested calls.
2) that you are calling the function with the right parameters. Generally, some progress must be made, with each execution through the recursive function.
5. so what you think must be the right recursive solution for this problem? shall I replace the while loop with another recursive call to says rec1?
6. thinking more recursively, I changed the while(1) loop into another recursive call:
Code:
```/*prime factor by recursion*/
#include<stdio.h>
int rec (int,int);
int rec1 (int, int); /* this semi colon is must*/
int main(void)
{
int n;
printf("Enter any positive integer for prime factorization ? ");
scanf("%d",&n);
rec(n,2);
return (0);
}
int rec(int n, int l)
{
n=rec1(n,l);
if (n>l)
rec(n,l+1);
return 0;
}
int rec1(int n, int l)
{
if (n%l==0)
{
printf("%d ",l);
n=n/l;
rec1(n,l);
}
return n;
return 0;
}```
But the code is wrong answer for n=288
any help possible?
7. what is a "prime" factor?
A prime number less than n, the greatest common denominator, any factor at all, or ??
Complimentary recursive functions are quite elegant, but more than you require, I believe.
Edit: I just looked up what prime factorization was. We use that fundamental theorem of Arithmetic for some programs that find primes, but I've never heard of this term before.
8. you are right.. I finally found a pure recursive solution with just one recursive function.
Code:
```/*prime factor by recursion*/
#include<stdio.h>
int rec (int,int); /* this semi colon is must*/
int main(void)
{
int n;
printf("Enter any positive integer for prime factorization ? ");
scanf("%d",&n);
rec(n,2);
return (0);
}
int rec (int n, int l)
{
if (n%l==0)
{
printf("%d ",l);
n=n/l;
rec(n,l);
}
else
{
if (n<l)
return 0;
else
{
l=l+1;
rec(n,l);
}
}
return 0;
}```
9. for learning purpose, How could we possibly use an recursive call from master recursive call? would be interesting to know..
10. May I note that using l's and 1's in the same loops is asking for errors? Try using i, or j, or k, or m, or anything else except O ("oh"), and l ("el"), and capital I.
I haven't seen them arranged as recursive functions with master and slave varieties. I've seen complimentary or symbiotic recursive functions that called each other, in old style mini-max functions. (used in chess, checkers, and other two party games of perfect information, in the early days). When the move was for white, one function was called recursively, and when the move changed one ply (half-move, where each players move is a half of a whole move), deeper, the other recursive function was called. This is a look ahead function, so each time the computer is thinking, both white and black moves have to be considered. Thus the back and forth calls.
Nowadays, they just use the same function, but scores are multiplied by -1 at odd or even ply depths, depending on which color the computer is, and how scoring is set up.
11. I found this code working fine. it uses another recursive call from a recursive call.. I don't know if calling them master-slave is valid or not.. but it is good learning about recursive functions and calling them from each other..
Code:
```/*prime factor by recursion*/
/* calling another recursive function from a recursive function*/
#include<stdio.h>
int rec (int,int);
int rec1 (int,int);
int main(void)
{
int n;
printf("Enter any positive integer for prime factorization ? ");
scanf("%d",&n);
rec(n,2);
return (0);
}
int rec(int n, int l)
{
int x;
x=rec1(n,l);
l=l+1;
if (x>=l)
rec(x,l);
else
return (0);
}
int rec1(int n, int l)
{
if (n%l!=0)
return (n);
else
{
printf("%d ",l);
n=n/l;
rec1(n,l);
}
}``` | 1,399 | 5,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2016-40 | latest | en | 0.825317 |
https://www.thefreedictionary.com/Newton's+law+of+gravitation | 1,623,718,096,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487614006.8/warc/CC-MAIN-20210614232115-20210615022115-00059.warc.gz | 947,078,156 | 11,654 | # Newton's law of gravitation
Also found in: Thesaurus, Medical, Encyclopedia, Wikipedia.
## Newton's law of gravitation
n.
The law proposed by Sir Isaac Newton that expresses the force of gravitational attraction between two bodies as a function of their masses and their distance. Expressed mathematically, F = Gm1m2/d2 where F is the force in Newtons, m1 and m2 are the masses of the bodies in kilograms, G is the gravitational constant, and d is the distance between the bodies in meters.
[After Isaac Newton.]
## Newton's law of gravitation
n
(General Physics) the principle that two particles attract each other with forces directly proportional to the product of their masses divided by the square of the distance between them
Collins English Dictionary – Complete and Unabridged, 12th Edition 2014 © HarperCollins Publishers 1991, 1994, 1998, 2000, 2003, 2006, 2007, 2009, 2011, 2014
## Newton's law of gravitation
The principle that two bodies exert a gravitational attraction for each other that increases as their masses increase and as the distance between them decreases. In mathematical terms, the force equals the product of the two masses multiplied by the gravitational constant and divided by the square of the distance. Also called law of gravitation, law of universal gravitation. See Note at gravity. | 304 | 1,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-25 | latest | en | 0.93033 |
https://astarmathsandphysics.com/university-physics-notes/thermal-physics/1676-alternative-formulation-of-the-second-law-of-thermodynamics.html?tmpl=component&print=1&page= | 1,519,318,216,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814140.9/warc/CC-MAIN-20180222160706-20180222180706-00394.warc.gz | 606,144,215 | 73,581 | ## Alternative Formulation of the Second Law of Thermodynamics
The thermodynamic state of a substance is defined by the quantities– volume, – pressure,– Temperature and– internal energy.
These four quanties are related by the equationsand one equation for the internal energy eg
This means we can choose any two of these quantities as independent andf express the other two – dependent – variables in terms of the independent variables.
The second law of thermodynamics can be written(1) whereandare independent.
Or aswhereandare independent (2)
Or aswhereandare independent (3)
Or aswhereandare independent (4)
Or aswhere U and p are independent (5)
Or aswhereandare independent. (6)
To show (1) write
Then(7)
Now let
Then(8)
Henceand
Equate this expression withandfrom (7) to give
and
Hence we can write (1) asor
he proof for (3) is similar.
To prove (4) takeandas independent, then
Then
Rearrange the second of these to give
Substitute into the first to give
Compare these two equations with equations (7) to give
and
Substitute these into (1) to obtain
(5) and (6) can be similarly proved. | 274 | 1,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-09 | longest | en | 0.860823 |
https://math.stackexchange.com/questions/1144136/is-euclidean-geometry-studied-at-all/1144258 | 1,726,657,550,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00363.warc.gz | 351,247,245 | 39,034 | # Is Euclidean Geometry studied at all?
Is there a place for Euclidean geometry in the hearts or minds of any mathematicians? I personally find it to be the most beautiful mathematics I have yet encountered but I see little of it on sites such as these, leading me to believe that it somehow disappears from the minds of all college mathematics students unless it is to calculate a distance or angle, which is not at all what appeals to me.
The only modern day type of Euclidean geometry research, etc. that I see being done at a college level would be the Forum Geometricorum. (Highly recommend it)
In short, will anyone see or study Euclidean geometry after high school?
• @Workaholic Are most mathematicians even interested in the decidability of the theories they use/study?
– Did
Commented Feb 11, 2015 at 21:37
• @Workaholic You mean, an algorithm that could in principle identify all theorems from non-theorems. The relevance of this remark to decide of the interest of mathematicians seems nearly null.
– Did
Commented Feb 11, 2015 at 21:57
This "comment" is a little too long for the comments.
I think, without diving too deep in to the link that you posted, that what you're lamenting is the lack of what I would call synthetic geometry, in particular as opposed to analytic geometry or other approaches to geometry ( eg algebraic geometry ). This would be things like ruler/compass constructions. To this complaint, I think many many mathematicians consider synthetic geometry very beautiful and it has a special place in their hearts, but it's kind of like owning a horse that you ride for pleasure and exercise, but you still drive a car to get places; we typically have better tools now. Synthetic geometric intuitions still play a big part and occasionally I've seen proofs that were more enlightening when done synthetically, but for the most part, practicing mathematicians use the more modern tools.
If it's synthetic geometry you're actually interested in, you shouldn't restrict yourself to Euclidean geometry! There's actually a lot of fun to be had trying to do some of the same things in non-euclidean geometry that were done for centuries in Euclidean geometry. I recommend the book "Euclidean and non-Euclidean Geometry". For more of a similar flavor, you could study projective geometry. I've never read it, but Robin Hartshorne has a very popular book that is of a synthetic flavor ( Note: His more famous algebraic geometry books are decidedly not of this flavor ).
As far as "Euclidean" geometry goes, though, that is still quite actively studied. It may not be the hottest topic, and the tools used are often pretty far removed the days of the Greeks or even Descartes, but there is plenty of geometry still done that studies good old Euclidean space.
• Yeah the geometry I was thinking about was synthetic, IMO style geometry with cyclic quads and triangle centers and all that good stuff. I like the analogy. Commented Feb 12, 2015 at 11:45
The area of circle packings is currently pretty wide open. Here's a specific easy example: take a circle of radius $r$ and place $n$ circles around it so that they are tangent to the center circle and to each other. The question is, what is the ratio of radii of the center circle to the smallest outer circle? So if $r_c$ is the radius of the center circle, and $r_1$ is the radius of the smallest circle, then how big can $r_c/r_1$ be? Here's an image:
Center circle, with 7 tangent outer circles. The left/right images are not allowed:
The answer is surprising, and is known as the Ring Lemma: $r_c/r_1$ is bounded above by a constant that only depends on $n$: the number of outer circles. In other words, the outer tangent circles can be arbitrarily big (this is obvious, just make one big) but they cannot be arbitrarily small for a given $n$. The proof of this is not trivial! Also surprisingly, this idea was used in proving distributional limits of random graphs.
As far as I'm aware, these results have not been proven in dimensions 4 and higher. These questions become very difficult in higher dimensions!
These kinds of things also come up in complex analysis (Schramm Loewner theory in particular), where one is trying to prove regularity results about complicated limits of stochastic processes. It helps to think of your domain as a packing of circles, and then look at conformal maps of this. | 969 | 4,394 | {"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} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.964651 |
https://classroom.thenational.academy/lessons/1-5-6grkae?step=5&activity=completed | 1,709,502,315,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00035.warc.gz | 173,002,661 | 26,688 | # 1 - 5
In this lesson, we will develop our problem solving skills.
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# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.On a compass, what direction is opposite NE (north east)?
1/3
Q2.What is a carabiner?
2/3
Q3.Which of the following is a famous British adventurer?
3/3
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# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.On a compass, what direction is opposite NE (north east)?
1/3
Q2.What is a carabiner?
2/3
Q3.Which of the following is a famous British adventurer?
3/3
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
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# 1-5
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.What other subject is this challenge closely linked to?
1/3
Q2.What is the least number of moves it takes to complete this challenge?
2/3
Q3.This challenge is based on what?
3/3
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# 1-5
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.What other subject is this challenge closely linked to?
1/3
Q2.What is the least number of moves it takes to complete this challenge?
2/3
Q3.This challenge is based on what?
3/3
# Lesson summary: 1 - 5
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Climb stairs
On the spot:
Chair yoga | 616 | 2,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-10 | latest | en | 0.911995 |
https://b-ok.org/book/456331/908dd8 | 1,571,213,312,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00182.warc.gz | 390,273,544 | 35,014 | Main Algorithm for solving 2nd order linear DE
# Algorithm for solving 2nd order linear DE
Year: 1986
Language: english
Pages: 21
Series: JSC
File: DJVU, 490 KB
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1
### Factorization of functional differential equations
Language: english
File: DJVU, 67 KB
2
### Complexity of factoring and GCD of linear differential operators
Year: 1990
Language: english
File: DJVU, 603 KB
```Editor's Announcement Membership of the AAR is inexpensive, providing a convenient way to follow various developments in the field. To subscribe, send \$5.00 to the Treasurer: Larry Henschen, 780 S. Warrington Road, Des Plaines, Illinois 60016 U.S.A. Other societies in the area of symbolic computation may be given space for a similar announcement in this journal. I Symbolic Computation A986) 2, 3-43 An Algorithm for Solving Second Order Linear Homogeneous Differential Equations JERALD J. KOVACIC JYACC Inc., 919 Third Avenue, New York, NY 10022, U.S.A. (Received 8 May 1985) In this paper we present an algorithm for finding a "closed-form" solution of the differential equation /'+ay' + by, where a and b are rational functions of a complex variable x, provided a "closed-form" solution exists. The algorithm is so arranged that if no solution is found, then no solution can exist. 1. Introduction In this paper we present an algorithm for finding a "closed-form" solution of the differential equation y" + ay' +by, where a and b are rational functions of a complex variable x, provided a "closed-form" solution exists. The algorithm is so arranged that if no solution is found, then no solution can exist. The first section makes precise what is meant by "closed-form" and shows that there are four possible cases. The first three cases are discussed in sections 3, 4 and 5 respectively. The last case is the case in which the given equation has no "closed-form" solution. It holds precisely when the first three cases fail. In the second section we present conditions that are necessary for each of the three cases. Although this material could have been omitted, it seems desirable to know in advance which cases are possible. The algorithm in cases 1 and 2 is quite simple and can usually be carried out by hand, provided the given equation is relatively simple. However, the algorithm in case 3 involves quite extensive computations. It can be programmed on a computer for a specific differential equation with no difficulty. In fact, the author has worked through several examples using only a programmable calculator. Only in one example was a computer necessary, and this was because intermediate numbers grew to 20 decimal digits, more than the calculator could handle. Fortunately, the necessary conditions for case 3 are quite strong so this case can often be eliminated from consideration. The algorithm does require that the partial fraction expansion of the coefficients of the differential equation be known, thus one needs to factor a polynomial in one variable over the complex numbers into linear factors. Once the partial fraction expansions are known, only linear algebra is required. Using the MACSYMA computer algebra system, see, for example, Pavelle & Wang A985), Bob Caviness and David Saunders of Rensselear Polytechnic Institute programmed the entire algorithm (see Saunders A981)). Meanwhile, the algorithm has
J. J. Kovacic been implemented also in the maple computer algebra system, see, for example, Char al. A985), by Carolyn Smith A984). This paper is arranged so that the algorithm may be studied independently of ih( proofs. In section 1, parts 1 and 2 are necessary to understand the algorithm, parts 3 an< 4 are devoted to proofs. In the other sections, part 1 describes the algorithm, part j contains examples, and the remaining parts contain proofs. Since the first appearance of this paper as a technical report, a number of papers havj appeared on the same problem: Baldassarri A980), Baldassarri & Dwork A979), Singe A979, 1981, 1985). Special thanks are due to Bob Caviness and David Saunders of RPI for thcit| encouragement and assistance during the preparation of this paper. 1. The Four Cases In the first part of this section we define precisely what we mean by "closed-form' solution. In the second part we state the four possible cases that can occur. These case are treated individually in the latter sections. The third part is devoted to a brie description of the Galois theory of differential equations. This theory is used in the proof! of the theorems of the present chapter and those of sections 4 and 5. Part 4 contains a proof of the theorem stated in part 2. 1.1. LIOUVILLIAN EXTENSIONS The goal of this paper is to find "closed-form" solutions of differential equations. By t "closed-form" solution we mean, roughly, one that can be written down by a first-yeai calculus student. Such a solution may involve esponentials, indefinite integrals and solutions of polynomial equations. (As we are considering functions of a comply variable, we need not explicitly mention trigonometric functions, they can be written it terms of exponentials. Note that logarithms are indefinite integrals and hence are allowed.) A more precise definition involves the notion of Liouvillian field. Definition. Let F be a differential field of functions of a complex variable x that contains] C(x). (Thus F is a field and the derivation operator ' (=d/dx) carries F into itself). К isi said to be Liouvillian if there is a tower of differential fields C(x) = F0cFlo..cF, = F such that, for each i = 1,..., n, either F( = F,_1(a) where a'/aeF,_t (F, is generated by an exponential of an integral over F,^) or F, = F,_1(a) where a'eF,_! (F, is generated by an integral over F,_i) or Fj is finite algebraic over F,.!. A function is said to be Liouvillian if it is contained in some Liouvillian differential field' Suppose that ц is a (non-zero) Liouvillian solution of the dilTcrcntial equation Solving Homogeneous Differential Equations v" + ay' + by, where a, beC(x). It follows that every solution of this differential equation is Liouvillian. Indeed, the method of reduction of order produces a second solution, namely «[(e'^/n2). This second solution is evidently Liouvillian and the two solutions are linearly independent. Thus any solution, being a linear combination of these two, is Liouvillian. We may use a well-known change of variable to eliminate the term involving / from the differential equation. Set z = eiiay. Then z" + (Ь-\а2-^а')г = О. This new equation still has coefficients in C(x) and evidently у is Liouvillian if and only if z is Liouvillian. Thus no generality is lost by assuming that the term involving y' is missing from the differential equation. 1.2 THE FOUR CASES In the remainder of this paper we shall consider the equation y" = ry, reC(x). We shall refer to this equation as "the DE". To avoid triviality, we assume that гфС. By a solution of the DE is always meant a non-zero solution. Theorem. There are precisely four cases that can occur. Case 1. The DE has a solution of the form е'ш where шеС(х). Case 2. The DE has a solution of the form е'ш where со is algebraic over C(x) of degree 2, and case 1 does not hold. Case 3. All solutions of the DE are algebraic over C(x) and cases 1 and 2 do not hold. Case 4. The DE has no Liouvillian solution. It is evident that these cases are mutually exclusive, the theorem states that they are exhaustive. The proof of this theorem will be presented in part 1.4. 1.3. THE DIFFERENTIAL GALOIS GROUP Here we present a brief summary of the Picard-Vessiot theory of differential equations (see Kaplansky A957), or Chapter 6 of Kolchin A973)), which is tailored specifically to the DE /' = ry. Suppose that n, С is a fundamental system of solutions of the DE (where tj, { are functions of a complex variable x). Form the differential extension field G of C(z) generated by n, (, thus G = !г, О = Then the Galois group of G over C(x), denoted by G(G/C(x)), is the group of all differential automorphisms of G that leave C(x) invariant. (An automorphism a is differential if a(a') = (aa)' for every a e G.) We refer the reader to the references cited above for a proof that the Fundamental Theorem of Galois Theory holds in this context. There is an isomorphism of G(G/C(x)) with a subgroup of GLB), the group of Avertible 2x2 matrices with coefficients in C. Let aeG(G/C(x)). Then (an)" = a(n") = а(гц) = rcrr;.
J. J. Kovacic Hence, an is also a solution of the DE and so is a linear combination от/ = aarj + ca(, а„ с„еС, of^, С Similarly, oZ, = bari + dal, for some Ь„, daeC c:a ¦ is immediately seen to be an injective group homomorphism. This representation с: G(G/C(x)) -»GLB) certainly does depend on the choice of| fundamental system n, (. If rjlt ?, is another fundamental system, then there is a matrix XeGLB) such that fa,, ?,) = fa, QX. Therefore, G = C(jc)<ij,f> = C(x)<4l,{;1> and Ci{a) = X-lc{e)X. The representation G(G/C(x)) -»GLB) is determined by the DE only up to conjugation. By abuse of language, we allow ourselves to speak of any one of these conjugate groups as the Galois group of the DE. If a fundamental system r\, ? is fixed, then we refer to c(G(G/C(x))) s GLB) as the Galois group of the DE relative to n, ?. Fix a fundamental system r\, ? of solutions of the DE and let G ? GLB) be the Galois group relative to rj, ?. Let W = t]('-rjX be the Wronskian of n, ?. A simple computation, using the DE, shows that W = 0, so W is a (non-zero) constant and is left fixed by any element of G(G/C(x)). Let a e G(G/C(x)), then, using the notation above, Solving Homogeneous Differential Equations W=aW= - W. Thus G <= SLB), the group of 2 x 2 matrices with determinant 1. Recall that a subgroup G of GLB) is an algebraic group if there exist a finite number of polynomials Р1,...,Рге?1ХиХ2,Хъ,Х^ suchthat Г JeG \c dj if and only if P^a, b,c,d)=--- = Pr(a, b, c, d) = 0. One of the principal facts in the Picard-Vessiot theory is that the Galois group of a differential equation is an algebraic group. For a proof in all generality, see the references cited above. Here we sketch a proof in the special case that we are considering. Let У, Z, У,, Zj be indeterminates over C(x) and consider the substitution homomorphism CO, Y, Z, Yu ZJ -» C[x, n, С tf, Cl The kernel of this mapping is a prime ideal p. Any element fa b' A = С d of SLB) induces an automorphism of C[x, У, Z, Yu Z,] over C[x] by the formula (У, Z, У„ Z,) -* (aY+cZ, bY+dZ, aY^+cZ^, bY^dZJ. Moreover, A e G if and only if p is carried into itself. The ideal p is finitely generated, say p = (<j];..., qs), where qu .. ., q, are linearly independent over C. Let n be the maximum of the degrees o( q,,..., </, in x, У, Z, У,, Z, and let V be the vector space over С of all polynomials in C[x, Y, Z, У,, Z,] of degree n or less. Evidently the action of SLB) on ?гх, У, Z, V"i,Zi] restricts to V. If <jx,. .., <?s,<js+i,..., q, is a basis of V, then there exist polynomials PijeClXl, X2, X3, X^\ such that the result of the action of A on q, is ? Р{1{а,Ъ,сЛ)ч}- It follows that AeG if and only if Pi/fl, b, c, d) = 0 for i=l,...,s, j = s+l, Therefore G is an algebraic group. . .,t. сеСсфО}, 1.4. proof In this section we shall prove the theorem that was stated in 1.2. We shall use several facts about algebraic groups. Suitable references are Borel A956), Kaplansky A957), and Chapter 5 of Kolchin A973). The following result is contained in Kaplansky A957, p. 31). Lemma. Let G be an algebraic subgroup of SLB). Then one of four cases can occur. Case 1. G is triangulisable. Case 2. G is conjugate to a subgroup of ;^Ь"'Н(-«-о) and case 1 does not hold. Case 3. G is finite and cases 1 and 2 do not hold. Case 4. G = SLB). Proof. Denote the component of the identity of G by G°. First we note that any two- dimensional Lie algebra is solvable, hence either dim G = 3 (in which case G = SLB)) or else G° is solvable. In the latter case, G° is triangulisable by the Lie-Kolchin Theorem. Assume that G° is triangular. If G° is not diagonalisable, then G° contains a matrix of the form I I with а Ф 0 (since an algebraic group contains the unipotent and semi-simple parts of all of its /1 a\ elements). Since G° is normal in G, any matrix in G conjugates I I into a triangular matrix. A direct computation shows that only triangular matrices have this property. Thus G itself is triangular. This is case 1. Assume next that G° is diagonal and infinite, so G° contains a non-scalar diagonal matrix A. Because G° is normal in G, any element of G conjugates A into a diagonal matrix. A direct computation shows that any matrix with this property must be contained in D\ Therefore either G is diagonal, this being case 1, or else G is contained in D\ this being case 2. Finally we observe that if G° is finite (and therefore G° = {1}), then G must also be finite. This is case 3. This proves the lemma. We shall now prove the theorem of section 2. Let n, ? be a fundamental system of solutions of the DE and let G be the Galois group relative to ц, ?. Set G = C(x)<»j, ?>. Case 1. G is triangulisable. We may assume that G is triangular. Then, for every
J. J. Ко Therefore aa> = a>, where m = n'/n, which a6 G(G/C(X)), or\ = car\, where с„еС, с„ implies that ш е C(x). Case 2. G is conjugate to be a subgroup of D\ We may assume that G is a subgroup о D1. If a> = n'/n and ф = ?'/?, then, for every <reG(G/C(x)), either асо = ш, аф = ф о aw = ф, аф = w. Thus a> is quadratic over C(x). Case 3. G is finite. In this case G has only a finite number of differential automorphisms ai, ¦.., <V Since the elementary symmetric function of oln,..., а„п are invariant under G(G/C(x)), n is algebraic over C(x). Similarly, ( is algebraic over C(x). Because every solution of the DE is contained in G, every solution of the DE is algebraic. Case 4. G = SLB). Suppose that the DE had a Liouvillian solution. Then, as pointed out in 1.1, every solution of the DE is Liouvillian. Thus G is contained in a Liouvillian field. It follows that G° is solvable (Kolchin, 1973, p. 415). Since G° = SLB) is not solvable, the DE can have no Liouvillian solution. This proves the theorem. 2. Necessary Conditions In this section we discuss some easy conditions that are necessary for cases 1, 2, or 3 to hold. These conditions give a sufficient condition for case 4 to hold (namely when the necessary conditions for cases 1, 2, and 3 fail). Throughout, we shall consider the DH y" = ry геОД 2.1. THE NECESSARY CONDITIONS Since r is a rational function, we may speak of the poles of r, by which we shall always mean the poles in the finite complex plane C. If r — s/t, with s, t eC[x], relatively prime, then the poles of r are the zeros of t and the order of the pole is the multiplicity of the zero of t. By the order of r at oo we shall mean the order of oo as a zero of r, thus the order of r at oo is deg t—degs. Theorem. The following conditions are necessary for the respective cases to hold. Case 1. Every pole ofr must have even order or else have order 1. The order ofrat a must be even or else be greater than 2. Case 2. r must have at least one pole that either has odd order greater than 2 or else has order 2. Case 3. The order of a pole ofr cannot exceed 2 and the order of r at oo must be at least 2. If the partial fraction expansion ofr is a' , у ft -c,J ?x-dj' then v/l+4a,eQ, for each i, ? /?, = 0, and if then Solving Homogeneous Differential Equations 2.2. examples Airey's Equation y" = xy has no Liouvillian solution (i.e. case 4 holds). This is clear because the necessary conditions for cases 1, 2, and 3 all fail. More generally, y" = Py, where РеСПх] has odd degree, has no Liouvillian solution. For Bessel's Equation , 4(n2-x2)-l (in self-adjoint form), only cases 1, 2, and 4 are possible. For Weber's Equation only cases 1 and 4 are possible. 2.3. proof In this section we prove the theorem of Section 1. Case 1. In this case the DE has a solution of the form n = ef<a where сиеС(х). Since n" = rrj, it follows that a>' + w2 — r (the Riccatti Equation). Both a> and r have Laurent series expansions about any point с of the complex plane, for ease of notation we take c = 0. Say (The dots represent terms involving x raised to powers higher than that shown.) Using the Riccatti Equation, we find that +b 2x2" + = axv+ As we need to show that every pole of r either has order 1 or else has even order, we may assume that v<— 3. Since a#0, — 3 > v > min (ц— \,2ц). It follows that ц< — 1 and 2n<fi~\. Since Ь2ф0, 2ц = v, which implies that v is even. For use in section 3.3, we remark that if r has a pole of order 2ц > 4 at c, then со must have a pole of order ц at с Now consider the Laurent series expansions of r and со at oo. a) = bx"+ ¦¦ ¦, /iel, ЬфО r = axv+---, veZ, a Ф 0. (The dots represent terms involving x raised to a power lower than that shown. The order °f r at со is —v.) As we need to show that either the order of r at oo is ^3 or else is even, we may assume that v ^ — 1. Using the Riccatti Equation, we have цЬх" ¦+b2, '+¦¦¦ = oav Just as above, - 1 s? v s? max (ji-1, 2ц), ц > — 1, 2ц > ц-1. Since b2 Ф 0, 2/x = v, so v is even. For use in section 3,3, we remark that if r has a pole of order 2ц > 0 at oo, then w has a pole of order fi at oo. This verifies the necessary conditions for case 1. Case 2. We analyse this case by considering the differential Galois group that must obtain. By section 1.4 the group must be conjugate to a subgroup G of D\ which is not Wangulisable (otherwise case 1 would hold). Let n, ( be a fundamental system of
ш J. J. Rovacic solutions of the DE relative to the group G. For every <reG(G/C(x)), either ar\ = ai a( = c~lC or ar\ =-c^C, a[, = car}. Evidently ц2B is an invariant of C(G/C(x)) and therefore i;2BeC(x). Moreover, diagonal group, which is case 1. Writing y ц( (/()) d ), for otherwise G would be a subgroup of the we have that at least one exponent e{ is odd. Without loss of generality we may assume that and that e is odd. Let 9 = 0Ю70Ю = i(*2C WC2) = 1«- '+••¦, where the dots represent terms involving x to non-negative powers. Since r\" = rr\ and Let r — axv + • • • be the Laurent series expansion of r at 0, where a # 0 and v e Z. From the equation above we obtain If v>-2, then 0 = 8e-6e2 + e3 = e(e-2)(e-4). This contradicts the fact that e is odd. Therefore v ^ -2. If v < -2, then e + v = 0, so v is odd. This verifies the necessary conditions for case 2. Case 3. In this case the DE has a solution q that is algebraic over C(x). r\ has a Puiseaux series expansion about any point с in the complex plane, for ease of notation we take c=0. Then rj = ax"+ ¦ ¦ ¦, where aeC, аФЪ, цеп. Since reC(x), r = axv + • • •, where a /0 and veZ. The DE implies that It follows that v>-2, i.e. r has no pole of order greater than 2. If v = -2, then A*(a*—l) = a. Because /ie<Q, we must have y/l+4ae6. So far we have shown that the partial fraction expansion of r has the form ft ¦+P, where i>eC[x] and ,/1+4а,е<0 for each i. Next, we consider the series expansions about x, ц = ax"+ • • •, r = yxv + where the dots represent lower powers of x than those shown. From the DE we obtain /«О*- 1)я.т*-2+ • • • HuinugeiiEuus uineiEiiuai ьциании» just as above, we obtain v ^ — 2 and therefore P = 0. But r = ) ——r 'x-d, where у = Z ai + Z A4(- Therefore ^ ft = 0 and /Д/* -1) = У- Since ц e Q, i 1 j This completes the proof of the theorem stated in section 2.1. 3. The Algorithm for Case 1 The first part of this section is devoted to a description of the algorithm. It is somewhat complicated to describe in full generality, yet, as the examples in part 2 show, it is often quite easy to apply. The third part is devoted to a proof that the algorithm is correct. 3.1. DESCRIPTION OF THE ALGORITHM The goal of this algorithm is to find a solution of the DE of the form q = Pe^, where PeC[x] and иеС(х). Since r\ may be written as r\ = e№lp + °>\ this is of the form described in section 1.2. The first step on the algorithm consists of determining "parts" of the partial fraction expansion of w. In the second step we put these "parts" together to form a candidate for со. The maximum number of candidates possible is 2P+1 where p is the number of poles of r. If there are no candidates, then case 1 cannot hold. The third and last step is applied to each candidate for w and consists of searching for a suitable polynomial P. If one is found, then we have the desired solution of the DE. If, for each candidate for со, we fail to find a suitable P, then case 1 cannot hold. We assume that the necessary condition of section 2.1 for case 1 holds, and we denote by Г the set of poles of r. Step 1. For each сеГи {oo} we define a rational function [,/r]с and two complex numbers ac+, ac~ as described below. (ci) If ceF and с is a pole of order 1, then (c2) If с б Г and с is a pole of order 2, then [>/rl = 0. Let b be the coefficient of \/(x-cJ in the partial fraction expansion for r. Then (c3) If с е Г and с is a pole of order 2v > 4 (necessarily even by the conditions of section 2.1), then [v/r]c is the sum of terms involving l/(x-c)' for 2 ^ i ^ v in the Laurent series expansion of */г at с There are two possibilities for [^/г]с, one being the negative of the other, either one may be chosen. Thus d ¦(*-<•)' + ¦ ¦ • + >
u j. j. In practice, one would not form the Laurent series for y/r, but rather would determine [y/r~]c by using undetermined coefficients. Let b be the coefficient o| l/(x-c)v+I in r minus the coefficient of l/(x-c)v+1 in (iyfr]c). Then " 2\^a з t) If the order of r at oo is > 2, then (oo2) If the order of r at oo is 2, then Let b be the coefficient of l/x2 in the Laurent series expansion of r at x. (I r = s/t, where s, t в C[x] are relatively prime, then b is the leading coefficient of > divided by the leading coefficient of t.) Then (oo3) If the order of r at 00 is — 2v ^0 (necessarily even by the conditions of section 2.1), then [y/r~]„ is the sum of terms involving x' for O^i^v in the Laurent series for y/r at 00. (Either one of the two possibilities may be chosen.) Thus , = ax'+--- +d. Let b be the coefficient of xvi in r minus the coefficient of xv~' in Then Step 2. For each family s = (s(c))csru@0), where s(c) is + or -, let If d is a non-negative integer, then со = «*>' is a candidate for a>. If rf is not a non-negative integer, then the family s may be removed from consideration. Step 3. This step should be applied to each of the families retained from Step 2, until success is achieved or the supply of families has been exhausted. In the latter event, case I cannot hold. For each family, search for a monic polynomial P of degree d (as defined in Step 2) that satisfies the differential equation P" + 2wP' + (со' + w2- r)P = 0. This is conveniently done by using undetermined coefficients and is a simple problem in linear algebra, which may or may not have a solution. If such a polynomial exists, then iiuiiiugtntuus uiiititmmi _. pe}01 is a solution of the DE. If no such polynomial is found for any family retained from Step 2, then Case 1 cannot hold. 3.2. examples Example 1. Consider the DE y" = ry where 4x6 - 8x5 +12x4 + 4x3 - 20x + 4 5 Since r has a single pole (at 0) and the order there is 4, the necessary conditions of section 2.1 for case 2 do not hold. Evidently the necessary conditions for case 3 also do not hold. We apply the algorithm for case 1 to this DE. The order of r at the pole 0 is 2v = 4. Therefore [^/^Jo = a/x2, and a2 = 1. We choose a= 1, so isff\0= l/x2. b = -5-0 = -5, and therefore a? = i(±(— 5/1) + 2), which gives ao+ = -3/2 and ao- = 7/2. At 00, v= 1, and [4/r]oc = ax-t-d. Comparing r and [yfr~\2o = a2x2 + 2adx + d2 we see that a1 = 1 and 2ad = -2. We choose a = 1, d = - 1. Thus l^/r]x = x —1. b = 3—1=2, anda+0O=l/2, a^ = -3/2. There are four families to consider. s@) = + s@) = + s@) = - s@) = - s(oo) = + , 5@0) = - s(oo) = + s(oo) = - d=l/2-(-3/2) rf= -3/2 —(-3/2 d= l/l-l/l d = -3/2-7/2 = 2 ) = o = -3 = -5. Only the first two remain for consideration. We shall treat the second family first, since d = 0 in that case. The candidate for w is We now search for a monic polynomial P of degree 0 such that P" + 2coP' + (o)' + CO2- r)P = 0. Since P—l, the existence of P is a question of whether or not co' + co2 — r = 0. But the coefficient of l/x in to'+ w2-r is —6. Thus no such polynomial P can exist. The only remaining family is the first family. The candidate for со is со = We now search for a monic polynomial P of degree 2 that satisfies the linear differential e4Uation given above. Writing P = x2 + ax + b, we easily determine that a — 0, b — — 1 Provides a solution. Therefore a solution of the DE is given by ^ _ pe\m _ (x2_1)ef(l/x2-3/Bx) + *-l)
Example 2. In this example we begin the discussion of Bessel's Equation (An2-\ У = 4x2 -1 U, neC. The necessary conditions of section 2.1 imply that case 3 cannot hold. Here we considei case 1, case 2 is worked out in section 3.2. The only pole of r is at с = 0 and the order there is 2. Thus = 0, 6 = Dn2-l)/4, «? At oo, r has order 0 and [,/r]^ = i. Evidently b = 0 so a* = 0. There are four families to consider. 5@) = +, 5@0) = +, d = -l/2-n s@) = +, s(oo) = -, d = -l/2-n 5@) = -, 5(OO) = +, d = 5@) = -, 5@0) = -, d = A necessary condition that case 1 holds is that -1/2 + n be a non-negative integer, i.e. that n be half an odd integer. We claim that this condition is also sufficient. If n is negative, and half an odd integer, then m = - 1/2-neN. This corresponds to the first family, in which case со = —m/x + i. We need to find a polynomial P of degree m such that 0 = P" + 2coP' + A0' + to2- r)P It is straightforward to verify that Bm-;)! is the desired polynomial. A solution to Bessel's Equation is given by r\ = x~mPeix. If n is positive, then m= —1/2+ n is a non-negative integer. This corresponds to the third family. In this case w= —m/x + i, and we are back to the case considered above. Example 3. In this example we treat the general situation where r is a polynomial of degree 2. We may write r = (ax + dJ + b for some a, b, d e С (a and d are determined by 1 only up to sign, we choose either of the two possibilities). We claim that the DE has a Liouvillian solution if and only if b/a is an odd integer. The necessary condition of section 2.1 implies that only cases 1 and 4 are possible. We consider case 1. Evidently [4/r]c0 = ax + dand a* = i(±(b/a) — 1). There are no poles. Thus dequals a* or a", so one of these two numbers must be a non-negative integer for case 1 to hold. И follows that b/a must be an odd integer, which is the necessity part of our claim. For sufficiency, we may assume that b/a = 2и +1 is positive, since a may be replaced by — a. Case 1 will hold provided that there is a monic polynomial P of degree n such that 0 = P" + 2coP' + (со' + cd2 - r)P = P" + 2(ax + d)F -2naP. Solving Homogeneous Differential Equations !f we write P = t pixi 1=0 and substitute, we obtain a system of linear equations in Po , Pn- j (Pn = 1) that has a solution, namely Bn+l)('+l)D Pt- + l) 0 Pi+ where Pn+i=0 and Р„=1. A special case of this example is Weber's Equation y" = ({x2-\-n)y, neC. Here a = - 1/2, b = -1/2 - n, d = 0. Thus b/a = 2n + 1 is an odd integer if and only if n is an integer. 3.3. proof In case 1, the DE has a solution of the form f/ = e'e, with 0eC(x). Since t\" = щ, we have ff + в2 = г (Riccatti Equation). We shall determine the partial fraction expansion of в using the Laurent series expansion of r and this Riccatti Equation. For ceC, we denote the "component at c" of the partial fraction expansion of 9 by L"JC ' x-c M (x-cI ' x-c' In order to simplify the notation, we assume that c — 0 and drop the subscript ". We shall also need to consider the Laurent series expansion of в about 0 +0 Here 0 = * + *x + • • •, where the * denotes a complex number whose particular value is irrelevant to our discussion. We assume that the necessary conditions for case 1 (see section 2.1) are satisfied, in particular we assume that the poles of r are either of even order or else of order 1. We split our proof into parts, depending on the nature of r at 0. This parallels the division of Step 1 of the algorithm. (c0 Suppose that 0 is a pole of r of order 1, so r = */x+ • • •. The Riccatti equation becomes Since а2 Ф 0, v «: 1 and [0] = 0. Substituting в-а/х + в into the Riccatti Equation, we have a ¦ ¦-2 a2 2a ^ ~ * —¦ + —0 + P =- +
J. J. Rovacic Therefore -a + a2 = 0, so a = 0 or a = 1. Were a = 0, the left-hand side of this equatio would have 0 as an ordinary point; however, the right-hand side has a pole at 0. Wi conclude that a = 1 and the component of the partial fraction expansion of 0 at 0 is (in thi notation of the algorithm) —, where a* = 1. x (c2) Suppose that r has a pole at 0 of order 2, say _ b * X2 X As in (с,), [0] = 0 and - a + a2 = b. Thus the component of the partial fraction expansio of в at 0 is —, where a* = x П+4Ь. (c3) Suppose that r has a pole at 0 of order 2ц ^ 4. In section 2.3, we showed tha v = ju. Recall from section 3.1 that where we have dropped the subscript ". Let r = v/r-[y/r]. Then r = [v/r]2 + 2r[v/r] + r2. From the Riccatti Equation we obtain the following formula (&) An examination of the right-hand side of this equation determines that it is free of terms involving 1/x1 for i = v + 2,.. , 2 (since v > 1). This implies that the left-hand side is 0. Indeed, since at least one of the factors involves l/x\ Were the other factor non-zero, it would involve 1/x' for some i>2. The product would then involve l/xv+' for some i> 2, which is absurd. Hence [0] = ± Г^Д]. The coefficient of l/xv+1 in the right-hand side of (&) is ±va + 2aa + b, where b is the coefficient of l/x'+1 in 2г\_^г} + г2 = г-[ч/7]2. Therefore a±=U±b/a + v). We have shown that if 0 is a pole of r of order 2v > 4, then the component of the partial fraction expansion of в at 0 is ±lyj~r]+—, where a±=-(±-+v). (c4) Finally, we must consider what happens when 0 is an ordinary point of r. As in (Ci), [0]=O and -a + a2 = 0. Contrary to the situation in (cx), however, we cannoi conclude that а Ф 0. Hence the component of the partial fraction expansion of r at 0 is cither 0 or 1/x. Solving Homogeneous Differential Equations 17 We collect together what we have proven so far. Let Г be the set of poles of r. Then where ReC[x], s(c) = + or -, and [_-Jr~]c, a*1 are as in the statement of the algorithm. Next we consider the Laurent series about oo. Suppose that (oo,) If r has order v > 2 at oo, then The Riccatti Equation implies that R = 0 and - am + a2m = 0, so a^ = 0 or (oo2) If r has order 2 at oo, then b * X2 XJ The Riccatti Equation implies that R = 0 and -«„ + oti = b, hence (oc3) In the other cases, the order of r at oo must be even, by the necessary conditions of section 2. Following an argument similar to that used in (c3) we find that and b is the where -2v is the order of r at oo, a is the leading coefficient of coefficient of l/xv~' in г-С/г]2»- We now know that the partial fraction expansion of в has the form Moreover, the coefficient of 1/x in the Laurent series expansion of в at oo is a^00'. Thus Let and -*( Р=П(*-< Then в = a) + P'/P. Again, using the Riccatti Equation, we obtain P" + 2wP' + (со' + w2 - r)P = 0. The converse, namely that if P is a solution of this equation, then в satisfies the Riccatti Equation, is a simple verification. It follows that if P is a solution of this equation, then П = Pef<° is a solution of the DF. y" = ry. This proves that the algorithm lor case 1 is correct.
