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http://www.ck12.org/book/CK-12-Middle-School-Math-Concepts-Grade-6/r4/section/5.0/ | 1,454,770,753,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701146550.16/warc/CC-MAIN-20160205193906-00244-ip-10-236-182-209.ec2.internal.warc.gz | 356,989,417 | 27,530 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 6 Go to the latest version.
# Chapter 5: Number Patterns and Fractions
Difficulty Level: At Grade Created by: CK-12
## Introduction
In Math 6, the learning content is divided into Concepts. Each Concept is complete and whole providing focused learning on an indicated objective. Theme-based Concepts provide students with experiences that integrate the content of each Concept. Students are given opportunities to practice the skills of each Concept through real-world situations, examples, guided practice and independent practice sections. In this fifth chapter, Number Patterns and Fractions, students will engage in many Concepts including factor pairs, divisibility, prime and composite numbers, prime factorization, greatest common factors, equivalent fractions, simplifying fractions, multiples, and writing fractions in different forms.
## Summary
Having completed this chapter, students are now ready to move on to Chapter Six. Each Concept has provided students with an opportunity to engage and practice skills in many Concepts including factor pairs, divisibility, prime and composite numbers, prime factorization, greatest common factors, equivalent fractions, simplifying fractions, multiples, and writing fractions in different forms.
Oct 29, 2012 | 305 | 1,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | longest | en | 0.900004 |
https://ambitiousbaba.com/reasoning-quiz-for-sbi-ibps-rrb-rbi-2021-25-march-2021/ | 1,674,971,192,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499700.67/warc/CC-MAIN-20230129044527-20230129074527-00069.warc.gz | 131,902,379 | 51,351 | # Reasoning Quiz for SBI, IBPS, RRB & RBI 2021| 25 March 2021
## Reasoning Quiz for SBI, IBPS, RRB & RBI 2021
Reasoning Quiz to improve your Reasoning Aptitude for SBI Po & SBI clerk exam IBPS PO Reasoning , IBPS Clerk Reasoning , IBPS RRB Reasoning, LIC AAO ,LIC Assistant and other competitive exam.
Direction (1 – 5): Study the following sequence and answer the given questions.
Q1. Which of the following element is twelfth to the left of the twentieth element from the left end of the given arrangement?
1. 6
2. &
3. M
4. \$
5. None of these
Q2. If all the symbols are dropped from the series, which element will be fourth to the right of the one which is twelfth from the right end?
1. 9
2. O
3. R
4. 7
5. None of these
Q3. How many such numbers are there in the given series which are immediately preceded by a symbol and followed by a letter?
1. None
2. One
3. Two
4. Three
5. Four
Q4. Four of the following five are alike in a certain way and forms a group find the one that does not belongs to that group?
1.
2.
3.
4.
5.
Q5. What should come in place of question mark (?) in the following series based on the above arrangement?
34% N\$M 6DL 8Q6 ?
1. %OR
2. 7Z%
3. O%R
4. R%O
5. R%7
Direction (6 – 10): Study the following information carefully and answer the question given below-
Seven people viz. P, Q, R, S, T, U and V are sitting around a circular table having equal distance between them. All of them are facing inside.
P sits immediate right of Q. Only one person sits between P and S (either from left or right). U sits third to the right of S. T is an immediate neighbor of U. R sits second to the left of V.
Q6. If all the persons are arranged according to the alphabetical order in anticlockwise direction starting from P, then how many persons position will remain unchanged (except P)?
1. Three
2. One
3. Two
4. None
5. None of these
Q7. How many persons sits between Q and U, if counted from the left of Q?
1. One
2. Two
3. Three
4. None
5. None of these
Q8. Who sits second to the right of T?
1. P
2. Q
3. R
4. S
5. None of these
Q9. Four of the following five belongs to a group find the one that does not belongs to that group?
1. VQ
2. PV
3. RT
4. SU
5. TQ
Q10. Who among the following sits second to the left of the one who sits 4th to the right of V?
1. U
2. T
3. R
4. S
5. None of these
## Solutions
Q1. Ans(3)
Q2. Ans(3)
Q3. Ans(4)
Q4. Ans(2)
Q5. Ans(1)
Sol. (6 -10):
Q6. Ans(3)
Q7. Ans(4)
Q8. Ans(2)
Q9. Ans(5)
Q10. Ans(1)
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3 | 1,010 | 3,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.869014 |
https://www.eeeguide.com/starting-induction-motor/ | 1,638,102,030,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358520.50/warc/CC-MAIN-20211128103924-20211128133924-00629.warc.gz | 690,457,564 | 22,768 | ## Starting Induction Motor:
At the time of Starting Induction Motor slip being unity, the load resistance with reference to the approximate circuit model of Fig. 9.23 ismeaning thereby that short-circuit conditions prevail. Therefore, the motor current at starting can be as large as five to six times the full-load current. In comparison the exciting current in the shunt branch of the circuit model can be neglected reducing the circuit to that of Fig. 9.31.
Assuming for simplicity that as a rough approximation
i.e., the magnetizing current is neglected even under full-load conditions. Then
The starting torque expressed as ratio of the full-load torque is
Squirrel-Cage Motors
Direct starting When a squirrel-cage motor is started ‘direct-on-line,’
i.e. starting current equals the short-circuit (blocked-rotor) current.
Then from Eq. (9.60)
This means that when the starting current is as large as five times the full-load current,
the Starting Induction Motor torque just equals the full-load torque. With such a large starting current, the motor must accelerate and reach normal speed quickly otherwise overheating may damage the motor. The load on the motor at the time of starting must, therefore, be very light or preferably the motor must be on no-load.
To protect their supply systems, the electrical utilities have regulations against short-time current peaks; the consumer is heavily penalized through suitable tarrif for such peaks. Therefore, it is only small-size motors that can be started direct online. A bulk consumer can start motors up to 10 kW direct-on-line as long as he arranges to stagger the starting of such motors.
Reduced voltage starting The starting current can be reduced to a tolerable level by reduced-voltage starting. This causes the starting torque to reduce heavily as it is proportional to the square of voltage. Such starting can only be carried out on no or light-load. Various methods of reduced voltage starting are discussed below.
Stator-impedance starting Inclusion of resistors or inductors in the three lines feeding the stator of the Starting Induction Motor reduces the stator terminal voltage to x V of the rated voltage V. The initial starting current is then
Substituting in Eq. (9.60)Thus, while the starting current reduces by a fraction x of the rated-voltage starting current (4c), the starting torque is reduced by a fraction x2 of that obtainable with direct switching. This method can be used for small motors such as those driving centrifugal pumps; but star-delta starting (given later) is cheaper with better starting torque.
Autotransformer starting Reduced voltage for starting can be obtained from three autotransformers connected in star as shown in the schematic diagram of Fig. 9.32. If the voltage is reduced to a fraction x of the rated voltage V, the motor starting current (initial) iswhere Isc = starting current (line) with full-voltage The current drawn from the supply is
It is found that while the starting torque is reduced to a fraction x2 of that obtainable by direct starting, the starting line current is also reduced by the same fraction. Compared to stator-impedance starting, the line current reduces further by a fraction x while torque remains the same. The autotransformer starting is much superior to the stator-impedance starting. Further, smooth starting and high acceleration are possible by gradually raising the voltage to the full line value.
After starting, the autotransformer is cut as shown in the wiring diagram of Fig. 9.33. It is to be observed that the autotransformer can be short-time rated. The use of an autotransformer is an expensive way of Starting Induction Motor and is warranted for large motors only.Star-delta starting The star-delta starting is an inexpensive two-step method of Starting Induction Motor. The motor designed for delta running is started across full-line voltage by connecting the phases in star as shown in the schematic diagram of Fig. 9.34.In direct delta starting:
In star starting:
Starting line (phase) current, IS (star)
Using Eq. (9.60)It is thus seen that star-delta starting reduces the starting torque to one-third that obtainable by direct-delta starting and also the starting line current to one-third. It just acts like autotransformer starting with
A star-delta starter is much cheaper than an autotransformer starter and is commonly employed for both small and medium-size motors. The wiring diagram of star-delta starting is shown in Fig. 9.35.
### Starting of Slip-ring Motors (Rotor-resistance Starting)
It has already been discussed that in slip-ring motors the starting current is reduced and the starting torque simultaneously increased by adding an external resistance in the rotor circuit. Figure 9.35 shows the motor circuit (magnetizing branch neglected) with external resistance added in the rotor circuit. For simplicity, the stator impedance is also neglected. Here
where Rext = actual external resistance in the rotor circuit
NowMaximum starting torque is achieved at
which equals the breakdown torque or
By a suitable choice of Rext the starting current and torque can be both adjusted to desirable levels. Since the maximum starting torque achievable is much more than the full-load torque, the use of slip-ring motors with rotor-resistance starting is ideal for starting on load. The external rotor resistance is arranged in steps which are gradually cut out during starting. | 1,137 | 5,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-49 | latest | en | 0.93036 |
https://zbmath.org/?q=an:0711.76083 | 1,713,935,380,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00132.warc.gz | 989,907,167 | 10,662 | ## Spectral and scattering theory for wave propagation in perturbed stratified media.(English)Zbl 0711.76083
Applied Mathematical Sciences, 87. New York etc.: Springer-Verlag. 188 p. DM 74.00 (1991).
A medium is called stratified if its physical properties depend on a single coordinate. The stratified medium is a mathematical model for ocean acoustics, integrated optics, geophysics.
In the book the author considers the same mathematical problems of wave propagation in media whose physical characteristics depend upon all of the coordinates and they stabilize at infinity to the functions which depend only on a single coordinate. This condition means these media are the perturbed stratified media. It should be noted that in the classical scattering theory the medium is the perturbation of the homogeneous medium.
In the first part of the book the author considers the acoustic equation in $${\mathbb{R}}^{n+1}:$$ $$(\partial^ 2u/\partial t^ 2)-c^ 2(x,y)\Delta u=0$$, $$x\in {\mathbb{R}}^ n$$, $$y\in {\mathbb{R}}$$, $$t\in {\mathbb{R}}^ 1$$, where c(x,y) is the sound speed, the fluid density $$\rho =const.$$; c(x,y) is a real valued measurable function on $${\mathbb{R}}^{n+1}$$ such that $$0<c_ m\leq c(x,y)\leq c_ M<\infty$$, for almost every (x,y) and some positive constant $$c_ m$$ and $$c_ M$$, and $$| c(x,y)-c_ 0(y)| \leq c(1+| (x,y)|)^{-1-\epsilon}$$, $$\epsilon >0$$, where $$c_ 0(y)$$ is a real valued measurable function on $${\mathbb{R}}$$ which satisfies the same conditions of stabilization at $$y\to \pm \infty.$$
The author considers the following problems: (1) the limiting absorption principle for acoustic operators in unperturbed (stratified) and perturbed media; 2) the absence of the positive eigenvalues on acoustic operators for perturbed media; 3) the acoustic scattering theory in which the unperturbed operator is an acoustic operator for the stratified media; 4) the generalized Fourier transforms; 5) the structure of the scattering matrix.
In the second part of the book the author considers the Maxwell system of equations
\left\{\begin{aligned} \nabla \times \bar E=-\mu (x,z)\partial \bar H/\partial t, \\ \nabla \times \bar H=\epsilon (x,z)\partial \bar E/\partial t, \\ \nabla \cdot (\epsilon (x,z)\bar E)=0,\\ \nabla \cdot (\mu (x,z)\bar H)=0,\end{aligned}\right.\tag{1} where $$x=(x_ 1,x_ 2)\in {\mathbb{R}}^ 2$$, $$z\in {\mathbb{R}}$$, $$t\in {\mathbb{R}}$$, $$\bar E=(E_ 1,E_ 2,E_ 3)$$, $$\bar H=(H_ 1,H_ 2,H_ 3)$$ are the functions from $${\mathbb{R}}^ 4$$ into $${\mathbb{R}}^ 3$$ that correspond, respectively, to the electric and magnetic fields, $$\epsilon(x,z)$$, $$\mu(x,z)$$ are real valued measurable and bounded function defined on $${\mathbb{R}}^ 3$$ that represent, respectively, the electric and magnetic susceptibilities.
The author assumes that $$0<c_ m\leq \epsilon (x,z)\leq c_ M<\infty$$, $$0<c_ m\leq \mu (x,z)\leq c_ M<\infty$$, and $$| \epsilon (x,z)- \epsilon_ 0(z)| \leq c(1+| (x,z)|)^{-1-\epsilon}$$, $$| \mu (x,z)-\mu_ 0(z)| \leq c(1+| (x,z)|)^{-1-\epsilon}$$, $$\epsilon >0$$, where $$\epsilon_ 0(z)$$, $$\mu_ 0(z)$$ are the real valued measurable functions defined on $${\mathbb{R}}$$, which satisfy the same conditions of stabilization at $$z\to \pm \infty.$$
The author considers the limiting absorption principle, the spectral properties and the scattering theory for the Maxwell system (1).
Reviewer’s remark: In the case when the fluid density $$\rho$$ (x,y) is the function depending upon all of the coordinates, similar problems for acoustic operators were considered in the works of M. Ben-Artzi, Y. Dermenjian and J.-C. Guillot [Commun. Partial Differ. Equations 14, No.4, 479-517 (1989; Zbl 0675.35065)] and V. S. Rabinovich [On the solvability of acoustic problems in an open waveguide (in Russian), Differ. Uravn. 26, No.12, 2178-2180 (1990)].
Reviewer: V.S.Rabinovich
### MSC:
76Q05 Hydro- and aero-acoustics 35P05 General topics in linear spectral theory for PDEs 76-02 Research exposition (monographs, survey articles) pertaining to fluid mechanics
Zbl 0675.35065 | 1,245 | 4,060 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-18 | latest | en | 0.780061 |
http://de.metamath.org/mpegif/axc5c711to11.html | 1,591,314,273,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348492295.88/warc/CC-MAIN-20200604223445-20200605013445-00052.warc.gz | 35,138,700 | 3,753 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > axc5c711to11 Structured version Unicode version
Theorem axc5c711to11 2252
Description: Re-derivation of ax-11 1843 from axc5c711 2249. Note that ax-c7 2217 and ax-11 1843 are not used by the re-derivation. The use of alimi 1634 (which uses ax-c5 2215) is allowed since we have already proved axc5c711toc5 2250. (Contributed by NM, 19-Nov-2006.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
axc5c711to11
Proof of Theorem axc5c711to11
StepHypRef Expression
1 axc5c711toc7 2251 . . 3
21con4i 130 . 2
3 pm2.21 108 . . . . . 6
4 axc5c711 2249 . . . . . 6
53, 4syl 16 . . . . 5
65alimi 1634 . . . 4
7 axc5c711toc7 2251 . . . 4
86, 7nsyl4 142 . . 3
98alimi 1634 . 2
102, 9syl 16 1
Colors of variables: wff setvar class Syntax hints: wn 3 wi 4 wal 1393 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1619 ax-4 1632 ax-11 1843 ax-c5 2215 ax-c4 2216 ax-c7 2217 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 474 | 1,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-24 | latest | en | 0.324102 |
https://ratva.noticiasalahora.info/3-set-venn-diagram-generator.php | 1,600,965,956,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219691.59/warc/CC-MAIN-20200924163714-20200924193714-00461.warc.gz | 564,950,425 | 7,875 | Online calculator to create venn diagram for three sets. Set Algebra Calculators (3.) Delete the "default" expression in the textbox of the calculator. (4.) Copy and paste the expression you typed, into the small textbox of the calculator. (5.) Click the "Submit" button. (6.) Check to make sure that it is the correct set you typed. (7.) Review the answer (Venn Diagram). Using the Venn Diagram Generator; Type: A ∪ B as A BioVenn - a web application for the comparison and Venn diagram generator. Please use it for the creation of area-proportional Venn diagrams for scientific publications and presentations. Give it a try! Shading Venn Diagrams (solutions, examples, videos)
## Venn diagram generator. Please use it for the creation of area-proportional Venn diagrams for scientific publications and presentations. Give it a try!
four circle four set venn diagram template. six circle A Venn diagram maker made for non-designers How to Use the Venn Diagram Maker in 3 Easy Steps: Online Venn Diagram Maker - Create Free Venn Diagrams in What should you do when you've got 3 unique data sets that need to be compared and contrasted quickly and plainly? In this visual world where catching Venn Diagram Maker Online. Create and download Create customizable venn diagrams online using our free tool. Just choose the amounts, the colors, the intersection and hit download! Venn Diagram Generator – GeoGebra Easy App to generate simple symmetric Venn Diagrams with 2 or 3 sets.
BioVenn - a web application for the comparison and Venn diagram generator. Please use it for the creation of area-proportional Venn diagrams for scientific publications and presentations. Give it a try! Shading Venn Diagrams (solutions, examples, videos) Shading Venn Diagrams Venn Diagrams: Shading Regions for Two Sets Basic Venn Diagram Shading 1) A ∪ B 2) A ∩ B 3) A' ∪ B 4) A ∩ B' 5) A ∪ B ∪ C
Three Circle Venn Diagrams | Passy's World of Mathematics Represent these results using a three circle Venn Diagram.” The type of three circle Venn Diagram we will need is the following: Image Source: Passy’s World of Mathematics. This three circle word problem is an easy one. All of the number values for each section of the diagram have been given to us in the question. Venn diagram creation bioinformatics tools | Next-generation
## Venn Diagram Generator | Academo.org - Free, interactive
3 Oct 2019 VennDiagram #3SetVenn #Venn Create presentation-ready Venn diagrams in 4 simple steps with Creately. Or choose from an array of Venn Diagram Template | Venn Diagrams | 5-Set Venn ConceptDraw DIAGRAM diagramming and vector drawing software presents the Venn Diagrams solution from "Diagrams" area which 3-set Venn diagram,. draw.triple.venn: Draw a Venn Diagram with Three Sets in In VennDiagram: Generate High-Resolution Venn and Euler Plots. Description A vector (length 3) of strings giving the category names of the sets. rotation. Venn Diagrams | Math Goodies
### Offers lots of ready-made printable 3 circles venn diagram examples and blank 3 Download 3-Set Numbers Venn Diagram Templates in Editable Format
22 May 2015 InteractiVenn allows set unions in Venn diagrams to be explored and BioVenn [3] that construct area-proportional diagrams, limited by the VennDiagram: a package for the generation of highly 26 Jan 2011 To fill this gap we introduce VennDiagram, an R package that enables Row 1, column 3: a two-set Euler diagram showing two distinct sets. Venn Diagram Examples - Creately
Represent these results using a three circle Venn Diagram.” The type of three circle Venn Diagram we will need is the following: Image Source: Passy’s World of Mathematics. This three circle word problem is an easy one. All of the number values for each section of the diagram have been given to us in the question. Venn diagram creation bioinformatics tools | Next-generation Assists users in comparing lists. Venny provides an interactive application with Venn's diagrams, a figure that shows all possible logical relations between a finite collection of different sets. It permits to compare 2, 3 or 4 lists of data. It also offers an offline version that consists on a single standard html file.
Online calculator to create venn diagram for three sets. | 982 | 4,249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-40 | latest | en | 0.757541 |
http://www.enotes.com/homework-help/using-ideal-gas-law-calculate-following-volume-0-426289 | 1,477,450,833,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720475.79/warc/CC-MAIN-20161020183840-00276-ip-10-171-6-4.ec2.internal.warc.gz | 433,818,114 | 9,858 | # Using the ideal gas law, calculate the density of CO2 at 4.00 atm pressure and -20.0 oC.
jerichorayel | College Teacher | (Level 2) Senior Educator
Posted on
Using the Ideal gas law we can solve for the density of gas. First, let us write down the Ideal gas equation and the formula for density.
`PV = nRT`
`rho = (mass)/(volume) = M/V`
We know that moles = mass (M)/ molar mass, so we can substitute it to the Ideal gas equation thus having:
`PV = (M)/(molar mass) * RT`
Further arrangement in the expression we can have:
`(P)/(RT) =(M)/(molar mass * V)`
Finally,
`(P* molar mass)/(RT) =(M)/(V)= rho`
`rho =(P* molar mass)/(RT)`
• P = 4.00 atm
• T = -20.0 + 273.15 = 253.15K
• R = 0.08206 atm-L/mol-K
• Molar mass of CO_2 = 44.01 grams/mol
`rho =(4.00* 44.01)/(0.08206* 253.15)`
`rho` = 8.47 g/L = 0.00847 g/ml
Sources: | 293 | 838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2016-44 | latest | en | 0.816321 |
https://www.sambuz.com/doc/adjustable-references-ppt-presentation-760935 | 1,695,915,154,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00811.warc.gz | 1,072,072,843 | 13,909 | # Adjustable References Viktor Vafeiadis Max Planck Institute for - PowerPoint PPT Presentation
## Adjustable References Viktor Vafeiadis Max Planck Institute for Software Systems (MPI-SWS) Viktor Vafeiadis Adjustable References Introduction Reasoning about imperative programs How to represent mutable state? Two standard approaches Deep
2. Introduction Reasoning about imperative programs How to represent mutable state? Two standard approaches Deep embedding Monadic What if state is only used as an optimisation? Viktor Vafeiadis Adjustable References
3. Memoization let memo f = let h = Hashtbl . create() in λ a . match Hashtbl . get h a with | Some b → b | None → let b = f a in Hashtbl . add h a b; b Viktor Vafeiadis Adjustable References
4. Memoization let memo f = let h = Hashtbl . create() in λ a . match Hashtbl . get h a with | Some b → b | None → let b = f a in Hashtbl . add h a b; b Intuitively, memo f = f. Viktor Vafeiadis Adjustable References
5. Memoization let memo f = let h = Hashtbl . create() in λ a . match Hashtbl . get h a with | Some b → b | None → let b = f a in Hashtbl . add h a b; b But with monads, f : A → B and memo f : A → State B. Viktor Vafeiadis Adjustable References
6. Adjustable References Like mutable references, but mutation is not observable. Internal representation type, R . Observation type, T . Observation function, f : R → T . Parameter aref : ∀ R T , ( R → T ) → Type . Viktor Vafeiadis Adjustable References
7. Adjustable References (2) Parameter aval : ∀ R T ( f : R → T ) , R → aref f . Parameter aget : ∀ R T ( f : R → T ) , aref f → T . Axiom aref inhabited : ∀ R T ( f : R → T ) ( r : aref f ) , ∃ v , r = aval f v . Axiom agetval : ∀ R T ( f : R → T ) v , aget(aval f v ) = f v . Viktor Vafeiadis Adjustable References
8. Unobservable Updates Naively, adjupd : ∀ R T ( f : R → T ) ( r : aref f ) ( v : R ) , f v = aget r → unit . Problems: 1 Evaluation order unknown & Coq can discard unused results 2 The new internal value cannot depend on the old internal value. 3 Not useful unless T is a function type. Viktor Vafeiadis Adjustable References
9. Adjusting Reads Parameter agetu : ∀ R A B ( f : R → A → B ) ( upd : R → A → R × B ) ( PF : ∀ x a , f ( fst ( upd x a )) = f x ) ( PF ′ : ∀ x a , snd ( upd x a ) = f x a ) , aref f → A → B . Axiom agetuE : ∀ R A B ( f : R → A → B ) upd PF PF ′ , agetu upd PF PF ′ = aget ( f := f ) . Viktor Vafeiadis Adjustable References
10. Allocations Parameter anew : ∀ R 1 T 1 ( f 1 : R 1 → T 1 ) ( r : aref f 1 ) R 2 T 2 ( f 2 : R 2 → T 2 ) ( g : ∀ v , aget r = f 1 v → R 2 ) ( PF : ∀ x p x y p y , f 2 ( g x p x ) = f 2 ( g y p y )) , aref f 2 . Axiom anewval : ∀ R 1 T 1 ( f 1 : R 1 → T 1 ) v R 2 T 2 ( f 2 : R 2 → T 2 ) g PF , anew (aval f 1 v ) g PF = aval f 2 ( g v (agetval f 1 v )) . Viktor Vafeiadis Adjustable References
11. Consistency Trivial; just ignore the adjustments. Definition aref R T f := R . Definition aval R T f v := v . Definition aget R T f r := f r . Definition agetu R A B f upd p p ′ r := f r . Definition anew R 1 T 1 f 1 r R 2 T 2 f 2 g p := g r eq refl . Viktor Vafeiadis Adjustable References
12. Imperative Implementation type ( ′ r , ′ t ) aref = ′ r ref let aval x = ref x let aget f r = f ! r let agetu u r a = let ( v , b ) = u ! r a in r := v ; b let anew r g = ref ( g ! r ()) Viktor Vafeiadis Adjustable References
13. Memoization Using Adjustable References (1) Section Memo . Variables ( AB : Type) ( f : A → B ) . Variable eqA : ∀ x y : A , { x = y } + { x � = y } . Variable hash : A → int . Definition cache := { c : Parray . t ( option ( A × B )) | ∀ x a b , Parray . get c x = Some ( a , b ) → b = f a } . Viktor Vafeiadis Adjustable References
14. Memoization Using Adjustable References (2) Program Definition mupd ( c : cache )( a : A ) := let h := hash a in match Parray . get c h with | None ⇒ let b := f a in ( Parray . set c h ( Some ( a , b )) , b ) | Some( a ′ , b ) ⇒ if eqA a a ′ then ( c , b ) else let b := f a in ( Parray . set c h ( Some ( a , b )) , b ) end . � ... � Program Definition memo := let r := aval ( λ c , f ) ( Parray . create 100 None ) in agetu mupd r . � ... � Viktor Vafeiadis Adjustable References
15. Memoization Using Adjustable References (3) Lemma memo eq : memo = f . Proof. by unfold memo ; rewrite arefgetuE , agetval . Qed. End Memo . Viktor Vafeiadis Adjustable References
16. Other examples Data structures that optimize their shape even when performing read-only operations. Union-find path compression (in the paper) Splay trees Viktor Vafeiadis Adjustable References
17. Limitation Cannot have any observable updates. But observable updates are fine... ...provided that nobody observes them. In a sequential setting, you can do a sequence of updates, whose total effect is unobservable. cf. Conchon and Filliˆ atre persistent arrays. Viktor Vafeiadis Adjustable References
18. Future work Overcome the limitation! runST : ∀ c : AdjStateMonad A . c is logically pure → A . Formally justify the imperative implementation Develop applications of adjustable references Viktor Vafeiadis Adjustable References
Recommend
More recommend | 1,550 | 5,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-40 | latest | en | 0.653011 |
https://forums.aat.org.uk/Forum/discussion/28097/pev-understanding-of-formulas | 1,716,253,940,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058342.37/warc/CC-MAIN-20240520234822-20240521024822-00727.warc.gz | 228,521,221 | 83,382 | # Pev - understanding of formulas
Options
Registered Posts: 60 Regular contributor ⭐
Hi
I will try to keep this short.
Simply - i know the formulas but can some one confirm my understanding of them.
Price variance
AQ x (SP-AP)
AQ = Actual Quantity of material purchaced
SP = Standard Price set for the cost/purchase of the material
AP = Actual Price cost of the material purchased
Usage Variance
SP x (SU-AU)
SP = same as earlier
SU = Standard Usage, being the amount of material that was issued from stores for actual production at the standard price.
AU = Actual Usage, being the actual amount of material used in production at the standard price.
Once i know this is ok then it will help with certain questions
• Registered Posts: 12 New contributor 🐸
Options
I am stuck on the exact same thing in my revision and its really annoying me, iv been taught to learn mine like this
Actual Qty x Actual Price =
Price Variance
Actual Qty x Standard Price =
Usage variance
Standard Qty x Standard Price =
But my answers dont seem to be matching up to the answers in the book :-(
• Registered Posts: 70 Regular contributor ⭐
Options
Are you adjusting the standard quantity for the actual production? because I sometimes forget to do that and only realise when my reconcilliation doesn't balance.
• Registered Posts: 63 Regular contributor ⭐
Options
S for Should
I don't know if this helps, but here is how I remember it:
S for Standard, S for Should've, so:
Standard Price is the price you should've paid,
Standard Usage (or Standard Quantity) is the amount you should've used (for the actual output you produced)
Standard Cost is the amount it should've cost (for the actual output produced.)
• Registered Posts: 122 Dedicated contributor 🦉
Options
We've been taught:
PAUS
Price variance use
Actual quantities
Usage variances use
Standard costing
Direct material price variance is therefore AQSC - AQAC using actual production unit
Direct material usage variance is therefore SQSC - AQSC using actual production units
SQ = quantity of material you would expect to use for actual production
AQ = actual quantity of material used
SC = what you budgeted to pay per unit of material
AC = what you actually paid per unit of material
I consistently got answers wrong early on in my revision because I kept using budgeted production units instead of actual production units, so what I do now is write out a standard cost card for one unit, which seems to help me.
• Registered Posts: 63 Regular contributor ⭐
Options
OK - this looks similar to what I've learned but what you call SC - Standard Cost, i call SP - standard price. So my formulas are:
Price variance: AQ x SP - AQ x AP
Usage variance: SQ x SP - AQ x SP
another thing I use when my brain turns to jelly is to remember that, for a price variance you fix the quantity and look at the difference in price (so AQ and AQ) and for the usage variance you fix the price and look at the difference in quantity (so SP and SP - or SC and SC in your case)
but maybe I'm just confusing you now!
• Registered Posts: 122 Dedicated contributor 🦉
Options
I understand you completely EAP!
• Registered Posts: 63 Regular contributor ⭐
Options
My tutor was useless at teaching this - he just wanted us to learn the formulas, so I sat and worked through this until my head hurt but I think I have finally got it. I wish I'd gone on the forums before as there is loads of useful stuff. I have to admit though, I am looking forward to never having to think about variances again!
• Registered Posts: 122 Dedicated contributor 🦉
Options
Our tutor completely flummoxed us with this too, using a graph.
Like you, I only really understood when I started revising and worked it out myself. Are you sitting PCR too? I haven't even started revising that yet!
• Registered Posts: 63 Regular contributor ⭐
Options
I'm doing PEV Monday, BTC Tuesday and PCR Wednesday. PEV I think will be OK, PCR is touch and go, and BTC I am trying not to care about coz I don't actually need it but deep down I can't bear the idea of failing it all the same.
• Registered Posts: 122 Dedicated contributor 🦉
Options
PEV Monday PCR Wednesday. Ducked out of Tax, so don't envy you that timetable. Good luck. Any tips for PCR revision?
• Registered Posts: 63 Regular contributor ⭐
Options
Do the last paper Dec 2009 first. It has opening and closing stocks of materials as well as opening and closing stocks of finished goods, so it's a pig of a paper but at least it won't freak you out if it comes up in the exam.
June 07 paper is a right can of worms, I'd be tempted to say avoid it all together. From the forums it seems like they changed the model answers a few times! It's section 1 that's the problem - they ask you to do an operating statement, but don't ask for opening or closing stocks (although apparently the original model answers did show opening and closing stocks). Nightmare!
Good luck for next week anyway!
• Registered Posts: 60 Regular contributor ⭐
Options
I am stuck on the exact same thing in my revision and its really annoying me, iv been taught to learn mine like this
Actual Qty x Actual Price =
Price Variance
Actual Qty x Standard Price =
Usage variance
Standard Qty x Standard Price =
But my answers dont seem to be matching up to the answers in the book :-(
Is it the Osbornes books - if you let me know what task then i will see if it works using my formulas and then i'll see if i can help you to use the formulas my way. To be honest i find it very difficult to work with any new versions given to me as i find mine works all the time. I actually use what twinmeister said PAUS and that works for me.
• Registered Posts: 60 Regular contributor ⭐
Options
I have just seen this from another thread and thaugh to post it here. I think it might just help us. Its been answered by Sandy Hood who is a PEV teacher so he knows his stuff.
#1 09-06-10, 19:57
sorcha
New Member Join Date: Nov 2009
Posts: 9
Standard v budgeted costs - what's the difference
Please, please can someone explain the difference between these two when calculating costings in MAC Unit 33
sorcha
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#2 09-06-10, 22:15
SandyHood
Well known Join Date: Nov 2007
Posts: 1,073
Budgeted costs are the standard costs of the planned output
Standard cost of actual production is the standard cost of the actual output
Standard cost of kgs purchased is the standard cost of 1 kg x the actual kgs purchased
__________________
sandy.hood@chichester.ac.uk | 1,528 | 6,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.866878 |
http://forum.allaboutcircuits.com/threads/inductance-and-superconductors.42262/ | 1,484,565,921,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279169.4/warc/CC-MAIN-20170116095119-00031-ip-10-171-10-70.ec2.internal.warc.gz | 107,267,196 | 16,354 | # Inductance and superconductors
Discussion in 'General Electronics Chat' started by wes, Aug 28, 2010.
1. ### wes Thread Starter Active Member
Aug 24, 2007
242
2
I have a question about inductors and superconductor's.
I have read that superconductor's expel all magnetic fields (or at least most )
so if a you had a inductor made out of a superconductor then how could it produce a inductance since each field generated by each wire could not induce a voltage in the next. if it would then why exactly since the field is expelled by the next wire?
Also what would the inductance be for it if say it's inductance would normally be 1 mh in a normal state but in a superconducting state what would it be?
2. ### marshallf3 Well-Known Member
Jul 26, 2010
2,358
201
Shoot, I used to know some of this stuff but I can tell some of the reading you've been doing hasn't explained things thoroughly enough.
If I recall the inductance still exists, it just has no resistive component to it.
Jun 1, 2009
499
37
I'm guessing that means it doesn't act as a classical inductor, since the magnetic fields are forced out of the super conducting region there is no interaction electrically but it still produces a magnetic field around the wire? That's my guess at least.
4. ### Wendy Moderator
Mar 24, 2008
20,772
2,540
Actually it will be the classic inductor squared, with no resistance there goes a lot of pesky negligible factors that aren't. It will also store energy indefinitely (IE, forever) if you short the leads while conducting current, thus forming a permanent electromagnet. Basically the Q of the circuit will be exceedingly high.
It will be fun to play with.
Remember, they are using superconducting electromagnets for many industrial and scientific uses. Electromagnets is just another word for inductor.
5. ### timrobbins Active Member
Aug 29, 2009
318
16
An ideal wire, or coil of wire, carrying a current will have a magnetic field in space dependant on the geometry of the wire. At the wire surface, the electric field doesn't penetrate the wire, because there is zero resistance throughout the wire corss-section. Hence there is no field in the wire itself - it is all external.
When the wire has resistance then the field extends in to the wire, reducing as you go in further, and the term 'skin depth' describes how far in the field extends. In normal wire the skin depth is frequency dependant. Superconductors have a very small 'skin depth' that is not frequency dependant, for normal superconducting conditions.
Skin depth is a term readily appreciated by the early HAM and rf transmitter community, and lately by switchmode designers - and maybe some other exotic applications (?).
Ciao, Tim
6. ### marshallf3 Well-Known Member
Jul 26, 2010
2,358
201
I have noticed an increase in availability of Litz wire lately at some of the parts websites.
7. ### retched AAC Fanatic!
Dec 5, 2009
5,201
313
That's funny you say that. I have had a funny feeling about seeing much more lately. It seems during times of war and major innovations, companies make more odd items. If you figure that they set up a factory tooled to produce litz, they would start marketing to sell inventory.
Dunno. Odd you noticed too.
8. ### wes Thread Starter Active Member
Aug 24, 2007
242
2
Ok so if anything then the superconductor inductor, lol would create a larger inductance for the same size inductor as a normal one
same number of turn
same size in all directions
everything is the same except the one is normal wire and the other is superconducting wire
also anyone heard of bifilar coil
they are supposed to create 0 inductance (well almost 0 )
is there still a net magnetic field produced by the coil even though each wire's current is in the opposite direction? I would think there would still be a small field since the very end wire's magnetic field on one side is not being opposed
Mar 24, 2008
20,772
2,540
10. ### marshallf3 Well-Known Member
Jul 26, 2010
2,358
201
Makes sense when you think about it. Switching PS ICs are operating at higher frequencies and producing more amps. Litz wire has another advantage than just minimizing the skin efect - it's easier to wind the equivalent AWG on a toroid form.
I assume you've seen this relatively new IC? LTC3788
11. ### retched AAC Fanatic!
Dec 5, 2009
5,201
313
No, no I have not.
Interesting. If I had time before work, I would play with it in LTspice...
I guess after work it is.
Very interesting.
12. ### marshallf3 Well-Known Member
Jul 26, 2010
2,358
201
2 separate outputs of up to 60V @ 10A each and no heat sinks required - even on the MOSFETS. | 1,150 | 4,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-04 | latest | en | 0.945836 |
https://math.stackexchange.com/questions/688985/how-to-know-that-the-slope-of-the-tangent-line-and-gradient-are-orthogonal | 1,585,454,711,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493684.2/warc/CC-MAIN-20200329015008-20200329045008-00153.warc.gz | 594,559,404 | 31,451 | # How to know that the slope of the tangent line and gradient are orthogonal?
Given a surface $f(x, y) = z$, a level set for the surface, and a point on that set, how to know that the slope of the tangent line to the level curve and the gradient vector are orthogonal?
• Welcome to Math.SE! This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Joe Tait Feb 24 '14 at 20:47
Consider any $C^1$ path $c$, defined on an non-empty interval $J$ such that $c(J)$ is a subset of the domain of the $C^1$-map $f\colon\boldsymbol R^n\to\boldsymbol R$. Then $f\mathop{\circ}c$ is a real-valued $C^1$-map defined on $J$. Its derivative is according to the chain rule $$(f\mathop{\circ}c)'=\langle\nabla (f\mathop{\circ}c),c'\rangle.$$ Now if $c$ is a level curve, $f\mathop{\circ}c$ is constant, hence $(f\mathop{\circ}c)'=0$. | 302 | 1,107 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-16 | latest | en | 0.885982 |
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Our flying geese calculator uses the flying geese formula to give you the required fabric pieces for your desired flying geese measurements!
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Scroll to Top | 1,000 | 4,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-38 | latest | en | 0.869415 |
http://mathhelpforum.com/advanced-statistics/226387-probabibility-problem-terms-binomial-coeficciants.html | 1,480,987,672,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541864.44/warc/CC-MAIN-20161202170901-00162-ip-10-31-129-80.ec2.internal.warc.gz | 169,061,831 | 10,462 | # Thread: Probabibility problem in terms of binomial coeficciants
1. ## Probabibility problem in terms of binomial coeficciants
From 6 positive integers and 8 negative integers, 4 are chosen at random without replacement. What is the probability that their product is positive? Express answer in terms of binomial coefficients
So I know the total is 14 integers, and for the product to be positive the selection has to be either all 4 are + , all 4 are - , or 2 are + & 2 are -
I'm confused as to how this would be expressed in terms of binomial coefficients. Would it be ((4 choose 6) + (8 choose 4)/(14 choose 4)) ?
2. ## Re: Probabibility problem in terms of binomial coeficciants
Hey crownvicman.
You are better off selecting the independent events and adding up their probabilities if they are disjoint. This avoids trying to use formulae that may not work.
3. ## Re: Probabibility problem in terms of binomial coeficciants
Originally Posted by crownvicman
From 6 positive integers and 8 negative integers, 4 are chosen at random without replacement. What is the probability that their product is positive? Express answer in terms of binomial coefficients
So I know the total is 14 integers, and for the product to be positive the selection has to be either all 4 are + , all 4 are - , or 2 are + & 2 are -
I'm confused as to how this would be expressed in terms of binomial coefficients. Would it be ((4 choose 6) + (8 choose 4)/(14 choose 4)) ?
Let $P =\displaystyle \sum\limits_{k = 0}^2 {\binom{6}{2k}\binom{8}{4-2k}}$ then find $\dfrac{P}{\binom{14}{4}}$. | 417 | 1,574 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2016-50 | longest | en | 0.91896 |
http://www.dictall.com/indu/155/1546054BE32.htm | 1,603,894,911,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107898577.79/warc/CC-MAIN-20201028132718-20201028162718-00464.warc.gz | 130,011,920 | 4,643 | 说明：双击或选中下面任意单词，将显示该词的音标、读音、翻译等；选中中文或多个词，将显示翻译。 您的位置：首页 -> 词典 -> 静电势能 1) electrostatic potential energy 静电势能 1. The author discusses respectively how the determination of potential-energy zero point influences geopotential energy,elastic potential energy and electrostatic potential energy. 讨论了势能零点的选取对重力势能、弹性势能和静电势能的影响，讨论了弹性势能与简谐振动的势能、静电势能与静电场能的关系。 2. Two simple methods for calculating the electrostatic potential energy between two infinite charged bodies are introduced,using Poisson equation and the basic concepts on electrostatic field respectively. 通过求解静电场的泊松方程以及静电场基本概念的运用 ,给出无限大带电体之间的静电势能的两种简便计算方法 ,与教程常用方法相比更易于掌握 。 2) electrostatic potential energy model 静电势能模型 1. Microtextural model and electrostatic potential energy model of Layered double hydroxides (LDHs) were built adopting theory of crystallography. 采用晶体学理论建立二元类水滑石(LDHs)微观结构模型与静电势能模型,将层板金属离子间距、层板电荷密度、层间阴离子间距等微观结构参数定量化,并将层间阴离子的静电势能表示成层板金属离子半径和物质的量之比、插层阴离子尺寸和电荷的函数。 3) electrostatic potential 静电势 1. Comparative investigation of the molecular electrostatic potential of fullerenes with density function theory and HF ab initio methods; 密度泛函理论和从头算方法对富勒烯分子静电势的比较研究 2. Theoretical study of the molecular electrostatic potential of finite armchair single-wall carbon nanotubes; 有限长扶手椅形单壁碳纳米管分子静电势的理论研究 3. Theoretical studies on the molecular electrostatic potential of the C_(100) fullerenes C_(100)富勒烯分子静电势的理论研究 4) electrostatic potential well 静电势阱 1. For long pulse times and especially for CW mode, the coupling between ions and electrostatic potential wells in the devices will lead to ion noise,which generally manifests itself as a slow phase fluctuation on the output signal. 在长脉冲或者连续波的工作状态下,这种离子化与设备中的静电势阱相互作用会引起离子噪声,通常表现为输出信号相位的缓慢波动。 5) rest potential 静态电势 6) electrostatic potential barrier 静电势垒 补充资料：LEP势能面 分子式：CAS号：性质：伦敦（London）最早把量子力学的玻恩-奥本海默近似用于化学反应。对X+Y—Z（三原子体系）→X—Y+Z反应，他提出了伦敦方程计算相互作用势能V：V=QXY+QYZ+QZX±{1/2[(JXY—JYZ)2+(JYZ—JZX)2+(JXZ—JXY)2]}=Q±J式中Qij为库伦能（积分）和Jij为交换能（积分）（i，j分别代表X、Y、Z），均为核间距的函数。上式远不能得到合乎要求的结果。艾林和波拉尼提出了一个半经验方法，根据XY、YZ、ZX的光谱数据得到Qij和Jij。如对Ha+HbHc选择QHH=0.14V可得与该反应活化能符合得很好的理论结果，据此方法制得的势能面即LEP势能面。但鞍点区出现一小谷，这是其不足之处。 说明：补充资料仅用于学习参考，请勿用于其它任何用途。 参考词条 ©2011 dictall.com | 994 | 2,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.338024 |
https://bedtimemath.org/fun-math-tortoise-mama/ | 1,524,680,644,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947939.52/warc/CC-MAIN-20180425174229-20180425194229-00103.warc.gz | 577,735,485 | 15,297 | # A Ride with Mama Turtle
Here's your nightly math! Just 5 quick minutes of number fun for kids and parents at home. Read a cool fun fact, followed by math riddles at different levels so everyone can jump in. Your kids will love you for it.
# A Ride with Mama Turtle
May 14, 2017
It’s Mother’s Day, the day we celebrate everything our moms do for us. But why not celebrate all moms, not just humans? Animal moms work just as hard to care for their kids. Look at this huge African tortoise mama carrying her babies. The African tortoise is the largest tortoise in Africa, with the grown-ups weighing nearly 180 pounds. Her 4-day-old babies, though, are only 2-3 inches long, and weigh only about 1 ounce (1/16th of pound). They’ll have to grow a lot before they can carry their own kids!
Wee ones: The African tortoise is the 3rd largest type of tortoise in the world. How many kinds of tortoises are bigger?
Little kids: If the 3rd tortoise slides off the shell, then the 4th tortoise, then the 5th…who’s next? Bonus: If the next 2 babies stay on, who slides off after them?
Big kids: If the mom and her 1-pound baby weigh 160 pounds together, how much does the mama weigh? Bonus: Once each baby weighs 1 1/2 pounds, how much do 2 of them weigh together?
The sky’s the limit: If 8 baby tortoises sit in a circle, and the mama kisses every 3rd one — the 1st, then the 4th, then the 7th, then looping back around to the 2nd, and so on…how many times does she have to go around to kiss all 8?
Wee ones: 2 types of tortoises.
Little kids: The 6th tortoise. Bonus: The 9th tortoise.
Big kids: 159 pounds. Bonus: 3 pounds.
The sky’s the limit: 3 times around, since she’s kissing about 1/3 of them each time. She’ll kiss the 1st, 4th and 7th baby, then the 2nd, 5th and 8th, then the 3rd and 6th.
### Laura Overdeck
Laura Bilodeau Overdeck is founder and president of Bedtime Math Foundation. Her goal is to make math as playful for kids as it was for her when she was a child. Her mom had Laura baking while still in diapers, and her dad had her using power tools at a very unsafe age, measuring lengths, widths and angles in the process. Armed with this early love of numbers, Laura went on to get a BA in astrophysics from Princeton University, and an MBA from the Wharton School of Business; she continues to star-gaze today. Laura’s other interests include her three lively children, chocolate, extreme vehicles, and Lego Mindstorms. | 653 | 2,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.953778 |
https://artsintegration.com/2012/11/08/equation-rhyme-time-math-through-music/ | 1,638,492,100,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00106.warc.gz | 200,049,633 | 20,695 | Susan Riley | November 2012
# Equation Rhyme Time: Math through Music
In today’s strategy post, we’ll be featuring two music techniques that can be directly tied to Common Core math and Common Core reading and writing. The simplicity of these strategies will (hopefully) provide you with the confidence you need to try making intentional connections between music and other content areas throughout the day.
## Equation Rhythms
Equation rhythms provide students with the time and safe space to share their understanding of equations, beginning algebraic concepts, and fractions when they might otherwise be nervous. When you use this method, students view it as a game and create rhythms that showcase not only their knowledge of these mathematical concepts, but also their ability to manipulate that knowledge into something new and creative.
### Here are the Steps:
1. Begin with simple rhythms the students know. Clap a 4-beat pattern of quarter notes (ta, ta, ta, ta) and then a 4-beat pattern of eighth notes (ti-ti, ti-ti, ti-ti, ti-ti). Ask students to repeat clap each of those lines.
2. Begin to mix up the two rhythms in your model clap (ta, ti-ti, ti-ti, ta) and have them echo back a few of these as well.
3. As students gain confidence, add two more rhythms to the mix: a half note (ta…aa) and a rest (just open up your hands for one beat) and have them echo back.
4. Once students have mastered variations of patterns with these 4 main rhythms, explain to the students that they are going to create an “answer” to your “question” rhythm.
5. Choose a model student who you feel would be good at this activity. Tell them that you are going to clap a rhythm while the rest of the class whisper-counts to 4. When you are finished, the student will clap an answer ,which must take place while the class whisper-counts to 4, and the answer must be different than the teacher’s question. IE: Teacher: ta, ta, ti-ti, ta Student: ti-ti, rest, ta, ta
6. Allow each student to practice this (may take several tries).
7. Write the beat counts on the board for a teacher rhythm (IE: ta, ta, ti-ti, ta would be written 1 + 1 + 1/2 + 1/2 + 1) and ask students to find the answer (4). Then, do the same thing for a student “answer” rhythm. Ask them to find the answer for that equation (should also be 4).
Extensions: You can extend this by sharing that the questions and the answers all create the same end result. Work backwards and have students create rhythmic compositions to traditional math equations.
## Active Listening
Active listening is a wonderful strategy that provides students with the opportunity to slow down and focus on specific elements of a piece of music. This is often an area of great comfort for many teachers because they themselves are not required to perform the music. I recommend choosing music that has a story behind it (ie: Saint-Saens “Danse Macbre” or Prokofiev’s “Peter and the Wolf”) so that you can listen for instrument choice, how feelings are portrayed, what dynamics were used, etc to tell the story.
### Here are the Steps:
1. Choose a piece of music that has a rich variety of expressive elements. Some recommendations include: Danse Macabre, Peter and the Wolf, Nimrod Symphony IX, In the Hall of the Mountain King, Beethoven’s Symphony number 3, Haydn’s Surprise Symphony.
2. Listen to the piece prior to your students. Allow yourself to write down any specific elements that pop out at you and what images that creates in your mind.
3. Beside that list, create a column with the header “what caused that?” at the top.
4. Listen to the piece again and when you hear an item on your list, write down what you hear this time that caused that initial reaction (was it the dynamics, the different use of instruments, the speed of the piece?) | 890 | 3,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-49 | latest | en | 0.939329 |
https://www.multiplication.com/our-blog/caycee-coffield/class-activity-teaching-4x6 | 1,670,651,275,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711712.26/warc/CC-MAIN-20221210042021-20221210072021-00596.warc.gz | 965,588,628 | 17,575 | # A Class Activity For Teaching 4x6
Posted by:
Teaching the multiplication facts with our pictures and story association is so much fun.
For the fact 4 (floor) x 6 (chicks) try this activity.
For this activity you will need:
-Candy or a fun treat. Using Peeps to represent 6 (chicks) is always fun!
- A mix of markers, crayons and colored pencils.
- White paper or small paper bags, enough for each student.
Begin by dividing one side of the bag into 4 equal parts.
Then label each portion: 1) Array 2) Grouping 3) Repeated Addition 4) Equation
Now have your students colorfully complete each section.
On the other side of the paper bag students will draw the picture story for 4 (door) x 6 (chicks) = 24 (denty floor)
Once the bags are complete fill them with a little treat.
We used peeps to represent the chicks, but you can use anything.
Also give each student a flash card with the story and equation, perfect for practicing at home.
The flashcards that go along with this activity can be found here.
Be sure to watch this quick video for 4 (door) x 6 (chicks) = 24 (denty floor). | 285 | 1,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.885833 |
https://www.atarimagazines.com/compute/issue74/readers_feedback_2.php | 1,723,148,285,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640740684.46/warc/CC-MAIN-20240808190926-20240808220926-00437.warc.gz | 508,214,134 | 4,566 | ` COMPUTE! ISSUE 74 / JULY 1986 / PAGE 10`
# Readers Feedback
### The Editors and Readers of COMPUTE!
If you have any questions, comments, or suggestions you would like to see addressed in this column, write to "Readers' Feedback," COMPUTE!, P.O. Box 5406, Greensboro, NC 27403. Due to the volume of mail we receive, we regret that we cannot provide personal answers to technical questions.
## Faster Fractals In Forth
I enjoyed reading Paul Carison's article on fractal graphics for the IBM PC/PCjr (COMPUTE!, March 1986). His explanations were very clear. But it must have been a real trial for him to develop the BASIC version of the "Eight Thousand Dragons" program. We would like to show the beauty of fractals when written with a language that supports recursion. Here is an example of Forth code that does the same thing. It's written for Mach1, our Forth compiler for the Apple Macintosh and Atari ST. The execution time for a fourteenth-degree dragon is only three minutes.
```:Drag RECURSIVE { x1 y1 x2 y2 x3 y3 k | x4 y4 x5 y5 }
PAUSE
k 0= (if k=0 just draw and return)
IF (else continue breaking lines down)
k 1 - -> x4
x1 x2 + 2 / y2 y1 - 2 / - -> x4
y1 y2 + 2 / x2 x1 - 2 / + -> y4
x2 x3 + 2 / y3 y2 - 2 / + -> x5
y2 y3 + 2 / x3 x2 - 2 / - -> y5
x1 y1 x4 y4 x2 y2 k Drag
x2 y2 x5 y5 x3 y3 k Drag
THEN;
:Dragon { iters -- } ('14 dragon' gives the best results)
CLS
100 190 CALL MoveTo (place pen at beginning)
100 190 228 62 356 190 iters Drag (start with initial seed)
```
Terry Noyes
Although recursive routines (program segments that call themselves) are ordinarily taboo in BASIC, they're not only feasible, but encouraged in other languages such as Logo and Forth. Besides speeding execution, recursion produces compact, elegant code, as this example shows. Thanks for the demonstration. | 532 | 1,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.881106 |
https://squamulation.xyz/blog/fdhnz/piecewise-function-worksheet.html | 1,638,319,832,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00600.warc.gz | 608,841,603 | 12,326 | # Piecewise Function Worksheet
Showing top 8 worksheets in the category piecewise functions. Name period worksheet piecewise functions evaluate the following for 2 35 4 4 xx fx xx.
Worksheet Piecewise Functions
### Worksheet piecewise functions answer keypdf.
Piecewise function worksheet. 0 0 t 5 if 2 4 if 2 xx fx x 2. Identify any points of discontinuity. Piecewise functions worksheet 2 part i.
Test how much you know about taking small bits and pieces of graphs and putting them together by. The extensive applications of piecewise functions are used in a variety of mathematical formulas. 0 0 t 2 1 if 1 2 3 if 1 xx fx xx.
This lesson quiz and worksheet form a complementary pair of resources. Worksheet piecewise functions answer keypdf. Cn q2l01s6 wkuufttaw mstoifhtjwgaarver vllwcgi zaalwlb rcisgshntksw srhesfelrpvceldro s zmjajdxel wninthq pinnlfjiinditmeu jpvrdepcjaflzcludluusy.
Learn how to translate piecewise functions. Before look at the worksheet if you would like to know the stuff related to piecewise functions please click here. Displaying worksheet piecewise functions answer keypdf.
Graph each of the following piecewise functions. Some of the worksheets displayed are piecewise functions date period math 2 name piecewise functions work 2 graph each work homework piecewise functions name mathematics ii unit 5 step and piecewise functions part 1 work piecewise functions absolute value and piecewise functions work piecewise functions. Worksheet given in this section is much useful to the students who would like to practice problems on piecewise defined functions.
Worksheet Piecewise Functions Holidayfu Com
Worksheet Piecewise Functions Algebra 2 Answers Unique
10 Best Images Of Algebra 2 Piecewise Function Worksheets
Piecewise Linear Functions Common Core Algebra 2 Homework Lovely
Homework Piecewise Functions Worksheet 1 8 Answers
Piecewise Function Limits Math Finding The Limit Functions And
Quiz Worksheet Finding The Domain Of Piecewise Functions Study Com
Piecewise Function Graphing Piecewise Functions Math Tutorvista Com
Free Algebra 2 Worksheets Piecewise Functions | 476 | 2,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.755294 |
https://www.solvedlib.com/n/algebra-and-trigonometry | 1,722,682,742,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00490.warc.gz | 791,756,032 | 11,201 | # Algebra and Trigonometry
All new and solved questions in Algebra and Trigonometry category
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##### Problem 1Leni plans to use a ramp to make it easier to move a piano out of the back of his truck. The back of the truck is 33 inches tall and the ramp is 65 inches long. What is the horizontal distance from the end of the ramp to the back of the truck?Problem 2BongBong is admiring a statue in Seaside Park. The statue is 12 feet taller than he is, and BongBong is standing 16 feet away. How far is it from the top of the statue to BongBong's headProblem 3At ice-skating lessons, Manny attempts
Problem 1 Leni plans to use a ramp to make it easier to move a piano out of the back of his truck. The back of the truck is 33 inches tall and the ramp is 65 inches long. What is the horizontal distance from the end of the ramp to the back of the truck? Problem 2 BongBong is admiring a statu...
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The floor in a computer lab has a width of 6x – 9 and a length of 6x + 9. Find the area of the floor. Each tile that will cover the floor has an area of 24x2- 12x – 36. Divide the two areas and tell how many tiles will be needed to cover the floor?...
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As a tennis ball is struck, it departs from the racket horizont- ally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. How far above the court is the tennis ball when leaves the racket?...
5 answers
##### II. Given the nth term formula, complete the first five terms of the sequence 1. an = 5n2 -2n for n ? 1 2. an = n/(n+1) for n?0 3. an = 4n2 –n - 1 for n ? 1 4. a_n= 2n^2+ n – 3 for n ? 0 5. a1 = 3, a2 = 7, and an = 2an-1 + an-2 for n ? 3III. Substitute the given values in the formula A = Pe^rt to find the missing quantity 1. P = 750,000, r = 18% per year, t = 8 years 2. A = 1,740,000, r =
II. Given the nth term formula, complete the first five terms of the sequence 1. an = 5n2 -2n for n ? 1 2. an = n/(n+1) for n?0 3. an = 4n2 –n - 1 for n ? 1 4. a_n= 2n^2+ n – 3 for n ? 0 5. a1 = 3, a...
5 answers
##### A space vehicle is coasting at a constant velocity of 21.0 m/s in the +y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.320 m/s in the +x direction. After 45.0 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle measured from the +y
A space vehicle is coasting at a constant velocity of 21.0 m/s in the +y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.320 m/s in the +x direction. After 45.0 s, the pilot shuts off the RCS thruster....
5 answers
##### A hot-air balloon is rising straight up with a speed of 3.0m/s. A ballast bag is released from rest relative to the balloon at 9.5m above the ground. How much... time elapses before the ballast bag hits the ground?
A hot-air balloon is rising straight up with a speed of 3.0m/s. A ballast bag is released from rest relative to the balloon at 9.5m above the ground. How much... time elapses before the ballast bag hits the ground?...
5 answers
##### A 12 sided die is rolled. The set of equally likely outcomes is {1,2,3,4,5,6,7,8,9,10,11,12}. Find the probability of rolling a 6.
A 12 sided die is rolled. The set of equally likely outcomes is {1,2,3,4,5,6,7,8,9,10,11,12}. Find the probability of rolling a 6....
-- 0.188559-- | 1,913 | 6,898 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-33 | latest | en | 0.912178 |
https://discusstest.codechef.com/t/help-for-problem-domsol-https-www-codechef-com-problems-domsol/21455 | 1,627,547,861,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00360.warc.gz | 216,117,192 | 3,952 | # help for problem 'DOMSOL' https://www.codechef.com/problems/DOMSOL
can someone tell me why my code is getting ‘wrong answer’ for the problem ‘DOMSOL’
``````/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int column = sc.nextInt();
int[][] matrix = new int[2][column];
for(int i = 0; i < 2; i++) {
for(int j = 0; j < column; j++) {
matrix[i][j] = sc.nextInt();
}
}
// for(int i = 0; i < 2; i++) {
// for(int j = 0; j < column; j++) {
// System.out.println(matrix[i][j]);
// }
// }
int type = 1; // vertical
long score = Math.abs(matrix[0][0] - matrix[1][0]);
// System.out.println(score);
for (int i = 1; i < column; i++) {
if (type == 1) {
if ((Math.abs(matrix[0][i-1] - matrix[1][i-1]) + Math.abs(matrix[0][i] - matrix[1][i])) >= (Math.abs(matrix[0][i] - matrix[0][i-1]) + Math.abs(matrix[1][i] - matrix[1][i-1]))) {
score += Math.abs(matrix[0][i] - matrix[1][i]);
} else {
type = 2;
score += - Math.abs(matrix[0][i-1] - matrix[1][i-1])
+ Math.abs(matrix[0][i] - matrix[0][i-1])
+ Math.abs(matrix[1][i] - matrix[1][i-1]);
}
} else {
int a = Math.abs(matrix[0][i] - matrix[1][i]) + Math.abs(matrix[0][i-1] - matrix[0][i-2]) + Math.abs(matrix[1][i-1] - matrix[1][i-2]);
int b = Math.abs(matrix[0][i-2] - matrix[1][i-2]) + Math.abs(matrix[0][i] - matrix[0][i-1]) + Math.abs(matrix[1][i] - matrix[1][i-1]);
if (a >= b) {
type = 1;
score += Math.abs(matrix[0][i] - matrix[1][i]);
} else {
score -= Math.abs(matrix[0][i-1] - matrix[0][i-2]) + Math.abs(matrix[1][i-1] - matrix[1][i-2]);
score += b;
}
}
}
System.out.println(score);
}
}``````
// | 647 | 1,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-31 | latest | en | 0.15173 |
https://www.got-it.ai/solutions/excel-chat/excel-tutorial/rank/rank-function-example | 1,725,956,863,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00074.warc.gz | 737,590,233 | 17,904 | Get instant live expert help with Excel or Google Sheets
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# How to use the Excel RANK function
We can use the RANK function to RANK VALUES from the highest to the lowest. The highest value is assigned as 1. The steps below will walk through the process.
