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1Department of Industrial and Production Engineering, The Federal University of Technology, Akure, Nigeria.
2Department of Physics, The Federal University of Technology, Akure, Nigeria.
3Department of Metallurgical and Materials Engineering, The Federal University of Technology, Akure, Nigeria.
The steam turbine is a prime mover that converts kinetic energy in steam into rotational mechanical energy through the impact or reaction of the steam against the blades. The aim of this study is to design a steam turbine for a small scale steam power plant with target of producing electricity. The turbine is driven by the heat energy from palm kernel shells as a renewable energy source obtained at a lower or no cost. The study was concentrated on design of turbine elements and its validation using computer packages. Specifically, the microturbine design was limited to design, modeling, simulation and analysis of the rotor, blades and nozzle under the palm kernel shell as fuel for the micro power plant. In blade design, stress failures, efficiency and blade angle parameters were considered. In casing volume design, the overall heat transfer and mean temperature, and different concepts were applied. The thermal distribution on stator and rotor was considered in order to determine its level of tolerance. The design software packages used for design validation were Solidworks and Comsol Multiphysics for analysis. Simulation results showed that the designed steam turbine can adequately tolerate change in stress/load, torsion/compression, temperature and speeds.
Palm Kernel Shell, Steam Turbine, Power Plant, Design Software, Renewable Energy
Kareem, B. , Ewetumo, T. , Adeyeri, M. , Oyetunji, A. and Olatunji, O. (2018) Design of Steam Turbine for Electric Power Production Using Heat Energy from Palm Kernel Shell. Journal of Power and Energy Engineering, 6, 111-125. doi: 10.4236/jpee.2018.611009.
{d}_{o}
{\text{N}}_{\text{b}}=\frac{2\pi Dm}{s}
{d}^{3}=\frac{16\text{T}}{0.27\pi {d}_{o}}
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Working With MDX · USACO Guide
HomeGeneralWorking With MDX
FrontmatterExample FrontmatterModule OrderingLinking to ModulesTable of ContentsMDX and Custom ComponentsSpoilersInfo BlockWarning BlockOptional ContentProblem ListsFocus ProblemResource ListsTooltips<TextTooltip><Asterisk>Incomplete SectionCode Blocks and Code SnippetsKattioC++ Long TemplateC++ Short TemplateLanguage-Specific ContentA heading that only appears in C++Quizzes
Author: Nathan Wang
Explanation of the frontmatter that precedes every module and solution, a list of custom components that may be used within modules or solutions.
We're using MDX, a superset of Markdown, using the XDM compiler. HTML and React components are supported, so it is possible to add interactivity / custom components to each module.
Frontmatter is the stuff in the beginning of each module that's surrounded by three dashes. Frontmatter is written in YAML. It stores the "metadata" for each module.
ID: Required. The ID of the module. Ex: getting-started, or containers. This ID is used to identify the module, so make sure it is unique and all lowercase with dashes only. The URL will be generated based off this.
Title: Required. The title of the module. Ex: Getting Started
Author: Required. The author of the module. Ex: Unknown
Contributors: Optional. The people who contributed code and/or short explanations to the module.
Description: Required. A short description of the module, similar to what codecademy has in their syllabus. Markdown/LaTeX does not work in the description field.
Prerequisites: Optional. Any prerequisites for this module. If you want to reference a module as a prerequisite, list it as a module ID. A link will be auto-generated.
Frequency: Optional. Takes a number 0-4 inclusive, where 0 = never shown up before and 4 = shows up ~once a contest. Leave this field out if you don't want to show the frequency.
Redirects: Optional. Takes a list of URLs that will redirect to the current module. Add a redirect whenever you change the module ID or move the module to a different division.
Example Frontmatter
id: getting-started
description: Welcome to the guide! We'll introduce what programming competitions are and how this guide is organized.
- Dummy prerequisite
- running-cpp
- /silver/bipartite
Located at content/ordering.ts, this file stores the ordering of the modules. The format should be self-explanatory (it matches based on the ID).
Linking to Modules
Linking to another module within the guide looks like this:
[insert text here](/general/practicing).
Don't use relative links like practicing, as that will break our link checker.
A table of contents will be auto-generated based off of the headings in the Markdown. Keep this in mind when formatting your module.
MDX and Custom Components
Optional: XDM and MDX
We use the XDM compiler, which has a few differences from MDX v1:
Markdown interleaved in JSX is fully supported; ie. this works as expected: <Info>some **markdown**</Info>
As an extension of (1), indentation is fully supported. You can indent the markdown nested in JSX tags. This also means that indenting text with four spaces doesn't make it a code block; explicitly wrap the code block with three backticks instead.
< and > need to be escaped with backslashes; ie. \<
Note that JSX comments ({/* ... */}) don't work well with Prettier, so use HTML comments instead. Internally we map HTML comments to JSX comments before passing the markdown to XDM. Don't worry if you don't understand all of this yet.
Some components are globally available in every module (without having to be imported):
<FocusProblem>
<TextTooltip>
<LanguageSection>
<CPPSection>
<JavaSection>
<PySection>
<IncompleteSection>
These are all documented below.
Spoilers are collapsible elements that only show themselves when the user clicks on it. It's useful when writing solutions to problems.
<Spoiler title="Show Solution">
- Insert solution here
<Info title="Insert Title Here">
**Markdown is Supported!!**
Markdown is Supported!!
<Warning title="Insert Title Here">
Fun fact: the title attribute is optional.
Warning: Insert Title Here
<Optional title="Insert Title Here">
Optional: Insert Title Here
Each module has two corresponding files, a .mdx file and a .problems.json file. The .problems.json file stores the focus problems and problem lists used in that module; it is also indexed by Algolia for problem search.
The .problems.json file holds an object, where keys are problem list names (or focus problem names) and values are arrays of ProblemMetadata objects. For focus problems, the array should have length exactly one. Additionally, the .problems.json file should have a MODULE_ID key with value equal to a string that represents the module ID.
For more information on problem definitions, refer to src/models/problem.ts.
<Problems problems="problems" />
[module].problems.json should have a key of problems that maps to an array of ProblemMetadata.
There is a distinction between ProblemInfo and ProblemMetadata. ProblemMetadata is what is stored in [module].problems.json. Gatsby takes ProblemMetadata and converts it into ProblemInfo at build time; React components use ProblemInfo when interacting with problem information. The documentation below is for ProblemMetadata, which is what content authors will be writing.
ProblemMetadata fields:
uniqueId -- The uniqueId of the problem. Problem progress is linked to this, so don't change this (otherwise problem progress will be lost). By convention, it's [source]-[SlugifiedProblemNameCamelCased].
If the problem name is only one word, the word is lower cased.
If the problem is USACO or CSES, the unique ID is instead usaco-[USACO URL Number] or cses-[CSES number].
If the problem is Codeforces, the unique ID is cf-[contestNumber][problemLetter]. If it's CF Gym, it's cfgym-[gymNumber][problemLetter].
If the problem is an OI with a year, the unique ID is [oiName]-[twodigityear]-[slugifiedName].
Here are some example unique ID's:
cses-2177
poi-08-blockade
apio-18-duathlon
dmoj-investment
infoarena-xortransform
usaco-949
kattis-chineseremainder
cfgym-102538F
spoj-LexicographicalStringSearch
ys-AssociativeArray
Problems with the same unique ID are expected to have identical names, sources, and URL's.
name -- The name of the problem. Should not include source.
Greedy Pie Eaters
2014 - The Stables of Genghis Khan
source -- The source of the problem. Restricted to: todo, refer to src/models/problem.ts contests and probSources
difficulty -- The difficulty of the problem relative to the module it is in. Valid options are Very Easy, Easy, Normal, Hard, Very Hard, Insane
isStarred -- Whether this problem should be starred or not.
tags -- List of tags for this problem.
solutionMetadata -- Information about the solution.
export type ProblemMetadata = Omit<ProblemInfo, 'solution'> & {
solutionMetadata:
// auto generate problem solution label based off of the given site
// For sites like CodeForces: "Check contest materials, located to the right of the problem statement."
kind: 'autogen-label-from-site';
// The site to generate it from. Sometimes this may differ from the source; for example, Codeforces could be the site while Baltic OI could be the source if Codeforces was hosting a Baltic OI problem.
// internal solution
kind: 'internal';
// URL solution
// Use this for links to PDF solutions, etc
kind: 'link';
// Competitive Programming Handbook
// Ex: 5.3 or something
kind: 'CPH';
// USACO solution, generates it based off of the USACO problem ID
// ex. 1113 is mapped to sol_prob1_gold_feb21.html
kind: 'USACO';
usacoId: string;
// IOI solution, generates it based off of the year
// ex. Maps year = 2001 to https://ioinformatics.org/page/ioi-2001/27
kind: 'IOI';
// no solution exists
kind: 'none';
// for focus problems, when the solution is presented in the module of the problem
kind: 'in-module';
kind: 'sketch';
sketch: string;
Editorials are also written in MDX. The frontmatter has four fields:
id: cses-1621
title: Distinct Numbers
The ID of the solution frontmatter must be the same as the unique ID of the problem. Make sure to also update the kind of the solutionMetadata to 'internal' for any associated problems. We assume that if there is an internal solution, we should use it; therefore, the build will throw an error if there is an internal solution but the solutionMetadata's kind isn't set to
'internal'. The Adding Solutions module describes how to add a new solution.
Displays a singular problem as a "focus problem."
<FocusProblem problem="genPermutations" />
[module].problems.json should have a key of genPermutations that maps to an array of length 1.
source="Errichto"
title="Video - How to test your solution"
url="https://www.youtube.com/watch?v=JXTVOyQpSGM"
using a script to stress test
Video - How to test your solution
Special functionality based on source:
If the source is a book, it'll automatically set the URL to point to the book.
Supported books:
GCP (Guide to Competitive Programming)
CPH (Competitive Programming Handbook)
PAPS (Principles of Algorithmic Problem Solving)
CP2 (Competitive Programming 2)
IUSACO (Darren's book; will auto-set URL based on user language; uses C++ for Python users)
Some sources will automatically have tooltips generated for them (listed here).
There are two main types of tooltips: text tooltips, which display a dotted underline under the text, and asterisk tooltips, which render an asterisk that can be hovered over.
<TextTooltip content="Popup text goes here">short text goes here</TextTooltip>
short text goes here
<Asterisk>Popup text goes here</Asterisk>
- this list is optional and can be used to specify what is missing
- missing 32-bit integer explanation
</IncompleteSection>
this list is optional and can be used to specify what is missing
missing 32-bit integer explanation
Code Blocks and Code Snippets
Code blocks are separated by three backticks, just like in normal markdown. Additionally, we have support for collapsible code snippets. An example for how to use them can be found below:
//BeginCodeSnip{Optional Code Snippet Title}
Code snippet goes here
You can indent the entire BeginCodeSnip block (including the BeginCodeSnip line) and it will function as expected
//EndCodeSnip
Code Snippet: Optional Code Snippet Title (Click to expand)
//BeginCodeSnip{}
Other code goes here
Code Snippet (Click to expand)
My non-snippet code goes here
Some common snippets have shorthand notations, as defined in src/mdx-plugins/rehype-snippets.js. They can be accessed using CodeSnip{Snip ID}.
CodeSnip{Kattio} gets replaced with an indented version (based off of indentation of CodeSnip):
Code Snippet: Kattio (Click to expand)
public static void main ...
C++ Long Template
CodeSnip{Benq Template} gets replaced with Benq's Long Template.
Code Snippet: Benq Template (Click to expand)
C++ Short Template
CodeSnip{CPP Short Template} is replaced with
Code Snippet: C++ Short Template (Click to expand)
Language-Specific Content
#### A heading that only appears in C++
</CPPSection>
#### A heading that only appears in Java
</JavaSection>
<PySection />
</LanguageSection>
In the example above, nothing will be rendered for Python.
A heading that only appears in C++
A heading that only appears in Java
<Quiz.Question>
<Quiz.Answer>
<Quiz.Explanation>
Almost. Prefer $\mathcal{O}$ over $O$.
</Quiz.Explanation>
</Quiz.Answer>
<Quiz.Answer correct>
$\mathcal{O}(\log n)$
That's not right. Latex is important...
</Quiz.Question>
// constant time code here
$O(100m)$
That's not correct. Constant factors are ignored.
$O(m)$
O(\log n)
\mathcal{O}(\log n)
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Solve this : Draw the sign scheme then find the solution a) x-22 x-3x-1 - Maths - Linear Inequalities - 12660851 | Meritnation.com
Draw the sign scheme then find the solution
a\right) \frac{{\left(x-2\right)}^{2} \left(x-3\right)}{x-1} > 0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}b\right) \frac{{\left(x-2\right)}^{2} \left(x-3\right)}{x-1} \ge 0
Romil Singh answered this
Which is not clear question or photo
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drule [Isabelle/HOL Support Wiki]
Trace: • drule
reference:drule
drule is a proof method. It applies a rule if possible.
\quad\bigwedge x_1 \dots x_k : [|\ A_1; \dots ; A_m\ |] \Longrightarrow C
and we want to use drule with rule
\quad[|\ P_1; \dots ; P_n\ |] \Longrightarrow Q
Then, drule does the following:
P_1
A_j
j
j
U
\quad\bigwedge x_1 \dots x_k : [|\ U(A_1); \dots ; U(A_{j-1}); U(A_{j+1}); \dots ; U(A_m)\ ; U(Q) |] \Longrightarrow U(C)
\quad\bigwedge x_1 \dots x_k : [|\ U(A_1); \dots ; U(A_{j-1}); U(A_{j+1}); \dots ; U(A_m)\ |] \Longrightarrow U(P_k)
k = 2, \dots, n
Note that drule is almost the same as frule. The only difference is that frule keeps the assumption “used” by the used rule
A_j
A_j
is needed to prove the new subgoals; drule is unsafe because of this.
\quad[|\ A \longrightarrow B; A\ |] \Longrightarrow B
\quad A \Longrightarrow A
\quad[|\ A; B\ |] \Longrightarrow B
which can both be solved by Assumption. Note that apply (drule(2) mp) is a shortcut for this and immediately solves the goal.
drule_tac
With drule_tac, you can force schematic variables in the used rule to take specific values. The extended syntax is:
apply (drule_tac ident1="expr1" and ident2="expr1" and ... in rule)
drule(k)
Oftentimes, a rule application results in several subgoals that can directly be solved by Assumption; see above for an example. Instead of applying assumption by hand, you can apply drule(k) which forces Isabelle to apply assumption
k
reference/drule.txt · Last modified: 2011/07/06 12:08 by 131.246.161.187
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Plane_stress Knowpia
In continuum mechanics, a material is said to be under plane stress if the stress vector is zero across a particular plane. When that situation occurs over an entire element of a structure, as is often the case for thin plates, the stress analysis is considerably simplified, as the stress state can be represented by a tensor of dimension 2 (representable as a 2×2 matrix rather than 3×3). [1] A related notion, plane strain, is often applicable to very thick members.
Figure 7.1 Plane stress state in a continuum.
Plane stress typically occurs in thin flat plates that are acted upon only by load forces that are parallel to them. In certain situations, a gently curved thin plate may also be assumed to have plane stress for the purpose of stress analysis. This is the case, for example, of a thin-walled cylinder filled with a fluid under pressure. In such cases, stress components perpendicular to the plate are negligible compared to those parallel to it.[1]
In other situations, however, the bending stress of a thin plate cannot be neglected. One can still simplify the analysis by using a two-dimensional domain, but the plane stress tensor at each point must be complemented with bending terms.
{\displaystyle \sigma ={\begin{bmatrix}\sigma _{11}&0&0\\0&\sigma _{22}&0\\0&0&0\end{bmatrix}}\equiv {\begin{bmatrix}\sigma _{x}&0&0\\0&\sigma _{y}&0\\0&0&0\end{bmatrix}}}
For example, consider a rectangular block of material measuring 10, 40 and 5 cm along the
{\displaystyle x}
{\displaystyle y}
{\displaystyle z}
, that is being stretched in the
{\displaystyle x}
direction and compressed in the
{\displaystyle y}
direction, by pairs of opposite forces with magnitudes 10 N and 20 N, respectively, uniformly distributed over the corresponding faces. The stress tensor inside the block will be
{\displaystyle \sigma ={\begin{bmatrix}500\mathrm {Pa} &0&0\\0&-4000\mathrm {Pa} &0\\0&0&0\end{bmatrix}}}
{\displaystyle \sigma ={\begin{bmatrix}\sigma _{11}&\sigma _{12}&0\\\sigma _{21}&\sigma _{22}&0\\0&0&0\end{bmatrix}}\equiv {\begin{bmatrix}\sigma _{x}&\tau _{xy}&0\\\tau _{yx}&\sigma _{y}&0\\0&0&0\end{bmatrix}}}
{\displaystyle \sigma _{ij}={\begin{bmatrix}\sigma _{11}&\sigma _{12}\\\sigma _{21}&\sigma _{22}\end{bmatrix}}\equiv {\begin{bmatrix}\sigma _{x}&\tau _{xy}\\\tau _{yx}&\sigma _{y}\end{bmatrix}}}
Constitutive equationsEdit
See Hooke's law#Plane_stress
Plane stress in curved surfacesEdit
In certain cases, the plane stress model can be used in the analysis of gently curved surfaces. For example, consider a thin-walled cylinder subjected to an axial compressive load uniformly distributed along its rim, and filled with a pressurized fluid. The internal pressure will generate a reactive hoop stress on the wall, a normal tensile stress directed perpendicular to the cylinder axis and tangential to its surface. The cylinder can be conceptually unrolled and analyzed as a flat thin rectangular plate subjected to tensile load in one direction and compressive load in another other direction, both parallel to the plate.
Plane strain (strain matrix)Edit
Figure 7.2 Plane strain state in a continuum.
If one dimension is very large compared to the others, the principal strain in the direction of the longest dimension is constrained and can be assumed as constant, that means there will be effectively zero strain along it, hence yielding a plane strain condition (Figure 7.2). In this case, though all principal stresses are non-zero, the principal stress in the direction of the longest dimension can be disregarded for calculations. Thus, allowing a two dimensional analysis of stresses, e.g. a dam analyzed at a cross section loaded by the reservoir.
{\displaystyle \varepsilon _{ij}={\begin{bmatrix}\varepsilon _{11}&\varepsilon _{12}&0\\\varepsilon _{21}&\varepsilon _{22}&0\\0&0&0\end{bmatrix}}\,\!}
and the corresponding stress tensor is:
{\displaystyle \sigma _{ij}={\begin{bmatrix}\sigma _{11}&\sigma _{12}&0\\\sigma _{21}&\sigma _{22}&0\\0&0&\sigma _{33}\end{bmatrix}}\,\!}
{\displaystyle \sigma _{33}\,\!}
term arises from the Poisson's effect. However, this term can be temporarily removed from the stress analysis to leave only the in-plane terms, effectively reducing the analysis to two dimensions.[1]
Stress transformation in plane stress and plane strainEdit
Consider a point
{\displaystyle P\,\!}
in a continuum under a state of plane stress, or plane strain, with stress components
{\displaystyle (\sigma _{x},\sigma _{y},\tau _{xy})\,\!}
and all other stress components equal to zero (Figure 8.1). From static equilibrium of an infinitesimal material element at
{\displaystyle P\,\!}
(Figure 8.2), the normal stress
{\displaystyle \sigma _{\mathrm {n} }\,\!}
{\displaystyle \tau _{\mathrm {n} }\,\!}
on any plane perpendicular to the
{\displaystyle x\,\!}
{\displaystyle y\,\!}
plane passing through
{\displaystyle P\,\!}
with a unit vector
{\displaystyle \mathbf {n} \,\!}
{\displaystyle \theta \,\!}
with the horizontal, i.e.
{\displaystyle \cos \theta \,\!}
is the direction cosine in the
{\displaystyle x\,\!}
direction, is given by:
{\displaystyle \sigma _{\mathrm {n} }={\frac {1}{2}}(\sigma _{x}+\sigma _{y})+{\frac {1}{2}}(\sigma _{x}-\sigma _{y})\cos 2\theta +\tau _{xy}\sin 2\theta \,\!}
{\displaystyle \tau _{\mathrm {n} }=-{\frac {1}{2}}(\sigma _{x}-\sigma _{y})\sin 2\theta +\tau _{xy}\cos 2\theta \,\!}
These equations indicate that in a plane stress or plane strain condition, one can determine the stress components at a point on all directions, i.e. as a function of
{\displaystyle \theta \,\!}
, if one knows the stress components
{\displaystyle (\sigma _{x},\sigma _{y},\tau _{xy})\,\!}
on any two perpendicular directions at that point. It is important to remember that we are considering a unit area of the infinitesimal element in the direction parallel to the
{\displaystyle y\,\!}
{\displaystyle z\,\!}
Figure 8.1 - Stress transformation at a point in a continuum under plane stress conditions.
Figure 8.2 - Stress components at a plane passing through a point in a continuum under plane stress conditions.
The principal directions (Figure 8.3), i.e., orientation of the planes where the shear stress components are zero, can be obtained by making the previous equation for the shear stress
{\displaystyle \tau _{\mathrm {n} }\,\!}
equal to zero. Thus we have:
{\displaystyle \tau _{\mathrm {n} }=-{\frac {1}{2}}(\sigma _{x}-\sigma _{y})\sin 2\theta +\tau _{xy}\cos 2\theta =0\,\!}
{\displaystyle \tan 2\theta _{\mathrm {p} }={\frac {2\tau _{xy}}{\sigma _{x}-\sigma _{y}}}\,\!}
This equation defines two values
{\displaystyle \theta _{\mathrm {p} }\,\!}
{\displaystyle 90^{\circ }\,\!}
apart (Figure 8.3). The same result can be obtained by finding the angle
{\displaystyle \theta \,\!}
which makes the normal stress
{\displaystyle \sigma _{\mathrm {n} }\,\!}
a maximum, i.e.