18 J. J. Kovacic 4. The Algorithm for Case 2 Following the pattern of section 3, we shall describe the algorithm in section 4.1, give examples in section 4.2 and the proof in section 4.3. The algorithm and its proof assume that case 1 is known to fail. 4.1. DESCRIPTION OF THE ALGORITHM Just as for case 1, we first collect data for each pole с of r and also for oo. The form of the data is a set Ec (or ?ш) consisting of from one to three integers. Next we consider families of elements of these sets, perhaps discarding some and retaining others. If no families are retained, case 2 cannot hold. For each family retained we search for a monii; polynomial that satisfies a certain linear differential equation. If no such polynomial exists for any family, then case 2 cannot hold. If such a polynomial does exist, then a solution to the DE has been found. Let Г be the set of poles of r. Step 1. For each сеГ we define Ec as follows. (ct) If с is a pole of r of order 1, then Ec = {4}. (c2) If с is a pole of r of order 2 and if b is the coefficient of l/(x — cJ in the partial fraction expansion of r, then Ec = {2 + feyi +4b\k = 0, ±2} n I. (c3) If с is a pole of r of order v > 2, then Ec = {v}. (оо^ If r has order >2 at oo, then ?„ = {0, 2, 4}. (co2) If r has order 2 at oo and b is the coefficient of x'2 in the Laurent series expansion of r at oc, then (oo3) If the order of г at oo is v < 2, then ?m = {v}. Step 2. We consider all families (ес)ееГи(а>] with eceEc. Those families all of whose coordinates are even may be discarded. Let I сеГ If d is a non-negative integer, the family should be retained, otherwise the family is discarded. If no families remain under consideration, case 2 cannot hold. Step 3. For each family retained from Step 2, we form the rational function Next we search for a monic polynomial P of degree d (as defined in Step 2) such that P'" + ЗвР" + C0* + 30' - 4r)P' + (в" + Звв' + в3- 4гв - 2/)Р = 0. If no such polynomial is found for any family retained from Step 2, then case 2 cannol hold. Suppose that such a polynomial is found. Let ф = в+Р'/Р and let ш be a solution of the equation a>2 2 - r) = 0. Then ц = ?>''" is a solution of the DE y" = ry. tuning HuiiiugEiieuus uiiieiEinmi ицшшш 4.2. examples Example 1. Consider the DE y" = ry where 1 3 The necessary conditions of section 2 show that cases 1 and 3 cannot hold. (The order of r at oo is 1.) The only pole of r is at 0 and the order there is 2. The coefficient of 1/x2 in the partial fraction expansion of r is Ь = —3/16. Since 2,уИ-4Ь = 1 is an integer, ?0 = {1,2,3}. The order of r at oo is 1 and?0O = {l}. We have three families to consider. e0 = 3, Only the third family need remain in consideration. For this family, в = 1/2х and we need to find a monic polynomial P, of degree 0, such that e=l, e=l, e= 1, d = -l/2 rf = -l d = 0. Evidently P must be 1, so the existence of P is a question of whether or not 0" + 38в' + в3-4г6-2г' is zero. That expression does happen to be 0, so P= 1 is the desired polynomial. Next we form The equation for со is The roots are It follows that = в + Р'/Р = 1 Ъс' 1 :47 f? = ' is a solution of the DE. (And xl'*e~2^'x is also a solution.) Example 2. In this example we finish consideration of Bessel's Equation У - l\y, that was started in section 3.2. In that section we observed that case 3 cannot hold and that case 1 holds if and only if n is half an odd integer. Here we treat case 2 and make the assumption that n is not half an odd integer. The only pole of r is at 0 and the order there is 2. Since = 2ч/1+4D«2-1)/4 = 4n, either ?0 = {2} or ?0 = {2, 2 + 4n, 2-4n}, depending on whether 4n is an integer or not. If 4" is not an integer, then there is only one case to consider. = 2, rf = — '
J. J. Rovacic Thus if An is not an integer, case 2 cannot hold. If An is an integer, there are three cases Ц consider. e0 = 2, e0 = 2 +An, e0 = 2-An, ^00 = 0, d = d = d = -1 -\-2n -l+2n. In order that d be a non-negative integer, it is necessary that n be half an integer. Since n is not half an odd integer, n must be half an even integer, that is n is an integer. But, for such n, both e0 and em are even. Hence all families are discarded and case 2 cannot hold. In this example, and in Example 2 of section 3.2, we have shown that Bessel's Equation has a Liouvillian solution if and only if n is half an odd integer. 4.3. proof For the proof of the algorithm for case 2 we shall rely heavily on the differential Galois group of the DE. In case 2, this group is (conjugate to) a subgroup of — c О с -1 0 ceC, Moreover, we may assume that case 1 does not hold, so the differential Galois group is not triangulisable. Let t\, f be a fundamental system of solutions of the DE corresponding to the subgroup of D\ For any differential automorphism a of C(x)(n, (> over C(x), either arj = cn, o[,=c~l[, or щ= — с-1{, <r? = cf/, for some ceC, c^O. Evidently a{n2^2)-n2C,2, therefore f/2?2eC(x). Moreover, ^f^C(x) since case 1 does not hold. We write сеГ (=1 where Г is the set of poles of r and the exponents ec,ft are integers. Our goal is to determine these exponents. Let Because ф - п'(ц + C'/C, it follows that t2) = i I — +i I — T- сеГ X—С /=1 л—и, We first determine ec for с е Г. In order to simplify the notation, we assume that с - 0. (с,) Suppose that 0 is a pole of r of order 1. The Laurent series expansions of r and </' at 0 are of the form -- (eeZ,/eC). Substituting these series into the equation (*) and retaining all those terms that involve x~3 and x'1, we obtain the following. = 2aex~2+ ¦ ¦ ¦ -<xx~2+ SSIVlltg HUtHUgEllEUUS LJIIlfcitntiaj Therefore e-fe2 + ?e3 =0, so e = 0,2,4. Also -fe/+|e2/=2ae-a. Because a^O, еф0,2. Hence, e must be 4. (c2) Suppose that 0 is a pole of r of order 2 and that b is the coefficient of 1/x2 in the Laurent series for r. That is r = bx~2+---, ф = ±ex~1+ ¦ • ¦ . Equating the coefficients of x~3 on the two sides of equation (*), we obtain e-le2+&3 = 2eb-Ab. The roots of this equation are e = 2, e = 2 + 2^1 +Ab. Of course, the latter two roots may be discarded in the case that they are non-integral. (e3) Finally we consider the possibility that 0 is a pole of r of order v > 2. Then /¦ = x~v+ ¦ • • and ф - \$ex~l+ ¦ • ¦. Equating the coefficients of x"v"' in (*) we obtain 0 = 2ae — 2av, hence e = v. In determining the exponents/, we may use the calculation of (cj) above if we replace a by 0 (since dt must be an ordinary point of r). We find that /, = 0, 2, or 4. We cannot exclude the possibility that/ = 2, but we can, of course, exclude the possibility / = 0. We have shown so far that where eceEc (as defined in section 4.1) and PeC|V|. Set 9 = |УЛ so ф^в + F/P. X — С The next step in our proof is to determine the degree d of P, which we do by examining the Laurent series expansion of ф at oo and using equation (*). (об,) Suppose that the order of r at oo is 2. As in (cx) we find that ea = 0, 2 or 4. @02) Suppose that the order of r at oo is 2 and that b is the coefficient of x ~2 in the Laurent series expansion of r at oo. Then, as in (c2), ex- 2, 2±24/T+4b and еш is integral. @03) Suppose that the order of r at oo is v < 2. As in (c3), it follows that eK = v. Note that at least one of the ec (с е Г) is odd, since nt, \$ <C(x). Using equation (*) and the equation ф = в + Р'/Р, we obtain P"> + №" + C02 + 30' - Ar)P' + (в" + 30^ + в3 - Агв - 2r')P = 0. This is a linear homogeneous differential equation for P, so there is a polynomial solution rf and only if there is a monic polynomial which is a solution. Now let со be a solution of the equation (**) со2-ф To complete the proof we need to show that n = ei<a is a solution of the DE y* = ry. From (**) we obtain (by differentiation) Bш-ф)ы' = ф'(о-1ф"-фф' + г'.
The factor Bco — ф) cannot be zero. Indeed, if ф = 2co, then со' + со2 — г = 0 (from (**)) so r\ = e^a is a solution of the DE. But со = \феC(x). This is case 1, which was assumed t0 fail. Using (**) and (*) we have 2Bы - ф){ы' + со2 - г) =-ф"- Ъфф' - ф3 + Агф + 2r' = 0. Thus to'+ to2 = r so t] = es<o is a solution of the DE. This completes the proof that the algorithm for case 2 is correct. 5. The Algorithm for Case 3 Following the pattern established in the previous two sections, we describe the algorithm in section 5.1 and give examples in section 5.2. The proof of the algorithm requires a knowledge of the finite subgroups of SLB) and their invariants, which is provided in section 5.3. The proof of the algorithm is presented in section 5.4. In case 3, the DE has only algebraic solutions and we assume that cases 1 and 2 are known to fail. (It is possible for the DE to have only algebraic solutions and for cases I or 2 to apply. For example, case 1 gives the solution j/ = xI/4 to the DE y" = —C/16л2)к, then reduction of order gives ( = x3!4 as a second solution.) 5.1. DESCRIPTION OF THE ALGORITHM Let ц be a solution of the DE /' = ry and set со = rj'/r]. Then, as we shall show in section 5.4, со is algebraic over C(x) of degree 4, 6 or 12. It is the minimal polynomial for со lhat we shall determine. We are unable to determine the minimal equation for ц (which would be of degree 24, 48 or 120). There are two possible methods for finding the minimal equation for со. We could find a polynomial of degree 12 and then factor it. We shall prove that if to is any solution of the 12th degree polynomial found by our method, then ц = е^т is a solution of the DE, hence any one of the irreducible factors may be used. This is the most direct method; however, the factorisation can be a formidable problem, even with the assistance of a computer. We illustrate this by example, in section 5.2. The alternative is to first attempt to find a 4th degree equation for со, then a 6th degree equation and finally a 12th degree equation. The advantage is that if an equation is found, then it is guaranteed to be irreducible. In our description of the algorithm, we shall combine the various possibilities, denoting by n the degree of the equation being sought. As before, we denote by Г the set of poles of r. Recall that, by the necessary conditions of section 2, r cannot have a pole of order > 2 Step 1. For each сеГи {оо} we define a set Ec of integers as follows. (c:) If с is a pole of r of order 1, then Ec = {12}. (c2) If с is a pole of r of order 2 and if a is the coefficient of l/(x — cJ in the partial fraction expansion of r, then (oo) If the Laurent series for r at oo is r = yx~2+ ¦ ¦ ¦ (уеГ, possibly 0), then Ex = h + ^ У1Т471 к = 0, ± 1, + 2,..., ± 5| n Z. 2. We consider all families (ес)сеГи{«) such that eceEc. For each such family, define Siep if d is a non-negative integer, the family is retained, otherwise the family is discarded. If no families are retained, then ш cannot satisfy a polynomial equation of degree n with coefficients in C(x). Step 3. For each family retained from step 2, form the rational function x-c Also define П (*~ Next search for a monic polynomial PeC[x] of degree d (as defined in step 2) such that when we define polynomials Р„, Pn-U ..., P_t recursively by the formulas below, then p_ l = 0 (identically). P,_i = -SF, + ((n-i)S'-Se)Pt-(n-i)(i+l)S2rPi+1 (i= n, n-1,. ..,0). This may be conveniently done by using undetermined coefficients for P. If no such polynomial P is found for any family retained from step 2, then со cannot satisfy a polynomial equation of degree n with coefficients in C(x). Assume that a family and its associated polynomial P has been found. Let to be a solution of the equation Then r[ = eia is a solution of the DE. 5.2. examples Example 1. Our first example illustrates the alternative technique mentioned at the beginning of the last section, namely to bypass the search for equations of degrees 4 and 6 for cu and proceed directly to the search for an equation of degree 12. We consider the hypergeometric equation y" = ry where r = — 16x2 The necessary conditions of section 2 show that all four cases are possible. APplying the algorithm for case 1, we find that «0+ = 3/4, = 2/3, = 2/3, «o = <*r = a» = 1/4 1/3 1/3, '"id (i = a±_a±_a± can never be a non-negative integer. Case 1 fails.
J. J. Rovacic Applying the algorithm for case 2, we find that йо = {2,3,1} Ei = m Ex = {2}, and d = ex-e0-e1 can never be a non-negative integer. Case 2 fails. We apply the algorithm for case 3, searching for an equation of degree 12 for со, thu n = 12 in the algorithm. At с = 0, a = -3/16 and v/l+4a = 1/2 (or - 1/2). Hence Eo = {3, 4, 5, 6, 7, 8, 9}. Ai с = 1, a = -2/9 and N/l+4a = 1/3. So ?j = {4, 5, 6, 7, 8}. At oo, у = -2/9 and ?m = {4. 5, 6, 7, 8}. Following the instructions of step 2, we now form the expression d — ex — e0 — e1 for every choice of ешб?„, еоеЕо, ejeE,. We discard those families for which d is a negative integer. Only four possibilities remain. «00 = «m = i «o = 3, e, =4, «,=4, «!=5, «,=4, d= 1 We now consider the first possibility, following step 3. We set в = 3/x + 4/(x — I), S = x2 — x, and search for a monic polynomial P of degree 1 that satisfies the conditions given in step 3. Of course, P=l. The computations are far too complicated to be accurately done by hand; however, they are easily programmed into a computer. Since P; is always a polynomial (/' = 12,..., — 1) whose degree is easily predicted (in this example deg Pt = 12 — /) arrays ol coefficients may be manipulated to carry through the computations. In order to avoid roundoff error, we computed 1212~'.Р, using 33 digit integer arithmetic. The results follow. Pn= P10 = P9 = Ps = P3 = 7x-3 (l/12)(-536x2+459x-99) C!/C-122))A8544x3-23799x2 + 10260x-1485) D!/A6-122))(-127488x4 + 217972x3-140879x2 + 40770x- E!/B'123))A74080x5-371748x4 + 320305x3-138975x2 + F!/125)(-8257536x6 + 21145136x5-22757500x4+13168377x3 -4318O83x2 + 76O347x-56133) G!/B-125))G929856x7-23673984xe + 30564708x5-22107287x4 + 9668646x3 - 255528Ox2 + 377622x - 24057) (8!/A6-126)X-26421152x8 + 900984832x7-1356734768x6 + 1177673400x5 - 644082327x* + 227124972x3 - 50398362x2 + 6429780x - 360855) (9!/C128))(l 74483046x9 - 6688997376x8 + 11509039440x7 - 11656902184x" + 7654170465x5-3376695033x4+1000183626x3-191681802x2 + 21552885x-1082565) Solving Homogeneous Differential Equations 25 p = (lO!/B129))(-2281701376xIO + 9713634848x9-18799438080x8 + 21766009616x7 -1668 3774768x6 + 8840413683x5 -3277319535x4 + 838780110x3-141739470x2 + 14270175x -649539) Pi =(ll!/12lo)A342177280xu-6282O18816xlo + 135O7531776x''-17598922384x8 +15426848952x7 - 9546427017x6 + 4252638672x5 —1362816657x4 + 307684656x3 - 46576539x2 + 4251528x - 177147) Po = A2!/1212)(-8589934592x12+43838865408xn-103681720320x10 + 150145637824x9- 148170380976x8 +104901110964x7 - 54596424249x6 - 21032969490x5 - 5948563455x4 + 1203654816x3-165278151x2+13817466x-531441) Therefore r\ = е^ш is a solution of the DE, where со is a solution of the equation 12 (x2 — xVP ,4 A2-0! , , Professors Caviness and Saunders of Rensselaer Polytechnic Institute kindly offered to attempt a factorisation of this polynomial for w. They used the exceedingly powerful system for algebraic manipulation called macsyma at MIT. The program took less than 5 minutes to write but took 3 minutes of CPU time to execute. The result is that the polynomial above is the cube of the following polynomial. (x2-xLco4-(l/3)(x2-xKGx- -A/432Кх2-х)C20х3-409х2 + + A/20736)B048х4-3484х3 Example 2. In this example we consider the DE /' = ry, where 5x + 27 Г~~36(х-1J' The necessary conditions of section 2 show that all four cases are possible. Note that the partial fraction expansion of r has the form 9(x+lJ and the Laurant series for r about oo is 9(x-lJ I , . , r = 36x2 Applying the algorithm for case 1 we find that ^ = 2/3, aZ, = 1/3 a^ = 2/3, a," = 1/3 a+ = 5/6, a' = 1/6. For hold no choice of signs is d = a* — a?! — a,1 a non-negative integer, thus case 1 cannot
26 J. J. Kovacic Applying the algorithm for case 2 we find that E^l = E1=E00 = {2}, and case 2 doi not hold. We now apply the algorithm for case 3, attempting to find an equation of degree 4 ov< C(x) that is satisfied by со. From step 1 we have that ?_1 = {4,4,6,7,8}, Ei = {4, 5, 6, 7, 8} and ?„ = {2,4,6,8,10}. There are four families with the property that d = ^(eco—e-1-e1) is a non-negati integer, namely еш=8, е-!=4, el = 4, d = 0, ee = 10, e_j=4, e,=6, d = 0, em = 10, e.1 = 5, ei=5, d = 0, e^ = 10, e_!=6, e1=4, rf = 0. The first possibility gives 8x Setting S = x2-l, we have S0 = fx, S2r= -^Eх2 + 27). We then have P4 = -l P3 = (8/3)x P2 = -(l/3)A5x2 + l) P1 = (l/9)E0x3 + 14x) P0 = -(l/54)A25x4+134x2-3) P_,-0. Let со be a solution of the equation If we make the substitution 6Sco = z + 4x, the equation simplifies to Z* = 6(x2-l)z2-8x(x2-l)z + 3(x2-lJ. Then r, = el°> = (x2-lI/3exp(J(z/(x2-l))dx) is a solution of the DE. 5.3. FINITE SUBGROUPS OF SLB) In this section we determine the finite subgroups of SLB), up to conjugation, and the' invariants. This work is classical, being found in the work of Klein, Jordan and other* For the sake of completeness we sketch the results here in the form needed in tl» subsequent section. Solving Homogeneous Differential Equations 27 -theorem 1. Let G be a finite subgroup ofSLB). Then either (/) G is conjugate to a subgroup of the group where D is the diagonal group, or (ii) the order of G is 24 (the "tetrahedral" case), or (Hi) the order of G is 48 (the "octahedral" case), or (iv) the order of G is 120 (the "icosahedral" case). In the last three cases G contains the scalar matrix — 1. The geometric names were used by Klein; however, our proof will be entirely algebraic. Proof We assume that G is not conjugate to a subgroup of D\ Let H be the set of scalar matrices in G, thus H = {1} or {1,-1}, so the order of H is 1 or 2. For any xeG-H(i.e. xeG and хфН) we denote by Z(x) the centraliser of x in G and by N(x) the normaliser of Z(x) in G, Let xeG-H. Since x is of finite order, x is diagonalisable. (The Jordan form of a non- diagonalisable matrix in SLB) must be ± I j.) Since the centraliser in SLB) of a diagonal non-scalar matrix is D (by direct computation) Z(x) must be the intersection of G and a conjugate of D. Hence Z(x) = Z(y) if and only if у е Z(x). Using this fact and the fact that Z(gxg~l) = gZ(x)g~x we may conclude that (for arbitrary x, у, д, д'е G) either дг(х)д^пд'г(у)д'-1 = H or gZ(x)g-1 = g'Z(y)q-' and in the latter case yEggZ(x)g'1g'. In addition gZ(x)g~l =g'Z(x)g'~1 if and only if g'~igeN(x). Therefore we may write G as a disjoint union G=UU (disjoint), ^ in G/N(x,), s is some natural number where the inner union is taken over all cosets andxj xseG-H. The group N(x,) is easy to describe since xf is diagonalisable. First note that the only matrices in SLB) that conjugate a diagonal non-scalar matrix into a diagonal matrix are the matrices in Df (by direct computation). It follows that ЛГ(х;) is the intersection of G and a conjugate of D\ in particular the index of Z(x,) in ЛГ(х,), [W(x,): Z(x,)], is either 1 Let M = ord (G/H) and e, = ord (Z(x,)/H). The representation of G as a disjoint union g!ves the following formulas. or or (#) M • ord Я = ? [G. ЛГ(х,)](е, • ord Я-ord H) + ord H, ., ^ м M --1 +1
28 J. J. Kovacic Certainly s # 0 since G ф H. If s = 1, then ):Z{xJ\el) = 1/ord H), so G = This contradicts the fact that G is not conjugate to a subgroup of i^. Since e,- > 2 (i = 1,. . ., s) we have so 1 ¦<2. Because [Ar(x,.):Z(x,)]=l or 2, there are only three solutions of this inequality. с — О TN(v \ ¦ 71 w W — 1 ГЛ/f v "i • У(\- \~\ — ~) .5 = 2, [^(x,): Z(x,)] = [iV(x2): Z(x2)] = 2, s = 3, f_N(Xl): Z{xJ] = [/V(x2): Z(x2)] = [/V(x3): Z(x3)] = 2. For all solutions [ЛГ(х2): Z(x2)] = 2. Thus G contains a conjugate of a matrix i D^-D, i.e. the conjugate of a matrix of the form . The square of such a matri: \-c 0 is — 1. Hence ord Я = 2. The first solution gives 1/M = 1/e, + l/Be2)-l/2, so «j = 3, e2 = 2 and M=12, so ord G = 24. (The point being that M > 2e2, since G is not conjugate to a subgroup of D and therefore g] > 3.) The second solution gives 1/M = l/Bea) + l/Be2), which is impossible since M > 2e2. The third solution gives 2 111, -- = — + — + 1. M et e2 e3 Assuming that ey < e2 ^ e3 we find that e, < 3 so et = 2 and 2 _ 1 1 1 M~e2 еъ 2' Also e2 = 3 since M > 2еъ. The solutions are e,=2, e2 = 3, e3 = 3, M = 12, ordG = 24, e3 = 4, M = 24, ord G = 48, e3 = 5, M = 60, ord G= 120. This proves the theorem. In the following sequence of theorems we shall explicitly determine the three "geometric" groups. To that end we need the following lemma. Lemma. Let G be a finite subgroup o/SLB, C) that is not conjugate to a subgroup ofDf. i-''1 H — {1, — 1}. Then G/H has no normal cyclic subgroup. Proof. If xH is a generator of a normal cyclic subgroup of G/H then the group general^ by x and — x is diagonalisable. Since this group would be normal in G, <7 would b« conjugate to a subcroun of/)*. Solving Homogeneous Differential Equations 29 Theorem 2. Let G be a subgroup ofSLB, C) of order 24 that is not conjugate to a subgroup oflf. Let H = {1, — 1}. Then G/H is isomorphic to A4, the alternating group on 4 letters. Moreover, G is conjugate to the group generated by the matrices 'Z о о г1 Ф 2 -1 where t, is a primitive 6th root of 1 and Ъф = 2?, — Prooh. Since ord G/H is 12, and because of the previous lemma, G/H has 4 Sylow 3-groups, and G/H acts by conjugation on the set of these Sylow 3-groups. This action induces a homomorphism G/H->S4 (the symmetric group on 4 letters). The subgroup of the image consisting of those permutations that leave a particular Sylow 3-group fixed must have index 4 since G/H acts transitively. Therefore the order of the image is divisible by 4. It follows that the order of the kernel is 1, 2 or 3. By the previous lemma, the order of the kernel must be 1, so G/H is isomorphic to a subgroup of S4. Now consider the composite homomorphism G/H->S4-»{1, -1}, with the last arrow being given by G->signum (a). By the previous lemma, G/H cannot have a normal subgroup of order 6 (since a subgroup of order 6 contains a unique subgroup of order 3 which would be normal in G/H). Therefore the composite homomorphism has trivial image and G/H is isomorphic to A4. Let t;G->A4 be a homomorphism with kernel H. Let Лет"'A23). We may conjugate G so that A is a diagonal matrix. Thus A = 'Z о о г1 Since тА3 = A), АъеН. However, тЛ #A) and тА2фA), thus АфН and А2фН. Replacing A by — A, if necessary, we may assume that { is a primitive 6th root of 1. Let ВетA2)C4). Since r(AB) ф х(ВА), В cannot be a diagonal matrix, i.e. not both B12 and B2i are zero. In fact neither is zero because if one were zero and the other non- nonzero, then В would have infinite order. We may conjugate G by I ,1 without affecting A. If we choose с = B21 and d = 2512, \0 d) then В has the form 0 d Ф Ф 2Ф -xj' Now xB1 = A) so B2eH. A direct computation shows that х~Ф- Next we observe that r(BA2) = r(ABJ so BA2 = ±(ABJ. We perform the computation and discover that ф(?2-1) = ±?* (using the fact that ффО). Replacing В by -B, if necessary, we may assume that фЦ? -1) = ?,*, hence 30 = 2?-l (using the relation ?2 = ?-l). Next we use the fact that detB=l to obtain the formula ф2 + 2ф2 = -1, or ¦V=±B?-l). If necessary, we conjugate G by ( _ J so that 3^ = 2^-1 =Ъф. This Proves the theorem. The group of this theorem is called the tetrahedral group.
Theorem 3. Let G be a subgroup ofSLB) of order 48 thai is not conjugate to a subgroup of Dt. Let H = {\, —1}. Then G/H is isomorphic to S4, the symmetric group on 4 letters. Moreover, G is conjugate to the group generated by the matrices 1 1 1 -1 where { is a primitive 8th root of 1 and 2ф = ?(?,2 +1). Proof. Since ord G/H = 24, and because of the previous lemma, G/H has 4 Sylow 3-groups. The action of G/H on the set of Sylow 3-groups (via conjugation) induces a homomorphism G/tf->S4. The image contains a subgroup of index 4, namely the subgroup of permutations leaving a particular Sylow 3-group fixed, since G/H act transitively on the set of Sylow 3-groups. Hence the order of the image is divisible by 4, so the order of the kernel is 1, 2, 3 or 6. Were the order of the kernel 6, then the kernel would contain a unique subgroup of order 3 which would be normal in G. This contradicts the lemma. Indeed, the lemma implies that ord ker= 1, so G/H is isomorphic toS4. Let t:G->S4 be a homomorphism with kernel H and let Лет~1A234). We may conjugate G so that A is a diagonal matrix Since тЛ* = A), ?4= ±1. However, were ?4 = 1, then ?2 = ±1 and A2eH. But хА1Ф(\). Hence { is a primitive 8th root of 1. Let Бет^). Since x(AB) ф x(BA), В cannot be a diagonal matrix, thus not both «,, and В2i are zero. In fact, neither is zero since В has finite order. We may conjugate G, without disturbing A, by (л л ), where с2 = B21 and d2 = Bl2. Then В has the form 0 d В Ф Ф -x Using the fact tB2 = A), i.e. B2eH, we obtain easily that х = ф. Because x(BA3) = x(ABJ, BA3 = ±(ABJ. Making this computation, and using the fact that ффО, we find that ф(?2-1) = ±?, or 2ф = ±Щ2+\). Replacing В by -B, if necessary, we may assume that 2ф = Щ2 +1). Then 2ф2 = - 1. Now we use the fact that 1 = det В = — ф2 — ф2 to conclude that 2ф2 = — 1. Conjugate G, if necessary, by ( so that ф = ф. Because xA, xB generate S4 and the group generated by А, В contains H, we can conclude that А, В generate G. This proves the theorem. The group of this theorem is called the octahedral group. Theorem 4. Let G be a subgroup ofSL{2) of order 120 that is not conjugate to a subgroup ofD\ Let H = {1, — 1}. Then G/H is isomorphic to A,, the alternating group on 5 letters- Moreover, G is conjugate to the group generated by the matrices where 0 {О Г1 ; is a primitive 10th root of 1, 50 ф ф' Ф -Ф, %3 - V + 4?-2, and 5ф = ?3 + 3?2-2f + 1. Proof. The proof that G/H is isomorphic to A5 may be found in Burnside A955, 127, p. 161-2). Let t:G-»A, be a homomorphism with kernel H and let Лет 'A2345). We may conjugate G so that A is a diagonal matrix 4 = 1 _,)• Since xA! = (l), ?5 = +l. Replacing A with —A, if necessary, we may assume that ?5 = -1. Evidently ? is a primitive 10th root of 1. Let 5бтA2)C4). As in the proof of Theorem 3, we may assume that В has the form Ф Ф Ф -Ф. Because x(A*B) = x(BAJ, АлВ=±(ВА)г. Making this computation we find that фA+?3) = ±?*, or 50 = +C?3-?2 + 4?-2). Replacing В by -B, if necessary, we may assume that the plus sign obtains. Now we use the fact that 1 = det В to conclude that 5i> = ±(?3 + 3?2-2?+1). Conjugate G by I I, if necessary, so that the plus sign obtains. Note that xA, xB generate A5. (This group generated by xA and xB contains an element of order 5, an element of order 2 and an element of order 3. Thus the order of this group is divisible by 30. Since A5 is simple, this group must be A5.) Also the group generated by А, В contains H. Therefore А, В generate G. This proves the theorem. The group described in this theorem is called the icosahedral group. For use in the next section, we also need to know the invariants of the three "geometric" groups. Theorem 5. Let G be the Galois group of the DE y" = ry and let n,{, be a fundamental system of solutions relative to the group G. (i) IfG is the tetrahedral group, then (f/4 + 8»/C3KeC(x). (it) IfG is the octahedral group, then fo'(->jt5JeC(jc). (ш) IfG is the icosahedral group, then ^Ч-П^-^еЦх). Pkoof. (i) Consider the tetrahedral group, using the notation of Theorem 2. Recall that CJ = -1, ?2 = ?-l and 3^ = 2^-1. |4 + 8»jC3 is carried into <*4(i/4 + 84C3) by the matrix Г ° Д The matrix ф Г j carries into )-(ч- 3<й-ДО?-1H,-2?О-4 = -3-ф*-BЦ-1У = -3 - (-1/3J ¦ (-3) Thus (,/4 + 8>/CK is an invariant of G and therefore is in C(x). (ii) Consider the octahedral group, using the notation of Theorem 3. Recall that
32 1 t- Rovacic f/5? — flC5 is carried into carries into by the matrix . The matrix ф 1\ 1 -1 Thus (ч5С-'/С5J is an invariant of G and therefore is in C(x). (iii) Consider the icosahedral group and use the notation of Theorem 4. First we collect some easily derivable formulas. The matrix j ) carries into -пС- С2) Thus пи?- l This proves the theorem. - Iln6F-n(n into itself. The matrix 2c2)-^2- ?V+?C2 is an invariant of G and therefore is in C(x). ix ( Ф -ф carries Solving Homogeneous Differential Equations 5.4. PROOF OF THE ALGORITHM We must prove the validity of four separate algorithms. We must show that the algorithms for finding a 4th, 6th and 12th degree equation for ы are correct for the tetrahedral, octahedral and icosahedral groups and that the equation obtained is irreducible, and finally that the algorithm for finding a 12th degree equation is all- inclusive (although the equation obtained need not be irreducible). In so far as is possible we carry out the proofs simultaneously. We begin by showing that the equations obtained for w in the tetrahedral, octahedral and icosahedral cases are minimal. Throughout we assume that the Galois group G of tlie DE y" = ry is the tetrahedral, octahedral or icosahedral group. We also fix a fundament'11 system of solutions n, ? of the DE relative to the group G and set со = rj'/n. Theorem 1. Let tj1 be any solution of the DE and let Wj = ц\1г\у. @ // G is the tetrahedral group, then degiU./rti 2»4 and deg, ыш = 4. octahedral group, then degc(x)w! > 6 and degcw« = 6. icosahedral group, then degcwco, 3s 12 and degcww = 12. /f 0 iii) If G IS (iii) If G is /f 0\ Proof. Since ш is left fixed by the group Gj generated by I x I, where t, is a primitive 6th 8th or 10th root of 1 in the tetrahedral, octahedral or icosahedral cases, the degree of 0 over C(x) is ^[G: GJ = 4, 6 or 12. The reverse inequality is proven more generally, as indicated in the statement of the theorem. Let G, be the subgroup of G that fixes t]x. Complete i/, toa fundamental system of solutions fh.Ci of the DE and conjugate G to XGX'1 so that XGX~l is the Galois group of the DE relative to nu d. Then XGXX'^ consists of matrices of the form c d | Since XG^X'1 is finite, d = 0 and cm=\, where mis the order of Gv Evidently oc'7 is a subgroup of the cyclic group с 0 P с" and therefore is cyclic. Hence GJH (where H is the centre of G) is isomorphic to a cyclic subgroup of A4 in the tetrahedral case, of S4 in the octahedral case, and of A5 in the icosahedral case. So ord GJH ^ 3, 4, 5 or ord Gt «: 6, 8, 10. Thus degcww1 = [G:G1]>4,6, 12. This proves the theorem. Throughout the remainder of this section we shall be considering a certain differential equation written recursively, namely «„ = -1 (#)„ a,.l=-a'i-zai-(n-i)(i+l)ral + 1 (i = n,.. ., 0) By a solution of (#)„ is meant a function z such that when а„,..., a_, are defined as above, then a.i is (identically) zero. Theorem 2. Let z be a solution о/(#)„, and let w be any solution of "~' a Then t] = eIra is a solution of the DE y" - ry. ''roof. Let i=-o(n-i)! where w is an indeterminate. We claim that dk + 1A „, ... . ,5M
For к = 0, we have дА , = naHwn+1 1-1 i/j П- 1 i +,4(n-o! ,4 1-0! UJ±1W' Our claim now follows by induction. To show that n = е'ш is a solution of the DE is equivalent to showing that co' + co2 = We assume that co' + co2 — г ф 0 and force a contradiction. Since Л (со) = 0, we have dA , Therefore Hence dw dA "dx" — (»)(«' + со2 -r) = - — (со) + (net) + z)A(w) + Ц- (to) = 0. 5- M = 0. 5 Assuming that we have Thus d dx (со). т (со) so = 0, Solving Homogeneous Differential Equations 35 The desired contradiction follows from the fact that (f A This proves the theorem. Theorem 3. (i) Suppose that (#L has a solution ze<C(x). Then the polynomial w4-?t/47w'6C(x)[>] is irreducible over C(x). (ii) Suppose that (#N /ws a solution zeC(x). Then the polynomial 5 n. is irreducible over C(x). Hi) Suppose that (#)l2 ^as a solution zeC(x) and that (#L and (#N do not solutions in C(x). Г/ien tfte polynomial is irreducible over C(x). Proof. By Theorems 1 and 2, any root of the polynomial w"- (л,еС(х)) must have degree 4, 6 or 12 over C(x). Statement (i) of the present theorem is clear. Statement (ii) follows from the fact that if a sextic is reducible, then one of the factors has degree ^3. To prove (iii) it suffices to show that if degC(x)co = n, then (#)„ has a solution zeC(x). Let ,4eC(x)[w] be the minimal polynomial for w over C(x). Let deg^A = n and write Consider the polynomial dw where The coefficient of w" + 1 in В is and the coefficient of w" in В is -(n-lK-r - " dx Л-1бС(х). |а„ + па„ = 0, ,_1+гя„ = а„-1-2 = 0,
.!Б J.J.R6vacic since а„ = — 1 and а„_ t = 2. Therefore deg^fl < n. But 5/1 Solving Homogeneous UiHerential Equations .* / This formula is easily checked for m = 1 and, for m > 1, = ^- (A(w)) + (nw + z)A(cd) = 0. Therefore В = 0. The coefficient of w' in В is '(п-1-Ol — + n a. (n+l-j)! (n-i)! («+!-/)! (n-i)! where a_, = 0. These are precisely the equations of (#)„. This proves the theorem. For any function b we denote by ISb = b'/b the "logarithmic derivative" of b. Theorem 4. Let F be any form (homogeneous polynomial) of degree n in solutions of the DK. Then z — ISF is a solution of(#)„. Proof. First we prove that if Fl and F2 are functions such that 15FX and 15F2 are solutions of (#)„, then W(c1F1+c2F2) is a solution of (#)„ for any c,,c2eC. Let a}, a?, a? denote the sequences determined by (#)„ for z = lSFl, ISF2, l6(c1Fl + c1F2) respectively. We claim that This is clear for 1 = n. Also (C1F1 +c2F2)a?^ = (c, f, clF1e,_1+c2F2a?_1 Therefore (c1F1+c2F2)al1 = CiFj which verifies our assertion. To prove the theorem, we may assume that ?.! = 0, i 1 where ^b .. .,г\„ are solutions of the DE. Let o)i = r}'i/r\i and denote by <rmk the fcth symmetric function of a>1,. amk = 0 for к < 0 or к > m, am0 — 1 and ..,a>m for 1 ^ /c < m. First we claim that -к)гатЛ.1 -ат + [(m + 1 - k)ram- lit_ 2 -<rm-,,, on _ x am_ i,t+j+(гга_ 1к which completes the induction. Next we use induction on 1 to prove that Evidently Using (#)„, we have = ? W, = 1=1 ()@<„1 „,( -(П-0(/+1)К-1)"-'(п-1-1)! Hence aM=(-D>+l)!»,,+ i=0. This completes the proof of the theorem. Theorem 5. @ If G is the tetrahedral group, then (ф)л has a solution z = Idu, where м3еС(х). (ii) If G is the octahedral group, then (#N has a solution z = Idu, where u2eC(x). (Hi) If G is either the tetrahedral group, the octahedral group or the icosahedral group, then (#)n has a solution 2 = Idu, where ueC(x). Proof. This theorem is a corollary of Theorem 3 of the present section and Theorem 3 of the previous section. For part (i) we may take u = nA + 9>nC, for part (ii) we may take " = VSC-^C5 and for part (iii) we may take и = (п4 + ЦС3)\ (f/sf-fffsJ or '/"C-iVc6-^11- We shall write *nere n = 4, 6 or 12 and eceZ. Our next step in the proof is to determine the various Possibilities for ec, as stated in step 1 of the algorithm. For ease of notation, we shall assume that с = 0. To this end we shall use the Laurent series for z = Ibu = ^ 15(u12!"),
38 J. J. Kovacic 2 = -r ex 1 + ¦ ¦ ¦ (e = e0 e 1, possibly 0) namely and for r, namely r = ax~2 + fix'1+ ••• (a, /JeC, possibly 0). (Note that, by the necessary conditions of section 2, r can have no pole of orde, exceeding 2.) First we consider the possibility that a = 0 and p ф 0, corresponding to (r t) of Step I 0| the algorithm. Theorem 6. If a = 0 and РФ0, then e = 12. Proof. We write 2 =y^e*~ and treat e and/as indeterminates. Then where At, B,, Ci are polynomials in e with coefficients in C. Using (#)„ we find that Solving Homogeneous Differential EqU3H8Bg The first equation implies that 12 / n for some / = 0 n. Suppose that / ф п. Then the second equation gives «-o "п сл—o. for i = n,. .., 0. We leave to the reader the verification that the solution to these equations is given by n-i-l / „ 4— Л [y-A- t=o \ 12 c'= <"-'¦> because and which implies that /} = 0. This contradiction shows that l = n and therefore e= 12. This proves the theorem. Next we consider the possibility that а ф 0. This corresponds to (c2) of Step 1 of the algorithm. As above we write at = /4,x''" + Lemma. /1,15 0 polynomial in e with coefficients in Q[a] vv/iose degree is n — i and whose leading coefficient is -(-(n/12))"'. Proof. Using (#)„ we have (n n-i-— The lemma is immediate from these formulas. The author did not succeed in finding a closed-form solution of these equations, thus we shall use an indirect argument. Assume that а Ф —1/4. Then the DE /' = ry has Puisseaux series solutions of the form By Theorem 4, /<5(^\»?5~0 is a solution of (#)„ for every i = 0,..., n. Since n (n the polynomial A _, must vanish for 12 n (n Theorem 7. /l+4a (i = 0,., @ /Issume tliat G is (йе tetrahedral group. Then e is an integer chosen from among 6 + к^/Г+4а,к = 0,±3,±6. ('') Assume that С is the octahedral group. Then e is an integer chosen from among l+4a fc = 0 ±2 ±4 ±6. . ...v. . ._,.. =0,±2, ±4, ±6. (iii) Assume that G is either the tetrahedral group, the octahedral group or the icosahedral group. Then e is an integer chosen from among 6 + ky/l+4a, /i = 0, +1 ±6.
40 J. J. Kovacic Proof, (i) In this case n = 4. If а Ф 1/4, then we may use the Lemma and the remark, following it to obtain following it to obtain Solving Homogeneous Differential Equations 41 a = —1/4, we compute directly. Using a programmable calculator we obtained the following. ¦ft i = 0 Thus e = 6 + ky/l+4<x, fe = O, ±3, ±6. If « = - 1/4, we compute directly, using the recurrence relations given above. A12=-\ An =e 148 5<>2 +2241^-1485 = - ^(e* -1 Se3 +135e2 - 459e + (ii) In this case n = 6. If а.ф —1/4, then we may use the Lemma and the remarks following it to obtain 0 = /t_, = Ц U -3 + C-«)v/H:4a). Thus е = 6 + Ц/Т+4а, fc = 0, ±2, ±4, ±6. If a = -1/4, we compute directly. 162378e-lu7fu A3 = e9- A2 = - e'° + 45e9 - 945e8 +12060e7 -103005e6 + 612927e5 - 2566620e4 — 7453620e3 — 2 A, =fn- 2 e ie + -74^r e- 92538045 -28176687e6+137179251e5-Mfi2§ilsie4+1240169535e3 4261026627 ,,2 .4446102717,. 4261026627 ~~ 2 e + 2 e~ 4 1_! =e13-78e12+2808en-61776e10 + 926640e9-10007712c8 + 80061696e7-480370176e6+2161665792e5-7205552640e4 + 17293326336e3-28298170368e2 + 28298170368e- 13060694016 Ш60е2-24336е + 22320) 7 - 42е6 + 756е5 - 756Ое* + 45360е3 - 163296е2 + 326592^ - 279936) (iii) In this case n= 12. If a^ — 1/4, then we may use the Lemma and the remarks following it to obtain Thus , /c = 0, ±1,.. ., ±6. This proves the theorem. Finally we consider what happens if a = ft = 0, i.e. at an ordinary point of r. Using the Previous theorem, we have that (и/12)е is an integer. Let Г denote the set of poles of r. We have proven the following. @ In the tetrahedral case, (#L has a solution z = ISu, where and ee is an integer chosen from among 6 + /с^/1 +4а, к = 0, ±3, +6. (>i) In the octahedral case, (#N has a solution 7 = I6u, where
42 J. J. Kovacic PeC[x] and ec in an integer chosen from among 6 + к^/\ + 4a, fc = 0, +2, +4, +6. (iii) In either the tetrahedral case, the octahedral case or the icosahedral case, (#), has a solution z = ISu, where Solving Homogeneous Uifferential Equations PeC[x] and ec is an integer chosen from among 6 + fc^/l +4a, fc = 0, ±1, Let d = dep P. Then the Laurent series for z at oo has the form .., +6. and the Laurent series for r at oo has the form r — yx+ ¦ • ¦. (By the necessar/'conditions of section 2, the order of r at oo is at least 2.) If we let 12 then, by a theorem analogous to Theorem 7, ет satisfies the same conditions as does each er. Also must be a non-negative integer. This is a restatement of step 2 of the algorithm. We shall complete the proof of the algorithm by showing that the recursive relations of step 3 are identical with (#)„. Let x-c Then z = ISu = Р'/Р + в. Also set Pt = S^'Pa^ Using (#)„, we have References assarri, F. A980). On second order linear differential equations with algebraic solutions on algebraic curves. Am. J. Math. 102, 517-535. oa|dassarri, F., Dwork, B. A979). On second order linear differential equations with algebraic solutions. Am. J. Math. 101,42-76. gorel. A. A956). Groupes lineaires algebriques. Ann. Math. 64, 20-82. gurnside, W. A955). Theory of Groups of Finite Order. 2nd edn. New York: Dover. Char. B. W., Fee, G. J., Geddes, K. O., Gonnet, G. H., Monagan, M. B. A986). A tutorial introduction to MAPLE. J. Symbolic Computation 2, (in press). Kapiansky, I. A957). An Introduction to Differential Algebra. Paris: Hermann. Kolchin, E. R. A973). Differential Algebra and Algebraic Groups. New York: Academic Press. Pavclle, R., Wang, P. S. A985). MACSYMA from F to G. /. Symbolic Compulation 1, 69-100. Saundcrs, B. D. A981). An implementation of Kovacic's algorithm for solving second order linear homogeneous differential equations. Proc. SYMSAC '81 (P. Wang, ed.), pp. 105-108. New York: ACM. Singer, M. F. A979). Algebraic solutions of nth order linear differential equations. Proceedings of the 1979 Queens University Conference on Number Theory. Queens Papers in Pure and Applied Mathematics. No. 54. Singer, M. F. A981). Liouvillian solutions of nth order homogeneous linear differential equations. Am. J. Math. 103, 661-682. Singer, M. F. A985). Solving homogeneous linear differential equations in terms of second order linear differential equations. Amer. J. Math, (in press). Smith, C. A984). A discussion and implementation of Kovacic's algorithm for ordinary differential equations. University of Waterloo Computer Science Department Research Report CS-84-35. Та,)' + (л - i)S"' 'S'Pa,- + S" This is precisely the equation of step 2 of the algorithm. Finally, the equation may be rewritten as This completes the proof of the algorithm.
``` | 23,237 | 73,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-43 | latest | en | 0.895186 |
https://www.physicsforums.com/threads/circular-motion.295023/ | 1,519,603,540,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817523.0/warc/CC-MAIN-20180225225657-20180226005657-00722.warc.gz | 910,800,304 | 14,624 | Circular motion
1. Feb 24, 2009
electricheart
1. The problem statement, all variables and given/known data
A 1.11 kg disc is whirled in a vertical circle with radius 0.680m about a fixed point. Find the tension at the top if the speed at the top is 8.76 m/s.
2. Relevant equations
F= ma
F= m* Acp
Acp= m(w^2*r)
3. The attempt at a solution
I figured centripetal force was acting on the mass so I used the equation w=v/r to find omega(w), then I found the Acp. But that answer is incorrect. If at the top of a circle will the tension be the same Acp.
2. Feb 24, 2009
rock.freak667
At the top of the circle, the forces acting are the normal reaction of the 1.11kg mass and the tension. Both acting downwards. The resultant of those two is the centripetal force mv2/r
3. Feb 24, 2009
electricheart
So F + M g = Acp.
Alright, thanks.
4. Feb 24, 2009
Dr.D
Centripetal force is not really a force at all, it is a mass*acceleration term.
You will get a much better understanding of this problem (as opposed to simply getting an answer quickly) if you draw a free body diagram and write Sum F = m*a where you recognize that for a body in uniform circular motion, a = - r * omega^2 * er, a vector pointing towards the center of the circle. | 351 | 1,247 | {"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} | 3.59375 | 4 | CC-MAIN-2018-09 | longest | en | 0.900327 |
https://paleozonerecipes.com/chickens/is-chicken-lighter-when-cooked.html | 1,638,502,175,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362589.37/warc/CC-MAIN-20211203030522-20211203060522-00523.warc.gz | 520,140,739 | 19,017 | # Is chicken lighter when cooked?
Contents
## Does chicken weigh less after cooking?
Cooked to raw: Cooked weight / 0.75 (example: 8oz cooked / 0.75 = 8oz raw) While it will vary slightly depending on the type of meat you use, you can safely assume that cooked meat will lose about 25% of its weight once it’s cooked. … A standard 8oz chicken breast will almost always cook down to 6oz.
## Is chicken lighter or heavier when cooked?
As a general rule of thumb, on average meat will lose about 25% of its weight when cooked. You still have to weigh out your meat in bulk when its raw, but you don’t need to re-weigh it cooked and figure out the math, just multiple the total raw weight by . 75 and that’s what your 1 oz logged will actually weigh.
## Should you weigh your food before or after cooking?
Enter in a recipe just like the chicken. If you don’t want to enter in a recipe, because it’s a pain, you can always weigh the whole pan of raw veggies, then weigh after it’s cooked and figure what it shrunk by. FOR EXAMPLE: I cooked 477 grams of this Delicata Squash, and when it came out it only weighed 270 grams.
## Are calories for meat cooked or raw?
Cooked items are often listed as having fewer calories than raw items, yet the process of cooking meat gelatinizes the collagen protein in meat, making it easier to chew and digest—so cooked meat has more calories than raw.
## Does chicken lose calories when cooked?
In general, meat, poultry and fish will shrink about 25 percent when cooked. … To help understand the 25 percent shrinkage rate, compare the calories of 4 ounces of raw chicken breast (134 calories) to 3 ounces of cooked chicken breast (139 calories).
## How do I know when my chicken is done?
For properly cooked chicken, if you cut into it and the juices run clear, then the chicken is fully cooked. If the juices are red or have a pinkish color, your chicken may need to be cooked a bit longer.