Figure 1: How to use the Excel RANK function
## Syntax
`=RANK(value,data)`
• Value is the number we want to rank
• Data is the named range of the values
## Formula
`=RANK(C4,Marks)`
## Setting up the Data
We have subjects offered by a student in a semester. We will rank the marks obtained from the highest to the lowest.
• We will input the SUBJECTS in Column B
• We will input the MARKS into Column C
• Column D is where we want the RANK FUNCTION to return the result
• NoteWe must highlight Cell C4 to Cell C11 and click on the drop-down arrow where there is C3 in figure 2. We will name the range as “Marks” and press enter
Figure 2: Setting up the Data
## RANK Function
• We will click on Cell D4
• We will insert the formula below into the cell
`=RANK(C4,Marks)`
• We will press the enter key
Figure 3: Output for Cell D4 with the RANK Function
• We will click on Cell D4 again
• We will double-click on the fill handle (the small plus sign at the bottom right of Cell D4) and drag down to copy the formula into the other cells
Figure 4: Result for Column D with the RANK function
## Explanation
• The RANK Function checks the Marks in Column C and assigns numbers to them beginning with 1 from the highest to the lowest mark.
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Trusted by people who work at | 643 | 2,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-38 | latest | en | 0.884712 |
http://www.differencebetween.net/science/mathematics-statistics/differences-between-the-taylor-and-maclaurin-series/ | 1,718,988,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862132.50/warc/CC-MAIN-20240621160500-20240621190500-00602.warc.gz | 41,282,364 | 25,809 | Difference Between Similar Terms and Objects
# Differences Between the Taylor and Maclaurin Series
Taylor vs Maclaurin Series
Aside from flying cockroaches, here is another thing that most people detest – math. We are often stricken with fear when we are facing math. The numbers seem like they are rattling our head, and it seems that math is eating up all of our life force. No matter what we do, we can’t escape the clutches of math. From counting to complex equations, we are always dealing with math. Nevertheless, we have to deal with it. Face your fear and learn to handle it. We have to meet Taylor and Maclaurin. Who are these people? These are not people. These are mathematical series.
In the field of mathematics, a Taylor series is defined as the representation of a function as an infinite sum of terms that are calculated from the values of the function’s derivatives at a single point. The Taylor series got its name from Brook Taylor. Brook Taylor was an English mathematician in 1715. It is all right to approximate the value of a function through utilizing the finite number of terms in the Taylor series. Approximating the value is already a common practice. In this approximation process, the Taylor series can yield quantitative estimates on the error. A Taylor polynomial is the term used to represent the finite number of the Taylor series’ initial function terms.
According to wikipedia.org, there are other uses of the Taylor series for determining analytic functions. The Taylor series can be used in obtaining the partial sums or the Taylor polynomials through using approximation techniques in the entire function. Another usage of the Taylor series is the differentiation and integration of the power series which can be done with each term. The Taylor series can also provide a complex analysis through integrating the analytic function with a holomorphic function in a complex plane. It can also be used to obtain and compute values numerically in a truncated series. This is done by applying the Chebyshev formula and Clenshaw algorithm. Another thing is that you can use the Taylor series in algebraic operations. An example of this is applying the Euler’s formula connecting with the Taylor series for the expansion of trigonometric and exponential functions. This can be used in the field of harmonic analysis. You can also use the Taylor series in the field of physics.
A Taylor series becomes a Maclaurin series if the Taylor series is centered at the point of zero. The Maclaurin series is named after Colin Maclaurin. Colin Maclaurin was a Scottish mathematician who had greatly used the Taylor series during the 18th century. A Maclaurin series is the expansion of the Taylor series of a function about zero. According to mathworld.wolfram.com, the Maclaurin series is a type of series expansion in which all terms are non-negative integer powers of the variable. Other more general types of series include the Laurent series and the Puiseux series. The Taylor and Maclaurin series have many uses in the mathematical field including the sciences.
Summary:
1. In the field of mathematics, a Taylor series is defined as the representation of a function as an infinite sum of terms that are calculated from the values of the function’s derivatives at a single point.
2. A Taylor series becomes a Maclaurin series if the Taylor series is centered at the point of zero. A Maclaurin series is the expansion of the Taylor series of a function about zero.
3. The Taylor series got its name from Brook Taylor. Brook Taylor was an English mathematician in 1715. The Maclaurin series is named after Colin Maclaurin. Colin Maclaurin was a Scottish mathematician who had greatly used the Taylor series during the 18th century.
Latest posts by Celine (see all)
Sharing is caring! | 791 | 3,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.943079 |
https://www.electricalclassroom.com/web-stories/lenzs-law/ | 1,619,097,346,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039610090.97/warc/CC-MAIN-20210422130245-20210422160245-00345.warc.gz | 839,174,172 | 11,830 | Lenz’s law - Explained
Lenz’s law – Statement
The direction of an induced e.m.f. is always such that it tends to set up a current opposing the motion or the change of flux responsible for inducing that e.m.f.
Lenz’s law equation
Lenz’s law is based on Faraday’s law of electromagnetic induction. The combined equation for these two laws are:
Experiment explaining Lenz law
Experiment explaining Lenz law
When a bar magnet is brought closer to the enclosed ring, it is repulsed by the magnet. In this case, the induced current resists the further change increase in magnet flux. When the magnet is pulled back, the ring is attracted by it. In this case, the induced current resists the decrease in magnet flux. | 175 | 722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | longest | en | 0.910621 |
https://www.physicsforums.com/threads/kibble-problem-linear-motion-chap-2.396184/ | 1,544,749,993,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825123.5/warc/CC-MAIN-20181214001053-20181214022553-00224.warc.gz | 1,000,316,492 | 13,769 | # Homework Help: Kibble problem (linear motion chap. 2)
1. Apr 18, 2010
### levilevi
1. The problem statement, all variables and given/known data
7. A particle of mass m moves (in the region x > 0) under a force F =
−kx+c/x, where k and c are positive constants. Find the corresponding
potential energy function. Determine the position of equilibrium, and
the frequency of small oscillations about it.
2. Relevant equations
dx/dt = [(2/m)(E - V(x)]^1/2
3. The attempt at a solution
dx/[(2/m)(E+ Clnx - (1/2)kx^2]^1/2 = dt
V = -Clnx + (1/2)kx^2]^1/2
i think at equilibrum point V(x) must be zero but then how can i continue? This is en exam question and i must solve it completely correct. Thank you for your answers. (my firt language is not english. Sorry for mistakes.)
2. Apr 18, 2010
### kuruman
V(x) is not zero at equilibrium. What does equilibrium mean to you?
3. Apr 18, 2010
### levilevi
Thanks, you are right. Total force, i think is zero at equilibrium point. So;
-ka + $$\frac{c}{a}$$ = 0 (a = equilibrium point.) k = c/a2
Is it true?
4. Apr 18, 2010
### kuruman
It is true. If this is an exam question, you have to finish it by yourself. I will not tell you how to solve it, but I can tell you if something is correct or not and the rest is up to you.
5. Apr 18, 2010
### levilevi
The exam was three days ago. Nobody could solve this question and a similar one (in which F = -kx +a/x^3) The teacher gave us as a homework. If i will solve maybe my exam mark will increase but i am not sure this. Maybe not. If you don not want to help it is up to you. I will study on it again and again...
Last edited: Apr 18, 2010
6. Apr 18, 2010
### levilevi
I solved at last. a=(c/k)^1/2 and w=(2k/m)^1/2 , w=(V''(a) /m)^1/2. Thanks. | 557 | 1,754 | {"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} | 3.6875 | 4 | CC-MAIN-2018-51 | latest | en | 0.93007 |
https://www.jiskha.com/questions/160816/how-do-i-graph-piecewise-functions-like-f-x-x-2-if-x-greater-equal-than-0-x-less-than-0 | 1,618,442,354,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078900.34/warc/CC-MAIN-20210414215842-20210415005842-00360.warc.gz | 932,563,716 | 5,210 | # Algebra ugh!
How do I graph piecewise functions?
Like f(x)= x+2 if x greater/equal than 0
x less than 0
Also, how do I write an equation of the line passing through (2,1) and is perpendicular to the line y= 1/2x+1
1. 👍
2. 👎
3. 👁
1. You graph it in two regions
1) less than zero
2) from zero toward +inf.
You know the slope of the perpendicular line is -2 (the negative reciprocal of 1/2).
y=mx+b
y=-2x+b
Put in the point 2,1, and solve for b.
then you have slope and b, so you know the equation.
1. 👍
2. 👎
👤
bobpursley
2. You helped a LOT on the second question, I forgot to change it to the negative reciprocal, so thanks! I can do the second one now. But I'm still not sure about the top question. I'm not sure how do graph piecewise functions. When you say from 0, do you mean from the origin?
1. 👍
2. 👎
3. f(x)={-x+3, if x <1
1. 👍
2. 👎
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https://www21.in.tum.de/~nipkow/pubs/Flyspeck/Isabelle/ListAux.html | 1,516,586,378,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890947.53/warc/CC-MAIN-20180122014544-20180122034544-00704.warc.gz | 1,025,864,063 | 16,583 | # Theory ListAux
Up to index of Isabelle/HOL/Tame
theory ListAux
imports Main EfficientNat
begin
```(* ID: \$Id: ListAux.thy,v 1.3 2006/01/11 19:40:44 nipkow Exp \$
Author: Gertrud Bauer, Tobias Nipkow
*)
header {* Basic Functions Old and New *}
theory ListAux
imports Main EfficientNat
begin
declare Let_def[simp]
declare comp_def[code unfold]
subsection {* HOL *}
lemma pairD: "(a,b) = p ==> a = fst p ∧ b = snd p"
by auto
lemmas conj_aci = conj_comms conj_assoc conj_absorb conj_left_absorb
lemma [code unfold]: "max x y == (let u = x; v = y in if u <= v then v else u)"
lemma [code unfold]: "min x y == (let u = x; v = y in if u <= v then u else v)"
lemmas[code] = lessThan_0 lessThan_Suc
subsection {* Lists *}
declare mem_iff[simp] list_all_iff[simp] list_ex_iff[simp]
subsubsection{* @{text length} *}
syntax "_length" :: "'a list => nat" ("|_|")
translations
"|xs|" == "length xs"
lemma length3D: "|xs| = 3 ==> ∃x y z. xs = [x, y, z]"
apply (cases xs) apply simp
apply (case_tac list) apply simp
apply (case_tac lista) by simp_all
lemma length4D: "|xs| = 4 ==> ∃ a b c d. xs = [a, b, c, d]"
apply (case_tac xs) apply simp
apply (case_tac list) apply simp
apply (case_tac lista) apply simp
apply (case_tac listb) by simp_all
subsubsection {* @{const filter} *}
lemma filter_emptyE[dest]: "(filter P xs = []) ==> x ∈ set xs ==> ¬ P x"
lemma filter_comm: "[x ∈ xs. P x ∧ Q x] = [x ∈ xs. Q x ∧ P x]"
lemma filter1: " [x∈xs . P x] = [] ==>
[x∈ xs. Q x ∧ P x] = []"
by (induct xs) (simp_all split: split_if_asm)
lemma filter_prop: "!!x. x ∈ set [u∈ys . P u] ==> P x"
proof (induct ys)
case Nil then show ?case by simp
next
case (Cons y ys) then show ?case
by (auto split: split_if_asm)
qed
lemma filter_Cons_prop: "[u∈ys . P u] = x # xs ==> P x"
proof -
assume "[u∈ys . P u] = x # xs"
then have "x ∈ set [u∈ys . P u]" by simp
then show "P x" by (rule filter_prop)
qed
lemma filter_compl1:
"([x ∈ xs. P x] = []) = ([x ∈ xs. ¬ P x] = xs)" (is "?lhs = ?rhs")
proof
show "?rhs ==> ?lhs"
proof (induct xs)
case Nil then show ?case by simp
next
case (Cons x xs)
have "[u∈xs . ¬ P u] ≠ x # xs"
proof
assume "[u∈xs . ¬ P u] = x # xs"
then have "|x # xs| = |[u∈xs . ¬ P u]|" by simp
also have "... ≤ |xs|" by simp
finally show False by simp
qed
with Cons show ?case by auto
qed
next
show "?lhs ==> ?rhs"
by (induct xs) (simp_all split: split_if_asm)
qed
lemma [simp]: "Not o (Not o P) = P"
by (rule ext) simp
lemma filter_compl2: "!!P. (filter (Not o P) xs = []) = (filter P xs = xs)"
lemma filter_eqI:
"(!!v. v ∈ set vs ==> P v = Q v) ==> [v ∈ vs . P v] = [v ∈ vs . Q v]"
by (induct vs) simp_all
lemma filter_simp: "(!!x. x ∈ set xs ==> P x) ==> [x ∈ xs. P x ∧ Q x] = [x ∈ xs. Q x]"
by (induct xs) auto
lemma filter_True_eq1:
"(length [y ∈ xs. P y] = length xs) ==> (!!y. y ∈ set xs ==> P y)"
proof (induct xs)
case Nil then show ?case by simp
next
case (Cons x xs)
have l: "length (filter P xs) ≤ length xs"
have hyp: "length (filter P (x # xs)) = length (x # xs)" .
then have "P x" by (simp split: split_if_asm) (insert l, arith)
moreover with hyp have "length (filter P xs) = length xs"
by (simp split: split_if_asm)
moreover have "y ∈ set (x#xs)" .
ultimately show ?case by (auto dest: Cons(1))
qed
lemma length_filter_True_eq:
"(length [y ∈ xs. P y] = length xs) = (∀y. y ∈ set xs --> P y)"
by (intro iffI allI impI, erule filter_True_eq1) simp_all
subsubsection {* @{const map} *}
syntax (xsymbols)
"@map" :: "[ 'b, pttrn, 'a list] => 'a list"("(1[_. _ ∈ _])")
syntax
"@map" :: "[ 'b, pttrn, 'a list] => 'a list"("(1[_./ _ : _])")
translations
"[f. x ∈ xs]"== "map (λx. f) xs"
"[f. x : xs]"== "map (λx. f) xs";
subsubsection {* @{const map_filter} *}
syntax (xsymbols)
"@map_filter" :: "['b, pttrn, 'a list, bool] => 'a list"("(1[_. _ ∈ _, _])")
syntax
"@map_filter" :: "['b, pttrn, 'a list, bool] => 'a list"("(1[_./ _ : _, _])")
translations
"[f. x ∈ xs, P]"== "map_filter (λx. f) P xs"
"[f. x : xs, P]"== "map_filter (λx. f) P xs"
lemma [simp]: "[f x. x ∈ xs, P] = [f x. x ∈ [x ∈ xs. P x]]"
by (induct xs) auto
subsubsection {* @{const concat} *}
syntax (xsymbols)
"@concat" :: "idt => 'a list => 'a list => 'a list" ("\<Squnion>_∈ _ _" 10)
translations "\<Squnion>x∈xs f" == "concat [f. x ∈ xs]"
subsubsection {* List product *}
constdefs listProd1 :: "'a => 'b list => ('a × 'b) list"
"listProd1 a bs ≡ [(a,b). b ∈ bs]"
constdefs listProd :: "'a list => 'b list => ('a × 'b) list" (infix "×" 50)
"as × bs ≡ \<Squnion>a ∈ as listProd1 a bs"
lemma(*<*)[simp]: (*>*) "set (xs × ys) = (set xs) × (set ys)"
by (auto simp: listProd_def listProd1_def)
subsubsection {* Minimum and maximum *}
(* FIXME combine minimal with min_list *)
(* more general:
foldl1 f (x#xs) = if xs=[] then x else f x (foldl1 f xs)
min_list = foldl1 min
max_list = foldl1 max
sum_list = foldl1 (op +)
Put in separate ExecList.thy
*)
consts minimal:: "('a => nat) => 'a list => 'a"
primrec
"minimal m (x#xs) =
(if xs=[] then x else
let mxs = minimal m xs in
if m x ≤ m mxs then x else mxs)"
lemma minimal_in_set[simp]: "xs ≠ [] ==> minimal f xs : set xs"
by(induct xs) auto
lemma minimal_Cons1:
"∀y ∈ set xs. f x ≤ f y ==> minimal f (x#xs) = x"
by (induct xs) auto
lemma minimal_append2:
"∀x ∈ set xs. f x > f y ==> minimal f (xs @ y # ys) = minimal f (y # ys)"
by (induct xs) simp_all
lemma minimal_neq_lowerbound:
"xs ≠ [] ==> ALL x: set xs. f x ≥ n ==> f(minimal f xs) ≠ n
==> ALL x: set xs. f x ≠ n"
apply(induct xs)
apply simp
apply (auto split:split_if_asm)
done
(* FIXME minList to min_list *)
consts minList :: "nat list => nat"
primrec
"minList (x#xs) = (if xs=[] then x else min x (minList xs))"
consts max_list :: "nat list => nat"
primrec
"max_list (x#xs) = (if xs=[] then x else max x (max_list xs))"
lemma minList_conv_Min[simp]:
"xs ≠ [] ==> minList xs = Min(set xs)"
by (induct xs) auto
lemma max_list_conv_Max[simp]:
"xs ≠ [] ==> max_list xs = Max(set xs)"
by (induct xs) auto
subsubsection {* replace *}
(* FIXME replace "remove1" by "replace1" in List.thy? *)
consts replace :: "'a => 'a list => 'a list => 'a list"
primrec
"replace x ys [] = []"
"replace x ys (z#zs) =
(if z = x then ys @ zs else z # (replace x ys zs))"
consts mapAt :: "nat list => ('a => 'a) => ('a list => 'a list)"
primrec
"mapAt [] f as = as"
"mapAt (n#ns) f as =
(if n < |as| then mapAt ns f (as[n:= f (as!n)])
else mapAt ns f as)"
lemma length_mapAt[simp]: "!!xs. length(mapAt vs f xs) = length xs"
by(induct vs) auto
lemma length_replace1[simp]: "length(replace x [y] xs) = length xs"
by(induct xs) simp_all
lemma replace_id[simp]: "replace x [x] xs = xs"
by(induct xs) simp_all
lemma len_replace_ge_same:
"length ys ≥ 1 ==> length(replace x ys xs) ≥ length xs"
by (induct xs) auto
lemma len_replace_ge[simp]:
"[| length ys ≥ 1; length xs ≥ length zs |] ==>
length(replace x ys xs) ≥ length zs"
apply(drule len_replace_ge_same[where x = x and xs = xs])
apply arith
done
lemma replace_append[simp]:
"replace x ys (as @ bs) =
(if x ∈ set as then replace x ys as @ bs else as @ replace x ys bs)"
by(induct as) auto
lemma filter_replace:
"¬ P y ==> filter P (replace x [y] xs) = remove1 x (filter P xs)"
by(induct xs) simp_all
lemma distinct_set_replace: "distinct xs ==>
set (replace x ys xs) =
(if x ∈ set xs then (set xs - {x}) ∪ set ys else set xs)"
apply(induct xs)
apply(simp)
apply simp
apply blast
done
lemma replace1:
"f ∈ set (replace f' fs ls ) ==> f ∉ set ls ==> f ∈ set fs"
proof (induct ls)
case Nil then show ?case by simp
next
case (Cons l ls) then show ?case by (simp split: split_if_asm)
qed
lemma replace2:
"f' ∉ set ls ==> replace f' fs ls = ls"
proof (induct ls)
case Nil then show ?case by simp
next
case (Cons l ls) then show ?case by (auto split: split_if_asm)
qed
lemma replace3[intro]:
"f' ∈ set ls ==> f ∈ set fs ==> f ∈ set (replace f' fs ls)"
by (induct ls) auto
lemma replace4:
"f ∈ set ls ==> oldF ≠ f ==> f ∈ set (replace oldF fs ls)"
by (induct ls) auto
lemma replace5: "f ∈ set (replace oldF newfs fs) ==> f ∈ set fs ∨ f ∈ set newfs"
by (induct fs) (auto split: split_if_asm)
lemma replace6: "distinct oldfs ==> x ∈ set (replace oldF newfs oldfs) =
((x ≠ oldF ∨ oldF ∈ set newfs) ∧ ((oldF ∈ set oldfs ∧ x ∈ set newfs) ∨ x ∈ set oldfs))"
by (induct oldfs) auto
lemma replace_delete_oldF:
"oldF ∉ set fs ==> distinct ls ==> oldF ∉ set (replace oldF fs ls)"
by (induct ls) auto
lemma distinct_replace:
"distinct fs ==> distinct newFs ==> set fs ∩ set newFs ⊆ {oldF} ==>
distinct (replace oldF newFs fs)"
proof (induct fs)
case Nil then show ?case by simp
next
case (Cons f fs)
then show ?case
proof (cases "f = oldF")
case True with Cons show ?thesis by simp blast
next
case False
moreover with Cons have "f ∉ set newFs" by simp blast
with Cons have "f ∉ set (replace oldF newFs fs)"
by simp (blast dest: replace1)
moreover from Cons have "distinct (replace oldF newFs fs)"
by (rule_tac Cons) auto
ultimately show ?thesis by simp
qed
qed
lemma replace_replace[simp]: "oldf ∉ set newfs ==> distinct xs ==>
replace oldf newfs (replace oldf newfs xs) = replace oldf newfs xs"
apply (induct xs) apply auto apply (rule replace2) by simp
lemma replace_distinct: "distinct fs ==> distinct newfs ==> oldf ∈ set fs --> set newfs ∩ set fs ⊆ {oldf} ==>
distinct (replace oldf newfs fs)"
apply (case_tac "oldf ∈ set fs") apply simp
apply (induct fs) apply simp
apply (auto simp: replace2) apply (drule replace1)
by auto
lemma filter_replace2:
"[| ¬ P x; ∀y∈ set ys. ¬ P y |] ==>
filter P (replace x ys xs) = filter P xs"
apply(cases "x ∈ set xs")
apply(induct xs)
apply simp
apply clarsimp
done
lemma length_filter_replace1:
"[| x ∈ set xs; ¬ P x |] ==>
length(filter P (replace x ys xs)) =
length(filter P xs) + length(filter P ys)"
apply(induct xs)
apply simp
apply fastsimp
done
lemma length_filter_replace2:
"[| x ∈ set xs; P x |] ==>
length(filter P (replace x ys xs)) =
length(filter P xs) + length(filter P ys) - 1"
apply(induct xs)
apply simp
apply auto
apply(drule split_list)
apply clarsimp
done
subsubsection {* @{const"distinct"} *}
lemma dist_filter_single:
"distinct ls ==> v ∈ set ls ==> [a∈ ls . a = v] = [v]"
proof (induct ls)
case Nil then show ?case by auto
next
case (Cons l ls) then show ?case
proof (cases "l = v")
case True with Cons
have "v ∉ set ls" by auto
then have "[a∈ls . a = v] = []" by (induct ls) auto
with Cons True show ?thesis by auto
next
case False with Cons show ?thesis by auto
qed
qed
lemma dist_at1: "!! c vs. distinct vs ==> vs = a @ r # b ==> vs = c @ r # d ==> a = c"
proof (induct a)
case Nil
assume dist: "distinct vs" and vs1: "vs = [] @ r # b" and vs2: "vs = c @ r # d"
from dist vs2 have rc: "r ∉ set c" by auto
from vs1 vs2 have "c @ r # d = r # b" by auto
then have "hd (c @ r # d) = r" by auto
then have "c ≠ [] ==> hd c = r" by auto
then have "c ≠ [] ==> r ∈ set c" by (induct c) auto
with rc have c: "c = []" by auto
then show ?case by auto
next
case (Cons x xs) then show ?case by (induct c) auto
qed
lemma dist_at: "distinct vs ==> vs = a @ r # b ==> vs = c @ r # d ==> a = c ∧ b = d"
proof -
assume dist: "distinct vs" and vs1: "vs = a @ r # b" and vs2: "vs = c @ r # d"
then have "a = c" by (rule_tac dist_at1) auto
with dist vs1 vs2 show ?thesis by simp
qed
lemma dist_at2: "distinct vs ==> vs = a @ r # b ==> vs = c @ r # d ==> b = d"
proof -
assume dist: "distinct vs" and vs1: "vs = a @ r # b" and vs2: "vs = c @ r # d"
then have "a = c ∧ b = d" by (rule_tac dist_at) auto
then show ?thesis by auto
qed
lemma distinct_split1: "distinct xs ==> xs = y @ [r] @ z ==> r ∉ set y"
apply auto done
lemma distinct_split2: "distinct xs ==> xs = y @ [r] @ z ==> r ∉ set z" apply auto done
lemma distinct_hd_not_cons: "distinct vs ==> ∃ as bs. vs = as @ x # hd vs # bs ==> False"
proof -
assume d: "distinct vs" and ex: "∃ as bs. vs = as @ x # hd vs # bs"
from ex have vsne: "vs ≠ []" by auto
with d ex show ?thesis apply (elim exE) apply (case_tac "as")
apply (subgoal_tac "hd vs = x") apply simp apply (rule sym) apply simp
apply (subgoal_tac "x = hd (x # (hd vs # bs))") apply simp apply (thin_tac "vs = x # hd vs # bs")
apply auto
apply (subgoal_tac "hd vs = a") apply simp
apply (subgoal_tac "a = hd (a # list @ x # hd vs # bs)") apply simp
apply (thin_tac "vs = a # list @ x # hd vs # bs") by auto
qed
subsubsection {* Misc *}
(* FIXME move to List *)
lemma drop_last_in: "!!n. n < length ls ==> last ls ∈ set (drop n ls)"
apply (frule_tac last_drop) apply(erule subst)
apply (case_tac "drop n ls" rule: rev_exhaust) by simp_all
lemma nth_last_Suc_n: "distinct ls ==> n < length ls ==> last ls = ls ! n ==> Suc n = length ls"
proof (rule ccontr)
assume d: "distinct ls" and n: "n < length ls" and l: "last ls = ls ! n" and not: "Suc n ≠ length ls"
then have s: "Suc n < length ls" by auto
def lls ≡ "ls!n"
with n have "take (Suc n) ls = take n ls @ [lls]" apply simp by (rule take_Suc_conv_app_nth)
then have "take (Suc n) ls @ drop (Suc n) ls = take n ls @ [lls] @ drop (Suc n) ls" by auto
then have ls: "ls = take n ls @ [lls] @ drop (Suc n) ls" by auto
with d have dls: "distinct (take n ls @ [lls] @ drop (Suc n) ls)" by auto
from lls_def l have "lls = (last ls)" by auto
with s have "lls ∈ set (drop (Suc n) ls)" apply simp by (rule_tac drop_last_in)
with dls show False by auto
qed
(****************************** rotate *************************)
subsubsection {* @{const rotate} *}
lemma plus_length1[simp]: "rotate (k+(length ls)) ls = rotate k ls "
proof -
have "!! k ls. rotate k (rotate (length ls) ls) = rotate (k+(length ls)) ls"
by (rule rotate_rotate)
then have "!! k ls. rotate k ls = rotate (k+(length ls)) ls" by auto
then show ?thesis by (rule sym)
qed
lemma plus_length2[simp]: "rotate ((length ls)+k) ls = rotate k ls "
proof -
def x ≡ "(length ls)+k"
then have "x = k+(length ls)" by auto
with x_def have "rotate x ls = rotate (k+(length ls)) ls" by simp
then have "rotate x ls = rotate k ls" by simp
with x_def show ?thesis by simp
qed
lemma rotate_minus1: "n > 0 ==> m > 0 ==>
rotate n ls = rotate m ms ==> rotate (n - 1) ls = rotate (m - 1) ms"
proof (cases "ls = []")
assume r: "rotate n ls = rotate m ms"
case True with r
have "rotate m ms = []" by auto
then have "ms = []" by auto
with True show ?thesis by auto
next
assume n: "n > 0" and m: "m > 0" and r: "rotate n ls = rotate m ms"
case False
then have lls: "length ls > 0" by auto
with r have lms: "length ms > 0" by auto
have mem1: "rotate (n - 1) ls = rotate ((n - 1) + length ls) ls" by auto
from n lls have "(n - 1) + length ls = (length ls - 1) + n" by arith
then have "rotate ((n - 1) + length ls) ls = rotate ((length ls - 1) + n) ls" by auto
with mem1 have mem2: "rotate (n - 1) ls = rotate ((length ls - 1) + n) ls" by auto
have "rotate ((length ls - 1) + n) ls = rotate (length ls - 1) (rotate n ls)" apply (rule sym)
by (rule rotate_rotate)
with mem2 have mem3: "rotate (n - 1) ls = rotate (length ls - 1) (rotate n ls)" by auto
from r have "rotate (length ls - 1) (rotate n ls) = rotate (length ls - 1) (rotate m ms)" by auto
with mem3 have mem4: "rotate (n - 1) ls = rotate (length ls - 1) (rotate m ms)" by auto
have "rotate (length ls - 1) (rotate m ms) = rotate (length ls - 1 + m) ms" by (rule rotate_rotate)
with mem4 have mem5: "rotate (n - 1) ls = rotate (length ls - 1 + m) ms" by auto
from r have "length (rotate n ls) = length (rotate m ms)" by simp
then have "length ls = length ms" by auto
with m lms have "length ls - 1 + m = (m - 1) + length ms" by arith
with mem5 have mem6: "rotate (n - 1) ls = rotate ((m - 1) + length ms) ms" by auto
have "rotate ((m - 1) + length ms) ms = rotate (m - 1) (rotate (length ms) ms)" by auto
then have "rotate ((m - 1) + length ms) ms = rotate (m - 1) ms" by auto
with mem6 show ?thesis by auto
qed
lemma rotate_minus1': "n > 0 ==> rotate n ls = ms ==>
rotate (n - 1) ls = rotate (length ms - 1) ms"
proof (cases "ls = []")
assume r: "rotate n ls = ms"
case True with r show ?thesis by auto
next
assume n: "n > 0" and r:"rotate n ls = ms"
then have r': "rotate n ls = rotate (length ms) ms" by auto
case False
with n r' show ?thesis apply (rule_tac rotate_minus1) by auto
qed
lemma rotate_inv1: "!! ms. n < length ls ==> rotate n ls = ms ==>
ls = rotate ((length ls) - n) ms"
proof (induct n)
case 0 then show ?case by auto
next
case (Suc n) then show ?case
proof (cases "ls = []")
case True with Suc
show ?thesis by auto
next
case False
then have ll: "length ls > 0" by auto
from Suc have nl: "n < length ls" by auto
from Suc have r: "rotate (Suc n) ls = ms" by auto
then have "rotate (Suc n - 1) ls = rotate (length ms - 1) ms" apply (rule_tac rotate_minus1') by auto
then have "rotate n ls = rotate (length ms - 1) ms" by auto
then have mem: "ls = rotate (length ls - n) (rotate (length ms - 1) ms)"
apply (rule_tac Suc) by (auto simp: nl)
have " rotate (length ls - n) (rotate (length ms - 1) ms) = rotate (length ls - n + (length ms - 1)) ms"
by (rule rotate_rotate)
with mem have mem2: "ls = rotate (length ls - n + (length ms - 1)) ms" by auto
from r have leq: "length ms = length ls" by auto
with False nl have "length ls - n + (length ms - 1) = length ms + (length ms - (Suc n))"
by arith
then have "rotate (length ls - n + (length ms - 1)) ms = rotate (length ms + (length ms - (Suc n))) ms"
by auto
with mem2 have mem3: "ls = rotate (length ms + (length ms - (Suc n))) ms" by auto
have "rotate (length ms + (length ms - (Suc n))) ms = rotate (length ms - (Suc n)) ms" by simp
with mem3 leq show ?thesis by auto
qed
qed
lemma rotate_conv_mod'[simp]: "rotate (n mod length ls) ls = rotate n ls"
lemma rotate_inv2: "rotate n ls = ms ==>
ls = rotate ((length ls) - (n mod length ls)) ms"
proof (cases "ls = []")
assume r: "rotate n ls = ms"
case True with r show ?thesis by auto
next
assume r: "rotate n ls = ms"
then have r': "rotate (n mod length ls) ls = ms" by auto
case False
then have "length ls > 0" by auto
with r' show ?thesis apply (rule_tac rotate_inv1) by auto
qed
lemma rotate_inv': "ls = rotate ((length ls) - (n mod length ls)) ms ==>
rotate n ls = ms"
proof (cases "ls = []")
assume r: "ls = rotate ((length ls) - (n mod length ls)) ms"
case True with r show ?thesis by auto
next
assume r: "ls = rotate ((length ls) - (n mod length ls)) ms"
case False
def len ≡ "length ls"
with r have r': "ls = rotate (len - (n mod len)) ms" by simp
with len_def have lms: "length ms = len" by auto
def y ≡ "n mod len"
from len_def False have ll: "len > 0" by auto
with y_def have small_y: "y < len" by auto
then have len_gz: "len > 0" by auto
from r' len_def have mem0: "rotate n ls = rotate n (rotate (len - (n mod len)) ms)" by auto
have "rotate n (rotate (len - (n mod len)) ms) = rotate (n + (len - (n mod len))) ms"
by (rule rotate_rotate)
with mem0 have mem1: "rotate n ls = rotate (n + (len - (n mod len))) ms" by auto
have "n mod len - n mod len = 0" by arith
then have "len + n mod len - n mod len = len" by arith
from y_def have ymod: "n mod len = y" by auto
with len_gz obtain r where n: "n = y + r*len" apply (drule_tac mod_eqD) by auto
then have "n + len = (r*len)+len + y" by arith
then have "n + len = ((r+1)*len) + y" by auto
with len_gz small_y have "n + (len - y) = ((r+1)*len)" by auto
then have zero: "(n + (len - y)) mod len = 0" by auto
have "rotate (n + (len - y)) ms = rotate ((n + (len - y)) mod length ms) ms" by (rule_tac rotate_conv_mod)
with lms zero have "rotate (n + (len - y)) ms = ms" by auto
with mem1 y_def show ?thesis by auto
qed
lemma rotate_id[simp]: "rotate ((length ls) - (n mod length ls)) (rotate n ls) = ls"
apply (rule sym) apply (rule rotate_inv2) by simp
lemma nth_rotate1_Suc: "Suc n < length ls ==> ls!(Suc n) = (rotate1 ls)!n"
apply (auto simp: rotate1_def) apply (cases ls) apply auto
lemma nth_rotate1_0: "ls!0 = (rotate1 ls)!(length ls - 1)" apply (cases ls) by (auto simp: rotate1_def)
lemma nth_rotate1: "0 < length ls ==> ls!((Suc n) mod (length ls)) = (rotate1 ls)!(n mod (length ls))"
proof (cases "0 < (Suc n) mod (length ls)")
assume lls: "0 < length ls"
case True
def m ≡ "(Suc n) mod (length ls) - 1"
with True have m: "Suc m = (Suc n) mod (length ls)" by auto
with lls have a: "(Suc m) < length ls" by auto
from lls m have "m = n mod (length ls)" by (simp add: mod_Suc split:split_if_asm)
with a m show ?thesis apply (drule_tac nth_rotate1_Suc) by auto
next
assume lls: "0 < length ls"
case False
then have a: "(Suc n) mod (length ls) = 0" by auto
with lls have "Suc (n mod (length ls)) = (length ls)" by (auto simp: mod_Suc split: split_if_asm)
then have "(n mod (length ls)) = (length ls) - 1" by arith
with a show ?thesis by (auto simp: nth_rotate1_0)
qed
lemma rotate_Suc2[simp]: "rotate n (rotate1 xs) = rotate (Suc n) xs"
apply (simp add:rotate_def) apply (induct n) by auto
lemma nth_rotate: "!! ls. 0 < length ls ==> ls!((n+m) mod (length ls)) = (rotate m ls)!(n mod (length ls))"
proof (induct m)
case 0 then show ?case by auto
next
case (Suc m)
def z ≡ "n + m"
then have "n + Suc m = Suc (z)" by auto
with Suc have r1: "ls ! ((Suc z) mod length ls) = rotate1 ls ! (z mod length ls)"
by (rule_tac nth_rotate1)
from Suc have "0 < length (rotate1 ls)" by auto
then have "(rotate1 ls) ! ((n + m) mod length (rotate1 ls))
= rotate m (rotate1 ls) ! (n mod length (rotate1 ls))" by (rule Suc)
with r1 z_def have "ls ! ((n + Suc m) mod length ls)
= rotate m (rotate1 ls) ! (n mod length (rotate1 ls))" by auto
then show ?case by auto
qed
lemma help1: "∃ n. filter f (rotate1 ls) = rotate n (filter f ls)"
proof (cases ls)
case Nil then show ?thesis by auto
next
case (Cons m ms)
then show ?thesis
proof (cases "f m")
case True with Cons
have "filter f (rotate1 (ls)) = rotate 1 (filter f (ls))" by auto
then show ?thesis apply (rule_tac exI) .
next
case False with Cons
have "filter f (rotate1 (ls)) = rotate 0 (filter f (ls))" by auto
then show ?thesis apply (rule_tac exI) .
qed
qed
lemma help3: "¬ f l ==> n < length ls ==> filter f (rotate n ls) = filter f (rotate n (rotate1 (l # ls)))"
proof -
assume fl: "¬ f l" and n: "n < length ls"
then have n2: "n - length ls = 0" by auto
from fl n show ?thesis apply simp
apply (subst rotate_drop_take)+ by auto
qed
lemma help4: "¬ f l ==> n < length ls ==> filter f (rotate n ls) = filter f (rotate (Suc n) (l # ls))"
proof -
assume fl: "¬ f l" and n: "n < length ls"
then have r1: "filter f (rotate n ls) = filter f (rotate n (rotate1 (l # ls)))" by (rule help3)
def ms ≡ "l # ls"
have "rotate n (rotate1 (ms)) = rotate (Suc n) (ms)" by simp
with ms_def have "rotate n (rotate1 (l # ls)) = rotate (Suc n) (l # ls)" by simp
with r1 show ?thesis by auto
qed
lemma help2: "∃ n. rotate1 (filter f ls) = filter f (rotate n ls)"
proof (induct ls)
case Nil then show ?case by auto
next
case (Cons m ms)
then show ?case
proof (cases "f m")
case True with Cons
have "filter f (m # ms) = m # filter f ms" by auto
from True Cons have
"rotate1 (filter f (m # ms)) = filter f (rotate 1 (m # ms))" by auto
then show ?thesis apply (rule_tac exI) .
next
case False with Cons
have f1: "filter f (m # ms) = filter f ms" by auto
then have mem: "rotate1 (filter f (m # ms)) = rotate1 (filter f ms)" by auto
from False have f2: "!! ks. filter f (ks @ [m]) = filter f ks" by auto
from Cons obtain n where n: "rotate1 (filter f ms) = filter f (rotate n ms)" by auto
with mem have mem1: "rotate1 (filter f (m # ms)) = filter f (rotate n ms)" by auto
def n' ≡ "n mod length ms"
then have "filter f (rotate n ms) = filter f (rotate n' ms)" by auto
with mem1 have mem2: "rotate1 (filter f (m # ms)) = filter f (rotate n' ms)" by auto
from n'_def have "ms ≠ [] ==> n' < length ms" by auto
with False have "filter f (rotate n' ms) = filter f (rotate (Suc n') (m # ms))"
apply (case_tac "ms = []") apply force apply (rule_tac help4) by auto
with mem2 have "rotate1 (filter f (m # ms)) = filter f ((rotate (Suc n') (m # ms)))" by auto
then show ?thesis apply (rule_tac exI) .
qed
qed
lemma rotate_help5: "∃ n. filter f (rotate m ls) = rotate n (filter f ls)"
proof (induct m)
case 0 then have "filter f (rotate 0 ls) = rotate 0 (filter f ls)" by auto
then show ?case by (rule exI)
next
case (Suc m)
def ks ≡ "rotate m ls"
then have mem0: "filter f (rotate (Suc m) ls) = filter f (rotate1 ks)" by auto
from ks_def Suc obtain n where n: "filter f ks = rotate n (filter f ls)" by auto
have "∃ n'. filter f (rotate1 ks) = rotate n' (filter f ks)" by (rule help1)
then obtain n' where "filter f (rotate1 ks) = rotate n' (filter f ks)" by auto
with n have "filter f (rotate1 ks) = rotate n' (rotate n (filter f ls))" by auto
with mem0 have mem1: "filter f (rotate (Suc m) ls) = rotate n' (rotate n (filter f ls))" by auto
have "rotate n' (rotate n (filter f ls)) = rotate (n'+n) (filter f ls)" by (rule rotate_rotate)
with mem1 have "filter f (rotate (Suc m) ls) = rotate (n'+n) (filter f ls)" by auto
then show ?case apply (rule_tac exI) .
qed
lemma rotate_help6: "∃ n. rotate m (filter f ls) = filter f (rotate n ls)"
proof (induct m)
case 0 then have "rotate 0 (filter f ls) = filter f (rotate 0 ls)" by auto
then show ?case by (rule exI)
next
case (Suc m)
have mem0: "rotate (Suc m) (filter f ls) = rotate1 (rotate m (filter f ls))" by auto
from Suc obtain n where n: "rotate m (filter f ls) = filter f (rotate n ls)" by auto
with mem0 have mem1: "rotate (Suc m) (filter f ls) = rotate1 (filter f (rotate n ls))" by auto
def ks ≡ "rotate n ls"
from help2 obtain n' where "rotate1 (filter f ks) = filter f (rotate n' ks)" by auto
with mem1 ks_def have mem2: "rotate (Suc m) (filter f ls) = filter f (rotate n' ks)" by auto
from ks_def have "rotate n' ks = rotate (n'+n) ls" apply simp by (rule rotate_rotate)
with mem2 have "rotate (Suc m) (filter f ls) = filter f(rotate (n'+n) ls)" by simp
then show ?case apply (rule_tac exI) .
qed
(************************* SplitAt *******************************************)
subsection {* @{text splitAt} *}
consts splitAtRec ::
"'a => 'a list => 'a list => 'a list × 'a list"
primrec
"splitAtRec c bs [] = (bs,[])"
"splitAtRec c bs (a#as) = (if a = c then (bs, as)
else splitAtRec c (bs@[a]) as)"
constdefs splitAt :: "'a => 'a list => 'a list × 'a list"
"splitAt c as ≡ splitAtRec c [] as"
subsubsection {* @{const splitAtRec} *}
lemma splitAtRec_conv: "!!bs.
splitAtRec x bs xs =
(bs @ takeWhile (%y. y≠x) xs, tl(dropWhile (%y. y≠x) xs))"
by(induct xs) auto
lemma splitAtRec_distinct_fst: "!! s. distinct vs ==> distinct s ==> (set s) ∩ (set vs) = {} ==> distinct (fst (splitAtRec ram1 s vs))"
by (induct vs) auto
lemma splitAtRec_distinct_snd: "!! s. distinct vs ==> distinct s ==> (set s) ∩ (set vs) = {} ==> distinct (snd (splitAtRec ram1 s vs))"
by (induct vs) auto
lemma splitAtRec_ram:
"!! us a b. ram ∈ set vs ==> (a, b) = splitAtRec ram us vs ==>
us @ vs = a @ [ram] @ b"
proof (induct vs)
case Nil then show ?case by simp
next
case (Cons v vs) then show ?case by (auto dest: Cons(1) split: split_if_asm)
qed
lemma splitAtRec_notRam:
"!! us. ram ∉ set vs ==> splitAtRec ram us vs = (us @ vs, [])"
proof (induct vs)
case Nil then show ?case by simp
next
case (Cons v vs) then show ?case by auto
qed
lemma splitAtRec_distinct: "!! s. distinct vs ==>
distinct s ==> (set s) ∩ (set vs) = {} ==>
set (fst (splitAtRec ram s vs)) ∩ set (snd (splitAtRec ram s vs)) = {}"
by (induct vs) auto
subsubsection {* @{const splitAt} *}
lemma splitAt_conv:
"splitAt x xs = (takeWhile (%y. y≠x) xs, tl(dropWhile (%y. y≠x) xs))"
lemma splitAt_no_ram[simp]:
"ram ∉ set vs ==> splitAt ram vs = (vs, [])"
by (auto simp: splitAt_def splitAtRec_notRam)
lemma splitAt_split:
"ram ∈ set vs ==> (a,b) = splitAt ram vs ==> vs = a @ ram # b"
by (auto simp: splitAt_def dest: splitAtRec_ram)
lemma splitAt_ram:
"ram ∈ set vs ==> vs = fst (splitAt ram vs) @ ram # snd (splitAt ram vs)"
by (rule_tac splitAt_split) auto
lemma fst_splitAt_last:
"[| xs ≠ []; distinct xs |] ==> fst (splitAt (last xs) xs) = butlast xs"
subsubsection {* Sets *}
lemma splitAtRec_union:
"!! a b s. (a,b) = splitAtRec ram s vs ==> (set a ∪ set b) - {ram} = (set vs ∪ set s) - {ram}"
apply (induct vs) by (auto split: split_if_asm)
lemma splitAt_union:
"(a,b) = splitAt ram vs ==> (set a ∪ set b) - {ram} = set vs - {ram}"
lemma splitAt_subset_ab:
"(a,b) = splitAt ram vs ==> set a ⊆ set vs ∧ set b ⊆ set vs"
apply (cases "ram ∈ set vs")
by (auto dest: splitAt_split simp: splitAt_no_ram)
lemma splitAt_subset_fst:
"set (fst (splitAt ram vs)) ⊆ set vs"
proof -
def a: a ≡ "fst (splitAt ram vs)"
def b: b ≡ "snd (splitAt ram vs)"
with a have "(a,b) = splitAt ram vs" by auto
with a show ?thesis by (auto dest: splitAt_subset_ab)
qed
lemma splitAt_subset_snd:
"set (snd (splitAt ram vs)) ⊆ set vs"
proof -
def a: a ≡ "fst (splitAt ram vs)"
def b: b ≡ "snd (splitAt ram vs)"
with a have "(a,b) = splitAt ram vs" by auto
with b show ?thesis by (auto dest: splitAt_subset_ab)
qed
lemma splitAt_in_fst[dest]: "v ∈ set (fst (splitAt ram vs)) ==> v ∈ set vs"
proof (cases "ram ∈ set vs")
assume v: "v ∈ set (fst (splitAt ram vs))"
def a ≡ "fst (splitAt ram vs)"
with v have vin: "v ∈ set a" by auto
def b ≡ "snd (splitAt ram vs)"
case True with a_def b_def have "vs = a @ ram # b" by (auto dest: splitAt_ram)
with vin show "v ∈ set vs" by auto
next
assume v: "v ∈ set (fst (splitAt ram vs))"
case False with v show ?thesis by (auto dest: splitAt_no_ram del: notI)
qed
lemma splitAt_not1:
"v ∉ set vs ==> v ∉ set (fst (splitAt ram vs))" by (auto dest: splitAt_in_fst)
lemma splitAt_in_snd[dest]: "v ∈ set (snd (splitAt ram vs)) ==> v ∈ set vs"
proof (cases "ram ∈ set vs")
assume v: "v ∈ set (snd (splitAt ram vs))"
def a ≡ "fst (splitAt ram vs)"
def b ≡ "snd (splitAt ram vs)"
with v have vin: "v ∈ set b" by auto
case True with a_def b_def have "vs = a @ ram # b" by (auto dest: splitAt_ram)
with vin show "v ∈ set vs" by auto
next
assume v: "v ∈ set (snd (splitAt ram vs))"
case False with v show ?thesis by (auto dest: splitAt_no_ram del: notI)
qed
subsubsection {* Distinctness *}
lemma splitAt_distinct_ab:
"distinct vs ==> (a,b) = splitAt ram vs ==> distinct a ∧ distinct b"
apply (cases "ram ∈ set vs") by (auto dest: splitAt_split simp: splitAt_no_ram)
lemma splitAt_distinct_a:
"distinct vs ==> (a,b) = splitAt ram vs ==> distinct a"
apply (cases "ram ∈ set vs") by (auto dest: splitAt_split simp: splitAt_no_ram)
lemma splitAt_distinct_b:
"distinct vs ==> (a,b) = splitAt ram vs ==> distinct b"
apply (cases "ram ∈ set vs") by (auto dest: splitAt_split simp: splitAt_no_ram)
lemma splitAt_distinct_fst[intro]:
"distinct vs ==> distinct (fst (splitAt ram vs))"
proof -
assume d: "distinct vs"
def b: b ≡ "snd (splitAt ram vs)"
def a: a ≡ "fst (splitAt ram vs)"
with b have "(a,b) = splitAt ram vs" by auto
with a d show ?thesis by (auto dest: splitAt_distinct_ab)
qed
lemma splitAt_distinct_snd[intro]:
"distinct vs ==> distinct (snd (splitAt ram vs))"
proof -
assume d: "distinct vs"
def b: b ≡ "snd (splitAt ram vs)"
def a: a ≡ "fst (splitAt ram vs)"
with b have "(a,b) = splitAt ram vs" by auto
with b d show ?thesis by (auto dest: splitAt_distinct_ab)
qed
lemma splitAt_distinct_ab:
"distinct vs ==> (a,b) = splitAt ram vs ==> set a ∩ set b = {}"
apply (cases "ram ∈ set vs") apply (drule_tac splitAt_split)
by (auto simp: splitAt_no_ram)
lemma splitAt_distinct_fst_snd:
"distinct vs ==> set (fst (splitAt ram vs)) ∩ set (snd (splitAt ram vs)) = {}"
by (rule_tac splitAt_distinct_ab) simp_all
lemma splitAt_distinct_ram_fst[intro]:
"distinct vs ==> ram ∉ set (fst (splitAt ram vs))"
apply (case_tac "ram ∈ set vs") apply (drule_tac splitAt_ram)
apply (rule distinct_split1) by (force dest: splitAt_in_fst)+
(* apply (frule splitAt_no_ram) by auto *)
lemma splitAt_distinct_ram_snd[intro]:
"distinct vs ==> ram ∉ set (snd (splitAt ram vs))"
apply (case_tac "ram ∈ set vs") apply (drule_tac splitAt_ram)
apply (rule distinct_split2) by (force dest: splitAt_in_fst)+
lemma splitAt_1[simp]:
"splitAt ram [] = ([],[])" by (simp add: splitAt_def)
lemma splitAt_2:
"v ∈ set vs ==> (a,b) = splitAt ram vs ==> v ∈ set a ∨ v ∈ set b ∨ v = ram "
apply (cases "ram ∈ set vs")
by (auto dest: splitAt_split simp: splitAt_no_ram)
lemma splitAt_or:
"v ∈ set vs ==> v ∈ set (fst (splitAt ram vs)) ∨ v ∈ set (snd (splitAt ram vs)) ∨ v = ram"
by (rule splitAt_2) auto
lemma splitAt_distinct_fst: "distinct vs ==> distinct (fst (splitAt ram1 vs))"
lemma splitAt_distinct_a: "distinct vs ==> (a,b) = splitAt ram vs ==> distinct a"
by (auto dest: splitAt_distinct_fst pairD)
lemma splitAt_distinct_snd: "distinct vs ==> distinct (snd (splitAt ram1 vs))"
lemma splitAt_distinct_b: "distinct vs ==> (a,b) = splitAt ram vs ==> distinct b"
by (auto dest: splitAt_distinct_snd pairD)
lemma splitAt_distinct: "distinct vs ==> set (fst (splitAt ram vs)) ∩ set (snd (splitAt ram vs)) = {}"
lemma splitAt_subset: "(a,b) = splitAt ram vs ==> (set a ⊆ set vs) ∧ (set b ⊆ set vs)"
apply (cases "ram ∈ set vs") by (auto dest: splitAt_split simp: splitAt_no_ram)
lemma splitAt_subset1: "(a,b) = splitAt ram vs ==> (set a ⊆ set vs)"
by (auto dest: splitAt_subset)
lemma splitAt_subset2: "(a,b) = splitAt ram vs ==> (set b ⊆ set vs)"
by (auto dest: splitAt_subset)
subsubsection {* @{const splitAt} composition *}
lemma set_help: "v ∈ set ( as @ bs) ==> v ∈ set as ∨ v ∈ set bs" by auto
lemma splitAt_elements: "ram1 ∈ set vs ==> ram2 ∈ set vs ==> ram2 ∈ set( fst (splitAt ram1 vs)) ∨ ram2 ∈ set [ram1] ∨ ram2 ∈ set( snd (splitAt ram1 vs))"
proof -
assume r1: "ram1 ∈ set vs" and r2: "ram2 ∈ set vs"
then have "ram2 ∈ set( fst (splitAt ram1 vs) @ [ram1]) ∨ ram2 ∈ set( snd (splitAt ram1 vs))"
apply (rule_tac set_help)
apply (drule_tac splitAt_ram) by auto
then show ?thesis by auto
qed
lemma splitAt_ram2: "ram2 ∉ set (snd (splitAt ram1 vs)) ==>
ram1 ∈ set vs ==> ram2 ∈ set vs ==> ram1 ≠ ram2 ==>
ram2 ∈ set (fst (splitAt ram1 vs))" by (auto dest: splitAt_elements)
lemma splitAt_ram3: "ram2 ∉ set (fst (splitAt ram1 vs)) ==>
ram1 ∈ set vs ==> ram2 ∈ set vs ==> ram1 ≠ ram2 ==>
ram2 ∈ set (snd (splitAt ram1 vs))" by (auto dest: splitAt_elements)
lemma splitAt_dist_ram: "distinct vs ==>
vs = a @ ram # b ==> (a,b) = splitAt ram vs"
proof -
assume dist: "distinct vs" and vs: "vs = a @ ram # b"
from vs have r:"ram ∈ set vs" by auto
with dist vs have "fst (splitAt ram vs) = a" apply (drule_tac splitAt_ram) by (rule_tac dist_at1) auto
then have first:"a = fst (splitAt ram vs)" by auto
from r dist have second: "b = snd (splitAt ram vs)" apply (drule_tac splitAt_ram) apply (rule dist_at2) apply simp
by (auto simp: vs)
show ?thesis by (auto simp: first second)
qed
lemma distinct_unique1: "distinct vs ==> ram ∈ set vs ==> EX! s. vs = (fst s) @ ram # (snd s)"
proof
assume d: "distinct vs" and r: "ram ∈ set vs"
def s ≡ "splitAt ram vs" with r show "vs = (fst s) @ ram # (snd s)"
by (auto intro: splitAt_ram)
next
fix s
assume d: "distinct vs" and vs1: "vs = fst s @ ram # snd s"
from d vs1 show "s = splitAt ram vs" apply (drule_tac splitAt_dist_ram) apply simp by simp
qed
lemma splitAt_dist_ram2: "distinct vs ==> vs = a @ ram1 # b @ ram2 # c ==>
(a @ ram1 # b, c) = splitAt ram2 vs"
by (auto intro: splitAt_dist_ram)
lemma splitAt_dist_ram20: "distinct vs ==> vs = a @ ram1 # b @ ram2 # c ==>
c = snd (splitAt ram2 vs)"
proof -
assume dist: "distinct vs" and vs: "vs = a @ ram1 # b @ ram2 # c"
then show "c = snd (splitAt ram2 vs)" apply (drule_tac splitAt_dist_ram2) by (auto dest: pairD)
qed
lemma splitAt_dist_ram21: "distinct vs ==> vs = a @ ram1 # b @ ram2 # c ==> (a, b) = splitAt ram1 (fst (splitAt ram2 vs))"
proof -
assume dist: "distinct vs" and vs: "vs = a @ ram1 # b @ ram2 # c"
then have "fst (splitAt ram2 vs) = a @ ram1 # b" apply (drule_tac splitAt_dist_ram2) by (auto dest: pairD)
with dist vs show ?thesis by (rule_tac splitAt_dist_ram) auto
qed
lemma splitAt_dist_ram22: "distinct vs ==> vs = a @ ram1 # b @ ram2 # c ==> (c, []) = splitAt ram1 (snd (splitAt ram2 vs))"
proof -
assume dist: "distinct vs" and vs: "vs = a @ ram1 # b @ ram2 # c"
then have "snd (splitAt ram2 vs) = c" apply (drule_tac splitAt_dist_ram2) by (auto dest: pairD)
with dist vs have "splitAt ram1 (snd (splitAt ram2 vs)) = (c, [])" by (auto intro: splitAt_no_ram)
then show ?thesis by auto
qed
lemma splitAt_dist_ram1: "distinct vs ==> vs = a @ ram1 # b @ ram2 # c ==> (a, b @ ram2 # c) = splitAt ram1 vs"
by (auto intro: splitAt_dist_ram)
lemma splitAt_dist_ram10: "distinct vs ==> vs = a @ ram1 # b @ ram2 # c ==> a = fst (splitAt ram1 vs)"
proof -
assume dist: "distinct vs" and vs: "vs = a @ ram1 # b @ ram2 # c"
then show "a = fst (splitAt ram1 vs)" apply (drule_tac splitAt_dist_ram1) by (auto dest: pairD)
qed
lemma splitAt_dist_ram11: "distinct vs ==> vs = a @ ram1 # b @ ram2 # c ==> (a, []) = splitAt ram2 (fst (splitAt ram1 vs))"
proof -
assume dist: "distinct vs" and vs: "vs = a @ ram1 # b @ ram2 # c"
then have "fst (splitAt ram1 vs) = a" apply (drule_tac splitAt_dist_ram1) by (auto dest: pairD)
with dist vs have "splitAt ram2 (fst (splitAt ram1 vs)) = (a, [])" by (auto intro: splitAt_no_ram)
then show ?thesis by auto
qed
lemma splitAt_dist_ram12: "distinct vs ==> vs = a @ ram1 # b @ ram2 # c ==> (b, c) = splitAt ram2 (snd (splitAt ram1 vs))"
proof -
assume dist: "distinct vs" and vs: "vs = a @ ram1 # b @ ram2 # c"
then have "snd (splitAt ram1 vs) = b @ ram2 # c" apply (drule_tac splitAt_dist_ram1) by (auto dest: pairD)
with dist vs show ?thesis by (rule_tac splitAt_dist_ram) auto
qed
lemma splitAt_dist_ram_all:
"distinct vs ==> vs = a @ ram1 # b @ ram2 # c
==> (a, b) = splitAt ram1 (fst (splitAt ram2 vs))
∧ (c, []) = splitAt ram1 (snd (splitAt ram2 vs))
∧ (a, []) = splitAt ram2 (fst (splitAt ram1 vs))
∧ (b, c) = splitAt ram2 (snd (splitAt ram1 vs))
∧ c = snd (splitAt ram2 vs)
∧ a = fst (splitAt ram1 vs)"
apply (rule_tac conjI) apply (rule_tac splitAt_dist_ram21) apply simp apply simp
apply (rule_tac conjI) apply (rule_tac splitAt_dist_ram22) apply simp apply simp
apply (rule_tac conjI) apply (rule_tac splitAt_dist_ram11 splitAt_dist_ram22) apply simp apply simp
apply (rule_tac conjI) apply (rule_tac splitAt_dist_ram12)apply simp apply simp
apply (rule_tac conjI) apply (rule_tac splitAt_dist_ram20) apply simp apply simp
by (rule_tac splitAt_dist_ram10) auto
subsubsection {* Mixed *}
lemma fst_splitAt_rev:
"distinct xs ==> x ∈ set xs ==>
fst(splitAt x (rev xs)) = rev(snd(splitAt x xs))"
lemma snd_splitAt_rev:
"distinct xs ==> x ∈ set xs ==>
snd(splitAt x (rev xs)) = rev(fst(splitAt x xs))"
lemma splitAt_take[simp]: "distinct ls ==> i < length ls ==> fst (splitAt (ls!i) ls) = take i ls"
proof -
assume d: "distinct ls" and si: "i < length ls"
then have ls1: "ls = take i ls @ ls!i # drop (Suc i) ls" by (rule_tac id_take_nth_drop)
from si have "ls!i ∈ set ls" by auto
then have ls2: "ls = fst (splitAt (ls!i) ls) @ ls!i # snd (splitAt (ls!i) ls)" by (auto dest: splitAt_ram)
from d ls2 ls1 have "fst (splitAt (ls!i) ls) = take i ls ∧ snd (splitAt (ls!i) ls) = drop (Suc i) ls" by (rule dist_at)
then show ?thesis by auto
qed
lemma splitAt_drop[simp]: "distinct ls ==> i < length ls ==> snd (splitAt (ls!i) ls) = drop (Suc i) ls"
proof -
assume d: "distinct ls" and si: "i < length ls"
then have ls1: "ls = take i ls @ ls!i # drop (Suc i) ls" by (rule_tac id_take_nth_drop)
from si have "ls!i ∈ set ls" by auto
then have ls2: "ls = fst (splitAt (ls!i) ls) @ ls!i # snd (splitAt (ls!i) ls)" by (auto dest: splitAt_ram)
from d ls2 ls1 have "fst (splitAt (ls!i) ls) = take i ls ∧ snd (splitAt (ls!i) ls) = drop (Suc i) ls" by (rule dist_at)
then show ?thesis by auto
qed
lemma fst_splitAt_upt:
"j <= i ==> i < k ==> fst(splitAt i [j..<k]) = [j..<i]"
using splitAt_take[where ls = "[j..<k]" and i="i-j"]
apply (simp del:splitAt_take) apply arith
done
lemma snd_splitAt_upt:
"j <= i ==> i < k ==> snd(splitAt i [j..<k]) = [i+1..<k]"
using splitAt_drop[where ls = "[j..<k]" and i="i-j"]
apply (simp del:splitAt_drop) apply arith
done
lemma local_help1: "!! a vs. vs = c @ r # d ==> vs = a @ r # b ==> r ∉ set a ==> r ∉ set b ==> a = c"
proof (induct c)
case Nil
then have ra: "r ∉ set a" and vs1: "vs = a @ r # b" and vs2: "vs = [] @ r # d"
by auto
from vs1 vs2 have "a @ r # b = r # d" by auto
then have "hd (a @ r # b) = r" by auto
then have "a ≠ [] ==> hd a = r" by auto
then have "a ≠ [] ==> r ∈ set a" by (induct a) auto
with ra have a: "a = []" by auto
then show ?case by auto
next
case (Cons x xs) then show ?case by (induct a) auto
qed
lemma local_help: "vs = a @ r # b ==> vs = c @ r # d ==> r ∉ set a ==> r ∉ set b ==> a = c ∧ b = d"
proof -
assume dist: "r ∉ set a" "r ∉ set b" and vs1: "vs = a @ r # b" and vs2: "vs = c @ r # d"
then have "a = c" apply (rule_tac local_help1) .