{\displaystyle {\frac {d\sigma _{\mathrm {n} }}{d\theta }}=0\,\!}
The principal stresses
{\displaystyle \sigma _{1}\,\!}
{\displaystyle \sigma _{2}\,\!}
, or minimum and maximum normal stresses
{\displaystyle \sigma _{\mathrm {max} }\,\!}
{\displaystyle \sigma _{\mathrm {min} }\,\!}
, respectively, can then be obtained by replacing both values of
{\displaystyle \theta _{\mathrm {p} }\,\!}
into the previous equation for
{\displaystyle \sigma _{\mathrm {n} }\,\!}
. This can be achieved by rearranging the equations for
{\displaystyle \sigma _{\mathrm {n} }\,\!}
{\displaystyle \tau _{\mathrm {n} }\,\!}
{\displaystyle \left[\sigma _{\mathrm {n} }-{\tfrac {1}{2}}(\sigma _{x}+\sigma _{y})\right]^{2}+\tau _{\mathrm {n} }^{2}=\left[{\tfrac {1}{2}}(\sigma _{x}-\sigma _{y})\right]^{2}+\tau _{xy}^{2}\,\!}
{\displaystyle (\sigma _{\mathrm {n} }-\sigma _{\mathrm {avg} })^{2}+\tau _{\mathrm {n} }^{2}=R^{2}\,\!}
{\displaystyle R={\sqrt {\left[{\tfrac {1}{2}}(\sigma _{x}-\sigma _{y})\right]^{2}+\tau _{xy}^{2}}}\quad {\text{and}}\quad \sigma _{\mathrm {avg} }={\tfrac {1}{2}}(\sigma _{x}+\sigma _{y})\,\!}
which is the equation of a circle of radius
{\displaystyle R\,\!}
{\displaystyle [\sigma _{\mathrm {avg} },0]\,\!}
, called Mohr's circle. But knowing that for the principal stresses the shear stress
{\displaystyle \tau _{\mathrm {n} }=0\,\!}
, then we obtain from this equation:
{\displaystyle \sigma _{1}=\sigma _{\mathrm {max} }={\tfrac {1}{2}}(\sigma _{x}+\sigma _{y})+{\sqrt {\left[{\tfrac {1}{2}}(\sigma _{x}-\sigma _{y})\right]^{2}+\tau _{xy}^{2}}}\,\!}
{\displaystyle \sigma _{2}=\sigma _{\mathrm {min} }={\tfrac {1}{2}}(\sigma _{x}+\sigma _{y})-{\sqrt {\left[{\tfrac {1}{2}}(\sigma _{x}-\sigma _{y})\right]^{2}+\tau _{xy}^{2}}}\,\!}
Figure 8.3 - Transformation of stresses in two dimensions, showing the planes of action of principal stresses, and maximum and minimum shear stresses.
{\displaystyle \tau _{xy}=0\,\!}
the infinitesimal element is oriented in the direction of the principal planes, thus the stresses acting on the rectangular element are principal stresses:
{\displaystyle \sigma _{x}=\sigma _{1}\,\!}
{\displaystyle \sigma _{y}=\sigma _{2}\,\!}
. Then the normal stress
{\displaystyle \sigma _{\mathrm {n} }\,\!}
and shear stress
{\displaystyle \tau _{\mathrm {n} }\,\!}
as a function of the principal stresses can be determined by making
{\displaystyle \tau _{xy}=0\,\!}
{\displaystyle \sigma _{\mathrm {n} }={\frac {1}{2}}(\sigma _{1}+\sigma _{2})+{\frac {1}{2}}(\sigma _{1}-\sigma _{2})\cos 2\theta \,\!}
{\displaystyle \tau _{\mathrm {n} }=-{\frac {1}{2}}(\sigma _{1}-\sigma _{2})\sin 2\theta \,\!}
Then the maximum shear stress
{\displaystyle \tau _{\mathrm {max} }\,\!}
{\displaystyle \sin 2\theta =1\,\!}
{\displaystyle \theta =45^{\circ }\,\!}
{\displaystyle \tau _{\mathrm {max} }={\frac {1}{2}}(\sigma _{1}-\sigma _{2})\,\!}
Then the minimum shear stress
{\displaystyle \tau _{\mathrm {min} }\,\!}
{\displaystyle \sin 2\theta =-1\,\!}
{\displaystyle \theta =135^{\circ }\,\!}
{\displaystyle \tau _{\mathrm {min} }=-{\frac {1}{2}}(\sigma _{1}-\sigma _{2})\,\!}
^ a b c Meyers and Chawla (1999): "Mechanical Behavior of Materials," 66-75.
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The Average Directional Index (ADX) Formulae
The ADX requires a sequence of calculations due to the multiple lines in the indicator.
\begin{aligned} &\text{+DI} = \left ( \frac{ \text{Smoothed +DM} }{ \text{ATR } } \right ) \times 100 \\ &\text{-DI} = \left ( \frac{ \text{Smoothed -DM} }{ \text{ATR } } \right ) \times 100 \\ &\text{DX} = \left ( \frac{ \mid \text{+DI} - \text{-DI} \mid }{ \mid \text{+DI} + \text{-DI} \mid } \right ) \times 100 \\ &\text{ADX} = \frac{ ( \text{Prior ADX} \times 13 ) + \text{Current ADX} }{ 14 } \\ &\textbf{where:}\\ &\text{+DM (Directional Movement)} = \text{Current High} - \text{PH} \\ &\text{PH} = \text{Previous High} \\ &\text{-DM} = \text{Previous Low} - \text{Current Low} \\ &\text{Smoothed +/-DM} = \textstyle{ \sum_{t=1}^{14} \text{DM} - \left ( \frac{ \sum_{t=1}^{14} \text{DM} }{ 14 } \right ) + \text{CDM} } \\ &\text{CDM} = \text{Current DM} \\ &\text{ATR} = \text{Average True Range} \\ \end{aligned}
+DI=(ATR Smoothed +DM)×100-DI=(ATR Smoothed -DM)×100DX=(∣+DI+-DI∣∣+DI−-DI∣)×100ADX=14(Prior ADX×13)+Current ADXwhere:+DM (Directional Movement)=Current High−PHPH=Previous High-DM=Previous Low−Current LowSmoothed +/-DM=∑t=114DM−(14∑t=114DM)+CDMCDM=Current DMATR=Average True Range
Calculating the Average Directional Movement Index (ADX)
Calculate +DM, -DM, and the true range (TR) for each period. Fourteen periods are typically used.
+DM = current high - previous high.
-DM = previous low - current low.
Use +DM when current high - previous high > previous low - current low. Use -DM when previous low - current low > current high - previous high.
TR is the greater of the current high - current low, current high - previous close, or current low - previous close.
Smooth the 14-period averages of +DM, -DM, and TR—the TR formula is below. Insert the -DM and +DM values to calculate the smoothed averages of those.
Next 14TR value = first 14TR - (prior 14TR/14) + current TR.
Next, divide the smoothed +DM value by the smoothed TR value to get +DI. Multiply by 100.
Divide the smoothed -DM value by the smoothed TR value to get -DI. Multiply by 100.
The directional movement index (DMI) is +DI minus -DI, divided by the sum of +DI and -DI (all absolute values). Multiply by 100.
To get the ADX, continue to calculate DX values for at least 14 periods. Then, smooth the results to get ADX.
First ADX = sum 14 periods of DX / 14.
After that, ADX = ((prior ADX * 13) + current DX) / 14.
What Does the Average Directional Index (ADX) Tell You?
The ADX, negative directional indicator (-DI), and positive directional indicator (+DI) are momentum indicators. The ADX helps investors determine trend strength, while -DI and +DI help determine trend direction.
The ADX identifies a strong trend when the ADX is over 25 and a weak trend when the ADX is below 20. Crossovers of the -DI and +DI lines can be used to generate trade signals. For example, if the +DI line crosses above the -DI line and the ADX is above 20, or ideally above 25, then that is a potential signal to buy. On the other hand, if the -DI crosses above the +DI, and the ADX is above 20 or 25, then that is an opportunity to enter a potential short trade.
Crosses can also be used to exit current trades. For example, if long, exit when the -DI crosses above the +DI. Meanwhile, when the ADX is below 20 the indicator is signaling that the price is trendless and that it might not be an ideal time to enter a trade.
The Average Directional Index (ADX) vs. The Aroon Indicator
The ADX indicator is composed of a total of three lines, while the Aroon indicator is composed of two.
The two indicators are similar in that they both have lines representing positive and negative movement, which helps to identify trend direction. The Aroon reading/level also helps determine trend strength, as the ADX does. The calculations are different though, so crossovers on each of the indicators will occur at different times.
Limitations of Using the Average Directional Index (ADX)
Crossovers can occur frequently, sometimes too frequently, resulting in confusion and potentially lost money on trades that quickly go the other way. These are called false signals and are more common when ADX values are below 25. That said, sometimes the ADX reaches above 25, but is only there temporarily and then reverses along with the price.
Like any indicator, the ADX should be combined with price analysis and potentially other indicators to help filter signals and control risk.
How is the Average Directional Index (ADX) calculated, and what is the formula?
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Home : Support : Online Help : Mathematics : Algebra : Expression Manipulation : Simplifying : RootOf
simplify expressions with the RootOf function
simplify(expr, RootOf)
literal name; RootOf
The simplify/RootOf function is used to simplify expressions which contain the RootOf function.
Polynomials and inverses of polynomials in RootOf are simplified. Any nth roots involving RootOf are resolved.
r≔\mathrm{RootOf}\left({x}^{2}-2=0,x\right):
\mathrm{simplify}\left({r}^{2},\mathrm{RootOf}\right)
\textcolor[rgb]{0,0,1}{2}
\mathrm{simplify}\left(\frac{1}{r},\mathrm{RootOf}\right)
\frac{\textcolor[rgb]{0,0,1}{\mathrm{RootOf}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{\mathrm{_Z}}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\right)}{\textcolor[rgb]{0,0,1}{2}}
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Non-Transitive Dice - Maple Help
Home : Support : Online Help : Math Apps : Probability and Statistics : Probability : Non-Transitive Dice
Imagine you are playing a game with a friend in which each of you chooses one die from the following set and roll them against each other to see who gets the higher face value.
\frac{5}{6}\cdot \left(3\right) + \frac{1}{6}\cdot \left(6\right) = \frac{21}{6}=3.5
\frac{3}{6}\cdot \left(2\right) + \frac{3}{6}\cdot \left(5\right)=\frac{21}{6}=3.5
\frac{1}{6}\cdot \left(1\right) + \frac{5}{6}\cdot \left(4\right) = \frac{21}{6}=3.5
\frac{1}{6}\cdot \left(1\right) + \frac{1}{6}\cdot \left(2\right)+\frac{1}{6}\cdot \left(3\right)+\frac{1}{6}\cdot \left(4\right)+\frac{1}{6}\cdot \left(5\right)+\frac{1}{6}\cdot \left(6\right)=\frac{21}{6} = 3.5
If you let your friend choose his or her die first, is there a particular die you could choose from the remaining three to give yourself a better chance of winning? Which die would you choose if you were able to pick first?
If your friend happens to choose one of the colored dice, A, B, or C, then yes, you can choose a certain die to give yourself the upper hand. If your friend happens to choose the normal die, N, then you will be playing with equal chances of winning, no matter which die you choose. The reason why competing with just the colored dice does not provide equal chances of rolling the higher number is because they are non-transitive dice —a special set of dice for which the property of "rolling the higher number more than half the time" is not transitive:
When rolling this set of dice repeatedly, you will see that A beats B most of the time, B beats C, and surprisingly, C beats A.
So, the fact that A beats B more than 50% of the time and B beats C more than 50% of the time DOES NOT ensure that A beats C more than 50% of the time!
A Look at the Probabilities
By observing all possible combinations found by rolling two of the dice, we can determine how often a certain die will roll a higher number than the competing die. In the case that we roll two of the non-transitive dice against one another, we will see that the probability of one of them winning is greater than 50%, meaning it should win more often. But, in the case that we roll a non-transitive die against a normal die, we will see that the probability of each of them winning is equal, meaning that they are equally matched and so should win the same amount of times.
Probability of A beating B =
\frac{21}{36}≈ 58.3 %
Probability of B beating A =
\frac{15}{36} ≈ 41.7 %
Therefore, A is more likely to roll the higher number.
Probability of B beating C =
\frac{21}{36}≈ 58.3 %
Probability of C beating B =
\frac{15}{36} ≈ 41.7 %
Therefore, B is more likely to roll the higher number.
Probability of A beating C =
\frac{11}{36}≈ 30.6 %
Probability of C beating A =
\frac{25}{36} ≈ 69.4 %
Therefore, C is more likely to roll the higher number.
Probability of A beating N =
\frac{15}{36}≈ 41.7 %
Probability of N beating A =
\frac{15}{36} ≈ 41.7 %
Probability of a tie =
\frac{6}{36}≈16.6 %
Therefore, A and N are equally likely to roll the higher number.
Probability of B beating N =
\frac{15}{36}≈ 41.7 %
Probability of N beating B =
\frac{15}{36} ≈ 41.7 %
\frac{6}{36}≈16.6 %
Therefore, B and N are equally likely to roll the higher number.
C vs. N
Probability of C beating N =
\frac{15}{36}≈ 41.7 %
Probability of N beating C =
\frac{15}{36} ≈ 41.7 %
\frac{6}{36}≈16.6 %
Therefore, C and N are equally likely to roll the higher number.
Use the check boxes below to choose two dice to roll, then press the "Roll!" button to roll them and see which one shows the higher face value. The Statistics plot will keep track of the number of "wins" for each competing die to allow you to observe whether or not one die is more likely to beat the other.
Alternatively, use the slider to choose the number of rounds to simulate and press the "Play" button to have the computer roll for you.
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Home : Support : Online Help : Programming : Input and Output : Translation : LaTeX : functions
When latex processes a Maple object of type function (i.e., an unevaluated function call), it checks to see if there exists a procedure by the name of latex/function_name, where function_name is the name of the function. If such a procedure exists, it is used to format the function call.
For instance, invoking latex(Int(exp(x), x=1..3)) causes latex to check to see if there is a procedure by the name of latex/Int. If such a procedure exists (not the case, but as an example), the above call to latex returns the result of `latex/Int`(exp(x), x=1..3) as its value. This allows LaTeX to produce standard mathematical output in most situations.
If such a procedure does not exist, latex formats the function call by recursively formatting the operands to the function call and inserting parentheses and commas at the appropriate points.
Since Maple 2021, there are no more pre-defined latex/ functions for anything. Instead, the latex command translates what the Maple Typesetting package produces as display. In this way everything you see on the screen can be translated to LaTeX as you see it; for details about that see latex.
Although latex produces a LaTeX version of the Bessel function of the first kind, BesselJ, and of the hypergeometric function based on the display for them by the Typesetting package, as an exercise consider a latex/BesselJ and latex/hypergeom routines to produce their mathematical notation. For BesselJ,latex/BesselJ would need to produce
{J}_{n}\left(z\right)
. This is how you can achieve that:
`latex/BesselJ` := proc(a,z)
sprintf("{ {\\rm J}_{%s}(%s) }",cat("", latex(a, output = string)), cat("", latex(z, output = string)))
With this procedure, the translation is:
\mathrm{latex}\left(\mathrm{BesselJ}\left(n,z+1\right)\right)
{ {\rm J}_{n}(z +1) }
A more sophisticated procedure, for the hypergeometric function pFq:
`latex/hypergeom` := proc(A0, B0, z0)
local nA, nB, pFq, p, q, A, B, z;
nA, nB := nops(A0), nops(B0);
pFq := cat('`\\mbox{$_`',nA,'`$F$_`',nB,'`$}`');
if nA = 0 then A, nA := [`\\ `], 1 else A := map(u -> latex(u, output = string), A0) end if;
if nB = 0 then B, nB := [`\\ `], 1 else B := map(u -> latex(u, output = string), B0) end if;
z := latex(z0, output = string);
p := op(map(u -> (u, '`,`'), A[1 .. -2])), A[-1];
q := op(map(u -> (u, '`,`'), B[1 .. -2])), B[-1];
cat(`{`, pFq, `(`, p, `;\\,`, q, `;\\,`, z, `)}`)
With this procedure, the translation of the three 2F1, 1F1, and 0F1 functions is as follows.
\mathrm{latex}\left(\mathrm{hypergeom}\left([a,b],[c],z\right)\right)
{\mbox{$_2$F$_1$}(a,b;\,c;\,z)}
\mathrm{latex}\left(\mathrm{hypergeom}\left([a],[c],z\right)\right)
{\mbox{$_1$F$_1$}(a;\,c;\,z)}
\mathrm{latex}\left(\mathrm{hypergeom}\left([],[c],z\right)\right)
{\mbox{$_0$F$_1$}(\ ;\,c;\,z)}
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Engineered Musings | Random thoughts from a random engineer.
K_p, K_i, K_id
dT
u
u
u
1.0
0.0
K_p, K_i, K_d
PB
PB/2
0
PB/2
1.0
T_i
T_d
K_p, K_i, K_d
0.5
Grandfather Clock - Test Setup
I finally got tired of having a movement, and not having it in action. However, a reasonable case is way to far away, so I put together a test rig. The rig consisted of some 2x4s, and a cedar top plate that allowed the cables for the weights to pass through. I picked up some lead shot (for making shot gun shells) for weight. The left two weights are 7.7 lb mason jars. The right weight is a 9.9 growler. The horizontal bar was required to keep the weights from interfering with the pendulum.
It actually took me more time than I care to admit to attach the pendulum (a 1x2) to the leader. Getting a rigid connection proved to be a challenge. The best solution ended up slotting the pendulum, and sliding it on the leader.
It runs! It ticks! It tocks! When I put hands on it, they even move. I spun the minute hand a little bit and the hammers started firing. A quick beat count with a stop watch yields it is 14% fast. Adding a clamp to the bottom of the pendulum got it to within 4%. Not bad for no calibration.
f = \frac{3960 beats}{hour} \cdot \frac{1 hour}{3600 seconds} = 1.1 \frac{beats}{second}
f = 1.1 \frac{beats}{second} \cdot \frac{1 cycle}{2 beats} = 0.55 Hz
T = \frac{1}{f} = \frac{1}{0.55Hz}
\boxed{T = 1.\overline{81} seconds}
Therefore, we are going to have to target a pendulum design that maintains a
1.8181 second
T \approx 2 \pi \sqrt{\frac{L}{g}} = 2 \pi \sqrt{\frac{94 cm}{981 cm/s^2}}
T = 1.95 seconds
Well, it looks like an ideal pendulum isn't going to do the trick at 94 cm. Therefore, what we really want is a pendulum with an effective length,
\bar{L}
T = 2 \pi \sqrt{\frac{\bar{L}}{g}}
\bar{L} = \left( \frac{T}{2 \pi} \right)^2 g
\bar{L} = \left( \frac{1.81818181s}{2 \pi} \right)^2 (980.665 cm/s^2)
\boxed{\bar{L} = 82.1175 cm = 32.3 in}
A bit arbitrarily, the bob was selected to be a 4" diameter, 1" thick, 6061 aluminum disc. Assuming a density of
0.0975 lb/in^3
, the bob will weight 1.23 lb.
The rod will consist of a 3/8" x 3/8" x 1/16" U-Shaped 6061 aluminum channel with a walnut inlay. Using the same assumed density of aluminum, and a density of walnut of
0.0243 lb/in^3
, the rod will weigh 0.012lb.
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Contributing · USACO Guide
Using This GuideIntroduction to Competitive ProgrammingChoosing a LanguageResources: Learning to CodeRunning Code OnlineData TypesInput & OutputExpected Knowledge
ContributingAdding SolutionsIntroducing ModulesWorking With MDX
How To DebugHow to PracticeContest StrategyResources: Competitive ProgrammingContestsOlympiads
Running Code LocallyC++ With the Command LineDebugging (Language-Specific)Fast Input & Output(Optional) C++ - Writing Generic Code(Optional) C++ - Lambda Expressions
USACO FAQsUSACO MonthliesUSACO CampResources: USA-Specific
HomeGeneralContributing
Quick StartUsing the Editor (recommended)Running Site LocallyWays to ContributeLessonProblemsMarkdown Resources
Authors: Nathan Wang, Benjamin Qi, Maggie Liu
General - Introducing Modules
All modules are written in MDX (Markdown + JSX) with the XDM compiler.
\LaTeX
is supported through KaTeX, and the Guide uses a number of custom MDX components. If you are confused about something, or if there's a certain feature that you want to add, reach out to Nathan Wang.
The USACO Guide has a public Github Repository.
If you're looking to add/modify modules, refer to the content/ folder.
If you're looking to add/modify problem solutions, refer to the solutions/ folder.
(Almost) all the other files and folders are related to the front-end code.
If you have any questions about how to contribute, feel free to just open an issue or a pull request and a team member will be able to assist you. If you want, you can also join our Discord server.
Using the Editor (recommended)
The easiest way to edit content is to use our live editor.
Running Site Locally
You can also run a local version of the site in order to view your changes. This is useful if you need to use Tailwind CSS classes, which don't work with /editor.
npm install -g yarn? might work
Via command line: git clone https://github.com/cpinitiative/usaco-guide.git
Or use Github Desktop
See the front end documentation for more information.
Convert lists of resources to tables (Plat / Advanced).
Add missing descriptions for sections, modules, or resources.
All resources should have descriptions.
Improve explanations for sample problems.
If starred resource or editorial already has a good explanation, no need to repeat it.
Improve implementations.
Make sure code compiles, remove excessive macros, add codesnips around templates.
Should be consistent across languages.
Adding modules!
If you add a substantial amount of text content to a module, add your name to the list of authors in the frontmatter.
If you add code implementations or improve existing implementations in a more significant way than simply refactoring code, or if you add a short explanation, add your name to the list of contributors.
Convert lists of problems to tables (Plat / Advanced).
Fix problem difficulties and tags.
Add problems that are good examples of the module topic (and remove those that are not).
Adding official editorial links.
Adding editorials.
If no editorial exists, or if existing editorial could be improved.
Or solution code in a different language, etc.
Using our online editor is probably best.
.md syntax highlighting is fine, is also ok for .mdx
You can open a .mdx file and set syntax highlighting to be the same as .md with View -> Syntax -> Open all with current extension as ... -> Markdown -> Markdown.
Automatically compiles
\LaTeX
(doesn't require an installation)
\LaTeX
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Point Update Range Sum · USACO Guide
HomeGoldPoint Update Range Sum
Segment TreeResourcesSolution - Dynamic Range Minimum QueriesSolution - Dynamic Range Sum QueriesBinary Indexed TreeResourcesSolution - Dynamic Range Sum Queries (With a BIT)Solution 1Solution 2Finding
k
-th ElementOrder Statistic TreeWith a BITWith a Segment TreeExample - Inversion CountingSolutionRange Sum ProblemsGeneralUSACO
Authors: Benjamin Qi, Dong Liu, Nathan Gong
Contributor: Andrew Wang
Silver - Introduction to Prefix Sums
k
Dynamic Range Sum Queries
Most gold range query problems require you to support following tasks in
\mathcal{O}(\log N)
time each on an array of size
N
Update the element at a single position (point).