## What is the weight difference between raw and cooked meat?
Here’s a rule of thumb to translate from raw to cooked portions of meats and poultry. Dubost suggests that for meats, it’s reasonable to estimate you’ll lose about a quarter of the weight in cooking. So four ounces of raw meat with no bones will serve up roughly three ounces cooked.
## How much does 6 ounces of raw chicken weigh cooked?
A raw chicken breast will lose roughly 25 percent of its weight and size when it’s cooked. So, it might be worth accounting for that in your cooking calculations. For example, one 6-oz (174 g) raw chicken breast will weigh around 4.5 oz (130.5 g) once cooked.
IT IS IMPORTANT: Best answer: Can you overcook pulled pork?
## Should you weigh potatoes before or after cooking?
If you start with a raw potato weighing 200g, the actual potato will still contain the same amount of calories after baking, but will weigh less. So if using the raw figure in your diary you would enter 200g = 150kcals. But because baking has caused water loss, the weight of the potato is now only 110g.
## Do foods lose calories when cooked?
When the food is cooked in water, the calorie count does not change, but it has a significant impact on the nutrient content. The nutrient content of the food changes depending on how fresh the food is and how long it is cooked. If the food is overcooked, then it will definitely have fewer nutrients.
## Does the weight of food matter?
When it comes to food, calories are king. As it turns out, most people eat about the same weight of food every day, and overweight or unhealthy people aren’t necessarily eating “more food” than healthier people. … | 823 | 3,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.933325 |
https://socratic.org/questions/how-do-you-find-the-equation-of-the-circle-centered-at-h-k-4-2-and-passing-throu | 1,726,290,325,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00542.warc.gz | 481,869,344 | 6,044 | # How do you find the equation of the circle centered at (h,k) = (-4,-2) and passing through the point (-1,2)?
Sep 15, 2016
${x}^{2} + {y}^{2} + 8 x + 4 x - 5 = 0$
#### Explanation:
The reqd. circle passes thro. the pt. $P \left(- 1 , 2\right)$ & has Centre
$C \left(- 4 , - 2\right)$.
Hence, its radius $r$ is given by the dist. $C P$, where,
${r}^{2} = {\left(- 1 + 4\right)}^{2} + {\left(2 + 2\right)}^{2} = 9 + 16 = 25$.
Therefore, the eqn. of circle is,
${\left(x + 4\right)}^{2} + {\left(y + 2\right)}^{2} = 25$
$\therefore {x}^{2} + {y}^{2} + 8 x + 4 x + 16 + 4 - 25 = 0 ,$" i.e.,
${x}^{2} + {y}^{2} + 8 x + 4 x - 5 = 0$ | 291 | 638 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-38 | latest | en | 0.689738 |
https://math.stackexchange.com/questions/3802225/given-x-1-x-2-independent-real-r-v-say-if-x-1-x-2-x-1-x-3-x-1-x-4 | 1,723,412,658,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00758.warc.gz | 289,796,864 | 40,073 | # Given $X_1, X_2..$ independent real r.v., say if $\{X_1 X_2, X_1 X_3, X_1 X_4, \ldots \}$ are always independent
I have a set of problems about independence of RV which I'm having quite a hard time solving. I think I know the theory, but when it's time to get the hands dirty, I do not know where to begin.
Here is one of them
Given $$X_1, X_2,\ldots$$ independent real r.v., say if the following proposition is always true or not $$\{X_1 X_2, X_1 X_3, X_1 X_4, \ldots\}$$ are independent.
I know that my objective is to prove that
$$P(\{X_1 X_2, X_1 X_3, X_1 X_4, \ldots\}) = P(X_1 X_2) \cdot P(X_1 X_2) \cdot P(X_1 X_3)\cdots$$ As the definition of independence. But I do not find the way.
What I tried so far:\
1. Conditioning on $$X_1$$ like $$P(a X_2\mid X_1=a, aX_3\mid X_1=a,\ldots) P(X_1=a)$$, given the $$X_is$$ are independent that factorizes, but i cannot put the "times $$X_1$$" back into each probability.\
2. Thinking as a cartesian product $$P(\{X_1\} \times \{X_2,X_3 \dots \})$$, and I couldn't actually elaborate much here.\
I am looking to both: A hint to advance with this one, and maybe some insight on how to address these kind of problems, since I have a few left.
• I assume the $X_i$ are independent to begin with, no? Commented Aug 24, 2020 at 23:30
• Also, by the way you are conditioning the $X_i$ it seems that these are discrete. Is that the case? Otherwise $P(X_i = a)$ will always be zero. (to clarify: I didn't downvote you) Commented Aug 24, 2020 at 23:31
• $P(X_1X_2)$ has no meaning. Why do you assume that all random variables are discrete? Sorry to say that there is too much of nonsense in what you have written. The question itself is absurd since there is no assumption of $X_i$'s. Commented Aug 24, 2020 at 23:33
• Let $X$ be a random variable with, say $N(0,1)$ distribution, and $X_i=X$ for all $i$. Do you see that this is a counter-example? Commented Aug 25, 2020 at 0:09
• omg, yes sorry, they are all independent. Commented Aug 25, 2020 at 0:49
As a hint, here is a particular random variable to think about: Let $$X_1$$ be Bernoulli(1/2), that is, it is $$0$$ with probability 1/2, and $$1$$ with probability 1/2. Now, construct some $$X_i$$s for yourself, and then ask questions like "What is the probability that $$X_1 X_i = 0$$ given $$X_1 X_j = 0?$$"
• Thanks a lot for your comment, really clear, and I take your recommendations and advice. I still have one final doubt:\ Sometimes I think I'm overthinking the problems. When you say $P(X_1 X_i =0 | X_1 X_j = 0)$ you mean think the $\frac{P(X_1 X_i =0 | X_1 X_j = 0)}{P(X_1 X_j = 0)}$ in some easy cases to have a counter-example, or is there a more direct way of thinking it? Commented Aug 25, 2020 at 13:04
This is obviously untrue. Consider the case of $$X_2 = X_3 = X_4 = ... = 1$$ and $$X_1 = N(0,1)$$. Then $$X_1, X_2, ...$$ are jointly independent, but the products are not. | 958 | 2,894 | {"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": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-33 | longest | en | 0.932305 |
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# KNOTS AND BRAIDS
REBECCA HOBERG
Abstract. In this paper we examine two physically-inspired objects, knots
and braids. The two are intimately related because when we connect the ends
of a braid, we end up with a knot or link. We show that braids can be defined
algebraically, geometrically, and topologically, and we determine when two
braids will yield the same knot. Finally, we prove that every knot or link is
the closure of some braid.
Contents
1. Introduction
2. Preliminary Definitions
3. Knots
4. The Braid Group
5. Knots to Braids
Acknowledgments
References
1
1
2
5
7
9
9
1. Introduction
A knot is a circle embedded in R3 . In the late 1800s Lord Kelvin suggested that
atoms might represent knots in the ether, with different elements corresponding to
different types of knots. Once this idea was shown to be false, knot theory remained
as a beautiful mathematical theory in its own right. Since then, several practical
applications of knot theory have come to light, including DNA knotting and other
topics in biology, chemistry and physics. For a further reference on some history,
see chapter one of Adams [1].
In this paper we focus on the connections between knot and braid theory. We
begin by defining knots and knot projections, and we give explicit examples of
knot invariants. We then define the braid group in three different ways and prove
the equivalence of these definitions. In particular, we show that a braid can be
defined in a purely algebraic manner, which makes it more convenient to write
out and manipulate than a knot. Finally, by introducing theorems of Markov and
Alexander, we show how related knots and braids really are.
2. Preliminary Definitions
Let X be a topological space.
Definition 2.1. A path in X is a continuous function f : [0, 1] → X.
Date: AUGUST 26, 2011.
1
In other words. such that: (1)The preimage of each point in P (K) contains at most two points. A knot is a smooth embedding of S 1 into R3 .2.2. (2) There are only finitely many points in P (K) with two preimages. 3. x0 ) or simply π1 (X) when independence from basepoint is clear.3.5. possibly linked to one another. We denote the fundamental group of X with basepoint x0 by π1 (X. See Figure 1 for an example of a projection. the fundamental group is independent of base point. If ft (1) = x0 for some x0 ∈ X and for all t ∈ [0. counting the number of times it winds around gives the integer multiple of the generator that represents this loop. To give an example. We call such points “crossings” and keep track of “over” and “under” strands. Two knots K1 and K2 are considered the same if there is an isotopy of R3 taking K1 to K2 . (Equivalently. Definition 3.3. Two paths f0 and f1 in X are homotopic if there exists a continuous family of paths ft (x) in X for t ∈ [0. The fundamental group of X at a point x0 ∈ X is the set of based homotopy classes of loops in X with base point x0 . given a path f in X and an isotopy φt of X.2 REBECCA HOBERG Definition 2. t < 1/2 . if f and g are loops. g(2t − 1). Intuitively. Definition 2. the fundamental group of the circle. A loop f in X with base point x0 is a path in X with f (0) = f (1) = x0 . . we have (f ◦ g)(t) = f (2t). Definition 2. 1]. we can think of a knot as a piece of infinitesimally thin string that is knotted and then has the ends glued together. with group operation defined by concatenation. a loop is a map from S 1 into X). 1]. If X is path connected. A projection of a knot K is the image of K under a projection P : R3 → R2 onto some affine plane. a link is a collection of disjoint knots. An isotopy of X is a continuous family of diffeomorphisms φt of X for t ∈ [0. A link is a smooth embedding of a disjoint union of circles into R3 . Definition 3.4. In particular. Knots Definition 3. 1] we say there is a based homotopy taking one path to the other. and the inverse of a loop is then simply the same loop in the reverse direction: f −1 (t) = f (1−t). In other words. is isomorphic to Z and is generated by one complete loop around the circle. we get a family of homotopic paths ft = φt ◦ f . t ≥ 1/2 The trivial loop is the constant path f (t) ≡ x0 . Definition 2. Given any loop. π1 (S 1 ). Two knots are the same if one can be deformed into the other without ungluing the ends. It is straightforward to check that these operations are well-defined on equivalence classes.1.
The knot which is represented by a projection with no crossings is the unknot. Type 1 Reidemeister move. However. One approach to this problem is to find knot invariants. A Reidemeister move is a local modification of a knot projection that changes one or two crossings. Definition 3. the result is a projection of an equivalent knot. determining when two knots are equivalent. One projection is shown in Figure 2. Figure 3. The Reidemeister moves are illustrated in figures 3. depicted below. A projection of the unknot. . A twist or untwist. One can show that after applying a Reidemeister move to a projection. Definition 3. which differentiate between non-equivalent knots. A fundamental problem in knot theory is classification.KNOTS AND BRAIDS 3 Figure 1. We’ll start our study of invariants with Reidemeister moves. there are also projections of the unknot with crossings.5. Figure 2. There are three possible moves. 4 and 5. that is.4. A projection of the trefoil knot.
The following theorem shows why Reidemeister moves are so important. Theorem 3. Definition 3. One strand passing over or under another strand. (Figure 7). A proof of this can be found in chapter one of Lickorish [5]. In particular. So in order to distinguish more knots.4 REBECCA HOBERG Figure 4. so must be nontrivial. so any knot which is tricolorable cannot be trivial. . The trefoil knot is tricolorable. A strand passing over or under a crossing. so it is a knot invariant. if any projection of a knot is tricolorable. The unknot is not tricolorable. One such is the knot group. A projection is tricolorable if it can be colored in such a way that at every crossing either three colors or one color meet at the crossing. Two knots are equivalent if and only if there is a series of Reidemeister moves taking one to another. For example. It turns out that the tricolorability of a knot is preserved under all three Reidemeister moves. Type 3 Reidemeister move. Color can only change at an undercrossing.6. Figure 5. the figure eight knot is not tricolorable. there are far more than two equivalence classes of knots. any property of a knot which is preserved by Reidemeister moves will be a knot invariant. The trefoil knot is tricolorable. However. This also lets us see that there exist nontrivial knots. Figure 6. then all projections must be. A knot is tricolorable if there exists a tricolorable projection of that knot. One such property is tricolorability. so there are many distinct knots with the same tricolorability status.7. Thus. we need stronger invariants. but also not trivial. Type 2 Reidemeister move. and three colors must be used.
2. (geometric) The n-strand braid group consists of equivalence classes of braids as defined above with the operation that connects the bottom of the first braid to the top of the second. This gives a shorthand way to express braids. The trefoil knot is not equivalent to its mirror image. For example. and makes the relations defining equivalent braids very clear. for example.8. . n. In other words.) The braid group also has a strictly algebraic definition. σn−1 } with the relations σi σj = σj σi . This is called a closed braid. two distinct knots can have isomorphic fundamental groups. the mirror image of the trefoil knot. Definition 4.. Definition 4. Consider. and rescales so that the new braid is still unit length. The knot group of a knot K is the fundamental group of the knot complement. n. 2.. This lets us define the braid group. there is no way to get it to satisfy tricolorability requirements. In R3 . An n-strand braid is a set of n non-intersecting smooth paths connecting the n points in A to the n points in B.. . for |i − j| ≥ 2 (∗) and σi σi+1 σi = σi+1 σi σi+1 .. However..KNOTS AND BRAIDS 5 Figure 7.3. Definition 4. The fundamental groups of two isotopic spaces are isomorphic. (algebraic) The n-strand braid group is the group given by the generators {σ1 . and B the set of points with y = 0. Definition 3. (Figure 8. Thus the knot group is a knot invariant.. z = 0. it is the group π1 (R3 \ K). . we will discuss in more detail how braids and knots are related in section 5. the knot group of the trivial knot is isomorphic to the integers. We say that two braids are equivalent if there is an isotopy of R3 taking one to the other. but their fundamental groups are isomorphic. and the inverse of a braid is its mirror image.1.. for 1 ≤ i ≤ n − 2 (∗∗) . z = 1. The Braid Group We will now define a braid and give three equivalent definitions of the braid group. and x = 1. Note that we can get a knot or link by connecting the corresponding top and bottom strands by paths in R3 . 4. One can think of a braid as a collection of strands in space with fixed endpoints that are braided around each other. let A be the set of points with coordinates y = 0... and x = 1. 3. No matter how one colors this knot. so the fundamental groups of equivalent knots are isomorphic. The trivial braid has n parallel strands. 2. A figure eight knot. While some connections between braids and knots will be immediately clear. 3.
Proposition 4.σn−1 } where σi represents strand i passing over strand i + 1. These two possibilities are illustrated in Figure 10. Without loss of generality. and σi−1 represents strand i passing under strand i + 1.4. For our third definition of the braid group. we have (∗∗). Also. Note that this particular braid yields a link of two components. A projection of a braid. A closed braid. the order doesn’t matter. For example. If the strands are adjacent.6 REBECCA HOBERG Figure 8. Figure 9. If the two crossings had the same z-value. Proof. because if strand j goes over strand j + 1 and then comes back over strand j + 1. One can check that different braid diagrams corresponding to the same braid differ precisely by the relations specified in the algebraic definition. the inverse relations hold. . a small perturbation would make them different heights. as in figure 9. we can arrange it so that only one crossing occurs in our projection for any value of z. The geometric and algebraic definitions of the braid group are equivalent. we will view braids as the fundamental group of a particular space called the n-configuration space . We can then describe the braid from top to bottom as a word in the generators {σ1 . this is isotopic to the trivial braid. In the same manner as a knot projection... if two crossings occur in a row at strands which are not adjacent. This particular projection can be described by σ1 σ2 . as we can shift one up or down along the braid. σ2 . any braid in R3 can be represented by a 2-dimensional projection onto the x − z plane. so σj σj−1 = e. This shows (∗).
In a similar way. There is clearly an isotopy between these two braids that only involves this segment of these three strands. each point will trace a curve in space that starts and ends at an integer in the set. the closure of a braid is always a knot or link. On the left we have σi σi+1 σi and on the right σi+1 σi σi+1 . as defined below. Definition 5. As we vary t.. This example shows that two distinct braids can yield the same knot. Consider the space R3 as C1 × R1 . with the natural topology. Proof... the top and bottom of the braid connect. Thus we have shown the equivalence of the definitions. many braids are not preserved under conjugation. Theorem 5. .5.. There are two types of Markov moves.. Two topological braids are isotopic if one can be continuously deformed into the other. So how do we know in general when two braids correspond to the same knot? Markov provided an answer to this question in the following theorem. 2. Proposition 4. The geometric and topological definitions of the braid group are equivalent. and therefore an element of the fundamental group. varying with the real variable time t. we can assume that the initial point in the topological definition is the set of integers {1. n − 1. (Markov) The closures of two braids B and B 0 represent the knot or link L if and only if one braid can be deformed into the other by a finite number of Markov moves or their inverses. .7. For example..6. 2. Knots to Braids As previously stated. 5. Definition 4. which is true if and only if their corresponding geometric braids are isotopic. this closure is not unique. Definition 4. n}.1. However.2. This gives a geometric realization of the braid. (topological) The n-strand braid group is the fundamental group of the unordered n-configuration space for R2 . every loop in the fundamental group can be thought of as the movement in the complex plane of these n points. When the braid is closed. and thus σi will cancel with σi−1 . Since R2 is path connected. A type one Markov move takes an n-strand braid to another nstrand braid via conjugation by σi for some i ∈ 1. 3. Now. every geometric braid can be considered as a loop in the unordered n-configuration space. The unordered n-configuration space of X is the space of all unordered sets of n distinct points of X. . However. consider conjugation by σi .KNOTS AND BRAIDS 7 Figure 10.
Remarkably. To every subarc oriented in the reverse direction. Figure 11. because a link that is entirely oriented around an axis is braided around that axis. Suppose we have a link L and its projection. The goal will be to manipulate L so that every component is oriented in a particular direction around this axis. In other words. Both of these knots are projections of the unknot. A type two Markov move takes an n-strand braid to an n + 1strand braid by adding σn or σn−1 to the end. choose some subarc S which is oriented opposite our chosen orientation. We then choose a point P on the projection such that P does not intersect the knot. we are doing a type one Reidemeister move. it is helpful to think of P as an axis extending through the projection plane. Though P is a point in the projection. We can continue doing this until we have a link which is all oriented around the axis. (i. Now consider an arbitrary Si . see [4]. But this is exactly what we want. it turns out that they all can. We pull it over all other parts of the knot. at every point where the orientation about P changes. Keeping the endpoints of the subarc fixed.e. Theorem 5. and thus we still get the same knot. We orient the components of L. except possibly the single crossing on Si . (Alexander) Every knot or link is isotopic to a closed braid. If we take a half-plane through the axis going . such that each subarc contains at most one crossing.8 REBECCA HOBERG Definition 5. Otherwise. Fix an orientation about P . (Figure 12). Figure 11 illustrates how adding σn to the end of an n-strand braid gives us the same knot with an extra twist. Proof. we can pull the subarc across our axis P to give it the correct orientation. We divide up S further into a finite number of subarcs {Si }.3. splitting it up into subarcs with at most one crossing and then pulling each piece over the knot and across the axis. which ensures that S is re-oriented in a finite number of steps.) If there are no subarcs oriented in the reverse direction. we put a point to separate the subarcs. As explained above. conjugation by σi is the same as multiplication by σi σi−1 once the braid is closed. The final question we wish to answer is which knots can be written as braids. We consider a finite number of subarcs of L such that each subarc is either oriented in our orientation or in the reverse orientation. In other words.4. we can apply the same procedure. Note that we can avoid adding another crossing to any part of S which is still oriented incorrectly. an n-strand braid B becomes Bσn or Bσn−1 . not both. For a proof that these are the only two moves necessary. then we’re done.
with the given orientation as “down”. If we regard the line L as the plane of the braid closure. (Figure 13). This motion is an isotopy of R3 . L Figure 13. for all their advice and guidance on this project.KNOTS AND BRAIDS 9 out to infinity. While at first S is going the opposite orientation from that of P . this half-plane necessarily passes through every strand. Acknowledgments. This completes the proof. If instead S had an undercrossing. William Lopes and Katie Mann. S would simply go under the other arc and then be pulled across P . It is a pleasure to thank my mentors. . and can be regarded as the plane of the braid closure. P S' Figure 12. on the right we see S 0 with the correct orientation. and they made the whole experience very enjoyable. They answered all the random questions that I had and gave me lots of things to think about in the future. then we see that this picture gives us the 2-strand braid σ1 σ1 . I couldn’t have asked for more helpful and enthusiastic mentors.
[5] W. Graduate Texts in Mathematics. An Introduction to Knot Theory. Vol. Nat. Springer-Verlag. Acad. [4] Kunio Murasugi and Bohdan I. 1997. Raymond Lickorish.10 REBECCA HOBERG References [1] Colin Adams. USA. The Knot Book. 1994. Kluwer Academic Publishers. Knot Theory. 2004.H. Alexander. 1923. . A Study of Braids. A Lemma on Systems of Knotted Curves. Freeman and Company. 1999. [2] Vassily Manturov. Proc. Sci. B. Kurpita. 9. [3] J. W. Chapman and Hall/CRC. W. | 4,431 | 17,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-51 | latest | en | 0.923703 |
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# Calculate Confidence Interval Standard Error Estimate
## Contents
Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. Each of these recent graduates is asked to indicate the amount of credit card debt they had at the time of graduation. For an upcoming national election, 2000 voters are chosen at random and asked if they will vote for candidate A or candidate B. Table 2 shows that the probability is very close to 0.0027. get redirected here
Next, consider all possible samples of 16 runners from the population of 9,732 runners. Some of these are set out in table 2. Retrieved 17 July 2014. Share Tweet Stats Calculator Sample SizeConfidence Interval Calculator forProportionsConfidence Interval Calculator forMeansZ-test for Proportions-IndependentGroupsIndependent T-testBinomial Test (for preferences) Top Newsletter Legal © 2016 McCallum Layton Respondent FAQ [email protected] Tel: +44 http://onlinestatbook.com/2/estimation/mean.html
## Calculate Confidence Interval From Standard Error In R
Here the size of the sample will affect the size of the standard error but the amount of variation is determined by the value of the percentage or proportion in the When the population standard deviation is unknown, like in this example, we can still get a good approximation by plugging in the sample standard deviation (s). Bookmark the permalink. ← Epidemiology - Attributable Risk (including AR% PAR +PAR%) Statistical Methods - Chi-Square and 2×2tables → Leave a Reply Cancel reply Enter your comment here... If you look closely at this formula for a confidence interval, you will notice that you need to know the standard deviation (σ) in order to estimate the mean.
As will be shown, the mean of all possible sample means is equal to the population mean. The standard deviation of the age was 3.56 years. Clearly, if you already knew the population mean, there would be no need for a confidence interval. Calculate Confidence Interval Variance As noted above, if random samples are drawn from a population, their means will vary from one to another.
This would give an empirical normal range . Calculate 95 Confidence Interval From Standard Error One of the children had a urinary lead concentration of just over 4.0 mmol /24h. The true standard error of the mean, using σ = 9.27, is σ x ¯ = σ n = 9.27 16 = 2.32 {\displaystyle \sigma _{\bar {x}}\ ={\frac {\sigma }{\sqrt Assuming a normal distribution, we can state that 95% of the sample mean would lie within 1.96 SEs above or below the population mean, since 1.96 is the 2-sides 5% point
Naming Colored Rectangle Interference Difference 17 38 21 15 58 43 18 35 17 20 39 19 18 33 15 20 32 12 20 45 25 19 52 33 17 31 Calculate Confidence Interval T Test This section considers how precise these estimates may be. Compare the true standard error of the mean to the standard error estimated using this sample. This formula is only approximate, and works best if n is large and p between 0.1 and 0.9.
## Calculate 95 Confidence Interval From Standard Error
A natural way to describe the variation of these sample means around the true population mean is the standard deviation of the distribution of the sample means. https://en.wikipedia.org/wiki/Standard_error These standard errors may be used to study the significance of the difference between the two means. Calculate Confidence Interval From Standard Error In R The ages in one such sample are 23, 27, 28, 29, 31, 31, 32, 33, 34, 38, 40, 40, 48, 53, 54, and 55. Calculate Confidence Interval Standard Deviation For example, the sample mean is the usual estimator of a population mean.
If one survey has a standard error of $10,000 and the other has a standard error of$5,000, then the relative standard errors are 20% and 10% respectively. The variation depends on the variation of the population and the size of the sample. Hutchinson, Essentials of statistical methods in 41 pages ^ Gurland, J; Tripathi RC (1971). "A simple approximation for unbiased estimation of the standard deviation". useful reference This often leads to confusion about their interchangeability. | 975 | 4,238 | {"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} | 3.78125 | 4 | CC-MAIN-2018-39 | latest | en | 0.851602 |
https://www.studyadda.com/ncert-solution/decimal_q33/541/44997 | 1,600,418,418,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187354.1/warc/CC-MAIN-20200918061627-20200918091627-00149.warc.gz | 1,474,641,880 | 18,256 | • # question_answer 33) 1. ind the value of (a) 9.756 ? 6.28 (b) 21.05 ? 15.27 (c) 18.5 ? 6.79 (d) 11.6 ? 9.847 TIPS Firstly, write the decimals in columns with decimal points directly below each other, then subtract same as whole numbers.
(a) We have, 9.756 ? 6.28 Now, $0.25=\frac{25}{100}=\frac{25\div 25}{100\div 25}=\frac{1}{4}$ $\because$ 9.756 ? 6.28 = 3.476 (b) We have, 21.05 ? 15.27 Now, $0.125=\frac{125}{1000}=\frac{125\div 125}{1000\div 125}=\frac{1}{8}$ $\because$ 21.05 ? 15.27 = 5.78 (c) We have, 185 ? 6.79 Now, $0.066=\frac{66}{1000}=\frac{66\div 2}{1000\div 2}=\frac{33}{500}$ $\because$ 18.50 ? 6.79 = 11.71 (d) We have, 11.6-9.847 Now, $\therefore$ $0.3=0+\frac{3}{10}$ 11.6 ? 9.847 = 1.753 | 336 | 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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2020-40 | latest | en | 0.456549 |
https://www.physicsforums.com/threads/possible-mistake-during-differentiation-please-check.552320/ | 1,701,552,872,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100452.79/warc/CC-MAIN-20231202203800-20231202233800-00780.warc.gz | 1,031,651,739 | 15,842 | # Possible mistake during differentiation? Please check
• Dominathan
Let's take a look at what you have so far:You are correct in that the quotient rule requires you to divide by the square of the denominator. However, in this case, you are dividing by 2, which is the square of 1.
#### Dominathan
For a function f(x), I have to determine intervals of increase/decrease, find local max(s)/min(s), and find intervals of concavity. The first thing I'm doing in this is to write out f'(x) and f''(x).
f(x) = $ln(x)/\sqrt{x}$
For f'(x), I used the quotient rule and received f'(x) = (($\frac{1}{x}$$\sqrt{x}$)-($\frac{-\sqrt{x}}{2}$ln(x))) / 2
However, I plugged f(x) into wolfram alpha and it gave me: $\frac{2-ln(x)}{2x^{3/2}}$
I don't understand the difference? I thought I had done this correctly but apparently not? Wolfram alpha used the product rule. Is there some kind of algebraic gymnastics I'm forgetting about? I really want to understand where my error was made, not just which is the correct answer. Thanks!
In using the quotient rule you are not differentiating $\sqrt{x}$ corretly and the quotient rule requires you to divide by the square of the denominator.
Dominathan said:
f(x) = $ln(x)/\sqrt{x}$
For f'(x), I used the quotient rule and received f'(x) = (($\frac{1}{x}$$\sqrt{x}$)-($\frac{-\sqrt{x}}{2}$ln(x))) / 2
I see a couple of issues here:
1) How did you get $\frac{-\sqrt{x}}{2}$ in the numerator? What is the derivative of $\sqrt{x}$ ?
2) How did you get 2 in the denominator? $(\sqrt{x})^{2} =$ ?
Using quotient rule, you should get $\frac{\frac{\sqrt{x}}{x} - \frac{\ln{x}}{2 \sqrt{x}}}{x}$ which simplifies to what you got from WA. It looks like you messed up on the derivative of $\sqrt{x}$ and on the bottom of the quotient rule. http://en.wikipedia.org/wiki/Quotient_rule
gb7nash said:
I see a couple of issues here:
1) How did you get $\frac{-\sqrt{x}}{2}$ in the numerator? What is the derivative of $\sqrt{x}$ ?
2) How did you get 2 in the denominator? $(\sqrt{x})^{2} =$ ?
1.
1. $\sqrt{x}$ = x$^{1/2}$
2. Using the power rule I bring the 1/2 out as a coefficient, and subtract one from the numerator : $\frac{1}{2}$$x^{-1/2}$
3. I simplified to : $\frac{-\sqrt{x}}{2}$ | 654 | 2,220 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-50 | latest | en | 0.916736 |
https://agda.github.io/agda-stdlib/Data.Vec.Base.html | 1,708,934,833,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474653.81/warc/CC-MAIN-20240226062606-20240226092606-00198.warc.gz | 84,171,134 | 19,233 | ```------------------------------------------------------------------------
-- The Agda standard library
--
-- Vectors, basic types and operations
------------------------------------------------------------------------
{-# OPTIONS --cubical-compatible --safe #-}
module Data.Vec.Base where
open import Data.Bool.Base using (Bool; if_then_else_)
open import Data.Nat.Base
open import Data.Fin.Base using (Fin; zero; suc)
open import Data.List.Base as List using (List)
open import Data.Product.Base as Prod using (∃; ∃₂; _×_; _,_; proj₁; proj₂)
open import Data.These.Base as These using (These; this; that; these)
open import Function.Base using (const; _∘′_; id; _∘_)
open import Level using (Level)
open import Relation.Binary.PropositionalEquality.Core using (_≡_; refl; trans; cong)
open import Relation.Nullary.Decidable.Core using (does)
open import Relation.Unary using (Pred; Decidable)
private
variable
a b c p : Level
A : Set a
B : Set b
C : Set c
m n : ℕ
------------------------------------------------------------------------
-- Types
infixr 5 _∷_
data Vec (A : Set a) : ℕ → Set a where
[] : Vec A zero
_∷_ : ∀ (x : A) (xs : Vec A n) → Vec A (suc n)
infix 4 _[_]=_
data _[_]=_ {A : Set a} : Vec A n → Fin n → A → Set a where
here : ∀ {x} {xs : Vec A n} → x ∷ xs [ zero ]= x
there : ∀ {i} {x y} {xs : Vec A n}
(xs[i]=x : xs [ i ]= x) → y ∷ xs [ suc i ]= x
------------------------------------------------------------------------
-- Basic operations
length : Vec A n → ℕ
length {n = n} _ = n
head : Vec A (1 + n) → A
head (x ∷ xs) = x
tail : Vec A (1 + n) → Vec A n
tail (x ∷ xs) = xs
lookup : Vec A n → Fin n → A
lookup (x ∷ xs) zero = x
lookup (x ∷ xs) (suc i) = lookup xs i
iterate : (A → A) → A → ∀ n → Vec A n
iterate s z zero = []
iterate s z (suc n) = z ∷ iterate s (s z) n
insertAt : Vec A n → Fin (suc n) → A → Vec A (suc n)
insertAt xs zero v = v ∷ xs
insertAt (x ∷ xs) (suc i) v = x ∷ insertAt xs i v
removeAt : Vec A (suc n) → Fin (suc n) → Vec A n
removeAt (x ∷ xs) zero = xs
removeAt (x ∷ xs@(_ ∷ _)) (suc i) = x ∷ removeAt xs i
updateAt : Vec A n → Fin n → (A → A) → Vec A n
updateAt (x ∷ xs) zero f = f x ∷ xs
updateAt (x ∷ xs) (suc i) f = x ∷ updateAt xs i f
-- xs [ i ]%= f modifies the i-th element of xs according to f
infixl 6 _[_]%=_ _[_]≔_
_[_]%=_ : Vec A n → Fin n → (A → A) → Vec A n
xs [ i ]%= f = updateAt xs i f
-- xs [ i ]≔ y overwrites the i-th element of xs with y
_[_]≔_ : Vec A n → Fin n → A → Vec A n
xs [ i ]≔ y = xs [ i ]%= const y
------------------------------------------------------------------------
-- Operations for transforming vectors
-- See README.Data.Vec.Relation.Binary.Equality.Cast for the reasoning
-- system of `cast`-ed equality.
cast : .(eq : m ≡ n) → Vec A m → Vec A n
cast {n = zero} eq [] = []
cast {n = suc _} eq (x ∷ xs) = x ∷ cast (cong pred eq) xs
map : (A → B) → Vec A n → Vec B n
map f [] = []
map f (x ∷ xs) = f x ∷ map f xs
-- Concatenation.
infixr 5 _++_
_++_ : Vec A m → Vec A n → Vec A (m + n)
[] ++ ys = ys
(x ∷ xs) ++ ys = x ∷ (xs ++ ys)
concat : Vec (Vec A m) n → Vec A (n * m)
concat [] = []
concat (xs ∷ xss) = xs ++ concat xss
-- Align, Restrict, and Zip.
alignWith : (These A B → C) → Vec A m → Vec B n → Vec C (m ⊔ n)
alignWith f [] bs = map (f ∘′ that) bs
alignWith f as@(_ ∷ _) [] = map (f ∘′ this) as
alignWith f (a ∷ as) (b ∷ bs) = f (these a b) ∷ alignWith f as bs
restrictWith : (A → B → C) → Vec A m → Vec B n → Vec C (m ⊓ n)
restrictWith f [] bs = []
restrictWith f (_ ∷ _) [] = []
restrictWith f (a ∷ as) (b ∷ bs) = f a b ∷ restrictWith f as bs
zipWith : (A → B → C) → Vec A n → Vec B n → Vec C n
zipWith f [] [] = []
zipWith f (x ∷ xs) (y ∷ ys) = f x y ∷ zipWith f xs ys
unzipWith : (A → B × C) → Vec A n → Vec B n × Vec C n
unzipWith f [] = [] , []
unzipWith f (a ∷ as) = Prod.zip _∷_ _∷_ (f a) (unzipWith f as)
align : Vec A m → Vec B n → Vec (These A B) (m ⊔ n)
align = alignWith id
restrict : Vec A m → Vec B n → Vec (A × B) (m ⊓ n)
restrict = restrictWith _,_
zip : Vec A n → Vec B n → Vec (A × B) n
zip = zipWith _,_
unzip : Vec (A × B) n → Vec A n × Vec B n
unzip = unzipWith id
-- Interleaving.
infixr 5 _⋎_
_⋎_ : Vec A m → Vec A n → Vec A (m +⋎ n)
[] ⋎ ys = ys
(x ∷ xs) ⋎ ys = x ∷ (ys ⋎ xs)
-- Pointwise application
infixl 4 _⊛_
_⊛_ : Vec (A → B) n → Vec A n → Vec B n
[] ⊛ [] = []
(f ∷ fs) ⊛ (x ∷ xs) = f x ∷ (fs ⊛ xs)
-- Multiplication
module CartesianBind where
infixl 1 _>>=_
_>>=_ : Vec A m → (A → Vec B n) → Vec B (m * n)
xs >>= f = concat (map f xs)
infixl 4 _⊛*_
_⊛*_ : Vec (A → B) m → Vec A n → Vec B (m * n)
fs ⊛* xs = fs CartesianBind.>>= λ f → map f xs
allPairs : Vec A m → Vec B n → Vec (A × B) (m * n)
allPairs xs ys = map _,_ xs ⊛* ys
-- Diagonal
diagonal : Vec (Vec A n) n → Vec A n
diagonal [] = []
diagonal (xs ∷ xss) = head xs ∷ diagonal (map tail xss)
module DiagonalBind where
infixl 1 _>>=_
_>>=_ : Vec A n → (A → Vec B n) → Vec B n
xs >>= f = diagonal (map f xs)
------------------------------------------------------------------------
-- Operations for reducing vectors
-- Dependent folds
module _ (A : Set a) (B : ℕ → Set b) where
FoldrOp = ∀ {n} → A → B n → B (suc n)
FoldlOp = ∀ {n} → B n → A → B (suc n)
foldr : ∀ (B : ℕ → Set b) → FoldrOp A B → B zero → Vec A n → B n
foldr B _⊕_ e [] = e
foldr B _⊕_ e (x ∷ xs) = x ⊕ foldr B _⊕_ e xs
foldl : ∀ (B : ℕ → Set b) → FoldlOp A B → B zero → Vec A n → B n
foldl B _⊕_ e [] = e
foldl B _⊕_ e (x ∷ xs) = foldl (B ∘ suc) _⊕_ (e ⊕ x) xs
-- Non-dependent folds
foldr′ : (A → B → B) → B → Vec A n → B
foldr′ _⊕_ = foldr _ _⊕_
foldl′ : (B → A → B) → B → Vec A n → B
foldl′ _⊕_ = foldl _ _⊕_
-- Non-empty folds
foldr₁ : (A → A → A) → Vec A (suc n) → A
foldr₁ _⊕_ (x ∷ []) = x
foldr₁ _⊕_ (x ∷ y ∷ ys) = x ⊕ foldr₁ _⊕_ (y ∷ ys)
foldl₁ : (A → A → A) → Vec A (suc n) → A
foldl₁ _⊕_ (x ∷ xs) = foldl _ _⊕_ x xs
-- Special folds
sum : Vec ℕ n → ℕ
sum = foldr _ _+_ 0
countᵇ : (A → Bool) → Vec A n → ℕ
countᵇ p [] = zero
countᵇ p (x ∷ xs) = if p x then suc (countᵇ p xs) else countᵇ p xs
count : ∀ {P : Pred A p} → Decidable P → Vec A n → ℕ
count P? = countᵇ (does ∘ P?)
------------------------------------------------------------------------
-- Operations for building vectors
[_] : A → Vec A 1
[ x ] = x ∷ []
replicate : (n : ℕ) → A → Vec A n
replicate zero x = []
replicate (suc n) x = x ∷ replicate n x
tabulate : (Fin n → A) → Vec A n
tabulate {n = zero} f = []
tabulate {n = suc n} f = f zero ∷ tabulate (f ∘ suc)
allFin : ∀ n → Vec (Fin n) n
allFin _ = tabulate id
------------------------------------------------------------------------
-- Operations for dividing vectors
splitAt : ∀ m {n} (xs : Vec A (m + n)) →
∃₂ λ (ys : Vec A m) (zs : Vec A n) → xs ≡ ys ++ zs
splitAt zero xs = [] , xs , refl
splitAt (suc m) (x ∷ xs) =
let ys , zs , eq = splitAt m xs in x ∷ ys , zs , cong (x ∷_) eq
take : ∀ m {n} → Vec A (m + n) → Vec A m
take m xs = proj₁ (splitAt m xs)
drop : ∀ m {n} → Vec A (m + n) → Vec A n
drop m xs = proj₁ (proj₂ (splitAt m xs))
group : ∀ n k (xs : Vec A (n * k)) →
∃ λ (xss : Vec (Vec A k) n) → xs ≡ concat xss
group zero k [] = ([] , refl)
group (suc n) k xs =
let ys , zs , eq-split = splitAt k xs in
let zss , eq-group = group n k zs in
(ys ∷ zss) , trans eq-split (cong (ys ++_) eq-group)
split : Vec A n → Vec A ⌈ n /2⌉ × Vec A ⌊ n /2⌋
split [] = ([] , [])
split (x ∷ []) = (x ∷ [] , [])
split (x ∷ y ∷ xs) = Prod.map (x ∷_) (y ∷_) (split xs)
uncons : Vec A (suc n) → A × Vec A n
uncons (x ∷ xs) = x , xs
------------------------------------------------------------------------
-- Operations involving ≤
-- Take the first 'm' elements of a vector.
truncate : ∀ {m n} → m ≤ n → Vec A n → Vec A m
truncate {m = zero} _ _ = []
truncate (s≤s le) (x ∷ xs) = x ∷ (truncate le xs)
-- Pad out a vector with extra elements.
padRight : ∀ {m n} → m ≤ n → A → Vec A m → Vec A n
padRight z≤n a xs = replicate _ a
padRight (s≤s le) a (x ∷ xs) = x ∷ padRight le a xs
------------------------------------------------------------------------
-- Operations for converting between lists
toList : Vec A n → List A
toList [] = List.[]
toList (x ∷ xs) = List._∷_ x (toList xs)
fromList : (xs : List A) → Vec A (List.length xs)
fromList List.[] = []
fromList (List._∷_ x xs) = x ∷ fromList xs
------------------------------------------------------------------------
-- Operations for reversing vectors
-- snoc
infixl 5 _∷ʳ_
_∷ʳ_ : Vec A n → A → Vec A (suc n)
[] ∷ʳ y = [ y ]
(x ∷ xs) ∷ʳ y = x ∷ (xs ∷ʳ y)
-- vanilla reverse
reverse : Vec A n → Vec A n
reverse = foldl (Vec _) (λ rev x → x ∷ rev) []
-- reverse-append
infix 5 _ʳ++_
_ʳ++_ : Vec A m → Vec A n → Vec A (m + n)
xs ʳ++ ys = foldl (Vec _ ∘ (_+ _)) (λ rev x → x ∷ rev) ys xs
-- init and last
initLast : ∀ (xs : Vec A (1 + n)) → ∃₂ λ ys y → xs ≡ ys ∷ʳ y
initLast {n = zero} (x ∷ []) = [] , x , refl
initLast {n = suc n} (x ∷ xs) =
let ys , y , eq = initLast xs in
x ∷ ys , y , cong (x ∷_) eq
init : Vec A (1 + n) → Vec A n
init xs = proj₁ (initLast xs)
last : Vec A (1 + n) → A
last xs = proj₁ (proj₂ (initLast xs))
------------------------------------------------------------------------
-- Other operations
transpose : Vec (Vec A n) m → Vec (Vec A m) n
transpose {n = n} [] = replicate n []
transpose {n = n} (as ∷ ass) = ((replicate n _∷_) ⊛ as) ⊛ transpose ass
------------------------------------------------------------------------
-- DEPRECATED
------------------------------------------------------------------------
-- Please use the new names as continuing support for the old names is
-- not guaranteed.
-- Version 2.0
remove = removeAt
{-# WARNING_ON_USAGE remove
"Warning: remove was deprecated in v2.0. | 3,592 | 9,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-10 | longest | en | 0.457956 |
http://www.ntrand.com/ntrandchisq/ | 1,503,229,341,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106465.71/warc/CC-MAIN-20170820112115-20170820132115-00001.warc.gz | 610,017,513 | 6,736 | # NTRANDCHISQ
Returns chi square pseudo random number(s) based on Mersenne Twister Algorithm which has long period (219937-1), high order of equidistribution and 623 dimensions.
## Syntax
NTRANDCHISQ(
• Size,
• N,
• Algorithm,
• Random seed1,
• Random seed2
)
#### Parameters
• Size is # of random numbers (Positive integer).
• N is a parameter of the distribution $N$ (Positive integer).
• Algorithm is a integer value that determines a method to generate uniform random number.
• 0: Mersenne Twister(2002)
• 1: Mersenne Twister(1998)
• 2: Numerical Recipesran2()
• Random seed1 is 1st. random seed.
• Random seed2 is 2nd. random seed.
## Remarks
• The distribution has a semi-infinite support $[0,+\infty)$.
• The distribution is leptokurtic.
## Example
• The example may be easier to understand if you copy it to a blank worksheet
How to copy an example
1. Create a blank workbook or worksheet.
2. Select the example in the Help topic.
Note Do not select the row or column headers.
Selecting an example from Help
3. Press CTRL+C.
4. In the worksheet, select cell A1, and press CTRL+V.
5. To switch between viewing the results and viewing the formulas that return the results, press CTRL+` (grave accent), or on the Tools menu, point to Formula Auditing, and then click Formula Auditing Mode.
1
2
3
4
A B
Data Description
9 Value of parameter N
Formula Description (Result)
=NTRANDCHISQ(100,A2,0) 100 chi square deviates based on Mersenne-Twister algorithm for which the parameters above
Note The formula in the example must be entered as an array formula. After copying the example to a blank worksheet, select the range A4:A103 starting with the formula cell. Press F2, and then press CTRL+SHIFT+ENTER. | 451 | 1,726 | {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.684822 |
https://jonathanfrech.wordpress.com/tag/iterations/ | 1,542,414,911,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743247.22/warc/CC-MAIN-20181116235534-20181117021534-00488.warc.gz | 652,514,195 | 16,257 | ## Bifurcation Diagram
Generating the famous fractal, which can be used to model populations with various cycles, generate pseudo-random numbers and determine one of nature’s fundamental constants, the Feigenbaum constant $\delta$.
The fractal nature comes from iteratively applying a simple function, $f(x) = \lambda \cdot x \cdot (1-x)$ with $0 \leq \lambda \leq 4$, and looking at its poles.
The resulting image looks mundane at first, when looking at $0 \leq \lambda \leq 3$, though the last quarter section is where the interesting things are happening (hence the image below only shows the diagram for $2 \leq \lambda \leq 4$).
From $\lambda = 3$ on, the diagram bifurcates, always doubling its number of poles, until it enters the beautiful realm of chaos and fractals.
For more on bifurcation, fractals and $\delta$, I refer to this Wikipedia entry and WolframMathworld.
# Python 2.7.7 Code
# Jonathan Frech, 24th of March 2017
## Haferman Carpet
The Haferman Carpet is a fractal, which kind of looks like a woven carpet. To generate it, you start with a single black pixel and apply in each cycle a set of rules.
In each generation every pixel in the carpet will be replaced by nine pixels according to the rules. A black pixel is represented by a 0, a white one by a 1.