with dist vs1 vs2 show ?thesis by simp
qed
lemma local_help': "a @ r # b = c @ r # d ==> r ∉ set a ==> r ∉ set b ==> a = c ∧ b = d"
by (rule local_help) auto
lemma splitAt_simp1: "ram ∉ set a ==> ram ∉ set b ==> fst (splitAt ram (a @ ram # b)) = a "
proof -
assume ramab: "ram ∉ set a" "ram ∉ set b"
def vs ≡ "a @ ram # b"
then have vs: "vs = a @ ram # b" by auto
then have "ram ∈ set vs" by auto
then have "vs = fst (splitAt ram vs) @ ram # snd (splitAt ram vs)" by (auto dest: splitAt_ram)
with vs ramab show ?thesis apply simp apply (rule_tac sym) apply (rule_tac local_help1) apply simp
apply (rule sym) apply assumption by auto
qed
lemma splitAt_simp2: "ram ∉ set b ==> fst (splitAt ram (ram # b)) = [] "
apply (subgoal_tac " fst (splitAt ram ([] @ ram # b)) = []") apply force
apply (rule splitAt_simp1) by auto
lemma splitAt_simp3: "ram ∉ set a ==> fst (splitAt ram (a @ [ram])) = a "
apply (subgoal_tac " fst (splitAt ram (a @ ram # [])) = a") apply force
apply (rule splitAt_simp1) by auto
lemma splitAt_simp4: "ram ∉ set a ==> ram ∉ set b ==> snd (splitAt ram (a @ ram # b)) = b "
proof -
assume ramab: "ram ∉ set a" "ram ∉ set b"
def vs ≡ "a @ ram # b"
then have vs: "vs = a @ ram # b" by auto
then have "ram ∈ set vs" by auto
then have "vs = fst (splitAt ram vs) @ ram # snd (splitAt ram vs)" by (auto dest: splitAt_ram)
with vs ramab show ?thesis by (simp add: splitAt_simp1)
qed
lemma help'''_in: "!! xs. ram ∈ set b ==> fst (splitAtRec ram xs b) = xs @ fst (splitAtRec ram [] b)"
proof (induct b)
case Nil then show ?case by auto
next
case (Cons b bs) show ?case using Cons(2)
apply (case_tac "b = ram") apply simp
apply simp
apply (subgoal_tac "fst (splitAtRec ram (xs @ [b]) bs) = (xs@[b]) @ fst (splitAtRec ram [] bs)") apply simp
apply (subgoal_tac "fst (splitAtRec ram [b] bs) = [b] @ fst (splitAtRec ram [] bs)") apply simp
apply (rule Cons) apply force
apply (rule Cons) by force
qed
lemma help'''_notin: "!! xs. ram ∉ set b ==> fst (splitAtRec ram xs b) = xs @ fst (splitAtRec ram [] b)"
proof (induct b)
case Nil then show ?case by auto
next
case (Cons b bs)
then have "ram ∉ set bs" by auto
then show ?case
apply (case_tac "b = ram") apply simp
apply simp
apply (subgoal_tac "fst (splitAtRec ram (xs @ [b]) bs) = (xs@[b]) @ fst (splitAtRec ram [] bs)") apply simp
apply (subgoal_tac "fst (splitAtRec ram [b] bs) = [b] @ fst (splitAtRec ram [] bs)") apply simp
apply (rule Cons) apply simp
apply (rule Cons) by simp
qed
lemma help''': "fst (splitAtRec ram xs b) = xs @ fst (splitAtRec ram [] b)"
apply (cases "ram ∈ set b")
apply (rule_tac help'''_in) apply simp
apply (rule_tac help'''_notin) apply simp done
lemma splitAt_simpA[simp]: "fst (splitAt ram (ram # b)) = []" by (simp add: splitAt_def)
lemma splitAt_simpB[simp]: "ram ≠ a ==> fst (splitAt ram (a # b)) = a # fst (splitAt ram b)" apply (simp add: splitAt_def)
apply (subgoal_tac "fst (splitAtRec ram [a] b) = [a] @ fst (splitAtRec ram [] b)") apply simp by (rule help''')
lemma splitAt_simpB'[simp]: "a ≠ ram ==> fst (splitAt ram (a # b)) = a # fst (splitAt ram b)" apply (rule splitAt_simpB) by auto
lemma splitAt_simpC[simp]: "ram ∉ set a ==> fst (splitAt ram (a @ b)) = a @ fst (splitAt ram b)"
apply (induct a) by auto
lemma help'''': "!! xs ys. snd (splitAtRec ram xs b) = snd (splitAtRec ram ys b)"
apply (induct b) by auto
lemma splitAt_simpD[simp]: "!! a. ram ≠ a ==> snd (splitAt ram (a # b)) = snd (splitAt ram b)" apply (simp add: splitAt_def)
by (rule help'''')
lemma splitAt_simpD'[simp]: "!! a. a ≠ ram ==> snd (splitAt ram (a # b)) = snd (splitAt ram b)" apply (rule splitAt_simpD) by auto
lemma splitAt_simpE[simp]: "snd (splitAt ram (ram # b)) = b" by (simp add: splitAt_def)
lemma splitAt_simpF[simp]: "ram ∉ set a ==> snd (splitAt ram (a @ b)) = snd (splitAt ram b) "
apply (induct a) by auto
lemma splitAt_rotate_pair_conv:
"!!xs. [| distinct xs; x ∈ set xs |]
==> snd (splitAt x (rotate n xs)) @ fst (splitAt x (rotate n xs)) =
snd (splitAt x xs) @ fst (splitAt x xs)"
apply(induct n) apply simp
apply(erule disjE) apply simp
apply(drule split_list)
apply clarsimp
done
subsection {* @{text between} *}
constdefs between :: "'a list => 'a => 'a => 'a list"
"between vs ram1 ram2 ≡
let (pre1, post1) = splitAt ram1 vs in
if ram2 mem post1
then let (pre2, post2) = splitAt ram2 post1 in pre2
else let (pre2, post2) = splitAt ram2 pre1 in post1 @ pre2"
lemma inbetween_inset:
"x ∈ set(between xs a b) ==> x ∈ set xs"
apply(blast dest:splitAt_in_snd)
apply(blast dest:splitAt_in_snd)
done
lemma notinset_notinbetween:
"x ∉ set xs ==> x ∉ set(between xs a b)"
by(blast dest:inbetween_inset)
lemma set_between_id:
"distinct xs ==> x ∈ set xs ==>
set(between xs x x) = set xs - {x}"
apply(drule split_list)
apply (clarsimp simp:between_def split_def Un_commute)
done
lemma split_between:
"[| distinct vs; r ∈ set vs; v ∈ set vs; u ∈ set(between vs r v) |] ==>
between vs r v =
(if r=u then [] else between vs r u @ [u]) @ between vs u v"
apply(cases "r = v")
apply(clarsimp)
apply(frule split_list[of v])
apply(clarsimp)
apply(erule disjE)
apply(frule split_list[of u])
apply(clarsimp)
apply(frule split_list[of u])
apply(clarsimp)
apply(clarsimp)
apply(frule split_list[of r])
apply(clarsimp)
apply(rename_tac as bs)
apply(erule disjE)
apply(frule split_list[of v])
apply(clarsimp)
apply(rename_tac cs ds)
apply(subgoal_tac "between (cs @ v # ds @ r # bs) r v = bs @ cs")
prefer 2 apply(simp add:between_def split_def split:split_if_asm)
apply simp
apply(erule disjE)
apply(frule split_list[of u])
apply(clarsimp simp:between_def split_def split:split_if_asm)
apply(frule split_list[of u])
apply(clarsimp simp:between_def split_def split:split_if_asm)
apply(frule split_list[of v])
apply(clarsimp)
apply(rename_tac cs ds)
apply(subgoal_tac "between (as @ r # cs @ v # ds) r v = cs")
prefer 2 apply(simp add:between_def split_def split:split_if_asm)
apply simp
apply(frule split_list[of u])
apply(clarsimp simp:between_def split_def split:split_if_asm)
done
subsection {* Tables *}
types ('a, 'b) table = "('a × 'b) list"
constdefs isTable :: "('a => 'b) => 'a list => ('a, 'b) table => bool"
"isTable f vs t ≡ ∀p. p ∈ set t --> snd p = f (fst p) ∧ fst p ∈ set vs"
lemma isTable_eq: "isTable E vs ((a,b)#ps) ==> b = E a"
lemma isTable_subset:
"set qs ⊆ set ps ==> isTable E vs ps ==> isTable E vs qs"
by (unfold isTable_def) auto
lemma isTable_Cons: "isTable E vs ((a,b)#ps) ==> isTable E vs ps"
by (unfold isTable_def) auto
constdefs
removeKey :: "'a => ('a × 'b) list => ('a × 'b) list"
"removeKey a ps ≡ [p ∈ ps. a ≠ fst p]"
consts removeKeyList :: "'a list => ('a × 'b) list => ('a × 'b) list"
primrec
"removeKeyList [] ps = ps"
"removeKeyList (w#ws) ps = removeKey w (removeKeyList ws ps)"
lemma removeKey_subset[simp]: "set (removeKey a ps) ⊆ set ps"
lemma length_removeKey[simp]: "|removeKey w ps| ≤ |ps|"
lemma length_removeKeyList:
"length (removeKeyList ws ps) ≤ length ps" (is "?P ws")
proof (induct ws)
show "?P []" by simp
fix w ws
have "length (removeKey w (removeKeyList ws ps))
≤ length (removeKeyList ws ps)"
by (rule length_removeKey)
also assume "?P ws"
finally show "?P (w#ws)" by simp
qed
lemma removeKeyList_subset[simp]: "set (removeKeyList ws ps) ⊆ set ps"
proof (induct ws)
case Nil then show ?case by simp
next
case (Cons w ws)
have "set (removeKey w (removeKeyList ws ps)) ⊆ set (removeKeyList ws ps)"
by (rule removeKey_subset)
with Cons show ?case by (simp add: removeKey_def)
qed
lemma notin_removeKey1: "(a, b) ∉ set (removeKey a ps)"
by (induct ps) (auto simp add: removeKey_def)
lemma notin_removeKey: "r ∉ fst ` set (removeKey r ps)"
by (induct ps) (auto simp add: removeKey_def)
lemma notin_removeKeyList1:
"!!a. a ∈ set rs ==> (a, b) ∉ set (removeKeyList rs ps)"
proof (induct rs)
case Nil then show ?case by simp
next
case (Cons r rs) then show ?case
proof cases
assume "a = r" then show ?thesis
by (auto simp add: split_beta Let_def notin_removeKey1 removeKey_subset)
next
assume a: "a ≠ r"
with Cons have"(a, b) ∉ set (removeKeyList rs ps)" by simp
moreover have "set (removeKey r (removeKeyList rs ps))
⊆ set (removeKeyList rs ps)"
by (rule removeKey_subset) (* fixme : finally *)
ultimately have "(a, b) ∉ set (removeKey r (removeKeyList rs ps))"
by blast
then show ?thesis by simp
qed
qed
lemma notin_removeKeyList: "!!r. r ∈ set rs ==> r ∉ fst ` set (removeKeyList rs ps)"
proof (induct rs)
case Nil then show ?case by simp
next
case (Cons r' rs) then show ?case
by (auto simp add: notin_removeKey1 split_paired_all removeKey_def)
qed
lemma removeKeyList_eq:
"removeKeyList as ps = [p ∈ ps. ∀a ∈ set as. a ≠ fst p]"
by (induct as) (simp_all add: filter_comm removeKey_def)
lemma removeKey_empty[simp]: "removeKey a [] = []"
lemma removeKeyList_empty[simp]: "removeKeyList ps [] = []"
by (induct ps) simp_all
lemma removeKeyList_cons[simp]:
"removeKeyList ws (p#ps)
= (if fst p ∈ set ws then removeKeyList ws ps else p#(removeKeyList ws ps))"
by (induct ws) (simp_all split: split_if_asm add: removeKey_def)
end
```
### HOL
lemma pairD:
` (a, b) = p ==> a = fst p ∧ b = snd p`
lemmas conj_aci:
` (P ∧ Q) = (Q ∧ P)`
` (P ∧ Q ∧ R) = (Q ∧ P ∧ R)`
` ((P ∧ Q) ∧ R) = (P ∧ Q ∧ R)`
` (A ∧ A) = A`
` (A ∧ A ∧ B) = (A ∧ B)`
lemmas conj_aci:
` (P ∧ Q) = (Q ∧ P)`
` (P ∧ Q ∧ R) = (Q ∧ P ∧ R)`
` ((P ∧ Q) ∧ R) = (P ∧ Q ∧ R)`
` (A ∧ A) = A`
` (A ∧ A ∧ B) = (A ∧ B)`
lemma
` max x y == let u = x; v = y in if u ≤ v then v else u`
lemma
` min x y == let u = x; v = y in if u ≤ v then u else v`
lemmas
` {..<0} = {}`
` {..<Suc k} = insert k {..<k}`
lemmas
` {..<0} = {}`
` {..<Suc k} = insert k {..<k}`
### Lists
#### @{text length}
lemma length3D:
` |xs| = 3 ==> ∃x y z. xs = [x, y, z]`
lemma length4D:
` |xs| = 4 ==> ∃a b c d. xs = [a, b, c, d]`
#### @{const filter}
lemma filter_emptyE:
` [| filter P xs = []; x ∈ set xs |] ==> ¬ P x`
lemma filter_comm:
` [x∈xs . P x ∧ Q x] = [x∈xs . Q x ∧ P x]`
lemma filter1:
` filter P xs = [] ==> [x∈xs . Q x ∧ P x] = []`
lemma filter_prop:
` x ∈ set (filter P ys) ==> P x`
lemma filter_Cons_prop:
` filter P ys = x # xs ==> P x`
lemma filter_compl1:
` (filter P xs = []) = ([x∈xs . ¬ P x] = xs)`
lemma
` Not o (Not o P) = P`
lemma filter_compl2:
` (filter (Not o P) xs = []) = (filter P xs = xs)`
lemma filter_eqI:
` (!!v. v ∈ set vs ==> P v = Q v) ==> filter P vs = filter Q vs`
lemma filter_simp:
` (!!x. x ∈ set xs ==> P x) ==> [x∈xs . P x ∧ Q x] = filter Q xs`
lemma filter_True_eq1:
` [| |filter P xs| = |xs|; y ∈ set xs |] ==> P y`
lemma length_filter_True_eq:
` (|filter P xs| = |xs|) = (∀y. y ∈ set xs --> P y)`
#### @{const map_filter}
lemma
` map_filter f P xs = map f (filter P xs)`
#### List product
lemma
` set (xs × ys) = set xs × set ys`
#### Minimum and maximum
lemma minimal_in_set:
` xs ≠ [] ==> minimal f xs ∈ set xs`
lemma minimal_Cons1:
` ∀y∈set xs. f x ≤ f y ==> minimal f (x # xs) = x`
lemma minimal_append2:
` ∀x∈set xs. f y < f x ==> minimal f (xs @ y # ys) = minimal f (y # ys)`
lemma minimal_neq_lowerbound:
` [| xs ≠ []; ∀x∈set xs. n ≤ f x; f (minimal f xs) ≠ n |] ==> ∀x∈set xs. f x ≠ n`
lemma minList_conv_Min:
` xs ≠ [] ==> minList xs = Min (set xs)`
lemma max_list_conv_Max:
` xs ≠ [] ==> max_list xs = Max (set xs)`
#### replace
lemma length_mapAt:
` |mapAt vs f xs| = |xs|`
lemma length_replace1:
` |replace x [y] xs| = |xs|`
lemma replace_id:
` replace x [x] xs = xs`
lemma len_replace_ge_same:
` 1 ≤ |ys| ==> |xs| ≤ |replace x ys xs|`
lemma len_replace_ge:
` [| 1 ≤ |ys|; |zs| ≤ |xs| |] ==> |zs| ≤ |replace x ys xs|`
lemma replace_append:
``` replace x ys (as @ bs) =
(if x ∈ set as then replace x ys as @ bs else as @ replace x ys bs)```
lemma filter_replace:
` ¬ P y ==> filter P (replace x [y] xs) = remove1 x (filter P xs)`
lemma distinct_set_replace:
``` distinct xs
==> set (replace x ys xs) =
(if x ∈ set xs then set xs - {x} ∪ set ys else set xs)```
lemma replace1:
` [| f ∈ set (replace f' fs ls); f ∉ set ls |] ==> f ∈ set fs`
lemma replace2:
` f' ∉ set ls ==> replace f' fs ls = ls`
lemma replace3:
` [| f' ∈ set ls; f ∈ set fs |] ==> f ∈ set (replace f' fs ls)`
lemma replace4:
` [| f ∈ set ls; oldF ≠ f |] ==> f ∈ set (replace oldF fs ls)`
lemma replace5:
` f ∈ set (replace oldF newfs fs) ==> f ∈ set fs ∨ f ∈ set newfs`
lemma replace6:
``` distinct oldfs
==> (x ∈ set (replace oldF newfs oldfs)) =
((x ≠ oldF ∨ oldF ∈ set newfs) ∧
(oldF ∈ set oldfs ∧ x ∈ set newfs ∨ x ∈ set oldfs))```
lemma replace_delete_oldF:
` [| oldF ∉ set fs; distinct ls |] ==> oldF ∉ set (replace oldF fs ls)`
lemma distinct_replace:
``` [| distinct fs; distinct newFs; set fs ∩ set newFs ⊆ {oldF} |]
==> distinct (replace oldF newFs fs)```
lemma replace_replace:
``` [| oldf ∉ set newfs; distinct xs |]
==> replace oldf newfs (replace oldf newfs xs) = replace oldf newfs xs```
lemma replace_distinct:
``` [| distinct fs; distinct newfs; oldf ∈ set fs --> set newfs ∩ set fs ⊆ {oldf} |]
==> distinct (replace oldf newfs fs)```
lemma filter_replace2:
` [| ¬ P x; ∀y∈set ys. ¬ P y |] ==> filter P (replace x ys xs) = filter P xs`
lemma length_filter_replace1:
``` [| x ∈ set xs; ¬ P x |]
==> |filter P (replace x ys xs)| = |filter P xs| + |filter P ys|```
lemma length_filter_replace2:
``` [| x ∈ set xs; P x |]
==> |filter P (replace x ys xs)| = |filter P xs| + |filter P ys| - 1```
#### @{const"distinct"}
lemma dist_filter_single:
` [| distinct ls; v ∈ set ls |] ==> [a∈ls . a = v] = [v]`
lemma dist_at1:
` [| distinct vs; vs = a @ r # b; vs = c @ r # d |] ==> a = c`
lemma dist_at:
` [| distinct vs; vs = a @ r # b; vs = c @ r # d |] ==> a = c ∧ b = d`
lemma dist_at2:
` [| distinct vs; vs = a @ r # b; vs = c @ r # d |] ==> b = d`
lemma distinct_split1:
` [| distinct xs; xs = y @ [r] @ z |] ==> r ∉ set y`
lemma distinct_split2:
` [| distinct xs; xs = y @ [r] @ z |] ==> r ∉ set z`
lemma distinct_hd_not_cons:
` [| distinct vs; ∃as bs. vs = as @ x # hd vs # bs |] ==> False`
#### Misc
lemma drop_last_in:
` n < |ls| ==> last ls ∈ set (drop n ls)`
lemma nth_last_Suc_n:
` [| distinct ls; n < |ls|; last ls = ls ! n |] ==> Suc n = |ls|`
#### @{const rotate}
lemma plus_length1:
` rotate (k + |ls|) ls = rotate k ls`
lemma plus_length2:
` rotate (|ls| + k) ls = rotate k ls`
lemma rotate_minus1:
``` [| 0 < n; 0 < m; rotate n ls = rotate m ms |]
==> rotate (n - 1) ls = rotate (m - 1) ms```
lemma rotate_minus1':
` [| 0 < n; rotate n ls = ms |] ==> rotate (n - 1) ls = rotate (|ms| - 1) ms`
lemma rotate_inv1:
` [| n < |ls|; rotate n ls = ms |] ==> ls = rotate (|ls| - n) ms`
lemma rotate_conv_mod':
` rotate (n mod |ls|) ls = rotate n ls`
lemma rotate_inv2:
` rotate n ls = ms ==> ls = rotate (|ls| - n mod |ls|) ms`
lemma rotate_inv':
` ls = rotate (|ls| - n mod |ls|) ms ==> rotate n ls = ms`
lemma rotate_id:
` rotate (|ls| - n mod |ls|) (rotate n ls) = ls`
lemma nth_rotate1_Suc:
` Suc n < |ls| ==> ls ! Suc n = rotate1 ls ! n`
lemma nth_rotate1_0:
` ls ! 0 = rotate1 ls ! (|ls| - 1)`
lemma nth_rotate1:
` 0 < |ls| ==> ls ! (Suc n mod |ls|) = rotate1 ls ! (n mod |ls|)`
lemma rotate_Suc2:
` rotate n (rotate1 xs) = rotate (Suc n) xs`
lemma nth_rotate:
` 0 < |ls| ==> ls ! ((n + m) mod |ls|) = rotate m ls ! (n mod |ls|)`
lemma help1:
` ∃n. filter f (rotate1 ls) = rotate n (filter f ls)`
lemma help3:
``` [| ¬ f l; n < |ls| |]
==> filter f (rotate n ls) = filter f (rotate n (rotate1 (l # ls)))```
lemma help4:
``` [| ¬ f l; n < |ls| |]
==> filter f (rotate n ls) = filter f (rotate (Suc n) (l # ls))```
lemma help2:
` ∃n. rotate1 (filter f ls) = filter f (rotate n ls)`
lemma rotate_help5:
` ∃n. filter f (rotate m ls) = rotate n (filter f ls)`
lemma rotate_help6:
` ∃n. rotate m (filter f ls) = filter f (rotate n ls)`
### @{text splitAt}
#### @{const splitAtRec}
lemma splitAtRec_conv:
``` splitAtRec x bs xs =
(bs @ takeWhile (λy. y ≠ x) xs, tl (dropWhile (λy. y ≠ x) xs))```
lemma splitAtRec_distinct_fst:
``` [| distinct vs; distinct s; set s ∩ set vs = {} |]
==> distinct (fst (splitAtRec ram1.0 s vs))```
lemma splitAtRec_distinct_snd:
``` [| distinct vs; distinct s; set s ∩ set vs = {} |]
==> distinct (snd (splitAtRec ram1.0 s vs))```
lemma splitAtRec_ram:
` [| ram ∈ set vs; (a, b) = splitAtRec ram us vs |] ==> us @ vs = a @ [ram] @ b`
lemma splitAtRec_notRam:
` ram ∉ set vs ==> splitAtRec ram us vs = (us @ vs, [])`
lemma splitAtRec_distinct:
``` [| distinct vs; distinct s; set s ∩ set vs = {} |]
==> set (fst (splitAtRec ram s vs)) ∩ set (snd (splitAtRec ram s vs)) = {}```
#### @{const splitAt}
lemma splitAt_conv:
` splitAt x xs = (takeWhile (λy. y ≠ x) xs, tl (dropWhile (λy. y ≠ x) xs))`
lemma splitAt_no_ram:
` ram ∉ set vs ==> splitAt ram vs = (vs, [])`
lemma splitAt_split:
` [| ram ∈ set vs; (a, b) = splitAt ram vs |] ==> vs = a @ ram # b`
lemma splitAt_ram:
` ram ∈ set vs ==> vs = fst (splitAt ram vs) @ ram # snd (splitAt ram vs)`
lemma fst_splitAt_last:
` [| xs ≠ []; distinct xs |] ==> fst (splitAt (last xs) xs) = butlast xs`
#### Sets
lemma splitAtRec_union:
` (a, b) = splitAtRec ram s vs ==> set a ∪ set b - {ram} = set vs ∪ set s - {ram}`
lemma splitAt_union:
` (a, b) = splitAt ram vs ==> set a ∪ set b - {ram} = set vs - {ram}`
lemma splitAt_subset_ab:
` (a, b) = splitAt ram vs ==> set a ⊆ set vs ∧ set b ⊆ set vs`
lemma splitAt_subset_fst:
` set (fst (splitAt ram vs)) ⊆ set vs`
lemma splitAt_subset_snd:
` set (snd (splitAt ram vs)) ⊆ set vs`
lemma splitAt_in_fst:
` v ∈ set (fst (splitAt ram vs)) ==> v ∈ set vs`
lemma splitAt_not1:
` v ∉ set vs ==> v ∉ set (fst (splitAt ram vs))`
lemma splitAt_in_snd:
` v ∈ set (snd (splitAt ram vs)) ==> v ∈ set vs`
#### Distinctness
lemma splitAt_distinct_ab:
` [| distinct vs; (a, b) = splitAt ram vs |] ==> distinct a ∧ distinct b`
lemma splitAt_distinct_a:
` [| distinct vs; (a, b) = splitAt ram vs |] ==> distinct a`
lemma splitAt_distinct_b:
` [| distinct vs; (a, b) = splitAt ram vs |] ==> distinct b`
lemma splitAt_distinct_fst:
` distinct vs ==> distinct (fst (splitAt ram vs))`
lemma splitAt_distinct_snd:
` distinct vs ==> distinct (snd (splitAt ram vs))`
lemma splitAt_distinct_ab:
` [| distinct vs; (a, b) = splitAt ram vs |] ==> set a ∩ set b = {}`
lemma splitAt_distinct_fst_snd:
` distinct vs ==> set (fst (splitAt ram vs)) ∩ set (snd (splitAt ram vs)) = {}`
lemma splitAt_distinct_ram_fst:
` distinct vs ==> ram ∉ set (fst (splitAt ram vs))`
lemma splitAt_distinct_ram_snd:
` distinct vs ==> ram ∉ set (snd (splitAt ram vs))`
lemma splitAt_1:
` splitAt ram [] = ([], [])`
lemma splitAt_2:
` [| v ∈ set vs; (a, b) = splitAt ram vs |] ==> v ∈ set a ∨ v ∈ set b ∨ v = ram`
lemma splitAt_or:
``` v ∈ set vs
==> v ∈ set (fst (splitAt ram vs)) ∨ v ∈ set (snd (splitAt ram vs)) ∨ v = ram```
lemma splitAt_distinct_fst:
` distinct vs ==> distinct (fst (splitAt ram1.0 vs))`
lemma splitAt_distinct_a:
` [| distinct vs; (a, b) = splitAt ram vs |] ==> distinct a`
lemma splitAt_distinct_snd:
` distinct vs ==> distinct (snd (splitAt ram1.0 vs))`
lemma splitAt_distinct_b:
` [| distinct vs; (a, b) = splitAt ram vs |] ==> distinct b`
lemma splitAt_distinct:
` distinct vs ==> set (fst (splitAt ram vs)) ∩ set (snd (splitAt ram vs)) = {}`
lemma splitAt_subset:
` (a, b) = splitAt ram vs ==> set a ⊆ set vs ∧ set b ⊆ set vs`
lemma splitAt_subset1:
` (a, b) = splitAt ram vs ==> set a ⊆ set vs`
lemma splitAt_subset2:
` (a, b) = splitAt ram vs ==> set b ⊆ set vs`
#### @{const splitAt} composition
lemma set_help:
` v ∈ set (as @ bs) ==> v ∈ set as ∨ v ∈ set bs`
lemma splitAt_elements:
``` [| ram1.0 ∈ set vs; ram2.0 ∈ set vs |]
==> ram2.0 ∈ set (fst (splitAt ram1.0 vs)) ∨
ram2.0 ∈ set [ram1.0] ∨ ram2.0 ∈ set (snd (splitAt ram1.0 vs))```
lemma splitAt_ram2:
``` [| ram2.0 ∉ set (snd (splitAt ram1.0 vs)); ram1.0 ∈ set vs; ram2.0 ∈ set vs;
ram1.0 ≠ ram2.0 |]
==> ram2.0 ∈ set (fst (splitAt ram1.0 vs))```
lemma splitAt_ram3:
``` [| ram2.0 ∉ set (fst (splitAt ram1.0 vs)); ram1.0 ∈ set vs; ram2.0 ∈ set vs;
ram1.0 ≠ ram2.0 |]
==> ram2.0 ∈ set (snd (splitAt ram1.0 vs))```
lemma splitAt_dist_ram:
` [| distinct vs; vs = a @ ram # b |] ==> (a, b) = splitAt ram vs`
lemma distinct_unique1:
` [| distinct vs; ram ∈ set vs |] ==> ∃!s. vs = fst s @ ram # snd s`
lemma splitAt_dist_ram2:
``` [| distinct vs; vs = a @ ram1.0 # b @ ram2.0 # c |]
==> (a @ ram1.0 # b, c) = splitAt ram2.0 vs```
lemma splitAt_dist_ram20:
``` [| distinct vs; vs = a @ ram1.0 # b @ ram2.0 # c |]
==> c = snd (splitAt ram2.0 vs)```
lemma splitAt_dist_ram21:
``` [| distinct vs; vs = a @ ram1.0 # b @ ram2.0 # c |]
==> (a, b) = splitAt ram1.0 (fst (splitAt ram2.0 vs))```
lemma splitAt_dist_ram22:
``` [| distinct vs; vs = a @ ram1.0 # b @ ram2.0 # c |]
==> (c, []) = splitAt ram1.0 (snd (splitAt ram2.0 vs))```
lemma splitAt_dist_ram1:
``` [| distinct vs; vs = a @ ram1.0 # b @ ram2.0 # c |]
==> (a, b @ ram2.0 # c) = splitAt ram1.0 vs```
lemma splitAt_dist_ram10:
``` [| distinct vs; vs = a @ ram1.0 # b @ ram2.0 # c |]
==> a = fst (splitAt ram1.0 vs)```
lemma splitAt_dist_ram11:
``` [| distinct vs; vs = a @ ram1.0 # b @ ram2.0 # c |]
==> (a, []) = splitAt ram2.0 (fst (splitAt ram1.0 vs))```
lemma splitAt_dist_ram12:
``` [| distinct vs; vs = a @ ram1.0 # b @ ram2.0 # c |]
==> (b, c) = splitAt ram2.0 (snd (splitAt ram1.0 vs))```
lemma splitAt_dist_ram_all:
``` [| distinct vs; vs = a @ ram1.0 # b @ ram2.0 # c |]
==> (a, b) = splitAt ram1.0 (fst (splitAt ram2.0 vs)) ∧
(c, []) = splitAt ram1.0 (snd (splitAt ram2.0 vs)) ∧
(a, []) = splitAt ram2.0 (fst (splitAt ram1.0 vs)) ∧
(b, c) = splitAt ram2.0 (snd (splitAt ram1.0 vs)) ∧
c = snd (splitAt ram2.0 vs) ∧ a = fst (splitAt ram1.0 vs)```
#### Mixed
lemma fst_splitAt_rev:
``` [| distinct xs; x ∈ set xs |]
==> fst (splitAt x (rev xs)) = rev (snd (splitAt x xs))```
lemma snd_splitAt_rev:
``` [| distinct xs; x ∈ set xs |]
==> snd (splitAt x (rev xs)) = rev (fst (splitAt x xs))```
lemma splitAt_take:
` [| distinct ls; i < |ls| |] ==> fst (splitAt (ls ! i) ls) = take i ls`
lemma splitAt_drop:
` [| distinct ls; i < |ls| |] ==> snd (splitAt (ls ! i) ls) = drop (Suc i) ls`
lemma fst_splitAt_upt:
` [| j ≤ i; i < k |] ==> fst (splitAt i [j..<k]) = [j..<i]`
lemma snd_splitAt_upt:
` [| j ≤ i; i < k |] ==> snd (splitAt i [j..<k]) = [i + 1..<k]`
lemma local_help1:
` [| vs = c @ r # d; vs = a @ r # b; r ∉ set a; r ∉ set b |] ==> a = c`
lemma local_help:
` [| vs = a @ r # b; vs = c @ r # d; r ∉ set a; r ∉ set b |] ==> a = c ∧ b = d`
lemma local_help':
` [| a @ r # b = c @ r # d; r ∉ set a; r ∉ set b |] ==> a = c ∧ b = d`
lemma splitAt_simp1:
` [| ram ∉ set a; ram ∉ set b |] ==> fst (splitAt ram (a @ ram # b)) = a`
lemma splitAt_simp2:
` ram ∉ set b ==> fst (splitAt ram (ram # b)) = []`
lemma splitAt_simp3:
` ram ∉ set a ==> fst (splitAt ram (a @ [ram])) = a`
lemma splitAt_simp4:
` [| ram ∉ set a; ram ∉ set b |] ==> snd (splitAt ram (a @ ram # b)) = b`
lemma help'''_in:
` ram ∈ set b ==> fst (splitAtRec ram xs b) = xs @ fst (splitAtRec ram [] b)`
lemma help'''_notin:
` ram ∉ set b ==> fst (splitAtRec ram xs b) = xs @ fst (splitAtRec ram [] b)`
lemma help''':
` fst (splitAtRec ram xs b) = xs @ fst (splitAtRec ram [] b)`
lemma splitAt_simpA:
` fst (splitAt ram (ram # b)) = []`
lemma splitAt_simpB:
` ram ≠ a ==> fst (splitAt ram (a # b)) = a # fst (splitAt ram b)`
lemma splitAt_simpB':
` a ≠ ram ==> fst (splitAt ram (a # b)) = a # fst (splitAt ram b)`
lemma splitAt_simpC:
` ram ∉ set a ==> fst (splitAt ram (a @ b)) = a @ fst (splitAt ram b)`
lemma help'''':
` snd (splitAtRec ram xs b) = snd (splitAtRec ram ys b)`
lemma splitAt_simpD:
` ram ≠ a ==> snd (splitAt ram (a # b)) = snd (splitAt ram b)`
lemma splitAt_simpD':
` a ≠ ram ==> snd (splitAt ram (a # b)) = snd (splitAt ram b)`
lemma splitAt_simpE:
` snd (splitAt ram (ram # b)) = b`
lemma splitAt_simpF:
` ram ∉ set a ==> snd (splitAt ram (a @ b)) = snd (splitAt ram b)`
lemma splitAt_rotate_pair_conv:
``` [| distinct xs; x ∈ set xs |]
==> snd (splitAt x (rotate n xs)) @ fst (splitAt x (rotate n xs)) =
snd (splitAt x xs) @ fst (splitAt x xs)```
### @{text between}
lemma inbetween_inset:
` x ∈ set (between xs a b) ==> x ∈ set xs`
lemma notinset_notinbetween:
` x ∉ set xs ==> x ∉ set (between xs a b)`
lemma set_between_id:
` [| distinct xs; x ∈ set xs |] ==> set (between xs x x) = set xs - {x}`
lemma split_between:
``` [| distinct vs; r ∈ set vs; v ∈ set vs; u ∈ set (between vs r v) |]
==> between vs r v =
(if r = u then [] else between vs r u @ [u]) @ between vs u v```
### Tables
lemma isTable_eq:
` isTable E vs ((a, b) # ps) ==> b = E a`
lemma isTable_subset:
` [| set qs ⊆ set ps; isTable E vs ps |] ==> isTable E vs qs`
lemma isTable_Cons:
` isTable E vs ((a, b) # ps) ==> isTable E vs ps`
lemma removeKey_subset:
` set (removeKey a ps) ⊆ set ps`
lemma length_removeKey:
` |removeKey w ps| ≤ |ps|`
lemma length_removeKeyList:
` |removeKeyList ws ps| ≤ |ps|`
lemma removeKeyList_subset:
` set (removeKeyList ws ps) ⊆ set ps`
lemma notin_removeKey1:
` (a, b) ∉ set (removeKey a ps)`
lemma notin_removeKey:
` r ∉ fst ` set (removeKey r ps)`
lemma notin_removeKeyList1:
` a ∈ set rs ==> (a, b) ∉ set (removeKeyList rs ps)`
lemma notin_removeKeyList:
` r ∈ set rs ==> r ∉ fst ` set (removeKeyList rs ps)`
lemma removeKeyList_eq:
` removeKeyList as ps = [p∈ps . ∀a∈set as. a ≠ fst p]`
lemma removeKey_empty:
` removeKey a [] = []`
lemma removeKeyList_empty:
` removeKeyList ps [] = []`
lemma removeKeyList_cons:
``` removeKeyList ws (p # ps) =
(if fst p ∈ set ws then removeKeyList ws ps else p # removeKeyList ws ps)``` | 23,014 | 66,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-05 | latest | en | 0.616088 |
https://www.calculateme.com/compare-fractions/5-6/3-2 | 1,669,602,363,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00748.warc.gz | 746,856,218 | 2,938 | # What's Bigger 5/6 or 3/2?
Is five sixths greater than three halves? Use this calculator to quickly compare the size of two fractions.
Fraction 1
/
Fraction 2
/
56
is smaller than
32
Fraction Decimal Value
56 ≈ 0.833
32 1.5
How much less?
56 | 77 | 244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-49 | latest | en | 0.894218 |
https://encyclopediaofmath.org/wiki/Cubic_hypersurface | 1,674,938,380,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00491.warc.gz | 260,930,492 | 9,512 | # Cubic hypersurface
A projective algebraic variety defined by a homogeneous equation $F _ {3} ( x _ {0}, \dots, x _ {n} ) = 0$ of degree three with coefficients in some ground field $k$.
## Cubic curves.
An irreducible cubic curve is either smooth (in which case its canonical class is 0, its genus 1) or has a unique singular double point (in which case it is rational). Cubic curves are the curves of lowest degree for which there exist moduli (cf. Moduli theory). Every smooth cubic curve $X$ over an algebraically closed field $k$ of characteristic $\neq 2$ or 3 can be reduced by birational transformations to Weierstrass form, which is, in terms of non-homogeneous coordinates on the $( x, y)$-plane,
$$y ^ {2} = \ 4x ^ {3} - g _ {2} x - g _ {3} ,$$
where $g _ {2} , g _ {3} \in k$, $g _ {2} ^ {3} - 27g _ {3} ^ {2} \neq 0$. Two cubic curves with coefficients $( g _ {2} , g _ {3} )$ and $( g _ {2} ^ \prime , g _ {3} ^ \prime )$ in Weierstrass form are isomorphic if and only if
$$\frac{g _ {2} ^ {3} }{g _ {2} ^ {3} - 27g _ {3} ^ {2} } = \ \frac{g _ {2} ^ {\prime 2} }{g _ {2} ^ {\prime 3} - 27g _ {3} ^ {\prime 2} } .$$
The function
$$j = \ \frac{1728g _ {2} ^ {3} }{g _ {2} ^ {3} - 27g _ {3} ^ {2} }$$
takes arbitrary values in $k$ and depends only on the curve $X$; it is called the absolute invariant of $X$.
One can define a binary composition law $( x _ {1} , x _ {2} ) \rightarrow x _ {1} \circ x _ {2}$ on the set of points $X ( k)$ of a cubic curve: $x _ {1} \circ x _ {2}$ is the third point of intersection of $X$ with the straight line through $x _ {1}$ and $x _ {2}$. If one fixes some point $x _ {0} \in X ( k)$, the composition
$$( x _ {1} , x _ {2} ) \rightarrow \ x _ {0} \circ ( x _ {1} \circ x _ {2} )$$
turns $X ( k)$ into an Abelian group with neutral element $x _ {0}$. A cubic curve endowed with this structure is a one-dimensional Abelian variety (an elliptic curve).
If $k = \mathbf C$ is the field of complex numbers, $X ( \mathbf C )$ is a Riemann surface of genus 1, i.e. a one-dimensional complex torus — a quotient group $\mathbf C / \Gamma ( X)$, where $\Gamma ( X)$ is a two-dimensional period lattice. The field $k$ of rational functions of the curve $X$ is then isomorphic to the field of elliptic functions on $\mathbf C$ with period lattice $\Gamma ( X)$. The coefficients $g _ {2} , g _ {3}$ are interpreted as modular forms of weight 4 and 6, respectively, that are identical, up to a constant factor, with the forms defined by Eisenstein series of lowest weights. In that case the function $f$ is none other than the modular invariant.
A rich arithmetic theory has also been developed for cubic curves over algebraically non-closed fields $k$ (see [2]). Significant achievements in that respect are the Mordell–Weil theorem, the theory of complex multiplication and the homology theory of principal homogeneous spaces. The main unsolved problems (as of 1982) are: boundedness of the rank over an algebraic number field; the finiteness conjecture for the group of principal homogeneous locally trivial spaces; the conjecture of Birch and Swinnerton-Dyer on the zeta-function; Weil's uniformization conjecture, etc. (See also Elliptic curve.)
## Cubic surfaces.
Over an algebraically closed field $k$, every irreducible cubic surface (that does not degenerate into a cone) is a rational surface. The class of a hyperplane section $h$ of a surface $F$ is precisely the canonical class $(- K _ {F} )$. Any smooth cubic surface can be obtained from the projective plane $P ^ {2}$ by blowing-up (i.e. performing a monoidal transformation) of 6 points, no three of which are collinear, which do not lie on a single conic. The appropriate birational mapping $\phi : P ^ {2} \rightarrow F$ is determined by the linear system of cubic curves passing through the 6 points. There are 27 straight lines on $F$, each of which is exceptional (see Exceptional subvariety); they are the only exceptional curves on $F$. The configuration of these 27 lines is rich in symmetries: The automorphism group of the corresponding graph is isomorphic to the Weil group of type $E _ {6}$. Cubic surfaces belong to the class of del Pezzo surfaces — projective surfaces with an ample anti-canonical class.
Over an algebraically non-closed field $k$, there are smooth cubic surfaces $F$ which are not birationally isomorphic to $P ^ {2}$ over $k$ (i.e. $F$ is not rational over $k$). Among these surfaces one finds surfaces possessing $k$-points, and these are unirational over $k$. Such cubic surfaces provide a counterexample for the Lüroth problem on surfaces over non-closed fields. There exist fields $k$ over which there are minimal cubic surfaces. Segre's minimality criterion [6]: $\mathop{\rm Pic} ( F) \simeq \mathbf Z$. The group of birational automorphisms of a minimal surface has been determined (in terms of its generators and defining relations) and an arithmetic theory of cubic surfaces has been developed [4]. In order to describe the set of points $F ( k)$ one appeals to non-associative structures, such as quasi-groups and Moufang loops.
## Cubic hypersurfaces of dimension 3.
All smooth cubic hypersurfaces of dimension $\geq 2$ over an algebraically closed field are unirational. As far back as the eighties of the 19th century, the following question was posed: Is a smooth three-dimensional cubic hypersurface rational? A negative answer has been obtained [3]. This also provides a negative solution to the Lüroth problem for three-dimensional varieties. For every smooth three-dimensional cubic hypersurface $V$ there exists a principal polarized five-dimensional Abelian variety — the intermediate Jacobian $J _ {3} ( V)$. If $k = \mathbf C$ it is defined as as the complex torus
$$H ^ {1,2} ( V, \mathbf C )/H ^ {3} ( V, \mathbf Z ),$$
where $H ^ {1,2} ( V, \mathbf C )$ is the corresponding Hodge component in the decomposition of the homology space $H ^ {3} ( V, \mathbf C )$. In order to prove that $V$ is non-rational, it was shown that $J _ {3} ( V)$ is not the Jacobian of any curve of genus 5. The fact that a cubic hypersurface over a field of finite characteristic is non-rational was established in [5].
A cubic hypersurface $V$ is uniquely determined by its Fano surface $\Phi ( V)$. For $\Phi ( V)$ one has the Torelli theorems (which are also valid for $V$ itself). The following problem is unsolved: Given a three-dimensional cubic hypersurface, describe its group of birational automorphisms.
It is not known (1987) whether every smooth cubic hypersurface of dimension $\geq 4$ is rational. Rationality has been proved in this case for certain hypersurfaces of a special type; for example:
$$\sum _ {i = 0 } ^ { {2m } + 1 } a _ {i} x _ {i} ^ {3} = 0,\ \ m \geq 2.$$
#### References
[1] A. Hurwitz, R. Courant, "Vorlesungen über allgemeine Funktionentheorie und elliptische Funktionen" , Springer (1964) MR0173749 Zbl 0135.12101 [2] J.W.S. Cassels, "Diophantine equations with special reference to elliptic curves" J. London Math. Soc. , 41 (1966) pp. 193–291 MR0199150 Zbl 0138.27002 [3] C.H. Clemens, P.A. Griffiths, "The intermediate Jacobian of the cubic threefold" Ann. of Math. , 95 (1972) pp. 281–356 MR0302652 Zbl 0245.14010 [4] Yu.I. Manin, "Cubic forms. Algebra, geometry, arithmetic" , North-Holland (1986) (Translated from Russian) MR0833513 Zbl 0582.14010 [5] J.P. Murre, "Reduction of the proof of the non-rationality of a non-singular cubic threefold to a result of Mumford" Comp. Math. , 27 (1973) pp. 63–82 MR0352088 MR0352089 Zbl 0271.14020 [6] B. Segre, "The non-singular cubic surfaces" , Clarendon Press (1942) MR0008171 Zbl 0061.36701 Zbl 68.0358.01 [7] A.N. Tyurin, "Five lectures on three-dimensional varieties" Russian Math. Surveys , 27 (1972) pp. 1–53 Uspekhi Mat. Nauk , 27 : 5 (1972) pp. 3–50 Zbl 0263.14012 [8] A.N. Tyurin, "The geometry of the Fano surface of a nonsingular cubic and Torelli theorems for Fano surfaces and cubics" Math. USSR-Izv. , 5 : 3 (1971) pp. 517–546 Izv. Akad. Nauk SSSR Ser. Mat. , 35 (1971) pp. 498–529 Zbl 0215.08201 [9] I.R. Shafarevich, "Basic algebraic geometry" , Springer (1977) (Translated from Russian) MR0447223 Zbl 0362.14001 | 2,423 | 8,180 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-06 | longest | en | 0.79043 |
http://www.jiskha.com/display.cgi?id=1190503092 | 1,498,320,716,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320264.42/warc/CC-MAIN-20170624152159-20170624172159-00704.warc.gz | 564,954,099 | 3,630 | # pre-algebra
posted by .
The perimeter of an equilateral triangle is 7 inches more than the perimeter of a square,and the side of the triangle is 5 inches longer than the side of square.Find the side of the triangle.
• pre-algebra -
Let triangle side length be T and square side length be S.
3T = 4S + 7
T = S + 5
4T = 4S + 20
Subtract the first equation from the last to get T | 106 | 385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-26 | latest | en | 0.869473 |
https://spider.wadsworth.org/spider_doc/spider/docs/man/cg.html | 1,632,123,720,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057033.33/warc/CC-MAIN-20210920070754-20210920100754-00130.warc.gz | 575,472,620 | 2,191 | ## CG - Center of Gravity
(5/04/13)
PURPOSE
Compute coordinates of center of gravity of an image/volume with respect to the SPIDER image/volume center: (NX/2 + 1, NY/2 + 1, NZ/2 + 1). Also computes radius of gyration for volumes. Example.
CG PH [Center of Gravity - Phase approximation ||] CENT PH [Center image/volume using phase approximation] SH [Shift - using bilinear/trilinear interpolation ||]
USAGE
.OPERATION: CG [x],[y],[z],[rg]
[This operation can return up to 4 optional register variables:
Variable Example Receives First [x] Sub-pixel X center of gravity relative to center Second [y] Sub-pixel Y center of gravity relative to center
Third (Volumes only) [z] Sub-pixel Z center of gravity relative to center
Fourth (Volumes only) [rg] Radius of gyration
.INPUT FILE: PIC001
[Enter name of image/volume.]
.THRESHOLD: 0.23
[Only data above this threshold will be used in computation.]
NOTES
1. The center of gravity determination is very poor with typical high noise cryo-em particle images without an appropriate threshold. This threshold is hard to determine apriori. Try 'CG PH' for such images, but even that alternative central measure may have problems. ]
2. For definition of the radius of gyration, see: International Tables of Crystallography, vol. III, p327.
3. COG Algorithm (1D analog): Sum product of all pixel values by their position, Then divide by sum of all pixel values.
4. USAGE EXAMPLE: To shift an image to its center of gravity computed for a threshold of T=0.23, one could use the following operation sequence:
CG [x],[y]
PIC001
.23
SH
PIC001
SHI001
-[x], -[y]
SUBROUTINES
CENGR3
CALLER
UTIL1 | 426 | 1,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-39 | latest | en | 0.787338 |
https://homeworkvendors.com/discussion-thread-chi-square-cross-tabulation-and-non-parametric-association/?doing_wp_cron=1656155968.7276749610900878906250 | 1,656,155,969,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00101.warc.gz | 360,081,498 | 9,643 | # Discussion Thread: Chi-Square, Cross Tabulation, and Non-parametric Association
In the thread for each short-answer discussion, the student will post short answers to the prompted questions. The answers must demonstrate course-related knowledge and support their assertions with scholarly citations in the latest APA format. The minimum word count for all short answers cumulatively is 200 words.
For each thread the student must include a title block with your name, class title, date, and the discussion forum number; write the question number and the question title as a level one heading (e.g. D1.1 Variables) and then provide your response; use Level Two headings for multi-part questions (e.g. D1.1 & D1.1.a, D1.1.b, etc.), and include a reference section.
RespondRespond to the following short answer questions in Chapter 7 from the Morgan, Leech, Gloeckner, & Barrett textbook:D.5.7.1 In Output 7.1: (a) What do the terms “count” and “expected count” mean? (b) What does the difference between them tell you?D.5.7.2 In Output 7.1: (a) Is the (Pearson) chi-square statistically significant? Explain what it means. (b) Are the expected values in at least 80% of the cells ≥ 5? How do you know? Why is this important?D.5.7.4 Because father’s and mother’s education revised are 3-level variables with at least ordinal data, which of the statistics used in ProblemD.5.7.3 is the most appropriate to measure the strength of the relationship: phi, Cramer’s V, or Kendall’s tau-b? Interpret the results. Why are tau-b and Cramer’s V different?D.5.7.5 In Output 7.4: (a) How do you know which is the appropriate value of eta? (b) Do you think it is high or low? Why? (c) How would you describe the results? | 431 | 1,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.896828 |
www.artmusicdance.com | 1,642,503,025,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300810.66/warc/CC-MAIN-20220118092443-20220118122443-00163.warc.gz | 81,424,616 | 13,216 | # Proof of the Existence of God The Ingenious Numeration of Three Constants
## Expanded Highlights of the Evidence Illustrated in Document Alpha
### Vasilios Gardiakos
#### Only one entity could manipulate these three constants to produce oddities that we can detect. Only one entity could do this specifically on these three constants and not on other numbers. It seems that God not only could but has formed the geometry of space and even defined the fundamentals of mathematics. He ingeniously manipulated the geometric fabric and the default mathematics of the universe so that we can detect and analyze the oddities and conclude that only a great intelligence can accomplish this and therefore we can know that God exists.
Note: Document A may look complicated but it is not! It is fairly simple to understand. Any one with a little knowledge of math can understand it. The many arrows on Document Alpha though they may look confusing or distracting are there to point or reference the numbers. The highlights below are discussed step by step in detail to explain the oddities. Once you understand Document Alpha you will realize its significance.
1) Pi (pi = 3.141592653...) starts with 3_4_5 which is the smallest Pythagorean Triplet possible forming the Pythagorean triangle. This triplet, 3_4_5 can be recognized as the representation of the Pythagorean Theorem. Also in pi the sequence 1_ _2_ _3. In √2 early the sequence1_2_3 and 4_ _3_ _2.
2) In pi, 31, 41 and 59 are primes (3.1 41 59...). No formula has been discovered that will produce all the prime numbers however 41 + n squared + n, 59 + 4n squared + 4n and 31 + 2n squared + 4n are three most productive formulas known to yield prime numbers. Both 41, 59 and 31 are used. See 5) bellow. Right at the start of pi there are three very productive prime number generators, 31, 41 and 59. What are the odds of this happening by chance?
The great mathematician's Leonhard Euler's formula 41 + n squared + n turns out to be surprisingly good for large values of n. A computer showed that for the primes under ten million, the formula generated primes 47.5 percent of the time. The formula works even better for lower values of n. For values of n under 2,398, there's' an even chance of getting a prime. And for values under 100 the formula yields 86 primes and only 14 composite or non-prime numbers.
For n=0, the formula yields the prime number 41. For n=2, the prime number 47. Indeed, as n takes on successive values from 0 to 39, Euler's formula yields nothing but primes.
Stanislaw Ulam, a well known and respected mathematician and his coworkers discovered another prime generating formula that is almost as good as Euler's. The formula 59 + 4 n squared + 4n has a success rate of 43.7 percent.
Steve Homewood a determined and ingenious mathematician based in the U.K. announce on March 7, 2009 that he had solved the A31 problem. He discovered the formula 31 + 2n squared + 4n generates prime numbers 43.5% of the time for all answers up to 10,000,000. So as he says “So, no, 31 has not been overlooked”.
3) In pi, 31, 41 and 59 oddly their prime order, being the 11th, 13th and 17th primes are themselves primes. More oddly 11, 13 and 17 are consecutive primes. Working with 11, 13 and 17, oddly 1+1+1+3+1+7=14 and 11+13+17= 41 (3.14159...). One more oddity, in 11, 13, and 17, 1+1=2, 1+3=4 and 1+7=8 a 2-4-8 progression. Also 31+ 41+ 59=131 and 3+1+4+1+5+9= 23, both sums are prime numbers.
4) Pi starts with the prime number 31 which is also a Mersenne Prime. The cube root of 31=3.141380652 which is within .00675% of pi. The first 8 digits of pi 3+1+4+1+5+9+2+6=31 also 1+4+1+5+9+2+6+3=31. There is no zero in the first 31 digits of pi. Note, 1+2+4+8+16=31 and (from 16=31) 16x31=496. The third perfect number is 496.
5) Odd that pi starts with 314159 and its mirror image 951413 are both prime numbers. About 1 in 7 primes is the reversal also a prime.
These are the four known pi forward primes:
3, 31, 314159 and 31415926535897932384626433832795028841
So in addition to 314159 being a reversible prime it is also a rare pi forward prime.
A lesser oddity, the first even number in pi, 314 divided by two equals 157 which is a prime. Also 14, an even number divided by two equals 7, a prime.
6) Pi = 3.14159265358979323... It is very odd that a group of eight small different contiguous primes: 3, 14159, 2, 653, 5, 89, 7, 9323 are right at the start of pi. Many and possibly infinite small (numbers with five or fewer digits) and large (greater than five digits) different contiguous primes may exist after 9323. As it turns out right after the 9323 prime the next contiguous prime is: 846264338327........303906979207, it is 3057 digits long. So if pi started with 8462... the first prime would be 3057 digits long. See: http://www.primepuzzles.net/problems/prob_018.htm
It will be interesting to see how many digits the average contiguous prime has. Perhaps more interesting may be to find how scarce are groups consisting of eight small contiguous primes of which none of the prime numbers are duplicated.
7) Starting with the first digits in √2, 1_1_2_3; the first digits in S, 8_6_2_9 and the first digits in pi, 2_5_4_3 (3_4_5_2 inverted) are prime numbers.
8) Pi and √2 have the 2653, 3562 cross-mirroring and the 141,414 cross-symmetry at DEF/2. (see alphanumeric locators on the sides of Document A) The mathematical block at E/3 reveals several vertical and diagonal oddities. The repeating "88" ties pi and √2 to S.
9) In √2 (√2= 1 .4 1 4 2 1 3 5 6 2 ...) early 1+_2+_3=_6. Not only does it equal 6, a perfect number but this is an example of how to derive the first perfect number which is 6. Again in √2 early14+14=28, 28 being the second perfect number. Furthermore sum of the digits 1+4+1+4 = 2+8 = 10 which is the fourth triangular number
10) √2 starts with the Fibonacci sequence 1_1_2_3 in which 11, 23 and 1123 are primes.
Also note:
√2 = 1 .4 1 4 2 1 3 5 6 2 3 7 3...
Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13...
In
√2 the Fibonacci sequence is included: 1, 1, 2, 3, 5, 8(6+2), 13(3+7+3). Very odd that seven numbers of the Fibonacci sequence are found at the start of the square root of 2.
11) The 1_1_2 in √2 is in an identical position right at the beginning as the 3_4_5 in pi. Both 3,4,5 and 1,1,√2 can be represented by a triangle. See Document A.
12) √2 has a strong 7 occurrence. Many consecutive numbers add up to multiples of 7. In √2 early 14-21-35 are multiples of 7. Note that 22/7 equals pi within 0.04025%. See G/1-9 on Document A.
13) S (Side=) starts with 88. Numbers 88, 8 and 8 are prominent in 62(6+2=8) and 26(2+6=8) add up to 88 and are in a sense self-referential. At DF/12 Pi and √2 also refer to 88: 62+26=88, the 62 from √2 and the 26 from pi oddly 6226 is found in S. Pi and √2 refer to 88: 35+53=88. See more 88 oddities on Document A.
14) The 22 oddity: 11, 22 and 88 figure prominently in pi, √2 and S. Also 88 is present in some way in all three constants and is a multiple of 11 and 22.
15) S=.886226925
886 22 6925
8+8+6=22=6+9+2+5
Also interesting: 88+6=94=69+25 also 69-25=44=2x22
16) S=.886226925, so that 886, 862, 622, 226, 692 (odd numbers are ignored):
abc-cba=d, d/x=22
886-688=198, 198/9=22
862-268=594, 594/27=22
622-226=396, 396/18=22
226-622=-396, -396/18=-22
692-296=396, 396/18=22
abcdef, abcdef/x=22
886226, 886226/40283=22
fedcba, fedcba/x=22
622688, 622688/28304=22
263538, 263538/11979=22
17) The initial consecutive digits in pi, √2 and S can add up to 41 (pi=3.141...):
pi: 1+4+1+5+9+2+6+5+3+5 = 41
√2: 4+1+4+2+1+3+5+6+2+3+7+3 = 41
S: 8+8+6+2+2+6+9 = 41
Multiply 41 by 3 (the three constants) which is equal to 123. Now 1+2+3=6 which is a perfect number.
18) Very strange the E-5 mathematical block is the result of the "1415" or 14-15 in Pi and "4142" or 41-42 at identical place relative to the decimal point on √2. This block contains several vertical and diagonal and prime number oddities.
===========================
Conclusion
The "oddities" explained above is
the evidence that proves that God
intentionally numerated the three
constants so that we can become
aware that He exists.
Check out the criticism: Page 1 Page 2 Page 3 To Page 4
===========================
And Now For The Fun Stuff
On a Humorous Note (S=886226925):
It takes 62 muscles to frown
and only 26 to smile.
(I am pushing things a bit here)
Photo V. Gardiakos - My 1973 trip to New Orleans
On a Telephone Note:
The telephone area code of St Louis
home of The Gateway Arch is 314
and has a zip code of 63141
(I am pushing things a lot here)
On a Musical Note (S=886226925):
88 represents the 88 keys of a piano.
(now this is ridiculous connection so I
include it only to extract a smile from you.
Notes:
A. All of the underlined hyperlinks in this page are not in the typical blue but are in the color of this font.
B. "_" The under score represents a digit in the sequence that is not shown. Sometimes I underscore a digit for emphasis. The underscore under a number does not indicate that it is a hyperlink.
C. Pi (π) is the ratio of the circumference of a circle to its diameter. Pi is the most well known non-integral mathematical constant. Pi is equal to 3.141592653... which is infinitely long. Pi shows up in many equations where you don't expect it. It is the 16th letter of the Greek alphabet.