Query the sum of some consecutive subarray.
Both segment trees and binary indexed trees can accomplish this.
A segment tree allows you to do point update and range query in
\mathcal{O}(\log N)
time each for any associative operation, not just summation.
You can skip more advanced applications such as lazy propagation for now. They will be covered in platinum.
EDU: Segment Tree Pt 1 Steps 1, 3, 4
basic operations, inversion counting
9.3 - Segment Trees
Same implementation as AICash below.
See slides after union-find. Also introduces sqrt bucketing.
Simplest form of a Segment Tree
"Advanced versions" are covered in Platinum.
Solution - Dynamic Range Minimum Queries
AICash - Efficient and easy segment trees
based off above
Note that st.init(n+1) allows us to update and query indices in the range [0,n+1)=[0,n].
public class DynamicRangeMinQueries {
int q = io.nextInt();
SegmentTree seg = new SegmentTree(n);
Solution - Dynamic Range Sum Queries
Compared to the previous problem, all we need to change are T, ID, and comb.
const T ID = 0; T comb(T a, T b) { return a+b; }
Compared to the previous problem, all we need to change is the way we aggregate values (change from Math.min() to summation) and the data type we use to store the query (int to long).
public class DynamicRangeSumQueries {
Implementation is shorter than segment tree, but maybe more confusing at first glance.
9.2, 9.4 - Binary Indexed Tree
also similar to above
Solution - Dynamic Range Sum Queries (With a BIT)
vector<ll> bit(MX), x(MX);
Writing a BIT this way has the advantage of generalizing to multiple dimensions.
mouse_wireless - Multi-dimensional BITs with Templates
* Description: range sum queries and point updates for $D$ dimensions
* Source: https://codeforces.com/blog/entry/64914
* Verification: SPOJ matsum
* Usage: \texttt{BIT<int,10,10>} gives 2D BIT
* Time: O((\log N)^D)
template <class T, int ...Ns> struct BIT {
T val = 0; void upd(T v) { val += v; }
static long[] bit;
k
-th Element
Suppose that we want a data structure that supports all the operations as a set in C++ in addition to the following:
order_of_key(x): counts the number of elements in the set that are strictly less than x.
find_by_order(k): similar to find, returns the iterator corresponding to the k-th lowest element in the set (0-indexed).
Luckily, such a built-in data structure already exists in C++.
adamant - Policy Based Data Structures
4.5 - Policy Based Data Structures
brief overview with find_by_order and order_of_key
Indexed Set
To use indexed set locally, you need to install GCC.
Assumes all updates are in the range
[1,N]
adamant - About Ordered Set
With a Segment Tree
Covered in Platinum.
Example - Inversion Counting
SPOJ - Easy
Range Sum Problems
If the coordinates are large (say, up to
10^9
), then you should apply coordinate compression before using a BIT or segment tree (though sparse segment trees do exist).
Easy Show Tags PURS
Easy Show Tags Inversions, PURS
Easy Show Tags Coordinate Compress, PURS
Increasing Subsequence II
Normal Show Tags Offline, PURS
Sleepy Cow Sort
Mincross
Hard Show Tags PURS
A hint regarding Sleepy Cow Sort: There is only one correct output.
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Exercise - Azur Lane Wiki
Exercises, sometimes called Operations, Mock Battles or PvP Battles, are the portion of the game where the player's fleet is pitted against a fleet constructed by another player, with the goal to accumulate Merit and progress in Rank.
2 Ranking and Merit
2.1 Total merit per season
7 Defense fleet strategies
Each player is given five additional attempts three times a day, and this number is shown at the right side of the screen as 'X/10':
The Daily Reset at 00:00 Server Time.
12 hours later, at 12:00 Server Time.
6 hours later, at 18:00 Server Time.
Server time for JP is UTC+9. Server time for EN is UTC-7. Server time always has the same UTC offset regardless of any country's daylight savings time.
These are added to the available attempts, up to a limit of 10, except at a new season's start. Each season lasts 2 weeks and players start with 5 attempts regardless of how many were held before the reset.
Ranks are reset every two weeks in tandem with the Weekly Tasks. The time until the next ranking reset is listed above the remaining number of attempts for the day. Time until the next refresh is displayed under the number of remaining attempts, as well as a button to switch the current opposing fleets with 4 new ones that are adjusted to the player's current Rank. This button can only be used 5 times per day, resetting at 00:00 JST(UTC+9). The number of available remaining resets is not displayed.
If the number of available PvP attempts is pushed over 10 by a reset, any attempts over 10 will be permanently lost. Thus, players are recommended to ensure they are at 5/10 or fewer attempts before a reset, and 0/10 before a maintenance that stretches through two resets. Note that the attempt number is deducted after a PvP battle ends, not after it begins - therefore, a match starting 1 minute before a reset but ending 1 minute after a reset will count as being deducted after the reset.
Ranking and Merit
The player will have a choice of four fleets, which are displayed in the middle of the screen, to battle against; these fleets are set up by other players whose Rank Position is close to the player's own. The highest-ranked opponent available is displayed on the left and the lowest-ranked on the right; maximum Rank Score is gained by choosing the highest-ranked (smallest number) opponent on the left side.
Rank information is displayed on the bottom left, which includes the icon for the player's current Rank, and from top to bottom, the following 4 numbers:
Total accumulated Merit this season
Remaining Rank Score to next Rank
Fleet selection icons are displayed on the bottom right. A defense fleet button, used to set up the defense fleet other players will battle against when the player is chosen as an opponent, can be seen, as well as a series of buttons at the bottom for an Introduction to PvP, a link to the Rank List, and a link to the Special Supply shop. The battles themselves are automated and end when all of the player's or the enemy's ships are defeated. In case of a stalemate, fights are limited to five minutes.
The amount of experience and Rank Score rewarded is determined by the opposing fleet, and the amount of merit gained is determined by the player's current Rank. A victory earns full experience, Rank Score and . A defeat earns halved experience and , and no Rank Score is gained.
Merit rewards per victory increase as the player progresses through Ranks. The amount of Merit awarded per Rank is listed in the table below.
Before the "Commodore" rank, a player's rank is determined solely by Rank Score; after that, the player will also need to reach certain Rank Position thresholds for promotion. While the 'remaining Rank Score to next rank' will be displayed as 0 if the relative rank (Top 100, etc) requirements are not met, players will still accumulate Rank Score through victories. Rank Position is determined by players' relative Rank Score to others in descending order (1 is the current top-ranked player), and in the event of a tie, the player who achieved that Rank Score earlier will have a higher Rank Position (lower number).
Every time a player ranks up, they will be rewarded with a mail which contains a certain amount of , which increases as the player progresses through Ranks. The amount per Rank is included in the table below.
Minimum Rank Score
Base Rank Score Gain (Estimated)
Private 軍曹 0 50 25 -- 25
Petty Officer 曹長 100 60 30 200 22
Ensign 少尉 200 70 35 600 20
Lieutenant Junior Grade 中尉 300 70 35 600 17
Lieutenant 大尉 400 70 35 600 15
Lieutenant Commander 少佐 550 80 40 1000 15
Commander 中佐 700 80 40 1000 15
Captain 大佐 850 80 40 1000 12
Rear Admiral Lower Half 准将 - Top 1000 1050 90 45 1500 10
Rear Admiral 少将 - Top 600 1250 90 45 1500 10
Vice Admiral 中将- Top 300 1450 90 45 1500 10
Admiral 大将 - Top 100 1650 90 45 1500 10
Fleet Admiral 上級大将 - Top 50 1900 90 45 1500 10
Admiral of the Navy 元帥 - Top 10 2200 100 50 2500 10
Rank Score awarded per victory depends on the difference between the defending player's Rank Score and the attacking player's Rank Score. While the exact calculations for Rank Score gain are unknown, a 30-35 Rank Score difference is estimated to have a difference of 1 on the Rank Score, with the base Rank Score gain dependent on rank as shown above. Challenging lower-ranked players will deduct Rank Score gained, so it is possible to get only 9 or fewer Rank Score per victory when challenging lower-ranked players.
Total merit per season
Total available Merit for a two-week season is as follows:
Ranks up to Captain (last rank achievable by Rank Score alone): 5000
Ranks past Captain: 10000
210 battles:
All losses: 5250
All wins: ~15000 (depending on rank)
Daily and Weekly tasks: 2000
All in all, a player can expect to earn about 17000 in a season if playing all the battles with a 50% win rate. A top player could earn about twice that.
A daily Task rewards Merit.
{\displaystyle {\begin{aligned}\mathrm {BaseExperience} &=\lfloor 1.5\times \sum _{i}^{N}\mathrm {EnemyLevel} \rfloor +171.\end{aligned}}}
For example, a full Level 125 enemy fleet gives 1296 experience and a two ship Level 1 enemy fleet gives 174 experience. It's based on the total level rather than the average level, so a full fleet of level 40 ships ends up giving the same experience as two level 120 ships.
The MVP receives x2 experience, the flagship receives x1.5 experience, and if the flagship is the MVP, she receives x3 experience. Morale and/or event experience modifiers do NOT apply in Exercises.
Defeat halves all experience values. For reference, losing to a full level 120 fleet gives equivalent XP to winning against a full level 50.5 fleet, and losing to a full level 125 fleet gives equivalent XP to winning against a full level 53 fleet.
Ships do not do exactly the same thing in exercises as they do in sorties. The following changes are applied:
Hitpoints All ship hitpoints are multiplied by a factor depending on the ship's level. Plane health is quadrupled.
{\displaystyle \mathrm {ShipExerciseHP} =\mathrm {ShipHP} \times \left(0.9+0.012\times \mathrm {ShipLevel} \right)}
Stat Modifiers Ships in PvP receive special effects depending on their hull class.
DD/DDG (Vanguard Mode)/AE: Receive 25% less damage and gain 5% evasion rate.
CL: Receive 25% less damage and deal 15% more damage.
CA/CB: Receive 15% less damage.
BB/BC/BBV/DDG (Main Mode)/CV: Deal 20% less damage.
CVL: Deal 10% less damage.
AR/BM: No changes. (Note: SS/SSV also have no modifier)
Attacker Bonus Attackers benefit from an additional 20% damage reduction factor.
Bonus Stacking As usual, all sources of damage reduction stack multiplicatively with each other and everything else. Damage Dealt modifiers stack additively with the above penalties, provided they normally stack additively with other Damage Dealt modifiers. (e.g. Tirpitz or Avrora damage bonuses)
Damage over Time As usual, Burn and Flood damage are not affected by Damage Dealt and Damage Received modifiers, so they are unaffected by special PvP rules.
Ships with barrages all have alternative versions that they use in exercises, most of which are equal in strength to their PvE counterparts, but the pattern changes to align with enemy backlines. When neither side has any remaining vanguards, the game reports that the "battle is accelerated," which gives each ship a massive RLD stat buff, equal to a 300% RLD buff plus an additional 300.0 RLD stat.
The rewards available from PvP are Ship Experience and Merit.
Therefore, there are two general trains of thought with regard to how to best use Exercises: One is to maximise Experience gained by ships, which involves using ships that are not at level 100 to challenge fleets they can defeat (which may not always be the highest-ranked player). Resets are used if there are no fleets the target fleet can defeat, and players may opt to challenge fleets they have no chance of winning in order to gain the 50% experience, should they run out of opponent resets.
The other is to maximise gain, which involves using a full level 120 fleet to defeat the highest-ranked defense fleet available, which requires significantly more effort.
In both cases players should always use all available PvP attempts for the season, but the ships used for PvP will differ, the opponents challenged will differ, and when the opponents are challenged will differ.
In order to maximise Merit gain, players will wish to reach the Admiral of the Fleet rank for the 2500 promotion reward, and do as many attempts at 90 per victory as possible. In addition, the available defense players are reset only when the player uses the opponent reset button, or when the player defeats one opponent from the current set.
Thus, in order to maximise Rank Score (and thereby maximise one's chances of reaching the Admiral of the Fleet rank), one should wait for an opponent in their list to do their own PvP, which increases their Rank Score, which thereby increases the Rank Score gained by the attacking player on victory. Resets are used in this scheme to swap opponents if none of the available 4 achieve a victory after a while (or if all 4 opponents use defense fleets the player cannot defeat). As one's opponents' current Rank Position is updated only when the application is refreshed, to check if an opponent has achieved a victory (and is now useable to gain a 11-point victory), players will wish to log out from the Options menu and log back in periodically.
However, as dropping below the Commodore Rank (top 1000) reduces Merit per victory to 80, one must also attack before dropping out of the top 1000 while waiting for opponents to do their PvP attacks to maximise Merit.
Maximising Merit is thus done mostly through maximising Rank, which is done through maximising Rank Score, which is done by balancing the wait for one's 4 opponents to do their battles first, while avoiding dropping below the top 1000. Players capable of reaching Admiral of the Fleet Rank will almost always be hiding at some point in the top 1000 but outside the top 100 until 2 days, 12 hours before the ranking season ends, which is the earliest point that any player will be able to attain the 2200 Rank Score minimum requirement for Admiral of the Fleet.
As different opponents do their PvP battles at different times, some players will gain more Rank Score than other players based on the opponents they get matched against. In a way, it can be said that being able to reach the Admiral of the Fleet Rank at all, as well as what Rank they hold at the end of the season, is determined by their luck with the matchmaking system.
Defense fleet strategies
To a player whose objective is maximised ship experience, there is no need to set any specific defense fleet, and therefore players may choose to set easy fleets to give other players an easier time, hard fleets to annoy other players, waifu/event-only fleets to show off, or meme fleets for jokes. Some leave the fleet as the default starter DD + Long Island combination because they can't be bothered to change the setting.
To a player whose objective is maximised Merit, defense fleet setting will matter. Essentially, because players are the opponent of other players themselves, one may wish to intentionally set easy fleets when they are awake and playing the game to increase the chances that others will defeat them, and if any of the victory-chains causes one of the 4 specified opponents to get PvP victories earlier, this will increase the chances of successfully attaining 5x 11 Rank Score victories per reset. However, they will also wish to intentionally set difficult fleets if they will be sleeping (especially sleep that causes one to go above 5/10 attempts), in order to reduce the chances that their Rank Position drops below the top 1000.
To a player whose objective is truly maximised score, strictly speaking, if your server has a wide variety of fleet strength or even if your server is at endgame (due to liquid meta, where an endgame player could defeat the others anyway), using weak defense fleet is NOT advised because in order to maximize your ranking, you would like to have opponents such as those who are capable of rushing PvP themselves (or at least capable of defeating fleets that are strong), which will be more difficult if you put weak defense fleet, as you have allowed more weaker opponents who cannot rush their PvP to fill up the gap, (More likely to roll into them), and also the chance of one of your 4 opponents meeting you is very unlikely (although it could occur) so if the only one who put weak defense fleet is you (or minority does as such), you will only disturb everyone because everyone can now roll into weak opponents who cannot rush PvP.
At later rank, due to the nature of when your score is relatively high, in order for other players to maximise their own score, they will only attack you if you are the leftmost opponent or could attack only IF you have spent your PvP charges so that they could maximise their own score even further, which mostly only occur if you rank higher than them, and even then the opponents they have are randomly chosen (they likely will attack someone else), your chance of dropping out of top 1000 is rather slim. Hence it could be said, again, that weak defense fleet is harmful to everyone if only a minority chooses to do as such.
Also, for Gensui Rush, referring to the hour at which players will be capable of reaching Admiral of the Fleet (or Gensui) rank, only top 10 will be able to get in, during this hour, it is advised to use the most difficult fleet you have, or the fleet which could stall as much time as possible, so that you could finish your PvP before they do and get your Gensui before they could. (Same score players will be ranked based on chronological order of reaching said score).
Retrieved from ‘https://azurlane.koumakan.jp/w/index.php?title=Exercise&oldid=177529’
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Transient Heat Transfer Measurements Using a Single Wide-Band Liquid Crystal Test | J. Turbomach. | ASME Digital Collection
Transient Heat Transfer Measurements Using a Single Wide-Band Liquid Crystal Test
Dragos N. Licu,
Dragos N. Licu
Matthew J. Findlay,
Matthew J. Findlay
Ian S. Gartshore,
Ian S. Gartshore
Contributed by the International Gas Turbine Institute and based on a paper presented at the 44th International Gas Turbine and Aeroengine Congress and Exhibition, Indianapolis, Indiana, June 7–10, 1999. Manuscript received by the International Gas Turbine Institute February 1999. Paper No. 99-GT-167. Review Chair: D. C. Wisler.
Licu, D. N., Findlay, M. J., Gartshore , I. S., and Salcudean, M. (February 1, 1999). "Transient Heat Transfer Measurements Using a Single Wide-Band Liquid Crystal Test ." ASME. J. Turbomach. July 2000; 122(3): 546–552. https://doi.org/10.1115/1.1303819
A technique using a thermochromic liquid crystal coating to measure film cooling effectiveness (η) and heat transfer coefficient
hf
has been developed so that both of these important parameters can be obtained, as a function of time, from a single transient test. The technique combines a real-time, true color (24 bit) imaging system with the use of a wide-band liquid crystal coating and multiple event sampling for the simultaneous determination of η and
hf
from the single test. To illustrate and validate this technique, the flow from compound-angle square jets in a crossflow is examined. The tests, in which the jet air was suddenly heated to about 40°C, lasted 30 seconds. The measured η is compared with measurements made in the same flow under steady-state conditions in a totally different way, using a mass/heat analogy and a flame ionization detector. Good agreement is obtained. Three different blowing ratios
M
of 0.5, 1.0, and 1.5 are investigated with a constant jet Reynolds number of about 5000. Detailed quantitative comparisons of the η measured in both ways are made for all blowing ratios, and plots of η and
hf
are presented. [S0889-504X(00)01403-3]
gas turbines, heat transfer, temperature measurement, liquid crystal devices, film flow, cooling, jets
Film cooling, Flow (Dynamics), Jets, Liquid crystals, Temperature, Transients (Dynamics), Heat transfer coefficients, Transient heat transfer, Flames, Ionization, Sensors
Vedula, R. J., and Metzger, D. E., 1991, “A Method for the Simultaneous Determination of Local Effectiveness and Heat Transfer Distribution in Three-Temperature Convection Situations,” ASME Paper No. 91-GT-345.
Influence of Gap Leakage Downstream of the Injection Holes on Film Cooling Performance
Chyu, M. K., and Hsing Y. C., 1996, “Use of Thermographic Phosphor Fluorescence Imaging System for Simultaneous Measurement of Film Cooling Effectiveness and Heat Transfer Coefficient,” ASME Paper No. 96-GT-430.
Licu, D. N., 1998, “Heat Transfer Characteristics in Film Cooling Applications,” Ph.D. thesis, Dept. Mech. Eng., The University of British Columbia, Vancouver, BC, Canada.
Findlay, M. J., 1998, “Experimental and Computational Investigation of Inclined Jets in a Crossflow,” Ph.D. thesis, Dept. Mech. Eng., The University of British Columbia, Vancouver, BC, Canada.
Thermax—Thermographic Measurements Ltd., 1998, http://www.thermax.com/.
Pitas, I., 1993, Digital Image Processing, Prentice Hall, Englewood Cliffs, NJ.
A Color Image Processing System for Transient Liquid Crystal Heat Transfer Experiments
Carslaw, H. S., and Jaeger J. C., 1959, Conduction of Heat in Solids, Oxford University Press, New York, 2nd ed.
Ajersch, P., 1995, “Detailed Measurements on a Row of Jets in a Crossflow—With Applications,” M.A.Sc. Thesis, Department of Mechanical Engineering, The University of British Columbia, Vancouver, BC, Canada.
Sun, Y., 1995, “An Experimental and Numerical Investigation of Heat Transfer Downstream of a Normal Film Cooling Injection Slot,” Ph.D. thesis, Department of Mechanical Engineering, The University of British Columbia, Vancouver, BC, Canada.
Kays, W. M., and Crawford, M. E., 1980, Convective Heat and Mass Transfer, McGraw-Hill, New York, 2nd ed.
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$L^p$ estimates for the wave equation and applications
{L}^{p}
estimates for the wave equation and applications
MR 94f:35076
title = {$L^p$ estimates for the wave equation and applications},
TI - $L^p$ estimates for the wave equation and applications
Sogge, Christopher D. $L^p$ estimates for the wave equation and applications. Journées équations aux dérivées partielles (1993), article no. 15, 12 p. http://www.numdam.org/item/JEDP_1993____A15_0/
1. M. Beals, Lp boundedness of Fourier integrals, Mem. Amer. Math. Soc. 264 (1982). | Zbl 0508.42020
2. M. Beals and M. Bezard, Low regularity local solutions for field equations, preprint. | Zbl 0852.35098
3. J. Bourgain, Averages in the plane over convex curves and maximal operators, J. Analyse Math. 47 (1986), 69-85. | MR 88f:42036 | Zbl 0626.42012
4. J. Bourgain, Besicovitch type maximal operators and applications to Fourier analysis, Geometric and Funct. Anal. 1 (1991), 69-85. | MR 92g:42010 | Zbl 0756.42014
5. J. Bourgain, A harmonic analysis approach to problems in nonlinear differential equations, preprint. | Zbl 0822.35116
6. M. Christ, Lectures on singular integral operators, C.B.M.S. Lecture Notes, no. 77, American Math. Soc., Providence, RI, 1990. | MR 92f:42021 | Zbl 0745.42008
7. M. Christ and M. Weinstein, Dispersion of low-amplitude solutions of the generalized Korteweg-de Vries equation, J. Funct. Anal. 100 (1991), 87-109. | MR 92h:35203 | Zbl 0743.35067
8. D. Grieser, Lp bounds for eigenfunctions and spectral projections of the Laplacian near concave boundaries, Thesis, UCLA (1992).