#### The rules
• $0 \rightarrow \left( \begin{array}{ccc}1&0&1\\0&1&0\\1&0&1\end{array} \right) \text{and } 1 \rightarrow \left( \begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array} \right)$
# Python 2.7.7 Code
# Pygame 1.9.1 (for Python 2.7.7)
# Jonathan Frech 26th of February, 2016 | 455 | 1,568 | {"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": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.815386 |
http://isabelle.in.tum.de/repos/isabelle/file/2c42d0a94f58/src/HOL/Finite_Set.thy | 1,581,902,880,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141460.64/warc/CC-MAIN-20200217000519-20200217030519-00547.warc.gz | 77,086,679 | 50,225 | src/HOL/Finite_Set.thy
author nipkow Tue Oct 04 23:30:46 2005 +0200 (2005-10-04) changeset 17761 2c42d0a94f58 parent 17589 58eeffd73be1 child 17782 b3846df9d643 permissions -rw-r--r--
new lemmas
``` 1 (* Title: HOL/Finite_Set.thy
```
``` 2 ID: \$Id\$
```
``` 3 Author: Tobias Nipkow, Lawrence C Paulson and Markus Wenzel
```
``` 4 with contributions by Jeremy Avigad
```
``` 5 *)
```
``` 6
```
``` 7 header {* Finite sets *}
```
``` 8
```
``` 9 theory Finite_Set
```
``` 10 imports Power Inductive Lattice_Locales
```
``` 11 begin
```
``` 12
```
``` 13 subsection {* Definition and basic properties *}
```
``` 14
```
``` 15 consts Finites :: "'a set set"
```
``` 16 syntax
```
``` 17 finite :: "'a set => bool"
```
``` 18 translations
```
``` 19 "finite A" == "A : Finites"
```
``` 20
```
``` 21 inductive Finites
```
``` 22 intros
```
``` 23 emptyI [simp, intro!]: "{} : Finites"
```
``` 24 insertI [simp, intro!]: "A : Finites ==> insert a A : Finites"
```
``` 25
```
``` 26 axclass finite \<subseteq> type
```
``` 27 finite: "finite UNIV"
```
``` 28
```
``` 29 lemma ex_new_if_finite: -- "does not depend on def of finite at all"
```
``` 30 assumes "\<not> finite (UNIV :: 'a set)" and "finite A"
```
``` 31 shows "\<exists>a::'a. a \<notin> A"
```
``` 32 proof -
```
``` 33 from prems have "A \<noteq> UNIV" by blast
```
``` 34 thus ?thesis by blast
```
``` 35 qed
```
``` 36
```
``` 37 lemma finite_induct [case_names empty insert, induct set: Finites]:
```
``` 38 "finite F ==>
```
``` 39 P {} ==> (!!x F. finite F ==> x \<notin> F ==> P F ==> P (insert x F)) ==> P F"
```
``` 40 -- {* Discharging @{text "x \<notin> F"} entails extra work. *}
```
``` 41 proof -
```
``` 42 assume "P {}" and
```
``` 43 insert: "!!x F. finite F ==> x \<notin> F ==> P F ==> P (insert x F)"
```
``` 44 assume "finite F"
```
``` 45 thus "P F"
```
``` 46 proof induct
```
``` 47 show "P {}" .
```
``` 48 fix x F assume F: "finite F" and P: "P F"
```
``` 49 show "P (insert x F)"
```
``` 50 proof cases
```
``` 51 assume "x \<in> F"
```
``` 52 hence "insert x F = F" by (rule insert_absorb)
```
``` 53 with P show ?thesis by (simp only:)
```
``` 54 next
```
``` 55 assume "x \<notin> F"
```
``` 56 from F this P show ?thesis by (rule insert)
```
``` 57 qed
```
``` 58 qed
```
``` 59 qed
```
``` 60
```
``` 61 lemma finite_ne_induct[case_names singleton insert, consumes 2]:
```
``` 62 assumes fin: "finite F" shows "F \<noteq> {} \<Longrightarrow>
```
``` 63 \<lbrakk> \<And>x. P{x};
```
``` 64 \<And>x F. \<lbrakk> finite F; F \<noteq> {}; x \<notin> F; P F \<rbrakk> \<Longrightarrow> P (insert x F) \<rbrakk>
```
``` 65 \<Longrightarrow> P F"
```
``` 66 using fin
```
``` 67 proof induct
```
``` 68 case empty thus ?case by simp
```
``` 69 next
```
``` 70 case (insert x F)
```
``` 71 show ?case
```
``` 72 proof cases
```
``` 73 assume "F = {}" thus ?thesis using insert(4) by simp
```
``` 74 next
```
``` 75 assume "F \<noteq> {}" thus ?thesis using insert by blast
```
``` 76 qed
```
``` 77 qed
```
``` 78
```
``` 79 lemma finite_subset_induct [consumes 2, case_names empty insert]:
```
``` 80 "finite F ==> F \<subseteq> A ==>
```
``` 81 P {} ==> (!!a F. finite F ==> a \<in> A ==> a \<notin> F ==> P F ==> P (insert a F)) ==>
```
``` 82 P F"
```
``` 83 proof -
```
``` 84 assume "P {}" and insert:
```
``` 85 "!!a F. finite F ==> a \<in> A ==> a \<notin> F ==> P F ==> P (insert a F)"
```
``` 86 assume "finite F"
```
``` 87 thus "F \<subseteq> A ==> P F"
```
``` 88 proof induct
```
``` 89 show "P {}" .
```
``` 90 fix x F assume "finite F" and "x \<notin> F"
```
``` 91 and P: "F \<subseteq> A ==> P F" and i: "insert x F \<subseteq> A"
```
``` 92 show "P (insert x F)"
```
``` 93 proof (rule insert)
```
``` 94 from i show "x \<in> A" by blast
```
``` 95 from i have "F \<subseteq> A" by blast
```
``` 96 with P show "P F" .
```
``` 97 qed
```
``` 98 qed
```
``` 99 qed
```
``` 100
```
``` 101 text{* Finite sets are the images of initial segments of natural numbers: *}
```
``` 102
```
``` 103 lemma finite_imp_nat_seg_image_inj_on:
```
``` 104 assumes fin: "finite A"
```
``` 105 shows "\<exists> (n::nat) f. A = f ` {i. i<n} & inj_on f {i. i<n}"
```
``` 106 using fin
```
``` 107 proof induct
```
``` 108 case empty
```
``` 109 show ?case
```
``` 110 proof show "\<exists>f. {} = f ` {i::nat. i < 0} & inj_on f {i. i<0}" by simp
```
``` 111 qed
```
``` 112 next
```
``` 113 case (insert a A)
```
``` 114 have notinA: "a \<notin> A" .
```
``` 115 from insert.hyps obtain n f
```
``` 116 where "A = f ` {i::nat. i < n}" "inj_on f {i. i < n}" by blast
```
``` 117 hence "insert a A = f(n:=a) ` {i. i < Suc n}"
```
``` 118 "inj_on (f(n:=a)) {i. i < Suc n}" using notinA
```
``` 119 by (auto simp add: image_def Ball_def inj_on_def less_Suc_eq)
```
``` 120 thus ?case by blast
```
``` 121 qed
```
``` 122
```
``` 123 lemma nat_seg_image_imp_finite:
```
``` 124 "!!f A. A = f ` {i::nat. i<n} \<Longrightarrow> finite A"
```
``` 125 proof (induct n)
```
``` 126 case 0 thus ?case by simp
```
``` 127 next
```
``` 128 case (Suc n)
```
``` 129 let ?B = "f ` {i. i < n}"
```
``` 130 have finB: "finite ?B" by(rule Suc.hyps[OF refl])
```
``` 131 show ?case
```
``` 132 proof cases
```
``` 133 assume "\<exists>k<n. f n = f k"
```
``` 134 hence "A = ?B" using Suc.prems by(auto simp:less_Suc_eq)
```
``` 135 thus ?thesis using finB by simp
```
``` 136 next
```
``` 137 assume "\<not>(\<exists> k<n. f n = f k)"
```
``` 138 hence "A = insert (f n) ?B" using Suc.prems by(auto simp:less_Suc_eq)
```
``` 139 thus ?thesis using finB by simp
```
``` 140 qed
```
``` 141 qed
```
``` 142
```
``` 143 lemma finite_conv_nat_seg_image:
```
``` 144 "finite A = (\<exists> (n::nat) f. A = f ` {i::nat. i<n})"
```
``` 145 by(blast intro: nat_seg_image_imp_finite dest: finite_imp_nat_seg_image_inj_on)
```
``` 146
```
``` 147 subsubsection{* Finiteness and set theoretic constructions *}
```
``` 148
```
``` 149 lemma finite_UnI: "finite F ==> finite G ==> finite (F Un G)"
```
``` 150 -- {* The union of two finite sets is finite. *}
```
``` 151 by (induct set: Finites) simp_all
```
``` 152
```
``` 153 lemma finite_subset: "A \<subseteq> B ==> finite B ==> finite A"
```
``` 154 -- {* Every subset of a finite set is finite. *}
```
``` 155 proof -
```
``` 156 assume "finite B"
```
``` 157 thus "!!A. A \<subseteq> B ==> finite A"
```
``` 158 proof induct
```
``` 159 case empty
```
``` 160 thus ?case by simp
```
``` 161 next
```
``` 162 case (insert x F A)
```
``` 163 have A: "A \<subseteq> insert x F" and r: "A - {x} \<subseteq> F ==> finite (A - {x})" .
```
``` 164 show "finite A"
```
``` 165 proof cases
```
``` 166 assume x: "x \<in> A"
```
``` 167 with A have "A - {x} \<subseteq> F" by (simp add: subset_insert_iff)
```
``` 168 with r have "finite (A - {x})" .
```
``` 169 hence "finite (insert x (A - {x}))" ..
```
``` 170 also have "insert x (A - {x}) = A" by (rule insert_Diff)
```
``` 171 finally show ?thesis .
```
``` 172 next
```
``` 173 show "A \<subseteq> F ==> ?thesis" .
```
``` 174 assume "x \<notin> A"
```
``` 175 with A show "A \<subseteq> F" by (simp add: subset_insert_iff)
```
``` 176 qed
```
``` 177 qed
```
``` 178 qed
```
``` 179
```
``` 180 lemma finite_Collect_subset: "finite A \<Longrightarrow> finite{x \<in> A. P x}"
```
``` 181 using finite_subset[of "{x \<in> A. P x}" "A"] by blast
```
``` 182
```
``` 183 lemma finite_Un [iff]: "finite (F Un G) = (finite F & finite G)"
```
``` 184 by (blast intro: finite_subset [of _ "X Un Y", standard] finite_UnI)
```
``` 185
```
``` 186 lemma finite_Int [simp, intro]: "finite F | finite G ==> finite (F Int G)"
```
``` 187 -- {* The converse obviously fails. *}
```
``` 188 by (blast intro: finite_subset)
```
``` 189
```
``` 190 lemma finite_insert [simp]: "finite (insert a A) = finite A"
```
``` 191 apply (subst insert_is_Un)
```
``` 192 apply (simp only: finite_Un, blast)
```
``` 193 done
```
``` 194
```
``` 195 lemma finite_Union[simp, intro]:
```
``` 196 "\<lbrakk> finite A; !!M. M \<in> A \<Longrightarrow> finite M \<rbrakk> \<Longrightarrow> finite(\<Union>A)"
```
``` 197 by (induct rule:finite_induct) simp_all
```
``` 198
```
``` 199 lemma finite_empty_induct:
```
``` 200 "finite A ==>
```
``` 201 P A ==> (!!a A. finite A ==> a:A ==> P A ==> P (A - {a})) ==> P {}"
```
``` 202 proof -
```
``` 203 assume "finite A"
```
``` 204 and "P A" and "!!a A. finite A ==> a:A ==> P A ==> P (A - {a})"
```
``` 205 have "P (A - A)"
```
``` 206 proof -
```
``` 207 fix c b :: "'a set"
```
``` 208 presume c: "finite c" and b: "finite b"
```
``` 209 and P1: "P b" and P2: "!!x y. finite y ==> x \<in> y ==> P y ==> P (y - {x})"
```
``` 210 from c show "c \<subseteq> b ==> P (b - c)"
```
``` 211 proof induct
```
``` 212 case empty
```
``` 213 from P1 show ?case by simp
```
``` 214 next
```
``` 215 case (insert x F)
```
``` 216 have "P (b - F - {x})"
```
``` 217 proof (rule P2)
```
``` 218 from _ b show "finite (b - F)" by (rule finite_subset) blast
```
``` 219 from insert show "x \<in> b - F" by simp
```
``` 220 from insert show "P (b - F)" by simp
```
``` 221 qed
```
``` 222 also have "b - F - {x} = b - insert x F" by (rule Diff_insert [symmetric])
```
``` 223 finally show ?case .
```
``` 224 qed
```
``` 225 next
```
``` 226 show "A \<subseteq> A" ..
```
``` 227 qed
```
``` 228 thus "P {}" by simp
```
``` 229 qed
```
``` 230
```
``` 231 lemma finite_Diff [simp]: "finite B ==> finite (B - Ba)"
```
``` 232 by (rule Diff_subset [THEN finite_subset])
```
``` 233
```
``` 234 lemma finite_Diff_insert [iff]: "finite (A - insert a B) = finite (A - B)"
```
``` 235 apply (subst Diff_insert)
```
``` 236 apply (case_tac "a : A - B")
```
``` 237 apply (rule finite_insert [symmetric, THEN trans])
```
``` 238 apply (subst insert_Diff, simp_all)
```
``` 239 done
```
``` 240
```
``` 241
```
``` 242 text {* Image and Inverse Image over Finite Sets *}
```
``` 243
```
``` 244 lemma finite_imageI[simp]: "finite F ==> finite (h ` F)"
```
``` 245 -- {* The image of a finite set is finite. *}
```
``` 246 by (induct set: Finites) simp_all
```
``` 247
```
``` 248 lemma finite_surj: "finite A ==> B <= f ` A ==> finite B"
```
``` 249 apply (frule finite_imageI)
```
``` 250 apply (erule finite_subset, assumption)
```
``` 251 done
```
``` 252
```
``` 253 lemma finite_range_imageI:
```
``` 254 "finite (range g) ==> finite (range (%x. f (g x)))"
```
``` 255 apply (drule finite_imageI, simp)
```
``` 256 done
```
``` 257
```
``` 258 lemma finite_imageD: "finite (f`A) ==> inj_on f A ==> finite A"
```
``` 259 proof -
```
``` 260 have aux: "!!A. finite (A - {}) = finite A" by simp
```
``` 261 fix B :: "'a set"
```
``` 262 assume "finite B"
```
``` 263 thus "!!A. f`A = B ==> inj_on f A ==> finite A"
```
``` 264 apply induct
```
``` 265 apply simp
```
``` 266 apply (subgoal_tac "EX y:A. f y = x & F = f ` (A - {y})")
```
``` 267 apply clarify
```
``` 268 apply (simp (no_asm_use) add: inj_on_def)
```
``` 269 apply (blast dest!: aux [THEN iffD1], atomize)
```
``` 270 apply (erule_tac V = "ALL A. ?PP (A)" in thin_rl)
```
``` 271 apply (frule subsetD [OF equalityD2 insertI1], clarify)
```
``` 272 apply (rule_tac x = xa in bexI)
```
``` 273 apply (simp_all add: inj_on_image_set_diff)
```
``` 274 done
```
``` 275 qed (rule refl)
```
``` 276
```
``` 277
```
``` 278 lemma inj_vimage_singleton: "inj f ==> f-`{a} \<subseteq> {THE x. f x = a}"
```
``` 279 -- {* The inverse image of a singleton under an injective function
```
``` 280 is included in a singleton. *}
```
``` 281 apply (auto simp add: inj_on_def)
```
``` 282 apply (blast intro: the_equality [symmetric])
```
``` 283 done
```
``` 284
```
``` 285 lemma finite_vimageI: "[|finite F; inj h|] ==> finite (h -` F)"
```
``` 286 -- {* The inverse image of a finite set under an injective function
```
``` 287 is finite. *}
```
``` 288 apply (induct set: Finites, simp_all)
```
``` 289 apply (subst vimage_insert)
```
``` 290 apply (simp add: finite_Un finite_subset [OF inj_vimage_singleton])
```
``` 291 done
```
``` 292
```
``` 293
```
``` 294 text {* The finite UNION of finite sets *}
```
``` 295
```
``` 296 lemma finite_UN_I: "finite A ==> (!!a. a:A ==> finite (B a)) ==> finite (UN a:A. B a)"
```
``` 297 by (induct set: Finites) simp_all
```
``` 298
```
``` 299 text {*
```
``` 300 Strengthen RHS to
```
``` 301 @{prop "((ALL x:A. finite (B x)) & finite {x. x:A & B x \<noteq> {}})"}?
```
``` 302
```
``` 303 We'd need to prove
```
``` 304 @{prop "finite C ==> ALL A B. (UNION A B) <= C --> finite {x. x:A & B x \<noteq> {}}"}
```
``` 305 by induction. *}
```
``` 306
```
``` 307 lemma finite_UN [simp]: "finite A ==> finite (UNION A B) = (ALL x:A. finite (B x))"
```
``` 308 by (blast intro: finite_UN_I finite_subset)
```
``` 309
```
``` 310
```
``` 311 lemma finite_Plus: "[| finite A; finite B |] ==> finite (A <+> B)"
```
``` 312 by (simp add: Plus_def)
```
``` 313
```
``` 314 text {* Sigma of finite sets *}
```
``` 315
```
``` 316 lemma finite_SigmaI [simp]:
```
``` 317 "finite A ==> (!!a. a:A ==> finite (B a)) ==> finite (SIGMA a:A. B a)"
```
``` 318 by (unfold Sigma_def) (blast intro!: finite_UN_I)
```
``` 319
```
``` 320 lemma finite_cartesian_product: "[| finite A; finite B |] ==>
```
``` 321 finite (A <*> B)"
```
``` 322 by (rule finite_SigmaI)
```
``` 323
```
``` 324 lemma finite_Prod_UNIV:
```
``` 325 "finite (UNIV::'a set) ==> finite (UNIV::'b set) ==> finite (UNIV::('a * 'b) set)"
```
``` 326 apply (subgoal_tac "(UNIV:: ('a * 'b) set) = Sigma UNIV (%x. UNIV)")
```
``` 327 apply (erule ssubst)
```
``` 328 apply (erule finite_SigmaI, auto)
```
``` 329 done
```
``` 330
```
``` 331 lemma finite_cartesian_productD1:
```
``` 332 "[| finite (A <*> B); B \<noteq> {} |] ==> finite A"
```
``` 333 apply (auto simp add: finite_conv_nat_seg_image)
```
``` 334 apply (drule_tac x=n in spec)
```
``` 335 apply (drule_tac x="fst o f" in spec)
```
``` 336 apply (auto simp add: o_def)
```
``` 337 prefer 2 apply (force dest!: equalityD2)
```
``` 338 apply (drule equalityD1)
```
``` 339 apply (rename_tac y x)
```
``` 340 apply (subgoal_tac "\<exists>k. k<n & f k = (x,y)")
```
``` 341 prefer 2 apply force
```
``` 342 apply clarify
```
``` 343 apply (rule_tac x=k in image_eqI, auto)
```
``` 344 done
```
``` 345
```
``` 346 lemma finite_cartesian_productD2:
```
``` 347 "[| finite (A <*> B); A \<noteq> {} |] ==> finite B"
```
``` 348 apply (auto simp add: finite_conv_nat_seg_image)
```
``` 349 apply (drule_tac x=n in spec)
```
``` 350 apply (drule_tac x="snd o f" in spec)
```
``` 351 apply (auto simp add: o_def)
```
``` 352 prefer 2 apply (force dest!: equalityD2)
```
``` 353 apply (drule equalityD1)
```
``` 354 apply (rename_tac x y)
```
``` 355 apply (subgoal_tac "\<exists>k. k<n & f k = (x,y)")
```
``` 356 prefer 2 apply force
```
``` 357 apply clarify
```
``` 358 apply (rule_tac x=k in image_eqI, auto)
```
``` 359 done
```
``` 360
```
``` 361
```
``` 362 text {* The powerset of a finite set *}
```
``` 363
```
``` 364 lemma finite_Pow_iff [iff]: "finite (Pow A) = finite A"
```
``` 365 proof
```
``` 366 assume "finite (Pow A)"
```
``` 367 with _ have "finite ((%x. {x}) ` A)" by (rule finite_subset) blast
```
``` 368 thus "finite A" by (rule finite_imageD [unfolded inj_on_def]) simp
```
``` 369 next
```
``` 370 assume "finite A"
```
``` 371 thus "finite (Pow A)"
```
``` 372 by induct (simp_all add: finite_UnI finite_imageI Pow_insert)
```
``` 373 qed
```
``` 374
```
``` 375
```
``` 376 lemma finite_UnionD: "finite(\<Union>A) \<Longrightarrow> finite A"
```
``` 377 by(blast intro: finite_subset[OF subset_Pow_Union])
```
``` 378
```
``` 379
```
``` 380 lemma finite_converse [iff]: "finite (r^-1) = finite r"
```
``` 381 apply (subgoal_tac "r^-1 = (%(x,y). (y,x))`r")
```
``` 382 apply simp
```
``` 383 apply (rule iffI)
```
``` 384 apply (erule finite_imageD [unfolded inj_on_def])
```
``` 385 apply (simp split add: split_split)
```
``` 386 apply (erule finite_imageI)
```
``` 387 apply (simp add: converse_def image_def, auto)
```
``` 388 apply (rule bexI)
```
``` 389 prefer 2 apply assumption
```
``` 390 apply simp
```
``` 391 done
```
``` 392
```
``` 393
```
``` 394 text {* \paragraph{Finiteness of transitive closure} (Thanks to Sidi
```
``` 395 Ehmety) *}
```
``` 396
```
``` 397 lemma finite_Field: "finite r ==> finite (Field r)"
```
``` 398 -- {* A finite relation has a finite field (@{text "= domain \<union> range"}. *}
```
``` 399 apply (induct set: Finites)
```
``` 400 apply (auto simp add: Field_def Domain_insert Range_insert)
```
``` 401 done
```
``` 402
```
``` 403 lemma trancl_subset_Field2: "r^+ <= Field r \<times> Field r"
```
``` 404 apply clarify
```
``` 405 apply (erule trancl_induct)
```
``` 406 apply (auto simp add: Field_def)
```
``` 407 done
```
``` 408
```
``` 409 lemma finite_trancl: "finite (r^+) = finite r"
```
``` 410 apply auto
```
``` 411 prefer 2
```
``` 412 apply (rule trancl_subset_Field2 [THEN finite_subset])
```
``` 413 apply (rule finite_SigmaI)
```
``` 414 prefer 3
```
``` 415 apply (blast intro: r_into_trancl' finite_subset)
```
``` 416 apply (auto simp add: finite_Field)
```
``` 417 done
```
``` 418
```
``` 419
```
``` 420 subsection {* A fold functional for finite sets *}
```
``` 421
```
``` 422 text {* The intended behaviour is
```
``` 423 @{text "fold f g z {x\<^isub>1, ..., x\<^isub>n} = f (g x\<^isub>1) (\<dots> (f (g x\<^isub>n) z)\<dots>)"}
```
``` 424 if @{text f} is associative-commutative. For an application of @{text fold}
```
``` 425 se the definitions of sums and products over finite sets.
```
``` 426 *}
```
``` 427
```
``` 428 consts
```
``` 429 foldSet :: "('a => 'a => 'a) => ('b => 'a) => 'a => ('b set \<times> 'a) set"
```
``` 430
```
``` 431 inductive "foldSet f g z"
```
``` 432 intros
```
``` 433 emptyI [intro]: "({}, z) : foldSet f g z"
```
``` 434 insertI [intro]:
```
``` 435 "\<lbrakk> x \<notin> A; (A, y) : foldSet f g z \<rbrakk>
```
``` 436 \<Longrightarrow> (insert x A, f (g x) y) : foldSet f g z"
```
``` 437
```
``` 438 inductive_cases empty_foldSetE [elim!]: "({}, x) : foldSet f g z"
```
``` 439
```
``` 440 constdefs
```
``` 441 fold :: "('a => 'a => 'a) => ('b => 'a) => 'a => 'b set => 'a"
```
``` 442 "fold f g z A == THE x. (A, x) : foldSet f g z"
```
``` 443
```
``` 444 text{*A tempting alternative for the definiens is
```
``` 445 @{term "if finite A then THE x. (A, x) : foldSet f g e else e"}.
```
``` 446 It allows the removal of finiteness assumptions from the theorems
```
``` 447 @{text fold_commute}, @{text fold_reindex} and @{text fold_distrib}.
```
``` 448 The proofs become ugly, with @{text rule_format}. It is not worth the effort.*}
```
``` 449
```
``` 450
```
``` 451 lemma Diff1_foldSet:
```
``` 452 "(A - {x}, y) : foldSet f g z ==> x: A ==> (A, f (g x) y) : foldSet f g z"
```
``` 453 by (erule insert_Diff [THEN subst], rule foldSet.intros, auto)
```
``` 454
```
``` 455 lemma foldSet_imp_finite: "(A, x) : foldSet f g z ==> finite A"
```
``` 456 by (induct set: foldSet) auto
```
``` 457
```
``` 458 lemma finite_imp_foldSet: "finite A ==> EX x. (A, x) : foldSet f g z"
```
``` 459 by (induct set: Finites) auto
```
``` 460
```
``` 461
```
``` 462 subsubsection {* Commutative monoids *}
```
``` 463
```
``` 464 locale ACf =
```
``` 465 fixes f :: "'a => 'a => 'a" (infixl "\<cdot>" 70)
```
``` 466 assumes commute: "x \<cdot> y = y \<cdot> x"
```
``` 467 and assoc: "(x \<cdot> y) \<cdot> z = x \<cdot> (y \<cdot> z)"
```
``` 468
```
``` 469 locale ACe = ACf +
```
``` 470 fixes e :: 'a
```
``` 471 assumes ident [simp]: "x \<cdot> e = x"
```
``` 472
```
``` 473 locale ACIf = ACf +
```
``` 474 assumes idem: "x \<cdot> x = x"
```
``` 475
```
``` 476 lemma (in ACf) left_commute: "x \<cdot> (y \<cdot> z) = y \<cdot> (x \<cdot> z)"
```
``` 477 proof -
```
``` 478 have "x \<cdot> (y \<cdot> z) = (y \<cdot> z) \<cdot> x" by (simp only: commute)
```
``` 479 also have "... = y \<cdot> (z \<cdot> x)" by (simp only: assoc)
```
``` 480 also have "z \<cdot> x = x \<cdot> z" by (simp only: commute)
```
``` 481 finally show ?thesis .
```
``` 482 qed
```
``` 483
```
``` 484 lemmas (in ACf) AC = assoc commute left_commute
```
``` 485
```
``` 486 lemma (in ACe) left_ident [simp]: "e \<cdot> x = x"
```
``` 487 proof -
```
``` 488 have "x \<cdot> e = x" by (rule ident)
```
``` 489 thus ?thesis by (subst commute)
```
``` 490 qed
```
``` 491
```
``` 492 lemma (in ACIf) idem2: "x \<cdot> (x \<cdot> y) = x \<cdot> y"
```
``` 493 proof -
```
``` 494 have "x \<cdot> (x \<cdot> y) = (x \<cdot> x) \<cdot> y" by(simp add:assoc)
```
``` 495 also have "\<dots> = x \<cdot> y" by(simp add:idem)
```
``` 496 finally show ?thesis .
```
``` 497 qed
```
``` 498
```
``` 499 lemmas (in ACIf) ACI = AC idem idem2
```
``` 500
```
``` 501 text{* Interpretation of locales: *}
```
``` 502
```
``` 503 interpretation AC_add: ACe ["op +" "0::'a::comm_monoid_add"]
```
``` 504 by(auto intro: ACf.intro ACe_axioms.intro add_assoc add_commute)
```
``` 505
```
``` 506 interpretation AC_mult: ACe ["op *" "1::'a::comm_monoid_mult"]
```
``` 507 apply -
```
``` 508 apply (fast intro: ACf.intro mult_assoc mult_commute)
```
``` 509 apply (fastsimp intro: ACe_axioms.intro mult_assoc mult_commute)
```
``` 510 done
```
``` 511
```
``` 512
```
``` 513 subsubsection{*From @{term foldSet} to @{term fold}*}
```
``` 514
```
``` 515 lemma image_less_Suc: "h ` {i. i < Suc m} = insert (h m) (h ` {i. i < m})"
```
``` 516 by (auto simp add: less_Suc_eq)
```
``` 517
```
``` 518 lemma insert_image_inj_on_eq:
```
``` 519 "[|insert (h m) A = h ` {i. i < Suc m}; h m \<notin> A;
```
``` 520 inj_on h {i. i < Suc m}|]
```
``` 521 ==> A = h ` {i. i < m}"
```
``` 522 apply (auto simp add: image_less_Suc inj_on_def)
```
``` 523 apply (blast intro: less_trans)
```
``` 524 done
```
``` 525
```
``` 526 lemma insert_inj_onE:
```
``` 527 assumes aA: "insert a A = h`{i::nat. i<n}" and anot: "a \<notin> A"
```
``` 528 and inj_on: "inj_on h {i::nat. i<n}"
```
``` 529 shows "\<exists>hm m. inj_on hm {i::nat. i<m} & A = hm ` {i. i<m} & m < n"
```
``` 530 proof (cases n)
```
``` 531 case 0 thus ?thesis using aA by auto
```
``` 532 next
```
``` 533 case (Suc m)
```
``` 534 have nSuc: "n = Suc m" .
```
``` 535 have mlessn: "m<n" by (simp add: nSuc)
```
``` 536 from aA obtain k where hkeq: "h k = a" and klessn: "k<n" by (blast elim!: equalityE)
```
``` 537 let ?hm = "swap k m h"
```
``` 538 have inj_hm: "inj_on ?hm {i. i < n}" using klessn mlessn
```
``` 539 by (simp add: inj_on_swap_iff inj_on)
```
``` 540 show ?thesis
```
``` 541 proof (intro exI conjI)
```
``` 542 show "inj_on ?hm {i. i < m}" using inj_hm
```
``` 543 by (auto simp add: nSuc less_Suc_eq intro: subset_inj_on)
```
``` 544 show "m<n" by (rule mlessn)
```
``` 545 show "A = ?hm ` {i. i < m}"
```
``` 546 proof (rule insert_image_inj_on_eq)
```
``` 547 show "inj_on (swap k m h) {i. i < Suc m}" using inj_hm nSuc by simp
```
``` 548 show "?hm m \<notin> A" by (simp add: swap_def hkeq anot)
```
``` 549 show "insert (?hm m) A = ?hm ` {i. i < Suc m}"
```
``` 550 using aA hkeq nSuc klessn
```
``` 551 by (auto simp add: swap_def image_less_Suc fun_upd_image
```
``` 552 less_Suc_eq inj_on_image_set_diff [OF inj_on])
```
``` 553 qed
```
``` 554 qed
```
``` 555 qed
```
``` 556
```
``` 557 lemma (in ACf) foldSet_determ_aux:
```
``` 558 "!!A x x' h. \<lbrakk> A = h`{i::nat. i<n}; inj_on h {i. i<n};
```
``` 559 (A,x) : foldSet f g z; (A,x') : foldSet f g z \<rbrakk>
```
``` 560 \<Longrightarrow> x' = x"
```
``` 561 proof (induct n rule: less_induct)
```
``` 562 case (less n)
```
``` 563 have IH: "!!m h A x x'.
```
``` 564 \<lbrakk>m<n; A = h ` {i. i<m}; inj_on h {i. i<m};
```
``` 565 (A,x) \<in> foldSet f g z; (A, x') \<in> foldSet f g z\<rbrakk> \<Longrightarrow> x' = x" .
```
``` 566 have Afoldx: "(A,x) \<in> foldSet f g z" and Afoldx': "(A,x') \<in> foldSet f g z"
```
``` 567 and A: "A = h`{i. i<n}" and injh: "inj_on h {i. i<n}" .
```
``` 568 show ?case
```
``` 569 proof (rule foldSet.cases [OF Afoldx])
```
``` 570 assume "(A, x) = ({}, z)"
```
``` 571 with Afoldx' show "x' = x" by blast
```
``` 572 next
```
``` 573 fix B b u
```
``` 574 assume "(A,x) = (insert b B, g b \<cdot> u)" and notinB: "b \<notin> B"
```
``` 575 and Bu: "(B,u) \<in> foldSet f g z"
```
``` 576 hence AbB: "A = insert b B" and x: "x = g b \<cdot> u" by auto
```
``` 577 show "x'=x"
```
``` 578 proof (rule foldSet.cases [OF Afoldx'])
```
``` 579 assume "(A, x') = ({}, z)"
```
``` 580 with AbB show "x' = x" by blast
```
``` 581 next
```
``` 582 fix C c v
```
``` 583 assume "(A,x') = (insert c C, g c \<cdot> v)" and notinC: "c \<notin> C"
```
``` 584 and Cv: "(C,v) \<in> foldSet f g z"
```
``` 585 hence AcC: "A = insert c C" and x': "x' = g c \<cdot> v" by auto
```
``` 586 from A AbB have Beq: "insert b B = h`{i. i<n}" by simp
```
``` 587 from insert_inj_onE [OF Beq notinB injh]
```
``` 588 obtain hB mB where inj_onB: "inj_on hB {i. i < mB}"
```
``` 589 and Beq: "B = hB ` {i. i < mB}"
```
``` 590 and lessB: "mB < n" by auto
```
``` 591 from A AcC have Ceq: "insert c C = h`{i. i<n}" by simp
```
``` 592 from insert_inj_onE [OF Ceq notinC injh]
```
``` 593 obtain hC mC where inj_onC: "inj_on hC {i. i < mC}"
```
``` 594 and Ceq: "C = hC ` {i. i < mC}"
```
``` 595 and lessC: "mC < n" by auto
```
``` 596 show "x'=x"
```
``` 597 proof cases
```
``` 598 assume "b=c"
```
``` 599 then moreover have "B = C" using AbB AcC notinB notinC by auto
```
``` 600 ultimately show ?thesis using Bu Cv x x' IH[OF lessC Ceq inj_onC]
```
``` 601 by auto
```
``` 602 next
```
``` 603 assume diff: "b \<noteq> c"
```
``` 604 let ?D = "B - {c}"
```
``` 605 have B: "B = insert c ?D" and C: "C = insert b ?D"
```
``` 606 using AbB AcC notinB notinC diff by(blast elim!:equalityE)+
```
``` 607 have "finite A" by(rule foldSet_imp_finite[OF Afoldx])
```
``` 608 with AbB have "finite ?D" by simp
```
``` 609 then obtain d where Dfoldd: "(?D,d) \<in> foldSet f g z"
```
``` 610 using finite_imp_foldSet by iprover
```
``` 611 moreover have cinB: "c \<in> B" using B by auto
```
``` 612 ultimately have "(B,g c \<cdot> d) \<in> foldSet f g z"
```
``` 613 by(rule Diff1_foldSet)
```
``` 614 hence "g c \<cdot> d = u" by (rule IH [OF lessB Beq inj_onB Bu])
```
``` 615 moreover have "g b \<cdot> d = v"
```
``` 616 proof (rule IH[OF lessC Ceq inj_onC Cv])
```
``` 617 show "(C, g b \<cdot> d) \<in> foldSet f g z" using C notinB Dfoldd
```
``` 618 by fastsimp
```
``` 619 qed
```
``` 620 ultimately show ?thesis using x x' by (auto simp: AC)
```
``` 621 qed
```
``` 622 qed
```
``` 623 qed
```
``` 624 qed
```
``` 625
```
``` 626
```
``` 627 lemma (in ACf) foldSet_determ:
```
``` 628 "(A,x) : foldSet f g z ==> (A,y) : foldSet f g z ==> y = x"
```
``` 629 apply (frule foldSet_imp_finite [THEN finite_imp_nat_seg_image_inj_on])
```
``` 630 apply (blast intro: foldSet_determ_aux [rule_format])
```
``` 631 done
```
``` 632
```
``` 633 lemma (in ACf) fold_equality: "(A, y) : foldSet f g z ==> fold f g z A = y"
```
``` 634 by (unfold fold_def) (blast intro: foldSet_determ)
```
``` 635
```
``` 636 text{* The base case for @{text fold}: *}
```
``` 637
```
``` 638 lemma fold_empty [simp]: "fold f g z {} = z"
```
``` 639 by (unfold fold_def) blast
```
``` 640
```
``` 641 lemma (in ACf) fold_insert_aux: "x \<notin> A ==>
```
``` 642 ((insert x A, v) : foldSet f g z) =
```
``` 643 (EX y. (A, y) : foldSet f g z & v = f (g x) y)"
```
``` 644 apply auto
```
``` 645 apply (rule_tac A1 = A and f1 = f in finite_imp_foldSet [THEN exE])
```
``` 646 apply (fastsimp dest: foldSet_imp_finite)
```
``` 647 apply (blast intro: foldSet_determ)
```
``` 648 done
```
``` 649
```
``` 650 text{* The recursion equation for @{text fold}: *}
```
``` 651
```
``` 652 lemma (in ACf) fold_insert[simp]:
```
``` 653 "finite A ==> x \<notin> A ==> fold f g z (insert x A) = f (g x) (fold f g z A)"
```
``` 654 apply (unfold fold_def)
```
``` 655 apply (simp add: fold_insert_aux)
```
``` 656 apply (rule the_equality)
```
``` 657 apply (auto intro: finite_imp_foldSet
```
``` 658 cong add: conj_cong simp add: fold_def [symmetric] fold_equality)
```
``` 659 done
```
``` 660
```
``` 661 lemma (in ACf) fold_rec:
```
``` 662 assumes fin: "finite A" and a: "a:A"
```
``` 663 shows "fold f g z A = f (g a) (fold f g z (A - {a}))"
```
``` 664 proof-
```
``` 665 have A: "A = insert a (A - {a})" using a by blast
```
``` 666 hence "fold f g z A = fold f g z (insert a (A - {a}))" by simp
```
``` 667 also have "\<dots> = f (g a) (fold f g z (A - {a}))"
```
``` 668 by(rule fold_insert) (simp add:fin)+
```
``` 669 finally show ?thesis .
```
``` 670 qed
```
``` 671
```
``` 672
```
``` 673 text{* A simplified version for idempotent functions: *}
```
``` 674
```
``` 675 lemma (in ACIf) fold_insert_idem:
```
``` 676 assumes finA: "finite A"
```
``` 677 shows "fold f g z (insert a A) = g a \<cdot> fold f g z A"
```
``` 678 proof cases
```
``` 679 assume "a \<in> A"
```
``` 680 then obtain B where A: "A = insert a B" and disj: "a \<notin> B"
```
``` 681 by(blast dest: mk_disjoint_insert)
```
``` 682 show ?thesis
```
``` 683 proof -
```
``` 684 from finA A have finB: "finite B" by(blast intro: finite_subset)
```
``` 685 have "fold f g z (insert a A) = fold f g z (insert a B)" using A by simp
```
``` 686 also have "\<dots> = (g a) \<cdot> (fold f g z B)"
```
``` 687 using finB disj by simp
```
``` 688 also have "\<dots> = g a \<cdot> fold f g z A"
```
``` 689 using A finB disj by(simp add:idem assoc[symmetric])
```
``` 690 finally show ?thesis .