D. 2 = 1.414213562373... ("√2" is the mathematical representation of square root of two)
E. S=0.886226925... which is 1/2 times the square root of pi. It is gamma(3/2), and is sometimes also called (1/2)! the factorial of 1/2. Herein "S" is used for "side" for the sake of convenience.
F. Mersenne Primes: 2, 3, 5, 7, 13, 17, 19, 31, 67, 127, 257 ... They become more rare as the value increases.
G. 22/7=3.142857... = pi within 0.04025%. The first theoretical calculation of a value of pi was that of Archimedes of Syracuse (287-212 BCE), one of the most brilliant mathematicians of all time. Archimedes worked out that 223/71 < π < 22/7. Interestingly 22/7 is also a member of the continued-fraction series for pi. Continued fraction approximations can be used to get "optimal" rational approximations for any number. By "optimal" we mean the closest approximation that is possible with a given-size numerator and denominator. The continued fraction approximations for π are: 3/1, 22/7, 333/106, 355/113, 103993/33102, 104348/33215, 208341/66317, 312689/99532, 833719/265381, 1146408/364913, ...
H. 7 is the smallest number whose reciprocal has a pattern of more than one repeating digits: 1/7 = 0.142857142857... It is also the smallest number for which the digit sequence is of length N-1 (the longest such a sequence can be). The next such numbers are 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, ... (Sloane's integer sequence A006883)
I. 7 is considered "lucky" by many people and given much spiritual significance. The early religious and cultural use of the 7-day week almost certainly arose from the fact that the moon goes through its 4 phases in a bit over 28 days, which divides nicely into 7 days per phase. These man made numeral references may be part of numerology however they are excluded as being significant or relevant in the oddities of Document A.
J. What is a prime number?
A prime number is a positive integer that has exactly two positive integer factors, 1 and itself. For example, if we list the factors of 28, we have 1, 2, 4, 7, 14, and 28. That's six factors. If we list the factors of 29, we only have 1 and 29. That's 2. So we say that 29 is a prime number, but 28 isn't. Another way of saying this is that a prime number is a whole number that is not the product of two smaller numbers.
Note that the definition of a prime number doesn't allow 1 to be a prime number: 1 only has one factor, namely 1. Prime numbers have exactly two factors, not "at most two" or anything like that. When a number has more than two factors it is called a composite number.
Here are the first few prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, etc.
Prime numbers sequences are useful as they do not occur in nature. Only intelligent beings can create a prime number sequence. In the search for extra terrestrial intelligence any detection of a prime number sequence will be recognized as being created by a intelligent beings.
Mirror prime or reversible prime number or emirp (prime spelled backwards) is a prime that gives you a different prime when its digits are reversed.
K. A31
Prior to March 6, 2009 only Euler's (1707 - 1783) 41 + n squared + n and Ulam's (1909 - 1984) 59 + 4n squared + 4n simple polynomial formulas were known to be very productive in generating prime numbers. Since pi starts with 3.14159... and if God intentionally numerated pi with 41 and 59 as prime number generators then it is logical that 31 could also be used to do likewise.
Steve Homewood a very talented mathematician based in the U.K. took up the painstaking task to solve the A31 problem and March 6, 2009 announce that he had done so! Referring to my question "Has 31 been overlooked in the search for other productive numbers?" Homewood replied with “So, no, 31 has not been overlooked”. He discovered the formula 31 + 2n squared + 4n generates prime numbers 43.5% of the time for all answers up to 10,000,000. There are other simple polynomial prime number generators but are not as productive as these three.
Nothing to do with A31 and thus not significant but nevertheless interesting oddity: 31, 331, 3331, 33331, 333331, 3333331, and 33333331 are all prime numbers. But the pattern is broken with the next number in this sequence: 333333331 is not prime. Back to 2)
L. What is the Fibonacci Sequence?
Fibonacci Sequence:1, 1, 2, 3, 5, 8, 13, 21, 34 ... The Fibonacci series is formed by starting with 0 and 1 and then adding the latest two numbers to get the next one. Each number is the sum of the previous two. Surprisingly, Fibonacci numbers can be used to calculate pi.
M. What is a perfect number?
A perfect number is a number that it is equal to the sum of its own divisors (other than itself). Examples:
6 = 1 + 2 + 3
28 = 1 + 2 + 4 + 7 + 14
N. What is a triangular number?
A triangular number or Tetraktys is the number of dots that may be arranged in an equilateral triangle: 1, 3, 6, 10, 15, 21, 28, 36... In general n(n + 1)/2.
O. What is the definition of "oddity" when referring to Document A?
Oddity defined: abnormal unusual strange atypical extraordinary surprising amazing weird uncommon unexpected unusual unpredicted unforeseen remarkable notable noteworthy.
In Document A "oddities" refers to simple unexpected occurrences that include prime numbers, Fibonacci sequence, perfect numbers, Pythagorean triplet etc. One could write countless unique equations to fit every irrational or random number and call it an oddity. This force to fit technique is not used in Document A. It is important to find simple oddities as there are limits to simplicity like counting, adding, subtracting and multiplying. Complexity has not limits. I suspect it is more difficult to numerate simple oddities as opposed to complex ones as complexity allows you an unlimited course of action. All the oddities found in Document A are of the simple type. No complex equations, complex algorithms or complex formulas are used to find the oddities. Perhaps natural rather than contrived may be a way to describe the 19 or so oddities in the webpage.
Another reason simple oddities are more relevant is because they are easier to be spotted and recognized as such. If the oddities are intentional then they should be easy to spot or what would be the sense of numerating them in the first place?
One oddity may be noteworthy but additional oddities make it significant. The more oddities the more noteworthy and significant it is. The 19 or more oddities in Document A is enough evidence to conclude that a powerful intelligence namely God numerated pi.
By the way one equation that many mathematicians would agree is an oddity is e^(i*pi)=1. To further clarify, I use "oddity" with a similar meaning.
Comment:
God herein means the Pythagorean God and not any of the Gods of the various religions.
______________________________________________________ | 4,744 | 16,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-05 | latest | en | 0.918847 |
https://askinglot.com/what-is-ellipse-in-autocad | 1,632,848,920,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00348.warc.gz | 154,602,447 | 7,558 | asked in category: General Last Updated: 22nd May, 2020
# What is ellipse in AutoCAD?
The AutoCAD ELlipse command is easy to use. Mathematically, an ellipse is defined by a major (long) axis and a minor (short) axis. These axes determine the ellipse's length, width, and degree of curvature. An elliptical arc is an arc cut from an ellipse.
Likewise, how do you draw an ellipse in AutoCAD?
Draw an Elliptical Arc Using Start and End Angles
1. Click Home tab Draw panel Ellipse drop-down Elliptical Arc. Find.
2. Specify endpoints for the first axis (1 and 2).
3. Specify a distance to define half the length of the second axis (3).
4. Specify the start angle (4).
5. Specify the end angle (5).
Also, what is AutoCAD Pline? Creates a 2D polyline, a single object that is composed of line and arc segments. You can create straight line segments, arc segments, or a combination of the two.
Similarly one may ask, how do you make an ellipse a polyline in AutoCAD?
1. Set the PELLIPSE system variable to 0 (to draw true ellipses)
2. Draw an ellipse on the screen.
3. Set the PELLIPSE system variable to 1 (to draw polyline ellipses).
4. Set the PLINETYPE system variable to 0 (polylines will not be altered when they are placed into the drawing)
5. Enter dxfout on the command line.
What does the Ellipse command do?
The AutoCAD ELlipse command provides a straightforward way to draw an ellipse: You specify the two endpoints of one of its axes and then specify an endpoint on the other axis. Like the Arc command, however, the ELlipse command offers several other options: Arc: Generates an elliptical arc, not a full ellipse.
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22nd May, 2020
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Users | 425 | 1,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-39 | latest | en | 0.791197 |
https://math.stackexchange.com/questions/1273555/show-that-creates-a-partition-of-m-2-mathbbr | 1,563,598,729,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526446.61/warc/CC-MAIN-20190720045157-20190720071157-00212.warc.gz | 462,608,349 | 37,177 | # Show that ~ creates a partition of $M_2(\mathbb{R})$
Let $M_2 (\mathbb{R})$ be the set of 2x2 matrices over $\mathbb{R}$: $$M_2 (\mathbb{R}) = \biggl\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \; \biggm| \; \text{with } a,b,c,d \in \mathbb{R} \biggr\}.$$ For $M_1, M_2 \in M_2(\mathbb{R})$, we say that $M_1 \sim M_2$ if $\det(M_1) = \det(M_2)$, where $$\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad-bc.$$
• Show that $\sim$ creates a partition of $M_2(\mathbb{R})$.
• What are the representative elements of each partition?
• Are there countably or uncountably many distinct equivalence classes?
I am stuck and I could really use some help. I know that in order to have a partition, the union of all the partitions $X_i$ should be $X$ and $X_i \cap X_j = \varnothing$ for $i \ne j$.
Hints:
• It suffices to show that $\sim$ is an equivalence relation on $M_2(\mathbb R)$. So simply show that it's reflexive, symmetric, and transitive. This follows immediately, since $\sim$ is defined in terms of $=$.
• Try some examples. Pick a matrix at random, and find its equivalence class. Then try going backwards: pick a determinant at random, then try to find a matrix that has that as a determinant.
• From the above bullet, try to show that each $x \in \mathbb R$ corresponds to a distinct equivalence class $[M]_\sim$.
Whenever we have a function $f\colon X\to Y$, we can define a relation $\sim_f$ over $X$ by declaring $$a\sim_f b\qquad\text{if and only if}\qquad f(a)=f(b)$$ This is readily seen to be an equivalence relation:
• for all $a\in X$, $a\sim_f a$: indeed $f(a)=f(a)$
• for all $a,b\in X$, if $a\sim_f b$, then $b\sim_f a$: indeed $f(a)=f(b)$ implies $f(b)=f(a)$
• for all $a,b,c\in X$, if $a\sim_f b$ and $b\sim_f c$, then $a\sim_f c$: indeed $f(a)=f(b)$ and $f(b)=f(c)$ implies $f(a)=f(c)$
The equivalence class of $a\in X$ is $a/{\sim_f}=\{x\in X:f(x)=f(a)\}$.
Your case is with $X=M_2(\mathbb{R})$, $Y=\mathbb{R}$ and $f$ is the determinant function.
“The” representative is not well defined. What you need is, given finding one representative, but there is no “canonical” choice. Since any real number can be the determinant of a suitable matrix, given $r\in\mathbb{R}$ you just need a matrix $A$ such that $\det A=r$.
Note also that, in the general case, one can define a map $$\bar{f}\colon X/{\sim_f}\to Y$$ by declaring that $\bar{f}(a/{\sim_f})=f(a)$. This map is injective and has the same image as $f$. So, in your particular case, the set of equivalence classes is in 1-1 correspondence with $\mathbb{R}$.
Compute $$\det\pmatrix{a&0\\0&1}$$ What do you notice?
The partition property is inherent because $\det$ is a function and $x \sim_f y :\Leftrightarrow f(x) = f(y)$ is an equivalence relation for any function $f$. | 907 | 2,772 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-30 | latest | en | 0.759862 |
http://www.upworthy.com/see-this-teachers-amazing-response-to-the-question-but-when-are-we-gonna-have-to-use-this?c=hpstream | 1,427,672,797,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131298819.92/warc/CC-MAIN-20150323172138-00257-ip-10-168-14-71.ec2.internal.warc.gz | 864,449,397 | 16,705 | # See This Teacher's Amazing Response To The Question, 'But When Are We Gonna Have To Use This?'
Middle-school math class: pre-algebra, permutations, combinations, and simply trying to learn anything in the midst of one of the most awkward stages of life.
A few months ago, we posted a video of a teacher using a counting puzzle to engage his students and teach them that math can be pretty cool. They had so much fun and were so excited that we picked up the story that they made a sequel.
First, a few words from Justin:
Q: What was it like to have your story on Upworthy?
"When we were filming the first video, I jokingly told my students that I thought it could go viral. When we put it up and got a thousand views just by word of mouth, we were really pleased — that was about as viral as any of us expected. So after Upworthy picked it up and we got 25,000 views in 24 hours, my students were blown away. It helped get them jazzed to take on another puzzle!"
Q: Why do you do this exercise with your students?
"Every year I get the question in class, 'When are we gonna have to use this?' And my answer to that question is, 'You're not.' That's not the point. By doing math, we are carving neural pathways that otherwise wouldn't have been there. Grappling with problems like this makes us better problem solvers, and by extension, better human beings."
And finally, the counting puzzle, take 2:
Transcript:
Show Transcript Hide Transcript
My name is Justin Solonynka. I still teach Math at Abington Friends School.
So, last year Justin found this puzzle.
We got a puzzle, called the All Aboard train puzzle.
And he brought it in to our class.
He said that the children can arrange the puzzle in countless ways.
We were trying to find out how many different ways we can connect them together.
We decided to figure out if it really was countless.
And then this year, we had the puzzle, Around the World by the same company and they sent it to us.
A puzzle with a circular shape rather than just a straight line.
And also said it can be arranged in countless ways.
Puzzle pieces fit together to create countless configurations.
We don't think its actually countless.
We don't think its actually countless.
We don't think its actually countless.
We don't think its actually countless.
[music]
Do you have a decent theory here?
No, no.
What makes it so much complicated?
If you flip one piece it makes a whole different...
Right.
[Inaudible]
Well, I'm not gonna tell you.
[music]
So, if you pull out an out, then the rest of the outs have seven factorial ways to fit in, is it eight factorial... Ok, there's a lot of ins and outs.
[music]
This is just, it's two times two, then add one more, then it's two to the third power, because you can flip it.
[music]
In the end, my class did find the number of ways to put the puzzle together and they got it right. It was never really about the answer, it was about the process. It was about working out our brains. And we had fun too, which isn't such a bad thing.
[music]
There may be small errors in this transcript.
This video was made by Justin Solonynka and uploaded by Abington Friends School. Don't forget to check out the original counting puzzle if you missed it!
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Discussion in 'Off-Topic' started by Mouthwash, Apr 4, 2014.
1. ### BvBPLPour Decision Maker
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The Fermi paradox rests upon Drake’s equation. Drake’s equation reads: N = (N*) (fp) (ne) (fl) (fi) (fc) (fL)
fl is the percent of life supporting planets on which life develops. This value is obviously not zero, but beyond that we don’t know what the value is.
Fermi paradox adherents hold that fl is a relatively large. There is no support for a high fl. In fact, work in abiogenesis suggests that the development of life is highly unlikely. Regardless, there’s no way to currently know the correct value of fl at this time.
The same is true of other values in Drake’s equation. fi is the percent of life bearing planets that develop intelligent life and ne is the percent of potential life bearing planets per star with planets. These values are still obviously non-zero, but beyond that it is a total guess as to what the values actually are.
The consequences of all these unknowns is that any end value determined via the equation can’t be verified. The Fermi paradox is only a paradox if those presently unknowable values are sufficiently high, but there’s no proof that those values are high at all. You might as well read the entrails of a bird, it would have just as much validity as picking out high numbers for Drake’s equation out of the ether to create a paradox where one may not exist. The fact that we do not see any other extraterrestrial civilizations strongly suggests that the values for Drake’s equation are not high at all and therefore the Fermi paradox doesn't exist.
The Great Filter hypothesis is an attempt to reconcile the Fermi paradox with Drake’s equation by theorizing that there is some sort of near universal mechanism that reduces the number the intelligent civilizations in the galaxy. The hypothesis is unnecessary if the Fermi paradox doesn’t exist because Drake’s equation values are sufficiently low.
Furthermore, the likelihood that any particular experience is (nearly) universal reduces as other variables enter the equation. It would be hard to imagine a more disparate environment then planets alien to each other and their species. Therefore it seems unlikely that there is some universal experience would be shared across nearly every form of life on every planet on which it arises.
2. ### onejayhawkAfflicted with reason
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I said was that we have the technology to build a viable starship, but not the desire to commit the financial, material and human resources necessary. What is non rational about that?
If you get the spaghetti monster, I'll take a bowl of the loin.
J
3. ### Mr. DictatorA Chain-Smoking Fox
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How much energy does it take to die, go to hell, then fly off into space though?
4. ### Aleksey_aka_alSmiley
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This is not enough for a logical statement quoted in the OP. The supposed capability was never tested or observed.
5. ### El_MachinaeColour vision since 2018Retired Moderator
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It is akin to saying that human civilization cannot send people to Mars. In some ways, you're right. In some ways, it's not a true statement. Also if you say 'cannot' meaning 'now and forevermore', it's most likely wrong.
Pending energy or ecological collapse, I mean.
6. ### Aleksey_aka_alSmiley
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Once again:
The original Earth has not yet produced such civilization. Speculations about what's possible or what will happen in the future do not matter. The whole reasoning in the OP is based on this one unverified assumption. More logically is to claim that if there are Earth-like civilizations somewhere nearby, they behave the same way as this one: keep to their planet.
7. ### El_MachinaeColour vision since 2018Retired Moderator
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That would be one of the filters.
8. ### onejayhawkAfflicted with reason
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So it is a rational statement with which you disagree. That is quite different from a non-rational statement.
The capacity has been both observed and tested, but on a small scale. No breakthroughs are needed to build a multi-generation ship. The issues are purely enigeering from well understood science.
If you believed I was speaking of faster than light travel, please reread the posts above.
Some of this is simply wrong. We can do it, but we do not think it is worth the effort. Soiit is dubious to think that the original Earth has not produced such a civilization.
Certainly assumptions are unverified. However, if we are to assume that other cultures behave like ours, then we should draw extrapolations from our own history. From such extrapolations, you would assume that interstellar travel is used as soon as practical, maybe a bit earlier.
J
9. ### Core ImposterDeity
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One good candidate for Great Filter is democracy.
People get paid for their votes. We finished up the Moon Project just about the time of the implementation of the Great Society. Since that time no serious projects have been attempted nor are they likely in the foreseeable future.
We haven't even figured out how to pay for the votes that have been cast in the past. All governments are indebted, corporations and businesses are indebted, individuals are indebted. We are bringing decades of economic activity forward via reckless credit. This is a global phenomena.
All as a result of the fruit of democracy. Everyone has a vote and thus is entitled to consume.
You can't do interstellar missions under those constraints any more than the Pharaohs could do pyramids without slave labor. We stress over the concentration of wealth in the hands of the 1%. To do serious spacefaring we'd have to beggar the common man completely.
We dream of an evolved human society but its total fantasy. About 50% of humanity will always be dead weight and no amount of education and social programing can fix that.
But they get to vote and have to get paid. Leaving no way to marshal the necessary resources for space exploration.
Therefore to slip that Great Filter we have to have a model like China succeed. No liberty, grand corruption and mass slave labor on a global scale. Personally, I don't think it can happen. The Communist Party of China will collapse and with them we will likely lose the last, best hope of humanity reaching the stars.
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There's also the posibility that "they" know of us and don't want to contact us for some reason or perhaps they use different forms of communication - something other than radio waves, high powered light beam modulation perhaps ? telepathy ? ... everything is possible I guess.
11. ### onejayhawkAfflicted with reason
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J
12. ### Core ImposterDeity
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Well, I am using 20 July 1969 as the end of the moon project so my correlation is thereby adjusted to fit seamlessly.
13. ### El_MachinaeColour vision since 2018Retired Moderator
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In the alternative, you could say that we're already mostly slaves of the 1%, and so the necessary concentration of resources already exists to get us space borne.
14. ### onejayhawkAfflicted with reason
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What? Deprive us of a Tom Hanks movie. Fiend!
Even that date is 3-5 years after.
J
15. ### peter grimes...
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Well, not exactly. Things are only possible within the framework of the laws of physics. We haven't yet come to understand the full picture, but we've got a deep enough grasp on most of it that we can safely rule some things out. So not everything. | 1,956 | 8,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.930922 |
http://stackoverflow.com/questions/12459724/matlab-interpolation-for-calculating-volume-within-a-given-area | 1,440,757,882,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644062760.2/warc/CC-MAIN-20150827025422-00298-ip-10-171-96-226.ec2.internal.warc.gz | 222,151,258 | 17,274 | # matlab interpolation for calculating volume within a given area
I have some depth measurements and corresponding area at each depth of a water column:
``````depth = [0,2,4,6,7,9,10];
Area = [2000,1890,1873,1654,1523,1433,1200];
figure(1);
plot(Area,depth);set(gca,'ydir','reverse');
``````
I want to calculate the volume of water between each depth indicated and then calculating the entire volume of water from this. Firstly I have interpolated the depth and area values to improve the accuracy of the calculation:
``````dz = 0.1;
newD = min(depth):dz:max(depth);
newA = interp1(depth,Area,newD);
figure(2);
plot(newA,newD);set(gca,'ydir','reverse');
``````
Where would I go from here for calculating the volume of water between each of the depths?
-
To get volume you multiply area by height. Hence, since your dz is constant, to get water volumes at chosen depths you need to multiply
``````newA*dz
``````
To get the total volume you sum it up
``````sum(newA*dz)
``````
Your calculations are done assuming linear change of water area with depth (that's what you achieve with interp1). I am sure you can manage to integrate a linear function analytically instead of using your current approach. You can easily get an exact number.
-
You actually want to multiply the areas by the total depth, not the depth step:
``````volume = newA.*newD;
totalVolume = sum(volume);
``````
-
Your answer is not correct. Kate wants to compute the water volume "the volume of water between each depth indicated", which I assume means for every dz. Your answer is wrong because you multiply water area at given depth by that depth. This does not take into account the fact that the water area changes with depth. So effectively you compute integrated water volume from the surface to a given depth with the assumption that the water area is constant. It is not, hence your answer is wrong. – angainor Sep 19 '12 at 11:05 | 472 | 1,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-35 | latest | en | 0.907995 |
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# Rotary Reciprocating Stirling
It goes round and round and up and down, and hopefully along.
(-1) [vote for, against]
A single cylinder split into segments. Each segment is a piston with its own crank shaft at the top, which has a gear at the outside end that goes along a toothed track around the circumference of the top of the cylinder.
One thin strip of the cylinder wall is hot, the opposite side has a larger cold area, and the areas in between are conduct heat around the remaining wall area to act as a heat exchanger.
As the gas is each section is heated, the piston extends and the cylinder rotates. Heat is then removed by the exchanger wall and then by the cold wall, compressing the piston. Heat is then returned to the gas by the heat exchanger before the cylinder rotates back to the heating position again.
I think this would be more efficient if there were quite a few segments. With 8, the wall could be divided into Hot, HE1, HE2, HE3, Cold, HE3, HE2, HE1 where HE = heat exchanger, with both HE1 segments connected (HE1 would be the hottest and HE3 the coldest).
BTW, I think the Stirling category should be in Engine rather than car: engine as that is the worst application for a Stirling.
— marklar, Mar 25 2010
[link]
I'm having a difficult time visualizing this. Can you draw it?
— AutoMcDonough, Mar 26 2010
First of all, picture it from the top - a circle divided into let's say 3 segments. There is a hot side and a cold side. When the segments rotate, they will heat and cool as they pass the hot and cold sides.
The heat will increase the pressure, so now picture each segment from the side as a piston. Each piston will be attached to a crank shaft like a normal engine, however, as the the pistons are not inline, they will each have their own shaft. These shafts will need to rotate the cylinders, so they will have a gear at the end.
— marklar, Mar 29 2010
I think I must have been drunk when I wrote this idea. I only vaguely remember it and the spelling is pretty bad. I may try to explain it more clearly and with a diagram or two when I have more time.
Yes, I know there's a comment by me a few days later. I guess I got drunk twice that week.
— marklar, Oct 07 2010
//I guess I got drunk twice that week.//
And maybe once this week?
— Boomershine, Oct 07 2010
[annotate]
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There has been a murder on the Stevenson estate. Use your analytical chemistry skills to assess the crime scene and identify the cause of death...
### Cobalt Decay
##### Stage: 5 Challenge Level:
Investigate the effects of the half-lifes of the isotopes of cobalt on the mass of a mystery lump of the element.
### Clear as Crystal
##### Stage: 5 Challenge Level:
Unearth the beautiful mathematics of symmetry whilst investigating the properties of crystal lattices
### Smoke and Daggers
##### Stage: 5 Challenge Level:
We all know that smoking poses a long term health risk and has the potential to cause cancer. But what actually happens when you light up a cigarette, place it to your mouth, take a tidal breath. . . .
### Mathematical Issues for Chemists
##### Stage: 5
A brief outline of the mathematical issues faced by chemistry students.
### What Salt?
##### Stage: 5 Challenge Level:
Can you deduce why common salt isn't NaCl_2?
### Reaction Rates!
##### Stage: 5
Fancy learning a bit more about rates of reaction, but don't know where to look? Come inside and find out more...
##### Stage: 5
Read about the mathematics behind the measuring devices used in quantitative chemistry
### Bond Angles
##### Stage: 5 Challenge Level:
Think about the bond angles occurring in a simple tetrahedral molecule and ammonia.
### Stereoisomers
##### Stage: 5 Challenge Level:
Put your visualisation skills to the test by seeing which of these molecules can be rotated onto each other.
### Gassy Information
##### Stage: 5 Challenge Level:
Do each of these scenarios allow you fully to deduce the required facts about the reactants?
### Coordinated Crystals
##### Stage: 5 Challenge Level:
Explore the lattice and vector structure of this crystal.
### Stemnrich - Applied Mathematics
##### Stage: 3 and 4 Challenge Level:
This is the area of the stemNRICH site devoted to the core applied mathematics underlying the sciences. | 1,690 | 7,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-30 | longest | en | 0.820717 |
https://metanumbers.com/1666667 | 1,720,877,452,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00811.warc.gz | 336,148,716 | 7,531 | # 1666667 (number)
1666667 is an odd seven-digits composite number following 1666666 and preceding 1666668. In scientific notation, it is written as 1.666667 × 106. The sum of its digits is 38. It has a total of 2 prime factors and 4 positive divisors. There are 1,631,160 positive integers (up to 1666667) that are relatively prime to 1666667.
## Basic properties
• Is Prime? no
• Number parity odd
• Number length 7
• Sum of Digits 38
• Digital Root 2
## Name
Name one million six hundred sixty-six thousand six hundred sixty-seven
## Notation
Scientific notation 1.666667 × 106 1.666667 × 106
## Prime Factorization of 1666667
Prime Factorization 47 × 35461
Composite number
Distinct Factors Total Factors Radical ω 2 Total number of distinct prime factors Ω 2 Total number of prime factors rad 1666667 Product of the distinct prime numbers λ 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1666667 is 47 × 35461. Since it has a total of 2 prime factors, 1666667 is a composite number.
## Divisors of 1666667
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ 4 Total number of the positive divisors of n σ 1.70218e+06 Sum of all the positive divisors of n s 35509 Sum of the proper positive divisors of n A 425544 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 1290.99 Returns the nth root of the product of n divisors H 3.91656 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1666667 can be divided by 4 positive divisors (out of which none is even, and 4 are odd). The sum of these divisors (counting 1666667) is 1702176, the average is 425544.
## Other Arithmetic Functions (n = 1666667)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 1631160 Total number of positive integers not greater than n that are coprime to n λ 815580 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 125540 Total number of primes less than or equal to n r2 0 The number of ways n can be represented as the sum of 2 squares
There are 1,631,160 positive integers (less than 1666667) that are coprime with 1666667. And there are approximately 125,540 prime numbers less than or equal to 1666667.
## Divisibility of 1666667
m n mod m
2 1
3 2
4 3
5 2
6 5
7 2
8 3
9 2
1666667 is not divisible by any number less than or equal to 9.
## Classification of 1666667
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion 1666667
Base System Value
2 Binary 110010110111001101011
3 Ternary 10010200020102
4 Quaternary 12112321223
5 Quinary 411313132
6 Senary 55420015
8 Octal 6267153
10 Decimal 1666667
12 Duodecimal 68460b
20 Vigesimal a86d7
36 Base36 zq0b
## Basic calculations (n = 1666667)
### Multiplication
n×y
n×2 3333334 5000001 6666668 8333335
### Division
n÷y
n÷2 833334 555556 416667 333333
### Exponentiation
ny
n2 2777778888889 4629632407407962963 7716055555557407407654321 12860095164614197531893004218107
### Nth Root
y√n
2√n 1290.99 118.563 35.9304 17.5537
## 1666667 as geometric shapes
### Circle
Diameter 3.33333e+06 1.0472e+07 8.72665e+12
### Sphere
Volume 1.93926e+19 3.49066e+13 1.0472e+07
### Square
Length = n
Perimeter 6.66667e+06 2.77778e+12 2.35702e+06
### Cube
Length = n
Surface area 1.66667e+13 4.62963e+18 2.88675e+06
### Equilateral Triangle
Length = n
Perimeter 5e+06 1.20281e+12 1.44338e+06
### Triangular Pyramid
Length = n
Surface area 4.81125e+12 5.45607e+17 1.36083e+06
## Cryptographic Hash Functions
md5 0ec127d8815e03e75749257595257e7c e0d50714b38fc0eca41014fb677d13df9645e10b b06bbeaff3250b78b1e39c680834a9650f024a8e6005b875a619bbeaf6b5d0b8 2ef4a5e8ef739b8c95e8fc1996c76e13fd488b9040be45e815fb9dc532c37e630a46da8f364d407bc59852bfe466d210b8ec36ceb417a0c8d59a339edaf07a9c 923667801a3a946a0899edff96c269b302972b08 | 1,472 | 4,206 | {"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} | 3.40625 | 3 | CC-MAIN-2024-30 | latest | en | 0.748276 |
http://mathhelpforum.com/algebra/53359-solve-b.html | 1,480,796,026,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541134.5/warc/CC-MAIN-20161202170901-00213-ip-10-31-129-80.ec2.internal.warc.gz | 169,653,933 | 9,787 | 1. ## solve A&B
hi guys help me to solve the equations..
x1 = 0.65A*(1-B)
X2 = 0.8 A*(1-B)
x3 = 1*A*(1-B)
2. Originally Posted by sanmo4
hi guys help me to solve the equations..
x1 = 0.65A*(1-B)
X2 = 0.8 A*(1-B)
x3 = 1*A*(1-B)
more information is needed. what do you want to solve for? A and B separately? do you want to solve for them in terms of x1, x2, and x3?
3. i want to solve for A and B separately. u can take X1, X2 and X3 any value.
okay i will keep as
x1 = 1000
x2 = 2000
x3 = 3000 | 193 | 499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2016-50 | longest | en | 0.927917 |
http://forum.onlineconversion.com/showthread.php?p=69696 | 1,369,134,585,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699899882/warc/CC-MAIN-20130516102459-00018-ip-10-60-113-184.ec2.internal.warc.gz | 101,223,084 | 8,583 | Welcome to OnlineConversion.com Forums
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Convert and Calculate Post any conversion related questions and discussions here. If you're having trouble converting something, this is where you should post.* Guest Posting is allowed.
#1
04-10-2010, 01:45 PM
irenairena78 Junior Member Join Date: Apr 2010 Posts: 4 Rep Power: 0
conversion
Can someone convert me mg/l to mmol/l.Please give me the answer as soon as possible
Thank you
#2
04-10-2010, 04:02 PM
JohnS Double Ultimate Supreme Member Join Date: Dec 2007 Location: SE Michigan, USA Posts: 8,705 Rep Power: 17
Re: conversion
Quote:
Originally Posted by irenairena78 Can someone convert me mg/l to mmol/l.Please give me the answer as soon as possible Thank you
Sure, just divide by the molar mass in grams per mole. You have to know the substance and either look up its molecular weight or calculate it from chemical formula. The molecular weight is numerically equal to the molar mass in grams per mole. In general, it is unique to the substance.
#3
04-11-2010, 03:47 AM
irenairena78 Junior Member Join Date: Apr 2010 Posts: 4 Rep Power: 0
Re: conversion
Thank you!
So,if I need to convert 220 mg/l CaCo3 to mmol/l it is gonna be 220mg/l * 100 g/mol*1g/1000mg.Is it ok my calculation?
#4
04-11-2010, 03:51 AM
irenairena78 Junior Member Join Date: Apr 2010 Posts: 4 Rep Power: 0
Re: conversion
Sorry,as i can see i have to divide 220 mg/l with molar mass.So,220mg/l/100 g/mol*1g/1000mg cause i need mg.
#5
04-11-2010, 01:40 PM
Mrs X can't count, can't spell! Join Date: Feb 2006 Location: New Zealand Posts: 2,294 Rep Power: 11
Re: conversion
Quote:
Originally Posted by irenairena78 Sorry,as i can see i have to divide 220 mg/l with molar mass.So,220mg/l/100 g/mol*1g/1000mg cause i need mg.
220mg/L CaCO3 = 2.2mmol/L
For CaCO3, there are 100.1g/L for a 1M solution, there are 100.1mg/L for a 1mM solution.
#6
04-11-2010, 11:10 PM
irenairena78 Junior Member Join Date: Apr 2010 Posts: 4 Rep Power: 0
Re: conversion
Thank you Mrs X
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All times are GMT -8. The time now is 03:09 AM. | 737 | 2,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.839381 |
https://too-meta.neocities.org/anki/inf-ml-ai/1552960400719/back/ | 1,679,659,504,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945282.33/warc/CC-MAIN-20230324113500-20230324143500-00453.warc.gz | 670,726,523 | 3,390 | Show Question
What is the derivative of negative log MLE (MLE used as a negative cost) when the variables are passed through softmax activations?
The derivative of the cost function is needed for back-propagation.
When the output layer uses a MLE/cross-entropy cost and softmax activation, the derivative of the cost function combined with the activation function simplifies to a very simple expression. This note covers this computation.
Consider the last layer of a neural network shown below:
For the given training example $$x_1$$, the correct category is 2 (out of all categories 0,1 and 2). Only the output corresponding to category 2 is used in the calculation of the loss contributed by the example. The loss is given by:
$Loss_{ex_1} = -ln(p_3) = -ln(\frac{e^{f_3}}{e^{f_1} + e^{f_2} + e^{f_3}})$
While only the probability for the correct category is used to calculate the loss for the example, the other outputs, $$p_1$$ and $$p_2$$ still affect the loss, as $$p_3$$ depends on these values (all sum to 1).
The partial derivatives will be:
\begin{align*} \frac{\partial L}{\partial f_1} &= p_1 \\ \frac{\partial L}{\partial f_2} &= p_2 \\ \frac{\partial L}{\partial f_3} &= -(1-p_3) = p_3 - 1\end{align*}
So if we have output probabilities [0.1, 0.3, 0.6], then the derivative of the loss with respect to last layer's output before applying the softmax activation will be [0.1, 0.3, -0.4].
### Derivations
Formulating the derivative of loss wrt different variables is simplified if the log and exponential are canceled at the beginning:
\begin{align*} Loss_{ex_1} = -ln(p_3) &= -ln(\frac{e^{f_3}}{e^{f_1} + e^{f_2} + e^{f_3}}) \\ &= ln(\frac{e^{f_1} + e^{f_2} + e^{f_3}}{e^{f_3}}) \\ &= ln(e^{f_1} + e^{f_2} + e^{f_3}) - ln(e^{f_3}) \\ &= ln(e^{f_1} + e^{f_2} + e^{f_3}) - f_3 \end{align*}
From here, the partial derivates are obvious. For example:
\begin{align*} \frac{\partial L}{\partial f_1} &= \frac{\partial}{\partial f_1}(ln(e^{f_1} + e^{f_2} + e^{f_3}) - f_3) \\ &= \frac{\partial}{\partial x}(ln(x)) \frac{\partial}{\partial f_1}(e^{f_1} + e^{f_2} + e^{f_3}) \\ &= \frac{1}{x} e^{f_1} \\ &= \frac{e^{f_1}}{e^{f_1} + e^{f_2} + e^{f_3}} \\ &= p_1 \end{align*}
The below image is part of a separate card, but I'm including it here, as it is quite relevant.
Old image: | 768 | 2,300 | {"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": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | longest | en | 0.710922 |
https://www.studentmovementusa.org/is-linear-algebra-very-hard/ | 1,685,645,220,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648000.54/warc/CC-MAIN-20230601175345-20230601205345-00144.warc.gz | 1,081,654,181 | 17,137 | # Is Linear Algebra very hard?
Is Linear Algebra Very Hard? – Tips and Tricks for Students
# Is Linear Algebra Very Hard? – Tips and Tricks for Students
Linear Algebra Discrete Mathematics
Level Second Year First Year
Skills Required Robust Reasoning, Analytical Skills Logical Thinking, Problem Solving
Focus Linear Equations, Matrices, Vector Spaces Combinatorics, Graph Theory, Number Theory
Applications Physics, Engineering, Computer Science Cryptography, Algorithms, Optimization
## FAQs About Linear Algebra Difficulty
### 1. What is linear algebra?
Linear algebra is a branch of mathematics that deals with linear equations, matrices, and vector spaces. It is an important subject in many fields of science, including physics, engineering, and computer science.
### 2. Why is linear algebra considered difficult?
Many students regard linear algebra as a difficult study due to its abstract nature and the fact that it requires robust reasoning and analytical skills. Linear algebra is more challenging than discrete mathematics, which is usually a first-year program taught in most STEM majors.
### 3. What are the main topics covered in linear algebra?
The main topics covered in linear algebra include linear equations, matrices, determinants, vector spaces, linear transformations, eigenvalues, and eigenvectors. These topics form the foundation of linear algebra and are used to solve various problems in science and engineering.
### 4. How can I improve my understanding of linear algebra?
Here are some tips to improve your understanding of linear algebra:
• Practice solving problems: Linear algebra requires a lot of practice. Try solving as many problems as possible to develop your analytical skills.
• Focus on understanding the concepts: Instead of memorizing formulas, try to understand the underlying concepts of linear algebra. This will make it easier to apply the concepts to new problems.
• Use visual aids: Linear algebra involves a lot of matrices and vectors, which can be difficult to visualize. Use graphic software or online resources to help you understand the geometric interpretations of linear algebra.
• Collaborate with others: Study in groups or join an online forum to discuss linear algebra concepts and problems. This will help you gain different perspectives and learn from others.
### 5. How important is linear algebra?
Linear algebra is an important subject in many fields of science and engineering. It provides a framework for solving a variety of problems, including those related to physics, computer graphics, cryptography, and machine learning.
## Conclusion
Overall, linear algebra is considered a difficult subject by many students due to its abstract nature and the analytical skills required to understand it. However, with enough practice and a focus on understanding the concepts, it is possible to master this subject and apply it to solve real-world problems in various fields of science and engineering.
Sources: | 555 | 2,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-23 | latest | en | 0.907945 |
https://www.reference.com/web?q=Centimeter%20to%20Millimeter&qo=pagination&o=600605&l=dir&sga=1&qsrc=998&page=3 | 1,632,028,749,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00070.warc.gz | 978,675,811 | 27,176 | Related Search
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quizlet.com/132557858/centimeter-and-millimeter-conversion-flash-cards
Start studying Centimeter and Millimeter Conversion. Learn vocabulary, terms, and more with flashcards, games, and other study tools. | 544 | 2,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-39 | latest | en | 0.770633 |
https://era.library.ualberta.ca/search?direction=asc&facets%5Ball_contributors_sim%5D%5B%5D=Kouritzin%2C+Michael&facets%5Bitem_type_with_status_sim%5D%5B%5D=article_published&facets%5Bmember_of_paths_dpsim%5D%5B%5D=e06d8668-5460-47ba-a842-ba7aa1ec4021%2F4c68e92a-605c-400d-8ab2-1a0faac571a3&facets%5Bmember_of_paths_dpsim%5D%5B%5D=e06d8668-5460-47ba-a842-ba7aa1ec4021&sort=sort_year | 1,659,889,415,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00491.warc.gz | 251,367,250 | 11,156 | # Search
• ### Rates of convergence in a central limit theorem for stochastic processes defined by differential equations with a small parameter
1992
Let μ be a positive finite Borel measure on the real line R. For t ≥ 0 let et · E1 and E2 denote, respectively, the linear spans in L2(R, μ) of {eisx, s > t} and {eisx, s < 0}. Let θ: R → C such that ∥θ∥ = 1, denote by αt(θ, μ) the angle between θ · et · E1 and E2. The problems considered here...
• ### A law of the iterated logarithm for stochastic processes defined by differential equations with a small parameter
1994
Consider the following random ordinary differential equation: X˙ϵ(τ)=F(Xϵ(τ),τ/ϵ,ω)subject toXϵ(0)=x0, where {F(x,t,ω),t≥0} are stochastic processes indexed by x in Rd, and the dependence on x is sufficiently regular to ensure that the equation has a unique solution Xϵ(τ,ω) over the interval...
• ### Strong Convergence in the Stochastic Averaging Principle.
1994
In this note we consider the almost sure convergence (as ϵ→0) of solution Xϵ(·), defined over the interval 0 ≤ τ ≤ 1, of the random ordinary differential equation View the MathML source Here {F(x, t, ω), t ≥ 0} is a strong mixing process for each x and (x, t) → F(x, t, ω) is subject to regularity...
• ### Strong approximation for cross-covariances of linear variables with long-range dependence.
1995
Suppose {εk, −∞ < k < ∞} is an independent, not necessarily identically distributed sequence of random variables, and {cj}∞j=0, {dj}∞j=0 are sequences of real numbers such that Σjc2j < ∞, Σjd2j < ∞. Then, under appropriate moment conditions on {εk, −∞ < k < ∞}, View the MathML source, View the...
• ### Invariance Principles for Parabolic Equations with Random Coefficients
1997
A general Hilbert-space-based stochastic averaging theory is brought forth herein for arbitrary-order parabolic equations with (possibly long range dependent) random coefficients. We use regularity conditions onView the MathML sourcewhich are slightly stronger than those required to prove...
• ### Averaging for Fundamental Solutions of Parabolic Equations.
1997
Herein, an averaging theory for the solutions to Cauchy initial value problems of arbitrary order,ε-dependent parabolic partial differential equations is developed. Indeed, by directly developing bounds between the derivatives of the fundamental solution to such an equation and derivatives of the...
• ### Convergence of Markov chain approximations to stochastic reaction diffusion equations.
2002
In the context of simulating the transport of a chemical or bacterial contaminant through a moving sheet of water, we extend a well-established method of approximating reaction-diffusion equations with Markov chains by allowing convection, certain Poisson measure driving sources and a larger...
• ### Markov chain approximations to filtering equations for reflecting diffusion processes.
2004
Herein, we consider direct Markov chain approximations to the Duncan–Mortensen–Zakai equations for nonlinear filtering problems on regular, bounded domains. For clarity of presentation, we restrict our attention to reflecting diffusion signals with symmetrizable generators. Our Markov chains are...
• ### Rates for branching particle approximations of continuous discrete filters.
2005
Herein, we analyze an efficient branching particle method for asymptotic solutions to a class of continuous-discrete filtering problems. Suppose that t→Xt is a Markov process and we wish to calculate the measure-valued process t→μt(⋅)≐P{Xt∈⋅|σ{Ytk, tk≤t}}, where tk=kɛ and Ytk is a distorted,...
• ### On extending classical filtering equations.
2008
In this paper, we give a direct derivation of the Duncan–Mortensen–Zakai filtering equation, without assuming right continuity of the signal, nor its filtration, and without the usual finite energy condition. As a consequence, the Fujisaki–Kallianpur–Kunita equation is also derived. Our results...
1 - 10 of 13 | 958 | 3,958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.825404 |
http://serc.carleton.edu/eslabs/hurricanes/5.html | 1,418,961,627,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768205.146/warc/CC-MAIN-20141217075248-00125-ip-10-231-17-201.ec2.internal.warc.gz | 247,920,591 | 13,004 | Hurricanes > Lab 5: All About Air Pressure
# All About Air Pressure
For this lab, you and your team will be moving through a series of stations completing hands-on experiments. These experiments will help you understand the basic principles that describe how air pressure responds to and is responsible for various phenomena. As you work through the activities, keep in mind that your eventual goal will be to understand the connection between air pressure and hurricanes.
At each station, read the procedure and use the materials to perform the experiment or demonstration.
## Station 1: Air pressure on a pop can
#### Materials
• Aluminum pop/soda can
• Hot plate or Bunsen burner and striker
• Tongs or claw holder (the kind that can be attached to ring stands)
• Bucket of cold water
• Oven mitt (or equivalent) and safety glasses
• Metric rulers
#### Procedure
1. Measure and record the height and diameter of the can.
2. Put a small amount of water in the bottom of the can, just enough to cover the bottom.
3. Wearing goggles and an oven mitt, place the can on the hot plate or use the tongs to hold the can over the heat source. Do so until there is a good chimney of steam coming out the opening in the can. This might take a couple of minutes.
4. Quickly invert the can into the bucket of cold water and watch the results.
5. Complete the Stop and Think questions.
## Stop and Think
1. Follow these three steps to calculate the total force that air pressure is exerting on the can:
• Use the chart to find the average barometric pressure at your altitude. This value is the pressure.
• Use the dimensions of the can to calculate its approximate surface area.
Surface Area = 2π (radius) (height) + π (radius)2
• Multiply the pressure by the can's surface area.
2. If all that force is being exerted on the outside of the can all the time, why didn't it collapse before you put it in the cold water?
## Station 2: Balloon in a Bell Jar
#### Materials
• Bell jar
• Vacuum pump
• 2 Small balloons, partially inflated to the same size.
#### Procedure
1. Tape one of the balloons to the top inside of the bell jar. Leave the other on the outside for comparison. (This step won't be necessary for subsequent groups.)
2. Connect the vacuum pump and evacuate some air out of the bell jar.
3. Complete questions on activity sheet.
4. Release the vacuum so that the apparatus is ready for the next group.
## Stop and Think
3. What happened to the balloon as you pumped air out of the jar?
4. Brainstorm an explanation for why this happened.
## Station 3: Ruler and Newspaper
#### Materials
• Sheets of newspaper
• Wooden rulers or flat pieces of wood
#### Procedure
Part A
1. Place the ruler on a bench top with about a quarter of its length hanging over the edge.
2. Make sure the area around you is clear of people, then give the overhanging piece a quick "karate chop" with your hand.
3. Retrieve the ruler and replace it in the same position on the bench.
4. Part B
5. Lay one full sheet of newspaper over the part of the ruler that is on the bench.
6. Repeat your chop to the overhanging part of the ruler.
7. Record your observations and answer the questions on the activity sheet.
## Stop and Think
5. What did you expect to happen during each part of the activity?
6. Is it the weight of the newspaper that causes the difference? Back up your answer with your observations of the experiment and materials.
## Station 4: Egg in a Bottle
#### Materials
• Hard boiled eggs (with shells removed).
• A bottle or flask with an opening that is just small enough to prevent the egg from entering the bottle. Each group should have their own bottle as one of the students will have to put it to their mouth.
• Matches
#### Procedure
Part A
1. Drop a burning match into the bottom of the bottle.
2. After a few seconds, place the egg onto the mouth of the bottle.
3. Compete the relevant areas on the activity sheet.
4. Part B
5. Now that the egg is in the bottle, turn the bottle upside down so that the egg is resting in the neck of the bottle.
6. Tip back your head, place your mouth over the bottle opening and blow vigorously into the bottle.
7. Quickly remove your lips from the bottle hold it over the bench.
8. Complete this section of the activity sheet.
9. Cleanup
10. Wash the bottle with soap and hot water so that it's ready for the next class.
## Stop and Think
7. What causes the egg to be pushed into the bottle? What does the match have to do with it?
8. Why does blowing into the bottle cause the egg to pop back out?
## Station 5: Soda Bottle and Ping Pong Ball Barometer
Based on work by Patrick Cooney of Millersville University
#### Materials
• Soda bottle (Sobe 20oz bottles work well)
• 1 Ping pong ball
• Metric ruler
• Beaker to collect water
• A barometer (ideally, one that reads in kPa; if not available, convert pressure units to kPa)
#### Procedure
Image of the final state of Station 5. The ping pong ball is being held in place by the balanced forces inside and outside the bottle. Image by John McDaris
1. Obtain a reading of atmospheric pressure from the barometer at the station. Record this on the activity sheet.
2. Fill the bottle all the way to the top with water.
3. Push the ping pong ball onto the top to squeeze out a small amount of water.
4. Now, pour about one third of the water from the bottle into the graduated cylinder. Measure and record the amount of water in the graduated cylinder as V0 (pronounced "vee sub zero").
5. Hold the ping pong ball on top of the bottle and invert it over the beaker. Hold the ball loosely against the opening so that some water is allowed to leak past the ball into the beaker. Don't jiggle or rotate the ball during this process.
6. Eventually, enough water will leak out that the pressure on both sides of the ball will be the sameyou can take your hand away and the ball will stay in place. Add the water that leaked out into the graduated cylinder. Measure and record this total amount as V1 (pronounced "vee sub one") on the activity sheet.
7. Measure the distance from the mouth of the bottle to the top of the water it encloses and record this distance as D on the activity sheet.
8. Complete the calculations called for on your activity sheet to determine the atmospheric pressure in your classroom. Compare your calculated value to the reading you took off the barometer and answer the questions on the activity sheet.
When you turn the bottle upside down and allow a little bit of water to leak out, you are increasing the "empty" space inside the bottle without letting in any more air. In other words, the same amount of air that was in the bottle before now has more room to "spread out" and the pressure of the air in the bottle is lower than the pressure of the air outside the bottle.
When you reach the point where the ball stays in place without you needing to hold it, the pressure on both sides of the ball is the same. This fact allows us to do a few calculations and figure out what the atmospheric pressure is. Calculation of Atmospheric Pressure: Because the pressure is the same on both sides of the ball after the water has leaked out, that means that: P1 + ρgD = Atmospheric Pressure Where P1 ("P sub 1") is the air pressure in the bottle, ρ (rho – pronounced like "row") is the density of water, g is the acceleration due to gravity, and D is the height of water in the bottle after some has leaked out (as shown above).
The term ρgD represents the pressure that the water exerts on the ball. We also know that, originally, the pressure inside the bottle was equal to atmospheric pressure. So, now we can say that: P1 + ρgD = P0 Since air is nearly an ideal gas, we can make the approximation that P1 V1 = P0 V0 and that means that P1 = (P0 V0)/V1. If we put this result in the equation above and solve for P0, we get a relation that will allow us to calculate what the atmospheric pressure is. P0 = ρgD [V1 / (V1 – V0)]
## Stop and Think
9. Substitute your measurements into the following equation and calculate P0, the atmospheric pressure. Show your work.
P0 = ρgD [V1 / (V1 – V0)]
10. Compare your result to the atmospheric pressure you read off the barometer at the beginning. Are they the same? What are some factors that could affect how closely your result matches the measurement?
## Station 6: News Article Review
#### Materials
Copies of these news articles:
#### Procedure
1. Everyone in the group should pick an article to read. Everyone should take a different one unless there are more group members than articles.
2. Spend the first few minutes reading your article and then write a paragraph summary (on your activity sheet) of what the main points of the article were and what you learned from it.
3. When everyone is finished, each person should spend 1 minute telling the rest of the group about the article and fielding any questions their group-mates might have about the material.
4. Keep an eye on the time so that everyone gets a chance to share what they learned!
5. In your own words, write a couple of sentences based on the summary that your group-mates give of their articles.
6. Leave the articles for the other groups to use when you move on to your next lab station.
## Extending Your Learning
To learn more about the connections between air pressure and hurricanes, check out the additional resources listed below. You'll also be working with air pressure in Lab 6: Why Keep and Eye on the Barometer?
Pressure differences get the wind going
This short page from USA Today talks explicitly about how differences in atmospheric pressure between places creates wind as air moves from areas of high pressure to areas of low pressure.
Global Atmospheric Sea Level Pressure during Hurricane Frances
This visualization from the NASA Scientific Visualization Studio shows the global sea level atmospheric pressure for a period of 4 days in September 2004. This is the period when Hurricane Frances was approaching the East Coast of North America and Typhoon Songda was spinning in the western Pacific Ocean.
## Checking In Questions
• Can you pick the two storms out in the visualization? What do they look like?
Hurricane Frances and Typhoon Songda show up as dark blue circles on the general green background indicating that they have lower pressure inside them than the surrounding atmosphere.
NDBC Hurricane Information Page
This page from the National Data Buoy Center talks about hurricanes and provides data collected from a buoy during Hurricane Bertha in 1996. There is a map showing where the buoy was located and a graph of several types of data that it gathered. Pay special attention to the red line on the graph - this represents the barometric (atmospheric) pressure measured by the buoy as the storm passed over it. | 2,402 | 10,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | longest | en | 0.904727 |
https://stats.stackexchange.com/questions/461366/probability-of-a-given-b-or-c | 1,723,182,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00249.warc.gz | 425,587,924 | 41,464 | # Probability of A given B or C
I know some formulae related to conditional probabilities of events conditional on intersection of two events.
But I have been unable to find any formula for the case where the condition is union of two or more events.
That is, how can I find
$$P(A | B\text{ or }C)$$
And in general
$$P(A | B\text{ or }C\text{ or }\dots\text{ or }X)$$
And what can be the continuous analog?
$$p(x | y \in [a,b])$$
When I already know $$p(x | y=t), \forall t \in [a,b]$$
Edit: An approach I tried
$$P(A |B\text{ or }C) = \frac{P ((A\text{ and }B)\text{ or }(A\text{ and }C))}{P(B\text{ or }C)}$$
$$=\frac {P(A,B)+P(A,C)-P(A,B,C)} {P(B\text{ or }C)}$$
For the continuous case I asked, I assume that $$y$$ can take only one value at a time, so the $$P(A,B,C)$$ term can be ignored.
I have not tried a derivation, but just by analogy, I guess the continuous expression would be
$$p(x|y \in [a,b]) = \frac { \int_a^b p(x|y) p(y) dy} {P(y \in [a,b])}$$
Is the approach for the discrete case correct?
I have just used analogy for the continuous case, is the formula I guessed correct? If yes, how to prove that? If not, what is the correct formula?
• just define the event $D = B$ union $C$ and then calculate $P(A|D )$. Commented Apr 19, 2020 at 3:48
• I have edited the question and added an approach I tried, please comment on the continuous part. And I am sorry, I tried several times but I was unable to fix the formatting error. Commented Apr 19, 2020 at 16:47
• Isn't this just a logical follow on from some simpler rules? As in the probability of B union C is P(B) + P(C) - P(B intersection C), and for a sequence of events, that is the union of this result and the next possible event, applied as many times as necessary. Commented Apr 26, 2020 at 21:38
1) $$P(A | B \text{ or } C)=P(A|B\cup C)=\frac{P(A\cap(B\cup C))}{P(B\cup C)}$$
2)$$P(A | B \text{ or } C \text{ or } \dots \color{red}{\text{or }X})$$
what is $$B \text{ or } X$$? If $$X$$ is a random variable, I think it is only valid if we use it like $$B\cup \{X\in E\}=\{\omega \in \Omega \mid \omega \in B \text{ or } x(\omega)\in E\}$$.
so $$P(A | B \text{ or } C \text{ or } \cdots \text{ or } \{X \in E\})$$ can be easily calculated by defining $$D=B \cup C \cup \cdots \cup \{X \in E\}$$.
3) $$P(X= x|Y\in [a,b])$$ for the case $$Y$$ is a continues random variable You can easily calculate it if you knowing $$P(X\leq x|Y\in [a,b])$$.
$$P(X\leq x|Y\in [a,b])=P(\{X \leq x\}|\{Y\in [a,b]\})=\frac{P(\{X \leq x\} \cap \{Y\in [a,b]\})}{P(\{Y\in [a,b]\})}=\frac{\int_{-\infty}^{x}\int_{y\in [a,b]}f_{(X,Y)}(t , y) dy dt}{P(\{Y\in [a,b]\})}=\frac{\int_{-\infty}^{x} \int_{y\in [a,b]}p(t | y)p(y) dy dt}{P(\{Y\in [a,b]\})}$$.
• Thank you for the answer. Commented Apr 27, 2020 at 16:27
• You are welcome! Commented Apr 27, 2020 at 16:29
• And regarding the second point you raised, I am sorry if I used confusing (or wrong) notation, by X, I simply meant another event and not a random variable when I wrote $P(A|B or C or ..... or X)$, just like B and C, and not a random variable. Commented Apr 27, 2020 at 16:30 | 1,095 | 3,115 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-33 | latest | en | 0.882609 |
http://fractionslearningpathways.ca/tasks/curriculumConnectionsOpE | 1,585,933,203,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370515113.54/warc/CC-MAIN-20200403154746-20200403184746-00229.warc.gz | 65,494,931 | 4,113 | Operation E
# Curriculum Connections
OP E Add and subtract fractions with unlike denominators (e.g., 2 and 7) using models and symbols
7
• use a variety of mental strategies to solve problems involving the addition and subtraction of fractions and decimals;
7
• add and subtract fractions with simple like and unlike denominators, using a variety of tools and algorithms;
8
• use estimation when solving problems involving operations with whole numbers, decimals, percents, integers, and fractions, to help judge the reasonableness of a solution;
8
• solve problems involving addition, subtraction, multiplication, and division with simple fractions.
9D
• simplify numerical expressions involving integers and rational numbers, with and without the use of technology;
9D
• solve problems requiring the manipulation of expressions arising from applications of percent, ratio, rate, and proportion;
9P
• solve problems requiring the expression of percents, fractions, and decimals in their equivalent forms
9P
• simplify numerical expressions involving integers and rational numbers, with and without the use of technology;* | 220 | 1,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-16 | latest | en | 0.878356 |
http://isabelle.systems/zulip-archive/stream/238552-Beginner-Questions/topic/About.20the.20code.20generation-.20Wellsortedness.20error.html | 1,709,030,892,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00882.warc.gz | 17,149,956 | 2,468 | Stream: Beginner Questions
Topic: About the code generation- Wellsortedness error
Hongjian Jiang (Jul 27 2023 at 13:47):
When I run my example, I meet below hints:
``````Wellsortedness error:
Type nat not of sort {enum,equal,order}
No type arity nat :: enum
``````
I am curious about the principal behind this.
The part of snippet could be:
``````definition check_eqv :: "'a::order rexp ⇒ 'a rexp ⇒ 'a list ⇒ bool" where
"check_eqv r s as =
(let nr = norm r; ns = norm s
in case closure as (nr, ns) of
Some([],_) ⇒ True | _ ⇒ False)"
fun f :: "nat ⇒ bool" where
"f n = (n < 3)"
lemma "check_eqv (Plus One (Times ((Sym (Predicate.Pred f))) (Star (Sym (Predicate.Pred f))))) (Star(Sym (Predicate.Pred f))) [1,2,3,4]"
by eval
``````
Any helps would be appreciated.
Mathias Fleury (Jul 27 2023 at 13:50):
Well the error message basically says that the code generator is not able to generate code, because the only way is to enumerate all solutions…
Hongjian Jiang (Jul 27 2023 at 13:51):
how to improve such case
Mathias Fleury (Jul 27 2023 at 13:55):
by looking at piece of the code, find out exactly which part is not working (I assume it is closure but I am not sure). Once you know that, you have to replace the code by something executable… (See https://isabelle.in.tum.de/dist/Isabelle2022/doc/codegen.pdf)
Hongjian Jiang (Jul 28 2023 at 13:45):
Hi, Mathisas. I thought I found the problem. I use the set comprehension to calculate the value.
Like`value "{a. f a}"`
Hongjian Jiang (Jul 28 2023 at 13:46):
But it is mandatory in my example I think.
Yong Kiam (Jul 31 2023 at 00:57):
like Mathias said, if you must use sets, you would have to provide rewrites for that in a more "executable" way that doesn't require searching over all numbers `a`
Last updated: Feb 27 2024 at 08:17 UTC | 532 | 1,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-10 | latest | en | 0.833138 |
http://www.sosmath.com/algebra/fraction/frac3/frac34/frac342/frac34214/frac34214.html | 1,542,184,164,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741764.35/warc/CC-MAIN-20181114082713-20181114104713-00541.warc.gz | 408,889,568 | 2,505 | # MULTIPLICATION: Problem 4
Rule 12: To multiply a whole number and a fraction, complete the following steps.
1.
Convert the whole number to a fraction. (See Rule 8)
2.
Multiply the numerators.
3.
Multiply the denominators.
4.
Reduce the results. (See Rule 10)
a.
Factor the product of the numerators.
b.
Factor the product of the denominators.
c.
Look for the fractions that have a value of 1.
Problem 4: Reduce .
Solution. Factor the numerator, factor the denominator, and locate the fraction(s) that has(have) a value of 1.
Although we have reduced the problem, it can be reduced further. Follow the same steps
You can check the answer with your calculator. If 72 divided by 90 yields the same answer as 4 divided by 5, is equivalent to .