9. M.G. Grillakis, Regularity for the wave equation with a critical nonlinearity, Comm. Pure and Appl. Math. 45 (1992), 749-774. | MR 93e:35073 | Zbl 0785.35065
10. J. Harmse, On Lebesgue space estimates for the wave equation, Indiana Math. J. 39 (1990), 229-248. | MR 91j:35158 | Zbl 0683.35008
11. L. Hörmander, The spectral function of an elliptic operator, Acta Math. 121 (1968), 193-218. | MR 58 #29418 | Zbl 0164.13201
L. Hörmander, Non-linear hyperbolic differential equations, Lund lecture notes, 1988.
13. F. John, The ultrahyperbolic equation with 4 independent variables, Duke J. Math. 4 (1938), 300-322. | JFM 64.0497.04 | Zbl 0019.02404
14. J.-L. Journé, A. Soffer and C.D. Sogge, Decay estimates for Schrödinger operators, Comm. Pure and Appl. Math. 44 (1991), 573-604. | MR 93d:35034 | Zbl 0743.35008
15. L. Kapitanski, Weak and yet weaker solutions of semilinear wave equations, Brown Univ. preprint. | Zbl 0831.35109
16. C.E. Kenig, A. Ruiz and C.D. Sogge, Uniform Sobolev inequalities and unique continuation for second order constant coefficient differential operators, Duke Math. J. 55 (1987), 329-349. | MR 88d:35037 | Zbl 0644.35012
17. S. Klainerman and M. Machedon, The null condition and global existence for nonlinear waves, Comm. Pure and Appl. Math. (to appear).
18. H. Lindblad, A sharp counter example to local existence of low regularity solutions to nonlinear wave equations, Duke Math. J. 72 (1993) (to appear). | MR 94h:35165 | Zbl 0797.35123
19. H. Lindblad and C.D. Sogge, Minimal regularity for local existence of solutions to for semilinear Lorentz-invariant wave equations, in preparation.
20. W. Littman, Lp → Lq estimates for singular integrals, Proc. Symp. Pure and Appl. Math., vol. 23, Amer. Math. Soc., 1973, pp. 479-481. | MR 50 #10909 | Zbl 0263.44006
21. G. Mockenhaupt, A. Seeger and C. D. Sogge, Wave front sets, local smoothing and Bourgain's circular maximal theorem, Annals of Math. 136 (1992), 207-218. | MR 93i:42009 | Zbl 0759.42016
22. G. Mockenhaupt, A. Seeger and C.D. Sogge, Local smoothing of Fourier integrals and Carleson-Sjölin estimates, J. Amer. Math. Soc. 6 (1993), 65-130. | MR 93h:58150 | Zbl 0776.58037
23. J. Peral, Lp estimates for the wave equation, J. Funct. Anal. 36 (1980), 114-145. | MR 81k:35089 | Zbl 0442.35017
24. J. Rauch, The u5-Klein-Gordan equation, Nonlinear PDE's and applications, vol. 53, Pitman Research Notes in Math., pp. 335-364. | MR 83a:35066 | Zbl 0473.35055
25. J. Shatah and M. Struwe, Regularity results for nonlinear wave equations, preprint. | Zbl 0836.35096
26. A. Seeger, C.D. Sogge and E.M. Stein, Regularity properties of Fourier integral operators, Annals of Math 134 (1991), 231-251. | MR 92g:35252 | Zbl 0754.58037
27. H. Smith and C.D. Sogge, Lp regularity for the wave equation with strictly convex obstacles (to appear). | Zbl 0805.35169
28. C.D. Sogge, Uniqueness in Cauchy problems for hyperbolic differential operators, Trans. Amer. Math. Soc. 333 (1992), 821-833. | MR 92m:35006 | Zbl 0763.35012
29. C.D. Sogge, Propagation of singularities and maximal functions in the plane, Invent. Math. 104 (1991), 349-376. | MR 92i:58192 | Zbl 0754.35004
30. C.D. Sogge, Fourier integrals in classical analysis, Cambridge Univ. Press, Cambridge, New York, 1993. | MR 94c:35178 | Zbl 0783.35001
31. E.M. Stein, Harmonic analysis real-variable methods, orthogonality, and oscillatory integrals, Princeton Univ. Press, Princeton, 1993. | MR 95c:42002 | Zbl 0821.42001
32. W. Strauss, Nonlinear wave equations, C.B.M.S. Lecture Notes, no. 73, American Math. Soc., Providence, RI, 1989. | MR 91g:35002 | Zbl 0714.35003
33. R. Strichartz, A priori estimates for the wave equation and some applications, J. Funct. Analysis 5 (1970), 218-235. | MR 41 #2231 | Zbl 0189.40701
34. R. Strichartz, Restriction of Fourier transform to quadratic surfaces and decay of solutions to the wave equation, Duke Math. J. 44 (1977), 705-714. | MR 58 #23577 | Zbl 0372.35001
35. M. Struwe, Semilinear wave equations, Bull. Amer. Math. Soc. 26 (1992), 53-85. | MR 92e:35112 | Zbl 0767.35045
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A googol is the lairge nummer 10100; that is, the digit 1 follaed bi 100 zeroes:
10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000.
The term wis coined in 1938[1] bi 9-year-auld Milton Sirotta, nephew o American mathematician Edward Kasner. Kasner popularized the concept in his 1940 beuk Mathematics and the Imagination.
Ither names for googol include ten duotrigintillion on the short scale, ten thoosand sexdecillion on the long scale, or ten sexdecilliard on the Peletier lang scale.
A googol haes na particular signeeficance in mathematics, but is uiseful when comparin wi ither very lairge quantities such as the nummer o subatomic pairticles in the veesible universe or the nummer o hypothetically possible chess genmes. Edward Kasner uised it tae illustrate the difference atween an unimaginably lairge nummer an infinity, an in this role it is sometimes uised in teachin mathematics.
A googol is approximately 70! (factorial o 70). In the binary numeral seestem, ane would need 333 bits tae represent a googol, i.e., 1 googol ≈ 2332.2, or exactly
{\displaystyle 2^{(100/\mathrm {log} _{10}2)}}
Googol is notable for bein the subject o the £1 million question in the infamous episode o Who Wants to Be a Millionaire?, when contestant Charles Ingram cheatit his way through the shaw bi gettin help frae his wife, who wis in the audience, an fellae contestant Tecwen Whittock. It is an aa the namesake o the internet company Google, wi the name "Google" bein a misspellin o "googol" bi the company's founders.[2]
Tae gie a sense o hou big a googol really is, the mass o an electron, juist unner 1×10-30 kg, can be compared tae the mass o the entire universe, estimatit at atween 1×1050kg an 1×1060 kg.[3] It is a ratio in the order o aboot 1080 tae 1090, still much smawer than the value o a googol.
↑ Kasner, Edward and Newman, James R. (1940). Mathematics and the Imagination. Simon and Schuster, New York. ISBN 0-486-41703-4. freemit airtin in |title= (help)CS1 maint: multiple names: authors leet (link)
↑ QI: Quite Interesting facts about 100 Archived 2012-12-28 at the Wayback Machine, telegraph.co.uk
↑ McPherson, Kristine (2006). Elert, Glenn (ed.). "Mass of the universe". The Physics Factbook. Retrieved 24 August 2019.
Taen frae "https://sco.wikipedia.org/w/index.php?title=Googol&oldid=848470"
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Guild - Azur Lane Wiki
The Guild menu allows for players to form their own groups/guilds where they can communicate and see each other's stats.
New players will be presented with a choice upon first selecting the Guild menu: To create their own Guild, to join an existing Guild or to join the Public Guild.
There are currently 2 main factions:
Crimson Axis (Red)
Led by Iron Blood and Sakura Empire, they utilized Siren technology to obtain powers capable of changing the world.
Azur Lane (Blue)
Represented by Eagle Union and Royal Navy, they advocate natural evolution, and seek for true freedom and peace.
3 Public Guild
4 Guild Screen
6.1 Fleet Mission
6.2 Provision Mission
6.3 Fleet Supply
8.3 Battle Support
8.4.1 Check Support Fleet
The first step to creating a Guild is to pick an 'Ideology'. The effects of the choices are currently unknown and are purely aesthetic choice. Once you choose your ideology, you'll be led to the creation screen. Creating a Guild will cost 300.
The creation screen will be the same, regardless of the ideology you choose, only the color scheme will change.
Your Guild's ideology, objective, and guild declaration can be changed for free at any time. Changing your Guild's name, however, will cost another 100 each time. You can also disband your Guild, if you choose to.
This will display a screen showing currently active Guilds, with an emblem for their Political System and their level. A box at the top allows players to search for specific Guilds with their name or ID number. After applying to a Guild, you will need to be approved by an Officer of that Guild before you join it.
Public Guild
This is where you get if you join the Public Guild. This is a system that allows you to enjoy a few benefits of being in a Guild without joining one. It only has 2 Tabs:
Logistics, where you have access to Provision Missions, although you don't get any Fleet Missions or Supplies and also not gain any funds or tech progress from the provisions provided.
Tech, without the Guild Roster Expansion Tech since there's no limit on members in the Public Guild. Upgrading it is 50% more expensive than in a guild that has the respective Tech researched. Public Guild Tech levels are the same across all servers. Also Public Guild generates 60 Tech points for a "random" Guild Tech every day.
This is the Tech as of 10th January 2022:
The Public Guild always belongs to the Azur Lane faction.
Guild Screen
After joining or creating a Guild, selection of the Guild menu will bring a player to their Guild's homepage. This will display the Guild's stats on the left hand side, and a Guild chat on the right. Selecting the second button on the left will display a list of the current Guild members. Guild members of ranks Master and Officer will also see a third button to review pending applications (see previous section).
Guild XP chart
XP to Next
10 Current limit 157000 40
This tab shows the guild's muster roll, with info for each member. When the player selects a guild member, they will see four options depending on their rank:
Rank: Guild members can hold the rank of Master, Officer, Elite or Recruit. The guild's founder will automatically be assigned the rank of Master, every new member will start out as Recruit. The Master and Officers can use this option to promote or demote members, but keep in mind that there can only be one Master.
Impeach: After 10 days of inactivity, Officers get the option to impeach the Master, taking over their position. If the Master does not log in within 24 hours, then leadership will be transferred to the member with the most contributions to the Guild.
Expel: The Master and Officers can use this option to expel members.
Info: This option shows the member's stats as shown in their profile page.
Sources of Activity include Daily and Weekly Missions and Guild Activities.
Complete a Guild Mission 2
Contribute materials 10
Participate in a Guild Event 1
Participate in a Guild Operation 10
You can view Leaderboards for Contributions, Missions, and Operations for Weekly, Monthly, and All Time.
The Guild's Master can select each week which mission the Guild members will have to complete. Members completing the mission for the first time each week will receive Guild Tokens. In addition, each completion will earn Guild Funds.
Siren Subjugation I/II/III Defeat 60/180/300 enemies 1500/2500/3500 Guild Supplies 80/240/400
Sea Sector Sweep I/II/III Obtain 15/45/75 victories
Materials Contribution I/II/III Contribute materials 3/9/15 times
Guild Operation I/II/III Participate in 1/3/5 Guild events
The mission can only be completed a total of 100 times per week, counted across all Guild members.
Provision Mission
Members can contribute resources, from a randomly chosen list of three options, to the Guild. Gold is always at least one of the options available. Doing so will grant the member 75 Guild Tokens and earn the Guild 1 Fund and 2 Tech Points. This can be done thrice a day. Possible contributions are:
20 Oxy-cola
10 Secret Coolant
20 T1 Gear Parts
5 T3 Gear Parts
Fleet Supply is an additional reward mechanism that can be bought by the guild's leaders for 3000 Guild Funds. During its runtime of 14 days, once per day, the Guild will distribute 30 Guild Tokens as supplies to each member. In addition, there's a chance to receive more Guild Tokens, Prototype Cores, Gems, and Prototype Gear Upgrade Parts.
In this tab, the Guild's Master and Officers can set the research for guild bonuses and check its progress. Apart from Guild Roster Size, these bonuses are applied to the individual members of the guild. For the tech to take effect for a member it needs to be purchased by that member. Techs that the Guild hasn't researched, but that are available to the Public Guild can be bought at 150% the price. Available bonuses are:
Guild Roster Size: +1 Player per level, 20 levels
Coins Storage Cap: +300 per level, 30 levels
Oil Storage Cap: +50 per level, 30 levels
Warehouse Slots: +5 Depot space per level, 10 levels
Dock Slots: +2 Dock slots per level, 10 levels
Cat Box Discount: -15 Cat Box cost per level, 20 levels
The research target can be changed once per day.
Coin/Oil Storage Expansion
Guild Roster Expansion
Cat Box Discount
Additional Dock/Warehouse Slots
Masters and Officers can initiate an Operation using Guild funds.
Every member can participate in Guild Operations twice a month. This limit is for the member and the count will carry over when changing guilds and will not reset until the next month.
Pacific Base Patrol
Northern Shipping Escort
Peninsular Raid
Solomon Air-Sea Battle
Rec. Participants
x? x8 x388 x12 x438 x15 x498 x18
→ Main Article: Guild Events
Once an Operation is initiated, Events will occur. Guild members can dispatch ships to assist in the Events. A total of 4 dispatches can be made in an Event. Additional dispatches are available 6, 12, 18, and 21 hours after daily reset.
Each event has a base time to complete of between 12.6 and 42.0 hours. The Total Efficiency of all participants increases the speed by approximately:
{\displaystyle {\text{speed bonus}}=200\%\times {\frac {\text{Total Efficiency}}{{\text{Total Efficiency}}+7000}}}
Thus, with infinite efficiency, the event would be completed in one-third of the base time.
The speed bonus is retroactive. For example, if the guild let the event run for half of the base time with low efficiency, and suddenly increased the efficiency to 7000 (+100% speed), the event would be completed instantly.
Prior to the Boss Event, members can assign up to two ships to Battle Support. These ships will be available to all Guild Members to use against the Event Boss.
The boss appears at the end of the Events. The boss can be fought once per day. Up to 3 ships can be used from the Support Fleet formed previously by members. However, the Vanguard and Main lines must each contain at least 1 of the member's own ships.
When battling the boss, you can exit out without consuming your daily attempt, although oil will still be consumed.
Check Support Fleet
The Guild Master can recommend up to 9 ships. Recommended ships will be tagged and display first in the list.
All Guild Members can Check Gear to view the gear of ships assigned to Battle Support.
Retrieved from ‘https://azurlane.koumakan.jp/w/index.php?title=Guild&oldid=165286’
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Correspondence to: *E-mail: ksw0020@kpetro.or.kr
Fuel economy, DQE index, EER, ASCR, RMSSE, IWR
에너지소비효율, 모드추종성 지표, 에너지효율변화율, 절대 속도 변화율, 제곱평균 속도오차, 관성일률
F.E.=\frac{734}{0.866×HC+0.429×CO+0.273×C{O}_{2}}
\text{EER=}\left[\text{1-}\frac{\text{DR/100+1}}{\text{ER/100+1}}\right]\text{×100}
\text{DR=}\frac{{\text{D}}_{\text{D}}\text{-}{\text{D}}_{\text{T}}}{{\text{D}}_{\text{T}}}\text{×100}
\text{ER=}\frac{{\text{CE}}_{\text{D}}\text{-}{\text{CE}}_{\text{T}}}{{\text{CE}}_{\text{T}}}\text{×100}
{\text{CE}}_{\text{J}}\text{=}\sum _{\text{i=1}}^{\text{N}}{\left[\text{1.105ET}{\text{Wa}}_{\text{Ji}}\text{+}{\text{F}}_{\text{0}}\text{+}{\text{F}}_{\text{1}}{\text{V}}_{\text{Ji}}\text{+}{\text{F}}_{\text{2}}{\text{V}}_{\text{Ji}}^{\text{2}}\right]}^{\text{+}}
{\text{V}}_{\text{Ji}}\text{=}\frac{\text{1}}{\text{5}}\sum _{\text{t=i-2}}^{\text{t=i+1}}{\text{V}}_{\text{t}}
\text{ASCR=}\frac{\text{AS}{\text{C}}_{\text{D}}\text{-AS}{\text{C}}_{\text{T}}}{\text{AS}{\text{C}}_{\text{T}}}\text{×100}
{\text{ASC}}_{\text{J}}\text{=0.1}\sum _{\text{i=1}}^{\text{N}}\left|{\text{a}}_{\text{Ji}}\right|
{\text{a}}_{\text{Ji}}\text{=}\frac{{\text{V}}_{\text{Ji+1}}\text{-}{\text{V}}_{\text{Ji-1}}}{\text{0.2}}
\text{RMSSE=2.237}\sqrt{\frac{\sum _{\text{i=1}}^{\text{N}}{\left({\text{V}}_{\text{Di}}\text{-}{\text{V}}_{\text{Ti}}\right)}^{\text{2}}}{\text{n}}}
\text{IWR=}\frac{\text{I}{\text{W}}_{\text{D}}\text{-I}{\text{W}}_{\text{T}}}{\text{I}{\text{W}}_{\text{T}}}
{\text{IW}}_{\text{J}}\text{=}{\sum _{\text{i=1}}^{\text{N}}\left[{\text{W}}_{\text{I-Ji}}\right]}^{\text{+}}
{\text{W}}_{\text{I-Ji}}\text{=}{\text{F}}_{\text{I-Ji}}{\text{d}}_{\text{Ji}}\text{=1.015ETW}{\text{a}}_{\text{Ji}}{\text{V}}_{\text{Ji}}\text{0.1}
H. Wi, and J. Park, “Analyzing Uncertainty in Evaluation of Vehicle Fuel Economy Using FTP-75”, Int. J. Automotive Technology, Vol.14(No.3), p471-477, (2013). [https://doi.org/10.1007/s12239-013-0051-x]
H. Kim, “Eco-Friendly Technology for Improving Fuel Economy”, Auto Journal, KSAE, Vol.34(No.1), p32-38, (2012).
U. S. Environmental Protection Agency, “Fuel Economy Procedures”, EPA/Industry Light Duty Vehicle Compliance Meeting, p50-63, (2013).
SAE J2951, Drive Quality Evaluation for Chassis Dynamometer Testing, SAE International, Surface Vehicle Recommended Practice, (2014).
WLTP Trace Index Task Force(TF), Progress Report Speed Trace Violations / Drive Trace Index, (2015).
S. Kim, K. Kim, J. Ha, S. Kwon, and Y. Seo, “A Study about Impact of Battery SOC on Fuel Economy of Conventional Diesel Vehicle”, Transactions of KSAE, Vol.24(No.4), p480-486, (2016). [https://doi.org/10.7467/ksae.2016.24.4.480]
G. Cho, H. Wi, J. Lee, J. Park, and K. Park, “Effect of Alternator Control on Vehicle Fuel Economy”, Transactions of KSAE, Vol.17(No.2), p20-25, (2009).
J. Lee, and J. Ryu, “Research for RVC Validation Method”, KSAE Annual Conference Proceedings, p1667-1670, (2011).
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Modeling | Bean Machine
Declarative Style
Bean Machine allows you to express models declaratively, in a way that closely follows the notation that statisticians use in their everyday work. Consider our example from the Quick Start. We could express this mathematically as:
n_\text{init}
: known constant
\texttt{reproduction\_rate} \sim \text{Exponential}(10.0)
n_\text{new} \sim \text{Poisson}(\texttt{reproduction\_rate} \cdot n_\text{init})
Let's take a look at the model again:
def num_new(num_current):
return dist.Poisson(reproduction_rate() * num_current)
You can see how the Python code maps almost one-to-one to the mathematical definition. When building models in Bean Machine's declarative syntax, we encourage you to first think of the model mathematically, and then to evolve the code to fit to that definition.
Importantly, note that there is no formal class delineating your model. This means you're maximally free to build models that feel organic with the rest of your codebase and compose seamlessly with models found elsewhere in your codebase. Of course, you're also free to consolidate related modeling functionality within a class, which can help keep your model appropriately scoped!
Random Variable Functions
Python functions annotated with @bm.random_variable, or random variable functions for short, are the building blocks of models in Bean Machine. This decorator denotes functions which should be treated by the framework as random variables to learn about.
A random variable function must return a PyTorch distribution representing the probability distribution for that random variable, conditional on sample values for any other random variable functions that it depends on. For the most part, random variable functions can contain arbitrary Python code to model your problem! However, please do not depend on mutable external state (such as Python's random module), since Bean Machine will not be aware of it and your inference results may be invalid.
As outlined in the next two sections, calling random variable functions has different behaviors depending upon the callee's context.
Calling a Random Variable from Another Random Variable Function
When calling a random variable function from within another random variable function, you should treat the return value as a sample from its underlying distribution. Bean Machine intercepts these calls, and will perform inference-specific operations in order to draw a sample from the underlying distribution that is consistent with the available observation data. Working with samples therefore decouples your model definition from the mechanics of inference going on under the hood.
Calls to random variable functions are effectively memoized during a particular inference iteration. This is a common pitfall, so it bears repeating: calls to the same random variable function with the same arguments will receive the same sampled value within one iteration of inference. This makes it easy for multiple components of your model to refer to the same logical random variable. This means that the common statistical notation discussed previously in Declarative Style can easily map to your code: a programmatic definition like reproduction_rate() will always map to its corresponding singular statistical concept of
n_\text{new}
, no matter how many times it is invoked within a single model. This can also be appreciated from a consistency point of view: if we define a new random variable tautology to be equal to reproduction_rate() <= 3.0 or reproduction_rate() > 3.0, the probability of tautology being True should be
1
, but if each invocation of reproduction_rate produced a different value, this would not hold. In Defining Random Variable Families, we'll see how to control this memoization behavior with function parameters.
Calling a Random Variable from an Ordinary Function
It is valid to call random variable functions from ordinary Python functions. In fact, you've seen it a few times in the Quick Start already! We've used it to bind data, specify our queries, and access samples once inference has been completed. Under the hood, Bean Machine transforms random variable functions so that they act like function references. Here's an example, which we just call from the Python toplevel scope:
num_new()
RVIdentifier(function=<function num_new at 0x7ff00372d290>, arguments=())
As you can see, the call to this random variable function didn't return a distribution, or a sample from a distribution. Rather, it resulted in an RVIdentifier object, which represents a reference to a random variable function. You as the user can't do much with this object on its own, but Bean Machine will use this reference to access and re-evaluate different parts of your model.
Defining Random Variable Families
As discussed in Calling a Random Variable from Another Random Variable Function, calls to a random variable function are memoized during a particular iteration of inference. How, then, can we create models with many random variables which have related but distinct distributions?