```
``` 691 qed
```
``` 692 next
```
``` 693 assume "a \<notin> A"
```
``` 694 with finA show ?thesis by simp
```
``` 695 qed
```
``` 696
```
``` 697 lemma (in ACIf) foldI_conv_id:
```
``` 698 "finite A \<Longrightarrow> fold f g z A = fold f id z (g ` A)"
```
``` 699 by(erule finite_induct)(simp_all add: fold_insert_idem del: fold_insert)
```
``` 700
```
``` 701 subsubsection{*Lemmas about @{text fold}*}
```
``` 702
```
``` 703 lemma (in ACf) fold_commute:
```
``` 704 "finite A ==> (!!z. f x (fold f g z A) = fold f g (f x z) A)"
```
``` 705 apply (induct set: Finites, simp)
```
``` 706 apply (simp add: left_commute [of x])
```
``` 707 done
```
``` 708
```
``` 709 lemma (in ACf) fold_nest_Un_Int:
```
``` 710 "finite A ==> finite B
```
``` 711 ==> fold f g (fold f g z B) A = fold f g (fold f g z (A Int B)) (A Un B)"
```
``` 712 apply (induct set: Finites, simp)
```
``` 713 apply (simp add: fold_commute Int_insert_left insert_absorb)
```
``` 714 done
```
``` 715
```
``` 716 lemma (in ACf) fold_nest_Un_disjoint:
```
``` 717 "finite A ==> finite B ==> A Int B = {}
```
``` 718 ==> fold f g z (A Un B) = fold f g (fold f g z B) A"
```
``` 719 by (simp add: fold_nest_Un_Int)
```
``` 720
```
``` 721 lemma (in ACf) fold_reindex:
```
``` 722 assumes fin: "finite A"
```
``` 723 shows "inj_on h A \<Longrightarrow> fold f g z (h ` A) = fold f (g \<circ> h) z A"
```
``` 724 using fin apply induct
```
``` 725 apply simp
```
``` 726 apply simp
```
``` 727 done
```
``` 728
```
``` 729 lemma (in ACe) fold_Un_Int:
```
``` 730 "finite A ==> finite B ==>
```
``` 731 fold f g e A \<cdot> fold f g e B =
```
``` 732 fold f g e (A Un B) \<cdot> fold f g e (A Int B)"
```
``` 733 apply (induct set: Finites, simp)
```
``` 734 apply (simp add: AC insert_absorb Int_insert_left)
```
``` 735 done
```
``` 736
```
``` 737 corollary (in ACe) fold_Un_disjoint:
```
``` 738 "finite A ==> finite B ==> A Int B = {} ==>
```
``` 739 fold f g e (A Un B) = fold f g e A \<cdot> fold f g e B"
```
``` 740 by (simp add: fold_Un_Int)
```
``` 741
```
``` 742 lemma (in ACe) fold_UN_disjoint:
```
``` 743 "\<lbrakk> finite I; ALL i:I. finite (A i);
```
``` 744 ALL i:I. ALL j:I. i \<noteq> j --> A i Int A j = {} \<rbrakk>
```
``` 745 \<Longrightarrow> fold f g e (UNION I A) =
```
``` 746 fold f (%i. fold f g e (A i)) e I"
```
``` 747 apply (induct set: Finites, simp, atomize)
```
``` 748 apply (subgoal_tac "ALL i:F. x \<noteq> i")
```
``` 749 prefer 2 apply blast
```
``` 750 apply (subgoal_tac "A x Int UNION F A = {}")
```
``` 751 prefer 2 apply blast
```
``` 752 apply (simp add: fold_Un_disjoint)
```
``` 753 done
```
``` 754
```
``` 755 text{*Fusion theorem, as described in
```
``` 756 Graham Hutton's paper,
```
``` 757 A Tutorial on the Universality and Expressiveness of Fold,
```
``` 758 JFP 9:4 (355-372), 1999.*}
```
``` 759 lemma (in ACf) fold_fusion:
```
``` 760 includes ACf g
```
``` 761 shows
```
``` 762 "finite A ==>
```
``` 763 (!!x y. h (g x y) = f x (h y)) ==>
```
``` 764 h (fold g j w A) = fold f j (h w) A"
```
``` 765 by (induct set: Finites, simp_all)
```
``` 766
```
``` 767 lemma (in ACf) fold_cong:
```
``` 768 "finite A \<Longrightarrow> (!!x. x:A ==> g x = h x) ==> fold f g z A = fold f h z A"
```
``` 769 apply (subgoal_tac "ALL C. C <= A --> (ALL x:C. g x = h x) --> fold f g z C = fold f h z C")
```
``` 770 apply simp
```
``` 771 apply (erule finite_induct, simp)
```
``` 772 apply (simp add: subset_insert_iff, clarify)
```
``` 773 apply (subgoal_tac "finite C")
```
``` 774 prefer 2 apply (blast dest: finite_subset [COMP swap_prems_rl])
```
``` 775 apply (subgoal_tac "C = insert x (C - {x})")
```
``` 776 prefer 2 apply blast
```
``` 777 apply (erule ssubst)
```
``` 778 apply (drule spec)
```
``` 779 apply (erule (1) notE impE)
```
``` 780 apply (simp add: Ball_def del: insert_Diff_single)
```
``` 781 done
```
``` 782
```
``` 783 lemma (in ACe) fold_Sigma: "finite A ==> ALL x:A. finite (B x) ==>
```
``` 784 fold f (%x. fold f (g x) e (B x)) e A =
```
``` 785 fold f (split g) e (SIGMA x:A. B x)"
```
``` 786 apply (subst Sigma_def)
```
``` 787 apply (subst fold_UN_disjoint, assumption, simp)
```
``` 788 apply blast
```
``` 789 apply (erule fold_cong)
```
``` 790 apply (subst fold_UN_disjoint, simp, simp)
```
``` 791 apply blast
```
``` 792 apply simp
```
``` 793 done
```
``` 794
```
``` 795 lemma (in ACe) fold_distrib: "finite A \<Longrightarrow>
```
``` 796 fold f (%x. f (g x) (h x)) e A = f (fold f g e A) (fold f h e A)"
```
``` 797 apply (erule finite_induct, simp)
```
``` 798 apply (simp add:AC)
```
``` 799 done
```
``` 800
```
``` 801
```
``` 802 subsection {* Generalized summation over a set *}
```
``` 803
```
``` 804 constdefs
```
``` 805 setsum :: "('a => 'b) => 'a set => 'b::comm_monoid_add"
```
``` 806 "setsum f A == if finite A then fold (op +) f 0 A else 0"
```
``` 807
```
``` 808 text{* Now: lot's of fancy syntax. First, @{term "setsum (%x. e) A"} is
```
``` 809 written @{text"\<Sum>x\<in>A. e"}. *}
```
``` 810
```
``` 811 syntax
```
``` 812 "_setsum" :: "pttrn => 'a set => 'b => 'b::comm_monoid_add" ("(3SUM _:_. _)" [0, 51, 10] 10)
```
``` 813 syntax (xsymbols)
```
``` 814 "_setsum" :: "pttrn => 'a set => 'b => 'b::comm_monoid_add" ("(3\<Sum>_\<in>_. _)" [0, 51, 10] 10)
```
``` 815 syntax (HTML output)
```
``` 816 "_setsum" :: "pttrn => 'a set => 'b => 'b::comm_monoid_add" ("(3\<Sum>_\<in>_. _)" [0, 51, 10] 10)
```
``` 817
```
``` 818 translations -- {* Beware of argument permutation! *}
```
``` 819 "SUM i:A. b" == "setsum (%i. b) A"
```
``` 820 "\<Sum>i\<in>A. b" == "setsum (%i. b) A"
```
``` 821
```
``` 822 text{* Instead of @{term"\<Sum>x\<in>{x. P}. e"} we introduce the shorter
```
``` 823 @{text"\<Sum>x|P. e"}. *}
```
``` 824
```
``` 825 syntax
```
``` 826 "_qsetsum" :: "pttrn \<Rightarrow> bool \<Rightarrow> 'a \<Rightarrow> 'a" ("(3SUM _ |/ _./ _)" [0,0,10] 10)
```
``` 827 syntax (xsymbols)
```
``` 828 "_qsetsum" :: "pttrn \<Rightarrow> bool \<Rightarrow> 'a \<Rightarrow> 'a" ("(3\<Sum>_ | (_)./ _)" [0,0,10] 10)
```
``` 829 syntax (HTML output)
```
``` 830 "_qsetsum" :: "pttrn \<Rightarrow> bool \<Rightarrow> 'a \<Rightarrow> 'a" ("(3\<Sum>_ | (_)./ _)" [0,0,10] 10)
```
``` 831
```
``` 832 translations
```
``` 833 "SUM x|P. t" => "setsum (%x. t) {x. P}"
```
``` 834 "\<Sum>x|P. t" => "setsum (%x. t) {x. P}"
```
``` 835
```
``` 836 text{* Finally we abbreviate @{term"\<Sum>x\<in>A. x"} by @{text"\<Sum>A"}. *}
```
``` 837
```
``` 838 syntax
```
``` 839 "_Setsum" :: "'a set => 'a::comm_monoid_mult" ("\<Sum>_" [1000] 999)
```
``` 840
```
``` 841 parse_translation {*
```
``` 842 let
```
``` 843 fun Setsum_tr [A] = Syntax.const "setsum" \$ Abs ("", dummyT, Bound 0) \$ A
```
``` 844 in [("_Setsum", Setsum_tr)] end;
```
``` 845 *}
```
``` 846
```
``` 847 print_translation {*
```
``` 848 let
```
``` 849 fun setsum_tr' [Abs(_,_,Bound 0), A] = Syntax.const "_Setsum" \$ A
```
``` 850 | setsum_tr' [Abs(x,Tx,t), Const ("Collect",_) \$ Abs(y,Ty,P)] =
```
``` 851 if x<>y then raise Match
```
``` 852 else let val x' = Syntax.mark_bound x
```
``` 853 val t' = subst_bound(x',t)
```
``` 854 val P' = subst_bound(x',P)
```
``` 855 in Syntax.const "_qsetsum" \$ Syntax.mark_bound x \$ P' \$ t' end
```
``` 856 in
```
``` 857 [("setsum", setsum_tr')]
```
``` 858 end
```
``` 859 *}
```
``` 860
```
``` 861 lemma setsum_empty [simp]: "setsum f {} = 0"
```
``` 862 by (simp add: setsum_def)
```
``` 863
```
``` 864 lemma setsum_insert [simp]:
```
``` 865 "finite F ==> a \<notin> F ==> setsum f (insert a F) = f a + setsum f F"
```
``` 866 by (simp add: setsum_def)
```
``` 867
```
``` 868 lemma setsum_infinite [simp]: "~ finite A ==> setsum f A = 0"
```
``` 869 by (simp add: setsum_def)
```
``` 870
```
``` 871 lemma setsum_reindex:
```
``` 872 "inj_on f B ==> setsum h (f ` B) = setsum (h \<circ> f) B"
```
``` 873 by(auto simp add: setsum_def AC_add.fold_reindex dest!:finite_imageD)
```
``` 874
```
``` 875 lemma setsum_reindex_id:
```
``` 876 "inj_on f B ==> setsum f B = setsum id (f ` B)"
```
``` 877 by (auto simp add: setsum_reindex)
```
``` 878
```
``` 879 lemma setsum_cong:
```
``` 880 "A = B ==> (!!x. x:B ==> f x = g x) ==> setsum f A = setsum g B"
```
``` 881 by(fastsimp simp: setsum_def intro: AC_add.fold_cong)
```
``` 882
```
``` 883 lemma strong_setsum_cong[cong]:
```
``` 884 "A = B ==> (!!x. x:B =simp=> f x = g x)
```
``` 885 ==> setsum (%x. f x) A = setsum (%x. g x) B"
```
``` 886 by(fastsimp simp: simp_implies_def setsum_def intro: AC_add.fold_cong)
```
``` 887
```
``` 888 lemma setsum_cong2: "\<lbrakk>\<And>x. x \<in> A \<Longrightarrow> f x = g x\<rbrakk> \<Longrightarrow> setsum f A = setsum g A";
```
``` 889 by (rule setsum_cong[OF refl], auto);
```
``` 890
```
``` 891 lemma setsum_reindex_cong:
```
``` 892 "[|inj_on f A; B = f ` A; !!a. a:A \<Longrightarrow> g a = h (f a)|]
```
``` 893 ==> setsum h B = setsum g A"
```
``` 894 by (simp add: setsum_reindex cong: setsum_cong)
```
``` 895
```
``` 896 lemma setsum_0[simp]: "setsum (%i. 0) A = 0"
```
``` 897 apply (clarsimp simp: setsum_def)
```
``` 898 apply (erule finite_induct, auto)
```
``` 899 done
```
``` 900
```
``` 901 lemma setsum_0': "ALL a:A. f a = 0 ==> setsum f A = 0"
```
``` 902 by(simp add:setsum_cong)
```
``` 903
```
``` 904 lemma setsum_Un_Int: "finite A ==> finite B ==>
```
``` 905 setsum g (A Un B) + setsum g (A Int B) = setsum g A + setsum g B"
```
``` 906 -- {* The reversed orientation looks more natural, but LOOPS as a simprule! *}
```
``` 907 by(simp add: setsum_def AC_add.fold_Un_Int [symmetric])
```
``` 908
```
``` 909 lemma setsum_Un_disjoint: "finite A ==> finite B
```
``` 910 ==> A Int B = {} ==> setsum g (A Un B) = setsum g A + setsum g B"
```
``` 911 by (subst setsum_Un_Int [symmetric], auto)
```
``` 912
```
``` 913 (*But we can't get rid of finite I. If infinite, although the rhs is 0,
```
``` 914 the lhs need not be, since UNION I A could still be finite.*)
```
``` 915 lemma setsum_UN_disjoint:
```
``` 916 "finite I ==> (ALL i:I. finite (A i)) ==>
```
``` 917 (ALL i:I. ALL j:I. i \<noteq> j --> A i Int A j = {}) ==>
```
``` 918 setsum f (UNION I A) = (\<Sum>i\<in>I. setsum f (A i))"
```
``` 919 by(simp add: setsum_def AC_add.fold_UN_disjoint cong: setsum_cong)
```
``` 920
```
``` 921 text{*No need to assume that @{term C} is finite. If infinite, the rhs is
```
``` 922 directly 0, and @{term "Union C"} is also infinite, hence the lhs is also 0.*}
```
``` 923 lemma setsum_Union_disjoint:
```
``` 924 "[| (ALL A:C. finite A);
```
``` 925 (ALL A:C. ALL B:C. A \<noteq> B --> A Int B = {}) |]
```
``` 926 ==> setsum f (Union C) = setsum (setsum f) C"
```
``` 927 apply (cases "finite C")
```
``` 928 prefer 2 apply (force dest: finite_UnionD simp add: setsum_def)
```
``` 929 apply (frule setsum_UN_disjoint [of C id f])
```
``` 930 apply (unfold Union_def id_def, assumption+)
```
``` 931 done
```
``` 932
```
``` 933 (*But we can't get rid of finite A. If infinite, although the lhs is 0,
```
``` 934 the rhs need not be, since SIGMA A B could still be finite.*)
```
``` 935 lemma setsum_Sigma: "finite A ==> ALL x:A. finite (B x) ==>
```
``` 936 (\<Sum>x\<in>A. (\<Sum>y\<in>B x. f x y)) = (\<Sum>(x,y)\<in>(SIGMA x:A. B x). f x y)"
```
``` 937 by(simp add:setsum_def AC_add.fold_Sigma split_def cong:setsum_cong)
```
``` 938
```
``` 939 text{*Here we can eliminate the finiteness assumptions, by cases.*}
```
``` 940 lemma setsum_cartesian_product:
```
``` 941 "(\<Sum>x\<in>A. (\<Sum>y\<in>B. f x y)) = (\<Sum>(x,y) \<in> A <*> B. f x y)"
```
``` 942 apply (cases "finite A")
```
``` 943 apply (cases "finite B")
```
``` 944 apply (simp add: setsum_Sigma)
```
``` 945 apply (cases "A={}", simp)
```
``` 946 apply (simp)
```
``` 947 apply (auto simp add: setsum_def
```
``` 948 dest: finite_cartesian_productD1 finite_cartesian_productD2)
```
``` 949 done
```
``` 950
```
``` 951 lemma setsum_addf: "setsum (%x. f x + g x) A = (setsum f A + setsum g A)"
```
``` 952 by(simp add:setsum_def AC_add.fold_distrib)
```
``` 953
```
``` 954
```
``` 955 subsubsection {* Properties in more restricted classes of structures *}
```
``` 956
```
``` 957 lemma setsum_SucD: "setsum f A = Suc n ==> EX a:A. 0 < f a"
```
``` 958 apply (case_tac "finite A")
```
``` 959 prefer 2 apply (simp add: setsum_def)
```
``` 960 apply (erule rev_mp)
```
``` 961 apply (erule finite_induct, auto)
```
``` 962 done
```
``` 963
```
``` 964 lemma setsum_eq_0_iff [simp]:
```
``` 965 "finite F ==> (setsum f F = 0) = (ALL a:F. f a = (0::nat))"
```
``` 966 by (induct set: Finites) auto
```
``` 967
```
``` 968 lemma setsum_Un_nat: "finite A ==> finite B ==>
```
``` 969 (setsum f (A Un B) :: nat) = setsum f A + setsum f B - setsum f (A Int B)"
```
``` 970 -- {* For the natural numbers, we have subtraction. *}
```
``` 971 by (subst setsum_Un_Int [symmetric], auto simp add: ring_eq_simps)
```
``` 972
```
``` 973 lemma setsum_Un: "finite A ==> finite B ==>
```
``` 974 (setsum f (A Un B) :: 'a :: ab_group_add) =
```
``` 975 setsum f A + setsum f B - setsum f (A Int B)"
```
``` 976 by (subst setsum_Un_Int [symmetric], auto simp add: ring_eq_simps)
```
``` 977
```
``` 978 lemma setsum_diff1_nat: "(setsum f (A - {a}) :: nat) =
```
``` 979 (if a:A then setsum f A - f a else setsum f A)"
```
``` 980 apply (case_tac "finite A")
```
``` 981 prefer 2 apply (simp add: setsum_def)
```
``` 982 apply (erule finite_induct)
```
``` 983 apply (auto simp add: insert_Diff_if)
```
``` 984 apply (drule_tac a = a in mk_disjoint_insert, auto)
```
``` 985 done
```
``` 986
```
``` 987 lemma setsum_diff1: "finite A \<Longrightarrow>
```
``` 988 (setsum f (A - {a}) :: ('a::ab_group_add)) =
```
``` 989 (if a:A then setsum f A - f a else setsum f A)"
```
``` 990 by (erule finite_induct) (auto simp add: insert_Diff_if)
```
``` 991
```
``` 992 lemma setsum_diff1'[rule_format]: "finite A \<Longrightarrow> a \<in> A \<longrightarrow> (\<Sum> x \<in> A. f x) = f a + (\<Sum> x \<in> (A - {a}). f x)"
```
``` 993 apply (erule finite_induct[where F=A and P="% A. (a \<in> A \<longrightarrow> (\<Sum> x \<in> A. f x) = f a + (\<Sum> x \<in> (A - {a}). f x))"])
```
``` 994 apply (auto simp add: insert_Diff_if add_ac)
```
``` 995 done
```
``` 996
```
``` 997 (* By Jeremy Siek: *)
```
``` 998
```
``` 999 lemma setsum_diff_nat:
```
``` 1000 assumes finB: "finite B"
```
``` 1001 shows "B \<subseteq> A \<Longrightarrow> (setsum f (A - B) :: nat) = (setsum f A) - (setsum f B)"
```
``` 1002 using finB
```
``` 1003 proof (induct)
```
``` 1004 show "setsum f (A - {}) = (setsum f A) - (setsum f {})" by simp
```
``` 1005 next
```
``` 1006 fix F x assume finF: "finite F" and xnotinF: "x \<notin> F"
```
``` 1007 and xFinA: "insert x F \<subseteq> A"
```
``` 1008 and IH: "F \<subseteq> A \<Longrightarrow> setsum f (A - F) = setsum f A - setsum f F"
```
``` 1009 from xnotinF xFinA have xinAF: "x \<in> (A - F)" by simp
```
``` 1010 from xinAF have A: "setsum f ((A - F) - {x}) = setsum f (A - F) - f x"
```
``` 1011 by (simp add: setsum_diff1_nat)
```
``` 1012 from xFinA have "F \<subseteq> A" by simp
```
``` 1013 with IH have "setsum f (A - F) = setsum f A - setsum f F" by simp
```
``` 1014 with A have B: "setsum f ((A - F) - {x}) = setsum f A - setsum f F - f x"
```
``` 1015 by simp
```
``` 1016 from xnotinF have "A - insert x F = (A - F) - {x}" by auto
```
``` 1017 with B have C: "setsum f (A - insert x F) = setsum f A - setsum f F - f x"
```
``` 1018 by simp
```
``` 1019 from finF xnotinF have "setsum f (insert x F) = setsum f F + f x" by simp
```
``` 1020 with C have "setsum f (A - insert x F) = setsum f A - setsum f (insert x F)"
```
``` 1021 by simp
```
``` 1022 thus "setsum f (A - insert x F) = setsum f A - setsum f (insert x F)" by simp
```
``` 1023 qed
```
``` 1024
```
``` 1025 lemma setsum_diff:
```
``` 1026 assumes le: "finite A" "B \<subseteq> A"
```
``` 1027 shows "setsum f (A - B) = setsum f A - ((setsum f B)::('a::ab_group_add))"
```
``` 1028 proof -
```
``` 1029 from le have finiteB: "finite B" using finite_subset by auto
```
``` 1030 show ?thesis using finiteB le
```
``` 1031 proof (induct)
```
``` 1032 case empty
```
``` 1033 thus ?case by auto
```
``` 1034 next
```
``` 1035 case (insert x F)
```
``` 1036 thus ?case using le finiteB
```
``` 1037 by (simp add: Diff_insert[where a=x and B=F] setsum_diff1 insert_absorb)
```
``` 1038 qed
```
``` 1039 qed
```
``` 1040
```
``` 1041 lemma setsum_mono:
```
``` 1042 assumes le: "\<And>i. i\<in>K \<Longrightarrow> f (i::'a) \<le> ((g i)::('b::{comm_monoid_add, pordered_ab_semigroup_add}))"
```
``` 1043 shows "(\<Sum>i\<in>K. f i) \<le> (\<Sum>i\<in>K. g i)"
```
``` 1044 proof (cases "finite K")
```
``` 1045 case True
```
``` 1046 thus ?thesis using le
```
``` 1047 proof (induct)
```
``` 1048 case empty
```
``` 1049 thus ?case by simp
```
``` 1050 next
```
``` 1051 case insert
```
``` 1052 thus ?case using add_mono
```
``` 1053 by force
```
``` 1054 qed
```
``` 1055 next
```
``` 1056 case False
```
``` 1057 thus ?thesis
```
``` 1058 by (simp add: setsum_def)
```
``` 1059 qed
```
``` 1060
```
``` 1061 lemma setsum_strict_mono:
```
``` 1062 fixes f :: "'a \<Rightarrow> 'b::{pordered_cancel_ab_semigroup_add,comm_monoid_add}"
```
``` 1063 assumes fin_ne: "finite A" "A \<noteq> {}"
```
``` 1064 shows "(!!x. x:A \<Longrightarrow> f x < g x) \<Longrightarrow> setsum f A < setsum g A"
```
``` 1065 using fin_ne
```
``` 1066 proof (induct rule: finite_ne_induct)
```
``` 1067 case singleton thus ?case by simp
```
``` 1068 next
```
``` 1069 case insert thus ?case by (auto simp: add_strict_mono)
```
``` 1070 qed
```
``` 1071
```
``` 1072 lemma setsum_negf:
```
``` 1073 "setsum (%x. - (f x)::'a::ab_group_add) A = - setsum f A"
```
``` 1074 proof (cases "finite A")
```
``` 1075 case True thus ?thesis by (induct set: Finites, auto)
```
``` 1076 next
```
``` 1077 case False thus ?thesis by (simp add: setsum_def)
```
``` 1078 qed
```
``` 1079
```
``` 1080 lemma setsum_subtractf:
```
``` 1081 "setsum (%x. ((f x)::'a::ab_group_add) - g x) A =
```
``` 1082 setsum f A - setsum g A"
```
``` 1083 proof (cases "finite A")
```
``` 1084 case True thus ?thesis by (simp add: diff_minus setsum_addf setsum_negf)
```
``` 1085 next
```
``` 1086 case False thus ?thesis by (simp add: setsum_def)
```
``` 1087 qed
```
``` 1088
```
``` 1089 lemma setsum_nonneg:
```
``` 1090 assumes nn: "\<forall>x\<in>A. (0::'a::{pordered_ab_semigroup_add,comm_monoid_add}) \<le> f x"
```
``` 1091 shows "0 \<le> setsum f A"
```
``` 1092 proof (cases "finite A")
```
``` 1093 case True thus ?thesis using nn
```
``` 1094 apply (induct set: Finites, auto)
```
``` 1095 apply (subgoal_tac "0 + 0 \<le> f x + setsum f F", simp)
```
``` 1096 apply (blast intro: add_mono)
```
``` 1097 done
```
``` 1098 next
```
``` 1099 case False thus ?thesis by (simp add: setsum_def)
```
``` 1100 qed
```
``` 1101
```
``` 1102 lemma setsum_nonpos:
```
``` 1103 assumes np: "\<forall>x\<in>A. f x \<le> (0::'a::{pordered_ab_semigroup_add,comm_monoid_add})"
```
``` 1104 shows "setsum f A \<le> 0"
```
``` 1105 proof (cases "finite A")
```
``` 1106 case True thus ?thesis using np
```
``` 1107 apply (induct set: Finites, auto)
```
``` 1108 apply (subgoal_tac "f x + setsum f F \<le> 0 + 0", simp)
```
``` 1109 apply (blast intro: add_mono)
```
``` 1110 done
```
``` 1111 next
```
``` 1112 case False thus ?thesis by (simp add: setsum_def)
```
``` 1113 qed
```
``` 1114
```
``` 1115 lemma setsum_mono2:
```
``` 1116 fixes f :: "'a \<Rightarrow> 'b :: {pordered_ab_semigroup_add_imp_le,comm_monoid_add}"
```
``` 1117 assumes fin: "finite B" and sub: "A \<subseteq> B" and nn: "\<And>b. b \<in> B-A \<Longrightarrow> 0 \<le> f b"
```
``` 1118 shows "setsum f A \<le> setsum f B"
```
``` 1119 proof -
```
``` 1120 have "setsum f A \<le> setsum f A + setsum f (B-A)"
```
``` 1121 by(simp add: add_increasing2[OF setsum_nonneg] nn Ball_def)
```
``` 1122 also have "\<dots> = setsum f (A \<union> (B-A))" using fin finite_subset[OF sub fin]
```
``` 1123 by (simp add:setsum_Un_disjoint del:Un_Diff_cancel)
```
``` 1124 also have "A \<union> (B-A) = B" using sub by blast
```
``` 1125 finally show ?thesis .
```
``` 1126 qed
```
``` 1127
```
``` 1128 lemma setsum_mono3: "finite B ==> A <= B ==>
```
``` 1129 ALL x: B - A.
```
``` 1130 0 <= ((f x)::'a::{comm_monoid_add,pordered_ab_semigroup_add}) ==>
```
``` 1131 setsum f A <= setsum f B"
```
``` 1132 apply (subgoal_tac "setsum f B = setsum f A + setsum f (B - A)")
```
``` 1133 apply (erule ssubst)
```
``` 1134 apply (subgoal_tac "setsum f A + 0 <= setsum f A + setsum f (B - A)")
```
``` 1135 apply simp
```
``` 1136 apply (rule add_left_mono)
```
``` 1137 apply (erule setsum_nonneg)
```
``` 1138 apply (subst setsum_Un_disjoint [THEN sym])
```
``` 1139 apply (erule finite_subset, assumption)
```
``` 1140 apply (rule finite_subset)
```
``` 1141 prefer 2
```
``` 1142 apply assumption
```
``` 1143 apply auto
```
``` 1144 apply (rule setsum_cong)
```
``` 1145 apply auto
```
``` 1146 done
```
``` 1147
```
``` 1148 (* FIXME: this is distributitivty, name as such! *)
```
``` 1149 (* suggested name: setsum_right_distrib (CB) *)
```
``` 1150
```
``` 1151 lemma setsum_mult:
```
``` 1152 fixes f :: "'a => ('b::semiring_0_cancel)"
```
``` 1153 shows "r * setsum f A = setsum (%n. r * f n) A"
```
``` 1154 proof (cases "finite A")
```
``` 1155 case True
```
``` 1156 thus ?thesis
```
``` 1157 proof (induct)
```
``` 1158 case empty thus ?case by simp
```
``` 1159 next
```
``` 1160 case (insert x A) thus ?case by (simp add: right_distrib)
```
``` 1161 qed
```
``` 1162 next
```
``` 1163 case False thus ?thesis by (simp add: setsum_def)
```
``` 1164 qed
```
``` 1165
```
``` 1166 lemma setsum_left_distrib:
```
``` 1167 "setsum f A * (r::'a::semiring_0_cancel) = (\<Sum>n\<in>A. f n * r)"
```
``` 1168 proof (cases "finite A")
```
``` 1169 case True
```
``` 1170 then show ?thesis
```
``` 1171 proof induct
```
``` 1172 case empty thus ?case by simp
```
``` 1173 next
```
``` 1174 case (insert x A) thus ?case by (simp add: left_distrib)
```
``` 1175 qed
```
``` 1176 next
```
``` 1177 case False thus ?thesis by (simp add: setsum_def)
```
``` 1178 qed
```
``` 1179
```
``` 1180 lemma setsum_divide_distrib:
```
``` 1181 "setsum f A / (r::'a::field) = (\<Sum>n\<in>A. f n / r)"
```
``` 1182 proof (cases "finite A")
```
``` 1183 case True
```
``` 1184 then show ?thesis
```
``` 1185 proof induct
```
``` 1186 case empty thus ?case by simp
```
``` 1187 next
```
``` 1188 case (insert x A) thus ?case by (simp add: add_divide_distrib)
```
``` 1189 qed
```
``` 1190 next
```
``` 1191 case False thus ?thesis by (simp add: setsum_def)
```
``` 1192 qed
```
``` 1193
```
``` 1194 lemma setsum_abs[iff]:
```
``` 1195 fixes f :: "'a => ('b::lordered_ab_group_abs)"
```
``` 1196 shows "abs (setsum f A) \<le> setsum (%i. abs(f i)) A"
```
``` 1197 proof (cases "finite A")
```
``` 1198 case True
```
``` 1199 thus ?thesis
```
``` 1200 proof (induct)
```
``` 1201 case empty thus ?case by simp
```
``` 1202 next
```
``` 1203 case (insert x A)
```
``` 1204 thus ?case by (auto intro: abs_triangle_ineq order_trans)
```
``` 1205 qed
```
``` 1206 next
```
``` 1207 case False thus ?thesis by (simp add: setsum_def)
```
``` 1208 qed
```
``` 1209
```
``` 1210 lemma setsum_abs_ge_zero[iff]:
```
``` 1211 fixes f :: "'a => ('b::lordered_ab_group_abs)"
```
``` 1212 shows "0 \<le> setsum (%i. abs(f i)) A"
```
``` 1213 proof (cases "finite A")
```
``` 1214 case True
```
``` 1215 thus ?thesis
```
``` 1216 proof (induct)
```
``` 1217 case empty thus ?case by simp
```
``` 1218 next
```
``` 1219 case (insert x A) thus ?case by (auto intro: order_trans)
```
``` 1220 qed
```
``` 1221 next
```
``` 1222 case False thus ?thesis by (simp add: setsum_def)
```
``` 1223 qed
```
``` 1224
```
``` 1225 lemma abs_setsum_abs[simp]:
```
``` 1226 fixes f :: "'a => ('b::lordered_ab_group_abs)"
```
``` 1227 shows "abs (\<Sum>a\<in>A. abs(f a)) = (\<Sum>a\<in>A. abs(f a))"
```
``` 1228 proof (cases "finite A")
```
``` 1229 case True
```
``` 1230 thus ?thesis
```
``` 1231 proof (induct)
```
``` 1232 case empty thus ?case by simp
```
``` 1233 next
```
``` 1234 case (insert a A)
```
``` 1235 hence "\<bar>\<Sum>a\<in>insert a A. \<bar>f a\<bar>\<bar> = \<bar>\<bar>f a\<bar> + (\<Sum>a\<in>A. \<bar>f a\<bar>)\<bar>" by simp
```
``` 1236 also have "\<dots> = \<bar>\<bar>f a\<bar> + \<bar>\<Sum>a\<in>A. \<bar>f a\<bar>\<bar>\<bar>" using insert by simp
```
``` 1237 also have "\<dots> = \<bar>f a\<bar> + \<bar>\<Sum>a\<in>A. \<bar>f a\<bar>\<bar>"
```
``` 1238 by (simp del: abs_of_nonneg)
```
``` 1239 also have "\<dots> = (\<Sum>a\<in>insert a A. \<bar>f a\<bar>)" using insert by simp
```
``` 1240 finally show ?case .
```
``` 1241 qed
```
``` 1242 next
```
``` 1243 case False thus ?thesis by (simp add: setsum_def)
```
``` 1244 qed
```
``` 1245
```
``` 1246
```
``` 1247 text {* Commuting outer and inner summation *}
```
``` 1248
```
``` 1249 lemma swap_inj_on:
```
``` 1250 "inj_on (%(i, j). (j, i)) (A \<times> B)"
```
``` 1251 by (unfold inj_on_def) fast
```
``` 1252
```
``` 1253 lemma swap_product:
```
``` 1254 "(%(i, j). (j, i)) ` (A \<times> B) = B \<times> A"
```
``` 1255 by (simp add: split_def image_def) blast
```
``` 1256
```
``` 1257 lemma setsum_commute:
```
``` 1258 "(\<Sum>i\<in>A. \<Sum>j\<in>B. f i j) = (\<Sum>j\<in>B. \<Sum>i\<in>A. f i j)"
```
``` 1259 proof (simp add: setsum_cartesian_product)
```
``` 1260 have "(\<Sum>(x,y) \<in> A <*> B. f x y) =
```
``` 1261 (\<Sum>(y,x) \<in> (%(i, j). (j, i)) ` (A \<times> B). f x y)"
```
``` 1262 (is "?s = _")
```
``` 1263 apply (simp add: setsum_reindex [where f = "%(i, j). (j, i)"] swap_inj_on)
```
``` 1264 apply (simp add: split_def)
```
``` 1265 done
```
``` 1266 also have "... = (\<Sum>(y,x)\<in>B \<times> A. f x y)"
```
``` 1267 (is "_ = ?t")
```
``` 1268 apply (simp add: swap_product)
```
``` 1269 done
```
``` 1270 finally show "?s = ?t" .
```
``` 1271 qed
```
``` 1272
```
``` 1273
```
``` 1274 subsection {* Generalized product over a set *}
```
``` 1275
```
``` 1276 constdefs
```
``` 1277 setprod :: "('a => 'b) => 'a set => 'b::comm_monoid_mult"
```
``` 1278 "setprod f A == if finite A then fold (op *) f 1 A else 1"
```
``` 1279
```
``` 1280 syntax
```
``` 1281 "_setprod" :: "pttrn => 'a set => 'b => 'b::comm_monoid_mult" ("(3PROD _:_. _)" [0, 51, 10] 10)
```
``` 1282 syntax (xsymbols)
```
``` 1283 "_setprod" :: "pttrn => 'a set => 'b => 'b::comm_monoid_mult" ("(3\<Prod>_\<in>_. _)" [0, 51, 10] 10)
```
``` 1284 syntax (HTML output)
```
``` 1285 "_setprod" :: "pttrn => 'a set => 'b => 'b::comm_monoid_mult" ("(3\<Prod>_\<in>_. _)" [0, 51, 10] 10)
```
``` 1286
```
``` 1287 translations -- {* Beware of argument permutation! *}
```
``` 1288 "PROD i:A. b" == "setprod (%i. b) A"
```
``` 1289 "\<Prod>i\<in>A. b" == "setprod (%i. b) A"
```
``` 1290
```
``` 1291 text{* Instead of @{term"\<Prod>x\<in>{x. P}. e"} we introduce the shorter
```
``` 1292 @{text"\<Prod>x|P. e"}. *}
```
``` 1293
```
``` 1294 syntax
```
``` 1295 "_qsetprod" :: "pttrn \<Rightarrow> bool \<Rightarrow> 'a \<Rightarrow> 'a" ("(3PROD _ |/ _./ _)" [0,0,10] 10)
```
``` 1296 syntax (xsymbols)
```
``` 1297 "_qsetprod" :: "pttrn \<Rightarrow> bool \<Rightarrow> 'a \<Rightarrow> 'a" ("(3\<Prod>_ | (_)./ _)" [0,0,10] 10)
```
``` 1298 syntax (HTML output)
```
``` 1299 "_qsetprod" :: "pttrn \<Rightarrow> bool \<Rightarrow> 'a \<Rightarrow> 'a" ("(3\<Prod>_ | (_)./ _)" [0,0,10] 10)
```
``` 1300
```
``` 1301 translations
```
``` 1302 "PROD x|P. t" => "setprod (%x. t) {x. P}"
```
``` 1303 "\<Prod>x|P. t" => "setprod (%x. t) {x. P}"
```
``` 1304
```
``` 1305 text{* Finally we abbreviate @{term"\<Prod>x\<in>A. x"} by @{text"\<Prod>A"}. *}
```
``` 1306
```
``` 1307 syntax
```
``` 1308 "_Setprod" :: "'a set => 'a::comm_monoid_mult" ("\<Prod>_" [1000] 999)
```
``` 1309
```
``` 1310 parse_translation {*
```
``` 1311 let
```
``` 1312 fun Setprod_tr [A] = Syntax.const "setprod" \$ Abs ("", dummyT, Bound 0) \$ A
```
``` 1313 in [("_Setprod", Setprod_tr)] end;
```
``` 1314 *}
```
``` 1315 print_translation {*
```
``` 1316 let fun setprod_tr' [Abs(x,Tx,t), A] =
```
``` 1317 if t = Bound 0 then Syntax.const "_Setprod" \$ A else raise Match
```
``` 1318 in
```
``` 1319 [("setprod", setprod_tr')]
```
``` 1320 end
```
``` 1321 *}
```
``` 1322
```
``` 1323
```
``` 1324 lemma setprod_empty [simp]: "setprod f {} = 1"
```
``` 1325 by (auto simp add: setprod_def)
```
``` 1326
```
``` 1327 lemma setprod_insert [simp]: "[| finite A; a \<notin> A |] ==>
```
``` 1328 setprod f (insert a A) = f a * setprod f A"
```
``` 1329 by (simp add: setprod_def)
```
``` 1330
```
``` 1331 lemma setprod_infinite [simp]: "~ finite A ==> setprod f A = 1"
```
``` 1332 by (simp add: setprod_def)
```
``` 1333
```
``` 1334 lemma setprod_reindex:
```
``` 1335 "inj_on f B ==> setprod h (f ` B) = setprod (h \<circ> f) B"
```
``` 1336 by(auto simp: setprod_def AC_mult.fold_reindex dest!:finite_imageD)
```
``` 1337
```
``` 1338 lemma setprod_reindex_id: "inj_on f B ==> setprod f B = setprod id (f ` B)"
```
``` 1339 by (auto simp add: setprod_reindex)
```
``` 1340
```
``` 1341 lemma setprod_cong:
```
``` 1342 "A = B ==> (!!x. x:B ==> f x = g x) ==> setprod f A = setprod g B"
```
``` 1343 by(fastsimp simp: setprod_def intro: AC_mult.fold_cong)
```
``` 1344
```
``` 1345 lemma strong_setprod_cong:
```
``` 1346 "A = B ==> (!!x. x:B =simp=> f x = g x) ==> setprod f A = setprod g B"
```
``` 1347 by(fastsimp simp: simp_implies_def setprod_def intro: AC_mult.fold_cong)
```
``` 1348
```
``` 1349 lemma setprod_reindex_cong: "inj_on f A ==>
```
``` 1350 B = f ` A ==> g = h \<circ> f ==> setprod h B = setprod g A"
```
``` 1351 by (frule setprod_reindex, simp)
```
``` 1352
```
``` 1353
```
``` 1354 lemma setprod_1: "setprod (%i. 1) A = 1"
```
``` 1355 apply (case_tac "finite A")
```
``` 1356 apply (erule finite_induct, auto simp add: mult_ac)
```
``` 1357 done
```
``` 1358
```
``` 1359 lemma setprod_1': "ALL a:F. f a = 1 ==> setprod f F = 1"
```
``` 1360 apply (subgoal_tac "setprod f F = setprod (%x. 1) F")
```
``` 1361 apply (erule ssubst, rule setprod_1)
```
``` 1362 apply (rule setprod_cong, auto)
```
``` 1363 done
```
``` 1364
```
``` 1365 lemma setprod_Un_Int: "finite A ==> finite B
```
``` 1366 ==> setprod g (A Un B) * setprod g (A Int B) = setprod g A * setprod g B"
```
``` 1367 by(simp add: setprod_def AC_mult.fold_Un_Int[symmetric])
```
``` 1368
```
``` 1369 lemma setprod_Un_disjoint: "finite A ==> finite B
```
``` 1370 ==> A Int B = {} ==> setprod g (A Un B) = setprod g A * setprod g B"
```
``` 1371 by (subst setprod_Un_Int [symmetric], auto)
```
``` 1372
```
``` 1373 lemma setprod_UN_disjoint:
```
``` 1374 "finite I ==> (ALL i:I. finite (A i)) ==>
```
``` 1375 (ALL i:I. ALL j:I. i \<noteq> j --> A i Int A j = {}) ==>
```
``` 1376 setprod f (UNION I A) = setprod (%i. setprod f (A i)) I"
```
``` 1377 by(simp add: setprod_def AC_mult.fold_UN_disjoint cong: setprod_cong)
```
``` 1378
```
``` 1379 lemma setprod_Union_disjoint:
```
``` 1380 "[| (ALL A:C. finite A);
```
``` 1381 (ALL A:C. ALL B:C. A \<noteq> B --> A Int B = {}) |]
```
``` 1382 ==> setprod f (Union C) = setprod (setprod f) C"
```
``` 1383 apply (cases "finite C")
```
``` 1384 prefer 2 apply (force dest: finite_UnionD simp add: setprod_def)
```
``` 1385 apply (frule setprod_UN_disjoint [of C id f])
```
``` 1386 apply (unfold Union_def id_def, assumption+)
```
``` 1387 done
```
``` 1388
```
``` 1389 lemma setprod_Sigma: "finite A ==> ALL x:A. finite (B x) ==>
```
``` 1390 (\<Prod>x\<in>A. (\<Prod>y\<in> B x. f x y)) =
```
``` 1391 (\<Prod>(x,y)\<in>(SIGMA x:A. B x). f x y)"
```
``` 1392 by(simp add:setprod_def AC_mult.fold_Sigma split_def cong:setprod_cong)
```
``` 1393
```
``` 1394 text{*Here we can eliminate the finiteness assumptions, by cases.*}
```
``` 1395 lemma setprod_cartesian_product:
```
``` 1396 "(\<Prod>x\<in>A. (\<Prod>y\<in> B. f x y)) = (\<Prod>(x,y)\<in>(A <*> B). f x y)"
```
``` 1397 apply (cases "finite A")
```
``` 1398 apply (cases "finite B")
```
``` 1399 apply (simp add: setprod_Sigma)
```
``` 1400 apply (cases "A={}", simp)
```
``` 1401 apply (simp add: setprod_1)
```
``` 1402 apply (auto simp add: setprod_def
```
``` 1403 dest: finite_cartesian_productD1 finite_cartesian_productD2)
```
``` 1404 done
```
``` 1405
```
``` 1406 lemma setprod_timesf:
```
``` 1407 "setprod (%x. f x * g x) A = (setprod f A * setprod g A)"
```
``` 1408 by(simp add:setprod_def AC_mult.fold_distrib)
```
``` 1409
```
``` 1410
```
``` 1411 subsubsection {* Properties in more restricted classes of structures *}
```
``` 1412
```
``` 1413 lemma setprod_eq_1_iff [simp]:
```
``` 1414 "finite F ==> (setprod f F = 1) = (ALL a:F. f a = (1::nat))"
```
``` 1415 by (induct set: Finites) auto
```
``` 1416
```
``` 1417 lemma setprod_zero:
```
``` 1418 "finite A ==> EX x: A. f x = (0::'a::comm_semiring_1_cancel) ==> setprod f A = 0"
```
``` 1419 apply (induct set: Finites, force, clarsimp)
```
``` 1420 apply (erule disjE, auto)
```
``` 1421 done
```
``` 1422
```
``` 1423 lemma setprod_nonneg [rule_format]:
```
``` 1424 "(ALL x: A. (0::'a::ordered_idom) \<le> f x) --> 0 \<le> setprod f A"
```
``` 1425 apply (case_tac "finite A")
```
``` 1426 apply (induct set: Finites, force, clarsimp)
```
``` 1427 apply (subgoal_tac "0 * 0 \<le> f x * setprod f F", force)
```
``` 1428 apply (rule mult_mono, assumption+)
```
``` 1429 apply (auto simp add: setprod_def)
```
``` 1430 done
```
``` 1431
```
``` 1432 lemma setprod_pos [rule_format]: "(ALL x: A. (0::'a::ordered_idom) < f x)
```
``` 1433 --> 0 < setprod f A"
```
``` 1434 apply (case_tac "finite A")
```
``` 1435 apply (induct set: Finites, force, clarsimp)
```
``` 1436 apply (subgoal_tac "0 * 0 < f x * setprod f F", force)
```
``` 1437 apply (rule mult_strict_mono, assumption+)
```
``` 1438 apply (auto simp add: setprod_def)
```
``` 1439 done
```
``` 1440
```
``` 1441 lemma setprod_nonzero [rule_format]:
```
``` 1442 "(ALL x y. (x::'a::comm_semiring_1_cancel) * y = 0 --> x = 0 | y = 0) ==>
```
``` 1443 finite A ==> (ALL x: A. f x \<noteq> (0::'a)) --> setprod f A \<noteq> 0"
```
``` 1444 apply (erule finite_induct, auto)
```
``` 1445 done
```
``` 1446
```
``` 1447 lemma setprod_zero_eq:
```
``` 1448 "(ALL x y. (x::'a::comm_semiring_1_cancel) * y = 0 --> x = 0 | y = 0) ==>
```
``` 1449 finite A ==> (setprod f A = (0::'a)) = (EX x: A. f x = 0)"
```
``` 1450 apply (insert setprod_zero [of A f] setprod_nonzero [of A f], blast)
```
``` 1451 done
```
``` 1452
```
``` 1453 lemma setprod_nonzero_field:
```
``` 1454 "finite A ==> (ALL x: A. f x \<noteq> (0::'a::field)) ==> setprod f A \<noteq> 0"
```
``` 1455 apply (rule setprod_nonzero, auto)
```
``` 1456 done
```
``` 1457
```
``` 1458 lemma setprod_zero_eq_field:
```
``` 1459 "finite A ==> (setprod f A = (0::'a::field)) = (EX x: A. f x = 0)"
```
``` 1460 apply (rule setprod_zero_eq, auto)
```
``` 1461 done
```
``` 1462
```
``` 1463 lemma setprod_Un: "finite A ==> finite B ==> (ALL x: A Int B. f x \<noteq> 0) ==>
```
``` 1464 (setprod f (A Un B) :: 'a ::{field})
```
``` 1465 = setprod f A * setprod f B / setprod f (A Int B)"
```
``` 1466 apply (subst setprod_Un_Int [symmetric], auto)
```
``` 1467 apply (subgoal_tac "finite (A Int B)")
```
``` 1468 apply (frule setprod_nonzero_field [of "A Int B" f], assumption)
```
``` 1469 apply (subst times_divide_eq_right [THEN sym], auto simp add: divide_self)
```
``` 1470 done
```
``` 1471
```
``` 1472 lemma setprod_diff1: "finite A ==> f a \<noteq> 0 ==>
```
``` 1473 (setprod f (A - {a}) :: 'a :: {field}) =
```
``` 1474 (if a:A then setprod f A / f a else setprod f A)"
```
``` 1475 apply (erule finite_induct)
```
``` 1476 apply (auto simp add: insert_Diff_if)
```
``` 1477 apply (subgoal_tac "f a * setprod f F / f a = setprod f F * f a / f a")
```
``` 1478 apply (erule ssubst)
```
``` 1479 apply (subst times_divide_eq_right [THEN sym])
```
``` 1480 apply (auto simp add: mult_ac times_divide_eq_right divide_self)
```
``` 1481 done
```
``` 1482
```
``` 1483 lemma setprod_inversef: "finite A ==>
```
``` 1484 ALL x: A. f x \<noteq> (0::'a::{field,division_by_zero}) ==>
```
``` 1485 setprod (inverse \<circ> f) A = inverse (setprod f A)"
```
``` 1486 apply (erule finite_induct)
```
``` 1487 apply (simp, simp)
```
``` 1488 done
```
``` 1489
```
``` 1490 lemma setprod_dividef:
```
``` 1491 "[|finite A;
```
``` 1492 \<forall>x \<in> A. g x \<noteq> (0::'a::{field,division_by_zero})|]
```
``` 1493 ==> setprod (%x. f x / g x) A = setprod f A / setprod g A"
```
``` 1494 apply (subgoal_tac
```
``` 1495 "setprod (%x. f x / g x) A = setprod (%x. f x * (inverse \<circ> g) x) A")
```
``` 1496 apply (erule ssubst)
```
``` 1497 apply (subst divide_inverse)
```
``` 1498 apply (subst setprod_timesf)
```
``` 1499 apply (subst setprod_inversef, assumption+, rule refl)
```
``` 1500 apply (rule setprod_cong, rule refl)
```
``` 1501 apply (subst divide_inverse, auto)
```
``` 1502 done
```
``` 1503
```
``` 1504 subsection {* Finite cardinality *}
```
``` 1505
```
``` 1506 text {* This definition, although traditional, is ugly to work with:
```
``` 1507 @{text "card A == LEAST n. EX f. A = {f i | i. i < n}"}.