[Previous Problem] [Next Problem] [Menu Back] | 198 | 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} | 3.765625 | 4 | CC-MAIN-2018-47 | latest | en | 0.866854 |
http://www.velocityreviews.com/forums/t956979-looking-for-advice-python.html | 1,394,747,507,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678682243/warc/CC-MAIN-20140313024442-00069-ip-10-183-142-35.ec2.internal.warc.gz | 589,823,857 | 9,178 | Velocity Reviews > looking for advice python
twiztidtrees@gmail.com
Guest
Posts: n/a
01-26-2013
Hey I'm new to programming and I have been working on calculating miles per gallon. iv posted below what I have and constructive criticism would be wonderful. Thanks
#This is a program used to calculate miles per gallon
#variable used to gather miles driven
string_miles = input('How many miles did you drive?')
#variable used to gather the amount of gallons used
string_gas = input('How many gallons of gas did you use?')
#used to convert the miles input
miles = int(string_miles)
#used to convert the gas input
gas = int(string_gas)
#used to calculate mpg through division
mpg = miles/gas
print(float(string_miles))
print(float(string_gas))
print('Your miles per gallon is', format(mpg,'.2f'))
Chris Angelico
Guest
Posts: n/a
01-26-2013
On Sun, Jan 27, 2013 at 9:26 AM, <(E-Mail Removed)> wrote:
> miles = int(string_miles)
> gas = int(string_gas)
>
> #used to calculate mpg through division
> mpg = miles/gas
>
> print(float(string_miles))
> print(float(string_gas))
> print('Your miles per gallon is', format(mpg,'.2f'))
Welcome aboard!
You turn your inputs into integers, then display them as floats...
from the original strings. (Though I guess this is probably debugging
code?) I would advise against doing this; at very least, it's
confusing. I would recommend simply using floats everywhere, and thus
allowing non-integer inputs:
How many miles did you drive?60.9
How many gallons of gas did you use?11.9
Traceback (most recent call last):
File "123.py", line 11, in <module>
miles = int(string_miles)
ValueError: invalid literal for int() with base 10: '60.9'
Small additional point: It's common to put a space at the end of your
prompt, to avoid the "crammed" look of "drive?60.9".
Are there any particular areas that you'd like comments on?
ChrisA
Dave Angel
Guest
Posts: n/a
01-26-2013
On 01/26/2013 05:26 PM, http://www.velocityreviews.com/forums/(E-Mail Removed) wrote:
> Hey I'm new to programming and I have been working on calculating miles per gallon. iv posted below what I have and constructive criticism would be wonderful. Thanks
>
A good post for the python-tutor mailing list.
If you want help with a program, the first thing you should specify is
what version of Python you're using. And usually which OS you're
running, but in this case it doesn't matter much.
I don't see either a shebang line nor a coding line.
> #This is a program used to calculate miles per gallon
>
>
> #variable used to gather miles driven
> string_miles = input('How many miles did you drive?')
>
> #variable used to gather the amount of gallons used
> string_gas = input('How many gallons of gas did you use?')
>
Why do you bother to have separate variables for the string versions?
Why not just miles = int( input("xxxx")) ? For that matter, what if
the user enters a decimal value for the gallons? Perhaps you'd better
use gas = float( input("yyy") )
> #used to convert the miles input
> miles = int(string_miles)
>
> #used to convert the gas input
> gas = int(string_gas)
>
> #used to calculate mpg through division
> mpg = miles/gas
This isn't portable to Python 2.x. In 2.x, it would truncate the result.
>
> print(float(string_miles))
> print(float(string_gas))
> print('Your miles per gallon is', format(mpg,'.2f'))
>
What if the user enters something that isn't a valid number, either int
or float? Where's the try/catch to handle it?
Is this a class assignment? If not, why would you have a comment and a
blank line between every line of useful code?
--
DaveA
twiztidtrees@gmail.com
Guest
Posts: n/a
01-27-2013
I am in a class and was just looking for different advice. This is the first time iv ever tried to do this. That's all that iv taken from two chapters and wondering how bad I did. I also like to learn. Thanks for everyones input
twiztidtrees@gmail.com
Guest
Posts: n/a
01-28-2013
On Sunday, January 27, 2013 1:57:47 PM UTC-5, (E-Mail Removed) wrote:
> I am in a class and was just looking for different advice. This is the first time iv ever tried to do this. That's all that iv taken from two chapters and wondering how bad I did. I also like to learn. Thanks for everyones input
This is what I have now thanks for the advice. It did seem a lot easier this way, but any criticism would be nice thanks.
windows 7 and python 3.3.0
while True:
try:
miles = float(input("How many miles did you drive?"))
break
except ValueError:
print("That is not a valid number. Please try again.")
while True:
try:
gas = float(input("How many gallons of gas did you use?"))
break
except ValueError:
print("That is not a valid number. Please try again.")
mpg = miles/gas
print('Your miles per gallons is', format(mpg,'.2f')) | 1,247 | 4,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | latest | en | 0.851598 |
https://www.ukessays.com/essays/business/impact-of-quality-management-on-value-chain-performance-business-essay.php | 1,481,430,778,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544097.11/warc/CC-MAIN-20161202170904-00287-ip-10-31-129-80.ec2.internal.warc.gz | 1,017,136,944 | 14,634 | # Impact of Quality Management on Value Chain Performance
Published:
This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.
## 3.1 Method of Data collection
Questionnaire was used to collect the data. Internet was used to collect the data. Personal visits were also made for some questionnaires.
## 3.2 Sampling Technique
The technique for my study was convenience, because it was most suitable technique for me to collect the data.
## 3.3 Sample Size
The size of population for the study is 30.
## 3.4 Instrument of Data Collection
We used the questionnaire as an instrument to collect the data.
## 3.5 Research Model developed
Quality management
Information technology Value chain performance
Technological diversity
Customer service
## 3.6 Statistical techniques
Simple regression techniques were used for the statistical test through the Spss software.
## 4.1 Finding and interpretation of results
H1: Has a significant Quality management impact on value chain performance?
## TABLE 4.1
The above given table shows the model summary of the applied model. Enter method is used to developed the results of this study. The R square value shows .352 or 35% variation. It means that a one unit change in the independent variable (Quality management) is bringing 35% change in the dependent variable that is value chain performance.
## TABLE 4.2
The above given table shows that the model is applicable for test because of significant value of the model is less than 0.05. Our regression model is suitable to be applied for this test.
In the coefficient table we see the dependent variable and independent variable relationship. The significance value is below .05. We can observe from this table that the quality management has a significant impact on the value chain activities. The above results show the acceptance of our hypothesis. The beta value 3.970 shows that there is positive relation between the value chain activities and quality management.
## H2
Technological diversity and introduction of modern technology will have positive effect on the value chain performance
## TABLE 4.4
The above given table shows the model summary of the applied model. Enter method is used to developed the results of this test. The R square value shows .171 or 17% variation. It means that a one unit change in the independent variable (Technological diversity) is bringing 17.1% change in the dependent variable that is value chain performance.
## TABLE 4.5
. The above given table shows that the model is applicable for test because of significant value of the model is less than 0.05. Our regression model is suitable to be applied for this test
## TABLE 4.6
In the coefficient table we see the dependent variable and independent variable relationship. The significance value is below .05. We can observe from this table that the Technological diversity has a significant impact on the value chain activities. The above results show the acceptance of our hypothesis. The beta value 2.445 shows that there is positive relation between the value chain activities and quality management.
## H3:
Information technology is has significant impact on value chain performance
## TABLE 4.7
The R square value is .302 or 30.2%. It shows that there is 30.2% change in the dependent variable due to one unit change in the independent variable.
## TABLE 4.8
The above given table shows that the model is applicable for test because of significant value of the model is less than 0.05. Our regression model is suitable to be applied for this test.
## TABLE 4.9
.In the coefficient table we see the dependent variable and independent variable relationship.The significance value is below .05. We can observe from this table that the information technology has a significant impact on the value chain activities. The above results show the acceptance of our hypothesis. The beta value 3.539 shows that there is positive relation between the value chain activities and quality management.
## H4:
Customer service has significant impact on value chain performance
## TABLE 4.10
The above given table shows the model summary of the applied model. Enter method is used to developed the results of this test. The R square value shows .257 or 25.7% variation. It means that a one unit change in the independent variable (Customer service) is bringing 25.7% change in the dependent variable that is value chain performance.
## TABLE 4.11
The above given table shows that the model is applicable for test because of significant value of the model is less than 0.05. Our regression model is suitable to be applied for this test.
## TABLE 4.12
In the coefficient table we see the dependent variable and independent variable relationship.The significance value is below .05. We can observe from this table that the Customer service has a significant impact on the value chain activities. The above results show the acceptance of our hypothesis. The beta value 3.539 shows that there is positive relation between the value chain activities and quality management.
No.
## HYPOTHSIS
Beta
SIG.
RESULT
H0
Quality management has a significant impact on value chain performance
.593
.000
Accepted
H1
Technological diversity and introduction of modern technology will have positive effect on the value chain performance
. 413
.021
Accepted
H2
Information technology is has significant impact on value chain performance
.549
.497
Accepted
H3
Customer service has significant impact on value chain performance
.507
.004
Accepted
## 5.1 Discussion
This research project has provided illuminating guidance for improving value chain performance for the textile sector. The acceptance of our hypotheses informs us that the textile garment sector will have to focus on the deterministic parameters studied in this project viz. Quality management, customer service, Technological excellence and information technology.
It is clear that the culture of customer service has lagged behind in Pakistan. As a result, many manufacturers have not recognized the importance of quality. A few unscrupulous exporters burnt their hands as whole lots of material marked for exports were rejected due to poor quality. On occasions customers were provided high quality sample, while material supplied was of poor quality resulting in rejection and loss of reputation.
This kind of practices are gradually disappearing as internal pressure from exporting organizations, competitive pressures from global competitors have made close compliance to customer requirement essential to success.
The importance’s of high quality machinery to produce consistently high quality products and supply chain compatibility with the customers are also recognized as indispensable for the value chain. The textile exporters have invested in state of the art machinery and customer relationship management, supplier relationship management and enterprise resource planning software to help improve communications as well as supply chain efficiency.
The acceptance of our hypotheses shows that the textile sector has recognized the importance of these factors the value chain parameters assessed in this research project are being recognized by textile exporters as essential for business.
## 5.2 Conclusion:
In the conclusion, this study identified that quality management, customer service, technological diversity and information technology have a critical role in the value chain activities. Quality management which came up with the highest beta shows the high significance. Textile exporters must recognize quality as a critical determinant of the success in the competitive global business. Information integration, access to sophisticated production machinery and above all customer service dimension must be recognized as critical to their success.
## 5.2 Implications and Recommendation
Implication of this research is obvious. In order to survive in a world market that focuses on efficiency and not on regional quota opens both challenges and opportunities. Pakistan’s textile sector can benefit from developing high quality indigenous textile machinery. Access to information technology such as ERP is essential to manage business operations as well as communications with the global partner. High cost of proprietary ERP has created many obstacles in acquiring the required information technology. Locally developed software promises much needed assistance in this area and needs to be supported by both government and textile exporters.
## 5.3 Future Research
The information collected and analyzed for the textile sector is applicable to any sector interested in exporting its products such as sports equipment, leather products, surgical equipment etc. The parameters investigated and identified in this research could be verified for their importance to various Export oriented industries.
Major research and development effort is clearly required in developing indigenous resources valuable to the supply chain. The prohibitive cost of information technology software, could be controlled by developing local software that could seamlessly integrate with proprietary software needs to be developed and research in this area would be truly helpful to the supply chain. | 1,770 | 9,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.911369 |
https://upokary.com/write-a-program-to-defuse-the-bomb/ | 1,721,702,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00270.warc.gz | 522,379,277 | 12,158 | # Write a Program to Defuse the Bomb
# 1652. Defuse the Bomb
You have a bomb to defuse, and your time is running out! Your informer will provide you with a circular array code of length of n and a key k.
To decrypt the code, you must replace every number. All the numbers are replaced simultaneously.
If k > 0, replace the ith number with the sum of the next k numbers.
If k < 0, replace the ith number with the sum of the previous k numbers. If k == 0, replace the ith number with 0. As code is circular, the next element of code[n-1] is code[0], and the previous element of code[0] is code[n-1].Given the circular array code and an integer key k, return the decrypted code to defuse the bomb!
```### Example 1:
“`
Input: code = [5,7,1,4], k = 3
Output: [12,10,16,13]
Explanation: Each number is replaced by the sum of the next 3 numbers. The decrypted code is [7+1+4, 1+4+5, 4+5+7, 5+7+1]. Notice that the numbers wrap around.
“`
```
```### Example 2:
```
Input: code = [1,2,3,4], k = 0
Output: [0,0,0,0]
Explanation: When k is zero, the numbers are replaced by 0.
```
```
```### Example 3:
```
Input: code = [2,4,9,3], k = -2
Output: [12,5,6,13]
Explanation: The decrypted code is [3+9, 2+3, 4+2, 9+4]. Notice that the numbers wrap around again. If k is negative, the sum is of the previous numbers.
```
```
```Constraints:
```
n == code.length
1 <= n <= 100
1 <= code[i] <= 100
-(n - 1) <= k <= n - 1
```
```
```/**
* @param {number[]} code
* @param {number} k
* @return {number[]}
*/
const decrypt = function(code, k) {
const len = code.length;
const out = new Array(len).fill(0)
const sign = k > 0 ? 1 : -1;
const limit = k > 0 ? len : -1;
for(let i = 0; i < len; i++) {
let count = 0, j = i + sign;
while(k > 0 ? count < k : count > k) {
if (j === limit) {
j = k > 0 ? 0 : len - 1;
}
out[i] += code[j];
count = count + sign;
j = j + sign;
}
}
return out;
};
``` | 634 | 1,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-30 | latest | en | 0.72853 |
http://brilliant.org/explorations/ace-the-amc/trigonometry-2/b-premium/?element_id=clicked_locked_chapter_btn_from_explorations_topic_page | 1,498,219,514,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320057.96/warc/CC-MAIN-20170623114917-20170623134917-00679.warc.gz | 62,646,590 | 9,965 | Back to all chapters
# Trigonometry
The basics, the laws and relationships, and roots of unity.
# A preview of "Trigonometry"Join Brilliant Premium
Right triangle $$\triangle ABC$$ has a right angle at $$B$$, and $$\angle BAC = 30^{\circ} = \frac{\pi}{6}$$. Find $\sin\angle BAC + \cos\angle BCA.$
In triangle $$\triangle ABC$$, $$AB = 3, AC = 4$$, and $$\angle ABC = \frac{\pi}{2}$$. What is $$BC$$?
Let $$ABC$$ be an equilateral triangle. Point $$D$$ lies on $$BC$$ so that $$\angle BAD = 15^{\circ}$$. Find $$\frac{[BAD]}{[CAD]}$$.
× | 179 | 543 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-26 | latest | en | 0.712978 |
https://answers.yahoo.com/question/index?qid=20120221201558AAkEkrS | 1,561,595,963,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000609.90/warc/CC-MAIN-20190626234958-20190627020958-00059.warc.gz | 348,663,953 | 25,114 | # Stats and Probabiltiy?
The director of student health at a small high school is studying how many calories students at her high school consume in a day. She takes a random sample of 10 students from the high school and records the number of calories they consume in a 24 hour period. From the sample she found the average number of... show more The director of student health at a small high school is studying how many calories students at her high school consume in a day. She takes a random sample of 10 students from the high school and records the number of calories they consume in a 24 hour period. From the sample she found the average number of calories consumed was 1873 and that the standard deviation was 330. However before she finds the confidence interval for how many calories students consume in one day she goes to the statistics teacher for advice. What should the statistics teacher's advice be?
Select all that apply.
Choose at least one answer. a. The standard devaition is not large enough to compute the confidence interval.
b.The population may not be normal, but the Central Limit Theorem applies. Thus she should compute the confidence interval as normal.
c.She can calculate the confidence interval but should use a t-distribution since it deals with an average.
d.She did not take a simple random sample of the students, thus she should not compute the confidence interval.
e.The population may not be normal, and the Central Limit Theorem does not apply. Thus she should not compute the confidence interval. | 309 | 1,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-26 | latest | en | 0.957018 |
http://kochanski.org/gpk/code/speechresearch/gmisclib/gmisclib.mcmc_big-pysrc.html | 1,508,818,391,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828134.97/warc/CC-MAIN-20171024033919-20171024053919-00407.warc.gz | 185,495,491 | 25,989 | [frames] | no frames]
# Source Code for Module gmisclib.mcmc_big
``` 1
2 """An extension of mcmc that includes new stepping algorithms.
3 """
4
5 from __future__ import with_statement
6
7 import math
8 import random
9 import numpy
10
11 import die
12 import mcmc
13 from mcmc import Debug
14 import gpk_lsq
15
16 SIGFAC = 3.0
17
18 NotGoodPosition = mcmc.NotGoodPosition
19
20 -def N_maximum(a):
21 return a[numpy.argmax(a)].item()
22
23
24 -def _parab_interp_guts(xylist):
25 ND = 3
26 assert len(xylist) >= 3
27
28 yy = numpy.zeros((len(xylist), 1))
29 a = numpy.zeros((len(xylist), ND))
30 for (i, (x, y)) in enumerate(xylist):
31 # Order of parameter is a Constant, linear slope, then quadratic.
32 yy.itemset(i, 0, y)
33 a.itemset(i, 0, 1.0)
34 a.itemset(i, 1, x)
35 a.itemset(i, 2, x*x)
36
37 lls = gpk_lsq.linear_least_squares(a, yy, copy=False)
38
39 try:
40 sv = lls.sv()
41 x = lls.x()
42 except (ValueError, numpy.linalg.linalg.LinAlgError), ex:
43 raise mcmc.NoBoot, "stepParab: %s in _compute_move" % str(ex)
44 if not (sv[-1] > 1e-6*sv[0]):
45 raise mcmc.NoBoot, "Sv ratio too extreme: %g/%g" % (sv[-1], sv[0])
46 c = x.item(0,0)
47 b = x.item(1,0)
48 a = x.item(2,0)
49 # print 'c=', c
50 # print 'cc=', cc
51 return (a, b, c)
52
53
54 -def _parab_interp(points):
55 xlist = [x for (x, pos) in points]
56 tmp = [pos.logp_nocompute() for (x, pos) in points]
57 tmax = max(tmp)
58 ylist = [q-tmax for q in tmp]
59 # print 'XY=', (x0, y0), (x1, y1), (x2, y2)
60
61 # print 'xy=', (x0, y0), (x1, y1), (x2, y2)
62 a, b, c = _parab_interp_guts(zip(xlist, ylist))
63 # print 'abc=', a, b, c
64 mn = min(xlist)
65 mx = max(xlist)
66 w = mx - mn
67 if a >= 0.0:
68 raise mcmc.NoBoot, "parab: positive curvature"
69 else:
70 # This looks like a nice parabola with a maximum. We step
71 # into the vicinity of the maximum.
72 xmin = -b/(2*a)
73 sigma = math.sqrt(-SIGFAC/(2*a))
74
75 TOOCLOSE = 0.02
76 TOOFAR = 2.0
77 if xmin < -TOOFAR*w:
78 xmin = -TOOFAR*w
79 elif xmin > TOOFAR*w:
80 xmin = TOOFAR*w
81 if sigma > TOOFAR*w:
82 sigma = TOOFAR*w
83 elif sigma < 2*TOOCLOSE*w:
84 sigma = 2*TOOCLOSE*w
85
86 xtmp = random.normalvariate(xmin, sigma)
87 while min([abs(xtmp-x) for x in xlist]) < TOOCLOSE*w:
88 xtmp = random.normalvariate(xtmp, sigma)
89 # The minimum is too nearly a replication of an existing point,
90 # so we try again to make sure we get some new information.
91 return xtmp
92
93
94 -def test():
95 a, b, c = _parab_interp_guts([(0.0, 1.0), (1.0, 0.0), (2.0, 1.0)])
96 assert abs(a-1.0) < 0.001
97 assert abs(-b/(2*a) - 1.0) < 0.001
98 assert abs(c-1.0) < 0.001
99
100
101
102 -def find_closest_p(v, vp):
103 """Searches a list of (v,p) pairs and finds the one whose v
104 is closest to the first argument. Returns
105 (v,p) of the closest pair.
106 """
107 assert len(vp) > 0
108 vc, pc = vp[0]
109 for (vi, pi) in vp[1:]:
110 if abs(vi-v) < abs(vc-v):
111 vc = vi
112 pc = pi
113 return (vc, pc)
114
115
116
117 -def _fast_adjust(moveB, moveV, vfac, aV, aB, accepted):
118 """@note: this breaks the Markov assumption."""
119 SLOP = 5.0
120 # If the moveB completely dominates moveV, then we can apply our knowledge of
121 # whether the step was accepted or not to adjusting the size of the Bootstrap step.
122 if numpy.greater(numpy.absolute(moveB), SLOP*numpy.absolute(moveV)).all():
123 aB.inctry(accepted)
124 # Likewise if moveV dominates:
125 if numpy.greater(numpy.absolute(moveV), SLOP*numpy.absolute(moveB)).all():
126 aV.inctry(accepted)
127 # Otherwise, we assume that the bootstrap steps are fairly successful,
128 # so the bootstrap size is approximately correct, and we
129 # adjust the stepV step size to approximately match the stepBoot step size.
130 if numpy.greater(numpy.absolute(moveB), SLOP*numpy.absolute(moveV)/vfac).all():
131 aV.inctry(1) # moveV is too small: pretend stepV succeeded so it gets bigger.
132 if numpy.greater(numpy.absolute(moveV)/vfac, SLOP*numpy.absolute(moveB)).all():
133 aV.inctry(0) # moveV is too big: pretend stepV failed so it gets smaller.
134
135
136
137 -def _has_enough(points, dP):
138 if len(points) < 3:
139 return False
140 if len(points) > 10: # Something is wrong. Stop wasting time on a parabolic approximation.
141 return True
142 lp = [pos.logp_nocompute() for (x, pos) in points]
143 spread = max(lp) - min(lp)
145
146
147 -class BootStepper(mcmc.BootStepper):
148 - def __init__(self, lop, v, strategy=mcmc.BootStepper.SSAUTO,
149 maxArchSize=None, parallelSizeDiv=1):
150 mcmc.BootStepper.__init__(self, lop, v, strategy=strategy,
151 maxArchSize=maxArchSize,
152 parallelSizeDiv=parallelSizeDiv)
153
154
155 - def step(self):
156 mcmc.stepper.step(self)
157 Wboot = max(self.archive.distinct_count()-2, 0)
158 WV = self.np
159 Wmixed = math.sqrt(Wboot * WV)
160 # Parabolic steps are a bootstrap method, so they needs a large archive.
161 # But, Step_parab is not really Markov, so we only want to do it
162 if self.archive.sorted:
163 Wparab = max(self.archive.distinct_count()-2, 0)
164 else:
165 Wparab = 0
166 Wprobe = (Wboot+WV+Wmixed+1.5*Wparab) * self.F
167 W = float(Wboot + Wmixed + Wparab + WV + Wprobe)
168 Pboot = Wboot/W
169 Pmixed = Pboot + Wmixed/W
170 Pparab = Pmixed + Wparab/W
171 PV = Pparab + WV/W
172
173 again = True
174 while again:
175 P = random.random()
176 try:
177 if P < Pboot:
178 accepted = self.step_boot()
179 elif P < Pmixed:
180 accepted = self.step_mixed()
181 elif P < Pparab:
182 # Note! This violates MCMC requirements!
183 # The resulting probability distribution will not be correct
184 # when step_parab() is used!
185 accepted = self.step_parab()
186 elif P < PV:
187 accepted = self.stepV()
188 else:
189 accepted = self.step_probe()
190 except mcmc.NoBoot, x:
191 if Debug > 0:
192 die.info('NoBoot: %s' % str(x))
193 again = True
194 else:
195 again = False
196 return accepted
197
198
199 - def step_mixed(self):
200 vfac = math.exp(random.random()*math.log(1e-5))
201 self.steptype = 'step_mixed'
202 if len(self.archive) <= 2:
203 # print 'NoBoot', len(self.archive)
204 raise mcmc.NoBoot, "mixed short archive"
205 p1 = self.archive.choose()
206 p2 = self.archive.choose()
207 if p1.uid() == p2.uid():
208 # print 'NoBoot dup'
209 raise mcmc.NoBoot, "mixed duplicate"
210
211 vsV = self.aV.vs() * vfac
212 if self.archive.distinct_count() >= min(5, self.np_eff):
213 vsV *= (self.archive.variance()/self.v.diagonal())**(1./4.)
214
215 moveB = (p1.vec() - p2.vec()) * self.aB.vs()
216 moveV = vsV * self.V.sample()
217 move = moveB + moveV
218 try:
219 tmp = self.current().new(move)
220 delta = tmp.logp() - self.current().logp()
221 except mcmc.NotGoodPosition:
222 die.warn('NotGoodPosition')
223 # Should I call self.aM.inctry(accepted) ?
224 return 0
225 if self.acceptable(delta):
226 accepted = 1
227 self._set_current(tmp)
228 if Debug>2:
229 die.info('StepMixed: Accepted logp=%g' % tmp.logp_nocompute())
230 else:
231 self._set_failed( tmp )
232 accepted = 0
233 self._set_current(self.current())
234 if Debug>2:
235 die.info('StepMixed: Rejected logp=%g vs. %g, T=%g'
236 % (tmp.logp_nocompute(), self.current().logp_nocompute(), self.acceptable.T()))
237
238 # Only do this after a reset or when we don't care about breaking Markov assumptions:
239 if self.archive.sorted:
240 _fast_adjust(moveB, moveV, vfac, self.aV, self.aB, accepted)
241 return accepted
242
243
244
245 - def step_parab(self):
246 self.steptype = 'step_parab'
247 # The parabolic fit will be degenerate if we pick vs0 too close
248 # to either of the existing points.
249 if len(self.archive) <= 2:
250 raise mcmc.NoBoot, "parab short archive"
251
252 p1 = self.current()
253 p2 = self.archive.choose()
254 if p1.uid() == p2.uid():
255 raise mcmc.NoBoot, "parab duplicate"
256 points = [(0.0, p1), (1.0, p2)]
257 pp = []
258 vsB = self.aB.vs()
259 while True:
260 vs0 = random.normalvariate(0.0, vsB)
261 # Don't pick a point too close to the existing points.
262 tooclose = False
263 for (x, pos) in points:
264 if abs(vs0-x) < 0.5/len(points):
265 tooclose = True
266 break
267 if tooclose:
268 continue
269
270 vbase, pbase = find_closest_p(vs0, points)
271 move = (p2.vec() - p1.vec()) * (vs0-vbase)
272 # This is an approximation to a MCMC step, at least when the
273 # archive is large and well-annealed. So, we might as well add
274 # it to the archive.
275 try:
276 tmp = pbase.new(move)
277 pp.append( (tmp, tmp.logp()) )
278 except mcmc.NotGoodPosition:
279 die.warn('NotGoodPosition2a at %.3f' % vs0)
280 return 0
281
282 points.append( (vs0, tmp) )
283 if _has_enough(points, self.acceptable.jitter()):
284 break
285 vsB *= 1.2
286 if Debug > 0:
287 die.info("step_parab: %d points at %s" % (len(points), [q for (q,pos) in points]))
288
289 # Now, we take one of the (potentially many) new points and treat that as
290 # a MCMC step and add it to the archive. We don't want to add them all,
291 # because we might build up strong linear structures in the archive and
292 # make it more likely that the optimization gets stuck in a subspace.
293 tmp, tmp_logp = random.choice(pp)
294 try:
295 delta = tmp_logp - self.current().logp()
296 except mcmc.NotGoodPosition:
297 die.warn('NotGoodPosition2a at %.3f' % vs0)
298 return 0
299 if self.acceptable(delta):
300 if Debug > 2:
301 die.info('StepParab Accepted1 at %.3f, logp=%g' % (vs0, tmp_logp))
302 self._set_current(tmp)
303 else:
304 self._set_failed( tmp )
305 self._set_current(self.current())
306 # Return from here? No. It still can be used to
307 # define the interpolation parabola, even if logP
308 # isn't particularly good.
309
310 try:
311 vsnew = _parab_interp(points)
312 except mcmc.NotGoodPosition:
313 die.warn('NotGoodPosition preparing for _parab_interp().')
314 return 0
315 vbase, pbase = find_closest_p(vsnew, points)
316 move = (p2.vec() - p1.vec()) * (vsnew-vbase)
317
318 try:
319 tmp = pbase.new(move)
320 delta = tmp.logp() - self.current().logp()
321 except mcmc.NotGoodPosition:
322 die.warn('NotGoodPosition')
323 accepted = 0
324 if self.acceptable(delta):
325 if Debug > 2:
326 die.info('StepParab Accepted2 at %.3f, logp=%g' % (vsnew, tmp.logp_nocompute()))
327 accepted = 1
328 self._set_current(tmp)
329 else:
330 self._set_failed( tmp )
331 accepted = 0
332 self._set_current(self.current())
333 if Debug>2:
334 die.info('StepParab: Rejected logp=%g vs. %g, T=%g'
335 % (tmp.logp_nocompute(), self.current().logp_nocompute(), self.acceptable.T()))
336 return accepted
337
338
339
340
341 -def bootstepper(logp, x, v, c=None, strategy=BootStepper.SSAUTO, fixer=None, repeatable=True):
342 """This is (essentially) another interface to the class constructor.
343 It's really there for backwards compatibility.
344 """
345 pd = mcmc.problem_definition_F(logp_fcn=logp, c=c, fixer=fixer)
346 position_constructor = [mcmc.position_nonrepeatable, mcmc.position_repeatable][repeatable]
347 return BootStepper(mcmc.make_list_of_positions(x, position_constructor, pd), v, strategy=strategy)
348
349 diag_variance = mcmc.diag_variance
350 stepper = mcmc.stepper
351 problem_definition = mcmc.problem_definition
352 problem_definition_F = mcmc.problem_definition_F
353 problem_definition = mcmc.problem_definition
354 position_repeatable = mcmc.position_repeatable
355 position_nonrepeatable = mcmc.position_nonrepeatable
356
357 -def test2d(stepper):
358 import dictops
359 start = numpy.array([9, 1])
360 c = numpy.array([1.0, 0.1])
361 V = numpy.identity(len(c), numpy.float)
362 # V = numpy.array([[ 20.69626808, 20.6904984 ], [ 20.6904984, 20.69477235]])
363
364 x = stepper(mcmc._logp1, start, V, c=c)
365 v = numpy.zeros((x.np, x.np))
366 n_per_steptype = dictops.dict_of_averages()
367 for i in range(1000):
368 accepted = x.step()
370 lpsum = 0.0
371 lps2 = 0.0
372 na = 0
373 psum = numpy.zeros((x.np,))
374 N = 30000
375 nreset = 0
376 nsorted = 0
377 rid = x.reset_id()
378 for i in range(N):
379 accepted = x.step()
380 na += accepted
382 nsorted += x.archive.sorted
383 if x.reset_id() != rid:
384 rid = x.reset_id()
385 nreset += 1
386 lp = x.current().logp()
387 lpsum += lp
388 lps2 += lp**2
389 p = x.current().vec()
391 v += numpy.outer(p, p)
392 for steptype in n_per_steptype.keys():
393 print 'Step %s has %.0f successes out of %.0f trials: %.2f' % (
394 steptype, n_per_steptype.get_both(steptype)[0],
395 n_per_steptype.get_both(steptype)[1],
396 n_per_steptype.get_avg(steptype)
397 )
398 # assert N*x.PBootLim**1.5 < nboot < N*x.PBootLim**0.7
399 assert nsorted < N//30
400 assert N//8 < na < N//2
401 assert nreset < 30
402 lpsum /= N
403 assert abs(lpsum+0.5*x.np) < 0.15
404 lps2 /= N-1
405 lpvar = lps2 - lpsum**2
406 assert abs(lpvar-0.5*x.np) < 1.0
407 numpy.divide(psum, N, psum)
408 assert numpy.alltrue(numpy.less(numpy.absolute(psum), 6.0/numpy.sqrt(c)))
409 numpy.divide(v, N-1, v)
410 for i in range(x.np):
411 for j in range(x.np):
412 if i == j:
413 # print '2*v[ii]*c=', 2*v[i,j]*c[i]**2
414 assert abs(math.log(2*v[i,j]*c[i]*c[j])) < 0.1
415 else:
416 assert abs(v[i,j]) < 20/(c[i]*c[j])
417
418
419 if __name__ == '__main__':
420 # mcmc.test(stepper=bootstepper)
421 Debug = 3
422 test()
423 mcmc.test_(stepper=bootstepper)
424
<!--
expandto(location.href);
// -->
```
Generated by Epydoc 3.0.1 on Thu Sep 22 04:25:13 2011 http://epydoc.sourceforge.net | 4,674 | 19,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-43 | latest | en | 0.407659 |
http://mathhelpforum.com/business-math/220931-two-asset-portfolio-variance.html | 1,481,350,877,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542939.6/warc/CC-MAIN-20161202170902-00270-ip-10-31-129-80.ec2.internal.warc.gz | 175,646,117 | 9,932 | 1. ## two-asset portfolio variance
Hi,
I'm having problems sorting out whether a segment of a youtube video on portfolio theory is correct. The issue relates to calculating the variance of a two-asset portfolio. Since there is no linearity the addition of a second asset cannot be assumed to simply add its variance to the variance of the first asset, due to the fact that the two assets, treated as random variables, are not necessarily uncorrelated. Therefore, the formula would be something like:
Var (p) = [Weighting (Asset 1)2 * Var (Asset 1)] + [W(Asset 2)2 * Var (Asst 2)] + [ 2 * W1* W2* cov (Ass1, Ass2) ]
However, the equation in the youtube video is:
Var (p) = [Weighting (Asset 1)2 * Var (Asset 1)] + [W(Asset 2)2 * Var (Asst 2)] + [ 2 * W1* W2* SD (Ass1) * SD (Ass2) ]
Further, the author of the video claims that [ 2 * W1* W2* SD (Ass1) * SD (Ass2) ] is the covariance.
What am I missing? | 272 | 909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2016-50 | longest | en | 0.888627 |
http://wingware.com/psupport/python-manual/3.3/library/heapq.html | 1,516,722,321,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891980.75/warc/CC-MAIN-20180123151545-20180123171545-00548.warc.gz | 379,792,999 | 8,611 | # 8.5. heapq — Heap queue algorithm¶
Source code: Lib/heapq.py
This module provides an implementation of the heap queue algorithm, also known as the priority queue algorithm.
Heaps are binary trees for which every parent node has a value less than or equal to any of its children. This implementation uses arrays for which heap[k] <= heap[2*k+1] and heap[k] <= heap[2*k+2] for all k, counting elements from zero. For the sake of comparison, non-existing elements are considered to be infinite. The interesting property of a heap is that its smallest element is always the root, heap[0].
The API below differs from textbook heap algorithms in two aspects: (a) We use zero-based indexing. This makes the relationship between the index for a node and the indexes for its children slightly less obvious, but is more suitable since Python uses zero-based indexing. (b) Our pop method returns the smallest item, not the largest (called a “min heap” in textbooks; a “max heap” is more common in texts because of its suitability for in-place sorting).
These two make it possible to view the heap as a regular Python list without surprises: heap[0] is the smallest item, and heap.sort() maintains the heap invariant!
To create a heap, use a list initialized to [], or you can transform a populated list into a heap via function heapify().
The following functions are provided:
heapq.heappush(heap, item)
Push the value item onto the heap, maintaining the heap invariant.
heapq.heappop(heap)
Pop and return the smallest item from the heap, maintaining the heap invariant. If the heap is empty, IndexError is raised.
heapq.heappushpop(heap, item)
Push item on the heap, then pop and return the smallest item from the heap. The combined action runs more efficiently than heappush() followed by a separate call to heappop().
heapq.heapify(x)
Transform list x into a heap, in-place, in linear time.
heapq.heapreplace(heap, item)
Pop and return the smallest item from the heap, and also push the new item. The heap size doesn’t change. If the heap is empty, IndexError is raised.
This one step operation is more efficient than a heappop() followed by heappush() and can be more appropriate when using a fixed-size heap. The pop/push combination always returns an element from the heap and replaces it with item.
The value returned may be larger than the item added. If that isn’t desired, consider using heappushpop() instead. Its push/pop combination returns the smaller of the two values, leaving the larger value on the heap.
The module also offers three general purpose functions based on heaps.
heapq.merge(*iterables)
Merge multiple sorted inputs into a single sorted output (for example, merge timestamped entries from multiple log files). Returns an iterator over the sorted values.
Similar to sorted(itertools.chain(*iterables)) but returns an iterable, does not pull the data into memory all at once, and assumes that each of the input streams is already sorted (smallest to largest).
heapq.nlargest(n, iterable, key=None)
Return a list with the n largest elements from the dataset defined by iterable. key, if provided, specifies a function of one argument that is used to extract a comparison key from each element in the iterable: key=str.lower Equivalent to: sorted(iterable, key=key, reverse=True)[:n]
heapq.nsmallest(n, iterable, key=None)
Return a list with the n smallest elements from the dataset defined by iterable. key, if provided, specifies a function of one argument that is used to extract a comparison key from each element in the iterable: key=str.lower Equivalent to: sorted(iterable, key=key)[:n]
The latter two functions perform best for smaller values of n. For larger values, it is more efficient to use the sorted() function. Also, when n==1, it is more efficient to use the built-in min() and max() functions.
## 8.5.1. Basic Examples¶
A heapsort can be implemented by pushing all values onto a heap and then popping off the smallest values one at a time:
```>>> def heapsort(iterable):
... 'Equivalent to sorted(iterable)'
... h = []
... for value in iterable:
... heappush(h, value)
... return [heappop(h) for i in range(len(h))]
...
>>> heapsort([1, 3, 5, 7, 9, 2, 4, 6, 8, 0])
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
```
Heap elements can be tuples. This is useful for assigning comparison values (such as task priorities) alongside the main record being tracked:
```>>> h = []
>>> heappush(h, (5, 'write code'))
>>> heappush(h, (7, 'release product'))
>>> heappush(h, (1, 'write spec'))
>>> heappush(h, (3, 'create tests'))
>>> heappop(h)
(1, 'write spec')
```
## 8.5.2. Priority Queue Implementation Notes¶
A priority queue is common use for a heap, and it presents several implementation challenges:
• Sort stability: how do you get two tasks with equal priorities to be returned in the order they were originally added?
• Tuple comparison breaks for (priority, task) pairs if the priorities are equal and the tasks do not have a default comparison order.
• If the priority of a task changes, how do you move it to a new position in the heap?
• Or if a pending task needs to be deleted, how do you find it and remove it from the queue?
A solution to the first two challenges is to store entries as 3-element list including the priority, an entry count, and the task. The entry count serves as a tie-breaker so that two tasks with the same priority are returned in the order they were added. And since no two entry counts are the same, the tuple comparison will never attempt to directly compare two tasks.
The remaining challenges revolve around finding a pending task and making changes to its priority or removing it entirely. Finding a task can be done with a dictionary pointing to an entry in the queue.
Removing the entry or changing its priority is more difficult because it would break the heap structure invariants. So, a possible solution is to mark the entry as removed and add a new entry with the revised priority:
```pq = [] # list of entries arranged in a heap
entry_finder = {} # mapping of tasks to entries
REMOVED = '<removed-task>' # placeholder for a removed task
counter = itertools.count() # unique sequence count
'Add a new task or update the priority of an existing task'
if task in entry_finder:
count = next(counter)
entry = [priority, count, task]
heappush(pq, entry)
'Mark an existing task as REMOVED. Raise KeyError if not found.'
entry[-1] = REMOVED
'Remove and return the lowest priority task. Raise KeyError if empty.'
while pq:
priority, count, task = heappop(pq)
if task is not REMOVED:
raise KeyError('pop from an empty priority queue')
```
## 8.5.3. Theory¶
Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for all k, counting elements from 0. For the sake of comparison, non-existing elements are considered to be infinite. The interesting property of a heap is that a[0] is always its smallest element.
The strange invariant above is meant to be an efficient memory representation for a tournament. The numbers below are k, not a[k]:
``` 0
1 2
3 4 5 6
7 8 9 10 11 12 13 14
15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
```
In the tree above, each cell k is topping 2*k+1 and 2*k+2. In an usual binary tournament we see in sports, each cell is the winner over the two cells it tops, and we can trace the winner down the tree to see all opponents s/he had. However, in many computer applications of such tournaments, we do not need to trace the history of a winner. To be more memory efficient, when a winner is promoted, we try to replace it by something else at a lower level, and the rule becomes that a cell and the two cells it tops contain three different items, but the top cell “wins” over the two topped cells.
If this heap invariant is protected at all time, index 0 is clearly the overall winner. The simplest algorithmic way to remove it and find the “next” winner is to move some loser (let’s say cell 30 in the diagram above) into the 0 position, and then percolate this new 0 down the tree, exchanging values, until the invariant is re-established. This is clearly logarithmic on the total number of items in the tree. By iterating over all items, you get an O(n log n) sort.
A nice feature of this sort is that you can efficiently insert new items while the sort is going on, provided that the inserted items are not “better” than the last 0’th element you extracted. This is especially useful in simulation contexts, where the tree holds all incoming events, and the “win” condition means the smallest scheduled time. When an event schedules other events for execution, they are scheduled into the future, so they can easily go into the heap. So, a heap is a good structure for implementing schedulers (this is what I used for my MIDI sequencer :-).
Various structures for implementing schedulers have been extensively studied, and heaps are good for this, as they are reasonably speedy, the speed is almost constant, and the worst case is not much different than the average case. However, there are other representations which are more efficient overall, yet the worst cases might be terrible.
Heaps are also very useful in big disk sorts. You most probably all know that a big sort implies producing “runs” (which are pre-sorted sequences, which size is usually related to the amount of CPU memory), followed by a merging passes for these runs, which merging is often very cleverly organised [1]. It is very important that the initial sort produces the longest runs possible. Tournaments are a good way to that. If, using all the memory available to hold a tournament, you replace and percolate items that happen to fit the current run, you’ll produce runs which are twice the size of the memory for random input, and much better for input fuzzily ordered.
Moreover, if you output the 0’th item on disk and get an input which may not fit in the current tournament (because the value “wins” over the last output value), it cannot fit in the heap, so the size of the heap decreases. The freed memory could be cleverly reused immediately for progressively building a second heap, which grows at exactly the same rate the first heap is melting. When the first heap completely vanishes, you switch heaps and start a new run. Clever and quite effective!
In a word, heaps are useful memory structures to know. I use them in a few applications, and I think it is good to keep a ‘heap’ module around. :-)
Footnotes
[1] The disk balancing algorithms which are current, nowadays, are more annoying than clever, and this is a consequence of the seeking capabilities of the disks. On devices which cannot seek, like big tape drives, the story was quite different, and one had to be very clever to ensure (far in advance) that each tape movement will be the most effective possible (that is, will best participate at “progressing” the merge). Some tapes were even able to read backwards, and this was also used to avoid the rewinding time. Believe me, real good tape sorts were quite spectacular to watch! From all times, sorting has always been a Great Art! :-) | 2,596 | 11,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-05 | longest | en | 0.887691 |
https://www.jagranjosh.com/articles/ncert-cbse-class-10th-science-chapter-12-electricity-1450241705-1 | 1,670,628,705,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711552.8/warc/CC-MAIN-20221209213503-20221210003503-00500.warc.gz | 878,331,111 | 34,902 | # NCERT Class 10 Science Chapter 12: Electricity
Get NCERT Class 10 Science Chapter 12: Electricity in its latest edition. Read this chapter thoroughly to make effective preparation for board exam 2020.
NCERT Class 10th Science Chapter-Electricity
Read latest NCERT Class 10 Science Chapter 12: Electricity to prepare for board exam 2020. NCERT books explain each and every topic in detail which helps to understand that topic and the concepts used in a very clear way. So, be consistent with the practice of topics and questions given in NCERT to perform well in your exams.
About NCERT Class 10 Science Chapter 12
This chapter explains electric current, its applications and various effects related to it. Go through the following major topics before starting with the NCERT questions and solutions:
• Introduction to electric current
• Potential difference
• Ohm’s law and its graphical representation
• Factors on which the Resistance of the conductor depends – Resistivity
• Resistors in series and parallel: Calculation of resultant resistance in the series, current and voltage across each resistor
• Advantage of parallel combination over the series combination
• Heating effects of electric circuit: Joule’s law of heating effect of electric current
• Electric fuse
• Power
NCERT Class 10 Science Chapter 12: Electricity
We are also providing here the accurate and exclusive NCERT solutions for chapter 12 of Class 10 Science NECRT book. Students must go through these detailed solutions to understand all the concepts clearly and manage their exam preparations in an efficient manner. All our NCERT books and solutions are reviewed by the subject experts to bring the reliable and error free content for students. Download all the NCERT solutions for questions asked in chapter 12, from the following link:
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खेलें हर किस्म के रोमांच से भरपूर गेम्स सिर्फ़ जागरण प्ले पर | 807 | 3,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-49 | latest | en | 0.912555 |
https://www.rampfesthudson.com/is-60kg-a-healthy-weight-for-160cm/ | 1,642,576,527,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301264.36/warc/CC-MAIN-20220119064554-20220119094554-00667.warc.gz | 973,251,362 | 21,208 | # Is 60kg a healthy weight for 160cm?
## Is 60kg a healthy weight for 160cm?
According to BMI chart and their classification, BMI score of 23.44 for 60 kilograms weight & 160 cm tall indicates that your weight is Normal. The Body Mass Index of 23.44 is derived from the weight of 60 kilograms divided by the square of height 160 centimeters.
## What should be the weight for 160 cm height?
Metric Measurement
158 cm 50-62 kg
160 cm 51-64 kg
162 cm 52-66 kg
164 cm 54-67 kg
## Is 157 cm a good height for girl?
Is 157 cm a good height for girl? The women of 157 centimeter height with weight of around 109.49 pounds or 49.67 kilograms considered to be healthy.
## How to calculate the BMI of a 64 year old?
BMI = 64 ÷ 1.9 2 BMI = 64 ÷ 3.61 BMI = 17.7 Example 2 (Calculating BMI in US units)
## What should your height and weight be according to your age?
This Age, Height, and Weight Chart helped to understand what height and weight are perfect according to your Age. In these days we read and fined many Charts like average male weight by age in kg and many more.
## How are height and weight measured on a BMI Calculator?
Height can be entered in inches, feet and inches, meters, or centimeters. This calculator will also calculate and display the normal weight range for the height entered. BMI is a way of measuring a person’s weight relative to his or her height.
## How to find out if your child is a healthy weight?
If you are over 18 years, you can use an adult calculator to find out whether you are a healthy weight. Please fix the issues above and try again. based on a n year old at kg and cm BMI-for-age is at the percentile. The result suggests your child is below a healthy weight for their age and height.
## What is the body mass index for 60 kg?
The Body Mass Index of 23.44 is derived from the weight of 60 kilograms divided by the square of height 160 centimeters. BMI chart for 60 kg weight & 160 cm height and its nearest weights to let you know how your height & weight describes you.
## How tall is 60 kg in weight and height?
Is 60 kg Weight & 160 cm Height Obese or Overweight? getcalc.com’s BMI calculator to find if a male or female of 60 kg weight & 160 cm height is obese, extreme obese, overweight, underweight or ideal weight.
## How tall is 70 kg in feet and in?
Current Weight Target Weight Curent BMI Target BMI 70 kg 63.74 kg 27.34 24.9
## What should my BMI be for 70 kg?
According to BMI chart and their classification, BMI score of 27.34 for 70 kilograms weight & 160 cm tall indicates that your weight is Overweight. The Body Mass Index of 27.34 is derived from the weight of 70 kilograms divided by the square of height 160 centimeters. | 679 | 2,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-05 | latest | en | 0.921428 |
https://docs.juliadiffeq.org/v4.0.1/solvers/sde_solve.html | 1,582,212,339,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144979.91/warc/CC-MAIN-20200220131529-20200220161529-00318.warc.gz | 354,519,902 | 5,955 | SDE Solvers
# SDE Solvers
## Recommended Methods
For most Ito diagonal and scalar noise problems where a good amount of accuracy is required and mild stiffness may be an issue, the SRIW1 algorithm should do well. If the problem has additive noise, then SRA1 will be the optimal algorithm. For commutative noise, RKMilCommute is a strong order 1.0 method which utilizes the commutivity property to greatly speed up the Wiktorsson approximation and can choose between Ito and Stratonovich. For non-commutative noise, EM and EulerHeun are the choices (for Ito and Stratonovich interpretations respectively).
For stiff problems with diagonal noise, ImplicitRKMil is the most efficient method and can choose between Ito and Stratonovich. If the noise is non-diagonal, ImplicitEM and ImplicitEulerHeun are for Ito and Stratonovich respectively. For each of these methods, the parameter theta can be chosen. The default is theta=1/2 which will not dampen numerical oscillations and thus is symmetric (and almost symplectic) and will lead to less error when noise is sufficiently small. However, theta=1/2 is not L-stable in the drift term, and thus one can receive more stability (L-stability in the drift term) with theta=1, but with a tradeoff of error efficiency in the low noise case. In addition, the option symplectic=true will turns these methods into an implicit Midpoint extension which is symplectic in distribution but has an accuracy tradeoff.
## Mass Matrices and Stochastic DAEs
The ImplicitRKMil, ImplicitEM, and ImplicitEulerHeun methods can solve stochastic equations with mass matrices (including stochastic DAEs written in mass matrix form) when either symplectic=true or theta=1.
## Special Noise Forms
Some solvers are for specialized forms of noise. Diagonal noise is the default setup. Non-diagonal noise is specified via setting noise_rate_prototype to a matrix in the SDEProblem type. A special form of non-diagonal noise, commutative noise, occurs when the noise satisfies the following condition:
$\sum_{i=1}^d g_{i,j_1}(t,u) \frac{\partial g_{k,j_2}}{\partial x_i} = \sum_{i=1}^d g_{i,j_2}(t,x) \frac{\partial g_{k,j_1}}{\partial x_i}$
for every $j_1,j_2$ and $k$. Additive noise is when $g(t,u)=g(t)$, i.e. is independent of u. Multiplicative noise is $g_i(t,u)=a_i u$.
## Special Keyword Arguments
• save_noise: Determines whether the values of W are saved whenever the timeseries is saved. Defaults to true.
• delta: The delta adaptivity parameter for the natural error estimator. Determines the balance between drift and diffusion error. For more details, see the publication.
# Full List of Methods
## StochasticDiffEq.jl
Each of the StochasticDiffEq.jl solvers come with a linear interpolation. Orders are given in terms of strong order.
### Nonstiff Methods
• EM- The Euler-Maruyama method. Strong Order 0.5 in the Ito sense. Can handle all forms of noise, including non-diagonal, scalar, and colored noise.†
• EulerHeun - The Euler-Heun method. Strong Order 0.5 in the Stratonovich sense. Can handle all forms of noise, including non-diagonal, scalar, and colored noise.†
• RKMil - An explicit Runge-Kutta discretization of the strong Order 1.0 Milstein method. Defaults to solving the Ito problem, but RKMil(interpretation=:Stratonovich) makes it solve the Stratonovich problem. Only handles scalar and diagonal noise.†
• RKMilCommute - An explicit Runge-Kutta discretization of the strong Order 1.0 Milstein method for commutative noise problems. Defaults to solving the Ito problem, but RKMilCommute(interpretation=:Stratonovich) makes it solve the Stratonovich problem.†
• SRA - The strong Order 1.5 methods for additive Ito and Stratonovich SDEs due to Rossler. Default tableau is for SRA1. Can handle non-diagonal and scalar additive noise.
• SRI - The strong Order 1.5 methods for diagonal/scalar Ito SDEs due to Rossler. Default tableau is for SRIW1.
• SRIW1 - An optimized version of SRIW1. Strong Order 1.5 for diagonal/scalar Ito SDEs.†
• SRA1 - An optimized version of SRA1. Strong Order 1.5 for additive Ito and Stratonovich SDEs. Can handle non-diagonal and scalar additive noise.†
Example usage:
sol = solve(prob,SRIW1())
### Tableau Controls
For SRA and SRI, the following option is allowed:
• tableau: The tableau for an :SRA or :SRI algorithm. Defaults to SRIW1 or SRA1.
### Stiff Methods
• ImplicitEM - An order 0.5 Ito implicit method. This is a theta method which defaults to theta=1/2 or the Trapezoid method on the drift term. This method defaults to symplectic=false, but when true and theta=1/2 this is the implicit Midpoint method on the drift term and is symplectic in distribution. Can handle all forms of noise, including non-diagonal, scalar, and colored noise.
• ImplicitEulerHeun - An order 0.5 Stratonovich implicit method. This is a theta method which defaults to theta=1/2 or the Trapezoid method on the drift term. This method defaults to symplectic=false, but when true and theta=1/2 this is the implicit Midpoint method on the drift term and is symplectic in distribution. Can handle all forms of noise, including non-diagonal, scalar, and colored noise.
• ImplicitRKMil - An order 1.0 implicit method. This is a theta method which defaults to theta=1/2 or the Trapezoid method on the drift term. Defaults to solving the Ito problem, but ImplicitRKMil(interpretation=:Stratonovich) makes it solve the Stratonovich problem. This method defaults to symplectic=false, but when true and theta=1/2 this is the implicit Midpoint method on the drift term and is symplectic in distribution. Handles diagonal and scalar noise.
These methods interpret the mass matrix equation as:
$Mu' = f(t,u)dt + Mg(t,u)dW_t$
i.e. with no mass matrix inversion applied to the g term. Thus these methods apply noise per dependent variable instead of on the combinations of the dependent variables and this is designed for phenomenological noise on the dependent variables (like multiplicative or additive noise)
## StochasticCompositeAlgorithm
One unique feature of StochasticDiffEq.jl is the StochasticCompositeAlgorithm, which allows you to, with very minimal overhead, design a multimethod which switches between chosen algorithms as needed. The syntax is StochasticCompositeAlgorithm(algtup,choice_function) where algtup is a tuple of StochasticDiffEq.jl algorithms, and choice_function is a function which declares which method to use in the following step. For example, we can design a multimethod which uses EM() but switches to RKMil() whenever dt is too small:
choice_function(integrator) = (Int(integrator.dt<0.001) + 1)
alg_switch = StochasticCompositeAlgorithm((EM(),RKMil()),choice_function)
The choice_function takes in an integrator and thus all of the features available in the Integrator Interface can be used in the choice function.
## BridgeDiffEq.jl
Bridge.jl is a set of fixed timestep algorithms written in Julia. These methods are made and optimized for out-of-place functions on immutable (static vector) types. Note that this setup is not automatically included with DifferentialEquaitons.jl. To use the following algorithms, you must install and use BridgeDiffEq.jl: | 1,745 | 7,177 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-10 | latest | en | 0.863848 |
https://www.investmentguide.co.uk/how-to-calculate-compound-annual-growth-rate-cagr-with-examples/ | 1,726,618,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00779.warc.gz | 753,856,259 | 14,057 | Introduction
Compound Annual Growth Rate (CAGR) is a useful tool for measuring the growth of an investment over a period of time. It is calculated by taking the average annual growth rate of an investment over a specified period of time. CAGR is a useful metric for investors to compare the performance of different investments and to assess the potential of an investment. In this article, we will discuss how to calculate CAGR, with examples to illustrate the process. We will also discuss the advantages and disadvantages of using CAGR as a metric for measuring investment performance.
What is Compound Annual Growth Rate (CAGR) and How to Calculate it?
Compound Annual Growth Rate (CAGR) is a measure of the rate of return of an investment over a period of time. It is calculated by taking the average annual growth rate of an investment over a specified period of time. CAGR is a useful tool for investors to compare the performance of different investments over time.
To calculate CAGR, you need to know the beginning and ending values of the investment, as well as the number of years the investment was held. First, calculate the total return of the investment by subtracting the beginning value from the ending value. Then, divide the total return by the beginning value. Finally, take the result and raise it to the power of 1 divided by the number of years the investment was held. The result is the CAGR.
For example, if you invested \$10,000 in a stock and it was worth \$15,000 after five years, the CAGR would be calculated as follows:
Total return = \$15,000 – \$10,000 = \$5,000
Divide total return by beginning value = \$5,000/\$10,000 = 0.5
Raise result to the power of 1/5 = (0.5)^(1/5) = 1.03
CAGR = 1.03 or 3%
CAGR is a useful tool for investors to compare the performance of different investments over time. It is important to note that CAGR does not take into account any fees or taxes associated with the investment, so it is important to consider these factors when evaluating an investment.
How to Use Compound Annual Growth Rate (CAGR) to Measure Investment Performance
Compound Annual Growth Rate (CAGR) is a useful tool for measuring the performance of an investment over a period of time. It is a measure of the average rate of return on an investment over a given period of time, expressed as a percentage.
To calculate CAGR, you need to know the beginning and ending values of the investment, as well as the length of the period. First, subtract the beginning value from the ending value. Then, divide that number by the beginning value. Finally, divide the result by the number of years in the period and multiply by 100 to get the CAGR.
For example, if you invested \$10,000 in a stock five years ago and it is now worth \$15,000, the CAGR would be calculated as follows:
(15,000 – 10,000) / 10,000 = 0.5
0.5 / 5 = 0.1
0.1 x 100 = 10%
Therefore, the CAGR of the investment is 10%.
READ ALSO: Day Order: definition and how it works in trading
CAGR is a useful tool for measuring the performance of an investment over a period of time because it takes into account the effects of compounding. Compounding is the process of earning interest on interest, which can significantly increase the value of an investment over time.
CAGR is also useful because it allows you to compare the performance of different investments over the same period of time. For example, if you had invested \$10,000 in two different stocks five years ago, you could use CAGR to compare the performance of each stock.
In conclusion, CAGR is a useful tool for measuring the performance of an investment over a period of time. It takes into account the effects of compounding and allows you to compare the performance of different investments over the same period of time.
A Step-by-Step Guide to Calculating Compound Annual Growth Rate (CAGR)
Calculating the Compound Annual Growth Rate (CAGR) is a great way to measure the performance of an investment over a period of time. CAGR is the average rate of return for an investment over a certain period of time, taking into account the effects of compounding. It is a useful tool for investors to compare the performance of different investments.
Here is a step-by-step guide to calculating CAGR:
Step 1: Gather the necessary information.
You will need to know the initial investment amount, the final investment amount, and the length of time the investment was held.
Step 2: Calculate the total return.
The total return is the difference between the final investment amount and the initial investment amount.
Step 3: Calculate the CAGR.
To calculate the CAGR, use the following formula:
CAGR = (Final Investment Amount / Initial Investment Amount)^(1/Number of Years) – 1
Step 4: Interpret the results.
The CAGR is expressed as a percentage. A positive CAGR indicates that the investment has grown over the period of time, while a negative CAGR indicates that the investment has declined.
By following these steps, you can easily calculate the CAGR of an investment. CAGR is a useful tool for investors to measure the performance of their investments over time.
How to Interpret Compound Annual Growth Rate (CAGR) Results
Interpreting Compound Annual Growth Rate (CAGR) results can be a great way to measure the performance of an investment over a period of time. CAGR is a measure of the average rate of return for an investment over a certain period of time, usually expressed as a percentage.
To calculate CAGR, you need to know the beginning and ending values of the investment, as well as the number of years it was held. The formula for CAGR is:
CAGR = (Ending Value / Beginning Value)^(1/Number of Years) – 1
For example, if you had an investment that started at \$100 and ended at \$150 after two years, the CAGR would be calculated as follows:
CAGR = (150 / 100)^(1/2) – 1 = 0.25 = 25%
This means that the investment had an average annual return of 25% over the two-year period.
READ ALSO: What is a fixed annuity?
When interpreting CAGR results, it is important to remember that CAGR is an average rate of return and does not take into account any fluctuations in the investment’s value over the period of time. It is also important to note that CAGR does not take into account any fees or taxes associated with the investment.
In conclusion, CAGR is a useful tool for measuring the performance of an investment over a period of time. It is important to remember that CAGR is an average rate of return and does not take into account any fluctuations in the investment’s value or any fees or taxes associated with the investment.
How to Use Compound Annual Growth Rate (CAGR) to Compare Investment Options
Compound Annual Growth Rate (CAGR) is a useful tool for comparing different investment options. It is a measure of the average rate of return on an investment over a period of time. It takes into account the effects of compounding, which is when the returns from an investment are reinvested and generate additional returns.
To use CAGR to compare investment options, you need to calculate the CAGR for each option. To do this, you need to know the initial investment amount, the final value of the investment, and the number of years the investment was held. You can then use the following formula to calculate the CAGR:
CAGR = (Final Value / Initial Value)^(1/Number of Years) – 1
Once you have calculated the CAGR for each investment option, you can compare them to determine which option is the most profitable. Generally, the higher the CAGR, the better the investment option.