Let's dive into this by extending our model. In the previous example, we were modeling the number of new cases on a given day as a function of the number of infected individuals on the previous day. What if we wanted to model the spread of disease over multiple days? This might correspond to the following mathematical model:
n_i-n_{i-1} \sim \text{Poisson}(\texttt{reproduction\_rate} \cdot n_{i-1})
n_i
represents the number of cases on day
i
n_0=n_\text{init}
It is common for statistical models to group random variables together into a random variable family as you see here. In Bean Machine, the ability of indexing into random variable families is generalized to arbitrary serializable Python objects. As an example, we could use a discrete time domain, here represented as a list of datetime.date objects,
in order to re-index the random varialble num_new() in our previous model:
Note how this allows us to express a more complex dependency structure: where we previously relied on the argument num_current to describe the infections at some unspecified "current time", we can now use a more precise notion of (for example) "the day before today". This knowledge is in turn represented in another part of our probabilistic generative model, namely in the function num_total:
# WARNING: INCORRECT COUNTER-EXAMPLE
Transforming Random Variables
The problem in the above code is that we can't decorate num_total() with @bm.random_variable. The reason we cannot is that it doesn't return a PyTorch elementary probability distribution. But, without a @bm.random_variable decorator on this function, Bean Machine won't know that it should treat num_new() inside its body as a random variable function. As we discussed in Calling a Random Variable from an Ordinary Function, this call to num_new() would merely return an RVIdentifier, which is not what we want.
What do we do then? What we need here, and what is also the last important construct in Bean Machine's modeling toolkit, is the @bm.functional decorator. This decorator behaves like @bm.random_variable, except that it does require the function it is decorating to return only elementary distributions. As such, it can be used to deterministically transform the results of one or more other @bm.random_variable or @bm.functional functions. With this construct we can now write this model as follows:
One last note: while a @bm.functional can be queried during inference, it can't have observations bound to it.
Next, we'll look at how you can use Inference to fit data to your model.
Random Variable Functions
Calling a Random Variable from Another Random Variable Function
Calling a Random Variable from an Ordinary Function
Defining Random Variable Families
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2015 Monotone and Concave Positive Solutions to Three-Point Boundary Value Problems of Higher-Order Fractional Differential Equations
Wenyong Zhong, Lanfang Wang
We study the three-point boundary value problem of higher-order fractional differential equations of the form
{Dc}_{0+}^{\rho }u\left(t\right)+f\left(t, u\left(t\right)\right)=0
0<t<1
2⩽n-1<\rho <n
{u}^{\prime }\left(0\right)={u}^{\prime \prime }\left(0\right)=\cdots ={u}^{\left(n-1\right)}\left(0\right)=0
u\left(1\right)+p{u}^{\prime }\left(1\right)=q{u}^{\prime }\left(\mathrm{\xi }\right)
{\mathrm{ }}^{c}{D}_{0+}^{\rho }
\rho
f:\left[0,1\right]×\left[0,\mathrm{\infty }\right)↦\left[0,+\mathrm{\infty }\right)
is continuously differentiable. Here,
0⩽q⩽p
0<\xi <1
2⩽n-1<\rho <n
. By virtue of some fixed point theorems, some sufficient criteria for the existence and multiplicity results of positive solutions are established and the obtained results also guarantee that the positive solutions discussed are monotone and concave.
Wenyong Zhong. Lanfang Wang. "Monotone and Concave Positive Solutions to Three-Point Boundary Value Problems of Higher-Order Fractional Differential Equations." Abstr. Appl. Anal. 2015 (SI05) 1 - 9, 2015. https://doi.org/10.1155/2015/728491
Wenyong Zhong, Lanfang Wang "Monotone and Concave Positive Solutions to Three-Point Boundary Value Problems of Higher-Order Fractional Differential Equations," Abstract and Applied Analysis, Abstr. Appl. Anal. 2015(SI05), 1-9, (2015)
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Boost drive shaft speed - Simulink - MathWorks Deutschland
ExtTrq
Additional torque input
Shaft inertia, J_shaft
Initial shaft speed, w_0
Min shaft speed, w_min
Max shaft speed, w_max
Turbine mechanical efficiency, eta_mech
Boost drive shaft speed
Powertrain Blockset / Propulsion / Combustion Engine Components / Boost
The Boost Drive Shaft block uses the compressor, turbine, and external torques to calculate the drive shaft speed. Use the block to model turbochargers and superchargers in an engine model.
You can specify these configurations:
Turbocharger — Connect the compressor to the turbine
Two-way ports for turbine and compressor connections
Option to add an externally applied input torque
Compressor only — Connect the drive shaft to the compressor
Two-way port for compressor connection
Externally applied input torque
Turbine only — Connect the drive shaft to the turbine
Two-way port for turbine connection
Externally applied load torque
For the Turbine only and Turbocharger configurations, the block modifies the turbine torque with a mechanical efficiency.
The Boost Drive Shaft block applies Newton’s Second Law for Rotation. Positive torques cause the drive shaft to accelerate. Negative torques impose a load and decelerate the drive shaft.
The block also calculates the power loss due to mechanical inefficiency.
Shaft dynamics
\frac{d\omega }{dt}=\frac{1}{{J}_{shaft}}\left({\eta }_{mech}{\tau }_{turb}+{\tau }_{comp}+{\tau }_{ext}\right)
with initial speed ω0
{\omega }_{min}\le \omega \le {\omega }_{max}
{\stackrel{˙}{W}}_{loss}=\omega {\tau }_{turb}\left(1-{\eta }_{mech}\right)
PwrCmpsr
Shaft power from compressor
{\tau }_{comp}\omega
PwrTurb
Shaft power from turbine
{\tau }_{turb}\omega
PwrExt Externally applied power
{\tau }_{ext}\omega
PwrMechLoss Mechanical power loss
-{\stackrel{˙}{W}}_{turb}
PwrStoredDriveshft Rate change in rotational kinetic energy
\left({\eta }_{mech}{\tau }_{turb}+{\tau }_{comp}+{\tau }_{ext}\right)\omega
Initial drive shaft speed
Minimum drive shaft speed
Maximum drive shaft speed
Jshaft
Mechanical efficiency of turbine
τcomp
Compressor torque
τext
Externally applied torque.
{\stackrel{˙}{W}}_{loss}
Power loss due to mechanical inefficiency
Cmprs — Compressor torque
Compressor torque, τcomp, in N·m.
To create this port, for the Configuration parameter, select Turbocharger or Compressor only.
Turb — Turbine torque
Turbine torque, τturb, in N·m.
To create this port, for the Configuration parameter, select Turbocharger or Turbine only.
ExtTrq — Externally applied torque
Externally applied torque, τext, in N·m.
For turbocharger configurations, to create this port, set Additional torque input to External torque input.
MechPwrLoss
Applied external torque
PwrInfo PwrTrnsfrd PwrCmpsr
PwrExt Externally applied power W
PwrNotTrnsfrd PwrMechLoss Mechanical power loss W
PwrStored PwrStoredDriveshft Rate change in rotational kinetic energy W
Cmprs — Compressor speed
Compressor speed, ω, in rad/s.
Turb — Turbine speed
Turbine speed, ω, in N·m.
Configuration — Specify configuration
Turbocharger (default) | Turbine only | Compressor only
Selecting Turbocharger or Compressor only creates the Cmprs port.
Selecting Turbocharger or Turbine only creates the Turb port.
Additional torque input — Specify external torque input
External torque input (default) | No external torque
To enable this parameter, select a Turbocharger configuration.
To create the Trq port, select External torque input.
Shaft inertia, J_shaft — Inertia
Shaft inertia, Jshaft, in kg·m^2.
Initial shaft speed, w_0 — Speed
Initial drive shaft speed, ω0, in rad/s.
Min shaft speed, w_min — Speed
Minimum drive shaft speed, ωmin, in rad/s.
Max shaft speed, w_max — Speed
Maximum drive shaft speed, ωmax, in rad/s.
Turbine mechanical efficiency, eta_mech — Efficiency
Mechanical efficiency of turbine ηmax.
To enable this parameter, select the Turbocharger or Turbine only configuration.
Compressor | Turbine
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Mini-Workshop: Zeta Functions, Index and Twisted K-Theory; Interactions with Physics | EMS Press
Simon G. Scott
This mini-workshop brought together number theorists, analysts, geometers and mathematical physicists to discuss current issues at the common boundary of mathematics and physics. Topics covered included the number theoretic and algebraic structures underlying renormalization, twisted K-theory and higher algebraic structures, modular forms, and arithmetic and spectral zeta functions. A particular theme was around developing interconnections between arithmetic (multiple) zeta functions, spectral zeta functions associated with elliptic operators (and related spectral invariants such as spectral flow) and current issues in physics such as renormalization and mirror symmetry. Multiple zeta functions appear in index theory and
K
-theory via their relation to anomalies, in number theory in their relation to polylogarithms, in renormalization questions in perturbative quantum field theory and Hopf algebras, in duality issues and in twisted
K
-theory for index theorems for projective families of elliptic operators, thereby providing a rich set of overlapping topics with common analytical issues.
This meeting was organized around one hour talks, four each day, with plenty of time between talks for informal discussion and a 45 minute talk in the afternoon for students; three graduate students were among the 16 participants. Some participants lectured for two hours in order to have time to introduce the audience to the subject before entering the technical details.
The organizers and participants would like to thank the {\em Mathematisches For\-schungs\-institut Oberwolfach} for providing a pleasant and stimulating enviroment for this meeting.
Sylvie Paycha, Steven Rosenberg, Simon G. Scott, Mini-Workshop: Zeta Functions, Index and Twisted K-Theory; Interactions with Physics. Oberwolfach Rep. 3 (2006), no. 2, pp. 1245–1284
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{\displaystyle A}
{\displaystyle A}
. A square matrix is full rank if all of its columns are independent. That is, a full rank matrix has no column vector
{\displaystyle v_{i}}
{\displaystyle A}
that can be expressed as a linear combination of the other column vectors
{\displaystyle v_{j}\neq \Sigma _{i=0,i\neq j}^{n}a_{i}v_{i}}
A simple test for determining if a matrix is full rank is to calculate its determinant. If the determinant is zero, there are linearly dependent columns and the matrix is not full rank. Prof. John Doyle also mentioned during lecture that one can perform the singular value decomposition of a matrix, and if the lowest singular value is near or equal to zero the matrix is likely to be not full rank ("singular").
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Error, invalid input: `simpl/abs` expects its 2nd argument, a2, to be of type algebraic, but received Array(1..2, [-1,-1]) - Maple Help
Home : Support : Online Help : Error, invalid input: `simpl/abs` expects its 2nd argument, a2, to be of type algebraic, but received Array(1..2, [-1,-1])
\mathrm{sin}\left(\left[1,2, 3\right]\right);
\mathrm{whattype}\left(\left[1,2,3\right]\right);
\textcolor[rgb]{0,0,1}{\mathrm{list}}
\left[1,2,3\right]
\mathrm{Describe}\mathit{}\left(\mathrm{sin}\right)
x
\mathrm{sin}
\mathrm{sin}\left(1\right);\mathrm{sin}\left(2\right);\mathrm{sin}\left(3\right)
\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\right)
\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2}\right)
\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\right)
\mathrm{sin}~\left(\left[1,2,3\right]\right)
\left[\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\right)\right]
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Minimal Surface Problem - MATLAB & Simulink
This example shows how to solve the minimal surface equation
-\nabla \cdot \left(\frac{1}{\sqrt{1+{|\nabla \mathit{u}|}^{2}}}\nabla \mathit{u}\right)=0
\Omega =\left\{\left(\mathit{x},\mathit{y}\right)|{\mathit{x}}^{2}+{\mathit{y}}^{2}\le 1\right\}
\mathit{u}\left(\mathit{x},\mathit{y}\right)={\mathit{x}}^{2}
\partial \Omega
. An elliptic equation in the toolbox form is
-\nabla \cdot \left(\mathit{c}\nabla \mathit{u}\right)+\mathit{au}=\mathit{f}
Therefore, for the minimal surface problem, the coefficients are as follows:
\mathit{c}=\frac{1}{\sqrt{1+{|\nabla \mathit{u}|}^{2}}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathit{a}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathit{f}=0
Because the coefficient c is a function of the solution u, the minimal surface problem is a nonlinear elliptic problem.
To solve the minimal surface problem using the programmatic workflow, first create a PDE model with a single dependent variable.
Create the geometry and include it in the model. The circleg function represents this geometry.
specifyCoefficients(model,'m',0,'d',0,'c',cCoef,'a',a,'f',f);
Specify the boundary conditions using the function
\mathit{u}\left(\mathit{x},\mathit{y}\right)={\mathit{x}}^{2}
bcMatrix = @(region,~)region.x.^2;
applyBoundaryCondition(model,'dirichlet',...
'Edge',1:model.Geometry.NumEdges,...
'u',bcMatrix);
Solve the problem by using the solvepde function. Because the problem is nonlinear, solvepde invokes a nonlinear solver. Observe the solver progress by setting the SolverOptions.ReportStatistics property of the model to 'on'.
model.SolverOptions.ReportStatistics = 'on';
Iteration Residual Step size Jacobian: Full
0 1.8540e-02
1 2.8715e-04 1.0000000
pdeplot(model,'XYData',u,'ZData',u);
zlabel 'u(x,y)'
title 'Minimal Surface'
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New Insights Into Learning With Correntropy-Based Regression | Neural Computation | MIT Press
Department of Mathematics and Statistics, State University of New York at Albany, Albany, NY 12222, U.S.A. ylfeng@albany.edu
Yunlong Feng; New Insights Into Learning With Correntropy-Based Regression. Neural Comput 2021; 33 (1): 157–173. doi: https://doi.org/10.1162/neco_a_01334
Stemming from information-theoretic learning, the correntropy criterion and its applications to machine learning tasks have been extensively studied and explored. Its application to regression problems leads to the robustness-enhanced regression paradigm: correntropy-based regression. Having drawn a great variety of successful real-world applications, its theoretical properties have also been investigated recently in a series of studies from a statistical learning viewpoint. The resulting big picture is that correntropy-based regression regresses toward the conditional mode function or the conditional mean function robustly under certain conditions. Continuing this trend and going further, in this study, we report some new insights into this problem. First, we show that under the additive noise regression model, such a regression paradigm can be deduced from minimum distance estimation, implying that the resulting estimator is essentially a minimum distance estimator and thus possesses robustness properties. Second, we show that the regression paradigm in fact provides a unified approach to regression problems in that it approaches the conditional mean, the conditional mode, and the conditional median functions under certain conditions. Third, we present some new results when it is used to learn the conditional mean function by developing its error bounds and exponential convergence rates under conditional (
1+ε
)-moment assumptions. The saturation effect on the established convergence rates, which was observed under (
1+ε
)-moment assumptions, still occurs, indicating the inherent bias of the regression estimator. These novel insights deepen our understanding of correntropy-based regression, help cement the theoretic correntropy framework, and enable us to investigate learning schemes induced by general bounded nonconvex loss functions.
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networks(deprecated)/allpairs - Maple Help
Home : Support : Online Help : networks(deprecated)/allpairs
allpairs(G)
allpairs(G, v)
name used to return a table of parents
Important: The networks package has been deprecated. Use the superseding command GraphTheory[AllPairsDistance] instead.
This procedure is an implementation of Floyd's allpairs shortest path algorithm.
The result, T, is a table of distances between any two vertices. Thus
{T}_{u,v}
is the shortest distance from u to v.
The optional extra parameter (eg. parents) is used to supply a name for a table of ancestors. Thus
{\mathrm{parents}}_{u,v}
is the ancestor of v in the shortest path tree rooted at u.
Edge weights are assumed to be lengths or distances. Undirected edges are assumed to be bidirectional.
Edge weights must be non-negative.
This routine is normally loaded via the command with(networks) but may also be referenced using the full name networks[allpairs](...).
\mathrm{with}\left(\mathrm{networks}\right):
G≔\mathrm{petersen}\left(\right):
T≔\mathrm{allpairs}\left(G,p\right):
T[1,3]
\textcolor[rgb]{0,0,1}{2}
p[1,3]
\textcolor[rgb]{0,0,1}{2}
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Modular Root - Maple Help
Home : Support : Online Help : Mathematics : Factorization and Solving Equations : Roots : Modular Root
ModularRoot(x, r, n)
The ModularRoot function computes a non-negative integer
y
{y}^{r}=x\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{mod}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}n
When x has more than one roots of order r, any one of them may be returned.
\mathrm{with}\left(\mathrm{NumberTheory}\right):
The following numbers have cube roots modulo
24
\mathrm{residues}≔{\mathrm{seq}\left({i}^{3}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{mod}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}24,i=0..23\right)}
\textcolor[rgb]{0,0,1}{\mathrm{residues}}\textcolor[rgb]{0,0,1}{≔}{\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{11}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{13}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{15}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{16}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{17}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{19}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{21}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{23}}
13
has a cube root modulo
24
\mathrm{evalb}\left(13\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{residues}\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{ModularRoot}\left(13,3,24\right)
\textcolor[rgb]{0,0,1}{13}
{13}^{3}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{mod}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}24
\textcolor[rgb]{0,0,1}{13}
12
does not have a cube root modulo
24
\mathrm{evalb}\left(12\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{residues}\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{ModularRoot}\left(12,3,24\right)
Error, (in NumberTheory:-ModularRoot) 12 is a 3rd order non-residue modulo 24
The NumberTheory[ModularRoot] command was introduced in Maple 2016.
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Theory of relativity/Special relativity/momentum - Wikiversity
Theory of relativity/Special relativity/momentum
< Theory of relativity | Special relativity(Redirected from Special relativity/momentum)
This article presumes that the reader has read Special relativity/space, time, and the Lorentz transform.
This article will derive the relativity-correct formula for momentum on theoretical grounds, using the Lorentz transform of Special Relativity and the requirement that momentum be conserved in all frames of reference.
The thought experiment is a simple collision between two identical particles, A and B. The collision is perfectly elastic, that is, energy is conserved. The particles have the same mass, and their speeds, before and after the collision, are the same. By setting up the collision in this way, we don't have to be concerned with issues such as mass change or energy change, that is, E = mc².
The collision as seen in the primed frame.
The collision takes place at x' = y' = t' = 0 in the primed frame of reference. The x and y components of the speed are ±r and ±s, respectively. The equations of the particles' motion in the primed frame are:
A before collision (t' < 0):
{\displaystyle x'=rt'\qquad y'=st'}
A after collision (t' > 0):
{\displaystyle x'=rt'\qquad \ y'=-st'}
B before collision (t' < 0):
{\displaystyle x'=-rt'\quad \ y'=-st'}
B after collision (t' > 0):
{\displaystyle x'=-rt'\qquad y'=st'}
Now choose an unprimed frame of reference moving to the left with speed r. This will cancel out the horizontal motion of particle B. The Lorentz transforms both primed-to-unprimed and unprimed-to-primed, are
{\displaystyle x={\frac {x'+rt'}{\sqrt {1-r^{2}/c^{2}}}}\qquad \qquad x'={\frac {x-rt}{\sqrt {1-r^{2}/c^{2}}}}}
{\displaystyle y=y'\qquad \qquad \qquad \qquad y'=y}
{\displaystyle t={\frac {t'+rx'/c^{2}}{\sqrt {1-r^{2}/c^{2}}}}\qquad \qquad t'={\frac {t-rx/c^{2}}{\sqrt {1-r^{2}/c^{2}}}}}
Solving these, we get, in the unprimed frame:
A before collision (t < 0):
{\displaystyle x={\frac {2rt}{1+r^{2}/c^{2}}}\qquad y={\frac {st{\sqrt {1-r^{2}/c^{2}}}}{1+r^{2}/c^{2}}}}
A after collision (t > 0):
{\displaystyle x={\frac {2rt}{1+r^{2}/c^{2}}}\qquad \ y={\frac {-st{\sqrt {1-r^{2}/c^{2}}}}{1+r^{2}/c^{2}}}}
B before collision (t < 0):
{\displaystyle x=0\qquad y={\frac {-st}{1+r^{2}/c^{2}}}}
B after collision (t > 0):
{\displaystyle x=0\qquad \ y={\frac {st}{1+r^{2}/c^{2}}}}
We now analyze the collision in the unprimed frame. In this frame, B has no horizontal speed. It goes straight down, collides, and comes straight back up.
The collision as seen in the unprimed frame.
A's speed is the square root of the sum of the squares of the x and y components of its speed:
{\displaystyle A_{\textrm {speed}}={\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}}
B's speed is simpler:
{\displaystyle B_{\textrm {speed}}={\frac {s}{\sqrt {1-r^{2}/c^{2}}}}}
We know that the momentum vector must point in the same direction as the particle's motion, and its magnitude must be some function of the mass and speed. It must also be exactly proportional to the mass—two protons have twice the momentum of one. Also, in the non-relativistic limit, the momentum is given by:
{\displaystyle {\vec {p}}=m{\vec {v}}}
So we set the momentum formula to:
{\displaystyle {\vec {p}}=m{\vec {v}}f(v)}
where f is some function of the (magnitude of the) momentum.
Sneak preview: Those who have seen the formula know that
{\displaystyle f(x)={\frac {1}{\sqrt {1-x^{2}/c^{2}}}}}
. The goal of this article is to derive that.