```
``` 1508 But now that we have @{text setsum} things are easy:
```
``` 1509 *}
```
``` 1510
```
``` 1511 constdefs
```
``` 1512 card :: "'a set => nat"
```
``` 1513 "card A == setsum (%x. 1::nat) A"
```
``` 1514
```
``` 1515 lemma card_empty [simp]: "card {} = 0"
```
``` 1516 by (simp add: card_def)
```
``` 1517
```
``` 1518 lemma card_infinite [simp]: "~ finite A ==> card A = 0"
```
``` 1519 by (simp add: card_def)
```
``` 1520
```
``` 1521 lemma card_eq_setsum: "card A = setsum (%x. 1) A"
```
``` 1522 by (simp add: card_def)
```
``` 1523
```
``` 1524 lemma card_insert_disjoint [simp]:
```
``` 1525 "finite A ==> x \<notin> A ==> card (insert x A) = Suc(card A)"
```
``` 1526 by(simp add: card_def)
```
``` 1527
```
``` 1528 lemma card_insert_if:
```
``` 1529 "finite A ==> card (insert x A) = (if x:A then card A else Suc(card(A)))"
```
``` 1530 by (simp add: insert_absorb)
```
``` 1531
```
``` 1532 lemma card_0_eq [simp]: "finite A ==> (card A = 0) = (A = {})"
```
``` 1533 apply auto
```
``` 1534 apply (drule_tac a = x in mk_disjoint_insert, clarify, auto)
```
``` 1535 done
```
``` 1536
```
``` 1537 lemma card_eq_0_iff: "(card A = 0) = (A = {} | ~ finite A)"
```
``` 1538 by auto
```
``` 1539
```
``` 1540 lemma card_Suc_Diff1: "finite A ==> x: A ==> Suc (card (A - {x})) = card A"
```
``` 1541 apply(rule_tac t = A in insert_Diff [THEN subst], assumption)
```
``` 1542 apply(simp del:insert_Diff_single)
```
``` 1543 done
```
``` 1544
```
``` 1545 lemma card_Diff_singleton:
```
``` 1546 "finite A ==> x: A ==> card (A - {x}) = card A - 1"
```
``` 1547 by (simp add: card_Suc_Diff1 [symmetric])
```
``` 1548
```
``` 1549 lemma card_Diff_singleton_if:
```
``` 1550 "finite A ==> card (A-{x}) = (if x : A then card A - 1 else card A)"
```
``` 1551 by (simp add: card_Diff_singleton)
```
``` 1552
```
``` 1553 lemma card_insert: "finite A ==> card (insert x A) = Suc (card (A - {x}))"
```
``` 1554 by (simp add: card_insert_if card_Suc_Diff1)
```
``` 1555
```
``` 1556 lemma card_insert_le: "finite A ==> card A <= card (insert x A)"
```
``` 1557 by (simp add: card_insert_if)
```
``` 1558
```
``` 1559 lemma card_mono: "\<lbrakk> finite B; A \<subseteq> B \<rbrakk> \<Longrightarrow> card A \<le> card B"
```
``` 1560 by (simp add: card_def setsum_mono2)
```
``` 1561
```
``` 1562 lemma card_seteq: "finite B ==> (!!A. A <= B ==> card B <= card A ==> A = B)"
```
``` 1563 apply (induct set: Finites, simp, clarify)
```
``` 1564 apply (subgoal_tac "finite A & A - {x} <= F")
```
``` 1565 prefer 2 apply (blast intro: finite_subset, atomize)
```
``` 1566 apply (drule_tac x = "A - {x}" in spec)
```
``` 1567 apply (simp add: card_Diff_singleton_if split add: split_if_asm)
```
``` 1568 apply (case_tac "card A", auto)
```
``` 1569 done
```
``` 1570
```
``` 1571 lemma psubset_card_mono: "finite B ==> A < B ==> card A < card B"
```
``` 1572 apply (simp add: psubset_def linorder_not_le [symmetric])
```
``` 1573 apply (blast dest: card_seteq)
```
``` 1574 done
```
``` 1575
```
``` 1576 lemma card_Un_Int: "finite A ==> finite B
```
``` 1577 ==> card A + card B = card (A Un B) + card (A Int B)"
```
``` 1578 by(simp add:card_def setsum_Un_Int)
```
``` 1579
```
``` 1580 lemma card_Un_disjoint: "finite A ==> finite B
```
``` 1581 ==> A Int B = {} ==> card (A Un B) = card A + card B"
```
``` 1582 by (simp add: card_Un_Int)
```
``` 1583
```
``` 1584 lemma card_Diff_subset:
```
``` 1585 "finite B ==> B <= A ==> card (A - B) = card A - card B"
```
``` 1586 by(simp add:card_def setsum_diff_nat)
```
``` 1587
```
``` 1588 lemma card_Diff1_less: "finite A ==> x: A ==> card (A - {x}) < card A"
```
``` 1589 apply (rule Suc_less_SucD)
```
``` 1590 apply (simp add: card_Suc_Diff1)
```
``` 1591 done
```
``` 1592
```
``` 1593 lemma card_Diff2_less:
```
``` 1594 "finite A ==> x: A ==> y: A ==> card (A - {x} - {y}) < card A"
```
``` 1595 apply (case_tac "x = y")
```
``` 1596 apply (simp add: card_Diff1_less)
```
``` 1597 apply (rule less_trans)
```
``` 1598 prefer 2 apply (auto intro!: card_Diff1_less)
```
``` 1599 done
```
``` 1600
```
``` 1601 lemma card_Diff1_le: "finite A ==> card (A - {x}) <= card A"
```
``` 1602 apply (case_tac "x : A")
```
``` 1603 apply (simp_all add: card_Diff1_less less_imp_le)
```
``` 1604 done
```
``` 1605
```
``` 1606 lemma card_psubset: "finite B ==> A \<subseteq> B ==> card A < card B ==> A < B"
```
``` 1607 by (erule psubsetI, blast)
```
``` 1608
```
``` 1609 lemma insert_partition:
```
``` 1610 "\<lbrakk> x \<notin> F; \<forall>c1 \<in> insert x F. \<forall>c2 \<in> insert x F. c1 \<noteq> c2 \<longrightarrow> c1 \<inter> c2 = {} \<rbrakk>
```
``` 1611 \<Longrightarrow> x \<inter> \<Union> F = {}"
```
``` 1612 by auto
```
``` 1613
```
``` 1614 (* main cardinality theorem *)
```
``` 1615 lemma card_partition [rule_format]:
```
``` 1616 "finite C ==>
```
``` 1617 finite (\<Union> C) -->
```
``` 1618 (\<forall>c\<in>C. card c = k) -->
```
``` 1619 (\<forall>c1 \<in> C. \<forall>c2 \<in> C. c1 \<noteq> c2 --> c1 \<inter> c2 = {}) -->
```
``` 1620 k * card(C) = card (\<Union> C)"
```
``` 1621 apply (erule finite_induct, simp)
```
``` 1622 apply (simp add: card_insert_disjoint card_Un_disjoint insert_partition
```
``` 1623 finite_subset [of _ "\<Union> (insert x F)"])
```
``` 1624 done
```
``` 1625
```
``` 1626
```
``` 1627 lemma setsum_constant [simp]: "(\<Sum>x \<in> A. y) = of_nat(card A) * y"
```
``` 1628 apply (cases "finite A")
```
``` 1629 apply (erule finite_induct)
```
``` 1630 apply (auto simp add: ring_distrib add_ac)
```
``` 1631 done
```
``` 1632
```
``` 1633
```
``` 1634 lemma setprod_constant: "finite A ==> (\<Prod>x\<in> A. (y::'a::recpower)) = y^(card A)"
```
``` 1635 apply (erule finite_induct)
```
``` 1636 apply (auto simp add: power_Suc)
```
``` 1637 done
```
``` 1638
```
``` 1639 lemma setsum_bounded:
```
``` 1640 assumes le: "\<And>i. i\<in>A \<Longrightarrow> f i \<le> (K::'a::{comm_semiring_1_cancel, pordered_ab_semigroup_add})"
```
``` 1641 shows "setsum f A \<le> of_nat(card A) * K"
```
``` 1642 proof (cases "finite A")
```
``` 1643 case True
```
``` 1644 thus ?thesis using le setsum_mono[where K=A and g = "%x. K"] by simp
```
``` 1645 next
```
``` 1646 case False thus ?thesis by (simp add: setsum_def)
```
``` 1647 qed
```
``` 1648
```
``` 1649
```
``` 1650 subsubsection {* Cardinality of unions *}
```
``` 1651
```
``` 1652 lemma of_nat_id[simp]: "(of_nat n :: nat) = n"
```
``` 1653 by(induct n, auto)
```
``` 1654
```
``` 1655 lemma card_UN_disjoint:
```
``` 1656 "finite I ==> (ALL i:I. finite (A i)) ==>
```
``` 1657 (ALL i:I. ALL j:I. i \<noteq> j --> A i Int A j = {}) ==>
```
``` 1658 card (UNION I A) = (\<Sum>i\<in>I. card(A i))"
```
``` 1659 apply (simp add: card_def del: setsum_constant)
```
``` 1660 apply (subgoal_tac
```
``` 1661 "setsum (%i. card (A i)) I = setsum (%i. (setsum (%x. 1) (A i))) I")
```
``` 1662 apply (simp add: setsum_UN_disjoint del: setsum_constant)
```
``` 1663 apply (simp cong: setsum_cong)
```
``` 1664 done
```
``` 1665
```
``` 1666 lemma card_Union_disjoint:
```
``` 1667 "finite C ==> (ALL A:C. finite A) ==>
```
``` 1668 (ALL A:C. ALL B:C. A \<noteq> B --> A Int B = {}) ==>
```
``` 1669 card (Union C) = setsum card C"
```
``` 1670 apply (frule card_UN_disjoint [of C id])
```
``` 1671 apply (unfold Union_def id_def, assumption+)
```
``` 1672 done
```
``` 1673
```
``` 1674 subsubsection {* Cardinality of image *}
```
``` 1675
```
``` 1676 text{*The image of a finite set can be expressed using @{term fold}.*}
```
``` 1677 lemma image_eq_fold: "finite A ==> f ` A = fold (op Un) (%x. {f x}) {} A"
```
``` 1678 apply (erule finite_induct, simp)
```
``` 1679 apply (subst ACf.fold_insert)
```
``` 1680 apply (auto simp add: ACf_def)
```
``` 1681 done
```
``` 1682
```
``` 1683 lemma card_image_le: "finite A ==> card (f ` A) <= card A"
```
``` 1684 apply (induct set: Finites, simp)
```
``` 1685 apply (simp add: le_SucI finite_imageI card_insert_if)
```
``` 1686 done
```
``` 1687
```
``` 1688 lemma card_image: "inj_on f A ==> card (f ` A) = card A"
```
``` 1689 by(simp add:card_def setsum_reindex o_def del:setsum_constant)
```
``` 1690
```
``` 1691 lemma endo_inj_surj: "finite A ==> f ` A \<subseteq> A ==> inj_on f A ==> f ` A = A"
```
``` 1692 by (simp add: card_seteq card_image)
```
``` 1693
```
``` 1694 lemma eq_card_imp_inj_on:
```
``` 1695 "[| finite A; card(f ` A) = card A |] ==> inj_on f A"
```
``` 1696 apply (induct rule:finite_induct, simp)
```
``` 1697 apply(frule card_image_le[where f = f])
```
``` 1698 apply(simp add:card_insert_if split:if_splits)
```
``` 1699 done
```
``` 1700
```
``` 1701 lemma inj_on_iff_eq_card:
```
``` 1702 "finite A ==> inj_on f A = (card(f ` A) = card A)"
```
``` 1703 by(blast intro: card_image eq_card_imp_inj_on)
```
``` 1704
```
``` 1705
```
``` 1706 lemma card_inj_on_le:
```
``` 1707 "[|inj_on f A; f ` A \<subseteq> B; finite B |] ==> card A \<le> card B"
```
``` 1708 apply (subgoal_tac "finite A")
```
``` 1709 apply (force intro: card_mono simp add: card_image [symmetric])
```
``` 1710 apply (blast intro: finite_imageD dest: finite_subset)
```
``` 1711 done
```
``` 1712
```
``` 1713 lemma card_bij_eq:
```
``` 1714 "[|inj_on f A; f ` A \<subseteq> B; inj_on g B; g ` B \<subseteq> A;
```
``` 1715 finite A; finite B |] ==> card A = card B"
```
``` 1716 by (auto intro: le_anti_sym card_inj_on_le)
```
``` 1717
```
``` 1718
```
``` 1719 subsubsection {* Cardinality of products *}
```
``` 1720
```
``` 1721 (*
```
``` 1722 lemma SigmaI_insert: "y \<notin> A ==>
```
``` 1723 (SIGMA x:(insert y A). B x) = (({y} <*> (B y)) \<union> (SIGMA x: A. B x))"
```
``` 1724 by auto
```
``` 1725 *)
```
``` 1726
```
``` 1727 lemma card_SigmaI [simp]:
```
``` 1728 "\<lbrakk> finite A; ALL a:A. finite (B a) \<rbrakk>
```
``` 1729 \<Longrightarrow> card (SIGMA x: A. B x) = (\<Sum>a\<in>A. card (B a))"
```
``` 1730 by(simp add:card_def setsum_Sigma del:setsum_constant)
```
``` 1731
```
``` 1732 lemma card_cartesian_product: "card (A <*> B) = card(A) * card(B)"
```
``` 1733 apply (cases "finite A")
```
``` 1734 apply (cases "finite B")
```
``` 1735 apply (auto simp add: card_eq_0_iff
```
``` 1736 dest: finite_cartesian_productD1 finite_cartesian_productD2)
```
``` 1737 done
```
``` 1738
```
``` 1739 lemma card_cartesian_product_singleton: "card({x} <*> A) = card(A)"
```
``` 1740 by (simp add: card_cartesian_product)
```
``` 1741
```
``` 1742
```
``` 1743
```
``` 1744 subsubsection {* Cardinality of the Powerset *}
```
``` 1745
```
``` 1746 lemma card_Pow: "finite A ==> card (Pow A) = Suc (Suc 0) ^ card A" (* FIXME numeral 2 (!?) *)
```
``` 1747 apply (induct set: Finites)
```
``` 1748 apply (simp_all add: Pow_insert)
```
``` 1749 apply (subst card_Un_disjoint, blast)
```
``` 1750 apply (blast intro: finite_imageI, blast)
```
``` 1751 apply (subgoal_tac "inj_on (insert x) (Pow F)")
```
``` 1752 apply (simp add: card_image Pow_insert)
```
``` 1753 apply (unfold inj_on_def)
```
``` 1754 apply (blast elim!: equalityE)
```
``` 1755 done
```
``` 1756
```
``` 1757 text {* Relates to equivalence classes. Based on a theorem of
```
``` 1758 F. Kammüller's. *}
```
``` 1759
```
``` 1760 lemma dvd_partition:
```
``` 1761 "finite (Union C) ==>
```
``` 1762 ALL c : C. k dvd card c ==>
```
``` 1763 (ALL c1: C. ALL c2: C. c1 \<noteq> c2 --> c1 Int c2 = {}) ==>
```
``` 1764 k dvd card (Union C)"
```
``` 1765 apply(frule finite_UnionD)
```
``` 1766 apply(rotate_tac -1)
```
``` 1767 apply (induct set: Finites, simp_all, clarify)
```
``` 1768 apply (subst card_Un_disjoint)
```
``` 1769 apply (auto simp add: dvd_add disjoint_eq_subset_Compl)
```
``` 1770 done
```
``` 1771
```
``` 1772
```
``` 1773 subsection{* A fold functional for non-empty sets *}
```
``` 1774
```
``` 1775 text{* Does not require start value. *}
```
``` 1776
```
``` 1777 consts
```
``` 1778 fold1Set :: "('a => 'a => 'a) => ('a set \<times> 'a) set"
```
``` 1779
```
``` 1780 inductive "fold1Set f"
```
``` 1781 intros
```
``` 1782 fold1Set_insertI [intro]:
```
``` 1783 "\<lbrakk> (A,x) \<in> foldSet f id a; a \<notin> A \<rbrakk> \<Longrightarrow> (insert a A, x) \<in> fold1Set f"
```
``` 1784
```
``` 1785 constdefs
```
``` 1786 fold1 :: "('a => 'a => 'a) => 'a set => 'a"
```
``` 1787 "fold1 f A == THE x. (A, x) : fold1Set f"
```
``` 1788
```
``` 1789 lemma fold1Set_nonempty:
```
``` 1790 "(A, x) : fold1Set f \<Longrightarrow> A \<noteq> {}"
```
``` 1791 by(erule fold1Set.cases, simp_all)
```
``` 1792
```
``` 1793
```
``` 1794 inductive_cases empty_fold1SetE [elim!]: "({}, x) : fold1Set f"
```
``` 1795
```
``` 1796 inductive_cases insert_fold1SetE [elim!]: "(insert a X, x) : fold1Set f"
```
``` 1797
```
``` 1798
```
``` 1799 lemma fold1Set_sing [iff]: "(({a},b) : fold1Set f) = (a = b)"
```
``` 1800 by (blast intro: foldSet.intros elim: foldSet.cases)
```
``` 1801
```
``` 1802 lemma fold1_singleton[simp]: "fold1 f {a} = a"
```
``` 1803 by (unfold fold1_def) blast
```
``` 1804
```
``` 1805 lemma finite_nonempty_imp_fold1Set:
```
``` 1806 "\<lbrakk> finite A; A \<noteq> {} \<rbrakk> \<Longrightarrow> EX x. (A, x) : fold1Set f"
```
``` 1807 apply (induct A rule: finite_induct)
```
``` 1808 apply (auto dest: finite_imp_foldSet [of _ f id])
```
``` 1809 done
```
``` 1810
```
``` 1811 text{*First, some lemmas about @{term foldSet}.*}
```
``` 1812
```
``` 1813 lemma (in ACf) foldSet_insert_swap:
```
``` 1814 assumes fold: "(A,y) \<in> foldSet f id b"
```
``` 1815 shows "b \<notin> A \<Longrightarrow> (insert b A, z \<cdot> y) \<in> foldSet f id z"
```
``` 1816 using fold
```
``` 1817 proof (induct rule: foldSet.induct)
```
``` 1818 case emptyI thus ?case by (force simp add: fold_insert_aux commute)
```
``` 1819 next
```
``` 1820 case (insertI A x y)
```
``` 1821 have "(insert x (insert b A), x \<cdot> (z \<cdot> y)) \<in> foldSet f (\<lambda>u. u) z"
```
``` 1822 using insertI by force --{*how does @{term id} get unfolded?*}
```
``` 1823 thus ?case by (simp add: insert_commute AC)
```
``` 1824 qed
```
``` 1825
```
``` 1826 lemma (in ACf) foldSet_permute_diff:
```
``` 1827 assumes fold: "(A,x) \<in> foldSet f id b"
```
``` 1828 shows "!!a. \<lbrakk>a \<in> A; b \<notin> A\<rbrakk> \<Longrightarrow> (insert b (A-{a}), x) \<in> foldSet f id a"
```
``` 1829 using fold
```
``` 1830 proof (induct rule: foldSet.induct)
```
``` 1831 case emptyI thus ?case by simp
```
``` 1832 next
```
``` 1833 case (insertI A x y)
```
``` 1834 have "a = x \<or> a \<in> A" using insertI by simp
```
``` 1835 thus ?case
```
``` 1836 proof
```
``` 1837 assume "a = x"
```
``` 1838 with insertI show ?thesis
```
``` 1839 by (simp add: id_def [symmetric], blast intro: foldSet_insert_swap)
```
``` 1840 next
```
``` 1841 assume ainA: "a \<in> A"
```
``` 1842 hence "(insert x (insert b (A - {a})), x \<cdot> y) \<in> foldSet f id a"
```
``` 1843 using insertI by (force simp: id_def)
```
``` 1844 moreover
```
``` 1845 have "insert x (insert b (A - {a})) = insert b (insert x A - {a})"
```
``` 1846 using ainA insertI by blast
```
``` 1847 ultimately show ?thesis by (simp add: id_def)
```
``` 1848 qed
```
``` 1849 qed
```
``` 1850
```
``` 1851 lemma (in ACf) fold1_eq_fold:
```
``` 1852 "[|finite A; a \<notin> A|] ==> fold1 f (insert a A) = fold f id a A"
```
``` 1853 apply (simp add: fold1_def fold_def)
```
``` 1854 apply (rule the_equality)
```
``` 1855 apply (best intro: foldSet_determ theI dest: finite_imp_foldSet [of _ f id])
```
``` 1856 apply (rule sym, clarify)
```
``` 1857 apply (case_tac "Aa=A")
```
``` 1858 apply (best intro: the_equality foldSet_determ)
```
``` 1859 apply (subgoal_tac "(A,x) \<in> foldSet f id a")
```
``` 1860 apply (best intro: the_equality foldSet_determ)
```
``` 1861 apply (subgoal_tac "insert aa (Aa - {a}) = A")
```
``` 1862 prefer 2 apply (blast elim: equalityE)
```
``` 1863 apply (auto dest: foldSet_permute_diff [where a=a])
```
``` 1864 done
```
``` 1865
```
``` 1866 lemma nonempty_iff: "(A \<noteq> {}) = (\<exists>x B. A = insert x B & x \<notin> B)"
```
``` 1867 apply safe
```
``` 1868 apply simp
```
``` 1869 apply (drule_tac x=x in spec)
```
``` 1870 apply (drule_tac x="A-{x}" in spec, auto)
```
``` 1871 done
```
``` 1872
```
``` 1873 lemma (in ACf) fold1_insert:
```
``` 1874 assumes nonempty: "A \<noteq> {}" and A: "finite A" "x \<notin> A"
```
``` 1875 shows "fold1 f (insert x A) = f x (fold1 f A)"
```
``` 1876 proof -
```
``` 1877 from nonempty obtain a A' where "A = insert a A' & a ~: A'"
```
``` 1878 by (auto simp add: nonempty_iff)
```
``` 1879 with A show ?thesis
```
``` 1880 by (simp add: insert_commute [of x] fold1_eq_fold eq_commute)
```
``` 1881 qed
```
``` 1882
```
``` 1883 lemma (in ACIf) fold1_insert_idem [simp]:
```
``` 1884 assumes nonempty: "A \<noteq> {}" and A: "finite A"
```
``` 1885 shows "fold1 f (insert x A) = f x (fold1 f A)"
```
``` 1886 proof -
```
``` 1887 from nonempty obtain a A' where A': "A = insert a A' & a ~: A'"
```
``` 1888 by (auto simp add: nonempty_iff)
```
``` 1889 show ?thesis
```
``` 1890 proof cases
```
``` 1891 assume "a = x"
```
``` 1892 thus ?thesis
```
``` 1893 proof cases
```
``` 1894 assume "A' = {}"
```
``` 1895 with prems show ?thesis by (simp add: idem)
```
``` 1896 next
```
``` 1897 assume "A' \<noteq> {}"
```
``` 1898 with prems show ?thesis
```
``` 1899 by (simp add: fold1_insert assoc [symmetric] idem)
```
``` 1900 qed
```
``` 1901 next
```
``` 1902 assume "a \<noteq> x"
```
``` 1903 with prems show ?thesis
```
``` 1904 by (simp add: insert_commute fold1_eq_fold fold_insert_idem)
```
``` 1905 qed
```
``` 1906 qed
```
``` 1907
```
``` 1908
```
``` 1909 text{* Now the recursion rules for definitions: *}
```
``` 1910
```
``` 1911 lemma fold1_singleton_def: "g \<equiv> fold1 f \<Longrightarrow> g {a} = a"
```
``` 1912 by(simp add:fold1_singleton)
```
``` 1913
```
``` 1914 lemma (in ACf) fold1_insert_def:
```
``` 1915 "\<lbrakk> g \<equiv> fold1 f; finite A; x \<notin> A; A \<noteq> {} \<rbrakk> \<Longrightarrow> g(insert x A) = x \<cdot> (g A)"
```
``` 1916 by(simp add:fold1_insert)
```
``` 1917
```
``` 1918 lemma (in ACIf) fold1_insert_idem_def:
```
``` 1919 "\<lbrakk> g \<equiv> fold1 f; finite A; A \<noteq> {} \<rbrakk> \<Longrightarrow> g(insert x A) = x \<cdot> (g A)"
```
``` 1920 by(simp add:fold1_insert_idem)
```
``` 1921
```
``` 1922 subsubsection{* Determinacy for @{term fold1Set} *}
```
``` 1923
```
``` 1924 text{*Not actually used!!*}
```
``` 1925
```
``` 1926 lemma (in ACf) foldSet_permute:
```
``` 1927 "[|(insert a A, x) \<in> foldSet f id b; a \<notin> A; b \<notin> A|]
```
``` 1928 ==> (insert b A, x) \<in> foldSet f id a"
```
``` 1929 apply (case_tac "a=b")
```
``` 1930 apply (auto dest: foldSet_permute_diff)
```
``` 1931 done
```
``` 1932
```
``` 1933 lemma (in ACf) fold1Set_determ:
```
``` 1934 "(A, x) \<in> fold1Set f ==> (A, y) \<in> fold1Set f ==> y = x"
```
``` 1935 proof (clarify elim!: fold1Set.cases)
```
``` 1936 fix A x B y a b
```
``` 1937 assume Ax: "(A, x) \<in> foldSet f id a"
```
``` 1938 assume By: "(B, y) \<in> foldSet f id b"
```
``` 1939 assume anotA: "a \<notin> A"
```
``` 1940 assume bnotB: "b \<notin> B"
```
``` 1941 assume eq: "insert a A = insert b B"
```
``` 1942 show "y=x"
```
``` 1943 proof cases
```
``` 1944 assume same: "a=b"
```
``` 1945 hence "A=B" using anotA bnotB eq by (blast elim!: equalityE)
```
``` 1946 thus ?thesis using Ax By same by (blast intro: foldSet_determ)
```
``` 1947 next
```
``` 1948 assume diff: "a\<noteq>b"
```
``` 1949 let ?D = "B - {a}"
```
``` 1950 have B: "B = insert a ?D" and A: "A = insert b ?D"
```
``` 1951 and aB: "a \<in> B" and bA: "b \<in> A"
```
``` 1952 using eq anotA bnotB diff by (blast elim!:equalityE)+
```
``` 1953 with aB bnotB By
```
``` 1954 have "(insert b ?D, y) \<in> foldSet f id a"
```
``` 1955 by (auto intro: foldSet_permute simp add: insert_absorb)
```
``` 1956 moreover
```
``` 1957 have "(insert b ?D, x) \<in> foldSet f id a"
```
``` 1958 by (simp add: A [symmetric] Ax)
```
``` 1959 ultimately show ?thesis by (blast intro: foldSet_determ)
```
``` 1960 qed
```
``` 1961 qed
```
``` 1962
```
``` 1963 lemma (in ACf) fold1Set_equality: "(A, y) : fold1Set f ==> fold1 f A = y"
```
``` 1964 by (unfold fold1_def) (blast intro: fold1Set_determ)
```
``` 1965
```
``` 1966 declare
```
``` 1967 empty_foldSetE [rule del] foldSet.intros [rule del]
```
``` 1968 empty_fold1SetE [rule del] insert_fold1SetE [rule del]
```
``` 1969 -- {* No more proves involve these relations. *}
```
``` 1970
```
``` 1971 subsubsection{* Semi-Lattices *}
```
``` 1972
```
``` 1973 locale ACIfSL = ACIf +
```
``` 1974 fixes below :: "'a \<Rightarrow> 'a \<Rightarrow> bool" (infixl "\<sqsubseteq>" 50)
```
``` 1975 assumes below_def: "(x \<sqsubseteq> y) = (x\<cdot>y = x)"
```
``` 1976
```
``` 1977 locale ACIfSLlin = ACIfSL +
```
``` 1978 assumes lin: "x\<cdot>y \<in> {x,y}"
```
``` 1979
```
``` 1980 lemma (in ACIfSL) below_refl[simp]: "x \<sqsubseteq> x"
```
``` 1981 by(simp add: below_def idem)
```
``` 1982
```
``` 1983 lemma (in ACIfSL) below_f_conv[simp]: "x \<sqsubseteq> y \<cdot> z = (x \<sqsubseteq> y \<and> x \<sqsubseteq> z)"
```
``` 1984 proof
```
``` 1985 assume "x \<sqsubseteq> y \<cdot> z"
```
``` 1986 hence xyzx: "x \<cdot> (y \<cdot> z) = x" by(simp add: below_def)
```
``` 1987 have "x \<cdot> y = x"
```
``` 1988 proof -
```
``` 1989 have "x \<cdot> y = (x \<cdot> (y \<cdot> z)) \<cdot> y" by(rule subst[OF xyzx], rule refl)
```
``` 1990 also have "\<dots> = x \<cdot> (y \<cdot> z)" by(simp add:ACI)
```
``` 1991 also have "\<dots> = x" by(rule xyzx)
```
``` 1992 finally show ?thesis .
```
``` 1993 qed
```
``` 1994 moreover have "x \<cdot> z = x"
```
``` 1995 proof -
```
``` 1996 have "x \<cdot> z = (x \<cdot> (y \<cdot> z)) \<cdot> z" by(rule subst[OF xyzx], rule refl)
```
``` 1997 also have "\<dots> = x \<cdot> (y \<cdot> z)" by(simp add:ACI)
```
``` 1998 also have "\<dots> = x" by(rule xyzx)
```
``` 1999 finally show ?thesis .
```
``` 2000 qed
```
``` 2001 ultimately show "x \<sqsubseteq> y \<and> x \<sqsubseteq> z" by(simp add: below_def)
```
``` 2002 next
```
``` 2003 assume a: "x \<sqsubseteq> y \<and> x \<sqsubseteq> z"
```
``` 2004 hence y: "x \<cdot> y = x" and z: "x \<cdot> z = x" by(simp_all add: below_def)
```
``` 2005 have "x \<cdot> (y \<cdot> z) = (x \<cdot> y) \<cdot> z" by(simp add:assoc)
```
``` 2006 also have "x \<cdot> y = x" using a by(simp_all add: below_def)
```
``` 2007 also have "x \<cdot> z = x" using a by(simp_all add: below_def)
```
``` 2008 finally show "x \<sqsubseteq> y \<cdot> z" by(simp_all add: below_def)
```
``` 2009 qed
```
``` 2010
```
``` 2011 lemma (in ACIfSLlin) above_f_conv:
```
``` 2012 "x \<cdot> y \<sqsubseteq> z = (x \<sqsubseteq> z \<or> y \<sqsubseteq> z)"
```
``` 2013 proof
```
``` 2014 assume a: "x \<cdot> y \<sqsubseteq> z"
```
``` 2015 have "x \<cdot> y = x \<or> x \<cdot> y = y" using lin[of x y] by simp
```
``` 2016 thus "x \<sqsubseteq> z \<or> y \<sqsubseteq> z"
```
``` 2017 proof
```
``` 2018 assume "x \<cdot> y = x" hence "x \<sqsubseteq> z" by(rule subst)(rule a) thus ?thesis ..
```
``` 2019 next
```
``` 2020 assume "x \<cdot> y = y" hence "y \<sqsubseteq> z" by(rule subst)(rule a) thus ?thesis ..
```
``` 2021 qed
```
``` 2022 next
```
``` 2023 assume "x \<sqsubseteq> z \<or> y \<sqsubseteq> z"
```
``` 2024 thus "x \<cdot> y \<sqsubseteq> z"
```
``` 2025 proof
```
``` 2026 assume a: "x \<sqsubseteq> z"
```
``` 2027 have "(x \<cdot> y) \<cdot> z = (x \<cdot> z) \<cdot> y" by(simp add:ACI)
```
``` 2028 also have "x \<cdot> z = x" using a by(simp add:below_def)
```
``` 2029 finally show "x \<cdot> y \<sqsubseteq> z" by(simp add:below_def)
```
``` 2030 next
```
``` 2031 assume a: "y \<sqsubseteq> z"
```
``` 2032 have "(x \<cdot> y) \<cdot> z = x \<cdot> (y \<cdot> z)" by(simp add:ACI)
```
``` 2033 also have "y \<cdot> z = y" using a by(simp add:below_def)
```
``` 2034 finally show "x \<cdot> y \<sqsubseteq> z" by(simp add:below_def)
```
``` 2035 qed
```
``` 2036 qed
```
``` 2037
```
``` 2038
```
``` 2039 subsubsection{* Lemmas about @{text fold1} *}
```
``` 2040
```
``` 2041 lemma (in ACf) fold1_Un:
```
``` 2042 assumes A: "finite A" "A \<noteq> {}"
```
``` 2043 shows "finite B \<Longrightarrow> B \<noteq> {} \<Longrightarrow> A Int B = {} \<Longrightarrow>
```
``` 2044 fold1 f (A Un B) = f (fold1 f A) (fold1 f B)"
```
``` 2045 using A
```
``` 2046 proof(induct rule:finite_ne_induct)
```
``` 2047 case singleton thus ?case by(simp add:fold1_insert)
```
``` 2048 next
```
``` 2049 case insert thus ?case by (simp add:fold1_insert assoc)
```
``` 2050 qed
```
``` 2051
```
``` 2052 lemma (in ACIf) fold1_Un2:
```
``` 2053 assumes A: "finite A" "A \<noteq> {}"
```
``` 2054 shows "finite B \<Longrightarrow> B \<noteq> {} \<Longrightarrow>
```
``` 2055 fold1 f (A Un B) = f (fold1 f A) (fold1 f B)"
```
``` 2056 using A
```
``` 2057 proof(induct rule:finite_ne_induct)
```
``` 2058 case singleton thus ?case by(simp add:fold1_insert_idem)
```
``` 2059 next
```
``` 2060 case insert thus ?case by (simp add:fold1_insert_idem assoc)
```
``` 2061 qed
```
``` 2062
```
``` 2063 lemma (in ACf) fold1_in:
```
``` 2064 assumes A: "finite (A)" "A \<noteq> {}" and elem: "\<And>x y. x\<cdot>y \<in> {x,y}"
```
``` 2065 shows "fold1 f A \<in> A"
```
``` 2066 using A
```
``` 2067 proof (induct rule:finite_ne_induct)
```
``` 2068 case singleton thus ?case by simp
```
``` 2069 next
```
``` 2070 case insert thus ?case using elem by (force simp add:fold1_insert)
```
``` 2071 qed
```
``` 2072
```
``` 2073 lemma (in ACIfSL) below_fold1_iff:
```
``` 2074 assumes A: "finite A" "A \<noteq> {}"
```
``` 2075 shows "x \<sqsubseteq> fold1 f A = (\<forall>a\<in>A. x \<sqsubseteq> a)"
```
``` 2076 using A
```
``` 2077 by(induct rule:finite_ne_induct) simp_all
```
``` 2078
```
``` 2079 lemma (in ACIfSL) fold1_belowI:
```
``` 2080 assumes A: "finite A" "A \<noteq> {}"
```
``` 2081 shows "a \<in> A \<Longrightarrow> fold1 f A \<sqsubseteq> a"
```
``` 2082 using A
```
``` 2083 proof (induct rule:finite_ne_induct)
```
``` 2084 case singleton thus ?case by simp
```
``` 2085 next
```
``` 2086 case (insert x F)
```
``` 2087 from insert(5) have "a = x \<or> a \<in> F" by simp
```
``` 2088 thus ?case
```
``` 2089 proof
```
``` 2090 assume "a = x" thus ?thesis using insert by(simp add:below_def ACI)
```
``` 2091 next
```
``` 2092 assume "a \<in> F"
```
``` 2093 hence bel: "fold1 f F \<sqsubseteq> a" by(rule insert)
```
``` 2094 have "fold1 f (insert x F) \<cdot> a = x \<cdot> (fold1 f F \<cdot> a)"
```
``` 2095 using insert by(simp add:below_def ACI)
```
``` 2096 also have "fold1 f F \<cdot> a = fold1 f F"
```
``` 2097 using bel by(simp add:below_def ACI)
```
``` 2098 also have "x \<cdot> \<dots> = fold1 f (insert x F)"
```
``` 2099 using insert by(simp add:below_def ACI)
```
``` 2100 finally show ?thesis by(simp add:below_def)
```
``` 2101 qed
```
``` 2102 qed
```
``` 2103
```
``` 2104 lemma (in ACIfSLlin) fold1_below_iff:
```
``` 2105 assumes A: "finite A" "A \<noteq> {}"
```
``` 2106 shows "fold1 f A \<sqsubseteq> x = (\<exists>a\<in>A. a \<sqsubseteq> x)"
```
``` 2107 using A
```
``` 2108 by(induct rule:finite_ne_induct)(simp_all add:above_f_conv)
```
``` 2109
```
``` 2110
```
``` 2111 subsubsection{* Lattices *}
```
``` 2112
```
``` 2113 locale Lattice = lattice +
```
``` 2114 fixes Inf :: "'a set \<Rightarrow> 'a" ("\<Sqinter>_" [900] 900)
```
``` 2115 and Sup :: "'a set \<Rightarrow> 'a" ("\<Squnion>_" [900] 900)
```
``` 2116 defines "Inf == fold1 inf" and "Sup == fold1 sup"
```
``` 2117
```
``` 2118 locale Distrib_Lattice = distrib_lattice + Lattice
```
``` 2119
```
``` 2120 text{* Lattices are semilattices *}
```
``` 2121
```
``` 2122 lemma (in Lattice) ACf_inf: "ACf inf"
```
``` 2123 by(blast intro: ACf.intro inf_commute inf_assoc)
```
``` 2124
```
``` 2125 lemma (in Lattice) ACf_sup: "ACf sup"
```
``` 2126 by(blast intro: ACf.intro sup_commute sup_assoc)
```
``` 2127
```
``` 2128 lemma (in Lattice) ACIf_inf: "ACIf inf"
```
``` 2129 apply(rule ACIf.intro)
```
``` 2130 apply(rule ACf_inf)
```
``` 2131 apply(rule ACIf_axioms.intro)
```
``` 2132 apply(rule inf_idem)
```
``` 2133 done
```
``` 2134
```
``` 2135 lemma (in Lattice) ACIf_sup: "ACIf sup"
```
``` 2136 apply(rule ACIf.intro)
```
``` 2137 apply(rule ACf_sup)
```
``` 2138 apply(rule ACIf_axioms.intro)
```
``` 2139 apply(rule sup_idem)
```
``` 2140 done
```
``` 2141
```
``` 2142 lemma (in Lattice) ACIfSL_inf: "ACIfSL inf (op \<sqsubseteq>)"
```
``` 2143 apply(rule ACIfSL.intro)
```
``` 2144 apply(rule ACf_inf)
```
``` 2145 apply(rule ACIf.axioms[OF ACIf_inf])
```
``` 2146 apply(rule ACIfSL_axioms.intro)
```
``` 2147 apply(rule iffI)
```
``` 2148 apply(blast intro: antisym inf_le1 inf_le2 inf_least refl)
```
``` 2149 apply(erule subst)
```
``` 2150 apply(rule inf_le2)
```
``` 2151 done
```
``` 2152
```
``` 2153 lemma (in Lattice) ACIfSL_sup: "ACIfSL sup (%x y. y \<sqsubseteq> x)"
```
``` 2154 apply(rule ACIfSL.intro)
```
``` 2155 apply(rule ACf_sup)
```
``` 2156 apply(rule ACIf.axioms[OF ACIf_sup])
```
``` 2157 apply(rule ACIfSL_axioms.intro)
```
``` 2158 apply(rule iffI)
```
``` 2159 apply(blast intro: antisym sup_ge1 sup_ge2 sup_greatest refl)
```
``` 2160 apply(erule subst)
```
``` 2161 apply(rule sup_ge2)
```
``` 2162 done
```
``` 2163
```
``` 2164
```
``` 2165 subsubsection{* Fold laws in lattices *}
```
``` 2166
```
``` 2167 lemma (in Lattice) Inf_le_Sup[simp]: "\<lbrakk> finite A; A \<noteq> {} \<rbrakk> \<Longrightarrow> \<Sqinter>A \<sqsubseteq> \<Squnion>A"
```
``` 2168 apply(unfold Sup_def Inf_def)
```
``` 2169 apply(subgoal_tac "EX a. a:A")
```
``` 2170 prefer 2 apply blast
```
``` 2171 apply(erule exE)
```
``` 2172 apply(rule trans)
```
``` 2173 apply(erule (2) ACIfSL.fold1_belowI[OF ACIfSL_inf])
```
``` 2174 apply(erule (2) ACIfSL.fold1_belowI[OF ACIfSL_sup])
```
``` 2175 done
```
``` 2176
```
``` 2177 lemma (in Lattice) sup_Inf_absorb[simp]:
```
``` 2178 "\<lbrakk> finite A; A \<noteq> {}; a \<in> A \<rbrakk> \<Longrightarrow> (a \<squnion> \<Sqinter>A) = a"
```
``` 2179 apply(subst sup_commute)
```
``` 2180 apply(simp add:Inf_def sup_absorb ACIfSL.fold1_belowI[OF ACIfSL_inf])
```
``` 2181 done
```
``` 2182
```
``` 2183 lemma (in Lattice) inf_Sup_absorb[simp]:
```
``` 2184 "\<lbrakk> finite A; A \<noteq> {}; a \<in> A \<rbrakk> \<Longrightarrow> (a \<sqinter> \<Squnion>A) = a"
```
``` 2185 by(simp add:Sup_def inf_absorb ACIfSL.fold1_belowI[OF ACIfSL_sup])
```
``` 2186
```
``` 2187
```
``` 2188 lemma (in Distrib_Lattice) sup_Inf1_distrib:
```
``` 2189 assumes A: "finite A" "A \<noteq> {}"
```
``` 2190 shows "(x \<squnion> \<Sqinter>A) = \<Sqinter>{x \<squnion> a|a. a \<in> A}"
```
``` 2191 using A
```
``` 2192 proof (induct rule: finite_ne_induct)
```
``` 2193 case singleton thus ?case by(simp add:Inf_def)
```
``` 2194 next
```
``` 2195 case (insert y A)
```
``` 2196 have fin: "finite {x \<squnion> a |a. a \<in> A}"
```
``` 2197 by(fast intro: finite_surj[where f = "%a. x \<squnion> a", OF insert(1)])
```
``` 2198 have "x \<squnion> \<Sqinter> (insert y A) = x \<squnion> (y \<sqinter> \<Sqinter> A)"
```
``` 2199 using insert by(simp add:ACf.fold1_insert_def[OF ACf_inf Inf_def])
```
``` 2200 also have "\<dots> = (x \<squnion> y) \<sqinter> (x \<squnion> \<Sqinter> A)" by(rule sup_inf_distrib1)
```
``` 2201 also have "x \<squnion> \<Sqinter> A = \<Sqinter>{x \<squnion> a|a. a \<in> A}" using insert by simp
```
``` 2202 also have "(x \<squnion> y) \<sqinter> \<dots> = \<Sqinter> (insert (x \<squnion> y) {x \<squnion> a |a. a \<in> A})"
```
``` 2203 using insert by(simp add:ACIf.fold1_insert_idem_def[OF ACIf_inf Inf_def fin])
```
``` 2204 also have "insert (x\<squnion>y) {x\<squnion>a |a. a \<in> A} = {x\<squnion>a |a. a \<in> insert y A}"
```
``` 2205 by blast
```
``` 2206 finally show ?case .
```
``` 2207 qed
```
``` 2208
```
``` 2209 lemma (in Distrib_Lattice) sup_Inf2_distrib:
```
``` 2210 assumes A: "finite A" "A \<noteq> {}" and B: "finite B" "B \<noteq> {}"
```
``` 2211 shows "(\<Sqinter>A \<squnion> \<Sqinter>B) = \<Sqinter>{a \<squnion> b|a b. a \<in> A \<and> b \<in> B}"
```
``` 2212 using A
```
``` 2213 proof (induct rule: finite_ne_induct)
```
``` 2214 case singleton thus ?case
```
``` 2215 by(simp add: sup_Inf1_distrib[OF B] fold1_singleton_def[OF Inf_def])
```
``` 2216 next
```
``` 2217 case (insert x A)
```
``` 2218 have finB: "finite {x \<squnion> b |b. b \<in> B}"
```
``` 2219 by(fast intro: finite_surj[where f = "%b. x \<squnion> b", OF B(1)])
```
``` 2220 have finAB: "finite {a \<squnion> b |a b. a \<in> A \<and> b \<in> B}"
```
``` 2221 proof -
```
``` 2222 have "{a \<squnion> b |a b. a \<in> A \<and> b \<in> B} = (UN a:A. UN b:B. {a \<squnion> b})"
```
``` 2223 by blast
```
``` 2224 thus ?thesis by(simp add: insert(1) B(1))
```
``` 2225 qed
```
``` 2226 have ne: "{a \<squnion> b |a b. a \<in> A \<and> b \<in> B} \<noteq> {}" using insert B by blast
```
``` 2227 have "\<Sqinter>(insert x A) \<squnion> \<Sqinter>B = (x \<sqinter> \<Sqinter>A) \<squnion> \<Sqinter>B"
```
``` 2228 using insert by(simp add:ACIf.fold1_insert_idem_def[OF ACIf_inf Inf_def])
```
``` 2229 also have "\<dots> = (x \<squnion> \<Sqinter>B) \<sqinter> (\<Sqinter>A \<squnion> \<Sqinter>B)" by(rule sup_inf_distrib2)
```
``` 2230 also have "\<dots> = \<Sqinter>{x \<squnion> b|b. b \<in> B} \<sqinter> \<Sqinter>{a \<squnion> b|a b. a \<in> A \<and> b \<in> B}"
```
``` 2231 using insert by(simp add:sup_Inf1_distrib[OF B])
```
``` 2232 also have "\<dots> = \<Sqinter>({x\<squnion>b |b. b \<in> B} \<union> {a\<squnion>b |a b. a \<in> A \<and> b \<in> B})"
```
``` 2233 (is "_ = \<Sqinter>?M")
```
``` 2234 using B insert
```
``` 2235 by(simp add:Inf_def ACIf.fold1_Un2[OF ACIf_inf finB _ finAB ne])
```
``` 2236 also have "?M = {a \<squnion> b |a b. a \<in> insert x A \<and> b \<in> B}"
```
``` 2237 by blast
```
``` 2238 finally show ?case .