It is important to note that CAGR does not take into account the risk associated with an investment. Therefore, it is important to consider other factors such as the volatility of the investment and the potential for losses when making an investment decision.
Using CAGR to compare investment options can help you make an informed decision about which option is best for you. It is a simple and effective way to compare different investment options and determine which one is most likely to generate the highest returns.
What Are the Benefits of Calculating Compound Annual Growth Rate (CAGR)?
Calculating Compound Annual Growth Rate (CAGR) is a useful tool for investors and business owners to measure the performance of their investments or businesses over a period of time. CAGR is a measure of the average rate of return over a period of time, taking into account the effects of compounding. It is a more accurate measure of performance than simple average returns, as it takes into account the effects of compounding.
The benefits of calculating CAGR include:
1. It provides a more accurate measure of performance: CAGR takes into account the effects of compounding, which can have a significant impact on the performance of an investment or business over time. This makes it a more accurate measure of performance than simple average returns.
READ ALSO: Bid price in trading: definition and importance
2. It allows for comparison of different investments or businesses: CAGR allows investors and business owners to compare the performance of different investments or businesses over a period of time. This can be useful for making decisions about which investments or businesses to invest in or pursue.
3. It allows for long-term planning: CAGR can be used to project the future performance of an investment or business over a long period of time. This can be useful for making decisions about how to allocate resources and plan for the future.
Overall, calculating CAGR can be a useful tool for investors and business owners to measure the performance of their investments or businesses over a period of time. It provides a more accurate measure of performance than simple average returns, allows for comparison of different investments or businesses, and allows for long-term planning.
Examples of Calculating Compound Annual Growth Rate (CAGR) for Different Investment Scenarios
Calculating the Compound Annual Growth Rate (CAGR) of an investment is a great way to measure the performance of your investments over time. CAGR is a measure of the average rate of return over a period of time, and it takes into account the effects of compounding. It’s a useful tool for comparing investments and understanding how your investments have grown over time.
Let’s look at a few examples of how to calculate CAGR for different investment scenarios.
Example 1: Calculating CAGR for a Single Investment
Let’s say you invested \$10,000 in a stock five years ago. Today, the value of your investment is \$15,000. To calculate the CAGR of this investment, you would use the following formula:
CAGR = (Ending Value / Beginning Value)^(1/Number of Years) – 1
In this example, the CAGR would be calculated as follows:
CAGR = (\$15,000 / \$10,000)^(1/5) – 1
CAGR = 1.14 – 1
CAGR = 0.14 or 14%
Example 2: Calculating CAGR for Multiple Investments
Let’s say you invested \$10,000 in two different stocks five years ago. Today, the value of your investments is \$15,000 and \$20,000 respectively. To calculate the CAGR of these investments, you would use the following formula:
CAGR = (Ending Value / Beginning Value)^(1/Number of Years) – 1
In this example, the CAGR would be calculated as follows:
CAGR = (\$15,000 / \$10,000)^(1/5) – 1
CAGR = (\$20,000 / \$10,000)^(1/5) – 1
CAGR = 1.14 – 1
CAGR = 1.32 – 1
CAGR = 0.14 or 14%
CAGR = 0.32 or 32%
As you can see, calculating the CAGR of your investments is a great way to measure the performance of your investments over time. It’s a useful tool for comparing investments and understanding how your investments have grown over time.
Conclusion
The Compound Annual Growth Rate (CAGR) is a useful tool for measuring the growth of an investment over a period of time. It is important to understand how to calculate CAGR in order to accurately assess the performance of an investment. By using the formula, examples, and tips provided in this article, you can easily calculate CAGR and use it to make informed decisions about your investments.
Helen Barklam
Helen Barklam is a journalist and writer with more than 25 years experience. Helen has worked in a wide range of different sectors, including health and wellness, sport, digital marketing, home design and finance. | 2,702 | 12,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-38 | latest | en | 0.957208 |
https://forums.wolfram.com/mathgroup/archive/1998/Sep/msg00284.html | 1,721,865,456,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00520.warc.gz | 221,372,244 | 7,613 | Re: Integral?
• To: mathgroup at smc.vnet.net
• Subject: [mg14148] Re: [mg14112] Integral?
• From: "Jens-Peer Kuska" <kuska at linmpi.mpg.de>
• Date: Mon, 28 Sep 1998 18:57:14 -0400
• Sender: owner-wri-mathgroup at wolfram.com
```Hi,
if You put the nested Integrate[] in one with
Integrate[Exp[x]*Exp[y]*p[x,y],{x,-Infinity,Infinity},{y,-Infinity,Infinity}
]
You get some messages and nice conditions for r and s but the If[]
expression
returns a result.
Hope that helps
Jens
-----Original Message-----
From: Torben Mikael Hansen <torhans at image.dk> To: mathgroup at smc.vnet.net
Subject: [mg14148] [mg14112] Integral?
>Hi
>How do I make Mathematica perform the double integral
>Integrate[Integrate[Exp[x]*Exp[y]*p[x,y],{x,-Infinity,Infinity}],{y,-Infini
ty,Infinity}]
>
>where
>p[x_,y_]:=1/(2 Pi s^2 Sqrt[1-r^2])*Exp[-(x^2-2 r s x y +y^2)/(2 s^2
>(1-r^2))]
>and I know
>r is a real, positive number less than 1 s is a real, positive number
>
>Another system gives
>Exp[(r+1) s^2]
>
>Regards
>Torben M. Hansen
>
>
>
```
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• Next by thread: Re: Integral? | 421 | 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} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.683134 |
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-9-cumulative-review-exercises-page-685/5 | 1,571,745,961,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987817685.87/warc/CC-MAIN-20191022104415-20191022131915-00293.warc.gz | 908,892,347 | 12,243 | ## Introductory Algebra for College Students (7th Edition)
The solution is $(3, 4)$.
We can use the substitution method for this problem because in the second equation, $y$ is already isolated. We can use the expression for $y$ to substitute into the first equation: $$3x - 2(10 - 2x) = 1$$ Use the distributive property to simplify: $$3x - (2)(10) - (2)(-2x) = 1$$ Multiply out the terms: $$3x - 20 + 4x = 1$$ Group like terms: $$(3x + 4x) - 20 = 1$$ Combine like terms: $$7x - 20 = 1$$ Add $20$ to each side to isolate the variable to one side of the equation and the constant to the other: $$7x = 21$$ Divide each side of the equation by $7$ to solve for $x$: $$x = 3$$ Now that we have the value for $x$, we can substitute this value into the second equation to find $y$: $$y = 10 - 2(3)$$ Divide first, according to order of operations: $$y = 10 - 6$$ Subtract to solve for $y$: $$y = 4$$ The solution is $(3, 4)$. | 290 | 920 | {"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} | 4.71875 | 5 | CC-MAIN-2019-43 | latest | en | 0.855094 |
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The International Physics Olympiad (IPhO) is an international physics competition for secondary school pupils. The first such competition was organised by Prof. Czesław Ścisłowski in Warsaw (Poland) in 1967. Since that era the International Physics Olympiads have been organised, with few variations that will be discussed later, in a different country every year.
## SYLLABUS:
1. Theoretical Part
The first column contains the main entries while the second column contains comments and remarks if necessary.
1. Mechanics
a) Foundation of kinematics of a point mass Vector description of the position of the point mass, velocity and acceleration as vectors b) Newton's laws, inertial systems Problems may be set on changing mass c) Closed and open systems, momentum and energy, work, power d) Conservation of energy, conservation of linear momentum, impulse e) Elastic forces, frictional forces, the law of gravitation, potential energy and work in a gravitational field Hooke's law, coefficient of friction (F/R = const), frictional forces, static and kinetic, choice of zero of potential energy f) Centripetal acceleration, Kepler's laws
1. Mechanics of Rigid Bodies
a) Statics, center of mass, torque Couples, conditions of equilibrium of bodies b) Motion of rigid bodies, translation, rotation, angular velocity, angular acceleration, conservation of angular momentum Conservation of angular momentum about fixed axis only c) External and internal forces, equation of motion of a rigid body around the fixed axis, moment of inertia, kinetic energy of a rotating body Parallel axes theorem (Steiner's theorem), additivity of the moment of inertia d) Accelerated reference systems, inertial forces Knowledge of the Coriolis force formula is not required
1. Hydromechanics
No specific questions will be set on this but students would be expected to know the elementary concepts of pressure, buoyancy and the continuity law.
1. Thermodynamics and Molecular Physics
• Oscillations and waves
• Electric Charge and Electric Field
• Current and Magnetic Field
• Electromagnetic waves
• Quantum Physics
• Relativity
• Matter
B. Practical PartThe Theoretical Part of the Syllabus provides the basis for all the experimental problems. The experimental problems given in the experimental contest should contain measurements
a) Internal energy, work and heat, first and second laws of thermodynamics Thermal equilibrium, quantities depending on state and quantities depending on process b) Model of a perfect gas, pressure and molecular kinetic energy, Avogadro's number, equation of state of a perfect gas, absolute temperature Also molecular approach to such simple phenomena in liquids and solids as boiling, melting etc. c) Work done by an expanding gas limited to isothermal and adiabatic processes Proof of the equation of the adiabatic process is not required d) The Carnot cycle, thermodynamic efficiency, reversible and irreversible processes, entropy (statistical approach), Boltzmann factor Entropy as a path independent function, entropy changes and reversibility, quasistatic processes a) Harmonic oscillations, equation of harmonic oscillation Solution of the equation for harmonic motion, attenuation and resonance -qualitatively b) Harmonic waves, propagation of waves, transverse and longitudinal waves, linear polarization, the classical Doppler effect, sound waves Displacement in a progressive wave and understanding of graphical representation of the wave, measurements of velocity of sound and light, Doppler effect in one dimension only, propagation of waves in homogeneous and isotropic media, reflection and refraction, Fermat's principle c) Superposition of harmonic waves, coherent waves, interference, beats, standing waves Realization that intensity of wave is proportional to the square of its amplitude. Fourier analysis is not required but candidates should have some understanding that complex waves can be made from addition of simple sinusoidal waves of different frequencies. Interference due to thin films and other simple systems (final formulae are not required), superposition of waves from secondary sources (diffraction) a) Conservation of charge, Coulomb's law b) Electric field, potential, Gauss' law Gauss' law confined to simple symmetric systems like sphere, cylinder, plate etc., electric dipole moment c) Capacitors, capacitance, dielectric constant, energy density of electric field a) Current, resistance, internal resistance of source, Ohm's law, Kirchhoff's laws, work and power of direct and alternating currents, Joule's law Simple cases of circuits containing non-ohmic devices with known V-I characteristics b) Magnetic field (B) of a current, current in a magnetic field, Lorentz force Particles in a magnetic field, simple applications like cyclotron, magnetic dipole moment c) Ampere's law Magnetic field of simple symmetric systems like straight wire, circular loop and long solenoid d) Law of electromagnetic induction, magnetic flux, Lenz's law, self-induction, inductance, permeability, energy density of magnetic field e) Alternating current, resistors, inductors and capacitors in AC-circuits, voltage and current (parallel and series) resonances Simple AC-circuits, time constants, final formulae for parameters of concrete resonance circuits are not required a) Oscillatory circuit, frequency of oscillations, generation by feedback and resonance b) Wave optics, diffraction from one and two slits, diffraction grating,resolving power of a grating, Bragg reflection, c) Dispersion and diffraction spectra, line spectra of gases d) Electromagnetic waves as transverse waves, polarization by reflection, polarizers Superposition of polarized waves e) Resolving power of imaging systems f) Black body, Stefan-Boltzmanns law Planck's formula is not required a) Photoelectric effect, energy and impulse of the photon Einstein's formula is required b) De Broglie wavelength, Heisenberg's uncertainty principle a) Principle of relativity, addition of velocities, relativistic Doppler effect b) Relativistic equation of motion, momentum, energy, relation between energy and mass, conservation of energy and momentum a) Simple applications of the Bragg equation b) Energy levels of atoms and molecules (qualitatively), emission, absorption, spectrum of hydrogen like atoms c) Energy levels of nuclei (qualitatively), alpha-, beta- and gamma-decays, absorption of radiation, halflife and exponential decay, components of nuclei, mass defect, nuclear reactions
70% | 1,341 | 6,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | longest | en | 0.863293 |
http://www.algebra.com/algebra/homework/Graphs.faq.question.173919.html | 1,386,186,805,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163037167/warc/CC-MAIN-20131204131717-00074-ip-10-33-133-15.ec2.internal.warc.gz | 254,485,807 | 4,353 | # SOLUTION: 2x + y -3 Can you show me how to graph this please?????
Algebra -> -> SOLUTION: 2x + y -3 Can you show me how to graph this please????? Log On
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Click here to see ALL problems on Graphs Question 173919: 2x + y -3 Can you show me how to graph this please?????Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!If this equation is 2x+y-3=0 then this is the graph if its 2x+y=-3 re write it.... | 168 | 598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2013-48 | latest | en | 0.853919 |
http://conversion.org/mass/kilogram/mite | 1,550,589,694,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247490225.49/warc/CC-MAIN-20190219142524-20190219164524-00571.warc.gz | 64,494,158 | 6,883 | Kilogram to mite conversion
Conversion number between Kilogram [kg] and mite is 308647.16705883. This means, that Kilogram is bigger unit than mite.
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• 1 Kilogram = (exactly) (1) / (3.2399455*10^-06) = 308647.16705883 mite
• 1 mite = (exactly) (3.2399455*10^-06) / (1) = 3.2399455 × 10-6 Kilogram
• ? Kilogram × (1 ("kg"/"Kilogram")) / (3.2399455*10^-06 ("kg"/"mite")) = ? mite
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Kilogrammite
00
103086471.6705883
206172943.3411766
309259415.0117649
4012345886.682353
5015432358.352941
6018518830.02353
7021605301.694118
8024691773.364706
9027778245.035295
10030864716.705883
11033951188.376471
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Kilogram
Definition of Kilogram unit: ≡ 1 kg. SI base unit. Mass of the prototype near Paris (≈ mass of 1 L of water)
Convert Mite to other unit:
mite
Definition of mite unit: ≡ 1⁄20 gr . = 1 grain / 20 = 3.2399455 mg
← Back to Mass units | 564 | 1,724 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-09 | latest | en | 0.554345 |
https://winter.group.shef.ac.uk/chemdex/chemistry/vn_cn_grid?element=g2 | 1,702,241,185,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102637.84/warc/CC-MAIN-20231210190744-20231210220744-00391.warc.gz | 674,255,216 | 6,883 | # Valence Number (VN) vs. Coordination Number (CN) grid
## Valence number (VN) vs. coordination number (CN) for oxygen (O)
Element chosen: O
The number of compounds found is 9141.
The intensity of the colour of each table cell is related to the percentage of compounds of O associated with the valence number (VN) and coordination number (CN) for that cell. Select any entry in the table to see compounds of with that VN and CN.
VN➝
% of each VN➝ 1 0 37.5 0 57.9 0 3.6 0 0 % of each CN↓ Count↓
CN = 0 0 0 0 0 0 0 0 0 0 0 0
CN = 1 1 0 11 0 22.1 0 0 0 0 34 3112
CN = 2 0 0 26.5 0 9.8 0 3.4 0 0 40 3634
CN = 3 0 0 0.011 0 26 0 0 0 0 26 2374
CN = 4 0 0 0 0 0.011 0 0 0 0 0.01 1
CN = 5 0 0 0 0 0 0 0 0 0 0 0
CN = 6 0 0 0 0 0 0 0.22 0 0 0.2 20
CN = 7 0 0 0 0 0 0 0 0 0 0 0
CN = 8 0 0 0 0 0 0 0 0 0 0 0
CN = 9 0 0 0 0 0 0 0 0 0 0 0
CN = 10 0 0 0 0 0 0 0 0 0 0 0
CN = 11 0 0 0 0 0 0 0 0 0 0 0
CN = 12 0 0 0 0 0 0 0 0 0 0 0
Average CN➝ 1 - 1.71 - 2.07 - 2.24 - - 1.93
Compound count➝ 90 0 3430 0 5291 0 330 0 0 9141
## Valence number (VN) vs. coordination number (CN) for sulfur (S)
Element chosen: S
The number of compounds found is 2402.
The intensity of the colour of each table cell is related to the percentage of compounds of S associated with the valence number (VN) and coordination number (CN) for that cell. Select any entry in the table to see compounds of with that VN and CN.
VN➝
% of each VN➝ 0.17 0 46.5 0 43.5 0 9.9 0 0 % of each CN↓ Count↓
CN = 0 0 0 0 0 0 0 0 0 0 0 0
CN = 1 0.17 0 20.1 0 3.3 0 0 0 0 24 565
CN = 2 0 0 26.4 0 15.6 0 0.042 0 0 42 1010
CN = 3 0 0 0 0 24.5 0 0.12 0 0 25 592
CN = 4 0 0 0 0 0.08 0 9.5 0 0 10 230
CN = 5 0 0 0 0 0 0 0 0 0 0 0
CN = 6 0 0 0 0 0 0 0.21 0 0 0.2 5
CN = 7 0 0 0 0 0 0 0 0 0 0 0
CN = 8 0 0 0 0 0 0 0 0 0 0 0
CN = 9 0 0 0 0 0 0 0 0 0 0 0
CN = 10 0 0 0 0 0 0 0 0 0 0 0
CN = 11 0 0 0 0 0 0 0 0 0 0 0
CN = 12 0 0 0 0 0 0 0 0 0 0 0
Average CN➝ 1 - 1.57 - 2.49 - 4.02 - - 2.21
Compound count➝ 4 0 1117 0 1044 0 237 0 0 2402
## Valence number (VN) vs. coordination number (CN) for selenium (Se)
Element chosen: Se
The number of compounds found is 546.
The intensity of the colour of each table cell is related to the percentage of compounds of Se associated with the valence number (VN) and coordination number (CN) for that cell. Select any entry in the table to see compounds of with that VN and CN.
VN➝
% of each VN➝ 3.7 0 63.4 0 28.2 0 4.8 0 0 % of each CN↓ Count↓
CN = 0 0 0 0.18 0 0 0 0 0 0 0.2 1
CN = 1 3.5 0 8.2 0 7.5 0 0 0 0 19 105
CN = 2 0.18 0 50.7 0 0.37 0 0.7 0 0 52 284
CN = 3 0 0 2.6 0 14.7 0 0 0 0 17 94
CN = 4 0 0 1.6 0 3.5 0 3.8 0 0 9 49
CN = 5 0 0 0 0 0.7 0 0.18 0 0 1 5
CN = 6 0 0 0 0 1.5 0 0 0 0 1 8
CN = 7 0 0 0 0 0 0 0 0 0 0 0
CN = 8 0 0 0 0 0 0 0 0 0 0 0
CN = 9 0 0 0 0 0 0 0 0 0 0 0
CN = 10 0 0 0 0 0 0 0 0 0 0 0
CN = 11 0 0 0 0 0 0 0 0 0 0 0
CN = 12 0 0 0 0 0 0 0 0 0 0 0
Average CN➝ 1.05 - 1.96 - 2.79 - 3.73 - - 2.24
Compound count➝ 20 0 346 0 154 0 26 0 0 546
## Valence number (VN) vs. coordination number (CN) for tellurium (Te)
Element chosen: Te
The number of compounds found is 453.
The intensity of the colour of each table cell is related to the percentage of compounds of Te associated with the valence number (VN) and coordination number (CN) for that cell. Select any entry in the table to see compounds of with that VN and CN.
VN➝
% of each VN➝ 2 0 44.2 0 41.9 0 11.9 0 0 % of each CN↓ Count↓
CN = 0 0 0 0.22 0 0 0 0 0 0 0.2 1
CN = 1 1.8 0 1.8 0 1.5 0 0 0 0 5 23
CN = 2 0.22 0 31.3 0 0.44 0 0 0 0 32 145
CN = 3 0 0 5.5 0 15.2 0 0 0 0 21 94
CN = 4 0 0 5.3 0 14.3 0 2.6 0 0 22 101
CN = 5 0 0 0 0 4.4 0 0.22 0 0 5 21
CN = 6 0 0 0 0 6 0 8.8 0 0 15 67
CN = 7 0 0 0 0 0 0 0.22 0 0 0.2 1
CN = 8 0 0 0 0 0 0 0 0 0 0 0
CN = 9 0 0 0 0 0 0 0 0 0 0 0
CN = 10 0 0 0 0 0 0 0 0 0 0 0
CN = 11 0 0 0 0 0 0 0 0 0 0 0
CN = 12 0 0 0 0 0 0 0 0 0 0 0
Average CN➝ 1.11 - 2.32 - 3.89 - 5.56 - - 3.34
Compound count➝ 9 0 200 0 190 0 54 0 0 453
## Valence number (VN) vs. coordination number (CN) for polonium (Po)
Element chosen: Po
The number of compounds found is 23.
The intensity of the colour of each table cell is related to the percentage of compounds of Po associated with the valence number (VN) and coordination number (CN) for that cell. Select any entry in the table to see compounds of with that VN and CN.
VN➝
% of each VN➝ 0 0 4.3 0 13 0 82.6 0 0 % of each CN↓ Count↓
CN = 0 0 0 0 0 0 0 0 0 0 0 0
CN = 1 0 0 0 0 0 0 0 0 0 0 0
CN = 2 0 0 0 0 0 0 0 0 0 0 0
CN = 3 0 0 0 0 0 0 0 0 0 0 0
CN = 4 0 0 0 0 0 0 13 0 0 13 3
CN = 5 0 0 0 0 0 0 0 0 0 0 0
CN = 6 0 0 0 0 13 0 65.2 0 0 78 18
CN = 7 0 0 0 0 0 0 0 0 0 0 0
CN = 8 0 0 4.3 0 0 0 4.3 0 0 9 2
CN = 9 0 0 0 0 0 0 0 0 0 0 0
CN = 10 0 0 0 0 0 0 0 0 0 0 0
CN = 11 0 0 0 0 0 0 0 0 0 0 0
CN = 12 0 0 0 0 0 0 0 0 0 0 0
Average CN➝ - - 8 - 6 - 5.79 - - 5.91
Compound count➝ 0 0 1 0 3 0 19 0 0 23
## Valence number (VN) vs. coordination number (CN) for livermorium (Lv)
Element chosen: Lv
The number of compounds found is 0. | 2,892 | 4,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-50 | longest | en | 0.714717 |
http://www.vias.org/physics/bk3_03_06_01.html | 1,544,747,317,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825123.5/warc/CC-MAIN-20181214001053-20181214022553-00306.warc.gz | 508,120,929 | 4,391 | Lectures on Physics has been derived from Benjamin Crowell's Light and Matter series of free introductory textbooks on physics. See the editorial for more information....
# The Doppler Effect
The pattern of waves made by a point source moving to the right across the water. Note the shorter wavelength of the forward-emitted waves and the longer wavelength of the backwardgoing ones. (PSSC Physics)
The figure shows the wave pattern made by the tip of a vibrating rod which is moving across the water. If the rod had been vibrating in one place, we would have seen the familiar pattern of concentric circles, all centered on the same point. But since the source of the waves is moving, the wavelength is shortened on one side and lengthened on the other. This is known as the Doppler effect.
Note that the velocity of the waves is a fixed property of the medium, so for example the forward-going waves do not get an extra boost in speed as would a material object like a bullet being shot forward from an airplane.
We can also infer a change in frequency. Since the velocity is constant, the equation v=fλ tells us that the change in wavelength must be matched by an opposite change in frequency: higher frequency for the waves emitted forward, and lower for the ones emitted backward. The frequency Doppler effect is the reason for the familiar dropping-pitch sound of a race car going by. As the car approaches us, we hear a higher pitch, but after it passes us we hear a frequency that is lower than normal.
The Doppler effect will also occur if the observer is moving but the source is stationary. For instance, an observer moving toward a stationary source will perceive one crest of the wave, and will then be surrounded by the next crest sooner than she otherwise would have, because she has moved toward it and hastened her encounter with it. Roughly speaking, the Doppler effect depends only the relative motion of the source and the observer, not on their absolute state of motion (which is not a well-defined notion in physics) or on their velocity relative to the medium.
Restricting ourselves to the case of a moving source, and to waves emitted either directly along or directly against the direction of motion, we can easily calculate the wavelength, or equivalently the frequency, of the Doppler-shifted waves. Let v be the velocity of the waves, and vs the velocity of the source. The wavelength of the forward-emitted waves is shortened by an amount vsT equal to the distance traveled by the source over the course of one period. Using the definition f=1/T and the equation v=fλ, we find for the wavelength λ' of the Doppler-shifted wave the equation
A similar equation can be used for the backward-emitted waves, but with a plus sign rather than a minus sign.
Doppler-shifted sound from a race car.
Doppler shift of the light emitted by a race car.
#### Optional Topic: A Note on Doppler Shifts of Light
If Doppler shifts depend only on the relative motion of the source and receiver, then there is no way for a person moving with the source and another person moving with the receiver to determine who is moving and who isn't. Either can blame the Doppler shift entirely on the other's motion and claim to be at rest herself. This is entirely in agreement with the principle stated originally by Galileo that all motion is relative.
On the other hand, a careful analysis of the Doppler shifts of water or sound waves shows that it is only approximately true, at low speeds, that the shifts just depend on the relative motion of the source and observer. For instance, it is possible for a jet plane to keep up with its own sound waves, so that the sound waves appear to stand still to the pilot of the plane. The pilot then knows she is moving at exactly the speed of sound. The reason this doesn't disprove the relativity of motion is that the pilot is not really determining her absolute motion but rather her motion relative to the air, which is the medium of the sound waves.
Einstein realized that this solved the problem for sound or water waves, but would not salvage the principle of relative motion in the case of light waves, since light is not a vibration of any physical medium such as water or air. Beginning by imagining what a beam of light would look like to a person riding a motorcycle alongside it, Einstein eventually came up with a radical new way of describing the universe, in which space and time are distorted as measured by observers in different states of motion. As a consequence of this Theory of Relativity, he showed that light waves would have Doppler shifts that would exactly, not just approximately, depend only on the relative motion of the source and receiver.
Discussion Questions
A
If an airplane travels at exactly the speed of sound, what would be the wavelength of the forward-emitted part of the sound waves it emitted? How should this be interpreted, and what would actually happen? What happens if it's going faster than the speed of sound. The figure shows a fighter jet that has just accelerated past the speed of sound. The sudden decompression of the air causes water droplets to condense, forming a cloud; why is the cloud in the shape of a cone?
B If bullets go slower than the speed of sound, why can a supersonic fighter plane catch up to its own sound, but not to its own bullets?
C If someone inside a plane is talking to you, should their speech be Doppler shifted?
Last Update: 2010-11-11 | 1,133 | 5,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-51 | latest | en | 0.956108 |
https://www.jobilize.com/course/section/refractive-index-geometrical-optics-grade-10-by-openstax?qcr=www.quizover.com | 1,621,350,551,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989637.86/warc/CC-MAIN-20210518125638-20210518155638-00591.warc.gz | 796,711,501 | 21,082 | # 0.1 Geometrical optics - grade 10 (Page 5/14)
Page 5 / 14
When light travels from one medium to another, it will be bent away from its original path. When it travels from an optically dense medium like water or glass to a less dense medium like air, it will be refracted away from the normal ( [link] ). Whereas, if it travels from a less dense medium to a denser one, it will be refracted towards the normal ( [link] ).
Just as we defined an angle of reflection in the previous section, we can similarly define an angle of refraction as the angle between the surface normal and the refracted ray. This is shown in [link] .
## Refractive index
Which is easier to travel through, air or water? People usually travel faster through air. So does light! The speed of light and therefore the degree of bending of the light depends on the refractive index of material through which the light passes. The refractive index (symbol $n$ ) is the ratio of the speed of light in a vacuum to its speed in the material.
Refractive Index
The refractive index of a material is the ratio of the speed of light in a vacuum to its speed in the medium.
## Interesting fact
The symbol $c$ is used to represent the speed of light in a vacuum.
$c=299\phantom{\rule{0.277778em}{0ex}}792\phantom{\rule{0.277778em}{0ex}}485\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}$
For purposes of calculation, we use $3×{10}^{8}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}$ . A vacuum is a region with no matter in it, not even air. However, the speed of light in air is very close to that in a vacuum.
Refractive Index
The refractive index (symbol $n$ ) of a material is the ratio of the speed of light in a vacuum to its speed in the material and gives an indication of how difficult it is for light to get through the material.
$n=\frac{c}{v}$
where
$n$ = refractive index (no unit) $c$ = speed of light in a vacuum ( $3,00×{10}^{8}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}$ ) $v$ = speed of light in a given medium ( $\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}$ )
## Refractive index and speed of light
Using
$n=\frac{c}{v}$
we can also examine how the speed of light changes in different media, because the speed of light in a vacuum ( $c$ ) is constant.
If the refractive index $n$ increases, the speed of light in the material $v$ must decrease. Light therefore travels slowly through materials of high $n$ .
[link] shows refractive indices for various materials. Light travels slower in any material than it does in a vacuum, so all values for $n$ are greater than 1.
Medium Refractive Index Vacuum 1 Helium 1,000036 Air* 1,0002926 Carbon dioxide 1,00045 Water: Ice 1,31 Water: Liquid ( ${20}^{\circ }$ C) 1,333 Acetone 1,36 Ethyl Alcohol (Ethanol) 1,36 Sugar solution (30%) 1,38 Fused quartz 1,46 Glycerine 1,4729 Sugar solution (80%) 1,49 Rock salt 1,516 Crown Glass 1,52 Sodium chloride 1,54 Polystyrene 1,55 to 1,59 Bromine 1,661 Sapphire 1,77 Glass (typical) 1,5 to 1,9 Cubic zirconia 2,15 to 2,18 Diamond 2,419 Silicon 4,01
## Snell's law
Now that we know that the degree of bending, or the angle of refraction, is dependent on the refractive index of a medium, how do we calculate the angle of refraction?
how can chip be made from sand
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
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## Statistics
Probability using discrete distributions.
a. If you roll a single fair die and count the number of dots on top, what is the sample space of all possible outcomes? Are the outcomes equally likely?
b. Assign probabilities to the outcomes of the sample space of part (a). Do probabilities add up to 1? Should they add upto 1? describe.
c. What is the likelihood of getting a number less than 5 on a single throw?
d. What is the possibility of getting 5 or 6 on a single throw?
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M920064
Have any Question?
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Create a provider database and related reports and queries to capture contact information for potential PC component pro | 964 | 4,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.942244 |
http://www.oldquestionpapers.net/2018/08/du-b-el-ed-entrance-question-paper-2019.html | 1,709,240,332,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474853.43/warc/CC-MAIN-20240229202522-20240229232522-00764.warc.gz | 56,353,602 | 17,278 | DUET B.EL.ED. Entrance Question Paper with Answer Keys 2019-20
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## DUET B.EL.ED. Entrance Question Paper with Answer Keys 2019-20
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#### EXAMPLE QUESTIONS
Q1. Rakesh rolls down a marble from a fixed height on an inclined plane. At the bottom of the inclined plane, the marble will cover the shortest distance if
• The surface is made of marble”
• The surface is made of sand
• The surface is covered with oil
• The surface is made of glass
Q2. In which of the following situations, largest force will be experienced by a current carrying conductor when placed in a magnetic field?
• when the direction of the electric current is 450to the magnetic field
• when the direction of the electric current is at 900 to the magnetic field
• when the direction of the electric current is at 1200 to the magnetic field
• when the direction of the electric current is at 1800 to the magnetic field
Q3. A person with short -sightedness (myopia) is recommended
• To use convex lens
• To use a combination of concave and convex lens
• To use concave lens
• To use sun protecting glasses
Q4. If the government of one state decides not to allow migration from other states, Which Fundamental Right will be violated?
• Right to Freedom
• Right against Exploitation
• Right to Constitutional Remedies
• Right to Equality
Q5. The minute and the hour hands of a clock are 5 cm and 3 cm respectively. What is the sum of the distances travelled by their tips in a day?
• 856
• 720
• 792
• 900
Q6. A number is chosen at random among the first 50 natural numbers. The probability of the number chosen being a multiple of6or 12 is:
• 4/25
• 3/25
• 1/25
• 2/25
Q7. The missing number in series: 24, 18, 27, 60.75, ….. , 683.4375 is:
• 540
• 127.75
• 182.25
• 98.5
Q8. Which of the following is not related to Akbar’s idea of ‘Sulh-i-kul’:
• followed by Shahjahan and Jahangir
• An attempt to spread Islam
• way of governance
• Universal peace and tolerance
Q9. Which of the following does not hold true for ‘Collateral asset’?
• It is an asset that the borrower owns and is used as a guarantee to a lender until the loan is repair
• It may not be needed by self-help groups of poor women
• It is a loan given by the government at cheap rate to buy assets
• It may be in the form of landtitle or livestock
Q10. 50% of a number is added to the number itself. The percentage of the result that needs to be subtracted to get the original number back is:
• 33.33%
• 50%
• 25%
• 40% | 933 | 3,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | latest | en | 0.850873 |
https://de.mathworks.com/matlabcentral/cody/problems/44314-a-simple-tide-gauge-with-matlab/solutions/1373048 | 1,590,655,656,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347398233.32/warc/CC-MAIN-20200528061845-20200528091845-00016.warc.gz | 316,671,822 | 15,637 | Cody
# Problem 44314. A Simple Tide Gauge with MATLAB
Solution 1373048
Submitted on 8 Dec 2017 by Smith
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = [5 8 10 12 8 13 14 10 10 15]; tide_correct = 1; assert(isequal(gauge(A),tide_correct))
2 Pass
A = [15 16 11 9 10 15 7 12 6 11 5 6]; tide_correct = 0; assert(isequal(gauge(A),tide_correct))
3 Pass
A = [9 15 3 9 5 18 4 17 18 19 8 13 12 21 17 24]; tide_correct = 1; assert(isequal(gauge(A),tide_correct))
4 Pass
A = [22 12 22 12 9 14 17 16 15 8 13 6 10 7 13 3]; tide_correct = 0; assert(isequal(gauge(A),tide_correct)) | 276 | 703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-24 | latest | en | 0.506812 |
https://respaper.com/icse/159/8791-pdf.html | 1,653,825,781,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662644142.66/warc/CC-MAIN-20220529103854-20220529133854-00192.warc.gz | 538,129,000 | 11,017 | Trending ▼ ResFinder
# ICSE Class X Board Specimen 2017 : Technical Drawing Applications
6 pages, 12 questions, 0 questions with responses, 0 total responses, 0 0
ICSE Indian Certificate of Secondary Education (ICSE), New Delhi
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TECHNICAL DRAWING APPLICATIONS (Three hours) Answers to this paper must be written on the paper provided separately. You will not be allowed to draw/write during the first 15 minutes. This time is to be spent in reading the question paper. The time given at the head of this paper is the time allowed for writing the answers. Attempt five questions in all. You must attempt three questions from Section A and two questions from Section B. Each Section should be answered on a separate paper. All questions must be answered in full scale. All construction lines must be shown. All dimensions are in millimetres unless specified otherwise. The intended marks for questions or parts of questions are given in brackets [ ]. SECTION A (48 Marks) Answer any three questions from this Section. Question 1 Construct a Parabola of 120 mm base and axis of 80 mm by the Tangent method. [16] Question 2. (a) Construct a parallelogram ABCD having its perimeter = 240 mm. AB : BC = 5 : 3 and angle BAD = 60 . On side AD draw an equilateral triangle ADE. (b) [8] Draw the front view, the top view and the left hand side view of a hexagonal right prism when its axis is parallel to the horizontal plane and parallel to the vertical plane. Two sides of the base are parallel to the horizontal plane. Use the THIRD ANGLE method of projection. [8] 111 ICSE Specimen Question Paper
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Tags : icse free download, icse papers, pdf download 2016 2017 2018 2019 2020 2021, icse sample papers, icse books, portal for icse india, icse question bank, indian certificate of secondary education, icse question papers with answers, icse model test papers, solved past board question papers of icse last year, previous years solved question papers, free online icse solved question paper, icse syllabus, india icse board sample questions papers, last 10 years icse papers, icse question papers 2017 - 2018, icse guess sample questions papers, icse important questions, class 10 specimen / guess / mock papers, icse pre board question papers, icse 2020 - 2021 pre-board examination | 570 | 2,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.86542 |
https://iotdesignpro.com/comment/97580 | 1,679,306,452,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00596.warc.gz | 384,065,335 | 32,349 | IoT Based Water Level Indicator Using Ultrasonic Sensor
Today we will be working on a Water level indicator whose data can be monitored through a webpage over a local area network. The water level will be detected by using an ultrasonic distance measurement sensor. We have previously built another IoT based tank water level monitoring system using a float sensor, but in this project, we will use an ultrasonic sensor for detecting the level of water.
You must have seen several water level indicators but most of them use special float sensors that are not easily available and neither simple to use. But in this tutorial, we will be using an ultrasonic sensor for measuring the water level. You will come to know how further in the article.
For making the project more user friendly, we will be integrating it with a local webserver through which you can monitor the data from any device connected to the same Wi-Fi as your ESP board.
Ultrasonic Sensor Work and Use
Before understanding the working of the project, let us first see how an ultrasonic sensor works. A typical HC-SR04 ultrasonic sensor used in this project is shown below.
Ultrasonic sensor is an electronic device that measures the distance of a target object by emitting ultrasonic sound waves and converts the reflected sound into an electrical signal.
Ultrasonic waves travel faster than the speed of audible sound (i.e. the sound that humans can hear). Ultrasonic sensors have two main components: the transmitter (which emits the sound using piezoelectric crystals) and the receiver (which encounters the sound after it has traveled to and from the target).
To calculate the distance between the sensor and the object, the sensor measures the time it takes between the emissions of the sound by the transmitter to its contact with the receiver. The formula for this calculation is-
`D = ½ T x C (where D is the distance, T is the time, and C is the speed of sound ~ 343 meters/second).`
So where can we use these sensors? We have previously used these sensors in Contactless Temperature measurementIoT based Inventory management, and many other projects, apart from this, it can also be used in Robot navigation, as well as factory automation. Water-level sensing is another good use and can be accomplished by positioning one sensor above a water surface. Another aquatic application is to use these sensors to “see” the bottom of a body of water, traveling through the water, but reflecting off the bottom surface below. We have used the same principle in our Flood Detection and Monitoring system as well.
Ultrasonic Sensor for Measuring Water Level
So, now as we know the working of the ultrasonic sensor, it is pretty straightforward to understand the working of the project. We just have to read the values from the sensor and convert it to CMs and after doing all that, we have to publish the data in a local webserver that will be created by our NodeMCU board after getting connected to Wi-Fi.
Required Components
For making this project, we will be needing a few basic components.
ESP8266 NodeMCU board: This will be the heart of our whole project.
HC-SR04 Ultrasonic Sensor: This will be used for sensing the level through ultrasonic sound waves as explained earlier.
Jumper wires: As we are using a breadboard, jumper or hookup wires are the way to go for connections.
Interfacing Ultrasonic Sensor with NodeMCU
As we are not using many components, the IoT water level indicator circuit diagram used in this project is fairly simple. We just need to connect the HC-SR04 Ultrasonic sensor to our NodeMCU module. The rest of the work will be done via the software part itself.
Here is the circuit diagram for the connection.
I used a breadboard to connect my ultrasonic sensor with the NodeMCU and then used connecting wires to make the connection. The testing set-up of mine looks like this below, later we will use longer jumper wires to mount the sensor over a demo water tank.
Programming NodeMCU to Read Water Level and Display on the Webserver
Now as we have finished making the connections. We will be proceeding to the coding part of the project. Let me clear the algorithm that we will be following while making the code. The complete code can be found at the bottom of this page.
What our Arduino code needs to do is, first of all, connect to the network whose SSID and password are mentioned in the code. After successfully getting, it needs to spit out the Local IP address to the serial monitor, so that we may access the webpage created by it. After doing so, it should start dealing with the ultrasonic sensor and after calculating the distance, it must send the data to the webpage that we have created.
Let us understand it part by part, so first of all, we are including the necessary libraries and the global variables that will be used further in the code. Along with that, we are also providing the SSID and passwords of our Wi-Fi router and initiating the webserver at port number 80.
```#include <ESP8266WiFi.h>
#include <WiFiClient.h>
#include <ESP8266WebServer.h>
int TRIGGER = D3;
int ECHO = D2;
// Replace with your network credentials
const char* ssid = "xxxxx";
ESP8266WebServer server(80); //instantiate server at port 80 (http port)
String page = "";
int data;
```
Further, in the setup part, we have set the pin mode of the pins as input and output and have stored our page in the page variable using HTML. We have kept a single line HTML code to make up our webpage. In the setup part itself, we get connected to the Wi-Fi router and print the IP address to the serial monitor.
The HTML page that we have created refreshes itself every three seconds. You may change it in the code itself.
```void setup(void){
pinMode(TRIGGER, OUTPUT);
pinMode(ECHO, INPUT);
delay(1000);
Serial.begin(115200);
Serial.println("");
// Wait for connection
while (WiFi.status() != WL_CONNECTED) {
delay(500);
Serial.print(".");
}
Serial.println("");
Serial.print("Connected to ");
Serial.println(ssid);
Serial.println(WiFi.localIP());
server.on("/", [](){
page = "<head><meta http-equiv=\"refresh\" content=\"3\"></head><center><h1>Web based Water Level monitor</h1><h3>Current water level is :-</h3> <h4>"+String(data)+"</h4></center>";
server.send(200, "text/html", page);
});
server.begin();
Serial.println("Web server started!");
}```
Most of the work is done, now in the loop part, we just calculate the distance from the ultrasonic sensor, and after storing it, we publish the data over our webpage. Our webpage has been made in such a way that, it refreshes every 3 seconds.
```void loop(void){
digitalWrite(TRIGGER, LOW);
delayMicroseconds(2);
digitalWrite(TRIGGER, HIGH);
delayMicroseconds(10);
digitalWrite(TRIGGER, LOW);
long duration = pulseIn(ECHO, HIGH);
data = (duration/2) / 29.09;
server.handleClient();
}```
This is pretty much about the code, Now just upload the code, and let us see the project in action.
IoT Based Water Level Indicator Testing and Working
This is the webpage that we have created using our NodeMCU board. It contains a heading that is, Web-based Water Level Indicator, below which is printing the live data coming through the HCSR04 Ultrasonic sensor. The water level values are printed in CMs, the less the value, the less empty the container, and vice versa.
Before directly testing the water level, we tested the distance measurement. It worked like charm as you may see from the above and below images.
After testing this, we set up a water level apparatus for demonstrating the working. We affixed the ultrasonic sensor on top of the jar and also placed a ruler for scale. This way, the monitor will show the data of how empty the jar is in CMs. Here is how the setup looks like
Now here is the working, firstly, we kept the jar empty, so it should display 20 in the server as the jar is 20cm deep and is empty as of now.
Now we have filled the jar by half so now it should display half the value of 20.
For seeing the working in detail, you may see the video we have attached below. Hope you enjoyed the article and learned something useful from it. If you have any questions, you can leave them in the comment section below.
Code
#include <ESP8266WiFi.h>
#include <WiFiClient.h>
#include <ESP8266WebServer.h>
int TRIGGER = D3;
int ECHO = D2;
// Replace with your network credentials
const char* ssid = "Comet";
ESP8266WebServer server(80); //instantiate server at port 80 (http port)
String page = "";
int data;
void setup(void){
pinMode(TRIGGER, OUTPUT);
pinMode(ECHO, INPUT);
delay(1000);
Serial.begin(115200);
Serial.println("");
// Wait for connection
while (WiFi.status() != WL_CONNECTED) {
delay(500);
Serial.print(".");
}
Serial.println("");
Serial.print("Connected to ");
Serial.println(ssid);
Serial.println(WiFi.localIP());
server.on("/", [](){
page = "<head><meta http-equiv=\"refresh\" content=\"3\"></head><center><h1>Web based Water Level monitor</h1><h3>Current water level is :-</h3> <h4>"+String(data)+"</h4></center>";
server.send(200, "text/html", page);
});
server.begin();
Serial.println("Web server started!");
}
void loop(void){
digitalWrite(TRIGGER, LOW);
delayMicroseconds(2);
digitalWrite(TRIGGER, HIGH);
delayMicroseconds(10);
digitalWrite(TRIGGER, LOW);
long duration = pulseIn(ECHO, HIGH);
data = (duration/2) / 29.09;
server.handleClient();
}
Video
Hi, I followed the above but…
Hi, I followed the above but in my case I get a constant result of 0cm no matter what I put in front of the sensor.
If I add the following bit of code the result I get is 0.96cm and 58 but again, no matter what I put in front of the sensor.
float test = (duration/2) / 29.09;
Serial.print(duration);
Serial.println(" \n");
Serial.print(test);
Serial.println(" cm");
Any idea what could be wrong?
Hi, Alfred... Same thing…
Hi, Alfred...
Same thing happened for me. 0cm.
Ever figure this out?
Thanks!
Rick
After I originally commented…
After I originally commented I appear to have clicked the -Notify me when new comments are added- checkbox and from now on each time a comment is added I receive 4 emails with the same comment. There has to be a way you can remove me from that service? Thanks a lot!|
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https://ar5iv.labs.arxiv.org/html/1009.3990 | 1,709,540,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476432.11/warc/CC-MAIN-20240304065639-20240304095639-00183.warc.gz | 104,902,772 | 35,861 | # A result of Lemmermeyer on class numbers
Paul Monsky Brandeis University, Waltham MA 02454-9110, USA
monsky@brandeis.edu
###### Abstract
I present Franz Lemmermeyerβs proof that if $p$ is a prime $\equiv 9\pod{16}$ then the class number of $\mathbb{Q}\left(p^{\frac{1}{4}}\right)$ is $\equiv 2\pod{4}$.
Let $p$ be a prime $\equiv 1\pod{4}$. Then the class number of $k=\mathbb{Q}\left(\sqrt{p}\right)$ is odd, and the fundamental unit of $\mathcal{O}_{k}$ has norm $-1$; this result in essence goes back to Gauss. Years ago I conjectured that if $p\equiv 9\pod{16}$ then the class number of $\mathbb{Q}\left(p^{\frac{1}{4}}\right)$ is $\equiv 2\pod{4}$. (Parry [2] had previously shown that itβs even, and that when 2 is not a fourth power in $\mathbb{Z}/p$ itβs $\equiv 2\pod{4}$.) I gave a proof of my conjecture assuming that the elliptic curve $y^{2}=x^{3}-px$ has positive rank, as the Birch Swinnerton-Dyer conjecture predicts.
Recently I asked on Mathoverflow whether the elliptic curve assumption could be eliminated. Franz Lemmermeyer responded with an unconditional proof that starts with Gaussβ result and continues with two applications of the ambiguous class number formula. His very nice argument deserves wider circulation, so Iβm writing it up here.
###### Theorem 1.
If $p\equiv 1\pod{8}$, $\mathbb{Q}\left(p^{\frac{1}{4}}\right)$ has even class number.
###### Proof (Lemmermeyer).
Let $F$ be the quartic subfield of $\mathbb{Q}(\mu_{p})$. Then $F\supset k=\mathbb{Q}(\sqrt{p})$. Since $p\equiv 1\pod{8}$, the infinite prime of $\mathbb{Q}$ is unramified in $F$, and the only prime of $\mathbb{Q}$ that ramifies in $F$ is $(p)$.
Since $F\left(p^{\frac{1}{4}}\right)$ is the compositum of $\mathbb{Q}\left(p^{\frac{1}{4}}\right)$ and $F$ it is a Galois extension of $k$ with Galois group $\mathbb{Z}/2\times\mathbb{Z}/2$. Since $p\neq 2$, the ramification is tame, and the prime above $p$ cannot ramify totally in the extension. It follows that $\left(p^{\frac{1}{4}}\right)$ cannot ramify from $\mathbb{Q}\left(p^{\frac{1}{4}}\right)$ to $F\left(p^{\frac{1}{4}}\right)$. So $\mathbb{Q}\left(p^{\frac{1}{4}}\right)$ has an everywhere unramified extension, $F\left(p^{\frac{1}{4}}\right)$, of degree 2, and class-field theory gives the result. β
###### Corollary 2.
Suppose $p\equiv 1\pod{8}$ and $F$ is as in Theorem 1. If $F\left(p^{\frac{1}{4}}\right)$ has odd class number then the class number of $\mathbb{Q}\left(p^{\frac{1}{4}}\right)$ is $\equiv 2\pod{4}$.
###### Proof.
Suppose on the contrary that 4 divides the class number. Then $\mathbb{Q}\left(p^{\frac{1}{4}}\right)$ admits an unramified abelian extension of degree 4. Translating by $F$ we get a degree 2 unramified extension of $F\left(p^{\frac{1}{4}}\right)$, contradicting the odd class number assumption. β
###### Lemma 3.
The $F$ of Theorem 1 is the unique degree 2 extension of $k$ unramified outside of $(\sqrt{p})$.
###### Proof.
Let $F^{\prime}$ be a second such extension. Since the class number of $k$ is odd, $k$ has no unramified extensions of degree 2 and $(\sqrt{p})$ must ramify in $F^{\prime}$. There is a third quadratic extension, $F^{\prime\prime}$, of $k$ contained in $FF^{\prime}$ and the same argument shows that $(\sqrt{p})$ ramifies in $F^{\prime\prime}$. So $(\sqrt{p})$ ramifies totally in $FF^{\prime}$, contradicting tameness. β
###### Theorem 4.
If $p\equiv 1\pod{8}$ and $\epsilon>0$ is a unit of norm $-1$ in the ring of integers of $k$, then $F=k\left(\sqrt{\epsilon\sqrt{p}}\right)$.
###### Proof.
$\epsilon\sqrt{p}$ and its $\mathbb{Q}$-conjugate $\epsilon^{-1}\sqrt{p}$ are both $>0$. So neither of the infinite primes of $k$ ramify in $k\left(\sqrt{\epsilon\sqrt{p}}\right)$. If $r^{2}-sp^{2}=-4$, $r$ and $s$ cannot both be odd. It follows that $\epsilon=a+b\sqrt{p}$ with $a$ and $b$ integers. Also, $0<\epsilon+\epsilon^{-1}=2b\sqrt{p}$, and $b>0$. Now $a^{2}-pb^{2}=-1$. Since $pb^{2}\equiv 1\pod{8}$, 8 divides $a^{2}$ and 4 divides $a$. Furthermore every prime that divides $b$ divides $a^{2}+1$, and so is $\equiv 1\pod{4}$. Since $b>0$, $b\equiv 1\pod{4}$. Let $P$ be a prime of $\mathcal{O}_{k}$ lying over $(2)$. Then in the $P$-completion of $\mathcal{O}_{k}$, $\epsilon\sqrt{p}=bp+a\sqrt{p}\equiv 1\pod{4}$. So $P$ does not ramify in $k\left(\sqrt{\epsilon\sqrt{p}}\right)$. It follows that the only prime that can ramify in $k\left(\sqrt{\epsilon\sqrt{p}}\right)$ is $(\sqrt{p})$, and we apply Lemma 3. β
###### Corollary 5.
$F\left(p^{\frac{1}{4}}\right)=F(\sqrt{\epsilon})$.
###### Proof.
Both fields are degree 2 extensions of $F$. Since $\sqrt{\epsilon\sqrt{p}}$ is in $F$, $\sqrt{\epsilon}$ is in $F\left(p^{\frac{1}{4}}\right)$. β
We now give the idea of Lemmermeyerβs proof. The class number of $k$ is known to be odd. Lemmermeyer uses the ambiguous class number formula to deduce that $k(\sqrt{\epsilon})$ has odd class number. Then assuming $p\equiv 9\pod{16}$ he uses it once more to show that $F(\sqrt{\epsilon})$ has odd class number. Corollaries 5 and 2 complete the proof.
We introduce some notation. Suppose $L\supset K$ is a degree 2 extension of number fields with Galois group $G=\{\mathrm{id},\sigma\}$. $U_{K}$ consists of the units of $\mathcal{O}_{K}$ while $h_{L}$ and $h_{K}$ are the class numbers of $L$ and $K$. $C_{L}$ is the class group of $L$, while $C_{L}^{G}$, the ambiguous class group, consists of the elements of $C_{L}$ fixed by $\sigma$. The following result is contained in Theorem 4.1 of [1].
###### Theorem 6.
$|C_{L}^{G}|=h_{K}\cdot(2^{t-1}/j)$ where $t$ is the number of primes of $K$, finite or infinite, that ramify in $L$, while $j$ is the index in $U_{K}$ of the subgroup consisting of elements that are norms from $L$. (Since this subgroup contains $U_{K}^{2}$, $j$ is a power of 2.) Furthermore, if $|C_{L}^{G}|$ is odd, $h_{L}$ is odd.
###### Lemma 7.
Just two primes of $k$ ramify in $k(\sqrt{\epsilon})$.
###### Proof.
$\epsilon>0$, and the $\mathbb{Q}$-conjugate $-\epsilon^{-1}$ of $\epsilon$ is $<0$. So one of the two infinite primes ramifies. Also $\epsilon=a+b\sqrt{p}$ with $a\equiv 0\pod{4}$, $b\equiv 1\pod{4}$. So $\epsilon\equiv 1\pod{4}$ in the completion of $\mathcal{O}_{k}$ at $\left(2,\frac{1-\sqrt{p}}{2}\right)$, and $\epsilon\equiv-1\pod{4}$ in the completion of $\mathcal{O}_{k}$ at $\left(2,\frac{1+\sqrt{p}}{2}\right)$. Finally no other primes can ramify. β
###### Theorem 8.
$k(\sqrt{\epsilon})$ has odd class number.
###### Proof.
Theorem 6 and Lemma 7 show that the ambiguous class number for the extension $k(\sqrt{\epsilon})\supset k$ is $\frac{2h_{k}}{j}$. So it suffices to show that $j>1$. Now $-1$ is in $U_{k}$. But as we saw above there is an infinite prime of $k$ ramifying in $k(\sqrt{\epsilon})$, and $-1$ evidently is not a local norm at that prime. β
###### Lemma 9.
$\epsilon$ represents a primitive fourth root of unity in $\mathcal{O}_{k}/(\sqrt{p})=\mathbb{Z}/p$. Furthermore the prime $(\sqrt{p})$ of $k$ splits in $k(\sqrt{\epsilon})$.
###### Proof.
$\epsilon=a+b\sqrt{p}$ with $a^{2}-pb^{2}=-1$. So mod $\sqrt{p}$, $\epsilon^{2}\equiv a^{2}\equiv-1$, giving the first result. Since $p\equiv 1\pod{8}$, any fourth root of unity in $(\mathbb{Z}/p)^{*}$ is a square, and the second result follows. β
###### Theorem 10(Lemmermeyer).
Suppose $p\equiv 9\pod{16}$. Then the ambiguous class number for the extension $F(\sqrt{\epsilon})\supset k(\sqrt{\epsilon})$ is odd. So $F(\sqrt{\epsilon})$ has odd class number, and Corollaries 5 and 2 show that the class number of $\mathbb{Q}\left(p^{\frac{1}{4}}\right)$ is $\equiv 2\pod{4}$.
###### Proof.
The only primes that can ramify are primes whose restriction to $k$ ramifies in $F$. In view of Lemma 9 the only possibilities are the 2 primes of $k(\sqrt{\epsilon})$ lying over $(\sqrt{p})$; itβs easy to see that they both ramify in $F(\sqrt{\epsilon})$. Furthermore $\sqrt{\epsilon}$ is not a local norm at either of these primes. (Because the prime ramifies it suffices to show that the image of $\sqrt{\epsilon}$ in the residue class field is not a square. But Lemma 9 shows that this image is a primitive eighth root of unity in $(\mathbb{Z}/p)^{*}$. And $p\not\equiv 1\pod{16}$. So in our quadratic extension, $t=2$ and $j$ is even. Since $k(\sqrt{\epsilon})$ has odd class number, Theorem 6 gives the desired result. β
## References
• [1] Serge Lang, Cyclotomic Fields II (1980), Springer Verlag, New York, Heidelberg, Berlin.
• [2] Charles J. Parry, A genus theory for quartic fields, J. Reine Angew.Β Math.Β 314 (1980), 40β71. | 2,898 | 8,586 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 186, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.742606 |
https://www.dataunitconverter.com/nibble-per-hour-to-exbibyte-per-minute | 1,718,897,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00434.warc.gz | 623,727,407 | 17,236 | # Nibble/Hr to EiB/Min → CONVERT Nibbles per Hour to Exbibytes per Minute
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Day
Nibble/Hr
## Nibbles per Hour (Nibble/Hr) Versus Exbibytes per Minute (EiB/Min) - Comparison
Nibbles per Hour and Exbibytes per Minute are units of digital information used to measure storage capacity and data transfer rate.
Nibbles per Hour is one of the very "basic" digital unit where as Exbibytes per Minute is a "binary" unit. One Nibble is equal to 4 bits. One Exbibyte is equal to 1024^6 bytes. There are 2,305,843,009,213,693,952 Nibble in one Exbibyte. Find more details on below table.
Nibbles per Hour (Nibble/Hr) Exbibytes per Minute (EiB/Min)
Nibbles per Hour (Nibble/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Nibbles that can be transferred in one Hour. Exbibytes per Minute (EiB/Min) is a unit of measurement for data transfer bandwidth. It measures the number of Exbibytes that can be transferred in one Minute.
## Nibbles per Hour (Nibble/Hr) to Exbibytes per Minute (EiB/Min) Conversion - Formula & Steps
The Nibble/Hr to EiB/Min Calculator Tool provides a convenient solution for effortlessly converting data rates from Nibbles per Hour (Nibble/Hr) to Exbibytes per Minute (EiB/Min). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Nibble) and target (Exbibyte) data units.
Source Data Unit Target Data Unit
Equal to 4 bits
(Basic Unit)
Equal to 1024^6 bytes
(Binary Unit)
The conversion from Data per Hour to Minute can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
The formula for converting the Nibbles per Hour (Nibble/Hr) to Exbibytes per Minute (EiB/Min) can be expressed as follows:
diamond CONVERSION FORMULA EiB/Min = Nibble/Hr x 4 ÷ (8x10246) / 60
Now, let's apply the aforementioned formula and explore the manual conversion process from Nibbles per Hour (Nibble/Hr) to Exbibytes per Minute (EiB/Min). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Exbibytes per Minute = Nibbles per Hour x 4 ÷ (8x10246) / 60
STEP 1
Exbibytes per Minute = Nibbles per Hour x 4 ÷ (8x1024x1024x1024x1024x1024x1024) / 60
STEP 2
Exbibytes per Minute = Nibbles per Hour x 4 ÷ 9223372036854775808 / 60
STEP 3
Exbibytes per Minute = Nibbles per Hour x 0.0000000000000000004336808689942017736029 / 60
STEP 4
Exbibytes per Minute = Nibbles per Hour x 0.0000000000000000000072280144832366962267
Example : By applying the previously mentioned formula and steps, the conversion from 1 Nibbles per Hour (Nibble/Hr) to Exbibytes per Minute (EiB/Min) can be processed as outlined below.
1. = 1 x 4 ÷ (8x10246) / 60
2. = 1 x 4 ÷ (8x1024x1024x1024x1024x1024x1024) / 60
3. = 1 x 4 ÷ 9223372036854775808 / 60
4. = 1 x 0.0000000000000000004336808689942017736029 / 60
5. = 1 x 0.0000000000000000000072280144832366962267
6. = 0.0000000000000000000072280144832366962267
7. i.e. 1 Nibble/Hr is equal to 0.0000000000000000000072280144832366962267 EiB/Min.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Nibbles per Hour to Exbibytes per Minute using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Nibble ?
A Nibble is a unit of digital information that consists of 4 bits. It is half of a byte and can represent a single hexadecimal digit. It is used in computer memory and data storage and sometimes used as a basic unit of data transfer in certain computer architectures.
arrow_downward
#### What is Exbibyte ?
An Exbibyte (EiB) is a binary unit of digital information that is equal to 1,152,921,504,606,846,976 bytes (or 9,223,372,036,854,775,808 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'exbi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'exabyte' (EB). It is widely used in the field of computing as it more accurately represents the storage size of high end servers and data storage arrays.
## Excel Formula to convert from Nibbles per Hour (Nibble/Hr) to Exbibytes per Minute (EiB/Min)
Apply the formula as shown below to convert from 1 Nibbles per Hour (Nibble/Hr) to Exbibytes per Minute (EiB/Min).