Using this formula with the unknown function f, we have:
{\displaystyle A_{\textrm {x\ momentum,\ before\ or\ after}}={\frac {2mr}{1+r^{2}/c^{2}}}\ \ f\left({\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}\right)}
{\displaystyle A_{\textrm {y\ momentum,\ before}}={\frac {ms{\sqrt {1-r^{2}/c^{2}}}}{1+r^{2}/c^{2}}}\ \ f\left({\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}\right)}
{\displaystyle A_{\textrm {y\ momentum,\ after}}={\frac {-ms{\sqrt {1-r^{2}/c^{2}}}}{1+r^{2}/c^{2}}}\ \ f\left({\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}\right)}
{\displaystyle B_{\textrm {x\ momentum,\ before\ or\ after}}=0\,}
{\displaystyle B_{\textrm {y\ momentum,\ before}}={\frac {-ms}{\sqrt {1-r^{2}/c^{2}}}}\ \ f\left({\frac {s}{\sqrt {1-r^{2}/c^{2}}}}\right)}
{\displaystyle B_{\textrm {y\ momentum,\ after}}={\frac {ms}{\sqrt {1-r^{2}/c^{2}}}}\ \ f\left({\frac {s}{\sqrt {1-r^{2}/c^{2}}}}\right)}
It is clear that the x component of the momentum is conserved:
{\displaystyle A_{\textrm {x,\ before}}+B_{\textrm {x,\ before}}=A_{\textrm {x,\ after}}+B_{\textrm {x,\ after}}}
The y component needs to be solved carefully:
{\displaystyle A_{\textrm {y,\ before}}+B_{\textrm {y,\ before}}=A_{\textrm {y,\ after}}+B_{\textrm {y,\ after}}}
Collecting terms in the above line, and dividing by 2ms:
{\displaystyle {\frac {\sqrt {1-r^{2}/c^{2}}}{1+r^{2}/c^{2}}}\ \ f\left({\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}\right)\ =\ {\frac {1}{\sqrt {1-r^{2}/c^{2}}}}\ \ f\left({\frac {s}{\sqrt {1-r^{2}/c^{2}}}}\right)}
{\displaystyle f\left({\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}\right)\ =\ {\frac {1+r^{2}/c^{2}}{1-r^{2}/c^{2}}}\ \ f\left({\frac {s}{\sqrt {1-r^{2}/c^{2}}}}\right)}
This is a rather formidable equation, involving two parameters, r and s, from which to deduce the function f.
We can simplify this by letting s approach zero:
{\displaystyle f\left({\frac {2r}{1+r^{2}/c^{2}}}\right)\ =\ {\frac {1+r^{2}/c^{2}}{1-r^{2}/c^{2}}}\ \ f(0)}
Since we have hypothesized that
{\displaystyle f(0)=1\,}
, that is, the momentum approaches
{\displaystyle {\vec {p}}=m{\vec {v}}}
in the classical limit:
{\displaystyle f\left({\frac {2r}{1+r^{2}/c^{2}}}\right)\ =\ {\frac {1+r^{2}/c^{2}}{1-r^{2}/c^{2}}}}
{\displaystyle z={\frac {2r}{1+r^{2}/c^{2}}}}
so (solving a quadratic equation):
{\displaystyle r={\frac {c^{2}}{z}}(1-{\sqrt {1-z^{2}/c^{2}}})}
{\displaystyle 1+{\frac {r^{2}}{c^{2}}}={\frac {2c^{2}}{z^{2}}}(1-{\sqrt {1-z^{2}/c^{2}}})}
{\displaystyle 1-{\frac {r^{2}}{c^{2}}}={\frac {2c^{2}}{z^{2}}}(z^{2}/c^{2}-1+{\sqrt {1-z^{2}/c^{2}}})}
{\displaystyle f(z)={\frac {1-{\sqrt {1-z^{2}/c^{2}}}}{{\sqrt {1-z^{2}/c^{2}}}\ -\ (1-z^{2}/c^{2})}}}
{\displaystyle \ \ ={\frac {1-{\sqrt {1-z^{2}/c^{2}}}}{{\sqrt {1-z^{2}/c^{2}}}\ (1-{\sqrt {1-z^{2}/c^{2}}})}}}
{\displaystyle \ \ ={\frac {1}{\sqrt {1-z^{2}/c^{2}}}}}
So the relativistically correct equation for momentum is:
{\displaystyle {\vec {p}}={\frac {m{\vec {v}}}{\sqrt {1-v^{2}/c^{2}}}}}
The Lorentz factor as a function of speed, from 0 to the speed of light.
The correction factor that we have derived is called the Lorentz factor, typically denoted with a lowercase gamma.
{\displaystyle \gamma ={\frac {1}{\sqrt {1-v^{2}/c^{2}}}}}
This same function appears prominently in the Lorentz transform. It determines the length contraction and time dilation. For non-relativistic speeds, it is nearly 1. As the speed approaches the speed of light, it grows without limit. At v=0.99c, it is about 7.
This means that an object's momentum grows unboundedly large as its speed approaches c.
The next article in this series is Special relativity/energy.
Special relativity/space, time, and the Lorentz transform
Special relativity/energy
Special relativity/E = mc²
Special relativity/spacetime diagrams and vectors
Retrieved from "https://en.wikiversity.org/w/index.php?title=Theory_of_relativity/Special_relativity/momentum&oldid=1834842"
|
Rms amplitude AGC - SEG Wiki
The rms amplitude AGC gain function is based on the rms amplitude within a specified time gate on an input trace. This gain function is computed as follows. The input trace is subdivided into fixed time gates. First, the amplitude of each sample in a gate is squared. Second, the mean of these values is computed and its square root is taken. This is the rms amplitude over that gate. The ratio of a desired rms amplitude (say 2000) to the actual rms value is assigned as the value of the gain function at the center of the gate. Hence, the scaling function g(t) at the gate center is given by
{\displaystyle g(t)={\frac {\text{desired rms}}{\sqrt {{\frac {1}{N}}\sum \nolimits _{i=1}^{N}{x_{i}^{2}}}}},}
where xi is the trace amplitude and N is the number of samples within the gate.
Figure 1.4-10 A portion of a CMP stack before and after application of two different rms AGC functions. Numbers on the top indicate the window sizes in milliseconds used in computing the AGC gain function described by equation ( 10 ).
Typically, we start out with a certain gate length at the shallow part of the trace. Gate length can be kept either constant or it can be increased systematically down the trace. At each gate center, the value of the gain function is computed as described above. Function g(t) then is interpolated between the gate centers. Note that the specified time gates are stationary — they do not slide down the trace.
Figure 1.4-10 shows the ungained data and two rms-gained sections. The gate lengths are indicated at the top of each panel. When the gate used in the computation is kept small, say 64 ms, then strong reflections become less distinct.
Retrieved from "https://wiki.seg.org/index.php?title=Rms_amplitude_AGC&oldid=17311"
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Home : Support : Online Help : Programming : Data Types : Conversion : truefalse
convert an expression to a value of type `truefalse'
convert( expr, 'truefalse' )
any Maple expression that can be evaluated as a boolean
The function convert( expr, 'truefalse' ) attempts to convert the expression expr to one of the two values true and false. This is intended to be used in composition with procedures that return a boolean literal that can include the value FAIL. The value FAIL is replaced by the value false.
The argument expression expr must be an expression that can be evaluated as a boolean, resulting in one of the values true, false, or FAIL. If
\mathrm{evalb}\left(\mathrm{expr}\right)
returns true, then the conversion also returns the value true. Otherwise, the conversion returns the value false.
\mathrm{convert}\left(\mathrm{true},'\mathrm{truefalse}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{convert}\left(\mathrm{false},'\mathrm{truefalse}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{convert}\left(\mathrm{FAIL},'\mathrm{truefalse}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{convert}\left(2<3,'\mathrm{truefalse}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{convert}\left(a<b,'\mathrm{truefalse}'\right)
Error, (in `convert/truefalse`) unable to convert a < b to type `truefalse'
\mathrm{sort}\left([\mathrm{posint},\mathrm{integer},\mathrm{numeric},\mathrm{string}],\mathrm{subtype}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{string}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{posint}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{integer}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{numeric}}]
\mathrm{sort}\left([\mathrm{posint},\mathrm{integer},\mathrm{numeric},\mathrm{string}],\mathrm{rcurry}\left(\mathrm{convert},'\mathrm{truefalse}'\right)@\mathrm{subtype}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{string}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{posint}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{integer}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{numeric}}]
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Tomáš Jakl - Interior spaces and frames
Interior spaces and frames
Today I attended a great talk by Ivan Di Liberti at YaMCATS and I relearned a nice little fact about interior spaces that Ivan showed me already some time ago. Let me write it down here before I forget again.
Ever since the beginning of topology, it was known that topological spaces can be equivalently represented as interior spaces, which are sets
X
equipped with an interior operator
\mathrm{int}\colon \mathcal P(X) \to \mathcal P(X)
\mathcal P(X)
is the powerset of
X
and the interior operator needs to satisfy the following axioms:
\mathrm{int}(M) \subseteq M
M \subseteq N
\mathrm{int}(M) \subseteq \mathrm{int}(N)
\mathrm{int}(\mathrm{int}(M)) = \mathrm{int}(M)
\mathrm{int}(M \cap N) = \mathrm{int}(M) \cap \mathrm{int}(N)
\mathrm{int}(X) = X
These look almost like the axioms of nuclei, except that the order in the first item is reversed and we have the extra assumption that
\mathrm{int}(X) = X
I am sure many people are familiar with with the adjunction between topological spaces and frames. However, what I haven’t seen before is the explicit construction of the adjunction between interior spaces and frames, and it’s rather nice!
Let us first show the mapping from frames to interior spaces. Recall that points
\mathrm{pt}(L)
of a frame
L
can be represented as frame homomorphisms
L \to 2
2
is the two element frame. Take the evaluation function:
L \times \mathrm{pt}(L) \to 2,\quad (a,p\colon L\to 2) \mapsto p(a)
By currying, we have a function
f\colon L \to 2^{\mathrm{pt}(L)} \cong \mathcal P(\mathrm{pt}(L))
It is a standard fact that this function is a frame homomorphism and it thus has a right adjoint
g
f
preserves finite meets, the composition
f \circ g
is an interior operator
\mathcal P(\mathrm{pt}(L)) \to \mathcal P(\mathrm{pt}(L))
(\mathrm{pt}(L), f \cdot g)
is an interior space. So neat!
Conversely, given an interior space
(X,\mathrm{int})
, we simply take the frame
\tau
of fixpoints of
\mathrm{int}
. The fact that this is well defined is precisely the same proof as when showing that
(X,\tau)
is a topological space, in the correspondence between interior spaces and topological spaces.
In fact, Ivan showed something much more sophisticated. Instead of frames he worked with Grothendieck toposes and instead of interior spaces he worked with certain (large) ionads, which are pairs
(\mathcal{X},\mathrm{Int})
\mathcal{X}
is a suitable locally small category and
\mathrm{Int}
is a comonad on the full subcategory
\overline{\mathcal P}(\mathcal{X})
[\mathcal{X},\mathrm{Set}]
, consisting of small copresheaves. The construction from Grothendieck toposes is basically as I described above (just replace
2
with the category
\mathrm{Set}
) and to go from Ionads to toposes just take the category of coalgebras for the comonad
\mathrm{Int}
In fact, when specialised to interior spaces and frames, we obtain precisely the same construction as described above. For details, see Ivan’s older slides here or his Ph.D. thesis.
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Divisibility · USACO Guide
HomeGoldDivisibility
ResourcesPrime FactorizationSolution - Counting DivisorsGCD & LCMGCDLCMProblems
Authors: Darren Yao, Michael Cao, Andi Qu, and Kevin Sheng
Contributor: Juheon Rhee
If you've never encountered any number theory before, AoPS is a good place to start.
practice problems, set focus to number theory!
13.1, 13.2 - Elementary Number Theory
21.1 - Primes & Factors
16.1, 16.2 - Number Theory
1, 3.1, and 3.2 - Divisors
nice proofs and problems
a
is called a divisor or a factor of a non-negative integer
b
b
a
, which means that there exists some integer
k
b = ka
n > 1
is prime if its only divisors are
1
n
. Integers greater than
1
that are not prime are composite.
Every positive integer has a unique prime factorization: a way of decomposing it into a product of primes, as follows:
n = {p_1}^{a_1} {p_2}^{a_2} \cdots {p_k}^{a_k}
p_i
a_i
Now, we will discuss how to find the prime factorization of any positive integer.
vector<int> factor(int n) {
if (n > 1) ret.push_back(n);
ArrayList<Integer> factor(int n) {
ArrayList<Integer> factors = new ArrayList<>();
if (n > 1) factors.add(n);
This algorithm runs in
\mathcal{O}(\sqrt{n})
time, because the for loop checks divisibility for at most
\sqrt{n}
values. Even though there is a while loop inside the for loop, dividing
n
i
quickly reduces the value of
n
, which means that the outer for loop runs less iterations, which actually speeds up the code.
Let's look at an example of how this algorithm works, for
n = 252
i
n
v
2
252
\{\}
2
126
\{2\}
2
63
\{2, 2\}
3
21
\{2, 2, 3\}
3
7
\{2, 2, 3, 3\}
At this point, the for loop terminates, because
is already 3 which is greater than
\lfloor \sqrt{7} \rfloor
. In the last step, we add
7
to the list of factors
v
, because it otherwise won't be added, for a final prime factorization of
\{2, 2, 3, 3, 7\}
Solution - Counting Divisors
The most straightforward solution is just to do what the problem asks us to do - for each
x
, find the number of divisors of
x
\mathcal{O}(\sqrt x)
for (int q = 0; q < n; q++) {
int div_num = 0;
public class Divisors {
int queryNum = Integer.parseInt(read.readLine());
for (int q = 0; q < queryNum; q++) {
div_num = 0
for i in range(1, int(sqrt(x) + 1)):
div_num = div_num + 1 if i ** 2 == x else div_num + 2
print(div_num)
This solution runs in
\mathcal{O}(n \sqrt x)
time, which is just fast enough to get AC. However, we can actually speed this up to get an
\mathcal{O}((x + n) \log x)
solution!
First, let's discuss an important property of the prime factorization. Consider:
x = {p_1}^{a_1} {p_2}^{a_2} \cdots {p_k}^{a_k}
Then the number of divisors of
x
(a_1 + 1) \cdot (a_2 + 1) \cdots (a_k + 1)
Why is this true? The exponent of
p_i
in any divisor of
x
must be in the range
[0, a_i]
and each different exponent results in a different set of divisors, so each
p_i
a_i + 1
to the product.
x
\mathcal{O}(\log x)
distinct prime factors, so if we can find the prime factorization of
x
efficiently, we can answer queries in
\mathcal{O}(\log x)
time instead of the previous
\mathcal{O}(\sqrt x)
Here's how we find the prime factorization of
x
\mathcal{O}(\log x)
\mathcal{O}(x \log x)
preprocessing:
k \leq 10^6
, find any prime number that divides
k
We can use the Sieve of Eratosthenes to find this efficiently.
x
, we can then find the prime factorization by repeatedly dividing
x
by a prime number that divides
x
x = 1
Alternatively, we can slightly modify the the prime factorization code above.
// max_div[i] contains the largest prime that goes into i
int max_div[MAX_N + 1];
for (int i = 2; i <= MAX_N; i++) {
private static final int MAX_N = (int) Math.pow(10, 6);
// maxDiv[i] contains the largest prime that can divide i
int[] maxDiv = new int[MAX_N + 1];
MAX_N = 10 ** 6
# max_div[i] contains the largest prime that can go into i
max_div = [0 for _ in range(MAX_N + 1)]
for i in range(2, MAX_N + 1):
if max_div[i] == 0:
for j in range(i, MAX_N + 1, i):
max_div[j] = i
Apply the linear sieve.
The greatest common divisor (GCD) of two integers
and
b
is the largest integer that is a factor of both
and
b
. In order to find the GCD of two non-negative integers, we use the Euclidean Algorithm, which is as follows:
\gcd(a, b) = \begin{cases} a & b = 0 \\ \gcd(b, a \bmod b) & b \neq 0 \\ \end{cases}
This algorithm is very easy to implement using a recursive function, as follows:
For C++14, you can use the built-in __gcd(a,b).
In C++17, there exists std::gcd and std::lcm in the <numeric> header, so there's no need to code your own GCD and LCM if you're using that.
This function runs in
\mathcal{O}(\log ab)
time because
a\le b \implies b\pmod a <\frac{b}{2}
The worst-case scenario for the Euclidean algorithm is when
and
b
are consecutive Fibonacci numbers
F_n
F_{n + 1}
. for an explanation). In this case, the algorithm will calculate
\gcd(F_n, F_{n + 1}) = \gcd(F_{n - 1}, F_n) = \dots = \gcd(0, F_1)
. This means that finding
\gcd(F_n, F_{n + 1})
n + 1
steps, which is proportional to
\log \left(F_n F_{n+1}\right)
The least common multiple (LCM) of two integers
and
b
is the smallest integer divisible by both
and
b
. The LCM can easily be calculated from the following property with the GCD:
\operatorname{lcm}(a, b) = \frac{a \cdot b}{\gcd(a, b)}=\frac{a}{\gcd(a,b)}\cdot b.
\text{lcm}
as a * b / gcd(a, b) might cause integer overflow if the value of a * b is greater than the max size of the data type of a * b (e.g. the max size of int is around 2 billion). Dividng a by gcd(a, b) first, then multiplying it by b will prevent integer overflow if the result fits in an int.
If we want to take the GCD or LCM of more than two elements, we can do so two at a time, in any order. For example,
\gcd(a_1, a_2, a_3, a_4) = \gcd(a_1, \gcd(a_2, \gcd(a_3, a_4))).
Div Game
Easy Show Tags Prime Factorization
Easy Show Tags Divisibility, Modular Arithmetic
Easy Show Tags NT
Diluc and Kavya
Easy Show Tags Divisibility
Orac and LCM
Normal Show Tags Prime Factorization
Hard Show Tags Divisibility
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Oil Cost - Azur Lane Wiki
Oil cost of ship is based on ship's rarity, hull class, level and limit break.
1 Oil cost equation
2 Max cost
2.1 Fuel Cost Limit
3 At Limit Break level caps
5 Other consumers of oil
Oil cost equation
Oil cost equation for surface ships is:
{\displaystyle {\text{OilCost}}=\left\lfloor {\text{MaxCost}}\cdot {\frac {100+\min \left({\text{Level}},99\right)}{200}}\right\rfloor +1}
For submarines it is:
{\displaystyle {\text{OilCost}}=\left\lfloor ({\text{MaxCost}}+1)\cdot {\frac {100+\min \left({\text{Level}},99\right)}{200}}\right\rfloor }
{\displaystyle MaxCost}
is maximum oil cost at the given limit break.
{\displaystyle Level}
is current level, capping at level 99.
{\displaystyle MaxCost}
is constant for PR/DR/UR/META Ships regardless of their development level/limit break/activation.
Max cost can be determined as follows:
1. Hull type.
SS, SSV CL, AE
AR, BM CA, CB
BB, BBV
2. Rarity.
Common Rare Elite Super Rare PR Ultra Rare DR
For META ships add 1 to the Oil Cost.
3. Limit Break.
For surface ships, add +2 for each Limit Break. PR, DR, UR and META ships are always considered to be at max limit break (+6 cost).
For submarines, add +1 for each Limit Break.
4. Extra modifier.
A few ships also have an extra Oil Cost modifier. Ships not in the table below can be assumed to have 0 extra modifier.
Ships from class
-2 Yuubari Yuubari
-1 County Dorsetshire
Kamikaze Hatakaze, Kamikaze, Matsukaze, Oite
Chao Ho Chao Ho, Ying Swei
+1 Amagi Amagi, Amagi-chan
Hiyou Hiyou, Junyou
Kongou Haruna, Hiei, Hiei-chan, Kirishima, Kongou
Brin Torricelli
Mogami Mikuma, Mogami
Fuel Cost Limit
At chapters 9 till 14 when Clearing Mode is on a maximum fuel (oil) consumption limit is enabled. This makes it so attacking escort nodes or boss nodes cannot take more oil than the map limit regardless of the regular oil cost of fleets. Submarine use also has a limiter.
Mob Node Limit
Boss Node Limit
Sub Fleet Limit
At Limit Break level caps
In this table, the "max cost" refers to the max cost at maximum limit break.
(CN only)
6 (sub) 3 4 5 6 3
This table shows the levels at which oil cost goes up. "Max cost" includes max cost from any limit breaks the ship has and excludes max cost from any limit breaks the ship doesn't have. The rows are arranged so that each limit break moves one row down.
Some cells are colored, indicating that ships outside of CN can only reach that level after limit breaks. This can prevent ships from actually reaching that oil cost. For example, both 13-max-cost LB1 ships and 14-max-cost LB0 ships will reach a maximum of 12 oil cost at their level cap.
Level at which cost is reached
3 (sub) 1 50
4 (sub) 1 20 60
6 (sub) 1 15 43 72
Other consumers of oil
If you leave a battle early, you pay only 1 Oil per ship instead of the full cost.
It costs 10 Oil to enter a map.
Some Commissions cost oil to run.
You can purchase Dorm food with Oil.
Operation Siren Data Logger
Retrieved from ‘https://azurlane.koumakan.jp/w/index.php?title=Oil_Cost&oldid=157825’
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How to Simplify an Improper Fraction: 12 Steps (with Pictures)
1 Using a Model
Fractions are numbers that represent parts of whole numbers. If a fraction has a numerator greater than its denominator it is termed an “improper fraction” and can be simplified as a mixed number (a number that combines a whole number and a fraction). There is nothing wrong with an improper fraction, and in fact in mathematics it is often easier to work with than a mixed number; however, in our daily lives we use mixed numbers more than improper fractions,[1] X Research source so it is helpful to know how to create them.
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Determine whether your fraction is improper. An improper fraction is a fraction that has a larger numerator than denominator.[2] X Research source
{\displaystyle {\frac {10}{4}}}
is an improper fraction, because ten is greater than 4.
Interpret the denominator. The denominator is the number below the fraction bar. It tells you how many equal pieces a whole is divided into.
For example, in the fraction
{\displaystyle {\frac {10}{4}}}
, 4 is the denominator, and it tells you that a whole is divided into four equal parts, or quarters.
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Interpret the numerator. The numerator is the number above the fraction bar. It tells you how many pieces you have.
{\displaystyle {\frac {10}{4}}}
, 10 is the numerator, and it tells you that you have 10 parts, or 10 quarters.
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Draw circles to represent the whole. Divide each whole according to the denominator of your fraction.
For example, if your denominator is 4, then divide each circle you draw into 4 equal pieces, or quarters.
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Shade in pieces according to your numerator. The number in the numerator tells you how many pieces you should shade in.
For example, if your fraction is
{\displaystyle {\frac {10}{4}}}
, you will shade in 10 quarters.
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Count how many whole circles you shaded in. To simplify an improper fraction, you must turn it into a mixed number, which includes a whole number and a fraction together. The number of whole circles you shaded in represents the whole number of your mixed fraction. Write this number down.
For example, for the fraction
{\displaystyle {\frac {10}{4}}}
, you should have shaded in 2 whole circles, so the whole number of your mixed number will be 2.
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Count how many parts of a whole you shaded in. The leftover shaded parts will represent the fraction in your mixed number. Write this fraction next to your whole number, and you have your mixed number.