```
``` 2239 qed
```
``` 2240
```
``` 2241
```
``` 2242 subsection{*Min and Max*}
```
``` 2243
```
``` 2244 text{* As an application of @{text fold1} we define the minimal and
```
``` 2245 maximal element of a (non-empty) set over a linear order. *}
```
``` 2246
```
``` 2247 constdefs
```
``` 2248 Min :: "('a::linorder)set => 'a"
```
``` 2249 "Min == fold1 min"
```
``` 2250
```
``` 2251 Max :: "('a::linorder)set => 'a"
```
``` 2252 "Max == fold1 max"
```
``` 2253
```
``` 2254
```
``` 2255 text{* Before we can do anything, we need to show that @{text min} and
```
``` 2256 @{text max} are ACI and the ordering is linear: *}
```
``` 2257
```
``` 2258 interpretation min: ACf ["min:: 'a::linorder \<Rightarrow> 'a \<Rightarrow> 'a"]
```
``` 2259 apply(rule ACf.intro)
```
``` 2260 apply(auto simp:min_def)
```
``` 2261 done
```
``` 2262
```
``` 2263 interpretation min: ACIf ["min:: 'a::linorder \<Rightarrow> 'a \<Rightarrow> 'a"]
```
``` 2264 apply(rule ACIf_axioms.intro)
```
``` 2265 apply(auto simp:min_def)
```
``` 2266 done
```
``` 2267
```
``` 2268 interpretation max: ACf ["max :: 'a::linorder \<Rightarrow> 'a \<Rightarrow> 'a"]
```
``` 2269 apply(rule ACf.intro)
```
``` 2270 apply(auto simp:max_def)
```
``` 2271 done
```
``` 2272
```
``` 2273 interpretation max: ACIf ["max:: 'a::linorder \<Rightarrow> 'a \<Rightarrow> 'a"]
```
``` 2274 apply(rule ACIf_axioms.intro)
```
``` 2275 apply(auto simp:max_def)
```
``` 2276 done
```
``` 2277
```
``` 2278 interpretation min:
```
``` 2279 ACIfSL ["min:: 'a::linorder \<Rightarrow> 'a \<Rightarrow> 'a" "op \<le>"]
```
``` 2280 apply(rule ACIfSL_axioms.intro)
```
``` 2281 apply(auto simp:min_def)
```
``` 2282 done
```
``` 2283
```
``` 2284 interpretation min:
```
``` 2285 ACIfSLlin ["min :: 'a::linorder \<Rightarrow> 'a \<Rightarrow> 'a" "op \<le>"]
```
``` 2286 apply(rule ACIfSLlin_axioms.intro)
```
``` 2287 apply(auto simp:min_def)
```
``` 2288 done
```
``` 2289
```
``` 2290 interpretation max:
```
``` 2291 ACIfSL ["max :: 'a::linorder \<Rightarrow> 'a \<Rightarrow> 'a" "%x y. y\<le>x"]
```
``` 2292 apply(rule ACIfSL_axioms.intro)
```
``` 2293 apply(auto simp:max_def)
```
``` 2294 done
```
``` 2295
```
``` 2296 interpretation max:
```
``` 2297 ACIfSLlin ["max :: 'a::linorder \<Rightarrow> 'a \<Rightarrow> 'a" "%x y. y\<le>x"]
```
``` 2298 apply(rule ACIfSLlin_axioms.intro)
```
``` 2299 apply(auto simp:max_def)
```
``` 2300 done
```
``` 2301
```
``` 2302 interpretation min_max:
```
``` 2303 Lattice ["op \<le>" "min :: 'a::linorder \<Rightarrow> 'a \<Rightarrow> 'a" "max" "Min" "Max"]
```
``` 2304 apply -
```
``` 2305 apply(rule Min_def)
```
``` 2306 apply(rule Max_def)
```
``` 2307 done
```
``` 2308
```
``` 2309
```
``` 2310 interpretation min_max:
```
``` 2311 Distrib_Lattice ["op \<le>" "min :: 'a::linorder \<Rightarrow> 'a \<Rightarrow> 'a" "max" "Min" "Max"]
```
``` 2312 .
```
``` 2313
```
``` 2314 text{* Now we instantiate the recursion equations and declare them
```
``` 2315 simplification rules: *}
```
``` 2316
```
``` 2317 (* Making Min or Max a defined parameter of a locale, suitably
```
``` 2318 extending ACIf, could make the following interpretations more automatic. *)
```
``` 2319
```
``` 2320 lemmas Min_singleton = fold1_singleton_def [OF Min_def]
```
``` 2321 lemmas Max_singleton = fold1_singleton_def [OF Max_def]
```
``` 2322 lemmas Min_insert = min.fold1_insert_idem_def [OF Min_def]
```
``` 2323 lemmas Max_insert = max.fold1_insert_idem_def [OF Max_def]
```
``` 2324
```
``` 2325 declare Min_singleton [simp] Max_singleton [simp]
```
``` 2326 declare Min_insert [simp] Max_insert [simp]
```
``` 2327
```
``` 2328
```
``` 2329 text{* Now we instantiate some @{text fold1} properties: *}
```
``` 2330
```
``` 2331 lemma Min_in [simp]:
```
``` 2332 shows "finite A \<Longrightarrow> A \<noteq> {} \<Longrightarrow> Min A \<in> A"
```
``` 2333 using min.fold1_in
```
``` 2334 by(fastsimp simp: Min_def min_def)
```
``` 2335
```
``` 2336 lemma Max_in [simp]:
```
``` 2337 shows "finite A \<Longrightarrow> A \<noteq> {} \<Longrightarrow> Max A \<in> A"
```
``` 2338 using max.fold1_in
```
``` 2339 by(fastsimp simp: Max_def max_def)
```
``` 2340
```
``` 2341 lemma Min_le [simp]: "\<lbrakk> finite A; A \<noteq> {}; x \<in> A \<rbrakk> \<Longrightarrow> Min A \<le> x"
```
``` 2342 by(simp add: Min_def min.fold1_belowI)
```
``` 2343
```
``` 2344 lemma Max_ge [simp]: "\<lbrakk> finite A; A \<noteq> {}; x \<in> A \<rbrakk> \<Longrightarrow> x \<le> Max A"
```
``` 2345 by(simp add: Max_def max.fold1_belowI)
```
``` 2346
```
``` 2347 lemma Min_ge_iff[simp]:
```
``` 2348 "\<lbrakk> finite A; A \<noteq> {} \<rbrakk> \<Longrightarrow> (x \<le> Min A) = (\<forall>a\<in>A. x \<le> a)"
```
``` 2349 by(simp add: Min_def min.below_fold1_iff)
```
``` 2350
```
``` 2351 lemma Max_le_iff[simp]:
```
``` 2352 "\<lbrakk> finite A; A \<noteq> {} \<rbrakk> \<Longrightarrow> (Max A \<le> x) = (\<forall>a\<in>A. a \<le> x)"
```
``` 2353 by(simp add: Max_def max.below_fold1_iff)
```
``` 2354
```
``` 2355 lemma Min_le_iff:
```
``` 2356 "\<lbrakk> finite A; A \<noteq> {} \<rbrakk> \<Longrightarrow> (Min A \<le> x) = (\<exists>a\<in>A. a \<le> x)"
```
``` 2357 by(simp add: Min_def min.fold1_below_iff)
```
``` 2358
```
``` 2359 lemma Max_ge_iff:
```
``` 2360 "\<lbrakk> finite A; A \<noteq> {} \<rbrakk> \<Longrightarrow> (x \<le> Max A) = (\<exists>a\<in>A. x \<le> a)"
```
``` 2361 by(simp add: Max_def max.fold1_below_iff)
```
``` 2362
```
``` 2363 subsection {* Properties of axclass @{text finite} *}
```
``` 2364
```
``` 2365 text{* Many of these are by Brian Huffman. *}
```
``` 2366
```
``` 2367 lemma finite_set: "finite (A::'a::finite set)"
```
``` 2368 by (rule finite_subset [OF subset_UNIV finite])
```
``` 2369
```
``` 2370
```
``` 2371 instance unit :: finite
```
``` 2372 proof
```
``` 2373 have "finite {()}" by simp
```
``` 2374 also have "{()} = UNIV" by auto
```
``` 2375 finally show "finite (UNIV :: unit set)" .
```
``` 2376 qed
```
``` 2377
```
``` 2378 instance bool :: finite
```
``` 2379 proof
```
``` 2380 have "finite {True, False}" by simp
```
``` 2381 also have "{True, False} = UNIV" by auto
```
``` 2382 finally show "finite (UNIV :: bool set)" .
```
``` 2383 qed
```
``` 2384
```
``` 2385
```
``` 2386 instance * :: (finite, finite) finite
```
``` 2387 proof
```
``` 2388 show "finite (UNIV :: ('a \<times> 'b) set)"
```
``` 2389 proof (rule finite_Prod_UNIV)
```
``` 2390 show "finite (UNIV :: 'a set)" by (rule finite)
```
``` 2391 show "finite (UNIV :: 'b set)" by (rule finite)
```
``` 2392 qed
```
``` 2393 qed
```
``` 2394
```
``` 2395 instance "+" :: (finite, finite) finite
```
``` 2396 proof
```
``` 2397 have a: "finite (UNIV :: 'a set)" by (rule finite)
```
``` 2398 have b: "finite (UNIV :: 'b set)" by (rule finite)
```
``` 2399 from a b have "finite ((UNIV :: 'a set) <+> (UNIV :: 'b set))"
```
``` 2400 by (rule finite_Plus)
```
``` 2401 thus "finite (UNIV :: ('a + 'b) set)" by simp
```
``` 2402 qed
```
``` 2403
```
``` 2404
```
``` 2405 instance set :: (finite) finite
```
``` 2406 proof
```
``` 2407 have "finite (UNIV :: 'a set)" by (rule finite)
```
``` 2408 hence "finite (Pow (UNIV :: 'a set))"
```
``` 2409 by (rule finite_Pow_iff [THEN iffD2])
```
``` 2410 thus "finite (UNIV :: 'a set set)" by simp
```
``` 2411 qed
```
``` 2412
```
``` 2413 lemma inj_graph: "inj (%f. {(x, y). y = f x})"
```
``` 2414 by (rule inj_onI, auto simp add: expand_set_eq expand_fun_eq)
```
``` 2415
```
``` 2416 instance fun :: (finite, finite) finite
```
``` 2417 proof
```
``` 2418 show "finite (UNIV :: ('a => 'b) set)"
```
``` 2419 proof (rule finite_imageD)
```
``` 2420 let ?graph = "%f::'a => 'b. {(x, y). y = f x}"
```
``` 2421 show "finite (range ?graph)" by (rule finite_set)
```
``` 2422 show "inj ?graph" by (rule inj_graph)
```
``` 2423 qed
```
``` 2424 qed
```
``` 2425
```
``` 2426 end
``` | 43,956 | 115,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-10 | latest | en | 0.646716 |
https://www.physicsforums.com/threads/electron-positron-collisions-and-decay-in-special-relativity.366059/ | 1,544,696,831,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824675.15/warc/CC-MAIN-20181213101934-20181213123434-00188.warc.gz | 1,018,000,295 | 12,805 | Homework Help: Electron-positron collisions and decay in Special Relativity
1. Dec 29, 2009
Keano16
1. The problem statement, all variables and given/known data
A particle physicist seeks to create a new fundamental particle with rest energy
200GeV by colliding electrons and positrons. What is the minimum positron energy
required when electrons and positrons travelling in opposite directions with equal speeds are
collided together?
The new particle is produced using this method with the minimum necessary
energy, and rapidly decays into two identical particles of rest energy 91.2GeV
2. Relevant equations
E2-p2c2=m2c4
3. The attempt at a solution
I am trying to do this question using 4-momentum. For a positron I have it as (E, p1, 0, 0) and for the electron I have it as once again (E, p2, 0, 0) (we are allowed to approximate the mass of a positron to that of an electron), the resultant being (2E, p1+p2, 0, 0).
I then equate (2E)2-(p1+p2)2=(200x103)2 (Working in MeV)
However, the problem I get is that as these particles are of essentially equal mass, and moving in opposite directions with the same speed, does p1+p2 in the 4-momentum effectively become 0?
If so, I get 4E2=(200x103)2
giving E=100GeV
I think there's something I am missing, as in the next part it says two particles of rest mass 91.2GeV are produced, which exceeds this energy. Or does this energy translate to kinetic energy for the 200GeV particle?
Any help would be appreciated.
2. Dec 29, 2009
Fightfish
There's no need to work so hard; simple Conservation of energy will solve the first part. Just 200/2 = 100GeV does the trick.
For the second part, the "missing energy" is indeed the kinetic energy of the decay products.
3. Dec 30, 2009
Keano16
Ah right I see now. Thanks! | 481 | 1,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.919042 |
https://www.jiskha.com/display.cgi?id=1528772225 | 1,529,306,145,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860089.13/warc/CC-MAIN-20180618070542-20180618090542-00484.warc.gz | 856,049,088 | 4,237 | # geometry
posted by jill
"We use different formulas to find the distance of a segment on a number line, d=|a−b|" role="presentation" style="display: inline; font-size: 16px; position: relative;">d=|a−b|, and the distance of a segment in the coordinate plane, d=(x2−x1)2+(y2−y1)2" role="presentation" style="display: inline; font-size: 16px; position: relative;">d=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√. Why is it necessary to use the absolute value of the difference when finding the distance on a number line, but not necessary when finding the differences of the coordinates in the coordinate plane?"
1. Steve
values on the number line can be measured in either direction. So, it makes a difference which one is subtracted. The distance though, is always positive.
In the xy distance formula, the differences are squared, so that always gives a positive result.
If you think about it, the same formula would work on the number line.
d = √(x1-x2)^2
will always give a positive result.
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More Similar Questions | 718 | 2,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.852677 |
https://forum.freecodecamp.org/t/sum-all-primes-in-java-script/299981 | 1,601,434,761,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402101163.62/warc/CC-MAIN-20200930013009-20200930043009-00339.warc.gz | 373,017,968 | 6,801 | # Sum All Primes in Java Script
Tell us what’s happening:
Hi, I’m totally confused by this challenge. As I understood, I only counted primes so far (is this correct, btw?). Could you also advise me how to count them all please?
``````
function sumPrimes(num) {
var primes=[];
if(num < 2)
return false;
for (var i = 2; i <= num; i++) {
if(num%i==0)
return false;
}
return true;
for(var j = 0; j <= num; j++){
if(isPrime(i))
primes.push(i);
}
return primes;
}
sumPrimes(10);
``````
Chrome Version 75.0.3770.142 (Official Build) (64-bit)
“sum” doesn’t mean count
Bit by bit here:
``````var primes=[];
``````
You’re looping over things in this function and adding up. I’m not sure why you feel you need to put them in an array, but anyway
``````if(num < 2) return false;
``````
Why `false`? This isn’t a particularly good thing to return, but I don’t think this gets tested so whatever.
``````for (var i = 2; i <= num; i++) {
if(num%i==0) return false;
}
return true;
``````
So when we get to this bit of code, this loop starts, and if it finds an even number, the function exits and returns false. Otherwise it exits as false.
I assume you meant to wrap that bit of code in a function called isPrime, because:
``````for(var j = 0; j <= num; j++){
if(isPrime(i)) primes.push(i);
}
return primes;
}
sumPrimes(10);
``````
Otherwise that piece of code doesn’t make sense.
Also, watch formatting consistently, the code is very difficult to read because the spacing and new lines seem random. And if you really really want to miss off brackets on if statements, at least put the expression on the same line. | 443 | 1,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | longest | en | 0.788419 |
http://www.q-chem.com/qchem-website/manual/qchem43_manual/sect-approx_hess.html | 1,506,164,780,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689624.87/warc/CC-MAIN-20170923104407-20170923124407-00117.warc.gz | 543,548,019 | 2,807 | Q-Chem 4.3 User’s Manual
# 9.7 Hessian-Free Transition State Search
Once a guess structure to the transition state is obtained, standard eigenvector-following methods such as the Partitioned-Rational Function Optimization (P-RFO) [432] can be employed to refine the guess to the exact transition state. The reliability of P-RFO depends on the quality of the Hessian input, which enables the method to distinguish between the reaction coordinate (characterized by a negative eigenvalue) and the remaining degrees of freedom. In routine calculations therefore, an exact Hessian is determined via frequency calculation prior to the P-RFO search. Since the cost of evaluating an exact Hessian typically scales one power of system size higher than the energy or the gradient, this step becomes impractical for systems containing large number of atoms.
The exact Hessian calculation can be avoided by constructing an approximate Hessian based on the output of FSM. The tangent direction at the transition state guess on the FSM string is a good approximation to the Hessian eigenvector corresponding to the reaction coordinate. The tangent is therefore used to calculate the correct eigenvalue and corresponding eigenvector by variationally minimizing the Rayleigh-Ritz ratio [433]. The reaction coordinate information is then incorporated into a guess matrix which, in turn, is obtained by transforming a diagonal matrix in delocalized internal coordinates [434] [435] to Cartesian coordinates. The resulting approximate Hessian, by design, has a single negative eigenvalue corresponding to the reaction coordinate. This matrix is then used in place of the exact Hessian as input to the P-RFO method.
An example of this one-shot, Hessian-free approach that combines the FSM and P-RFO methods in order to determine the exact transition state from reactant and product structures is shown below:
Example 9.199
```\$molecule
0 1
Si 1.028032 -0.131573 -0.779689
H 0.923921 -1.301934 0.201724
H 1.294874 0.900609 0.318888
H -1.713989 0.300876 -0.226231
H -1.532839 0.232021 0.485307
****
Si 0.000228 -0.000484 -0.000023
H 0.644754 -1.336958 -0.064865
H 1.047648 1.052717 0.062991
H -0.837028 0.205648 -1.211126
H -0.8556026 0.079077 1.213023
\$end
\$rem
jobtype fsm
fsm_nnode 18
fsm_mode 2
fsm_opt_mode 2
method b3lyp
basis 6-31g
symmetry false
sym_ignore true
\$end
@@@
\$rem
jobtype ts
max_scf_cycles 250
geom_opt_dmax 50
geom_opt_max_cycles 100
method b3lyp
basis 6-31g
symmetry false
sym_ignore true
\$end
\$molecule | 710 | 2,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-39 | latest | en | 0.874248 |
https://www.vedantu.com/question-answer/solubility-of-salt-a2b3-is-1-times-10-4its-class-11-chemistry-cbse-60a3d6997375ee051db107aa | 1,708,663,161,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00541.warc.gz | 1,094,415,879 | 30,476 | Courses
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# Solubility of salt ${A_2}{B_3}$, is $\;1 \times {10^{ - 4}}$its solubility product isOptionA.$\;1.08 \times {10^{ - 18}}$B.$\;1.08 \times {10^{ - 13}}$C.$\;1.08 \times {10^{ - 15}}$D.$\;1.08 \times {10^{ - 17}}$
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Hint:The ability of a substance called a solute to dissolve in a solvent and form a solution is known as solubility. Ionic compounds (which dissociate to form cations and anions) have a wide range of solubility in water. Some compounds are very soluble and can also absorb moisture from the air, while others are extremely insoluble.
The solubility product is a type of equilibrium constant whose value is temperature dependent. Due to enhanced solubility, ${K_{sp}}$ typically rises as the temperature rises. The ability of a substance called a solute to dissolve in a solvent and form a solution is known as solubility.
Solubility product is represented as
${K_{sp}} = {\left[ {{A^ + }} \right]^a}{\left[ {{B^ - }} \right]^b}$
${A_2}{B_3} \Leftrightarrow 2{A^{3 + }} + \quad 3{B^{2 - }}$
$2S \,\,\,\,\,\,\,\,\,\,\,\,\,\,\ 3S$
Substituting the value in ${K_{sp}}$
${K_{sp}} = {(2S)^2}{(3S)^3}$
2 x 2 x 3 x 3 x 3 = 108
Hence,
${K_{sp}} = 108{S^5}$
${K_{sp}} = 108 \times {\left( {{{10}^{ - 3}}} \right)^5}$
${K_{sp}} = 108 \times {10^{ - 15}}$
${K_{sp}} = 1.08 \times {10^{ - 13}}$
Hence option B Is correct. | 529 | 1,487 | {"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} | 3.546875 | 4 | CC-MAIN-2024-10 | longest | en | 0.828826 |
https://www.homebrewersassociation.org/forum/index.php?topic=24614.0;prev_next=prev | 1,670,358,284,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00296.warc.gz | 864,071,742 | 6,961 | ### Author Topic: Stupid calcium chloride solution question (Read 1040 times)
#### homoeccentricus
• Brewmaster General
• Posts: 2009
• A twerp from Antwerp
##### Stupid calcium chloride solution question
« on: October 10, 2015, 01:15:48 pm »
If a solution contains 33 percent calcium chloride, is this by weight, volume or a combination of the two? And so, how many grams of calcium chloride would there be in a liter of the solution?
Frank P.
Staggering on the shoulders of giant dwarfs.
#### a10t2
• Official Poobah of No Life. (I Got Ban Hammered by Drew)
• Posts: 4696
• Ask me why I don't like Chico!
##### Re: Stupid calcium chloride solution question
« Reply #1 on: October 10, 2015, 02:32:44 pm »
There's no real way to tell if it isn't explicitly stated. Conventions vary by industry, so you might be able to guess from context.
To actually answer the question, it would be 330 g/L if it's 33% w/v or 326 g/L if it's 33% w/w. The effect on volume is so small that it probably doesn't matter for your purposes.
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#### homoeccentricus
• Brewmaster General
• Posts: 2009
• A twerp from Antwerp
##### Re: Stupid calcium chloride solution question
« Reply #2 on: October 10, 2015, 02:37:42 pm »
Apparently this is on the label:
Calcium chloride content : > 33% (w/w) / > 430 g/l
How do you get to 326 grams per liter?
Frank P.
Staggering on the shoulders of giant dwarfs.
#### klickitat jim
• I must live here
• Posts: 8604
##### Re: Stupid calcium chloride solution question
« Reply #3 on: October 10, 2015, 02:40:56 pm »
Grams or grains?
#### a10t2
• Official Poobah of No Life. (I Got Ban Hammered by Drew)
• Posts: 4696
• Ask me why I don't like Chico!
##### Re: Stupid calcium chloride solution question
« Reply #4 on: October 10, 2015, 02:53:10 pm »
That checks out. I must have fat-fingered something.
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Refractometer Calculator | Batch Sparging Calculator | Two Mile Brewing Co. | 624 | 2,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.854722 |
https://www.scribd.com/doc/300319735/3-methodsofintegration | 1,579,429,420,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594391.21/warc/CC-MAIN-20200119093733-20200119121733-00534.warc.gz | 1,098,725,643 | 70,139 | You are on page 1of 8
# Math 1132 Worksheet 3
Name
Discussion Section
Methods of Integration
Solutions to these problems should show all of your work, not just a single
Part 1: Integration by parts. Do each problem as follows: (1) specify u and dv, (2)
compute du and v, (3) use integration by parts with your choice of u and dv. (4) If you need
integration by parts more than once, each time go through steps 1, 2, and 3 again.
Z
Example. Compute x2 ex dx.
Solution.
(1) Set u = x2 and dv = ex dx.
(2) We have du = 2x dx and v = ex .
Z
Z
Z
Z
Z
2 x
2 x
x
2 x
(3) Now x e dx = u dv = uv v du = x e e (2x) dx = x e 2 xex dx.
Z
## (4) To find xe dx, set u = x and dv = e dx, so du = dx and v = e . Then
Z
Z
Z
x
u dv = uv v du = xe ex dx = xex ex .
(5) Substituting the result of (4) into (3),
Z
x2 ex dx = x2 ex 2(xex ex ) + C = (x2 2x + 2)ex + C.
Z
1. Compute
x cos(5x) dx.
xex dx =
Z
2. Compute
## x2 2x dx. (Hint: You can find an antiderivative of 2x by recalling how to
differentiate 2x .)
## Part 2: Integration of rational functions.
Z
2x + 1
dx using partial fractions.
Example. Compute
x2 4
2x + 1
A
B
Solution. Write 2
=
+
for some A and B. Clearing the denominator,
x 4
x+2 x2
2x + 1 = A(x 2) + B(x + 2). Setting x = 2 we get 5 = 4B, so B = 5/4. Setting x = 2
2x + 1
3/4
5/4
we get 3 = 4A, so A = 3/4. Thus 2
=
+
, so
x 4
x+2 x2
Z
Z
2x + 1
3/4
5/4
3
5
dx =
+
dx = ln |x + 2| + ln |x 2| + C.
2
x 4
x+2 x2
5
4
Z
3. Compute
x3
Z
4. Compute
10
dx using partial fractions.
x2 6x
x2 + x + 1
dx using partial fractions.
x(x2 + 4)
## Part 3: Approximate Integration.
Z
Example. (a) Compute the trapezoid approximation to
x dx using n = 4 subin-
## tervals, rounding your approximation to 5 digits after the decimal point.
(b) Use the error bound for the trapezoid rule to determine an n such that the trapezoid
approximation is guaranteed by the error bound to be with .01 of the value of the integral.
Solution.
(a) The trapezoid approximation with n = 4 is
ba
2
(f (x0 ) + 2f (x1 ) + 2f (x2 ) + 2f (x3 ) + f (x4 )) =
(f (1) + 2f (1.5) + 2f (2) +
2n
8
2f (2.5) + f (3))
2.79306.
(b) An upper bound on the error from the trapezoid rule with n intervals is K(ba)
(x)2 ,
12
00
where x = (b
a)/n and K is an upper bound on |f (x)| for all x in [a, b]. In our
problem, f (x) = x, so f 00 (x) = 14 x3/2 . For 1 x 3, we have x3/2 1, so
|f 00 (x)| 1/4 when 1 x 3. Thus we can use K = 1/4, so the trapezoid error bound
is (1/4)(31)
( n2 )2 = 6n1 2 . Having the error be less than .01 means
12
1
1
< .01 n2 >
n >
2
6n
6(.01)
1
4.082,
.06
## so for n 5 the trapezoid approximation will be within .01 of the integral.
Z 3
5. (a) Compute the trapezoid approximation to
x sin x dx using n = 4 subintervals,
2
rounding your approximation to 5 digits after the decimal point. (Remember to set your
calculator to radian mode for trigonometric functions.)
(b) Use the error bound for the trapezoid rule to determine an n such that the trapezoid
approximation is guaranteed by the error bound to be with .01 of the value of the integral.
Z
6. (a) Compute the Simpsons rule approximation to
x dx using n = 4 subintervals,
## rounding your approximation to 5 digits after the decimal point.
(b) Use the error bound for Simpsons rule to determine an n such that the Simpsons
rule approximation is guaranteed by the error bound to be with 106 of the value of the
integral. (Remember n must be even.)
## Part 4: Improper Integrals.
Z
0
Z
8. Compute the improper integral
0
ax
## 7. For a > 0, compute the improper integrals
will be in terms of a.
Z
dx and
dx
using partial fractions.
(x + 2)(x + 5)
Z
9. Decide if the improper integral
gent, evaluate it.
x2
x
dx is convergent or divergent. If it is conver+1
Optional Question.
10. Vibrations show up in many places: civil engineering (oscillations in a bridge or the
reaction of a building to an earthquake), music (sound is a vibration of pressure waves),
and ski design (smaller vibrations make a smoother ride). The following computation is
fundamental in any mathematical study of vibrations: for all positive integers m and n,
use integration by parts to show
Z 2
sin(mx) cos(nx) dx = 0.
0
(Hint: Use the bounds of integration during the integration by parts, and treat m = n
and m 6= n separately. It may help to first try this for specific m and n, such as m = 2
and n = 3, and then m = 5 and n = 5.) There are two other integral formulas related
the one above, with products of two sines and two cosines:
(
(
Z 2
Z 2
, if m = n,
, if m = n,
cos(mx) cos(nx) dx =
sin(mx) sin(nx) dx =
0, if m 6= n.
0, if m 6= n
0
0
Here too m and n are positive integers. The optional question is only the first formula. | 1,647 | 4,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2020-05 | latest | en | 0.798888 |
http://www.jiskha.com/display.cgi?id=1298856047 | 1,493,501,822,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123590.89/warc/CC-MAIN-20170423031203-00193-ip-10-145-167-34.ec2.internal.warc.gz | 575,788,644 | 3,724 | posted by on .
Determine the values of a M and N , so that the polynomial 2x^3+mx^2+nx-3 and X^3-3mx^2+2nx+4 are both divisible by x-2.
I am so lost, i do not know how to start this problem! please help i have a test soon!
• MATH grade 12 - ,
Just change the variable x with the value x = 2. So, we get:
f(x) = 2x^3 + mx^2 + nx - 3
f(2) = 0 = 2(8) + 4m + 2n - 3
0 = 13 + 4m + 2n. ..(1)
g(x) = 3mx^2 + 2nx + 4
g(2) = 0 = 12m + 4n + 4
0 = 6m + 2n + 2. ..(2)
Using elimination and substitution for both equation (1) and (2), you'll get the values of m and n | 240 | 560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-17 | latest | en | 0.820352 |
https://www.organised-sound.com/how-to-calculate-equivalent-resistance-in-series-circuit/ | 1,660,394,722,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571950.76/warc/CC-MAIN-20220813111851-20220813141851-00378.warc.gz | 794,451,007 | 18,066 | # How To Calculate Equivalent Resistance In Series Circuit
By | July 26, 2022
Solved calculate the equivalent resistance for this circuit chegg com series and parallel circuits learn sparkfun 4 ways to wikihow calculation in a complex using matlab electrical academia of following network brainly capacitance inductance dummies l4 resistors physical computing combined objectives 1 cur voltage 2 ppt answered 18 shown r r2 bartleby each theory goal lab is test equations resistor applications guide experiment docsity calculating worksheet calculator engineering electronics tools online consider figure find b potential difference between points c 20 e 3 given as follow electric drop on total d below i ii physics shaalaa 11 42 measurement resistances connect figures ri are r3 r4 1ov fig 5 diagram how study
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Solved calculate the equivalent series and parallel circuits learn 4 ways to resistance calculation in a of capacitance l4 resistors objectives answered 18 circuit shown r lab is test equations resistor experiment calculating worksheet calculator ppt 3 given between total 42 diagram | 421 | 2,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-33 | longest | en | 0.723185 |
https://en.m.wikipedia.org/wiki/Analytic_capacity | 1,606,297,479,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181482.18/warc/CC-MAIN-20201125071137-20201125101137-00277.warc.gz | 272,671,245 | 14,357 | # Analytic capacity
In the mathematical discipline of complex analysis, the analytic capacity of a compact subset K of the complex plane is a number that denotes "how big" a bounded analytic function on C \ K can become. Roughly speaking, γ(K) measures the size of the unit ball of the space of bounded analytic functions outside K.
It was first introduced by Ahlfors in the 1940s while studying the removability of singularities of bounded analytic functions.
## Definition
Let KC be compact. Then its analytic capacity is defined to be
${\displaystyle \gamma (K)=\sup\{|f'(\infty )|;\ f\in {\mathcal {H}}^{\infty }(\mathbf {C} \setminus K),\ \|f\|_{\infty }\leq 1,\ f(\infty )=0\}}$
Here, ${\displaystyle {\mathcal {H}}^{\infty }(U)}$ denotes the set of bounded analytic functions UC, whenever U is an open subset of the complex plane. Further,
${\displaystyle f'(\infty ):=\lim _{z\to \infty }z\left(f(z)-f(\infty )\right)}$
${\displaystyle f(\infty ):=\lim _{z\to \infty }f(z)}$
Note that ${\displaystyle f'(\infty )=g'(0)}$ , where ${\displaystyle g(z)=f(1/z)}$ . However, usually ${\displaystyle f'(\infty )\neq \lim _{z\to \infty }f'(z)}$ .
If AC is an arbitrary set, then we define
${\displaystyle \gamma (A)=\sup\{\gamma (K):K\subset A,\,K{\text{ compact}}\}.}$
## Removable sets and Painlevé's problem
The compact set K is called removable if, whenever Ω is an open set containing K, every function which is bounded and holomorphic on the set Ω \ K has an analytic extension to all of Ω. By Riemann's theorem for removable singularities, every singleton is removable. This motivated Painlevé to pose a more general question in 1880: "Which subsets of C are removable?"
It is easy to see that K is removable if and only if γ(K) = 0. However, analytic capacity is a purely complex-analytic concept, and much more work needs to be done in order to obtain a more geometric characterization.
## Ahlfors function
For each compact KC, there exists a unique extremal function, i.e. ${\displaystyle f\in {\mathcal {H}}^{\infty }(\mathbf {C} \setminus K)}$ such that ${\displaystyle \|f\|\leq 1}$ , f(∞) = 0 and f′(∞) = γ(K). This function is called the Ahlfors function of K. Its existence can be proved by using a normal family argument involving Montel's theorem.
## Analytic capacity in terms of Hausdorff dimension
Let dimH denote Hausdorff dimension and H1 denote 1-dimensional Hausdorff measure. Then H1(K) = 0 implies γ(K) = 0 while dimH(K) > 1 guarantees γ(K) > 0. However, the case when dimH(K) = 1 and H1(K) ∈ (0, ∞] is more difficult.
### Positive length but zero analytic capacity
Given the partial correspondence between the 1-dimensional Hausdorff measure of a compact subset of C and its analytic capacity, it might be conjectured that γ(K) = 0 implies H1(K) = 0. However, this conjecture is false. A counterexample was first given by A. G. Vitushkin, and a much simpler one by John B. Garnett in his 1970 paper. This latter example is the linear four corners Cantor set, constructed as follows:
Let K0 := [0, 1] × [0, 1] be the unit square. Then, K1 is the union of 4 squares of side length 1/4 and these squares are located in the corners of K0. In general, Kn is the union of 4n squares (denoted by ${\displaystyle Q_{n}^{j}}$ ) of side length 4n, each ${\displaystyle Q_{n}^{j}}$ being in the corner of some ${\displaystyle Q_{n-1}^{k}}$ . Take K to be the intersection of all Kn then ${\displaystyle H^{1}(K)={\sqrt {2}}}$ but γ(K) = 0.
### Vitushkin's conjecture
Let KC be a compact set. Vitushkin's conjecture states that
${\displaystyle \gamma (K)=0\ \iff \ \int _{0}^{\pi }{\mathcal {H}}^{1}(\operatorname {proj} _{\theta }(K))\,d\theta =0}$
where ${\displaystyle \operatorname {proj} _{\theta }(x,y):=x\cos \theta +y\sin \theta }$ denotes the orthogonal projection in direction θ. By the results described above, Vitushkin's conjecture is true when dimHK ≠ 1.
Guy David published a proof in 1998 of Vitushkin's conjecture for the case dimHK = 1 and H1(K) < ∞. In 2002, Xavier Tolsa proved that analytic capacity is countably semiadditive. That is, there exists an absolute constant C > 0 such that if KC is a compact set and ${\displaystyle K=\bigcup _{i=1}^{\infty }K_{i}}$ , where each Ki is a Borel set, then ${\displaystyle \gamma (K)\leq C\sum _{i=1}^{\infty }\gamma (K_{i})}$ .
David's and Tolsa's theorems together imply that Vitushkin's conjecture is true when K is H1-sigma-finite. However, the conjecture is still open for K which are 1-dimensional and not H1-sigma-finite.
## References
• Mattila, Pertti (1995). Geometry of sets and measures in Euclidean spaces. Cambridge University Press. ISBN 0-521-65595-1.
• Pajot, Hervé (2002). Analytic Capacity, Rectifiability, Menger Curvature and the Cauchy Integral. Lecture Notes in Mathematics. Springer-Verlag.
• J. Garnett, Positive length but zero analytic capacity, Proc. Amer. Math. Soc. 21 (1970), 696–699
• G. David, Unrectifiable 1-sets have vanishing analytic capacity, Rev. Math. Iberoam. 14 (1998) 269–479
• Dudziak, James J. (2010). Vitushkin's Conjecture for Removable Sets. Universitext. Springer-Verlag. ISBN 978-14419-6708-4.
• Tolsa, Xavier (2014). Analytic Capacity, the Cauchy Transform, and Non-homogeneous Calderón–Zygmund Theory. Progress in Mathematics. Birkhäuser Basel. ISBN 978-3-319-00595-9. | 1,619 | 5,335 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 18, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-50 | latest | en | 0.863173 |
https://kopiandproperty.com/2021/03/09/affordable-increase/ | 1,627,446,660,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00601.warc.gz | 359,368,342 | 57,939 | # RM28,888 is an affordable increase. This is why.
What is an affordable increase? Actually this happens daily for all products we buy. The seller is always trying to maximise the profits by slowly increasing the price over the years. My personal term for it is affordable increase.
These days, affordable property is the rage. Transactions for properties continue to happen and that’s because people could afford them. This is despite the fact that total transactions are lower than pre-COVID period. In fact more and more of these affordable units would be built even in 2021 and private developers realise that they need to go for higher number of sales with a lower margin versus fewer sales but higher margins previously. In other words, everyone is working harder than previously just to fulfil the need for a home.
In the future, what are the chances than the home we bought today could have an affordable increase in price? Speaking of affordability, my personal thought is that for the next couple of years, RM400,000 would continue to be that ceiling for affordable home. Perhaps we look at an increase of RM28,888 every year? Let’s look at some calculations then.
Could we save RM113 every month if needed?
Think about it. 30 days in a month. RM113 divided by 30 days is RM3.77 per day. If we used to spend RM25 for a latte and a slice of cheesecake every week and we chose NOT to spend this RM25, then we have saved RM3.57 per day. Actually very close to the need to save RM3.77
For my friend who’s smoking, if he stopped smoking, then he could definitely save RM3.77 per day too.
For someone who had decided to buy that RM1,588 smartphone versus a RM3,000 phone, he would have saved RM1,412 and this comes to RM117.67 per month.
In other words, all these scenarios would have enabled us to save RM113 per month, if not more. Now, let’s look into why we are so focused on RM113 per month.
RM400,000 home, 10 percent downpayment, 30 years and interest rate of 3.2%
If we had bought this home (as per image above), the monthly repayment would come to RM1,556.88 per month. After paying for it for one whole year, what is the chance that we could sell this for RM28,888 higher than the price we paid? Well, if we love the home for the right reasons, there are definitely others who would love it for all the same reasons too. This person may be willing to pay you a higher price for it. Specifically RM28,888 higher. Let’s look at what this new buyer would pay you.
RM428,888 home, 10% downpayment, 30 years and interest rate of 3.2%
The total repayment is now RM1,669.32 per month and this is RM113 higher than what you were paying. In other words, the buyer who will be buying from you would have to fork out an additional RM113 and somehow save this amount of money from the scenarios we spoke about OR his salary increment was higher than this, so he could use some of it for this home. Meanwhile, for the owner, that’s you, you will get an extra RM28,888 for the home.
This is why RM28,888 increase in property price could still be affordable to that next buyer. The payment is stretched over 30 years for the buyer. This is also the reason why if we look at property price on a longer term basis, this is how the graph looks like, even in slow times. (Image below) Property prices would just slowly inch up. Do note that that anything above zero do mean that property prices are strill increasing every year yeah.
DO NOT buy homes for this RM28,888 price increase yeah. Please buy a home because you want to stay in it and many years later, you could be pleasantly surprised by the price which the next buyer is willing to pay for you. Actually it’s the same for you. When you are ready to upgrade, you will also think an extra RM28,888 looks affordable too. It’s just an extra RM113 per month.
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An article a day, keeps updated all the way. Subscribe for free! | 1,019 | 4,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-31 | latest | en | 0.970684 |
http://www.gradesaver.com/textbooks/math/other-math/basic-college-mathematics-9th-edition/chapter-6-percent-6-5-using-the-percent-equation-6-5-exercises-page-429/8 | 1,519,439,872,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815034.13/warc/CC-MAIN-20180224013638-20180224033638-00133.warc.gz | 457,246,023 | 13,093 | ## Basic College Mathematics (9th Edition)
Published by Pearson
# Chapter 6 - Percent - 6.5 Using the Percent Equation - 6.5 Exercises: 8
189.2 liters
#### Work Step by Step
Divide the percent by 100. Fill in the percent equation. part=(percent)(whole) $x=(0.44)(430)$ $x=189.2$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 119 | 444 | {"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} | 3.609375 | 4 | CC-MAIN-2018-09 | latest | en | 0.818583 |
electricbikebuilding.com | 1,701,730,917,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00800.warc.gz | 267,270,828 | 61,974 | # The Price of Charging Your Electric Bike
Some links may be affiliate links. You may receive compensation if you make a purchase or take action after clicking on one of these.
## How much electricity does an e-bike use?
The most effective way to address this query is by examining the usual energy consumption of e-bikes and comparing it to other modes of transportation in order to demonstrate their exceptional efficiency in terms of power usage – which is relatively minimal by most standards!
Comparing watts is the optimal method for measurement and comparison when addressing this question. Not only does it measure electrical energy, but it also encompasses energy consumed by any form of movement, ranging from walking to flying, and everything in between. Watt-hours, on the other hand, are an even more valuable measure. They simply represent the number of watts an e-bike (or any other device) consumes in an hour – hence the term watt hours (Wh).
Watt hours are commonly used to gauge the amount of electrical energy stored in an e-bike battery. Although 1000 Wh batteries are considered large for an e-bike, they are not uncommon. For the sake of comparison, let’s assume that a 1000Wh (or 1 kilowatt hour – kWh) e-bike battery can enable a single rider to travel 100km (approximately 60 miles). This assumption is based on the numerous real-world tests conducted by EBR as part of our electric bike reviews.
Now, let’s compare this measurement of covering 100km on 1 kWh of electricity to other methods of transportation in the following diagram. All the figures are based on data obtained from the Without Hot Air website.
Of course, many of the other modes of transportation depicted here do not rely on electricity, but they all consume energy obtained from various fuels, which can be equated to kWh. Therefore, an e-bike utilizes approximately one twenty-fifth of the electrical energy consumed by an electric car for mobility and one fiftieth of the energy consumed by an airplane. These figures, of course, are rough approximations based on averages and assumptions. Nonetheless, the key takeaway remains unchanged – e-bikes require a remarkably small amount of electrical energy when compared to alternative options.
Now, let’s delve into the cost of recharging your e-bike.
## How much does it cost to charge your e-bike?
Determining the cost of fully recharging a battery from empty is a relatively straightforward calculation:
READ MORE : Guide to Charging Stations for Electric Bikes
(Battery capacity in kWh x 1.25) x cost of electricity per kWh
The presence of the 1.25 multiplier accounts for the fact that not all the electricity that flows from your wall socket ends up being stored in your battery cells. Due to the inefficiencies of power transfer, up to approximately 20% can be lost – you may notice the transformer warming up during charging, as it requires electricity to generate heat!
What are the costs of filling a 500Wh e-bike battery in the US?
Let’s look at a worked example using a typical 500Wh e-bike battery, which costs \$0.106 per kWh of electricity. According to an article by Homeserve, the average cost of electricity for US consumers is 10.6 cents per kWh.
(0.5kWh battery size x 1.25) x \$0.106 = \$0.066. That’s right, it only costs just over 6.5 cents to fill a decent sized e-bike battery!
But keep in mind that the cost of electricity can vary significantly. To demonstrate this, let’s take examples from the opposite ends of the spectrum and apply them to the above calculation.
In Hawaii, electricity can cost as much as 38 cents per kWh. In Louisiana, it’s around 7.5 cents. This gives us the following example costs:
Hawaii
(0.5 x 1.25) x \$0.38 = \$0.0237, which is around 24 cents.
Louisiana
(0.5 x 1.25) x \$0.075 = \$0.047, which is less than 5 cents.
No matter what your tariff is, refilling your e-bike battery will always be a true bargain!
How does the cost per trip compare?
Sometimes, it’s more helpful to know the cost of a specific trip you take regularly, rather than the cost of a full recharge. If you already have a full battery, calculating the cost is quite simple. However, you will need a plug-in consumption meter, like the one available on Amazon, to measure the amount of electricity required to fully recharge the battery.
(Measured kWh) x (Electricity cost per kWh)
Let’s consider a typical example of a 15-mile round trip by e-bike, which requires 300Wh of electricity. At the average cost of 10.6 cents per kWh, the calculation would be:
0.3kWh x \$0.106 = \$0.032. That’s right, just over three cents worth of electricity for a typical commuting trip.
While it’s difficult to make an exact comparison with other means of transportation due to various factors, consider the examples provided in an article by The Points Guy, which highlights public transit passes in major US cities ranging from \$1.25 to \$2.75. There is no contest in terms of cost when compared to an e-bike.
READ MORE : Exploring London and South-East England on eBikes: Guide to Electric Bike Rentals/Hire
## What about other countries? Electricity prices, e-bike costs, and comparisons
Does the incredibly low cost of recharging your e-bike hold true in other countries? This question arises due to the disparity in electricity prices between countries. To illustrate this point, we selected the United Kingdom as an example. Firstly, because EBR has a considerable number of UK readers, and secondly, because electricity prices per kWh are considerably higher in the UK compared to the US. In fact, they have recently experienced a significant price hike, with the average figure now standing at around £0.35 (\$0.45) per kWh.
Let’s consider a basic example of fully recharging a typical 500Wh e-bike battery, based on the UK average cost of £0.35 per kWh:
(0.5kWh x 1.25) x £0.35 = £0.22 – which is approximately 22 pence (or around 30 cents) for a complete recharge. In comparison, the US, with an average electricity tariff of 6.5 cents (or 5 pence), would only cost a mere 6.5 cents for the same process.
Therefore, while the cost of refueling your e-bike in the UK and several other Western European countries may be higher than in the US, it still presents an astonishingly good deal. Let’s consider a personal experience shared by the author: for his regular hilly commuting route, a 500Wh battery would give him a range of approximately 30 miles. Utilizing public transport for the same journey would cost around £6 – nearly thirty times the cost of the electricity used by the e-bike!
## E-bikes vs Electric cars: Charging costs comparison
Electric cars are frequently championed as the solution to the climate crisis by many governments worldwide. However, e-bikes often receive less attention despite their significantly lower energy consumption compared to electric cars.