A B C
1 Nibbles per Hour (Nibble/Hr) Exbibytes per Minute (EiB/Min)
2 1 =A2 * 0.0000000000000000004336808689942017736029 / 60
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Nibbles per Hour (Nibble/Hr) to Exbibytes per Minute (EiB/Min) Conversion
You can use below code to convert any value in Nibbles per Hour (Nibble/Hr) to Nibbles per Hour (Nibble/Hr) in Python.
nibblesperHour = int(input("Enter Nibbles per Hour: "))
exbibytesperMinute = nibblesperHour * 4 / (8*1024*1024*1024*1024*1024*1024) / 60
print("{} Nibbles per Hour = {} Exbibytes per Minute".format(nibblesperHour,exbibytesperMinute))
The first line of code will prompt the user to enter the Nibbles per Hour (Nibble/Hr) as an input. The value of Exbibytes per Minute (EiB/Min) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Exbibytes(EiB) are there in a Nibble?expand_more
There are 0.0000000000000000004336808689942017736029 Exbibytes in a Nibble.
#### What is the formula to convert Nibble to Exbibyte(EiB)?expand_more
Use the formula EiB = Nibble x 4 / (8x10246) to convert Nibble to Exbibyte.
#### How many Nibbles are there in an Exbibyte(EiB)?expand_more
There are 2305843009213693952 Nibbles in an Exbibyte.
#### What is the formula to convert Exbibyte(EiB) to Nibble?expand_more
Use the formula Nibble = EiB x (8x10246) / 4 to convert Exbibyte to Nibble.
#### Which is bigger, Exbibyte(EiB) or Nibble?expand_more
Exbibyte is bigger than Nibble. One Exbibyte contains 2305843009213693952 Nibbles.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,958 | 6,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.754374 |
http://www.chegg.com/homework-help/questions-and-answers/ip-snowballs-thrown-speed-12-m-s-roof-60-m-ground-snowball-thrown-straight-downward-snowba-q929161 | 1,472,218,638,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982295494.5/warc/CC-MAIN-20160823195815-00198-ip-10-153-172-175.ec2.internal.warc.gz | 366,101,941 | 14,517 | IP Snowballs are thrown with a speed of 12 m/s from a roof 6.0 m above the ground. Snowball A is thrown straight downward; snowball B is thrown in a direction 30 above the horizontal.
Part A
Is the landing speed of snowball A greater than, less than, or the same as the landing speed of snowball B?
greater
less
the same
Part B
Explain.
Part C
Verify your answer to part (a) by calculating the landing speed of both snowballs. Find the speed of snowball A.
=
Part D
Find the speed of snowball B.
= | 132 | 501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-36 | latest | en | 0.906318 |
https://www.cemetech.net/projects/uti/viewtopic.php?t=490 | 1,601,008,472,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400221980.49/warc/CC-MAIN-20200925021647-20200925051647-00742.warc.gz | 774,320,749 | 6,611 | This is an archived, read-only copy of the United-TI subforum , including posts and topic from May 2003 to April 2012. If you would like to discuss any of the topics in this forum, you can visit Cemetech's z80 & ez80 Assembly subforum. Some of these topics may also be directly-linked to active Cemetech topics. If you are a Cemetech member with a linked United-TI account, you can link United-TI topics here with your current Cemetech topics.
Z80 & 68k Assembly => z80 & ez80 Assembly
Author Message
Jeffrey
Member
Joined: 12 Jun 2003
Posts: 212
Posted: 31 Aug 2003 08:58:42 am Post subject: This topic pertains to the other topic (Drawing Pics with ASM), but is an entire topic of its own (confusing, huh?). How do you load the real variable A into register A in an ASM program?
Tyraniek
Member
Joined: 07 Jun 2003
Posts: 133
Posted: 31 Aug 2003 10:41:27 am Post subject: First of all, you have to understand that you won't be able to load exactly the real number A in the register A, because the register is 8-bits (one byte), and the real variable is 18 bytes and can be decimal ! That means that while the variable A can be between -10^99 and 10^99 and can be decimal, the register will only be able to store a number between 0 and 255 !!! Then, I'm not sure getting the variable A is a good idea. You'd better try to get the variable X or even "Ans", so that you're able to use a ROM call. However, you could do the following (supposing you've stored A in X) : Code: ``` B_CALL(_rclX) ; X is in OP1 B_CALL(_convop1) ; Now you get the Less Signifiant Byte of OP1 in register A``` You could do the same with B_CALL(_rclAns), but 'm not sure there is any "_RclA".
aka Tianon
Know-It-All
Joined: 02 Jun 2003
Posts: 1874
Posted: 31 Aug 2003 02:23:28 pm Post subject: techinacally, no, but acutally, yes... Code: ```rclA: ;var A to ACC (Real A to Register a) bcall(_zeroop1) ld hl,op1+1 ld (hl),'A' bcall(_rclvarsym) bcall(_convop1) ret```does that make sence? it is just a simple little thing i learned from MV's ZBasic source Last edited by Guest on 31 Aug 2003 02:23:38 pm; edited 1 time in total
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Page 1 of 1 » All times are GMT - 5 Hours | 766 | 2,705 | {"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} | 2.59375 | 3 | CC-MAIN-2020-40 | latest | en | 0.874735 |
https://msdn.microsoft.com/en-us/library/hk63d62d(v=vs.110).aspx | 1,448,962,612,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398466178.24/warc/CC-MAIN-20151124205426-00158-ip-10-71-132-137.ec2.internal.warc.gz | 850,242,230 | 12,684 | Decimal Constructor (Int32)
# Decimal Constructor (Int32)
.NET Framework 4.6 and 4.5
Initializes a new instance of Decimal to the value of the specified 32-bit signed integer.
Namespace: System
Assembly: mscorlib (in mscorlib.dll)
## Syntax
```public Decimal(
int value
)
```
#### Parameters
value
Type: System.Int32
The value to represent as a Decimal.
## Examples
The following code example creates several Decimal numbers using the constructor overload that initializes a Decimal structure with an Int32 value.
```// Example of the decimal( int ) constructor.
using System;
class DecimalCtorIDemo
{
// Create a decimal object and display its value.
public static void CreateDecimal( int value, string valToStr )
{
decimal decimalNum = new decimal( value );
// Format the constructor for display.
string ctor = String.Format( "decimal( {0} )", valToStr );
// Display the constructor and its value.
Console.WriteLine( "{0,-30}{1,16}", ctor, decimalNum );
}
public static void Main( )
{
Console.WriteLine( "This example of the decimal( int ) " +
"constructor \ngenerates the following output.\n" );
Console.WriteLine( "{0,-30}{1,16}", "Constructor", "Value" );
Console.WriteLine( "{0,-30}{1,16}", "-----------", "-----" );
// Construct decimal objects from int values.
CreateDecimal( int.MinValue, "int.MinValue" );
CreateDecimal( int.MaxValue, "int.MaxValue" );
CreateDecimal( 0, "0" );
CreateDecimal( 999999999, "999999999" );
CreateDecimal( 0x40000000, "0x40000000" );
CreateDecimal( unchecked( (int)0xC0000000 ),
"(int)0xC0000000" );
}
}
/*
This example of the decimal( int ) constructor
generates the following output.
Constructor Value
----------- -----
decimal( int.MinValue ) -2147483648
decimal( int.MaxValue ) 2147483647
decimal( 0 ) 0
decimal( 999999999 ) 999999999
decimal( 0x40000000 ) 1073741824
decimal( (int)0xC0000000 ) -1073741824
*/
```
## Version Information
Universal Windows Platform
Available since 4.5
.NET Framework
Available since 1.1
Portable Class Library
Supported in: portable .NET platforms
Silverlight
Available since 2.0
Windows Phone Silverlight
Available since 7.0
Windows Phone
Available since 8.1 | 571 | 2,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | longest | en | 0.303376 |
https://discussions.unity.com/t/shader-function-to-adjust-texture-contrast/656195 | 1,723,284,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00207.warc.gz | 162,484,139 | 6,953 | Hey guys,
I’m just wondering if theres a built in function or some logic that can adjust the contrast within a mapped texture. Basically so that when I adjust the value the brights get brighter and the darks get darker?
For example I have Noise texture from grey to white and I want to increase the contrast. I can’t just add brightness because it increases the dark parts too? If that makes sense
Anyway help would be great!
1 Like
Depends on what you mean by “contrast”.
The easiest way to do it would be:
``````half3 AdjustContrast(half3 color, half contrast) {
return saturate(lerp(half3(0.5, 0.5, 0.5), color, contrast));
}
``````
That’ll work for gamma space rendering. Passing in a contrast value of 0.0 will give you middle grey. A contrast of 100.0 will make almost everything black or white (or solid colors), but it’ll crush the lights and darks entirely even at lower contrast values. The above is fairly straight forward to understand if you’re familiar with linear interpolation or even alpha blending, at least for contrast values between 0.0 and 1.0. This just blends between grey and the color, but linear interpolation has the handy behavior of working just fine with values >1.0, or even negatives if you want to invert the colors. For example a contrast of 2 will make color values of 63/255 black, and 192/255 white.
Another slightly more complex method is:
``````half3 AdjustContrastCurve(half3 color, half contrast) {
return pow(abs(color * 2 - 1), 1 / max(contrast, 0.0001)) * sign(color - 0.5) + 0.5;
}
``````
Unlike the above method the lights and darks won’t be immediately crushed, however very low contrasts will look kind of funny as the fully black and white areas will remain fully black and white. This has a bit more going on, but the essence is we’re applying a power curve to the color, but applying the power curve pivoted around “0.5”. There are a few curious things going on if you look closely, like using an abs() to the color after remapping it from 0.0~1.0 to -1.0~1.0, but that’s because the power might remove the sign sometimes (like pow(x, 2)) but not all the time, so we remove it before the power, then reapply it afterwards with the sign() function. The max() is there to prevent a divide by zero.
Both of these only really work “correctly” as is if you’re using the gamma color space, but with out tweaks it may still give you acceptable results depending on what you like the look of. If you need this to work properly in linear space as well you’ll need to convert the color back and forth between the linear and gamma space representations.
Here’s the first method updated account for the color space
``````half3 AdjustContrast(half3 color, half contrast) {
#if !UNITY_COLORSPACE_GAMMA
color = LinearToGammaSpace(color);
#endif
color = saturate(lerp(half3(0.5, 0.5, 0.5), color, contrast));
#if !UNITY_COLORSPACE_GAMMA
color = GammaToLinearSpace(color);
#endif
return color;
}
``````
I’ll leave updating the other method up to you.
21 Likes
Awesome that explains it really well. Thanks for the help!
I know this is an old post but it helped me out so I wanted to say thanks again bgolus for the in-depth explanation, you’re a life saver.
1 Like | 799 | 3,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-33 | latest | en | 0.878057 |
http://www.chegg.com/homework-help/elementary-differential-equations-with-boundary-value-problems-2nd-edition-chapter-9.6-solutions-9780321398505 | 1,455,266,805,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701163512.72/warc/CC-MAIN-20160205193923-00103-ip-10-236-182-209.ec2.internal.warc.gz | 342,429,372 | 15,872 | View more editions
Elementary Differential Equations With Boundary Value Problems
# TEXTBOOK SOLUTIONS FOR Elementary Differential Equations With Boundary Value Problems 2nd Edition
• 1947 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 10
Question:
Consider the initial value problem
• Step 2 of 10
• Step 3 of 10
(a) Solve this problem for the given parameter values and the given initial conditions
• Step 4 of 10
(b) Assume the solution represents the displacement at time t and position x. Determine the velocity, . (In Exercises 7-10, assume the series can be differentiated termwise.)
• Step 5 of 10
, , ,
• Step 6 of 10
Solution:
The general solution to the wave equation in a finite one-dimensional medium
is of the following forms, where ,
and
Additionally, the coefficients are determined by the following integrals
• Step 7 of 10
(a) From the problem, we have , , and . From this we have a solution of the form
where
• Step 8 of 10
By using inspection, we see that for all n, and for
• Step 9 of 10
Using these coefficients, we find that the Fourier series solution to this partial differential equation is
• Step 10 of 10
(b) Evaluating , where , we have
Corresponding Textbook
Elementary Differential Equations With Boundary Value Problems | 2nd Edition
9780321398505ISBN-13: 0321398505ISBN: Authors: | 385 | 1,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | latest | en | 0.824123 |
https://popflock.com/learn?s=0_(number) | 1,632,027,521,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00488.warc.gz | 495,145,091 | 50,328 | 0 (number)
Get 0 Number essential facts below. View Videos or join the 0 Number discussion. Add 0 Number to your PopFlock.com topic list for future reference or share this resource on social media.
0 Number
← -1 0 1 →
Cardinal0, zero, , nought, naught, nil
OrdinalZeroth, noughth, 0th
Binary02
Ternary03
Octal08
Duodecimal012
Arabic, Kurdish, Persian, Sindhi, Urdu?
Bengali?
Hindu Numerals?
Chinese?, ?
Khmer?
Thai?
0 (zero) is a number,[1] and the numerical digit used to represent that number in numerals. It fulfills a central role in mathematics as the additive identity[2] of the integers, real numbers, and many other algebraic structures. As a digit, 0 is used as a placeholder in place value systems. Names for the number 0 in English include zero, nought (UK), naught (US; ), nil, or--in contexts where at least one adjacent digit distinguishes it from the letter "O"--oh or o . Informal or slang terms for zero include zilch and zip.[3] Ought and aught ,[4] as well as cipher,[5] have also been used historically.[6][7]
## Etymology
The word zero came into the English language via French zéro from the Italian zero, a contraction of the Venetian zevero form of Italian zefiro via ?afira or ?ifr.[8] In pre-Islamic time the word ?ifr (Arabic ) had the meaning "empty".[9] Sifr evolved to mean zero when it was used to translate nya (Sanskrit: ) from India.[9] The first known English use of zero was in 1598.[10]
The Italian mathematician Fibonacci (c. 1170-1250), who grew up in North Africa and is credited with introducing the decimal system to Europe, used the term zephyrum. This became zefiro in Italian, and was then contracted to zero in Venetian. The Italian word zefiro was already in existence (meaning "west wind" from Latin and Greek zephyrus) and may have influenced the spelling when transcribing Arabic ?ifr.[11]
### Modern usage
Depending on the context, there may be different words used for the number zero, or the concept of zero. For the simple notion of lacking, the words "nothing" and "none" are often used. Sometimes, the word "nought" or "naught" is used. It is often called "oh" in the context of telephone numbers. Slang words for zero include "zip", "zilch", "nada", and "scratch."[12]
"Nil" is used for many sports in British English. Several sports have specific words for a score of zero, such as "love" in tennis – from French l'oeuf, "the egg" – and "duck" in cricket, a shortening of "duck's egg"; "goose egg" is another general slang term used for zero.[12]
## History
### Ancient Near East
nfr
heart with trachea
beautiful, pleasant, good
Ancient Egyptian numerals were of base 10.[13] They used hieroglyphs for the digits and were not positional. By 1770 BC, the Egyptians had a symbol for zero in accounting texts. The symbol nfr, meaning beautiful, was also used to indicate the base level in drawings of tombs and pyramids, and distances were measured relative to the base line as being above or below this line.[14]
By the middle of the 2nd millennium BC, the Babylonian mathematics had a sophisticated sexagesimal positional numeral system. The lack of a positional value (or zero) was indicated by a space between sexagesimal numerals. In a tablet unearthed at Kish (dating to as early as 700 BC), the scribe Bêl-bân-aplu used three hooks as a placeholder in the same Babylonian system. [15] By 300 BC, a punctuation symbol (two slanted wedges) was co-opted to serve as this placeholder.[16][17]
The Babylonian placeholder was not a true zero because it was not used alone, nor was it used at the end of a number. Thus numbers like 2 and 120 (2×60), 3 and 180 (3×60), 4 and 240 (4×60) looked the same, because the larger numbers lacked a final sexagesimal placeholder. Only context could differentiate them.[]
### Pre-Columbian Americas
The back of Epi-Olmec stela C from Tres Zapotes, the second oldest Long Count date discovered. The numerals 7.16.6.16.18 translate to September, 32 BC (Julian). The glyphs surrounding the date are thought to be one of the few surviving examples of Epi-Olmec script.
The Mesoamerican Long Count calendar developed in south-central Mexico and Central America required the use of zero as a placeholder within its vigesimal (base-20) positional numeral system. Many different glyphs, including this partial quatrefoil----were used as a zero symbol for these Long Count dates, the earliest of which (on Stela 2 at Chiapa de Corzo, Chiapas) has a date of 36 BC.[a]
Since the eight earliest Long Count dates appear outside the Maya homeland,[18] it is generally believed that the use of zero in the Americas predated the Maya and was possibly the invention of the Olmecs.[19] Many of the earliest Long Count dates were found within the Olmec heartland, although the Olmec civilization ended by the , several centuries before the earliest known Long Count dates.
Maya numeral zero.
Although zero became an integral part of Maya numerals, with a different, empty tortoise-like "shell shape" used for many depictions of the "zero" numeral, it is assumed to have not influenced Old World numeral systems.
Quipu, a knotted cord device, used in the Inca Empire and its predecessor societies in the Andean region to record accounting and other digital data, is encoded in a base ten positional system. Zero is represented by the absence of a knot in the appropriate position.
### Classical antiquity
The ancient Greeks had no symbol for zero (), and did not use a digit placeholder for it.[20] They seemed unsure about the status of zero as a number. They asked themselves, "How can nothing be something?", leading to philosophical and, by the medieval period, religious arguments about the nature and existence of zero and the vacuum. The paradoxes of Zeno of Elea depend in large part on the uncertain interpretation of zero.[21]
Example of the early Greek symbol for zero (lower right corner) from a 2nd-century papyrus
By AD 150, Ptolemy, influenced by Hipparchus and the Babylonians, was using a symbol for zero [22][23] in his work on mathematical astronomy called the Syntaxis Mathematica, also known as the Almagest.[24] This Hellenistic zero was perhaps the earliest documented use of a numeral representing zero in the Old World.[25] Ptolemy used it many times in his Almagest (VI.8) for the magnitude of solar and lunar eclipses. It represented the value of both digits and minutes of immersion at first and last contact. Digits varied continuously from as the Moon passed over the Sun (a triangular pulse), where twelve digits was the angular diameter of the Sun. Minutes of immersion was tabulated from , where 0?0? used the symbol as a placeholder in two positions of his sexagesimal positional numeral system,[b] while the combination meant a zero angle. Minutes of immersion was also a continuous function (a triangular pulse with convex sides), where d was the digit function and 31?20? was the sum of the radii of the Sun's and Moon's discs.[26] Ptolemy's symbol was a placeholder as well as a number used by two continuous mathematical functions, one within another, so it meant zero, not none.
The earliest use of zero in the calculation of the Julian Easter occurred before AD311, at the first entry in a table of epacts as preserved in an Ethiopic document for the years AD311 to 369, using a Ge'ez word for "none" (English translation is "0" elsewhere) alongside Ge'ez numerals (based on Greek numerals), which was translated from an equivalent table published by the Church of Alexandria in Medieval Greek.[27] This use was repeated in AD525 in an equivalent table, that was translated via the Latin nulla or "none" by Dionysius Exiguus, alongside Roman numerals.[28] When division produced zero as a remainder, nihil, meaning "nothing", was used. These medieval zeros were used by all future medieval calculators of Easter. The initial "N" was used as a zero symbol in a table of Roman numerals by Bede--or his colleagues around AD 725.[29]
### China
This is a depiction of zero expressed in Chinese counting rods, based on the example provided by A History of Mathematics. An empty space is used to represent zero.[30]
The S?nz? Suànj?ng, of unknown date but estimated to be dated from the 1st to , and Japanese records dated from the 18th century, describe how the Chinese counting rods system enabled one to perform decimal calculations. As noted in Xiahou Yang's Suanjing (425-468 AD) that states that to multiply or divide a number by 10, 100, 1000, or 10000, all one needs to do, with rods on the counting board, is to move them forwards, or back, by 1, 2, 3, or 4 places,[31] According to A History of Mathematics, the rods "gave the decimal representation of a number, with an empty space denoting zero."[30] The counting rod system is considered a positional notation system.[32]
In AD 690, Empress W? promulgated Zetian characters, one of which was "?"; originally meaning 'star', it subsequently [when?] came to represent zero.
Zero was not treated as a number at that time, but as a "vacant position".[33] Qín Ji?sháo's 1247 Mathematical Treatise in Nine Sections is the oldest surviving Chinese mathematical text using a round symbol for zero.[34] Chinese authors had been familiar with the idea of negative numbers by the Han Dynasty , as seen in The Nine Chapters on the Mathematical Art.[35]
### India
Pingala (c. 3rd/2nd century BC[36]), a Sanskrit prosody scholar,[37] used binary numbers in the form of short and long syllables (the latter equal in length to two short syllables), a notation similar to Morse code.[38] Pingala used the Sanskrit word nya explicitly to refer to zero.[36]
The concept of zero as a written digit in the decimal place value notation was developed in India, presumably as early as during the Gupta period , with the oldest unambiguous evidence dating to the 7th century.[39]
A symbol for zero, a large dot likely to be the precursor of the still-current hollow symbol, is used throughout the Bakhshali manuscript, a practical manual on arithmetic for merchants.[40] In 2017, three samples from the manuscript were shown by radiocarbon dating to come from three different centuries: from AD 224-383, AD 680-779, and AD 885-993, making it South Asia's oldest recorded use of the zero symbol. It is not known how the birch bark fragments from different centuries forming the manuscript came to be packaged together.[41][42][43]
The Lokavibh?ga, a Jain text on cosmology surviving in a medieval Sanskrit translation of the Prakrit original, which is internally dated to AD 458 (Saka era 380), uses a decimal place-value system, including a zero. In this text, nya ("void, empty") is also used to refer to zero.[44]
The Aryabhatiya (c. 500), states sth?n?t sth?na? da?agu?a? sy?t "from place to place each is ten times the preceding."[45][46][47]
Rules governing the use of zero appeared in Brahmagupta's Brahmasputha Siddhanta (7th century), which states the sum of zero with itself as zero, and incorrectly division by zero as:[48][49]
A positive or negative number when divided by zero is a fraction with the zero as denominator. Zero divided by a negative or positive number is either zero or is expressed as a fraction with zero as numerator and the finite quantity as denominator. Zero divided by zero is zero.
#### Epigraphy
The number 605 in Khmer numerals, from the Sambor inscription (Saka era 605 corresponds to AD 683). The earliest known material use of zero as a decimal figure.
There are numerous copper plate inscriptions, with the same small o in them, some of them possibly dated to the 6th century, but their date or authenticity may be open to doubt.[15]
A stone tablet found in the ruins of a temple near Sambor on the Mekong, Kratié Province, Cambodia, includes the inscription of "605" in Khmer numerals (a set of numeral glyphs for the Hindu-Arabic numeral system). The number is the year of the inscription in the Saka era, corresponding to a date of AD 683.[50]
The first known use of special glyphs for the decimal digits that includes the indubitable appearance of a symbol for the digit zero, a small circle, appears on a stone inscription found at the Chaturbhuj Temple, Gwalior, in India, dated 876.[51][52] Zero is also used as a placeholder in the Bakhshali manuscript, portions of which date from AD 224-383.[53]
### Middle Ages
#### Transmission to Islamic culture
The Arabic-language inheritance of science was largely Greek,[54] followed by Hindu influences.[55] In 773, at Al-Mansur's behest, translations were made of many ancient treatises including Greek, Roman, Indian, and others.
In AD 813, astronomical tables were prepared by a Persian mathematician, Mu?ammad ibn M?s? al-Khw?rizm?, using Hindu numerals;[55] and about 825, he published a book synthesizing Greek and Hindu knowledge and also contained his own contribution to mathematics including an explanation of the use of zero.[56] This book was later translated into Latin in the 12th century under the title Algoritmi de numero Indorum. This title means "al-Khwarizmi on the Numerals of the Indians". The word "Algoritmi" was the translator's Latinization of Al-Khwarizmi's name, and the word "Algorithm" or "Algorism" started to acquire a meaning of any arithmetic based on decimals.[55]
Muhammad ibn Ahmad al-Khwarizmi, in 976, stated that if no number appears in the place of tens in a calculation, a little circle should be used "to keep the rows". This circle was called ?ifr.[57]
#### Transmission to Europe
The Hindu-Arabic numeral system (base 10) reached Western Europe in the 11th century, via Al-Andalus, through Spanish Muslims, the Moors, together with knowledge of classical astronomy and instruments like the astrolabe; Gerbert of Aurillac is credited with reintroducing the lost teachings into Catholic Europe. For this reason, the numerals came to be known in Europe as "Arabic numerals". The Italian mathematician Fibonacci or Leonardo of Pisa was instrumental in bringing the system into European mathematics in 1202, stating:
After my father's appointment by his homeland as state official in the customs house of Bugia for the Pisan merchants who thronged to it, he took charge; and in view of its future usefulness and convenience, had me in my boyhood come to him and there wanted me to devote myself to and be instructed in the study of calculation for some days. There, following my introduction, as a consequence of marvelous instruction in the art, to the nine digits of the Hindus, the knowledge of the art very much appealed to me before all others, and for it I realized that all its aspects were studied in Egypt, Syria, Greece, Sicily, and Provence, with their varying methods; and at these places thereafter, while on business. I pursued my study in depth and learned the give-and-take of disputation. But all this even, and the algorism, as well as the art of Pythagoras, I considered as almost a mistake in respect to the method of the Hindus (Modus Indorum). Therefore, embracing more stringently that method of the Hindus, and taking stricter pains in its study, while adding certain things from my own understanding and inserting also certain things from the niceties of Euclid's geometric art. I have striven to compose this book in its entirety as understandably as I could, dividing it into fifteen chapters. Almost everything which I have introduced I have displayed with exact proof, in order that those further seeking this knowledge, with its pre-eminent method, might be instructed, and further, in order that the Latin people might not be discovered to be without it, as they have been up to now. If I have perchance omitted anything more or less proper or necessary, I beg indulgence, since there is no one who is blameless and utterly provident in all things. The nine Indian figures are: 9 8 7 6 5 4 3 2 1. With these nine figures, and with the sign 0 ... any number may be written.[58][59][60]
Here Leonardo of Pisa uses the phrase "sign 0", indicating it is like a sign to do operations like addition or multiplication. From the 13th century, manuals on calculation (adding, multiplying, extracting roots, etc.) became common in Europe where they were called algorismus after the Persian mathematician al-Khw?rizm?. The most popular was written by Johannes de Sacrobosco, about 1235 and was one of the earliest scientific books to be printed in 1488. Until the late 15th century, Hindu-Arabic numerals seem to have predominated among mathematicians, while merchants preferred to use the Roman numerals. In the 16th century, they became commonly used in Europe.
## Mathematics
0 is the integer immediately preceding 1. Zero is an even number[61] because it is divisible by 2 with no remainder. 0 is neither positive nor negative,[62] or both positive and negative.[63] Many definitions[64] include 0 as a natural number, in which case it is the only natural number that is not positive. Zero is a number which quantifies a count or an amount of null size. In most cultures, 0 was identified before the idea of negative things (i.e., quantities less than zero) was accepted.
As a value or a number, zero is not the same as the digit zero, used in numeral systems with positional notation. Successive positions of digits have higher weights, so the digit zero is used inside a numeral to skip a position and give appropriate weights to the preceding and following digits. A zero digit is not always necessary in a positional number system (e.g., the number 02). In some instances, a leading zero may be used to distinguish a number.
### Elementary algebra
The number 0 is the smallest non-negative integer. The natural number following 0 is 1 and no natural number precedes 0. The number 0 may or may not be considered a natural number, but it is an integer, and hence a rational number and a real number (as well as an algebraic number and a complex number).
The number 0 is neither positive nor negative, and is usually displayed as the central number in a number line. It is neither a prime number nor a composite number. It cannot be prime because it has an infinite number of factors, and cannot be composite because it cannot be expressed as a product of prime numbers (as 0 must always be one of the factors).[65] Zero is, however, even (i.e. a multiple of 2, as well as being a multiple of any other integer, rational, or real number).
The following are some basic (elementary) rules for dealing with the number 0. These rules apply for any real or complex number x, unless otherwise stated.
• Addition: x + 0 = 0 + x = x. That is, 0 is an identity element (or neutral element) with respect to addition.
• Subtraction: x - 0 = x and 0 - x = -x.
• Multiplication: x · 0 = 0 · x = 0.
• Division: 0/x = 0, for nonzero x. But x/0 is undefined, because 0 has no multiplicative inverse (no real number multiplied by 0 produces 1), a consequence of the previous rule.
• Exponentiation: x0 = x/x = 1, except that the case x = 0 may be left undefined in some contexts. For all positive real x, .
The expression 0/0, which may be obtained in an attempt to determine the limit of an expression of the form f(x)/g(x) as a result of applying the lim operator independently to both operands of the fraction, is a so-called "indeterminate form". That does not simply mean that the limit sought is necessarily undefined; rather, it means that the limit of f(x)/g(x), if it exists, must be found by another method, such as l'Hôpital's rule.
The sum of 0 numbers (the empty sum) is 0, and the product of 0 numbers (the empty product) is 1. The factorial 0! evaluates to 1, as a special case of the empty product.
### Related mathematical terms
• A zero of a function f is a point x in the domain of the function such that . When there are finitely many zeros these are called the roots of the function. This is related to zeros of a holomorphic function.
• The zero function (or zero map) on a domain D is the constant function with 0 as its only possible output value, i.e., the function f defined by for all x in D. The zero function is the only function that is both even and odd. A particular zero function is a zero morphism in category theory; e.g., a zero map is the identity in the additive group of functions. The determinant on non-invertible square matrices is a zero map.
• Several branches of mathematics have zero elements, which generalize either the property , or the property or both.
## Physics
The value zero plays a special role for many physical quantities. For some quantities, the zero level is naturally distinguished from all other levels, whereas for others it is more or less arbitrarily chosen. For example, for an absolute temperature (as measured in kelvins), zero is the lowest possible value (negative temperatures are defined, but negative-temperature systems are not actually colder). This is in contrast to for example temperatures on the Celsius scale, where zero is arbitrarily defined to be at the freezing point of water. Measuring sound intensity in decibels or phons, the zero level is arbitrarily set at a reference value--for example, at a value for the threshold of hearing. In physics, the zero-point energy is the lowest possible energy that a quantum mechanical physical system may possess and is the energy of the ground state of the system.
## Chemistry
Zero has been proposed as the atomic number of the theoretical element tetraneutron. It has been shown that a cluster of four neutrons may be stable enough to be considered an atom in its own right. This would create an element with no protons and no charge on its nucleus.
As early as 1926, Andreas von Antropoff coined the term neutronium for a conjectured form of matter made up of neutrons with no protons, which he placed as the chemical element of atomic number zero at the head of his new version of the periodic table. It was subsequently placed as a noble gas in the middle of several spiral representations of the periodic system for classifying the chemical elements.
## Computer science
The most common practice throughout human history has been to start counting at one, and this is the practice in early classic computer programming languages such as Fortran and COBOL. However, in the late 1950s LISP introduced zero-based numbering for arrays while Algol 58 introduced completely flexible basing for array subscripts (allowing any positive, negative, or zero integer as base for array subscripts), and most subsequent programming languages adopted one or other of these positions. For example, the elements of an array are numbered starting from 0 in C, so that for an array of n items the sequence of array indices runs from 0 to . This permits an array element's location to be calculated by adding the index directly to address of the array, whereas 1-based languages precalculate the array's base address to be the position one element before the first.[]
There can be confusion between 0- and 1-based indexing, for example Java's JDBC indexes parameters from 1 although Java itself uses 0-based indexing.[]
In databases, it is possible for a field not to have a value. It is then said to have a null value.[66] For numeric fields it is not the value zero. For text fields this is not blank nor the empty string. The presence of null values leads to three-valued logic. No longer is a condition either true or false, but it can be undetermined. Any computation including a null value delivers a null result.[]
A null pointer is a pointer in a computer program that does not point to any object or function. In C, the integer constant 0 is converted into the null pointer at compile time when it appears in a pointer context, and so 0 is a standard way to refer to the null pointer in code. However, the internal representation of the null pointer may be any bit pattern (possibly different values for different data types).[]
In mathematics both -0 and +0 represent exactly the same number, i.e., there is no "positive zero" or "negative zero" distinct from zero. However, in some computer hardware signed number representations, zero has two distinct representations, a positive one grouped with the positive numbers and a negative one grouped with the negatives; this kind of dual representation is known as signed zero, with the latter form sometimes called negative zero. These representations include the signed magnitude and one's complement binary integer representations (but not the two's complement binary form used in most modern computers), and most floating point number representations (such as IEEE 754 and IBM S/390 floating point formats).
In binary, 0 represents the value for "off", which means no electricity flow.[67]
Zero is the value of false in many programming languages.
The Unix epoch (the date and time associated with a zero timestamp) begins the midnight before the first of January 1970.[68][69][70]
The Classic Mac OS epoch and Palm OS epoch (the date and time associated with a zero timestamp) begins the midnight before the first of January 1904.[71]
Many APIs and operating systems that require applications to return an integer value as an exit status typically use zero to indicate success and non-zero values to indicate specific error or warning conditions.
Programmers often use a slashed zero to avoid confusion with the letter "O".[72]
## Other fields
• In comparative zoology and cognitive science, recognition that some animals display awareness of the concept of zero leads to the conclusion that the capability for numerical abstraction arose early in the evolution of species.[73]
• In telephony, pressing 0 is often used for dialling out of a company network or to a different city or region, and 00 is used for dialling abroad. In some countries, dialling 0 places a call for operator assistance.
• DVDs that can be played in any region are sometimes referred to as being "region 0"
• Roulette wheels usually feature a "0" space (and sometimes also a "00" space), whose presence is ignored when calculating payoffs (thereby allowing the house to win in the long run).
• In Formula One, if the reigning World Champion no longer competes in Formula One in the year following their victory in the title race, 0 is given to one of the drivers of the team that the reigning champion won the title with. This happened in 1993 and 1994, with Damon Hill driving car 0, due to the reigning World Champion (Nigel Mansell and Alain Prost respectively) not competing in the championship.
• On the U.S. Interstate Highway System, in most states exits are numbered based on the nearest milepost from the highway's western or southern terminus within that state. Several that are less than half a mile (800 m) from state boundaries in that direction are numbered as Exit 0.
## Symbols and representations
The modern numerical digit 0 is usually written as a circle or ellipse. Traditionally, many print typefaces made the capital letter O more rounded than the narrower, elliptical digit 0.[74] Typewriters originally made no distinction in shape between O and 0; some models did not even have a separate key for the digit 0. The distinction came into prominence on modern character displays.[74]
A slashed zero can be used to distinguish the number from the letter. The digit 0 with a dot in the center seems to have originated as an option on IBM 3270 displays and has continued with some modern computer typefaces such as Andalé Mono, and in some airline reservation systems. One variation uses a short vertical bar instead of the dot. Some fonts designed for use with computers made one of the capital-O-digit-0 pair more rounded and the other more angular (closer to a rectangle). A further distinction is made in falsification-hindering typeface as used on German car number plates by slitting open the digit 0 on the upper right side. Sometimes the digit 0 is used either exclusively, or not at all, to avoid confusion altogether.
## Year label
In the BC calendar era, the year 1 BC is the first year before AD 1; there is not a year zero. By contrast, in astronomical year numbering, the year 1 BC is numbered 0, the year 2 BC is numbered -1, and so forth.[75]
## Notes
1. ^ No long count date actually using the number 0 has been found before the 3rd century AD, but since the long count system would make no sense without some placeholder, and since Mesoamerican glyphs do not typically leave empty spaces, these earlier dates are taken as indirect evidence that the concept of 0 already existed at the time.
2. ^ Each place in Ptolemy's sexagesimal system was written in Greek numerals from , where 31 was written meaning 30+1, and 20 was written ? meaning 20.
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• Douglas Harper (2011), Zero, Etymology Dictionary, Quote="figure which stands for naught in the Arabic notation," also "the absence of all quantity considered as quantity," c. 1600, from French zéro or directly from Italian zero, from Medieval Latin zephirum, from Arabic sifr "cipher," translation of Sanskrit sunya-m "empty place, desert, naught";
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• Smithsonian Institution, Oriental Elements of Culture in the Occident, p. 518, at Google Books, Annual Report of the Board of Regents of the Smithsonian Institution; Harvard University Archives, Quote="Sifr occurs in the meaning of "empty" even in the pre-Islamic time. ... Arabic sifr in the meaning of zero is a translation of the corresponding India sunya.";
• Jan Gullberg (1997), Mathematics: From the Birth of Numbers, W.W. Norton & Co., ISBN 978-0-393-04002-9, p. 26, Quote = Zero derives from Hindu sunya - meaning void, emptiness - via Arabic sifr, Latin cephirum, Italian zevero.;
• Robert Logan (2010), The Poetry of Physics and the Physics of Poetry, World Scientific, ISBN 978-981-4295-92-5, p. 38, Quote = "The idea of sunya and place numbers was transmitted to the Arabs who translated sunya or "leave a space" into their language as sifr."
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48. ^ Algebra with Arithmetic of Brahmagupta and Bhaskara, translated to English by Henry Thomas Colebrooke (1817) London
49. ^ Kaplan, Robert (1999). The Nothing That Is: A Natural History of Zero. New York: Oxford University Press. pp. 68-75. ISBN 978-0-19-514237-2.
50. ^ Coedès, Georges, "A propos de l'origine des chiffres arabes," Bulletin of the School of Oriental Studies, University of London, Vol. 6, No. 2, 1931, pp. 323-328. Diller, Anthony, "New Zeros and Old Khmer," The Mon-Khmer Studies Journal, Vol. 25, 1996, pp. 125-132.
51. ^ Casselman, Bill. "All for Nought". ams.org. University of British Columbia), American Mathematical Society.
52. ^ Ifrah, Georges (2000), p. 400.
53. ^
54. ^ Pannekoek, A. (1961). A History of Astronomy. George Allen & Unwin. p. 165.
55. ^ a b c Will Durant (1950), The Story of Civilization, Volume 4, The Age of Faith: Constantine to Dante - A.D. 325-1300, Simon & Schuster, ISBN 978-0-9650007-5-8, p. 241, Quote = "The Arabic inheritance of science was overwhelmingly Greek, but Hindu influences ranked next. In 773, at Mansur's behest, translations were made of the Siddhantas - Indian astronomical treatises dating as far back as 425 BC; these versions may have the vehicle through which the "Arabic" numerals and the zero were brought from India into Islam. In 813, al-Khwarizmi used the Hindu numerals in his astronomical tables."
56. ^ Brezina, Corona (2006). Al-Khwarizmi: The Inventor of Algebra. The Rosen Publishing Group. ISBN 978-1-4042-0513-0.
57. ^ Will Durant (1950), The Story of Civilization, Volume 4, The Age of Faith, Simon & Schuster, ISBN 978-0-9650007-5-8, p. 241, Quote = "In 976, Muhammad ibn Ahmad, in his Keys of the Sciences, remarked that if, in a calculation, no number appears in the place of tens, a little circle should be used "to keep the rows". This circle the Mosloems called ?ifr, "empty" whence our cipher."
58. ^ Sigler, L., Fibonacci's Liber Abaci. English translation, Springer, 2003.
59. ^ Grimm, R.E., "The Autobiography of Leonardo Pisano", Fibonacci Quarterly 11/1 (February 1973), pp. 99-104.
60. ^ Hansen, Alice (9 June 2008). Primary Mathematics: Extending Knowledge in Practice. SAGE. ISBN 978-0-85725-233-3.
61. ^ Lemma B.2.2, The integer 0 is even and is not odd, in Penner, Robert C. (1999). Discrete Mathematics: Proof Techniques and Mathematical Structures. World Scientific. p. 34. ISBN 978-981-02-4088-2.
62. ^ W., Weisstein, Eric. "Zero". mathworld.wolfram.com. Retrieved 2018.
63. ^ Weil, Andre (6 December 2012). Number Theory for Beginners. Springer Science & Business Media. ISBN 978-1-4612-9957-8.
64. ^ Bunt, Lucas Nicolaas Hendrik; Jones, Phillip S.; Bedient, Jack D. (1976). The historical roots of elementary mathematics. Courier Dover Publications. pp. 254-255. ISBN 978-0-486-13968-5., Extract of pp. 254-255
65. ^ Reid, Constance (1992). From zero to infinity: what makes numbers interesting (4th ed.). Mathematical Association of America. p. 23. ISBN 978-0-88385-505-8. zero neither prime nor composite.
66. ^ Wu, X.; Ichikawa, T.; Cercone, N. (25 October 1996). Knowledge-Base Assisted Database Retrieval Systems. World Scientific. ISBN 978-981-4501-75-0.
67. ^ Chris Woodford 2006, p. 9.
68. ^ Paul DuBois. "MySQL Cookbook: Solutions for Database Developers and Administrators" 2014. p. 204.
69. ^ Arnold Robbins; Nelson Beebe. "Classic Shell Scripting". 2005. p. 274
70. ^ Iztok Fajfar. "Start Programming Using HTML, CSS, and JavaScript". 2015. p. 160.
71. ^ Darren R. Hayes. "A Practical Guide to Computer Forensics Investigations". 2014. p. 399
72. ^ "Font Survey: 42 of the Best Monospaced Programming Fonts". codeproject.com. 18 August 2010. Retrieved 2021.
73. ^ Cepelewicz, Jordana Animals Count and Use Zero. How Far Does Their Number Sense Go?, Quantam, August 9 2021
74. ^ a b Bemer, R. W. (1967). "Towards standards for handwritten zero and oh: much ado about nothing (and a letter), or a partial dossier on distinguishing between handwritten zero and oh". Communications of the ACM. 10 (8): 513-518. doi:10.1145/363534.363563. S2CID 294510.
75. ^ Steel, Duncan (2000). Marking time: the epic quest to invent the perfect calendar. John Wiley & Sons. p. 113. ISBN 978-0-471-29827-4. In the B.C./A.D. scheme there is no year zero. After 31 December 1 BC came 1 January AD 1. ... If you object to that no-year-zero scheme, then don't use it: use the astronomer's counting scheme, with negative year numbers.
## Bibliography
### Historical studies
• Bourbaki, Nicolas (1998). Elements of the History of Mathematics. Berlin, Heidelberg, and New York: Springer-Verlag. ISBN 3-540-64767-8.
• Diehl, Richard A. (2004). The Olmecs: America's First Civilization. London: Thames & Hudson.
• Ifrah, Georges (2000). The Universal History of Numbers: From Prehistory to the Invention of the Computer. Wiley. ISBN 0-471-39340-1.
• Kaplan, Robert (2000). The Nothing That Is: A Natural History of Zero. Oxford University Press.
• Seife, Charles (2000). Zero: The Biography of a Dangerous Idea. Penguin USA. ISBN 0-14-029647-6. | 10,484 | 41,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-39 | latest | en | 0.893894 |
https://zbmath.org/?q=an:0665.32011 | 1,611,256,539,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527224.75/warc/CC-MAIN-20210121163356-20210121193356-00739.warc.gz | 1,082,179,541 | 12,018 | # zbMATH — the first resource for mathematics
Twisted Picard-Lefschetz formulas. (English. Russian original) Zbl 0665.32011
Funct. Anal. Appl. 22, No. 1, 10-18 (1988); translation from Funkts. Anal. Prilozh. 22, No. 1, 12-22 (1988).
Let f: ($${\mathbb{C}}^ n,0)\to ({\mathbb{C}},0)$$ be the germ of a holomorphic function defining an isolated hypersurface singularity of multiplicity $$\mu$$ in $${\mathbb{C}}^ n$$. Furthermore, let F: ($${\mathbb{C}}^ n\times {\mathbb{C}}^{\mu},0\times 0)\to ({\mathbb{C}},0)$$ be a (mini-)versal deformation of f, $$F_{\lambda}:=F(-,\lambda)$$, and $$V_{\lambda}:=F_{\lambda}^{-1}(0)$$ for any $$\lambda\in {\mathbb{C}}$$. The discriminant of the versal deformation $$(V_{\lambda})_{\lambda \in {\mathbb{C}}^{\mu}}$$ of the given hypersurface singularity is defined to be the locus $$\Delta \subset {\mathbb{C}}^{\mu}$$ of the parameters $$\lambda$$ for which the hypersurface $$V_{\lambda}$$ is singular. Then one has a natural representation of the fundamental group $$\pi_ 1({\mathbb{C}}^{\mu}\setminus \Delta)$$ as a linear transformation group of the intermediate homology groups $$H_{n-1}(V_{\lambda},{\mathbb{Z}})\cong {\mathbb{Z}}^{\mu}$$, $$\lambda \in {\mathbb{C}}^{\mu}$$, just given by displacement of cycles along closed paths. The image of this representation of $$\pi_ 1({\mathbb{C}}^{\mu}\setminus \Delta)$$ in $$GL_{\mu}({\mathbb{Z}})$$ is called the monodromy group of the versal family. The monodromy group can be thought of as a subgroup of the isotropy group of the intersection form on the intermediate homology $$H_{n- 1}(V_{\lambda},Z)$$. The classical Picard-Lefschetz theory describes the generators of the monodromy group of a versal deformation (of isolated hypersurface singularities) in terms of a particular basis in $$H_{n- 1}(V_{\lambda},{\mathbb{Z}})$$, the so-called basis of vanishing cycles.
In the present paper, the author extends the classical Picard-Lefschetz theory, in that he investigates the action of the fundamental group $$\pi_ 1({\mathbb{C}}^{\mu}\setminus \Delta)$$ on the homology groups $$H_ n({\mathbb{C}}^ n\setminus V_{\lambda},{\mathcal Z}(q))$$ of the complement of a hypersurface $$V_{\lambda}$$ with values in a certain non-trivial locally constant sheaf $${\mathcal Z}(q)$$. This “twisted” sheaf, with stalk $${\mathbb{Z}}[q,q^{-1}]$$, involves the complex variable q and is characterized by the property that a closed path around $$V_{\lambda}$$ induces the multiplication by q on its stalks. The main result of the paper provides an explicit set of generators of the monodromy group of the given versal deformation in $$H_ n({\mathbb{C}}^ n\setminus V_{\lambda},{\mathcal Z}(q))$$, and that in terms of a suitable homology basis. In the concluding section of his paper, the author discusses examples, applications, and possible generalizations of his approach to the monodromy group. He shows how the Burau representation of the braid group $$\pi_ 1({\mathbb{C}}^{\mu}\setminus \Delta)$$ can be obtained, how the twisted cohomology of the complement of a non-singular hypersurface V in $${\mathbb{C}}^ n$$ can be interpreted as the De Rham cohomology of a certain complex of singular differential forms on $${\mathbb{C}}^ n\setminus V$$, and how the classical monodromy operator and the signature of the intersection form are calculated from his twisted Picard-Lefschetz formulae.
Reviewer: W.Kleinert
##### MSC:
32S30 Deformations of complex singularities; vanishing cycles 32Sxx Complex singularities 14D05 Structure of families (Picard-Lefschetz, monodromy, etc.) 32L10 Sheaves and cohomology of sections of holomorphic vector bundles, general results
Full Text:
##### References:
[1] V. I. Arnol’d, ”Normal forms of functions near degenerate critical points, the Weyl groups Ak, Dk, Ek, and Lagrangian singularities,” Funkts. Anal. Prilozhen.,5, No. 4, 3–25 (1972). [2] V. I. Arnol’d, ”Critical points of functions on manifolds with boundary, simple Lie groups Bk, Ck, F4, and singularities of evolutes,” Usp. Mat. Fiz.,33, No. 5, 91–105 (1978). [3] V. I. Arnol’d, A. N. Varchenko, and S. M. Gusein-Zade, Singularities of Differentiable Maps [in Russian], Vol. 1, Nauka, Moscow (1982). [4] V. I. Arnol’d, A. N. Varchenko, and S. M. Gusein-Zade, Singularities of Differentiable Maps [in Russian], Vol. 2, Nauka, Moscow (1984). [5] I. N. Bernshtein, ”Analytic continuation of generalized functions with respect to a parameter,” Funkts. Anal. Prilozhen.,6, No. 4, 26–40 (1972). [6] A. N. Varchenko, ”Semicontinuity of the spectrum and an upper bound for the number of singular points of a projective hypersurface,” Dokl. Akad. Nauk SSSR,270, No. 6, 1294–1297 (1983). · Zbl 0537.14003 [7] A. N. Varchenko, ”Local classification of volume forms in the presence of a hypersurface,” Funkts. Anal. Prilozhen.,19, No. 4, 23–31 (1985). · Zbl 0661.32017 [8] A. N. Varchenko, ”Combinatorics and topology of the disposition of affine hyperplanes in real space,” Funkts. Anal. Prilozhen.,21, No. 1, 11–22 (1987). · Zbl 0615.52005 [9] V. A. Vasil’ev, I. M. Gel’fand, and A. V. Zelevinskii, ”General hypergeometric functions on complex Grassmanians,” Funkts. Anal. Prilozhen.,21, No. 1, 23–38 (1987). · Zbl 0614.33008 [10] A. B. Givental’, ”Lagrangian manifolds with singularities and irreducible s2-modules,” Usp. Mat. Nauk,38, No. 6, 109–110 (1983). · Zbl 0535.58016 [11] A. B. Givental’ and V. V. Shekhtman, ”Monodromy groups and Hecke algebras,” Usp. Mat. Nauk,42, No. 4, 138 (1987). [12] V. V. Goryunov, ”Geometry of bifurcation diagrams of simple projections to a line,” Funkts. Anal. Prilozhen.,15, No. 2, 1–8 (1981). · Zbl 0463.30010 · doi:10.1007/BF01082373 [13] P. Griffiths and J. Harris, Principles of Algebraic Geometry [Russian translation], Mir, Moscow (1982). · Zbl 0531.14002 [14] V. P. Kostov, ”Versal deformations of differential forms of degree $$\alpha$$ on the line,” Funkts. Anal. Prilozhen.,18, No. 4, 81–82 (1984). · Zbl 0582.32041 · doi:10.1007/BF01076376 [15] S. K. Lando, ”Normal forms of powers of the volume form,” Funkts. Anal. Prilozhen.,19, No. 2, 78–79 (1985). · Zbl 0589.32048 [16] J. Milnor, Singular Points of Complex Hypersurfaces [Russian translation], Mir, Moscow (1971). · Zbl 0224.57014 [17] F. Fam, ”Generalized Picard–Lefschetz formulas and branched integrals,” Matematika,13, No. 4, 61–93 (1969). [18] K. Aomoto, ”On the structure of integrals of power product of linear functions,” Sc. Papers Coll. General Educ. Univ. Tokyo,27, 49–61 (1977). · Zbl 0384.35045 [19] T. Kohno, ”Linear representations of braid groups and classical Yang–Baxter equation,” in: Artin’s Braid Groups, Santa Cruz (1987). · Zbl 0634.58040 [20] J. Steenbrink, ”Mixed Hodge structure on the vanishing cohomology,” in: Real and Complex Singularities, Nordic Summer School, Oslo (1976). · Zbl 0373.14007 [21] J. Steenbrink, ”Semicontinuity of the singularity spectrum,” Invent. Math.,79, No. 3, 557–565 (1985). · Zbl 0568.14021 · doi:10.1007/BF01388523 [22] W. Ebeling, ”On the monodromy groups of singularities,” in: Proc. Symp. Pure Math., Vol. 40, Part 1 (1983), pp. 327–336. · Zbl 0518.32007 [23] S. V. Chmutov, ”Monodromy groups of singularities of functions of two variables,” Funkts. Anal. Prilozhen.,15, No. 1, 61–66 (1981). · Zbl 0464.51002 · doi:10.1007/BF01082383 [24] G. G. Il’yuta, ”Monodromy and vanishing cycles of boundary singularities,” Funkts. Anal. Prilozhen.,19, No. 3, 11–21 (1985).
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2,559 | 7,783 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-04 | latest | en | 0.673243 |
https://altair-viz.github.io/gallery/horizon_graph.html | 1,726,064,142,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00713.warc.gz | 76,803,420 | 6,303 | # Horizon Graph#
This example shows how to make a Horizon Graph with 2 layers. (See https://idl.cs.washington.edu/papers/horizon/ for more details on Horizon Graphs.)
```import altair as alt
import pandas as pd
source = pd.DataFrame([
{"x": 1, "y": 28}, {"x": 2, "y": 55},
{"x": 3, "y": 43}, {"x": 4, "y": 91},
{"x": 5, "y": 81}, {"x": 6, "y": 53},
{"x": 7, "y": 19}, {"x": 8, "y": 87},
{"x": 9, "y": 52}, {"x": 10, "y": 48},
{"x": 11, "y": 24}, {"x": 12, "y": 49},
{"x": 13, "y": 87}, {"x": 14, "y": 66},
{"x": 15, "y": 17}, {"x": 16, "y": 27},
{"x": 17, "y": 68}, {"x": 18, "y": 16},
{"x": 19, "y": 49}, {"x": 20, "y": 15}
])
area1 = alt.Chart(source).mark_area(
clip=True,
interpolate='monotone',
opacity=0.6
).encode(
alt.X('x').scale(zero=False, nice=False),
alt.Y('y').scale(domain=[0, 50]).title('y'),
).properties(
width=500,
height=75
)
area2 = area1.encode(
alt.Y('ny:Q').scale(domain=[0, 50])
).transform_calculate(
"ny", alt.datum.y - 50
)
area1 + area2
```
```import altair as alt
import pandas as pd
source = pd.DataFrame([
{"x": 1, "y": 28}, {"x": 2, "y": 55},
{"x": 3, "y": 43}, {"x": 4, "y": 91},
{"x": 5, "y": 81}, {"x": 6, "y": 53},
{"x": 7, "y": 19}, {"x": 8, "y": 87},
{"x": 9, "y": 52}, {"x": 10, "y": 48},
{"x": 11, "y": 24}, {"x": 12, "y": 49},
{"x": 13, "y": 87}, {"x": 14, "y": 66},
{"x": 15, "y": 17}, {"x": 16, "y": 27},
{"x": 17, "y": 68}, {"x": 18, "y": 16},
{"x": 19, "y": 49}, {"x": 20, "y": 15}
])
area1 = alt.Chart(source).mark_area(
clip=True,
interpolate='monotone'
).encode(
alt.X('x', scale=alt.Scale(zero=False, nice=False)),
alt.Y('y', scale=alt.Scale(domain=[0, 50]), title='y'),
opacity=alt.value(0.6)
).properties(
width=500,
height=75
)
area2 = area1.encode(
alt.Y('ny:Q', scale=alt.Scale(domain=[0, 50]))
).transform_calculate(
"ny", alt.datum.y - 50
)
area1 + area2
``` | 823 | 1,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.468027 |
https://mail.python.org/pipermail/edu-sig/2006-October/007387.html | 1,529,895,390,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867364.94/warc/CC-MAIN-20180625014226-20180625034226-00453.warc.gz | 665,025,081 | 2,523 | # [Edu-sig] Truth values and comparisons
John Zelle john.zelle at wartburg.edu
Mon Oct 30 18:07:04 CET 2006
```On Monday 30 October 2006 10:49 am, Arthur wrote:
> davelist at mac.com wrote:
> >I've not used .any or .all, but having just taught my CS1 students about
> > boolean operators, I was reminded that Python works as the following
> > example describes:
> >
> >x = a and b
> ># if both a and b are true, x is assigned b, otherwise x is assigned a
> >x = 2 and 3 # x is assigned 3
> >x = 0 and 2 # x is assigned 0
> >
> >x = a or b
> ># if a is true, x is assigned a, if a is true, x is assigned a, if a is
> > false and b is true, x is assigned b, if both a and b are false, x is
> > assigned False
> >
> >You need to determine what should make vertex_map[tp] be true...
>
> thanks, but having some trouble:
> >>> import Numeric as N
> >>> a=N.array([0,0,0])
> >>> b=N.array([0,0,1])
> >>> a and b
>
> array([0, 0, 0])
This tells me that a zero array is being treated as False (the logical
extension to arrays of 0 being false.
>
> >>> b and a
>
> array([0, 0, 0])
Right. In both cases, the answer is False, which Python gives to you by
handing you the first False expression. In either case the false part of the
and is the 0 array, while the other is True (a non-zero array).
You can check this out:
>>> not(a)
True
>>> not(b)
False
>
> Can this be? Either both a and b are true, or they are not, so can it
> be returning the "a" in both cases? Something screwy, other than my
> comprehension here?
Because a is False ;-). They're not both true.
> Same problem with
>
> >>> a or b
>
> array([0, 0, 1])
>
> >>> b or a
>
> array([0, 0, 1])
>
This works the same way, but returns the first True expression as the value
of "True."
> >>> any(a)
>
> False
>
> >>> all(a)
>
> False
>
> >>> any(b)
>
> True
>
> >>> all(b)
>
> False
>
> Though anyone growing up with the Python boolean operator might wonder
> why it is as it is - i.e. when 0 was the way to spell False this
> behavior is fairly well expected. Now that False is spelled "False",
> having "0" any less true than "1", when thinks one is dealing with
> numbers as numbers, is likely to catch the less geeky unaware, IMO.
This is why in teaching I prefer to use explicit tests:
if x != 0:
do something
Rather than
if x:
do something
--John
--
John M. Zelle, Ph.D. Wartburg College
Professor of Computer Science Waverly, IA
john.zelle at wartburg.edu (319) 352-8360
``` | 771 | 2,489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-26 | latest | en | 0.918294 |
http://heli-air.net/2016/02/15/boundary-layer-equations-3/ | 1,582,433,180,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145746.24/warc/CC-MAIN-20200223032129-20200223062129-00111.warc.gz | 66,184,217 | 16,536 | # Boundary-Layer Equations
The boundary layer can be considered as the phenomenological model of attached viscous flow. We derive in the following the boundary-layer equations for steady, compressible, three-dimensional flow past a flat surface. We assume laminar flow, but note that the resulting equations also hold for turbulent flows, if we treat them as Reynolds – or Favre-averaged flows [1, 6].
The derivation is made in Cartesian coordinates. We keep the notation introduced in Fig. 4.1 with x and z (z not indicated there) being the coordinates tangential to the body surface, and y the coordinate normal to it.
Fig. 7.2. Streamline curvature and cross-flow profiles of a three-dimensional boundary layer (schematically) [3]: a) negative cross-flow profile, b) s-shaped cross-flow profile, c) positive cross-flow profile. The coordinates t and n are tangential to the body surface, г is normal to the body surface.
Accordingly u and w are the tangential velocity components, and v is the component normal to the body surface.[86]
We try to keep the derivation as simple as possible in order to concentrate on the basic physical problems. (That is also the reason why we employ Cartesian coordinates.) Therefore, we do not include the description of the typical phenomena connected with hypersonic attached flow like the extra formulations for high-temperature real-gas effects, surface-radiation cooling, and slip-flow effects. Their introduction into the resulting system of equations is straight forward.
The boundary-layer equations are derived from the Navier-Stokes equations, together with the continuity equation and the energy equation, SubSection 4.3. They cannot be derived from first principles [1]. One has to introduce the observation—the boundary-layer assumption, originally conceived by L. Prandtl [7]—that the extension of the boundary layer in direction normal to the body surface (coordinate y and the involved boundary-layer thicknesses) is very small, like also the velocity component v in the boundary layer normal to the body surface. Actually the observation is that the different boundary-layer thicknesses and v in y-direction are inversely proportional to the square root of the Reynolds number.
We take care of this observation by introducing the so-called boundary – layer stretching, which brings y and v, non-dimensionalized with reference data Lref and vref, respectively, to O(1):
Reref = Preft’re/Lre/ . (7.3)
ftref
The prime above denotes variables, which were non-dimensionalized and stretched, however, we use it in the following also for variables, which are only non-dimensionalized.
All other variables are simply made dimensionless with appropriate reference data, and assumed then to be O(1): velocity components u and w with vref, lengths x and z with Lref, temperature T with Tref, density p with pref, pressure p with prefv? ff, the transport coefficients /л and к with /i, ref and kref, respectively, and finally the specific heat at constant pressure cp with cPref.4 Each resulting dimensionless variable is marked by a prime, for instance
u’ = —. (7.4)
vref
We introduce boundary-layer stretching and non-dimensionalization first into the continuity equation, eq. (4.83). We do this for illustration in full detail. In three dimensions and without the partial time derivative, Section 4.1, we replace и with u’vref, eq. (7.4), v with v’vrefjjReref, eq. (7.2), etc, and find
df/prefu’vref ^ dp’prefv’yref/-sjReref ^ dp’pref w’vref _ 0 ^ ^
Ox Lref dy’Lref/ уReref k)z Lrej
Because all reference parameters, and also Reref are constants, we find immediately the stretched and dimensionless continuity equation which has the same form—this does not hold for the other equations—as the original equation:
4 In classical boundary-layer theory the pressure is made dimensionless with Pref v2ref, which has the advantage that the equations describe in this form both compressible and incompressible flows. For general hypersonic viscous flows, however, we choose pref to make the pressure dimensionless, Sub-Section 7.1.7.