For the fraction
{\displaystyle {\frac {10}{4}}}
, you should have shaded in
{\displaystyle {\frac {2}{4}}}
of a circle, so the fraction part of your mixed number will be
{\displaystyle {\frac {2}{4}}}
{\displaystyle {\frac {10}{4}}}
{\displaystyle 2{\frac {2}{4}}}
Simplify your answer, if necessary. Sometimes the fraction of your mixed number will need to be reduced before you reach your final, simplified answer.[3] X Research source
For example, if your mixed number is
{\displaystyle 2{\frac {2}{4}}}
, you could simplify it to
{\displaystyle 2{\frac {1}{2}}}
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{\displaystyle {\frac {10}{4}}}
is an improper fraction, because
{\displaystyle 10>4}
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Divide the numerator by the denominator. Remember that the fraction bar can be interpreted as a division symbol.[5] X Research source To simplify an improper fraction, you must turn it into a mixed number, which includes a whole number and a fraction together. The number of times you can divide the numerator evenly by the denominator will be the whole number of your mixed number. Write this number down, and note the remainder.
The denominator will not divide evenly into the numerator. The remainder will be interpreted as the fraction part of your mixed number.
{\displaystyle {\frac {10}{4}}}
complete the calculation
{\displaystyle 10\div 4=2R2}
. So the whole number of your fraction will be
{\displaystyle 2}
Turn the remainder into a fraction. To do this, take the remainder, and place it over the denominator of the original improper fraction. Combine this new fraction with the whole number, and you have your mixed number.
{\displaystyle 10\div 4=2R2}
, so the fraction would be
{\displaystyle {\frac {2}{4}}}
{\displaystyle {\frac {10}{4}}}
{\displaystyle 2{\frac {2}{4}}}
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{\displaystyle 2{\frac {2}{4}}}
{\displaystyle 2{\frac {1}{2}}}
What if I have a big numerator, like 124?
As long as the denominator is less than 124 you have an improper fraction, and you can use the methods presented here to solve. The division method would be easiest, since drawing a model with 124 pieces would take some time.
What if I want to simplify it without making it a mixed number? Like simplifying it into a regular fraction?
You cannot convert an improper fraction into a proper fraction.
How do I reduce 42 by 18 5/9?
First change 42 to 41 9/9. Then subtract 18 from 41, and subtract 5/9 from 9/9. The difference is 23 4/9.
How do I simplify, to the lowest term, 15/6?
Divide both numerator and denominator by the largest factor that both numbers have in common. 15 = 5x3. 6 = 2x3. Therefore, 3 is the largest common factor. Divide 3 into both the numerator and denominator to simplify the fraction to its lowest terms.
How do I simplify 11 28/24?
Change 28/24 to 1 4/24. Combine the "1" with the "11" and we have 12 4/24, which simplifies to 12 1/6.
What is the simplest form of 15/12.
It reduces to 5/4 or 1¼.
How would I simplify 42/24?
First divide both the numerator and the denominator by 2 to get the fraction 21/12. Then divide both the new numerator and new denominator by 3 to get the fraction 7/4. (You can also arrive at the same result by dividing the original numerator and denominator by 6, which is 2 x 3.) That's as far as the fraction will reduce, because there is no whole number that will divide evenly into both the numerator and the denominator of this last fraction. So the final answer is 7/4 or 1¾.
If I multiply both terms of 2/3 by 4, what do I get?
If you multiply both numerator and denominator by 4, you get 8/12 or 2/3. In other words, you're multiplying 2/3 by 4/4, which is multiplying by 1.
How do I simplify 18/21?
Divide the numerator and denominator by 3. That makes 6/7.
To convert a mixed numbers back to improper fraction form, multiply the whole number by the denominator and add the product to the numerator.
Keep your denominator. For example,
{\displaystyle 2{\frac {1}{2}}}
{\displaystyle {\frac {5}{2}}}
{\displaystyle 2\times 2+1=5}
Some improper fractions can also represent whole numbers, such as
{\displaystyle {\frac {24}{3}}}
To simplify an improper fraction, start by turning it into a mixed number by dividing the numerator by the denominator. Then, turn the remainder into a fraction by placing it over the denominator of the original fraction. If necessary, simplify the final fraction to get your answer. For example, if your improper fraction was 10/4, you would start by dividing 10 by 4 to get 2 with a remainder of 2. Then, place the remainder over the denominator of the original fraction to get 2/4, and simplify to get 1/2, making the answer 2 1/2. To learn how to simplify an improper fraction by using a model, keep reading!
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Embeddable anticonformal automorphisms of Riemann surfaces | EMS Press
Embeddable anticonformal automorphisms of Riemann surfaces
Let S be a Riemann surface and f be an automorphism of finite order of S. We call f embeddable if there is a conformal embedding
e : S \to \bf {E}^3
e \circ f \circ e^{-1}
is the restriction to e(S) of a rigid motion. In this paper we show that an anticonformal automorphism of finite order is embeddable if and only if it belongs to one of the topological conjugation classes here described. For conformal automorphisms a similar result was known by R.A. Rüedy [R3].
Antonio F. Costa, Embeddable anticonformal automorphisms of Riemann surfaces. Comment. Math. Helv. 72 (1997), no. 2, pp. 203–215
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Helicity (particle physics) - Wikipedia
Projection of spin along the direction of momentum
This article is about helicity in physics. For other uses, see Helicity (disambiguation).
See also: Chirality (physics)
In physics, helicity is the projection of the spin onto the direction of momentum.
1.1 Comparison with chirality
2 Little group
The angular momentum J is the sum of an orbital angular momentum L and a spin S. The relationship between orbital angular momentum L, the position operator r and the linear momentum (orbit part) p is
{\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }
so L's component in the direction of p is zero. Thus, helicity is just the projection of the spin onto the direction of linear momentum. The helicity of a particle is positive (" right-handed") if the direction of its spin is the same as the direction of its motion and negative ("left-handed") if opposite.
Helicity is conserved.[1] That is, the helicity commutes with the Hamiltonian, and thus, in the absence of external forces, is time-invariant. It is also rotationally invariant, in that a rotation applied to the system leaves the helicity unchanged. Helicity, however, is not Lorentz invariant; under the action of a Lorentz boost, the helicity may change sign. Consider, for example, a baseball, pitched as a gyroball, so that its spin axis is aligned with the direction of the pitch. It will have one helicity with respect to the point of view of the players on the field, but would appear to have a flipped helicity in any frame moving faster than the ball (e.g. a bullet train, as both bullet trains and gyroballs are popular in Japan, while trains are popular in special relativity.)
Comparison with chirality[edit]
In this sense, helicity can be contrasted[2] to chirality, which is Lorentz invariant, but is not a constant of motion for massive particles. For massless particles, the two coincide: the helicity is equal to the chirality, and both are Lorentz invariant, and are constants of motion.
In quantum mechanics, angular momentum is quantized, and thus helicity is quantized as well. Because the eigenvalues of spin with respect to an axis have discrete values, the eigenvalues of helicity are also discrete. For a massive particle of spin S, the eigenvalues of helicity are S, S − 1, S − 2, ..., −S.[3]: 12 In massless particles, not all of these correspond to physical degrees of freedom: for example, the photon is a massless spin 1 particle with helicity eigenvalues −1 and +1, and the eigenvalue 0 is not physically present.[4]
All known spin 1⁄2 particles have non-zero mass; however, for hypothetical massless spin 1⁄2 particles (the Weyl spinors), helicity is equivalent to the chirality operator multiplied by 1⁄2ħ. By contrast, for massive particles, distinct chirality states (e.g., as occur in the weak interaction charges) have both positive and negative helicity components, in ratios proportional to the mass of the particle.
A treatment of the helicity of gravitational waves can be found in Weinberg.[5] In short, they come in only two forms: +2 and −2, while the +1, 0 and −1 helicities are non-dynamical (can be gauged away).
Little group[edit]
In 3 + 1 dimensions, the little group for a massless particle is the double cover of SE(2). This has unitary representations which are invariant under the SE(2) "translations" and transform as eihθ under a SE(2) rotation by θ. This is the helicity h representation. There is also another unitary representation which transforms non-trivially under the SE(2) translations. This is the continuous spin representation.
In d + 1 dimensions, the little group is the double cover of SE(d − 1) (the case where d ≤ 2 is more complicated because of anyons, etc.). As before, there are unitary representations which don't transform under the SE(d − 1) "translations" (the "standard" representations) and "continuous spin" representations.
Helicity basis
Gyroball, a macroscopic object (specifically a baseball) exhibiting an analogous phenomenon
^ Landau, L.D.; Lifshitz, E.M. (2013). Quantum mechanics. A shorter course of theoretical physics. Vol. 2. Elsevier. pp. 273–274. ISBN 9781483187228.
^ Chart in Robert Klauber (2013). Student Friendly Quantum Field Theory ISBN 978-0984513956
^ Troshin, S.M.; Tyurin, N.E. (1994). Spin phenomena in particle interactions. Singapore: World Scientific. ISBN 9789810216924.
^ Thomson (2011). "Handout 13" (PDF). High Energy Physics. Part III, Particles. U.K.: Cambridge U.
^ Steven Weinberg (1972). Gravitation and Cosmology: Principles and Application of the General Theory of Relativit Wiley & Sons. (See chapter 10.)
Povh, Bogdan; Lavelle, Martin; Rith, Klaus; Scholz, Christoph; Zetsche, Frank (2008). Particles and nuclei an introduction to the physical concepts (6th ed.). Berlin: Springer. ISBN 9783540793687.
Schwartz, Matthew D. (2014). "Chirality, helicity and spin". Quantum field theory and the standard model. Cambridge: Cambridge University Press. pp. 185–187. ISBN 9781107034730.
Taylor, John (1992). "Gauge theories in particle physics". In Davies, Paul (ed.). The new physics (1st pbk. ed.). Cambridge, [England]: Cambridge University Press. pp. 458–480. ISBN 9780521438315.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Helicity_(particle_physics)&oldid=1086504533"
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An object of mass 3 kg is moving with a velocity of 5 m/s along-Turito
f\left(x\right)=\left\{\begin{array}{l}\left[x\right] \text{ if }-3<x\le -1\\ |x| \text{ if }-1<x<1\\ |\left[-x\right]| \text{ if }1\le x\le 3\end{array}\text{ then }\left\{x:f\left(x\right)\ge 0\right\}=\right\
f\left(x\right)=\left\{\begin{array}{l}\left[x\right] \text{ if }-3<x\le -1\\ |x| \text{ if }-1<x<1\\ |\left[-x\right]| \text{ if }1\le x\le 3\end{array}\text{ then }\left\{x:f\left(x\right)\ge 0\right\}=\right\
A thin liquid film formed between a u-shaped wire and a light slider supports a weight of (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is.
Three masses 1 kg and 6 kg and 3 kg are connected to each other with threads and are placed on a table as shown in figure. The acceleration with which the system is moving is-ms2(g=10ms2)
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polynomial_series_sol - Maple Help
Home : Support : Online Help : Mathematics : Differential Equations : Slode : polynomial_series_sol
formal power series solutions with polynomial coefficients for a linear ODE
polynomial_series_sol(ode, var,opts)
polynomial_series_sol(LODEstr,opts)
The polynomial_series_sol command returns one formal power series solution or a set of formal power series solutions of the given linear ordinary differential equation with polynomial coefficients. The ODE must be either homogeneous or inhomogeneous with a right-hand side that is a polynomial, a rational function, or a "nice" power series (see LODEstruct) in the independent variable
x
x
x
x
\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{\sum }_{n=0}^{\mathrm{\infty }}v\left(n\right){P}_{n}\left(x\right)
{P}_{n}\left(x\right)
{\left(x-a\right)}^{n}
\frac{{\left(x-a\right)}^{n}}{n!}
\frac{1}{{x}^{n}}
\frac{1}{{x}^{n}n!}
a
v\left(n\right)
v\left(n\right)
v\left(n\right)
is a polynomial for all sufficiently large
n
\mathrm{\infty }
If this option is given, then the command returns one formal power series solution at a with polynomial coefficients if it exists; otherwise, it returns NULL. If a is not given, it returns a set of formal power series solutions with polynomial coefficients for all possible points that are determined by Slode[candidate_points](ode,var,'type'='polynomial').
\mathrm{with}\left(\mathrm{Slode}\right):
\mathrm{ode}≔\left(3{x}^{2}-6x+3\right)\mathrm{diff}\left(\mathrm{diff}\left(y\left(x\right),x\right),x\right)+\left(12x-12\right)\mathrm{diff}\left(y\left(x\right),x\right)+6y\left(x\right)
\textcolor[rgb]{0,0,1}{\mathrm{ode}}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}\right)\textcolor[rgb]{0,0,1}{}\left(\frac{{\textcolor[rgb]{0,0,1}{ⅆ}}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{ⅆ}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)\right)\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{12}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{12}\right)\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{ⅆ}}{\textcolor[rgb]{0,0,1}{ⅆ}\textcolor[rgb]{0,0,1}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)
\mathrm{polynomial_series_sol}\left(\mathrm{ode},y\left(x\right)\right)
{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{\sum }_{\textcolor[rgb]{0,0,1}{\mathrm{_n}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}}^{\textcolor[rgb]{0,0,1}{\mathrm{\infty }}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{_n}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_C}}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{\mathrm{_C}}}_{\textcolor[rgb]{0,0,1}{0}}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{\mathrm{_n}}}}
\mathrm{polynomial_series_sol}\left(\mathrm{ode}=\frac{6\left(180{x}^{2}-150x+25+3{x}^{4}-42{x}^{3}\right)}{{\left(x-5\right)}^{3}},y\left(x\right),'\mathrm{index}'=n\right)
{\textcolor[rgb]{0,0,1}{80}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{24}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{25}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{\sum }_{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}}^{\textcolor[rgb]{0,0,1}{\mathrm{\infty }}}\textcolor[rgb]{0,0,1}{}{\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\right)}^{\textcolor[rgb]{0,0,1}{n}}\right)}
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Metamath - Wikipedia
Formal language and associated computer program
For the study of mathematics using mathematical methods, see Metamathematics.
0.196[1] / 2 January 2021; 16 months ago (2 January 2021)
github.com/metamath/metamath-exe
Computer-assisted proof checking
GNU General Public License (Creative Commons Public Domain Dedication for databases)
Metamath is a formal language and an associated computer program (a proof checker) for archiving, verifying, and studying mathematical proofs.[2] Several databases of proved theorems have been developed using Metamath covering standard results in logic, set theory, number theory, algebra, topology and analysis, among others.[3]
As of February 2022[update], the set of proved theorems using Metamath is one of the largest bodies of formalized mathematics, containing in particular proofs of 74[4] of the 100 theorems of the "Formalizing 100 Theorems" challenge,[5] making it fourth after HOL Light, Isabelle, and Coq, but before Mizar, ProofPower, Lean, Nqthm, ACL2, and Nuprl. There are at least 17 proof verifiers for databases that use the Metamath format.[6]
This project is the first one of its kind that allows for interactive browsing of its formalized theorems database in the form of an ordinary website.[7]
1 Metamath language
2 Metamath proof checker
3 Metamath databases
3.1 Metamath Proof Explorer
3.2 Intuitionistic Logic Explorer
3.3 New Foundations Explorer
3.4 Higher-Order Logic Explorer
3.5 Databases without explorers
3.6 Older explorers
5 Other works connected to Metamath
5.1 Proof checkers
Metamath language[edit]
The Metamath language is a metalanguage, suitable for developing a wide variety of formal systems. The Metamath language has no specific logic embedded in it. Instead, it can simply be regarded as a way to prove that inference rules (asserted as axioms or proven later) can be applied. The largest database of proved theorems follows conventional ZFC set theory and classic logic, but other databases exist and others can be created.
The Metamath language design is focused on simplicity; the language, employed to state the definitions, axioms, inference rules and theorems is only composed of a handful of keywords, and all the proofs are checked using one simple algorithm based on the substitution of variables (with optional provisos for what variables must remain distinct after a substitution is made).[8]
Language basics[edit]
The set of symbols that can be used for constructing formulas is declared using $c (constant symbols) and $v (variable symbols) statements; for example:
The grammar for formulas is specified using a combination of $f (floating (variable-type) hypotheses) and $a (axiomatic assertion) statements; for example:
$( Define "wff" (part 1) $)
Axioms and rules of inference are specified with $a statements along with ${ and $} for block scoping and optional $e (essential hypotheses) statements; for example:
$( State axiom a1 $)
Using one construct, $a statements, to capture syntactic rules, axiom schemas, and rules of inference is intended to provide a level of flexibility similar to higher order logical frameworks without a dependency on a complex type system.
Theorems (and derived rules of inference) are written with $p statements; for example:
$( Prove a theorem $)
th1 $p |- t = t $=
$( Here is its proof: $)
tt tze tpl tt weq tt tt weq tt a2 tt tze tpl
tt weq tt tze tpl tt weq tt tt weq wim tt a2
tt tze tpl tt tt a1 mp mp
Note the inclusion of the proof in the $p statement. It abbreviates the following detailed proof:
tt $f term t
tze $a term 0
1,2 tpl $a term ( t + 0 )
3,1 weq $a wff ( t + 0 ) = t
1,1 weq $a wff t = t
1 a2 $a |- ( t + 0 ) = t
10,11 wim $a wff ( ( t + 0 ) = t -> t = t )
14,1,1 a1 $a |- ( ( t + 0 ) = t -> ( ( t + 0 ) = t -> t = t ) )
8,12,13,15 mp $a |- ( ( t + 0 ) = t -> t = t )
4,5,6,16 mp $a |- t = t
The "essential" form of the proof elides syntactic details, leaving a more conventional presentation:
a2 $a |- ( t + 0 ) = t
a1 $a |- ( ( t + 0 ) = t -> ( ( t + 0 ) = t -> t = t ) )
2,3 mp $a |- ( ( t + 0 ) = t -> t = t )
1,4 mp $a |- t = t
A step-by-step proof
All Metamath proof steps use a single substitution rule, which is just the simple replacement of a variable with an expression and not the proper substitution described in works on predicate calculus. Proper substitution, in Metamath databases that support it, is a derived construct instead of one built into the Metamath language itself.
The substitution rule makes no assumption about the logic system in use and only requires that the substitutions of variables are correctly done.
Here is a detailed example of how this algorithm works. Steps 1 and 2 of the theorem 2p2e4 in the Metamath Proof Explorer (set.mm) are depicted left. Let's explain how Metamath uses its substitution algorithm to check that step 2 is the logical consequence of step 1 when you use the theorem opreq2i. Step 2 states that ( 2 + 2 ) = ( 2 + ( 1 + 1 ) ). It is the conclusion of the theorem opreq2i. The theorem opreq2i states that if A = B, then (C F A) = (C F B). This theorem would never appear under this cryptic form in a textbook but its literate formulation is banal: when two quantities are equal, one can replace one by the other in an operation. To check the proof Metamath attempts to unify (C F A) = (C F B) with ( 2 + 2 ) = ( 2 + ( 1 + 1 ) ). There is only one way to do so: unifying C with 2, F with +, A with 2 and B with ( 1 + 1 ). So now Metamath uses the premise of opreq2i. This premise states that A = B. As a consequence of its previous computation, Metamath knows that A should be substituted by 2 and B by ( 1 + 1 ). The premise A = B becomes 2=( 1 + 1 ) and thus step 1 is therefore generated. In its turn step 1 is unified with df-2. df-2 is the definition of the number 2 and states that 2 = ( 1 + 1 ). Here the unification is simply a matter of constants and is straightforward (no problem of variables to substitute). So the verification is finished and these two steps of the proof of 2p2e4 are correct.
When Metamath unifies ( 2 + 2 ) with B it has to check that the syntactical rules are respected. In fact B has the type class thus Metamath has to check that ( 2 + 2 ) is also typed class.
Metamath proof checker[edit]
The Metamath program is the original program created to manipulate databases written using the Metamath language. It has a text (command line) interface and is written in C. It can read a Metamath database into memory, verify the proofs of a database, modify the database (in particular by adding proofs), and write them back out to storage.
It has a prove command that enables users to enter a proof, along with mechanisms to search for existing proofs.
The Metamath program can convert statements to HTML or TeX notation; for example, it can output the modus ponens axiom from set.mm as:
{\displaystyle \vdash \varphi \quad \&\quad \vdash (\varphi \rightarrow \psi )\quad \Rightarrow \quad \vdash \psi }
Many other programs can process Metamath databases, in particular, there are at least 17 proof verifiers for databases that use the Metamath format.[9]
Metamath databases[edit]
The Metamath website hosts several databases that store theorems derived from various axiomatic systems. Most databases (.mm files) have an associated interface, called an "Explorer", which allows one to navigate the statements and proofs interactively on the website, in a user-friendly way. Most databases use a Hilbert system of formal deduction though this is not a requirement.
Metamath Proof Explorer[edit]
A proof of the Metamath Proof Explorer
us.metamath.org/mpeuni/mmset.html
The Metamath Proof Explorer (recorded in set.mm) is the main and by far the largest database, with over 23,000 proofs in its main part as of July 2019. It is based on classical first-order logic and ZFC set theory (with the addition of Tarski-Grothendieck set theory when needed, for example in category theory). The database has been maintained for over twenty years (the first proofs in set.mm are dated August 1993). The database contains developments, among other fields, of set theory (ordinals and cardinals, recursion, equivalents of the axiom of choice, the continuum hypothesis...), the construction of the real and complex number systems, order theory, graph theory, abstract algebra, linear algebra, general topology, real and complex analysis, Hilbert spaces, number theory, and elementary geometry. This database was first created by Norman Megill, but as of 2019-10-04 there have been 48 contributors (including Norman Megill).[10]
The Metamath Proof Explorer references many text books that can be used in conjunction with Metamath.[11] Thus, people interested in studying mathematics can use Metamath in connection with these books and verify that the proved assertions match the literature.
Intuitionistic Logic Explorer[edit]
This database develops mathematics from a constructive point of view, starting with the axioms of intuitionistic logic and continuing with axiom systems of constructive set theory.
New Foundations Explorer[edit]
This database develops mathematics from Quine's New Foundations set theory.
Higher-Order Logic Explorer[edit]
This database starts with higher-order logic and derives equivalents to axioms of first-order logic and of ZFC set theory.