Now, let’s examine the charging costs for electric cars in comparison. Although the diagram presented in the previous section already conveyed that e-bikes require considerably less energy than electric cars, let’s delve into the specifics by analyzing a couple of examples using the most popular electric car models in the US and UK.
To calculate the charging costs, we apply the previously mentioned formula, utilizing the average kWh costs from the US and UK (10.6 cents / 8 pence and 35 pence / 45 cents per kWh). It’s important to note that these calculations are based on home charging rates, as public car charging points tend to be more expensive but provide faster charging.
READ MORE : Three Factors Demonstrating the Impracticality of Investing in an Inexpensive Replica Bicycle.
Tesla Model Y – US figures
(75kWh x 1.25) x \$0.106 = \$9.93 (£7.55)
Tesla Model 3 – UK figures
(50kWh x 1.25) x £0.35 = £21.87 (\$29)
Once again, the comparison clearly favors e-bikes in terms of cost-efficiency.
Additionally, it is crucial to consider that electric cars consume significantly more electricity to cover the same distance as an e-bike, ultimately resulting in more frequent recharging. A notable article provides consumption figures that indicate electric cars typically require 190-290 Wh per mile, whereas e-bikes have a consumption range of 10-25 Wh per mile. Even the least efficient e-bikes are approximately eight times more energy-efficient than the most efficient electric cars.
Undoubtedly, when compared to a purchase price of at least \$1000 (potentially even a few thousand dollars), the expenses involved in recharging an e-bike are truly insignificant. Naturally, there are also certain costs associated with the regular maintenance of an e-bike. While most of these can be considered relatively inexpensive in comparison to the initial purchase price, the cost of replacing batteries can amount to a few hundred dollars.
While it is beyond the scope of this article to delve into a detailed analysis of the long-term total cost of owning an e-bike, previous comparisons (such as those found in the book Electric Bicycles – although slightly outdated with current figures likely to be significantly higher) suggest that e-bikes are exceptionally cost-effective when compared to other modes of transportation:
### Comparative Running Costs:
• Conventional (non-electric) Bike: \$0.07 – \$0.09 per mile / £0.03 – £0.07 per mile
• Electric Bike: \$0.11 – \$0.16 p/m / £0.088 – £0.12 p/m
• Small Car: \$0.42 – \$1.91 p/m / £0.32 – £1.45 p/m
• Small Car – local use only: \$0.84 – \$3.37 p/m / £0.64 – £2.56 p/m
• Bus: ~ \$0.55 p/m / ~ £0.40 p/m
• Train: \$0.26 – \$0.80 p/m / £0.20 – £0.60 p/m
In summary, all the available evidence suggests that by embracing e-biking, you are not only enjoying the countless benefits of this form of transportation, but you are also significantly reducing your expenses! | 2,235 | 9,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-50 | longest | en | 0.949733 |
https://stacks.math.columbia.edu/tag/0B4E | 1,721,372,915,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00330.warc.gz | 457,726,971 | 6,242 | Lemma 68.22.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume
1. $X$ and $Y$ are decent and have finitely many irreducible components,
2. $f$ is integral and birational,
3. $Y$ is normal, and
4. $X$ is reduced.
Then $f$ is an isomorphism.
Proof. Let $V \to Y$ be an étale morphism with $V$ affine. It suffices to show that $U = X \times _ Y V \to V$ is an isomorphism. By Lemma 68.22.6 and its proof we see that $U$ and $V$ are decent and have finitely many irreducible components and that $U \to V$ is birational. By Properties, Lemma 28.7.5 $V$ is a finite disjoint union of integral schemes. Thus we may assume $V$ is integral. As $f$ is birational, we see that $U$ is irreducible and reduced, i.e., integral (note that $U$ is a scheme as $f$ is integral, hence representable). Thus we may assume that $X$ and $Y$ are integral schemes and the result follows from the case of schemes, see Morphisms, Lemma 29.54.8. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 350 | 1,172 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-30 | latest | en | 0.882613 |
https://socratic.org/questions/what-is-the-slope-and-intercept-of-x-5-0 | 1,722,659,859,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00813.warc.gz | 439,009,094 | 6,093 | # What is the slope and intercept of x-5 = 0?
Jun 6, 2016
$\text{y-intercept } \to \left(x , y\right) = \left(0 , - 5\right)$
Slope is 1
#### Explanation:
Write as:$\text{ } x - 5 = y$
Turn it round and we have the same thing
$y = x - 5$
Compare to $y = m x + c$
where m is the gradient. It is known that $1 \times x = x$ so the gradient m must be 1.
The slope is 1. This means that for every 1 along left to right the graph goes up 1.
The y intercept is at $x = 0$ so
$y = x - 5 \text{ becomes } y = 0 - 5$
So the y intercept is at $y = - 5$
To write this properly you would present it as
$\text{y-intercept } \to \left(x , y\right) = \left(0 , - 5\right)$ | 241 | 672 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-33 | latest | en | 0.899557 |
https://www.topperlearning.com/doubts-solutions/can-you-please-tell-me-most-13579 | 1,521,296,537,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645177.12/warc/CC-MAIN-20180317135816-20180317155816-00512.warc.gz | 897,512,823 | 40,832 | 1800-212-7858 (Toll Free)
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# Can you please tell me most important topic from both chapters
Asked by jayantvatsa2005 6th March 2017, 6:16 PM
Hi Nikita,
Answered by Expert 6th March 2017, 6:17 PM
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You have rated this answer /10
## blueprint of maths class 10
Asked by SNEHA 002 17th March 2018, 7:31 PM
## Four cubes of edge 5 cm each are placed side by side and stuck together . If the outer surface alone is painted at Rs 300 per meter square , find the cost of painting
Asked by deepikankur 17th March 2018, 6:55 PM
## A cylindrical roller is used to level a rectangular playground . The length of the playground is 3.5 m and its diameter is 2.8m . If the roller rolls over 200 times to completely cover the ground, find the area of the playground
Asked by deepikankur 17th March 2018, 6:53 PM | 334 | 1,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-13 | latest | en | 0.936666 |
http://www.edupil.com/question/difference-of-two-numbers-is-if-7-5-of-one-number-is-equal-to-12-5/ | 1,477,473,548,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720845.92/warc/CC-MAIN-20161020183840-00549-ip-10-171-6-4.ec2.internal.warc.gz | 418,765,628 | 17,285 | # The difference of two numbers is If 7.5% of one number is equal to 12.5%
The difference of two numbers is If 7.5% of one number is equal to 12.5% of the other number, then the two numbers are
(1) 2490, 4150
(2) 2510, 4170
(3) 2600, 4260
(4) 2700, 4360
Anurag Mishra Professor Asked on 19th April 2016 in
Solution:-
Let the one number be X
The difference of two numbers is 1660
Then, another number = X + 1660
If 7.5% of one number is equal to 12.5% of the other number.
Then, X x 7.5/100 = (X + 1660) x 12.5/100
75 X/1000 = (X + 1660) x 125/1000
75 X = 125 X + 1660 x 125
50 X = 1660 x 125
X = 1660 x 125/50
X = 830 x 5
X = 4150
Then, other number will be 4150 – 1660
= 2490
So, the two numbers are 2490, 4150.
Hence, the correct answer is option (1) 2490, 4150.
Anurag Mishra Professor Answered on 20th April 2016. | 342 | 834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | longest | en | 0.835356 |
http://mathhelpforum.com/trigonometry/179270-area-five-sided-coin.html | 1,481,117,281,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542112.77/warc/CC-MAIN-20161202170902-00163-ip-10-31-129-80.ec2.internal.warc.gz | 174,467,983 | 10,528 | # Thread: Area of five-sided coin
1. ## Area of five-sided coin
stuck on another problem from Abbotts "Teach Yourself" Trigonometry:
A new five-sided coin is to be made in the shape of figure /image attached/
Тhe point А on the cirсumferenсe of the coin is the сentre of the
arc CD, which has а radius of 2 сm. Similarly B is the сentre of
the arс DЕ, and so on. Find the area of one faсe of the сoin.
hope you can help me, thanks in advance!
2. Originally Posted by dmi3
stuck on another problem from Abbotts "Teach Yourself" Trigonometry:
A new five-sided coin is to be made in the shape of figure /image attached/
Тhe point А on the cirсumferenсe of the coin is the сentre of the
arc CD, which has а radius of 2 сm. Similarly B is the сentre of
the arс DЕ, and so on. Find the area of one faсe of the сoin.
hope you can help me, thanks in advance!
1. Draw a sketch (see attachment)
2. Draw a pentagon ABCDE. Determine the interior angles.
3. The area of the coin consists of
triangle
• ACD: $a_{ACD} = \frac12 \cdot 2 \cdot 2 \cdot \sin(36^\circ)$
• 2 triangles ABC: $a_{ABC} = \frac12 \cdot 2 \cdot h = \tan(36^\circ)$
• 5 segments: area of sector - area of triangle ACD: $a_{segm}= \frac{36^\circ}{360^\circ} \cdot \pi \cdot 2^2 - \frac12 \cdot 2 \cdot 2 \cdot \sin(36^\circ)$
4. I've got $a_{coin} \approx 3.034\ cm^2$
3. 3.034 is correct, thanks! | 446 | 1,366 | {"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": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.845228 |
http://slideplayer.com/slide/3581374/ | 1,524,223,369,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937440.13/warc/CC-MAIN-20180420100911-20180420120911-00371.warc.gz | 280,209,641 | 20,111 | # Factor Proportions and the Structure of Trade: HOS- Krugman-DFS Model The Explanation of International Trade: Differences across countries in relative.
## Presentation on theme: "Factor Proportions and the Structure of Trade: HOS- Krugman-DFS Model The Explanation of International Trade: Differences across countries in relative."— Presentation transcript:
Factor Proportions and the Structure of Trade: HOS- Krugman-DFS Model The Explanation of International Trade: Differences across countries in relative abundance of factors of production. Assumptions:Identical Technologies Identical Demand Patterns The HOS Model:
Relative Factor Intensity : Full employment: Y X A B* L constant K constant Structural Bias: The Transformation Curve( = ABC) shifts asymmetrically with unbalanced changes in K and L. A Rise in K, with no change in L, leads to an increase(fall) in X (Y)). C D B E F
AT POINT F 1) Labor is unemployed: W=0. (2) The X-industry is active The Y-industry is inactive. Therefore: AT POINT A 1) Capital is unemployed: R=0. (2) Y-industry is active X-industry is inactive. Therefore:
AT Point A (continue):
At Point B Relative Supply
Two Countries: H and F: H is more capital abundant. H’s Relative Supply is biased towards X:
Free Trade and Autarkic Equilibria 32 1 2=Free trade 1=autarky in H 3=autarky in F
Full Employment Supply of X and Y :
The Heckscher-Ohlin Proposition #1: Any country will export the good which makes intensive use in its production of relative abundant factor supply.
Full Employment Factor Prices:
Income Distribution and International Trade R W A B B’ D C ABC=factor price frontier A rise in (X is capital intensive) will raise R and decrease W. Industry X-Line Industry Y-Line
The Heckscher-Ohlin Proposition #2(dual to Proposition #1): Free trade causes an increase in the factor price of the factor of production which is used intensively in the export industry and a fall in the factor price used intensively in the import competing industry.
Factor Price Equalization: Failures Two ways to generate a failure of FPE: Assume that factor proportions are sufficiently different that they are outside the FPE set. Introduce costs to international trade, which could have strong effect on trade volume.
Romalis (AER, March 2004, 94, No.1, 67-97) Generalizes a Heckscher-Ohlin model of Dornbusch-Fischer-Samuelson framework, and explains trade structure; Assumes a many-country version of the Heckscher-Ohlin model; Integrates this with Krugman intra-industry trade; Allows for transportation costs.
The Model There are 2M countries, M each in the North and South. Southern variables are marked with an asterisk. There are two factors of production: skilled and unskilled labor. The proportion of skilled labor is Northern countries are abundant in skilled labor
Preferences
Monopolistic Competition = Production of variety i Number of of varieties in industry z Sub-utility function Fixed cost Dual Unit cost Number of countries
Transportation costs Units of a good must be shipped for 1 unit to arrive in any other country
Equilibrium in an industry Solve for the share of world production that each country commands, conditional on relative production costs. Countries with lower costs capture larger market shares. Consumer price Ideal Price index
National income and Spending A constant fraction of income b(z) is spent on industry z
World Demand
North-South relative price
If is low, (1) is the solution; if is high, (2) is the solution (1) (2)
General Equilibrium
Special Case The Dornbusch-Fischer- Samuelson Model is a special case with no transportation costs Perfect competition
Transport costs The addition of the transport costs leads a stark structure of production and trade: Share of industry Skill intensity of industry (z) Unskilled goods produced in south Non-traded goods Produced In North Non-traded goods Produced In South Skill-intensive goods Produced In North
Producer prices p=factory-gate price domestically; Sold in M-1 markets abroad | 895 | 4,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-17 | latest | en | 0.816015 |
https://proofwiki.org/wiki/Exponential_Generating_Function_for_Boubaker_Polynomials | 1,670,491,143,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711286.17/warc/CC-MAIN-20221208082315-20221208112315-00149.warc.gz | 485,412,693 | 11,769 | # Exponential Generating Function for Boubaker Polynomials
## Theorem
The Boubaker polynomials, defined as:
$\map {B_n} x = \begin {cases} 1 & : n = 0 \\ x & : n = 1 \\ x^2 + 2 & : n = 2 \\ x \map {B_{n - 1} } x - \map {B_{n - 2} } x & : n > 2 \end {cases}$
have as an exponential generating function:
$\ds \map {f_{B_n, \operatorname {EXP} } } {x, t} = \sum_{n \mathop = 0}^\infty \map {B_n} x \frac {t^n} {n!} = 4 x e^{t \frac x 2} \frac {\map \sin {t \sqrt {1 - \paren {\frac x 2}^2} } } {\sqrt {1 - \paren {\frac x 2}^2} } - 2 e^{t \frac x 2} \map \cos {t \sqrt {1 - \paren {\frac x 2}^2} } - 3$
## Proof
From the definition of the Boubaker polynomials, we have:
$\begin {cases} \map {B_0} x = 1 \\ \map {B_1} x = x \end {cases}$
We have also, for $T_n$, the Chebyshev polynomials of the first kind and $U_n$, the Chebyshev polynomials of the second kind:
$\begin {cases} \map {T_0} x = 1 \\ \map {T_1} x = x \end {cases}$
and:
$\begin {cases} \map {U_{-1} } x = 0 \\ \map {U_0} x = 1 \\ \map {U_1} x = 2 x \end {cases}$
Then, from the definition of the exponential generating functions of the Chebyshev polynomials of the first kind:
$\ds \map {f_{T_n, \operatorname {EXP} } } {x, t} = e^{t \frac x 2} \map \cos {t \sqrt {1 - \paren {\frac x 2}^2} }$
$\ds \map {f_{U_n, \operatorname {EXP} } } {x, t} = e^{t \frac x 2} \frac {\map \sin {t \sqrt {1 - \paren {\frac x 2}^2} } } {\sqrt {1 - \paren {\frac x 2}^2} }$
and from the relation:
$\map {B_n} x = 4 x \map {U_{n - 1} } {\dfrac x 2} - 2 \map {T_n} {\dfrac x 2}$
which is valid only for $n > 1$, we have:
$\ds \map {B_0} x \dfrac {t^0} {0!} + \map {B_1} x \dfrac {t^1} {1!} + \sum_{n \mathop = 2}^\infty \map {B_n} x \dfrac {t^n} {n!} = 4 x \sum_{n \mathop = 2}^\infty \map {U_{n - 1} } x \dfrac {t^n} {n!} - 2 \sum_{n \mathop = 2}^\infty \map {T_n} x \dfrac {t^n} {n!} + \map {B_0} x \dfrac {t^0} {0!} + \map {B_1} x \dfrac {t^1} {1!} - 4 x \sum_{n \mathop = 0}^1 \map {U_{n - 1} } x \dfrac {t^n} {n!} - 2 \sum_{n \mathop = 0}^1 \map {T_n} x \dfrac {t^n} {n!}$
Finally, by simplifying the two sides and considering the exponential generating functions and the values of the first terms of $T_n$ and $U_n$, we obtain:
$\ds \sum_{n \mathop = 0}^\infty \map {B_n} x \frac{t^n} {n!} = 4 x e^{t \frac x 2} \dfrac {\map \sin {t \sqrt {1 - \paren {\frac x 2}^2} } } {\sqrt {1 - \paren {\frac x 2}^2} } - 2 e^{t \frac x 2} \map \cos {t \sqrt {1 - \paren {\dfrac x 2}^2} } - 3$
Hence the result.
$\blacksquare$ | 1,101 | 2,482 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-49 | latest | en | 0.447777 |
https://www.jiskha.com/display.cgi?id=1319674049 | 1,503,245,131,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106779.68/warc/CC-MAIN-20170820150632-20170820170632-00003.warc.gz | 939,993,314 | 4,879 | # math
posted by .
if the ratio of rose bushes to trees in the yard is 3 to 1. Which of these shows possible numbers of roses to trees in the yard.
• math -
Which of WHAT?
• math -
a- 20 rose bushes, 7 trees
b- 7 rose bushes, 3 trees
c- 21 rose bushes, 7 trees
d- 7 roses bushes, 21 trees
• math -
• math -
if the ratio of girls to boys on the track team is 5 to 6 which of these shows possible numbers of girls to boys on the track team.
a- 24 girls, 20 boys
b-50 girls, 56 boys
c- 60 girls, 50 boys
d- 20 girls,24 boys
• math -
• math -
is the correct answer 20 girls to 24 boys
• math -
Yes. 5:6 = 20/24
• math -
is the correct on the rose bushes
7 roses bushes, 3 trees
• math -
• math -
if the ratio of red bows to green bows is 2 to 3 which of these shows possible numbers of red bows to green bows.
is this the correct answer 23 red bows, 32 green bows
• math -
is the correct answer on the rose bushes 21 rose bushes to 7 trees
• math -
2/3 = 4/6 = 6/9 = 8/12 = 10/15 . . .
Your rose bushes answer is now right >> 3 : 1 = 21/7
• math -
is this the correct answer for the bows
132 red bows,88green bows
• math -
No.
With a ratio of 2 to 3, you have fewer red bows than green.
• math -
is this correct 88 red bows, 132 green bows? thank you
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John has 1/4 yard of fabric. He cuts it into 3 equal pieces . What fraction of a yard is each piece of fabric? | 897 | 3,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-34 | latest | en | 0.916009 |
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# Picture a Pyramid ...
##### Stage: 2 Challenge Level:
You will probably have to write some numbers down.
BUT
Try not to draw anything.
Here we go ...
Imagine a step pyramid.
Each layer is made from a number of stone cubes.
There is a square number of cubes in each layer, which increases for each layer (increasing from top to bottom).
The cubes are numbered starting from the top layer so that, looking from above, each new layer carries on counting starting from the North West corner. (When the end of a row is reached you start at the West end of the next row.)
There are equal steps on all four sides of the pyramid (in other words, each successive layer "sits" exactly in the middle of the layer beneath it).
We'll imagine a six-layered pyramid that uses 91 cubes in total.
What numbered cubes are vertically underneath (in a straight line):
the cube that is number 1?
the cube that is number 4?
the cube that is number 8?
Now imagine stepping up two layers at a time, starting at the middle cube on the South side, second layer up. Which numbered cubes would you step on to end at the top, if you stepped up in a straight line?
Starting at the South West Corner and stepping up one layer at a time on the corner steps, which numbered cubes would you step on?
What if a second pyramid was built, numbering the cubes as they built the pyramid, so the numbering would start from the bottom North West corner and starting the second row as before.
So number 1 at the bottom and the last cube number 91.
What numbered cubes are vertically underneath (in a straight line):
the cube that is number 91?
the cube that is number 89?
the cube that is number 80?
Is there a cube that has the same number in both pyramids?
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https://luanvan68.com/how-many-16-9-oz-bottles-in-a-gallon/ | 1,713,599,936,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817491.77/warc/CC-MAIN-20240420060257-20240420090257-00120.warc.gz | 345,108,754 | 51,443 | # How Many 16.9 oz Bottles in a Gallon – A Comprehensive Guide
Discover how to calculate how many 16.9 oz bottles in a gallon with our comprehensive guide. Get accurate measurements for cooking, mixing, and more!
## Introduction
Do you ever find yourself wondering how many 16.9 oz bottles are in a gallon? Whether it’s for cooking, measuring liquids or even just out of curiosity, understanding this conversion is vital. In this comprehensive guide, we’ll explore everything you need to know about gallons and fluid ounces, why knowing the number of bottles per gallon is important, and how to calculate it accurately.
### What are Gallons and Fluid Ounces?
Before delving into the calculation of 16.9 oz bottles per gallon, let’s first understand what gallons and fluid ounces are. A gallon is an imperial unit of volume measurement commonly used in the United States and other countries that have not adopted the metric system. One US liquid gallon is equivalent to approximately 3.785 liters or 128 fluid ounces. On the other hand, fluid ounces measure smaller quantities of liquids such as milk, water, or juice.
### Importance of Knowing How Many Bottles are in a Gallon
Knowing how many bottles are in a gallon can be essential when preparing meals or mixing drinks. It ensures accuracy in measurements and simplifies recipe preparation by reducing guesswork. Additionally, it can help save money by determining whether buying larger quantity containers is more cost-effective than smaller ones.
## Conclusion
Understanding conversions between different units of measurements like gallons and fluid ounces is essential knowledge to have for everyday life. This guide has covered what gallons and fluid ounces are, emphasized the importance of knowing how many bottles are in a gallon, and provided an overview of the article’s content. Now that we’ve covered the basics let’s dive into calculating how many 16.9oz bottles there are in a gallon!
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## Understanding the Basics: Gallons and Fluid Ounces
### What are Gallons and Fluid Ounces?
Gallons and fluid ounces are both units of volume measurement commonly used in the United States. As previously mentioned, a gallon is an imperial unit that measures larger quantities of liquids such as gasoline or milk. In contrast, fluid ounces measure smaller amounts of liquid like water or juice.
### Conversion Factors between Gallons and Fluid Ounces
To convert gallons to fluid ounces, you need to multiply the number of gallons by 128. For example, if you have two gallons of liquid, it would be equivalent to 256 fluid ounces (2 x 128). Conversely, to convert fluid ounces to gallons, divide the number of fluid ounces by 128. For instance, if you had 384 fluid ounces, it would equate to three gallons (384 รท 128).
Understanding these conversion factors is crucial when calculating how many 16.9 oz bottles there are in a gallon accurately. Now that we’ve covered the basics let’s move on to determining how many bottles there are per gallon!
## Determining the Number of 16.9 oz Bottles in a Gallon
### Calculation Method for Determining How Many 16.9 oz Bottles are in a Gallon
To calculate how many 16.9 oz bottles make up one gallon, we need to know that one US liquid gallon is equivalent to 128 fluid ounces and that each bottle contains 16.9 fluid ounces of liquid.
Using these values, we can determine the number of bottles per gallon by dividing the total number of fluid ounces in a gallon (128) by the volume of each bottle (16.9). The calculation looks like this:
``````Number of Bottles = Total Fluid Ounces / Volume per Bottle
= 128 / 16.9
= 7.57``````
Therefore, there are approximately 7.57 bottles of 16.9 oz size in one gallon.
### Examples of How to Apply the Calculation Method
Let’s consider an example where you have a recipe that requires two gallons of water and uses sixteen-ounce cups to measure it out. Knowing that there are approximately seven and a half bottles in one gallon, you can calculate how many cups of water you will need as follows:
``````Total Cups Needed = Total Gallons x Cups per Gallon
= 2 x 128
= 256
Cups per Bottle = Volume per Bottle / Volume per Cup
= 16.9 / 8
= 2.11
Bottles Needed = Total Cups Needed / Cups per Bottle
= 256 / 2.11
โ 121
Gallons Equivalent= Bottles Needed * Volume Per Bottle/Volume Per Gallon
โ121*16.9/128
โ15.92
``````
So, you would need around 121 bottles of 16.9 oz size to make two gallons of water or approximately 15.92 gallons equivalent.
Knowing how to calculate the number of bottles per gallon is essential for precision in measurements, making sure you have the right amount of liquid for your recipe and avoiding waste.
## Factors That Can Affect the Number of Bottles in a Gallon
When calculating how many 16.9 oz bottles are in a gallon, it’s important to consider that several factors could affect the accuracy of your calculations. Here are some common factors you need to keep in mind:
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### Variations in Bottle Sizes
Bottle sizes can vary depending on the manufacturer or brand. Some bottles may have slightly different dimensions, which can cause variations in their volume capacity. As such, it is essential to ensure that all bottles being used have the same size and shape for accurate measurements.
### Differences in Liquid Density
Different liquids have varying densities, which means that the volume they occupy will differ even if they share similar measurements. For instance, one gallon of water weighs around 8 pounds while one gallon of honey weighs about 12 pounds. This difference in density affects how much liquid can fit into a bottle and ultimately impacts your calculation.
### Other Factors That Could Affect Calculations
The temperature of the liquid and atmospheric pressure can also impact calculations. For example, hot liquids tend to expand while cold liquids contract. Therefore, the temperature at which you measure liquid volumes should be consistent throughout the experiment. Similarly, atmospheric pressure changes with altitude; as you go higher up a mountain or building, air pressure decreases, impacting liquid measurement.
It’s crucial to recognize these factors when making conversions between units of measurements like gallons and fluid ounces. Failing to account for these variables might lead to inaccurate results that could impact recipe preparation or other tasks where precise measurements are necessary.
Now that we’ve covered the potential factors influencing bottle-to-gallon conversion let’s move onto some practical applications!
## Practical Applications of Knowing How Many Bottles Are in a Gallon
Knowing how many 16.9 oz bottles there are in a gallon can have several practical applications. Here are some use cases and benefits of understanding this conversion:
### Use Cases for Calculating the Number of Bottles per Gallon
• Cooking: Recipes often require specific quantities of liquids, and knowing how many bottles of a particular size make up a gallon can help ensure accuracy in cooking.
• Mixing drinks: Bartenders or mixologists need to know how many bottles it takes to fill a container or mixer accurately.
• Budgeting: Understanding the number of bottles per gallon can help determine whether purchasing smaller or larger quantity containers is more cost-effective.
### Benefits of Knowing This Information
• Saves time and effort by reducing guesswork when measuring liquids.
• Reduces waste by preventing overuse of ingredients due to inaccurate measurements.
• Helps prepare meals efficiently and with better precision, improving overall quality.
• Enables bartenders or mixologists to create consistent cocktails every time without wasting any excess liquid.
By knowing how many 16.9 oz bottles there are in a gallon, you can simplify your daily routine and improve your efficiency in various tasks that require accurate measurements.
### What factors affect the number of bottles in a gallon?
Several factors such as variations in bottle sizes, differences in liquid density, and other environmental conditions can affect the number of bottles that can fit in a gallon. It is important to note that understanding these factors can help you determine how much liquid you need to purchase for your project or recipe accurately.
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### How do I calculate the number of 16.9 oz bottles per gallon?
To calculate the number of 16.9 oz bottles per gallon, first divide the number of fluid ounces in a gallon (128) by the size of each bottle (16.9). The result is approximately 7.57 bottles per gallon.
### Can I use this conversion method for other bottle sizes?
Yes, this conversion method can be used with other bottle sizes as well. Simply replace the size of the bottle in fluid ounces within the calculation formula to get an accurate result.
### Why is it essential to know how many bottles are in a gallon?
Knowing how many bottles are in a gallon helps ensure accuracy when measuring liquids for cooking or mixing drinks. It simplifies recipe preparation and reduces guesswork while also helping save money when purchasing larger quantities.
### Are there any tools available that can help me perform these calculations easily?
Yes, several online calculators are available that make these calculations easy and straightforward, saving time and effort on manual calculations. These calculators take into account different variations such as bottle size and liquid density, providing accurate results every time.
## Conclusion
Converting between gallons and fluid ounces is easy once you understand the basic principles behind it. This section has answered some frequently asked questions related to converting bottles to gallons and vice versa, providing clear explanations for each query. With this information at hand, you’re now ready to tackle any conversion challenge thrown your way!
## Tips for Accurate Calculations
When calculating the number of 16.9 oz bottles in a gallon, accuracy is key. Here are some best practices to help you achieve accurate calculations.
### Best Practices for Accurate Calculations when Determining Number of Bottles per Gallon
1. Use a measuring cup with clear markings: Measuring cups with clear markings make it easier to read measurements and ensure an accurate calculation.
2. Double-check conversions: It’s important to double-check your calculations or use an online calculator to avoid errors that can cause inaccurate results.
3. Account for variations in bottle sizes: Not all bottles will be exactly 16.9 ounces, so it’s crucial to consider any variations in size when calculating the number of bottles in a gallon.
4. Weigh liquids: In some cases, weighing the liquid may be more accurate than using volume measurements, especially if there are differences in density between liquids.
### Tools That Can Help Make Calculations Easier
1. Online calculators: There are plenty of online calculators available that can quickly convert gallons to fluid ounces and vice versa.
2. Conversion charts: Printable conversion charts can be useful references when doing manual calculations.
3. Smart kitchen scales: Some smart kitchen scales come with built-in conversion features that can convert units of measurement automatically.
By following these tips and utilizing helpful tools, you can ensure that your calculations are as precise as possible when determining how many 16.9 oz bottles there are in a gallon.
## Conclusion
In conclusion, understanding how many 16.9 oz bottles are in a gallon is an essential skill for anyone who needs to measure liquids accurately. We’ve covered the basics of gallons and fluid ounces, learned how to calculate the number of bottles in a gallon, discussed factors that can affect the calculation, and explored practical applications for this knowledge.
Knowing the precise number of bottles per gallon can save time and money when purchasing or preparing drinks or meals. It also helps ensure consistent results in cooking recipes by providing accurate measurements.
Remember to keep in mind any variations in bottle sizes or liquid density that could impact your calculations and use best practices for accuracy. With this comprehensive guide, you now have everything you need to confidently convert between gallons and 16.9 oz bottles with ease.
So next time you’re wondering how many 16.9 oz bottles are in a gallon, remember these tips and tricks to make your calculations quick and effortless!
## How Many Pounds is 600 kg? – A Comprehensive Guide
Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature.
## How Many Pounds is 600 kg? – A Comprehensive Guide
Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature.
## How Many Ounces in 1.5 Pounds?
Learn how to convert pounds to ounces and vice versa accurately! Discover “how many ounces in 1.5 pounds” and more with this comprehensive guide.
## How Many Pounds Is 23 Kilos?
Learn how to convert 23 kilos to pounds and gain a better understanding of the conversion process between these two units of measurement.
## How Many M&Ms in a Pound: Everything You Need to Know
Discover the exact number of M&Ms in a pound and more with our comprehensive guide. Learn how to measure them by weight and find out what affects their quantity!
## How Many M&Ms in a Pound: Everything You Need to Know
Discover the exact number of M&Ms in a pound and more with our comprehensive guide. Learn how to measure them by weight and find out what affects their quantity! | 2,810 | 13,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-18 | latest | en | 0.904775 |
http://www.cprogrammingcode.com/2016/04/find-maximum-difference-between-two.html | 1,516,728,403,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892059.90/warc/CC-MAIN-20180123171440-20180123191440-00177.warc.gz | 439,655,671 | 20,810 | Find Maximum Difference between Two Elements of an Array - Java Program
Write a program to find maximum difference between two elements of an array. Given an unsorted array find maximum difference between two element of an array.
How to find maximum difference between two elements of an array
Maximum difference = Maximum value of an array - Minimum value of an array
Java programs
Find second largest number in array
So to solve this problem we need to find maximum and minimum value of an array. Then subtracting these two values. we can find the maximum difference.
Find Maximum Difference between Two Elements of an Array
METHOD 1 : Using two for loops
```package maximumdifference;
import java.util.*;
public class MaximumDifference {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
int size, i, j, max = 0;
Scanner in = new Scanner(System.in);
System.out.println("Enter the size of an array");
size = in.nextInt();
/* Declare an array */
int[] arr = new int[size];
System.out.println("Enter values in an array");
for (i = 0; i < size; i++) {
arr[i] = in.nextInt();
}
for (i = 0; i < size; i++) {
for (j = 0; j < size; j++) {
if ( (arr[i] - arr[j]) > max) {
max = arr[i] - arr[j];
}
}
}
System.out.println(" Max difference is "+ max);
}
}
```
Output
Enter the size of an array 5
Enter values in an array
3
1
6
7
9
Max difference is 8
METHOD 2 - By finding minimum and maximum value in an array
```package maximumdifference;
import java.util.*;
public class MaximumDifference {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
int size, i, j, max, min;
Scanner in = new Scanner(System.in);
System.out.println("Enter the size of an array");
size = in.nextInt();
/* Declare an array */
int[] arr = new int[size];
System.out.println("Enter values in an array");
for (i = 0; i < size; i++) {
arr[i] = in.nextInt();
}
/* Initialize min and max value */
min = arr[0];
max = arr[0];
/* Find min and max in an array */
for (i = 0; i < size; i++) {
if ( arr[i] > max) {
max = arr[i];
}
if ( arr[i] < min) {
min = arr[i];
}
}
/* Maximum difference is max - min */
System.out.println(" Max difference is "+ (max - min));
}
}
```
You can also use sorting algorithm to find minimum and maximum value in an array.
Insertion sort implementation in java
Bubble sort
Quick sort
time complexity of sorting algorithms . | 618 | 2,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | longest | en | 0.541232 |
http://betonvalue.net/sport-betting/sports-betting-programs/ | 1,701,411,326,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100276.12/warc/CC-MAIN-20231201053039-20231201083039-00187.warc.gz | 4,952,309 | 12,425 | # Sports Betting Programs
I suppose almost everyone already knows those legends about people that have invented and used sports betting systems so they can avoid loses and get great profits in the long run. Well, in today?s hi-tech world where sports betting industry is continually growing and making larger and larger profit annually, it really is almost impossible to believe that numerous people still live from betting.
Many have asserted maybe it?s due to the sports betting systems they normally use that they can still carry on and bet with the hope of winning bucks win-every-time.com. The truth of the matter is, there’s really no particular system that will promise you 100% profit and no risk of getting broke.
Nevertheless, many bettors believe you can still find certain sports betting systems that work pretty well for a number of people. These sports betting systems, although functional to some degree, however, imposes bigger risk for the players. Here are few of them:
Martingale System
Martingale System is fundamentally the most widely known sports betting system in the world. It has nothing to do with picking the winner as the whole story of this method is in picking the right stakes. Well, as outlined by some resources, the Martingale System of sports betting is based on the odds of losing infinite times repeatedly. It really is applied by beginning with one bet, and starting again in the event you win. However, should you lose, you double your bet, and each time you lose, you double your last lost bet. As believed, this may eventually enables you to win the betting and when you win you’d probably recover all your lost bets plus one unit profit against your initial wager. The most critical sentence when it comes to Martingale is “double you stake after you loose and begin again when you win”.
Parlay System
Parlay System is probably the most well-known sports betting systems that are widely used in horse racing. Experts have asserted unlike the other sports betting systems, the Parlay System provides the effect of pyramiding your profit blog here. Pyramiding is really a parlay wager whereby the initial wager plus its winnings are played on successive wagers. Following the principles of pyramiding, what’s basically involved in the Parlay System of sports betting is that you make a bet and if you win you re-invest the winnings for the next bet.
So in horse racing, for example, you merely allow it to ride. Also, unlike other sports betting systems, the Parlay System affords the least degree of risk of all wagers for the reason that the bettors are only concerned with either a win, place or show selection or maybe a combination of the three.
Paroli System
Of all the so-called well-known sports betting systems, the Paroli System is considered to be the exact opposite of Martingale. Well, the real difference lies on the notion that on Paroli, you start with one bet and increase the bet whenever you win rather than you lose. The main advantage of this particular one of the very popular sports betting systems is that you simply do not require a sizable bankroll. The system lets the profit run and cut short the losses. | 642 | 3,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-50 | latest | en | 0.965536 |
https://community.qlik.com/thread/126623 | 1,532,278,409,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593378.85/warc/CC-MAIN-20180722155052-20180722175052-00509.warc.gz | 621,363,552 | 22,775 | 5 Replies Latest reply: Jul 22, 2014 7:38 AM by Martyn Lloyd
# Aging of invoices
Hi All,
Can I ask a little help regarding my report.
I'd like to get the aging of all the unpaid invoices and segregate them according to aging, 30 days, 60 days , 90 days and beyond 90 days excluding weekends and holiday. Below is/are the following field name:
Entry Date - this is the starting date
07.22.2014 - current date or end date
Thanks!
-Ahyel
• ###### Re: Aging of invoices
Apart from QV script, practically I have never seen excluding off days for ageing.
• ###### Re: Aging of invoices
In Calculated Dimension create expression like following will create buckets
=if((num((num(vDateForAgeing)-[Due Date CLE]),'#######0') )<=0 ,'Not Due',
if(IsNull([Due Date CLE]) OR [Due Date CLE]='','Not Due',
if((num((num(vDateForAgeing)-[Due Date CLE]),'#######0'))>=0.1 and (num((num(vDateForAgeing)-[Due Date CLE]),'#######0'))<=30,'1-30',
if((num((num(vDateForAgeing)-[Due Date CLE]),'#######0'))>=30.1 and (num((num(vDateForAgeing)-[Due Date CLE]),'#######0'))<=60,'31-60',
if((num((num(vDateForAgeing)-[Due Date CLE]),'#######0'))>=60.1 and (num((num(vDateForAgeing)-[Due Date CLE]),'#######0'))<=90,'61-90',
if( (num((num(vDateForAgeing)-[Due Date CLE]),'#######0'))>=90.1 and (num((num(vDateForAgeing)-[Due Date CLE]),'#######0'))<=180,'91-180',
if(num((num(vDateForAgeing)-[Due Date CLE]),'#######0')>=180.1,'Above 181',
0)))))))
Where VDateForAgeing will be Variable -> =IF(ISNULL(GetFieldSelections(MonthName))=0 OR ISNULL(GetFieldSelections(FinancialYear))=0,DATE(MonthEnd(MAX(PostingDate))),DATE(Today()))
Using this you can get current ageing as well as as-off date ageing.
Hope this helps you.
Vikas
• ###### Re: Aging of invoices
I assume Starting Date = [Doc Date1] and Unpaid Amount= Receivable. You can add expressions as below:
1. Days Overdue - =if(Receivable=0,0,Today()-[Doc Date1])
2. Current - if([Days Overdue]=0,Receivable,0)
3. 0-30 - =IF([Days Overdue]>0 and [Days Overdue]<=30,Receivable,0)
4. 0-60 - =IF([Days Overdue]>30 and [Days Overdue]<=60,Receivable,0)
5. 0-90 - =IF([Days Overdue]>60 and [Days Overdue]<=90,Receivable,0)
6. 90+ - =IF([Days Overdue]>90,Receivable,0)
Hope it helps.
• ###### Re: Aging of invoices
Try Something like this:
if((Num(Today())-Num(Entry Date))<30,'< 30 Days',
if((Num(Today())-Num(Entry Date))>=30 AND (Num(Today())-Num(Entry Date))<60,'30-60 Days',
if((Num(Today())-Num(Entry Date))>=60 AND (Num(Today())-Num(Entry Date))<90,'60-90 Days',
if((Num(Today())-Num(Entry Date))>=90,'>=90 Days')))) AS InvoiceAgeingBuckets
Regards
Aviral Nag
• ###### Re: Aging of invoices
// First load a table of invoices
Invoices:
InvoiceID,
DueDate,
IF(OutstandingAmount=0, 0, Interval(CurrentDate- DueDate, 'D') as DebtorDays
From...
// Then age:
Left join (Invoices) | 927 | 2,861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-30 | latest | en | 0.74994 |
https://www.civilengineeringweb.com/2023/07/deep-beam-design-criteria.html | 1,709,455,849,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476211.69/warc/CC-MAIN-20240303075134-20240303105134-00699.warc.gz | 704,189,487 | 37,059 | # What is Deep Beam | Deep Beam Design Criteria as per IS456
In this article, we share the information regarding deep beam in structure, such as what is deep beam, deep beam design and reinforcement details criteria, etc.
## What is Deep Beam
As per IS 456 Clause no. 29.1, Deep beam is defined as a beam which effective span to depth ration is less than 2.0 or 2.5. The ratio is depends on types of beam as mentioned below.
• For simply supported beam : 2.0
• For continuous beam : 2.5
A beam becomes a deeper when the loads on it is too high which result the high shear force and bending moment into the section. To resist the such too much forces, we increases the size of section while the length of beam is constant and leads deep beams in structure.
Deep beam is associated with high forces within the section. Due to high force in short length, shear force become critical factor in design of deep beam. As a result, deep beams are designed to withstand shear forces. While, due to their shorter span, it can effectively resist bending moments.
Usually, Deep beams comes in building where the column are placed closer and the load on beam is too high. Also it is used in girder and shear wall to transfer the heavy loads.
Also Read: Ductile Detailing of Column as per IS 13920-2016
## Lever Arm Criteria For Deep Beam
The lever arm Z for deep beam shall be determined as below:
• a) For Simply Supported Beams:
• Z = 0.2 (L+2D), When L <= L/D < 2
• b) For Continuous Beams:
• Z = 0.5 L, When L/D<1
Where L is the effective span taken as center to center distance between supports or 1.15 times the clear span whichever is smaller, and D is the overall depth.
## Reinforcement Details for Deep Beam
### Positive Reinforcement details for deep beam
The tensile reinforcement required to resist positive bending moment in any span of deep beam shall:
• Extend without curtailment between supports:
• Be embedded beyond the face of each support, so that at the face of the support it shall have a development length not less than 0.8Ld; Where Ld is the development length for the design stress in the reinforcement, and
• Be placed within a zone of depth equal to 0.25D – 0.05L adjacent to the tension face of the beam where D is the overall depth and L is the effective span.
Also Read: Arrangement of Transverse Reinforcement in Column
### Negative Reinforcement Details in Deep Beam
• a) Termination of reinforcement: For tensile reinforcement required to resist negative bending moment over a support of a deep beam:
• It shall be permissible to terminate not more than half of the reinforcement at a distance of 0.5D from the face of the support.
• The remainder shall extend over the full span.
• b) Distribution: When the ratio of clear span to overall depth is in the range 1.0 to 2.5, tensile reinforcement over a support of a deep beam shall be placed in two zones comprising:
• A zone of depth 0.2D, adjacent to the tension face, which shall contain a proportion of the tension steel given by (0.5 (L/D – 0.5))
• A zone measuring 0.3D on either side of the mid-depth of the beam, which shall contain the remainder of the tension steel, evenly distributed.
• For span to depth ratios less than unity, the steel shall be evenly distributed over a depth of 0.8D measured from the tension face.
Also Read: 5 Best Structural Design and Analysis Software
### Vertical Reinforcement details in deep beam
If forces are applied to a deep beam in such a way that hanging action required, bars or suspension stirrups shall be provided to carry all the forces concerned.
## Deep Beams||IS:456 Code Clause-wise Detailed Explanations|Fundamentals |Design Specification|Part-56
#IS456CodeExplanations#Design For Online Test Series for all CE Exams | 878 | 3,770 | {"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} | 2.96875 | 3 | CC-MAIN-2024-10 | longest | en | 0.926365 |
https://math.answers.com/basic-math/What_is_the_least_common_factor_of_99_and_132 | 1,726,660,382,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00308.warc.gz | 359,204,442 | 47,840 | 0
# What is the least common factor of 99 and 132?
Updated: 4/28/2022
Wiki User
14y ago
The least common factor is a term often mistakenly given to either the greatest common factor (GCF) or the least common multiple (LCM). This term is not often used because it does not describe a useful relationship between numbers. Since 1 evenly divides all integers, 1 is technically the least common factor for any set of integers.
The smallest non-one common factor of 99 and 132 is 3.
Wiki User
14y ago
Earn +20 pts
Q: What is the least common factor of 99 and 132?
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Related questions
### What is the least common multiple of 99 and 132?
The least common multiple of 132 and 99 is 396.
### What is the greatest common factor of 99 and 132 165?
The Greatest Common Factor of 99, 132, 165: 33
### What is the least common multiplier of 132 and 99?
Least Common Multiple (LCM) for 132 99 is 396
The GCF is 33.
### What is the greatest common factor of 99 and 132?
The GCF of 99 and 132 is 33Definition: A factor is a divisor - a number that will evenly divide into another number. The greatest common factor of two or more numbers is the largest factor that the numbers have in common.Method:One way to determine the common factors and greatest common factor is to find all the factors of the numbers and compare them.The factors of 99 are 1, 3, 9, 11, 33, and 99.The factors of 132 are 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, and 132.The common factors are 1, 3, 11, and 33. Therefore, the greatest common factor is 33.
### What is the Highest Common Factor between 99 and 132?
99 = 3 * 3 * 11132 = 2 * 2* 3* 11hcf = 3 * 11hcf =33
### What is the highest common factor for 99 and 165?
What is the highest common factor of 33, 3 and 99
33
### What is the least common factor of 66 and 99?
The LCF is 1. The GCF is 33. The LCM is 198.
### What is the LCF of 99 and 100?
The least common factor of any two (or more) numbers is ' 1 '.
### What is the greatest common factor of 4 and 99?
The greatest common factor of 4 and 99 is 1.
### What is the greatest common factor of 99 and 90?
The greatest common factor of 99 and 90 is 9. | 637 | 2,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-38 | latest | en | 0.934204 |
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