Again we introduce non-dimensional and stretched variables, as we did above. We also write explicitly all terms of O(1), and bundle together all terms, which are of smaller order of magnitude, now except for two of the heat-conduction terms:and Eref the reference Eckert number:
Eref = (jref — l)Mref ■ (7.22)
(Both were introduced in Sub-Section 4.3.2.)
We have retained on purpose in this equation two terms, which are nominally of lower order of magnitude. They are the gradients of the heat – conduction terms in x and z direction, which are of O(l/Reref). The reason is that we in general consider radiation-cooled surfaces, where we have to take into account possible strong gradients of Tw in both x and z direction. They appear there on the one hand, because the thermal state of the surface changes strongly in the down-stream direction, usually the main-axis direction of a flight vehicle, Chapter 3. On the other hand, strong changes are present in both x and z direction, if laminar-turbulent transition occurs, see, e. g., Sub-Section 7.3.
The question now is, under what conditions can we drop the two terms, regarding the changes of the wall temperature. To answer it, we follow an argumentation given by Chapman and Rubesin [10]. We consider first (in two dimensions) the gradient of the heat-conduction term in direction normal to the surface in dimensional and non-stretched form, eq. (7.19), and introduce finite differences, as we did in Sub-Section 3.2.1:
with 5T being the thickness of the thermal boundary layer.
The gradient of the heat-conduction term in x-direction is written, assuming that (dT/dx)w is representative for it
The result is: provided, that eq. (7.26) holds, the gradient term of heat conduction in ж-direction can be neglected, because St/L ос 1/(/Pr/Re) ^ 1 in general means
d dT d dT
« Ж,(к 17’■
In [10] it is assumed, that the recovery temperature is representative for the wall temperature
Tw=Tr= TU1 + г^-ІАф. (7.29)
Introducing this into eq. (7.26), together with St ~ S « cL/^/Reref for laminar flow, we obtain
With r = y/Pr = a/0.72, Y = 1.4, c = 6 we arrive finally, after having introduced non-dimensional variables, at the Chapman-Rubesin criterion [10]. It says that the term eq. (7.24) can be neglected, if
dT’
— |ш ^ <U>3.l/:( x H,,. (7.31)
This means, that, for instance, for M^ = 1 and Reref = 106, the maximum permissible temperature gradient would be equivalent to a thirty-fold increase of T/T^ along a surface of length Lref. From this it can be concluded, that in general for high Mach-number and Reynolds-number flows the Chapman-Rubesin criterion is fulfilled, as long as the surface-temperature distribution is “reasonably smooth and continuous”. The situation can be different for low Mach numbers and Reynolds numbers.
With radiation-cooled surfaces, as we noted above, we do not necessarily have reasonably smooth and continuous surface-temperature distributions in both x and z direction. Moreover, the basic relation eq. (7.23) needs to be adapted, because it does not describe the situation at a radiation-cooled surface. For that situation we introduce a slightly different formulation for both directions:
(7.32)
(7.32) because, at least for laminar flow, Tr — Tw is the characteristic temperature difference, see Sub-Section 3.2.2. We also introduce the absolute values dT/dxw and dT/dzw, because the gradients will be negative downstream
of the forward stagnation point, Sub-Section 3.2.1, but may be positive or negative in laminar-turbulent transition regimes, Sub-Section 7.3, and in hotspot and cold-spot situations, Sub-Section 3.2.4.
The modified Chapman-Rubesin criterion is then: if eqs. (7.32) and (7.33) hold, the gradient term of heat conduction in both x and ^-direction can be neglected, because again St/L ос 1/(/Pr/Re)
__L___ d_
Reref dx’
1 d ‘ Reref dz
We refrain to propose detailed criteria, like the original Chapman-Rubesin criterion, eq. (7.31). This could be done for the region downstream of the forward stagnation point, but not in the other regimes. In practice the results of an exploration solution should show, if and where the modified Chapman- Rubesin criterion is violated or not and whether the two tangential heat conduction terms must be kept or not.[87]
Provided that the modified Chapman-Rubesin criterion is fulfilled, we arrive at the classical boundary-layer equations by neglecting all terms of O(1/Reref) and O(1/Re2ref) in eqs. (7.6), (7.16) to (7.18), and (7.20). We write the variables without prime, understanding that the equations can be read in either way, non-dimensional, stretched or non-stretched, and dimensional and non-stretched, then without the similarity parameters Prref and
Eref :
dpu dpv dx dy g о c§" ^ + = 0, (7.36) du du du dp d f du (7.37) fmd~x + ,n% + PWTz dx + % 1 V%) ’ 0 = – dp dy’ (7.38) dw dw dw dp d f dw (7.39) PU^ + ""% + pwlb dz + dy Ы) ’
du 2 / dw 2 1 dy) +dy) j
These equations are the ordinary boundary-layer equations which describe attached viscous flow fields on hypersonic flight vehicles. If thick boundary layers are present and/or entropy-layer swallowing occurs, they must be employed in second-order formulation, see below. For very large reference Mach numbers Mref the equations become fundamentally changed, see Sub-Section 7.1.7.[88]
With the above equations we can determine the unknowns u, v, w, and T. The unknowns density p, viscosity p, thermal conductivity k, and specific heat at constant pressure cp are to be found with the equation of state p = pRT, and the respective relations given in Chapters 4 and 5. If the boundary – layer flow is turbulent, the apparent transport properties must be introduced, [1] and Section 8.5. If high-temperature real-gas effects are present in the flow field under consideration, the respective formulations and laws must be incorporated.
Since dp/dy is zero, the pressure field of the external inviscid flow field, represented by dp/dx and dp/dz, is imposed on the boundary layer. This means, that in the boundary layer dp/dx and dp/dz are constant in y – direction. This holds for first-order boundary layers. If second-order effects are present, dp/dx and dp/dz in the boundary layer are implicitly corrected via dp/dy = 0 by centrifugal terms, see below.
The equations are first-order boundary-layer equations, based on Cartesian coordinates. In general locally monoclinic surface-oriented coordinates, factors and additional terms are added, which bring in the metric properties of the coordinate system [1]. It should be noted, that the equations for the general coordinates are formulated such that also the velocity components are transformed. This is in contrast to modern Euler and Navier – Stokes/RANS methods formulated for general coordinates. There only the geometry is transformed, Section A, and not the velocity components.
If locally the boundary-layer thickness is not small compared to the smallest radius of curvature of the surface, the pressure gradient in the boundary layer in direction normal to the surface, dp/dy, is no longer small of higher order, and hence no longer can be neglected.[89] This is a situation found typically in hypersonic flows, where also entropy-layer swallowing can oc
cur, Sub-Section 6.4.2. This situation is taken into account by second-order boundary-layer equations, which basically have the same form as the first – order equations [1, 12]. Information about the curvature properties of the surface is added. The y-momentum equation does not degenerate into dp/dy = 0. dp/dy is finite because centrifugal forces have to be taken into account. At the outer edge of the boundary layer the boundary conditions are determined by values from within the inviscid flow field, not from the surface as in first-order theory, see the discussion in Sub-Section 6.4.2. Also the first derivatives of the tangential velocity components, of temperature, density and pressure are continuous [12], which is not the case in first-order theory, see, e. g., Fig. 6.23. | 2,903 | 12,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-10 | latest | en | 0.915392 |
https://blackpineanimalpark.com/book-819 | 1,675,521,722,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500140.36/warc/CC-MAIN-20230204142302-20230204172302-00371.warc.gz | 146,363,250 | 3,582 | The formula of the radius can be simply derived by dividing the diameter of the circle by two. Radius $${\rm{ = }}\frac{{{\rm{Diameter}}}}{{\rm{2}}}$$ Circle: Tangent
Do mathematic problem
Free time to spend with your friends
## Radius Formula of a Circle Using Diameter, Area
Bohr Radius Formula [Click Here for Sample Questions] The formula of Bohr radius is. a 0 =4πε 0 (h/2π) 2 /m e e 2 =(h/2π)/m e cα. Where, a o = Bohr radius. m e =rest mass of
## Equation of a Circle
Formulas for Radius r is the radius of the circle d is the diameter of the circle C is the circumference of the circle A is the area of the circle
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Recurring customers | 194 | 675 | {"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} | 3.4375 | 3 | CC-MAIN-2023-06 | latest | en | 0.776047 |
https://www.mathblog.dk/tag/pathfinding/ | 1,638,881,213,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363376.49/warc/CC-MAIN-20211207105847-20211207135847-00331.warc.gz | 960,664,205 | 12,933 | # Pathfinding
## Project Euler 83: Find the minimal path sum from the top left to the bottom right by moving left, right, up, and down.
Problem 83 of Project Euler is the natural continuation of Problem 81 and 82 which we have already solved. The problem reads
NOTE: This problem is a significantly more challenging version of Problem 81.
In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by moving left, right, up, and down, is indicated in bold red and is equal to 2297.
131 673 234 103 18 201 96 342 965 150 630 803 746 422 111 537 699 497 121 956 805 732 524 37 331
Find the minimal path sum, in matrix.txt (right click and ‘Save Link/Target As…’), a 31K text file containing a 80 by 80 matrix, from the top left to the bottom right by moving left, right, up, and down.
Posted by Kristian in Project Euler, 11 comments | 242 | 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} | 2.78125 | 3 | CC-MAIN-2021-49 | latest | en | 0.884098 |
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# How can you calculate amortized loan payments with a formula?
A:
The formula for calculating amortized loan payments is A=P(r(1+r)^n)/((1+r)^n-1), states Vertex42. "A" is the payment amount per period, "P" is the beginning loan principal, "r" is the interest rate per period, and "n" is the number of payment periods.
## Keep Learning
To use this formula, add one to the interest rate per period, and multiply that number by itself "n" times, where "n" is the number of payment periods, according to Vertex42. Subtract one from this number to get the denominator of the formula. To calculate the numerator, multiply the previous multiplied number, without subtracting one, by the rate and then by the principal loan amount, Then divide the denominator from the numerator to get the payment amount per period.
Sources:
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### Messages - DuncanDoenitz
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1
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 28, 2021, 10:19:42 AM »
@Mark1986
Actually thought about this a bit this afternoon. What you describe is akin to the Doppler effect. The amount of frequency change differs depending upon whether you are travelling away from the sound source or toward the sound source.
Not going to put the effort into trying to come up with actual numbers but conceptually I see this in your argument. In RE, skydiver steps out of plane and falls x in time t. In this scenario, the diver has negligible impact on the position of the earth so for all intents and purposes only the diver is moving and the earth can be considered as stationary.
In FE, skydiver steps out of plane a distance x from the earth. However, in this case the earth pushing the atmosphere also pushes the diver. This means that in time t, the earth must travel not only x but also the distance that the atmosphere has pushed the skydiver in that time. Perhaps there's a FEer that can discuss this. I see it as an interesting anamoly in the FE equivalence argument.
I don't know enough about the doppler effect to comment. I'd have to think about it. But I don't necessarily see this as a problem with the equivalence argument in general so much as how they want to apply it. I get the whole "reverse the signs" thing, but you can't just pick and choose which ones you want to reverse. If gravity is up instead of down, you have to reverse the direction of any force that works with, against or depends on gravity. Technically, weight force would have negative value on FE. Specific to this scenario...air resistance would become air thrust.
The basic flaw in the FE reasoning is that air resistance, by definition, opposes the direction of motion. But they are trying to make it work with air resistance moving in the same direction as the skydiver...from the bottom up.
If you don't mind looking stupid, try this. Take one of those plastic grocery bags that have handles. Wrap a blow dryer around it, put it on the floor so it is blowing up and see what happens. To keep it from blowing away, you have to apply some resistance towards the floor. In order for it to move towards the floor, you have to pull it down, with more force than the air is blowing it up. There is absolutely no reason why the bag would ever move towards the floor while the air flow is up, unless there is a stronger force opposite pulling it down. The dryer doesn't make the bag move more slowly towards the floor, it keeps it from hitting the floor at all, ever. For as long as you sit there looking stupid blowing air into a plastic bag with a blow dryer, the bag will never hit the ground unless you pull it from the bottom (or push it from the top, too, I guess). Let go of the bag and it just blows away in the direction of the air flow.
In other words, if there is no force pulling him down, air flow from the bottom up would only propel the skydiver up. The force coming up from the ground and catching the parachute isn't "resistance", its thrust.
Now, I think you might be getting it.
2
##### Flat Earth Investigations / Re: Branson to go only 55 miles up !
« on: July 28, 2021, 09:36:37 AM »
Hi Pete,
I am not trolling and I'm certainly not handwaving away your arguments.
Here's what I did in order to make clear that the photo you posted does not show concave curvature in my opinion:
(For your reference: I used the free Inkscape drawing program. It is a vector based program)
- I drew a dashed line over the horizon. I used a dashed type of line, so that you are able to verify for yourself that I'm not selecting some arbitrary points in the picture
- Then I drew a straight line which connects both extremes of the visible horizon
- I added a horizontal line for reference
see this picture: https://imgur.com/N7vkwuA
- I exported this picture to png
- Then I resized the width from 100% to 20% without preserving the aspect ratio and saved it as a 2nd picture. This allows to see curved lines more pronounced.
Please note that the straight lines which I drew are still straight. Only the angle of the line which connects both extremes of the visible horizon has changed (more tilted), the shape has not changed.
See: https://imgur.com/iaiQf3E
I hope at least it explains why I'm not seeing any concaved curve. The line of the horizon is a bit fuzzy, so I would understand if you don't agree 100% with the dotted line I drew, but I'm willing to exchange the Inkscape csv file if you like.
Best regards
The original image in question was captured at short-haul airliner altitude during the airborne glide back to base. Covid-permitting, many of us will have an opportunity to take clearer photographs this summer.
3
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 27, 2021, 01:38:16 PM »
This is the difference between arithmetic and science. You are putting data into a calculator and expecting it to work it out for you. You can't do that if you haven't got your head around the science first. If we actually wanted to know the terminal velocity of the skydiver, or how long it takes him to impact, the calculator would be great. But we don't. If we had 2 identical spheres of pure gold, and wanted to know the weight one of them we could weigh it using some kind of scales to balance against a known mass. Or we could determine the volume of water it displaces, and multiply by the density of pure gold. But we don't, we only want to know if they are identical, so we can just balance them together on scales. No calculations. No numbers.
Lets see where we disagree on the science:
Round Earth
The skydiver has mass, and he will be accelerated downward by a force of 9.81 m/s/s multiplied by his mass.
At the instant of release, he will feel weightlessness, as the force of gravity is not being opposed.
As soon as he develops airspeed he will feel an opposing force upward due to aerodynamic drag.
The drag force is proportional to his size and shape, and to the square of his airspeed.
As soon as he develops airspeed, and hence drag, he will begin to feel some of the effect of gravity.
Because he is feeling a downward force (Gravity) opposed by an upward force (drag), his airspeed will increase at <9.81m/s/s.
As airspeed increases, the force of drag will increase.
At a particular airspeed, the increasing force due to drag will equal his mass multiplied by 9.81 due to gravity. He will stop accelerating. His airspeed will be constant thereafter, until impact. This is the Terminal Velocity (TV).
He is no longer weightless, as he is being supported by the force of drag so feels the effect of gravity.
The force of drag is not a constant; it increases with airspeed until stabilizing at terminal velocity.
If he was a styrofoam mannequin, but with the same aerodynamic size and shape, the upward drag-force would equal his gravitational down-force at a reduced airspeed, so his TV would be lower, so time-to-impact would be longer.
Flat Earth
Prior to jumping, the skydiver is at identical altitude to the RE case.
The human skydiver has identical mass and aerodynamics.
Prior to jumping, skydiver has the same upward velocity and (UA) acceleration as Earth. His airspeed is zero. (His speed relative to the universe is the same as Earth, a gazillion mph, but we are only interested in speed relative to the Earth/Atmoplane).
He jumps....
For a few milliseconds his airspeed remains zero, but he is no longer accelerating upward and his universal speed is constant. He experiences weightlessness.
At no point is he subjected to acceleration downward.
The Earth continues to accelerate upwards at 9.81 m/s/s. As the Atmoplane is static with respect to Earth, it also accelerates at 9.81m/s/s, unmitigated by any aerodynamic forces.
As the skydiver has constant Universal Velocity, but the atmoplane is accelerating upward, then he will begin to experience an upward airspeed.
As the airspeed increases at 9.81m/s/s, the skydiver will become subject to the force of drag, which will increase identically to the RE case, as his aerodynamic properties are identical.
As soon as he develops airspeed, and hence drag, he will begin to feel some of the effects of UA; no longer weightless.
This upward force will be opposed by the inertia of the skydiver due to his mass, which is identical to the RE case.
As he continues to accelerate upwards with respect to the universe, the rate of acceleration will increase, due to increasing airspeed, until the upward force of drag equals his mass multiplied by 9.81 (due to UA). This is TV.
At TV, his acceleration is now identical to the Earth/Atmoplane, but his Universal Velocity is lower, so the Earth will be catching up.
He is no longer weightless, as he is being accelerated at 9.81m/s/s by the atmosphere.
If he was a styrofoam mannequin , the upward drag force would equal his mass multiplied by UA at a lower airspeed, so he would have a lower TV. Because his TV is lower, his UV is higher, hence closer to that of Earth. Because he is now accelerating at the same rate as Earth, but spent less time un-accelerated, earth will still impact him, but time to impact would be longer.
In summary; same mass, same drag, same acceleration. Different pos/neg signs.
Please let us know if you (or anyone) have a problem any these statements.
4
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 26, 2021, 09:52:08 PM »
You found an online calculator for a Flat Earth Skydiver??
5
##### Flat Earth Investigations / Re: Branson to go only 55 miles up !
« on: July 26, 2021, 07:04:02 PM »
Personally, I'm not convinced of any shape to be determined from what are obviously short focal-length cameras on the Virgin craft. Face it, the cameras are not there for science, they are there to provide shiny publicity images for the Virgin operation. You are always going to get this when the camera is attached to the device it is photographing. The image from Felix Baumgartner's capsule is another example.
Having said that, the FE cause is hardly advanced by Pete giving us 3 screenshots from at least 2 different cameras, taken at airliner-altitude and purporting to show concavity (unless, of course, he is just seeking to demonstrate the futility of our attempting to draw such conclusions).
On the other hand, the RE cause is not advanced by making unsubstantiated claims about what we think the craft passengers saw through the enormous windows.
6
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 25, 2021, 11:57:08 AM »
First of all, your (light brown) "47 second" line is not valid because it is a straight line. It should be a curve as the object is accelerating, until it reaches terminal velocity, at which point it will become straight. What is the terminal velocity?; depends on the aerodynamic drag. You can't just interpolate linear points between zero and 47 seconds.
Second of all, you seem to be adding the acceleration of the (RE) skydiver to the acceleration of (FE) Earthing in coming to your 8 second and 11 second estimates.
Third of all, you are still including a hypothetical starting velocity of 100-somethings. The initial velocity is zero, because that is the relative velocity of Earth and the object/skydiver.
Look at it again:
Initial relative velocity of object to Round Earth-and-atmosphere = zero. Initial relative velocity of object to Flat-Earth-and-atmoplane = zero.
RE acceleration due to gravity = 9.81. FE acceleration due to magical UA = 9.81.
Drag-coefficient of object is identical in both scenarios.
On release, object and Earth initially accelerate towards each other at identical rates in both scenarios.
Airspeed of the objects develops identically in both scenarios.
As airspeed is identical, the force of drag is identical.
RE object will continue to accelerate due to gravity at 9.81, opposed by drag (which is increasing with airspeed) until the force of drag = force of gravity. At this point, Terminal Velocity, the object will stop accelerating and will continue to fall at constant velocity until impact.
Flat Earth-and-atmoplane will continue to accelerate at 9.81. The object is not accelerating due to UA, but begins to feel an upward force due to drag as the atmoplane develops airspeed around it. The drag increases with airspeed until its force = 9.81. At this point, Terminal Velocity, the object will have identical acceleration to the Earth-Atmoplane. Although its acceleration is identical, its universal velocity is lower, so the FE continues to approach it until impact.
It doesn't matter what the numbers are. For the same object in the same density air; the relative starting-velocity is zero, the acceleration of the object/air due to gravity/UA is identical, therefore the airspeed is identical, therefore the drag is identical, therefore terminal velocity is identical, therefore time-to-impact is identical.
7
##### Flat Earth Investigations / Re: Branson to go only 55 miles up !
« on: July 24, 2021, 09:23:28 AM »
The FAA know as we all do that this is a very 'cheap' claim from these billionaires who have trawled for the lowest bar possible and want to hand out the accolade of astronaut to anyone who pays up.
Now that's not really what NASA et al want. They want you to think astronauts are brave. And so when men were in rocket aircraft in the 1960s breaking speed and altitude records, they wanted to give them something for that. Little astronaut wings. But just sitting on a small private aircraft to receive the same honour cheapens it.
So they've got them on "You just sat there and did nothing". I'm sure they'd like to get them on "You're not in space" but then they have to remove the honours they handed out to men who actually deserved them.
But the thing about truth is that the FAAs or NASAs version of it is neither here nor there. At 85km ... Branson is still in the territory of clouds. I don't care what NASA say, you don't get clouds in space.
Will be interesting to see if the pilots of Branson's aircraft get given astronaut wings because they weren't sat there doing nothing. That would put Branson's nose right out of joint.
For the record, no, they didn't "get given" astronaut wings following this flight.
In fact, they didn't "get given" anything following this flight. Dave Mackay and Michael Masucci both already earned their FAA accredited astronaut wings as test pilots of VSS Unity in 2019.
And 100%, in my opinion, Sir Richard and the other passengers have been to space, but have not earned anything.
8
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 23, 2021, 09:03:35 AM »
Don't sweat it, its an interesting thought experiment, but I don't know what the drag coefficient of our hypothetical object is either. The thing is, its like a wind tunnel. Happy with that concept? Drag increases with velocity, doesn't matter whether our object is accelerating in still (RE) air, or it's stationary in accelerating Wind Tunnel/FE air; same result. And the relative acceleration is always going to 9.81 m/s/s.
9
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 23, 2021, 08:42:47 AM »
Quote
You are the one who introduced it to begin with, that's why I am asking you.
Sorry I wasn't more clear. I meant I don't have the answers to that question because it's not my theory. The earth has been accelerating according to FE, so at any given moment it has an "initial velocity". I just picked 1000 at random. And I don't know what it is supposed to be relative to because the story keeps changing. Is the relative velocity between the earth and a falling object 0? If so, and they are accelerating at the same rate, then they would never meet. So the velocity of the falling object must be different, but I don't know what that is supposed to be, so i just started with 9.8 at 1 second. If you think it should be something else, I can change it.
The velocity and acceleration of a SUSPENDED object are IDENTICAL to EARTH.
At the INSTANT of RELEASE, the velocity is identical, BUT THE OBJECT STOPS ACCELERATING.
In the seconds following release, the OBJECT MAINTAINS CONSTANT VELOCITY, BUT THE EARTH CONTINUES TO ACCELERATE, so it starts to catch up with the object.
10
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 22, 2021, 12:16:11 AM »
1. First, nice that you've shown some calculations. Thank you.
2. You seem to be using a mixture of SI, MPH and Ft/Sec which is confusing.
3. How did you calculate the aerodynamic drag? I don't see a Cd. What is its size and shape? How have you incorporated Velocity-squared? Why is it in units of "Kg/m"?
4. Why are you doubling the acceleration in the FE case?
5. Why have you specified a Mass? Its immaterial. But you haven't specified a size and shape (Surface area and Cd)
6. How have you calculated that "with air resistance it will take 14 seconds ......".
7. In FE (and why the f@@@ I am defending FE I'm not sure), the atmosphere is not providing any resistance to the Earth's acceleration, because there is no relative airspeed between the Earth and the atmosphere. The Earth and its atmosphere are essentially one unit that is being accelerated upwards by a magical force*. The acceleration of Earth serves to compress the atmosphere at its base, which is why its density decreases with altitude, but it is static, it has no inherent movement relative to the Earth so, so its velocity-squared equals zero, so its drag equals zero.
8. I'm with Clyde; why 1000 mph (or m/s on the spreadsheet)?
* As a side issue, we may like to ask ourselves why, in the Earthly realm, Magical Acceleration only acts directly on the Earth. Everything else is not accelerated, unless it is supported by the Earth directly (like houses, the atmosphere, you and me) or indirectly (like an aircraft supported by the atmosphere, or the 50 kg object in your hand). Odd thing is, that all this stuff (the houses, the object, the aircraft, molten lava) is all actually made out of Earth, but doesn't float. Weird.
11
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 21, 2021, 08:17:27 AM »
Quote
The Earth's initial velocity with respect to the about-to-be-free-falling object is 0mph. It seems like you aren't taking that into consideration. I don't know why you are choosing an initial velocity of 1000mph, what that velocity is supposed to be relative to, or how you are constructing this in your head, but I think you are adding some extraneous data points that are confusing you.
If the relative velocity between the object and the earth is 0 and rate of acceleration is the same, the earth and the object will never meet. The distance between them will always be the same.
Duncan suggested that the object would be subject to some air resistance, which would change the relative velocities. That’s reasonable. That would mean even though the initial velocities are the same, the rate of acceleration and the velocity of the object would be reduced and eventually the earth and the object would meet.
The problem is they would never meet in the same time frame that we see in reality.
With gravity at 9.8 and air resistance of 2.4 a 50 kg. object will fall 1000 ft. and meet the ground in 10 seconds. That makes the effective rate of acceleration 9.5 That’s what we would expect to see in reality, using standard method of calculation.
So if we have the earth accelerating up from 0 height at 1000 mph and 9.8 and the object accelerating up from an initial height of 1000 ft. at a rate of 9.5, at the end of 10 seconds the object would be at 17225 and the earth would be at 16274. They are still 951 feet apart.
(In the FE Model); Immediately upon release, the Earth&Atmosphere&Object have identical velocities. However, while the Earth&Atmosphere continue accelerating up at 9.81 m/s/s, there is now no force whatever acting upon the object so it is at constant velocity, so the Earth&Atmosphere will start to catch up with the Object.
In the seconds following release, as the accelerating atmosphere begins to have velocity relative to the object, the Object will become subject to aerodynamic drag, which will provide a force Up, beginning to accelerate the Object upwards.
It is important to note that this is upward force is not a fixed quantity. (I don't know how you came up with a figure of 2.4 (units?) for a 50 kg object). The upward force is directly proportional to the Object's size and shape (technically, its surface area and Drag Coefficient), to air density (so it is reduced at higher altitudes) and to the square-of-air-velocity. That "square" is important, as it means the drag force is very small at low velocities, but increases rapidly with higher airspeed. It is not a constant force, it increases as velocity increases.
Please note that the upwards force is completely unrelated to mass. Flat or Round Earth;
1. The "downward" force ("gravity" or "UA") is related only to mass and a constant rate of acceleration.
2. The "upward" force is completely unrelated to mass. It is related only to size and shape of the Object, air density, and the square of relative-velocity.
All of this is exactly like RE, applying Jack's pos/neg sign changes.
12
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 20, 2021, 02:21:00 PM »
Quote
But if you drop an object with initial velocity zero, then that initial velocity is with respect to the earth's surface, so both the object and the earth (and indeed the atmosphere) are moving at the same velocity. From the moment it leaves your hands, the earth and atmosphere would continue accelerating, but the object would not, other than a small acceleration caused by the atmosphere pushing on it.
If the object and atmosphere are accelerating up at the same rate and velocity as the earth, the earth and the object would never meet.
When released, the object is not accelerating up, but the Earth-and-atmosphere are. Once the object begins to feel airspeed, it will begin to accelerate up again, but not at the same rate as Earth-Atmosphere. The amount of acceleration it gets from aerodynamic drag is directly proportional to its size and shape, but it will always be less than the acceleration of Earth-Atmosphere.
13
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 20, 2021, 08:27:22 AM »
Just for the record, Jack appeared to agree with me. That is not necessarily the same thing as me agreeing with Jack.
However; I was stating that a Round-Earth case, where a "falling" object accelerates downward towards a constant-velocity planet whilst being opposed by aerodynamic drag, produces the same effect as a hypothetical disc-world accelerating upward towards a constant-velocity object, with the atmospheric air-cushion providing an equivalent upward aerodynamic force.
Yes, acceleration is a vector quantity. In RE it's (what we perceive as) "Down", and in FE it would be "Up". So yes, I would still agree with Jack that this would represent a simple change of pos/neg, and the magnitude remains identical.
In both cases, and again I think Jack would agree, the aerodynamic accelerating force is always "Up".
He is also correct in bringing up the Archimedes effect, which is independent of aerodynamics.
It is not exactly the same - on a hypothetical upwards accelerating disc world, gravity would be uniform world-wide. Whether you are standing on top of Mt. Everest, or sailing on the Indian Ocean, you should experience the same downward acceleration no matter where you are on the plane. Such would be the nature of an upwards accelerating plane.
But the reality is that you do not experience the same gravity everywhere on earth. Because the Earth is not a perfect sphere, you can have a variable distance to the Earth's center, which has a tiny, but measurable effect on the gravity you experience. Gravity is 9.7803 metres per second squared at the equator and 9.8322 m/s2 at the poles. Mount Nevado Huascarán in Peru has the lowest gravitational acceleration, at 9.7639 m/s2, while the highest is at the surface of the Arctic Ocean, at 9.8337 m/s2.
Agreed. I was answering the OP's question.
In practice, there are variations in Planet-Earth's surface g, just as there are variations in the atmospheric drag due to the vertical component of atmospheric currents (like thermals).
14
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 19, 2021, 10:30:49 PM »
Just for the record, Jack appeared to agree with me. That is not necessarily the same thing as me agreeing with Jack.
However; I was stating that a Round-Earth case, where a "falling" object accelerates downward towards a constant-velocity planet whilst being opposed by aerodynamic drag, produces the same effect as a hypothetical disc-world accelerating upward towards a constant-velocity object, with the atmospheric air-cushion providing an equivalent upward aerodynamic force.
Yes, acceleration is a vector quantity. In RE it's (what we perceive as) "Down", and in FE it would be "Up". So yes, I would still agree with Jack that this would represent a simple change of pos/neg, and the magnitude remains identical.
In both cases, and again I think Jack would agree, the aerodynamic accelerating force is always "Up".
He is also correct in bringing up the Archimedes effect, which is independent of aerodynamics.
15
##### Science & Alternative Science / Re: FE and ICBMs
« on: July 16, 2021, 02:10:20 PM »
And you keep using "that word," when you call out what is, quite obviously, TWO WORDS.
Have a nice day.
Since we're not advancing the argument at this stage, I think you quite obviously mean "..... what are, quite obviously, TWO WORDS".
Anywho; Has everyone heard of "muzzle velocity". Its the key parameter in the ballistics of firearms and artillery. Nothing to do with rates of acceleration. Nothing to do with average speed in the barrel.
Muzzle velocity. The instantaneous speed at the end of burn.
16
##### Science & Alternative Science / Re: FE and ICBMs
« on: July 15, 2021, 01:02:48 PM »
A Formula 1 race starts when the red lights go out, not when the car starts moving. If a car is moving when the red lights go out, it is disqualified.
Its velocity at the start is zero.
17
##### Flat Earth Investigations / Re: Branson to go only 55 miles up !
« on: July 14, 2021, 01:04:28 PM »
Where does London start? City of London? Greater London?
London is enormous. You missed another of its excellent airports. London Oxford Airport.
No, I didn't miss it. I was listing the destinations offered by a particular service provider. But you are correct in pointing out that other alternatives are offered by other providers.
Now, it calls itself Windsor castle, but I think it is pretty obvious that it is not Windsor castle.
With respect, its only obvious to you because you did your research. You would want to be sure that its a "Windsor Castle" that satisfies your requirements before you book an Uber. Would you want to finish up at the the big fortified house in Berkshire, for instance, when your mates are waiting for you down at the pub? If I was going to spend several hundred thousand dollars on a ticket from Sir Richard, I think I would first satisfy myself with where he was taking me. Its on the VG website:
APOGEE
Nearly 300,000 feet above Earth, the cabin becomes your playground to unbuckle and experience weightlessness.
18
##### Flat Earth Investigations / Re: Branson to go only 55 miles up !
« on: July 14, 2021, 10:26:22 AM »
Where does London start? City of London? Greater London?
Easyjet claims to be able to fly me to 4 airports in London; London Gatwick, London Luton, London Southend, and London Stansted. None of these are within any recognised boundary of the City, and Stansted is actually 47 Km from the Tower of London.
So are they mis-selling? I'm an intelligent kind of guy, and before I spend my £19.99 I'm going to do my research see if its where I want to go. I could pay more and get a helicopter into Hyde Park. Or maybe Luton is close enough to satisfy me. (Disclaimer; have you been to Luton? Don't). Can I buy a souvenir London Bus at Gatwick? You betcha!
Branson's marketing is quite clear on what you're getting for your money. You want more? Just like Easyjet, the developing market is filling up with alternative providers, who will fly you to London City, Earth-orbit, or (potentially) the Moon.
19
##### Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 12, 2021, 10:10:13 AM »
By "light", I assume you mean "less dense".
Same as Round Earth. In RE all matter with mass is attracted to the Earth at the same rate of acceleration, but as soon as it develops velocity it becomes subject to aerodynamic drag. The drag of less dense items is generally greater than denser ones due to the difference in surface area, so they feel more drag, so they move more slowly. eg; a one gram feather has more surface area than a one gram lead-pellet.
In FE, the atmosphere ("atmo-plane" in FE yuk-speak) is supported by the Earth so accelerates at the same rate. Hold the lead-pellet and the feather in your hand and they are feeling Universal acceleration because your feet are on the accelerating Earth. Release them, they stop accelerating, but the Earth and atmo-plane continue accelerating towards them. The pellet has a smaller surface area, so experiences less drag and remains relatively stationary. The feather, due to its larger surface area, feels more drag so starts accelerating upwards in the airstream so, to the observer, appears to fall less slowly.
If both items were in a vacuum chamber they would fall at the same rate, RE or FE.
20
##### Science & Alternative Science / Re: FE and ICBMs
« on: June 21, 2021, 03:05:40 PM »
Oh Jesus. 7.17km/s is a velocity. As he says in his quote. They have different units than acceleration. Slow down. Take a breath.
If you understand enough to correct what it should be to your own mind, then run with it.
This is you objectively fucking up and then doubling down on it. What will it take for you to understand that you aren’t currently knowledgeable enough to analyze this issue?
The rate I put forth is fine. I mistakenly put the word acceleration instead of the word velocity.
Kindly pat yourself on the back, take two if you care, my apology for using the wrong word in this case, and have a great rest of your day.
Before you go, if you can explain how a 16,038 mile per hour velocity at t+5 could possibly translate into an average rate of 3,000 mph over 5 minutes, that would be terrific.
ETA: To totally satisfy what appears to be a certain need for perfection in others, 0 - 16,038 mph translates to an acceleration of 23.9 m/s2.
You need to understand the difference between an average and an absolute. The average family has 2.4 children. So does an average family exist?
In physics, averages might make for interesting comparison, but are no basis for calculation. The only considerations in calculating the energy state at a particular instant are its instantaneous position and velocity.
Consider this; Urbanville and Townsville and are 60 miles apart. They are served by a train which does the journey in one hour. How the f*** are you supposed to get off the train in Townsville if it is doing an average of 60 mph?
Pages: [1] 2 3 ... 8 Next > | 7,538 | 31,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-31 | latest | en | 0.936426 |
https://doisinkidney.com/posts/2019-05-28-linear-phases.html | 1,723,415,470,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00581.warc.gz | 169,472,694 | 7,784 | ## Deriving a Linear-Time Applicative Traversal of a Rose Tree
Posted on May 28, 2019
Part 7 of a 10-part series on Breadth-First Traversals
Tags:
# The Story so Far
Currently, we have several different ways to enumerate a tree in breadth-first order. The typical solution (which is the usual recommended approach in imperative programming as well) uses a queue, as described by Okasaki (2000). If we take the simplest possible queue (a list), we get a quadratic-time algorithm, with an albeit simple implementation. The next simplest version is to use a banker’s queue (which is just a pair of lists). From this version, if we inline and apply identities like the following:
``foldr f b . reverse = foldl (flip f) b``
We’ll get to the following definition:
``````bfe :: Forest a -> [a]
bfe ts = foldr f b ts []
where
f (Node x xs) fw bw = x : fw (xs : bw)
b [] = []
b qs = foldl (foldr f) b qs []``````
We can get from this function to others (like one which uses a corecursive queue, and so on) through a similar derivation. I might some day write a post on each derivation, starting from the simple version and demonstrating how to get to the more efficient at each step.
For today, though, I’m interested in the traversal of a rose tree. Traversal, here, of course, is in the applicative sense.
Thus far, I’ve managed to write linear-time traversals, but they’ve been unsatisfying. They work by enumerating the tree, traversing the effectful function over the list, and then rebuilding the tree. Since each of those steps only takes linear time, the whole thing is indeed a linear-time traversal, but I hadn’t been able to fuse away the intermediate step.
# Phases
The template for the algorithm I want comes from the `Phases` applicative (Easterly 2019):
``````data Phases f a where
Lift :: f a -> Phases f a
(:<*>) :: f (a -> b) -> Phases f a -> Phases f b``````
We can use it to write a breadth-first traversal like so:
``````bft :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = runPhases . go
where
go (Node x xs) = liftA2 Node (Lift (f x)) (later (traverse go xs))``````
The key component that makes this work is that it combines applicative effects in parallel:
``````instance Functor f => Functor (Phases f) where
fmap f (Lift x) = Lift (fmap f x)
fmap f (fs :<*> xs) = fmap (f.) fs :<*> xs
instance Applicative f => Applicative (Phases f) where
pure = Lift . pure
Lift fs <*> Lift xs = Lift (fs <*> xs)
(fs :<*> gs) <*> Lift xs = liftA2 flip fs xs :<*> gs
Lift fs <*> (xs :<*> ys) = liftA2 (.) fs xs :<*> ys
(fs :<*> gs) <*> (xs :<*> ys) = liftA2 c fs xs :<*> liftA2 (,) gs ys
where
c f g ~(x,y) = f x (g y)``````
We’re also using the following helper functions:
``````runPhases :: Applicative f => Phases f a -> f a
runPhases (Lift x) = x
runPhases (fs :<*> xs) = fs <*> runPhases xs
later :: Applicative f => Phases f a -> Phases f a
later = (:<*>) (pure id)``````
The problem is that it’s quadratic: the `traverse` in:
``go (Node x xs) = liftA2 Node (Lift (f x)) (later (traverse go xs))``
Hides some expensive calls to `<*>`.
The problem with the `Phases` traversal is actually analogous to another function for enumeration: `levels` from Gibbons (2015).
``````levels :: Tree a -> [[a]]
levels (Node x xs) = [x] : foldr lzw [] (map levels xs)
where
lzw [] ys = ys
lzw xs [] = xs
lzw (x:xs) (y:ys) = (x ++ y) : lzw xs ys``````
`lzw` takes the place of `<*>` here, but the overall issue is the same: we’re zipping at every point, making the whole thing quadratic.
However, from the above function we can derive a linear time enumeration. It looks like this:
``````levels :: Tree a -> [[a]]
levels ts = f ts []
where
f (Node x xs) (q:qs) = (x:q) : foldr f qs xs
f (Node x xs) [] = [x] : foldr f [] xs``````
Our objective is clear, then: try to derive the linear-time implementation of `bft` from the quadratic, in a way analogous to the above two functions. This is actually relatively straightforward once the target is clear: the rest of this post is devoted to the derivation.
# Derivation
First, we start off with the original `bft`.
``````bft :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = runPhases . go
where
go (Node x xs) = liftA2 Node (Lift (f x)) (later (traverse go xs))``````
Inline `traverse`.
``````bft :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = runPhases . go
where
go (Node x xs) = liftA2 Node (Lift (f x)) (later (go' xs))
go' = foldr (liftA2 (:) . go) (pure [])``````
Factor out `go''`.
``````bft :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = runPhases . go
where
go (Node x xs) = liftA2 Node (Lift (f x)) (later (go' xs))
go' = foldr go'' (pure [])
go'' (Node x xs) ys = liftA2 (:) (liftA2 Node (Lift (f x)) (later (go' xs))) ys``````
Inline `go'` (and rename `go''` to `go'`)
``````bft :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = runPhases . go
where
go (Node x xs) = liftA2 Node (Lift (f x)) (later (foldr go' (pure []) xs))
go' (Node x xs) ys = liftA2 (:) (liftA2 Node (Lift (f x)) (later (foldr go' (pure []) xs))) ys``````
Definition of `liftA2`
``````bft :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = runPhases . go
where
go (Node x xs) = liftA2 Node (Lift (f x)) (later (foldr go' (pure []) xs))
go' (Node x xs) ys = liftA2 (:) (fmap Node (f x) :<*> (foldr go' (pure []) xs)) ys``````
Definition of `liftA2` (pattern-matching on `ys`)
``````bft :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = runPhases . go
where
go (Node x xs) = liftA2 Node (Lift (f x)) (later (foldr go' (pure []) xs))
go' (Node x xs) (Lift ys) = fmap (((:).) . Node) (f x) :<*> (foldr go' (pure []) xs) <*> Lift ys
go' (Node x xs) (ys :<*> zs) = fmap (((:).) . Node) (f x) :<*> (foldr go' (pure []) xs) <*> ys :<*> zs``````
Definition of `<*>`.
``````bft :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = runPhases . go
where
go (Node x xs) = liftA2 Node (Lift (f x)) (later (foldr go' (pure []) xs))
go' (Node x xs) (Lift ys) = liftA2 flip (fmap (((:).) . Node) (f x)) ys :<*> foldr go' (pure []) xs
go' (Node x xs) (ys :<*> zs) = liftA2 c (fmap (((:).) . Node) (f x)) ys :<*> liftA2 (,) (foldr go' (pure []) xs) zs
where
c f g ~(x,y) = f x (g y)``````
Fuse `liftA2` with `fmap`
``````bft :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = runPhases . go
where
go (Node x xs) = liftA2 Node (Lift (f x)) (later (foldr go' (pure []) xs))
go' (Node x xs) (Lift ys) = liftA2 (flip . (((:).) . Node)) (f x) ys :<*> foldr go' (pure []) xs
go' (Node x xs) (ys :<*> zs) = liftA2 (c . (((:).) . Node)) (f x) ys :<*> liftA2 (,) (foldr go' (pure []) xs) zs
where
c f g ~(x,y) = f x (g y)``````
Beta-reduction.
``````bft :: Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = go
where
go (Node x xs) = liftA2 Node (f x) (runPhases (foldr go' (pure []) xs))
go' (Node x xs) (Lift ys) = liftA2 (\y zs ys -> Node y ys : zs) (f x) ys :<*> foldr go' (pure []) xs
go' (Node x xs) (ys :<*> zs) = liftA2 c (f x) ys :<*> liftA2 (,) (foldr go' (pure []) xs) zs
where
c y g ~(ys,z) = Node y ys : g z``````
At this point, we actually hit a wall: the expression
``liftA2 (,) (foldr go' (pure []) xs) zs``
Is what makes the whole thing quadratic. We need to find a way to thread that `liftA2` along with the fold to get it to linear. This is the only real trick in the derivation: I’ll use polymorphic recursion to avoid the extra zip.
``````bft :: forall f a b. Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = go
where
go (Node x xs) = liftA2 (\y (ys,_) -> Node y ys) (f x) (runPhases (foldr go' (pure ([],())) xs))
go' :: forall c. Tree a -> Phases f ([Tree b], c) -> Phases f ([Tree b], c)
go' (Node x xs) ys@(Lift _) = fmap (\y -> first (pure . Node y)) (f x) :<*> foldr go' ys xs
go' (Node x xs) (ys :<*> zs) = liftA2 c (f x) ys :<*> foldr go' (fmap ((,) []) zs) xs
where
c y g ~(ys,z) = first (Node y ys:) (g z)``````
And that’s it!
# Avoiding Maps
We can finally write a slightly different version that avoids some unnecessary `fmap`s by basing `Phases` on `liftA2` rather than `<*>`.
``````data Levels f a where
Now :: a -> Levels f a
Later :: (a -> b -> c) -> f a -> Levels f b -> Levels f c
instance Functor f => Functor (Levels f) where
fmap f (Now x) = Now (f x)
fmap f (Later c xs ys) = Later ((f.) . c) xs ys
runLevels :: Applicative f => Levels f a -> f a
runLevels (Now x) = pure x
runLevels (Later f xs ys) = liftA2 f xs (runLevels ys)
bft :: forall f a b. Applicative f => (a -> f b) -> Tree a -> f (Tree b)
bft f = go
where
go (Node x xs) = liftA2 (\y (ys,_) -> Node y ys) (f x) (runLevels (foldr go' (Now ([],())) xs))
go' :: forall c. Tree a -> Levels f ([Tree b], c) -> Levels f ([Tree b], c)
go' (Node x xs) ys@(Now _) = Later (\y -> first (pure . Node y)) (f x) (foldr go' ys xs)
go' (Node x xs) (Later k ys zs) = Later id (liftA2 c (f x) ys) (foldr go' (fmap ((,) []) zs) xs)
where
c y g ~(ys,z) = first (Node y ys:) (k g z)``````
# References
Easterly, Noah. 2019. “Functions and newtype wrappers for traversing Trees: Rampion/tree-traversals.” https://github.com/rampion/tree-traversals. | 3,159 | 9,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.899912 |
https://betterlesson.com/lesson/resource/1988243/discovering-what-it-means-to-be-similar-volume-of-sphere-bellringer-mov | 1,511,139,263,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805881.65/warc/CC-MAIN-20171119234824-20171120014824-00746.warc.gz | 585,698,939 | 20,594 | ## Discovering What it Means to be Similar Volume of Sphere Bellringer.mov - Section 1: Opening Bellringer Activity
Discovering What it Means to be Similar Volume of Sphere Bellringer.mov
# Discovering What it Means to be Similar Triangles Continued
Unit 7: Linear Equations in two Variables
Lesson 6 of 23
## Big Idea: Imagine teaching similarity, angle-angle similarity for triangles, and dilations all at once! It's a three for one deal!
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Standards:
Subject(s):
Math, dilation (stretches and shrinks), Geometry, Transformations (Geom), Similarity and Congruence, similar triangles, 8th Math, hands on transformations, master teacher, tangram, projector
61 minutes
### Christa Lemily
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###### Developing Right and Straight Angle Intuition
8th Grade Math » Lines, Angles, and Algebraic Reasoning
Big Idea: Introduce and explore the connections between supplementary and complementary angles using algebra.
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New York, NY
Environment: Urban
###### PTA (Parallel Lines, Transversals and Angles)
Geometry » Line-sanity!
Big Idea: Students will draw and measure to discover relationships of angles formed by parallel lines cut by a transversal.
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Saratoga Springs, NY
Environment: Suburban
###### Angles and Parallel Lines (Day 1 of 2)
ALGEBRA /My Betterlesson Curriculum » Congruence and Similarity
Big Idea: Students recognize and use properties of lines and angle to determine angle measures.
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Windermere, FL
Environment: Urban | 365 | 1,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.806024 |
https://findthefactors.com/tag/1481/ | 1,680,280,771,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949644.27/warc/CC-MAIN-20230331144941-20230331174941-00382.warc.gz | 302,933,873 | 14,862 | # 1481 is the First Prime Number in the Fifth Prime Decade
### What Kind of Prime Number is 1481?
1481 is the 234th prime number. It is part of a twin prime pair, a prime triplet, a prime quadruplet, and even a prime quintuplet. All of those designations are nice, but to me, the most beautiful is the prime decade (the prime quadruplets not counting (5, 7, 11, 13)).
1481 is the first prime number in the fifth prime decade. The last digits of the numbers in every prime decade make a lovely pattern: 1, 3, 7, and 9. Once you know the first prime number in the decade, you also know the other three! That’s beautiful!
I found the numbers in the first five prime decades easy to memorize. We start with the teen decade followed by the first and last decades in the 100’s. Then199 is almost 200 and that helps me remember 821, and 821 helps me remember 1481.
### Factors of 1481:
• 1481 is a prime number.
• Prime factorization: 1481 is prime.
• 1481 has no exponents greater than 1 in its prime factorization, so √1481 cannot be simplified.
• The exponent in the prime factorization is 1. Adding one to that exponent we get (1 + 1) = 2. Therefore 1481 has exactly 2 factors.
• The factors of 1481 are outlined with their factor pair partners in the graphic below.
How do we know that 1481 is a prime number? If 1481 were not a prime number, then it would be divisible by at least one prime number less than or equal to √1481. Since 1481 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, or 37, we know that 1481 is a prime number.
### More about the Number 1481:
1481 is the sum of two squares:
35² + 16² = 1481.
1481 is the hypotenuse of a Pythagorean triple:
969-1120-1481, calculated from 35² – 16², 2(35)(16), 35² + 16².
Here’s another way we know that 1481 is a prime number: Since its last two digits divided by 4 leave a remainder of 1, and 35² + 16² = 1481 with 35 and 16 having no common prime factors, 1481 will be prime unless it is divisible by a prime number Pythagorean triple hypotenuse less than or equal to √1481. Since 1481 is not divisible by 5, 13, 17, 29, or 37, we know that 1481 is a prime number. | 635 | 2,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.920958 |
https://www.proprofs.com/discuss/c/kindergarten?utm_source=quiz_topic_page&utm_medium=discuss_button_click&utm_campaign=quiz_topic_page_discuss_button_click | 1,582,675,001,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146176.73/warc/CC-MAIN-20200225233214-20200226023214-00479.warc.gz | 861,679,679 | 43,548 | # Kindergarten Questions and Answers (Q&A)
L. Gibson, Science Professor
Answered: Aug 16, 2019
There are many correct answers to this question, for there are many living things that are around us every day. To answer this question accurately, one would need first to know the definition of...Read More
2 Answers
211 views
L. Gibson, Science Professor
Answered: Aug 16, 2019
The correct answer to this question is B, Every Good Boy Deserves Fudge. These lines are in the Trebel Clef of the staff. Staff is the set of lines in sheet music. The notes on the treble cleft...Read More
2 Answers
204 views
L. Gibson, Science Professor
Answered: Aug 16, 2019
The correct answer to this question is A, Orange. In the color spectrum, orange is the color between red and yellow. Looking at the umbrella, one can see the hints of red and yellow. Many...Read More
2 Answers
209 views
L. Gibson, Science Professor
Answered: Aug 16, 2019
The correct answer to this question is A, Green. Frogs can come in several different colors, and they come in just about every color of the rainbow. The image shown in this question reveals a...Read More
2 Answers
204 views
L. Gibson, Science Professor
Answered: Aug 16, 2019
The correct answer to this question is B, Yellow. The image shown is of a YIELD sign. The purpose of the sign is to alert the merging driver to stop to let another driver proceed. In other words,...Read More
2 Answers
216 views
L. Gibson, Science Professor
Answered: Aug 16, 2019
The correct answer to this question is C, 30. This type of question would be found on a test for early learners, such as kindergartners, who are just starting to learn math. It can test their...Read More
2 Answers
204 views
L. Gibson, Science Professor
Answered: Aug 16, 2019
The correct answer to this question is C, 25. To answer this question, one would need to know the primary sequence of numbers. One would have to be at least able to count to 26. Knowing the...Read More
2 Answers
204 views
L. Gibson, Science Professor
Answered: Aug 16, 2019
The correct answer to this question is A, 5. This type of question would be found on a test for early learners, such as kindergartners, who are just starting to learn numbers and the order of...Read More
2 Answers
200 views
M. Parker, Internet Researcher
Answered: Aug 14, 2019
The correct answer to this question is A, Cat. Answer B, CT, is mainly associated with CT scans or computed tomography scans. Answer C, CA, is primarily related to being an abbreviation for the...Read More
2 Answers
202 views
M. Parker, Internet Researcher
Answered: Aug 14, 2019
The correct answer to this question is B, Pre-K. This stands for Pre-Kindergarten. It is a preschool program that is in Turkey, Canada, and the United States. Pre-K is for students under the age...Read More
2 Answers
201 views
M. Parker, Internet Researcher
Answered: Aug 14, 2019
The correct answer to this question is B, Yellow. It would be easy for one to get this question wrong, for the bus shown and the colors of it differ than the context of the question. Like all...Read More
2 Answers
205 views
M. Parker, Internet Researcher
Answered: Aug 14, 2019
The correct answer to this question is B, Storytime. While most preschoolers are unable to read on their own, reading to them can help them learn about a lot of things about reading. This...Read More
2 Answers
204 views
L. Gibson, Science Professor
Answered: Aug 16, 2019
The correct answer to this question is B, Phonic songs. Kindergarten is early education for toddlers to prepare them for elementary school. Phonics is a way to teach kids how to read. It does so...Read More
2 Answers
201 views
L. Gibson, Science Professor
Answered: Aug 16, 2019
The correct answer to this question is C, Cognitive skills, and motor skills. Cognitive skills are when a person can perform activities associated with learning and solving problems. It deals...Read More
2 Answers
202 views
L. Gibson, Science Professor
Answered: Aug 16, 2019
The correct answer to this question is A, FACE. This format is used to help new learners understand music notes. FACE is found on the treble cleft, which is the top staff in sheet music. Starting...Read More
2 Answers
207 views
### Related Topics Of Kindergarten
Loading, please wait... | 1,077 | 4,293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-10 | latest | en | 0.951998 |
https://metanumbers.com/12990 | 1,624,114,198,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487648373.45/warc/CC-MAIN-20210619142022-20210619172022-00488.warc.gz | 359,237,386 | 10,840 | ## 12990
12,990 (twelve thousand nine hundred ninety) is an even five-digits composite number following 12989 and preceding 12991. In scientific notation, it is written as 1.299 × 104. The sum of its digits is 21. It has a total of 4 prime factors and 16 positive divisors. There are 3,456 positive integers (up to 12990) that are relatively prime to 12990.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 21
• Digital Root 3
## Name
Short name 12 thousand 990 twelve thousand nine hundred ninety
## Notation
Scientific notation 1.299 × 104 12.99 × 103
## Prime Factorization of 12990
Prime Factorization 2 × 3 × 5 × 433
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 12990 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 12,990 is 2 × 3 × 5 × 433. Since it has a total of 4 prime factors, 12,990 is a composite number.
## Divisors of 12990
1, 2, 3, 5, 6, 10, 15, 30, 433, 866, 1299, 2165, 2598, 4330, 6495, 12990
16 divisors
Even divisors 8 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 31248 Sum of all the positive divisors of n s(n) 18258 Sum of the proper positive divisors of n A(n) 1953 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 113.974 Returns the nth root of the product of n divisors H(n) 6.65131 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 12,990 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 12,990) is 31,248, the average is 1,953.
## Other Arithmetic Functions (n = 12990)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 3456 Total number of positive integers not greater than n that are coprime to n λ(n) 432 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1548 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 3,456 positive integers (less than 12,990) that are coprime with 12,990. And there are approximately 1,548 prime numbers less than or equal to 12,990.
## Divisibility of 12990
m n mod m 2 3 4 5 6 7 8 9 0 0 2 0 0 5 6 3
The number 12,990 is divisible by 2, 3, 5 and 6.
• Arithmetic
• Abundant
• Polite
• Square Free
## Base conversion (12990)
Base System Value
2 Binary 11001010111110
3 Ternary 122211010
4 Quaternary 3022332
5 Quinary 403430
6 Senary 140050
8 Octal 31276
10 Decimal 12990
12 Duodecimal 7626
20 Vigesimal 1c9a
36 Base36 a0u
## Basic calculations (n = 12990)
### Multiplication
n×i
n×2 25980 38970 51960 64950
### Division
ni
n⁄2 6495 4330 3247.5 2598
### Exponentiation
ni
n2 168740100 2191933899000 28473221348010000 369867145310649900000
### Nth Root
i√n
2√n 113.974 23.5073 10.6758 6.64847
## 12990 as geometric shapes
### Circle
Diameter 25980 81618.6 5.30113e+08
### Sphere
Volume 9.18155e+12 2.12045e+09 81618.6
### Square
Length = n
Perimeter 51960 1.6874e+08 18370.6
### Cube
Length = n
Surface area 1.01244e+09 2.19193e+12 22499.3
### Equilateral Triangle
Length = n
Perimeter 38970 7.30666e+07 11249.7
### Triangular Pyramid
Length = n
Surface area 2.92266e+08 2.58322e+11 10606.3
## Cryptographic Hash Functions
md5 f9c1b5bf25d49031367434c598ee9250 b63d3032e85da47e3a7cc31a5c57d2ffe4566409 d0bcba597a2caaf263d1463ed6866560389d45408881e444ed81d23ef0f14d83 e73d7b3fc8f2f41e030c4672f297e35ceeb03a8ae5ca712402bb32dea9cf998ca1611a12d17de3475894f6da8ec5c94d1e528b412f26bd97b4554e8c47b16c3f 458c7dff50eddbbac76bd0aeb0b8281a88ec2785 | 1,477 | 4,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.80239 |
http://www.rfwireless-world.com/Tutorials/radar-measurements.html | 1,558,583,300,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257002.33/warc/CC-MAIN-20190523023545-20190523045545-00052.warc.gz | 336,269,728 | 6,147 | # RF Wireless World
## Home of RF and Wireless Vendors and Resources
One Stop For Your RF and Wireless Need
This tutorial section on radar basics covers following sub topics:
This page describes radar measurements which is very useful during radar maintanance and calibration. We will cover both the transmitter part as well as receiver part measurements. The common tools used for radar maintenance and measurements are multimeter, spectrum analyzer, oscilloscope, power meter, frequency counter, signal generator, crystal detector, attenuator etc.
### Transmitter Measurements
The radar transmitter measurements include transmitted Power and Stability, PW, PRF, Duty Cycle/ Factor, Klystron Pulse, Klystron Current, Klystron RF Input Level, Transmitted frequency, Occupied BW and VSWR. Let us examine these measurement parameters one by one.
### Transmitted Power
The transmitted power is measured with the help of calibrated power meter. The value is calculated taking into consideration loss of cable as well as duty cycle factor.
Peak Power (dBm) = Average power (dBm) - Duty Factor(dB) + Loss(dB)
Where, Duty Factor = 10 log(Pulse Width(sec) * PRF(Hz))
Loss will be path loss between output flange of klystron and rf transmitter monitor connector.
### Power Stability
With Pulse Width of 1 µsec and PRF of 250Hz. Acceptable level for power stability is about +/-0.5 dB.
For example, if power output varies from -14.95dBm to -15.05dBm in a half an hour period. Difference in variation of power here is 14.95 - (-15.05) = 0.1 dB. This is less than acceptable level.
### PW,PRF,Duty cycle/duty factor
This measurement is carried out using oscilloscope. If oscilloscope is not capable of measuring high frequency than arrangement as shown in the figure is done. step attenuator and crystal detector is used for this purpose.
To measure PW we need 3dB attenuation, as it is the width at half power points. similarly PRF and duty factor is measured using oscilloscope.
### Klystron Pulse, Klystron current
Klystron pulse waveform is measured at the output of transmitter front panel using oscilloscope. Klystron current is calculated by multiplying oscilloscope value with 10. For example, oscilloscope reading for max. peak is 1.2V than current is 12Amp.
### Klystron RF input level
In this test, rf drive power level and rf frequency is measured to check whether they are within normal range or not. Frequency should be about 5625MHz. The same is measured using either spectrum analyzer or frequency counter.
Occupied BW is the width of band over which emitted mean power is about 99% of the total mean power. Difference between lower and upper frequency points is referred as occupied BW.
### VSWR measurement
It is the measure of how much power is transmitted and how much is reflected.
Return loss in general= Forward power - Reflected power,
Refer RL vs VSWR for more.
Radar receiver is very critical part in radar system as it has to detect and amplify the received weak signal from antenna. The radar receiver measurements include STALO level measurement, COHO level measurement, receiver Gain, Minimum Detectable Signal(MDS), Dynamic range, intensity check, velocity check etc.
### STALO Level Measurement
Radar usually will have different STALO frequencies and power levels. The main aim is to determine weak signal receiver power. The digital power meter is used for this purpose.
### COHO Level Measurement
COHO stands for COHerent Oscillator. It has frequency of about 30MHz.
Radar transmitter is kept off and signal generator frequency is set equal to radar receiver for this test.
Gain(dB) = output power (dB) - input power(dB).
Similarly dynamic range is calculated by finding MDS and saturated input points. Refer MDS vs SFDR for more.
### Tx IF Out and Exciter RF
Transmitter IF OUT is pulse modulated COHO output which is up converted to RF level. Exciter RF signal is passed to the klystron. The spectrum analyzer is used for this purpose.
### Intensity Check measurement
It is done to see how well the system performs the processing of the reflected signal after coming into LNA through Signal Processor.
### Velocity Accuracy
Velocity accuracy check measurement is done to see how well the system calculates the velocity of an echo. | 949 | 4,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.888404 |
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