Databases without explorers[edit]
The Metamath website hosts a few other databases which are not associated with explorers but are nonetheless noteworthy. The database peano.mm written by Robert Solovay formalizes Peano arithmetic. The database nat.mm[12] formalizes natural deduction. The database miu.mm formalizes the MU puzzle based on the formal system MIU presented in Gödel, Escher, Bach.
Older explorers[edit]
The Metamath website also hosts a few older databases which are not maintained anymore, such as the "Hilbert Space Explorer", which presents theorems pertaining to Hilbert space theory which have now been merged into the Metamath Proof Explorer, and the "Quantum Logic Explorer", which develops quantum logic starting with the theory of orthomodular lattices.
Natural deduction[edit]
Because Metamath has a very generic concept of what a proof is (namely a tree of formulas connected by inference rules) and no specific logic is embedded in the software, Metamath can be used with species of logic as different as Hilbert-style logics or sequents-based logics or even with lambda calculus.
However, Metamath provides no direct support for natural deduction systems. As noted earlier, the database nat.mm formalizes natural deduction. The Metamath Proof Explorer (with its database set.mm) instead use a set of conventions that allow the use of natural deduction approaches within a Hilbert-style logic.
Other works connected to Metamath[edit]
Proof checkers[edit]
Using the design ideas implemented in Metamath, Raph Levien has implemented very small proof checker, mmverify.py, at only 500 lines of Python code.
Ghilbert is a similar though more elaborate language based on mmverify.py.[13] Levien would like to implement a system where several people could collaborate and his work is emphasizing modularity and connection between small theories.
Using Levien seminal works, many other implementations of the Metamath design principles have been implemented for a broad variety of languages. Juha Arpiainen has implemented his own proof checker in Common Lisp called Bourbaki[14] and Marnix Klooster has coded a proof checker in Haskell called Hmm.[15]
Although they all use the overall Metamath approach to formal system checker coding, they also implement new concepts of their own.
Mel O'Cat designed a system called Mmj2, which provides a graphic user interface for proof entry.[16] The initial aim of Mel O'Cat was to allow the user to enter the proofs by simply typing the formulas and letting Mmj2 find the appropriate inference rules to connect them. In Metamath on the contrary you may only enter the theorems names. You may not enter the formulas directly. Mmj2 has also the possibility to enter the proof forward or backward (Metamath only allows to enter proof backward). Moreover Mmj2 has a real grammar parser (unlike Metamath). This technical difference brings more comfort to the user. In particular Metamath sometimes hesitates between several formulas it analyzes (most of them being meaningless) and asks the user to choose. In Mmj2 this limitation no longer exists.
There is also a project by William Hale to add a graphical user interface to Metamath called Mmide.[17] Paul Chapman in its turn is working on a new proof browser, which has highlighting that allows you to see the referenced theorem before and after the substitution was made.
Milpgame is a proof assistant and a checker (it shows a message only something gone wrong) with a graphic user interface for the Metamath language(set.mm),written by Filip Cernatescu, it is an open source(MIT License) Java application (cross-platform application: Window,Linux,Mac OS). User can enter the demonstration(proof) in two modes : forward and backward relative to the statement to prove. Milpgame checks if a statement is well formed (has a syntactic verifier). It can save unfinished proofs without the use of dummylink theorem. The demonstration is shown as tree, the statements are shown using html definitions (defined in typesetting chapter). Milpgame is distributed as Java .jar(JRE version 6 update 24 written in NetBeans IDE).
^ "Release 0.196". 2 January 2021. Retrieved 29 June 2021.
^ Megill, Norman; Wheeler, David A. (2019-06-02). Metamath: A Computer Language for Mathematical Proofs (Second ed.). Morrisville, North Carolina, US: Lulul Press. p. 248. ISBN 978-0-359-70223-7.
^ Megill, Norman. "What is Metamath?". Metamath Home Page.
^ Metamath 100.
^ https://www.cs.ru.nl/~freek/100/
^ Megill, Norman. "Known Metamath proof verifiers". Retrieved 14 July 2019.
^ TOC of Theorem List - Metamath Proof Explorer
^ Megill,Norman. "How Proofs Work". Metamath Proof Explorer Home Page.
^ Wheeler, David A. "Metamath set.mm contributions viewed with Gource through 2019-10-04". YouTube. Archived from the original on 2021-12-19.
^ Megill, Norman. "Reading suggestions". Metamath.
^ Liné, Frédéric. "Natural deduction based Metamath system". Archived from the original on 2012-12-28.
^ Levien,Raph. "Ghilbert".
^ Arpiainen, Juha. "Presentation of Bourbaki". Archived from the original on 2012-12-28.
^ Klooster,Marnix. "Presentation of Hmm". Archived from the original on 2012-04-02.
^ O'Cat,Mel. "Presentation of mmj2". Archived from the original on December 19, 2013.
^ Hale, William. "Presentation of mmide". Archived from the original on 2012-12-28.
Metamath: official website.
What do mathematicians think of Metamath: opinions on Metamath.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Metamath&oldid=1084582571"
Large-scale mathematical formalization projects
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Quiz 8: Hypothesis Testing Using the One-Sample T-Test | Quiz+
Hypothesis Testing Using the One-Sample T-Test
Looking for Finite Mathematics / Applied Calculus Homework Help?
When is a t-test used instead of a z-test?
S \bar{X}
Which of the following is one of the requirements of a one-sample t-test?
t_{o b t}
value indicate?
Which of the following requirements is common to both the z-test and the one-sample t-test?
Colin has computed the variance for his 10 raw scores and obtained = 16.37.He then computed
S_{X}^{2}
S \bar{X}
s \bar{x}=\sqrt{\frac{4.046}{10}}=0.636
What is wrong?
Which of the following is not one of the steps in a one-sample t-test?
How is the t-distribution defined?
For a given level of α, when does
t_{c r i t}=Z_{c r i t}
For a two-tailed test with α = 0.05, the
t_{\mathrm{crit}}
Unless we use the correct from the t-distribution for the appropriate N,
What happens to the absolute value of
t_{\mathrm{crit}}
as df increases? It
Which of the following would make it more likely your t-test will be significant?
After determining your one-tailed t-test is significant, what should you do?
The precise location on the dependent measure where we expect our population mean to fall refers to
Interval estimation suggests a ________ where we expect the ________ to fall.
Suppose a poll has been conducted on Americans' favorable attitudes towards a certain issue.If it is reported that Americans are 56%+4% in favor of the issue, which of the following is not a possible value represented within the margin of error?
When is a sample mean likely to represent a particular population mean?
To achieve statistical significance for a one-sample t-test,
Suppose we have a 95% confidence interval of 12.23 and 16.75.What does this mean?
|
Discrete_wavelet_transform Knowpia
Haar waveletsEdit
The first DWT was invented by Hungarian mathematician Alfréd Haar. For an input represented by a list of
{\displaystyle 2^{n}}
numbers, the Haar wavelet transform may be considered to pair up input values, storing the difference and passing the sum. This process is repeated recursively, pairing up the sums to prove the next scale, which leads to
{\displaystyle 2^{n}-1}
differences and a final sum.
Daubechies waveletsEdit
The dual-tree complex wavelet transform (DℂWT)Edit
The dual-tree complex wavelet transform (ℂWT) is a relatively recent enhancement to the discrete wavelet transform (DWT), with important additional properties: It is nearly shift invariant and directionally selective in two and higher dimensions. It achieves this with a redundancy factor of only
{\displaystyle 2^{d}}
, substantially lower than the undecimated DWT. The multidimensional (M-D) dual-tree ℂWT is nonseparable but is based on a computationally efficient, separable filter bank (FB).[4]
The Haar DWT illustrates the desirable properties of wavelets in general. First, it can be performed in
{\displaystyle O(n)}
operations; second, it captures not only a notion of the frequency content of the input, by examining it at different scales, but also temporal content, i.e. the times at which these frequencies occur. Combined, these two properties make the Fast wavelet transform (FWT) an alternative to the conventional fast Fourier transform (FFT).
Time issuesEdit
The discrete wavelet transform has a huge number of applications in science, engineering, mathematics and computer science. Most notably, it is used for signal coding, to represent a discrete signal in a more redundant form, often as a preconditioning for data compression. Practical applications can also be found in signal processing of accelerations for gait analysis,[13][14] image processing,[15][16] in digital communications and many others.[17][18][19]
Example in image processingEdit
{\displaystyle {\begin{bmatrix}1&1&1&1\\1&-i&-1&i\\1&-1&1&-1\\1&i&-1&-i\end{bmatrix}}}
{\displaystyle {\begin{bmatrix}1&1&1&1\\1&1&-1&-1\\1&-1&0&0\\0&0&1&-1\end{bmatrix}}}
{\displaystyle {\begin{aligned}(1,0,0,0)&={\frac {1}{4}}(1,1,1,1)+{\frac {1}{4}}(1,1,-1,-1)+{\frac {1}{2}}(1,-1,0,0)\qquad {\text{Haar DWT}}\\(1,0,0,0)&={\frac {1}{4}}(1,1,1,1)+{\frac {1}{4}}(1,i,-1,-i)+{\frac {1}{4}}(1,-1,1,-1)+{\frac {1}{4}}(1,-i,-1,i)\qquad {\text{DFT}}\end{aligned}}}
{\displaystyle {\begin{aligned}&\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)\\&\left({\frac {1}{2}},{\frac {1}{2}},0,0\right)\qquad {\text{2-term truncation}}\\&\left(1,0,0,0\right)\end{aligned}}}
{\displaystyle {\begin{aligned}&\left({\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}},{\frac {1}{4}}\right)\\&\left({\frac {3}{4}},{\frac {1}{4}},-{\frac {1}{4}},{\frac {1}{4}}\right)\qquad {\text{2-term truncation}}\\&\left(1,0,0,0\right)\end{aligned}}}
Notably, the middle approximation (2-term) differs. From the frequency domain perspective, this is a better approximation, but from the time domain perspective it has drawbacks – it exhibits undershoot – one of the values is negative, though the original series is non-negative everywhere – and ringing, where the right side is non-zero, unlike in the wavelet transform. On the other hand, the Fourier approximation correctly shows a peak, and all points are within
{\displaystyle 1/4}
of their correct value, though all points have error. The wavelet approximation, by contrast, places a peak on the left half, but has no peak at the first point, and while it is exactly correct for half the values (reflecting location), it has an error of
{\displaystyle 1/2}
for the other values.
One level of the transformEdit
The DWT of a signal
{\displaystyle x}
is calculated by passing it through a series of filters. First the samples are passed through a low-pass filter with impulse response
{\displaystyle g}
resulting in a convolution of the two:
{\displaystyle y[n]=(x*g)[n]=\sum \limits _{k=-\infty }^{\infty }{x[k]g[n-k]}}
The signal is also decomposed simultaneously using a high-pass filter
{\displaystyle h}
. The outputs give the detail coefficients (from the high-pass filter) and approximation coefficients (from the low-pass). It is important that the two filters are related to each other and they are known as a quadrature mirror filter.
However, since half the frequencies of the signal have now been removed, half the samples can be discarded according to Nyquist’s rule. The filter output of the low-pass filter
{\displaystyle g}
in the diagram above is then subsampled by 2 and further processed by passing it again through a new low-pass filter
{\displaystyle g}
and a high- pass filter
{\displaystyle h}
with half the cut-off frequency of the previous one, i.e.:
{\displaystyle y_{\mathrm {low} }[n]=\sum \limits _{k=-\infty }^{\infty }{x[k]g[2n-k]}}
{\displaystyle y_{\mathrm {high} }[n]=\sum \limits _{k=-\infty }^{\infty }{x[k]h[2n-k]}}
With the subsampling operator
{\displaystyle \downarrow }
{\displaystyle (y\downarrow k)[n]=y[kn]}
{\displaystyle y_{\mathrm {low} }=(x*g)\downarrow 2}
{\displaystyle y_{\mathrm {high} }=(x*h)\downarrow 2}
However computing a complete convolution
{\displaystyle x*g}
with subsequent downsampling would waste computation time.
Cascading and filter banksEdit
At each level in the above diagram the signal is decomposed into low and high frequencies. Due to the decomposition process the input signal must be a multiple of
{\displaystyle 2^{n}}
{\displaystyle n}
For example a signal with 32 samples, frequency range 0 to
{\displaystyle f_{n}}
and 3 levels of decomposition, 4 output scales are produced:
{\displaystyle 0}
{\displaystyle {f_{n}}/8}
{\displaystyle {f_{n}}/8}
{\displaystyle {f_{n}}/4}
{\displaystyle {f_{n}}/4}
{\displaystyle {f_{n}}/2}
{\displaystyle {f_{n}}/2}
{\displaystyle f_{n}}
Relationship to the mother waveletEdit
The filterbank implementation of wavelets can be interpreted as computing the wavelet coefficients of a discrete set of child wavelets for a given mother wavelet
{\displaystyle \psi (t)}
. In the case of the discrete wavelet transform, the mother wavelet is shifted and scaled by powers of two
{\displaystyle \psi _{j,k}(t)={\frac {1}{\sqrt {2^{j}}}}\psi \left({\frac {t-k2^{j}}{2^{j}}}\right)}
{\displaystyle j}
is the scale parameter and
{\displaystyle k}
is the shift parameter, both which are integers.
Recall that the wavelet coefficient
{\displaystyle \gamma }
{\displaystyle x(t)}
{\displaystyle x(t)}
onto a wavelet, and let
{\displaystyle x(t)}
be a signal of length
{\displaystyle 2^{N}}
. In the case of a child wavelet in the discrete family above,
{\displaystyle \gamma _{jk}=\int _{-\infty }^{\infty }x(t){\frac {1}{\sqrt {2^{j}}}}\psi \left({\frac {t-k2^{j}}{2^{j}}}\right)dt}
Now fix
{\displaystyle j}
at a particular scale, so that
{\displaystyle \gamma _{jk}}
{\displaystyle k}
only. In light of the above equation,
{\displaystyle \gamma _{jk}}
can be viewed as a convolution of
{\displaystyle x(t)}
with a dilated, reflected, and normalized version of the mother wavelet,
{\displaystyle h(t)={\frac {1}{\sqrt {2^{j}}}}\psi \left({\frac {-t}{2^{j}}}\right)}
, sampled at the points
{\displaystyle 1,2^{j},2^{2j},...,2^{N}}
. But this is precisely what the detail coefficients give at level
{\displaystyle j}
of the discrete wavelet transform. Therefore, for an appropriate choice of
{\displaystyle h[n]}
{\displaystyle g[n]}
, the detail coefficients of the filter bank correspond exactly to a wavelet coefficient of a discrete set of child wavelets for a given mother wavelet
{\displaystyle \psi (t)}
As an example, consider the discrete Haar wavelet, whose mother wavelet is
{\displaystyle \psi =[1,-1]}
. Then the dilated, reflected, and normalized version of this wavelet is
{\displaystyle h[n]={\frac {1}{\sqrt {2}}}[-1,1]}
, which is, indeed, the highpass decomposition filter for the discrete Haar wavelet transform.
Time complexityEdit
{\displaystyle g[n]}
{\displaystyle h[n]}
are both a constant length (i.e. their length is independent of N), then
{\displaystyle x*h}
{\displaystyle x*g}
each take O(N) time. The wavelet filterbank does each of these two O(N) convolutions, then splits the signal into two branches of size N/2. But it only recursively splits the upper branch convolved with
{\displaystyle g[n]}
(as contrasted with the FFT, which recursively splits both the upper branch and the lower branch). This leads to the following recurrence relation
{\displaystyle T(N)=2N+T\left({\frac {N}{2}}\right)}
As an example, the discrete Haar wavelet transform is linear, since in that case
{\displaystyle h[n]}
{\displaystyle g[n]}
are constant length 2.
{\displaystyle h[n]=\left[{\frac {-{\sqrt {2}}}{2}},{\frac {\sqrt {2}}{2}}\right]g[n]=\left[{\frac {\sqrt {2}}{2}},{\frac {\sqrt {2}}{2}}\right]}
Other transformsEdit
The multiplicative (or geometric) discrete wavelet transform [25] is a variant that applies to an observation model
{\displaystyle {\bf {y}}=f{\bf {X}}}
involving interactions of a positive regular function
{\displaystyle f}
and a multiplicative independent positive noise
{\displaystyle X}
{\displaystyle \mathbb {E} X=1}
{\displaystyle {\cal {W}}}
, a wavelet transform. Since
{\displaystyle f{\bf {X}}=f+{f({\bf {X}}-1)}}
, then the standard (additive) discrete wavelet transform
{\displaystyle {\cal {W^{+}}}}
{\displaystyle {\cal {W^{+}}}{\bf {y}}={\cal {W^{+}}}f+{\cal {W^{+}}}{f({\bf {X}}-1)},}
where detail coefficients
{\displaystyle {\cal {W^{+}}}{f({\bf {X}}-1)}}
cannot be considered as sparse in general, due to the contribution o{\displaystyle f}
in the latter expression. In the multiplicative framework, the wavelet transform is such that
{\displaystyle {\cal {W^{\times }}}{\bf {y}}=\left({\cal {W^{\times }}}f\right)\times \left({\cal {W^{\times }}}{\bf {X}}\right).}
This `embedding' of wavelets in a multiplicative algebra involves generalized multiplicative approximations and detail operators: For instance, in the case of the Haar wavelets, then up to the normalization coefficient
{\displaystyle \alpha }
, the standard
{\displaystyle {\cal {W^{+}}}}
approximations (arithmetic mean)
{\displaystyle c_{k}=\alpha (y_{k}+y_{k-1})}
and details (arithmetic differences)
{\displaystyle d_{k}=\alpha (y_{k}-y_{k-1})}
become respectively geometric mean approximations
{\displaystyle c_{k}^{\ast }=(y_{k}\times y_{k-1})^{\alpha }}
and geometric differences (details)
{\displaystyle d_{k}^{\ast }=\left({\frac {y_{k}}{y_{k-1}}}\right)^{\alpha }}
{\displaystyle {\cal {W^{\times }}}}
Example of above codeEdit
The wavelet transform is a multiresolution, bandpass representation of a signal. This can be seen directly from the filterbank definition of the discrete wavelet transform given in this article. For a signal of length
{\displaystyle 2^{N}}
, the coefficients in the range
{\displaystyle [2^{N-j},2^{N-j+1}]}
represent a version of the original signal which is in the pass-band
{\displaystyle \left[{\frac {\pi }{2^{j}}},{\frac {\pi }{2^{j-1}}}\right]}
. This is why zooming in on these ranges of the wavelet coefficients looks so similar in structure to the original signal. Ranges which are closer to the left (larger
{\displaystyle j}
in the above notation), are coarser representations of the signal, while ranges to the right represent finer details.
|
Dictionary:Gaussian distribution - SEG Wiki
(gaus' ē ∂n) A normal or bell-shaped distribution. A set of values so distributed about a mean value m that the probability
{\displaystyle \varepsilon (\Delta a)}
of a value lying within a small interval
{\displaystyle \Delta a}
centered at the point a is
{\displaystyle \varepsilon (\Delta a)={\text{erf}}(\Delta a)={\frac {e^{-\left(a-m\right)^{2}}\Delta a}{\sigma {\sqrt {2\pi }}}}}
{\displaystyle \sigma }
is the standard error and
{\displaystyle \varepsilon (\Delta a)}
is called the error function.
Retrieved from "https://wiki.seg.org/index.php?title=Dictionary:Gaussian_distribution&oldid=95819"
|
Accountancy Dk Goel 2018 for Class 11 Commerce Accountancy Chapter 4 - Accounting For Goods & Service Tax Gst
accounting for goods & service tax gst
Accountancy Dk Goel 2018 Solutions for Class 11 Commerce Accountancy Chapter 4 Accounting For Goods & Service Tax Gst are provided here with simple step-by-step explanations. These solutions for Accounting For Goods & Service Tax Gst are extremely popular among Class 11 Commerce students for Accountancy Accounting For Goods & Service Tax Gst Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Accountancy Dk Goel 2018 Book of Class 11 Commerce Accountancy Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Accountancy Dk Goel 2018 Solutions. All Accountancy Dk Goel 2018 Solutions for class Class 11 Commerce Accountancy are prepared by experts and are 100% accurate.
Pass entries in the books of Mukerjee & Sons. assuming all transactions have taken place within the state of Uttar Pradesh. Assume CGST @9% and SGST @ 9%.
March 1 Purchased goods for ₹ 5,00,000 from Mehta Bros.
March 10 Sold goods for ₹ 8,00,000 to Munjal & Co.
March 15 Paid for advertisement ₹ 40,000 by cheque.
March 18 Purchased furniture for office use ₹ 50,000 and payment made by cheque.
March 25 Paid for printing and stationery ₹ 8,000.
March 31 Payment made of balance amount of GST.
To Mehta Bros. A/c
(Purchased from Mehta Bros.)
Munjal & Co. A/c
(Sold goods to Munjal & Co.)
(Paid for advertisement)
(Purchased office furniture)
Printing & Stationery A/c
(Paid for printing and stationery)
(Input tax credit availed)
Output SGST A./c
(Balance tax paid to government)
Pass entries in the books of Devdhar & Bros. Odisha, assuming all transactions have been entered within the state, charging CGST and SGST @ 9% each:
March 4 Purchased goods for ₹ 5,00,000 from Sunil Bros.
7 Goods returned to Sunil Bros. for ₹ 20,000
10 Sold goods to Mehta & Co. for ₹ 8,00,000
12 Goods returned by Mehta & Co. for ₹ 30,000
20 Goods withdrawn by Proprietor for personal use ₹ 10,000
25 Goods distributed as free samples ₹ 5,000
26 Paid advertisement expenses by cheque ₹ 20,000
31 Payment made of balance amount of GST.
To Sunil Bros. A/c
(Purchased from Sunil Bros.)
Sunil Bros. A/c
(Goods returned to Sunil Bros.)
Mehta& Co. A/c
(Sold goods to Mehta& Co.)
To Mehta & Co. A/c
(Goods returned by Mehta & Co.)
Free Samples A/c
(Goods distributed as free samples)
Pass entries in the books of Ganguli & Sons. assuming all transactions have been entered in the state of West Bengal:
(i) Purchased goods for ₹ 2,00,000 and payment made by cheque.
(ii) Sold goods for ₹ 1,60,000 to Devki Nandan & Sons.
(iii) Purchased goods for ₹ 50,000 on credit.
(iv) Paid for printing and stationery ₹ 4,000.
(v) Received interest ₹ 5,000.
(vi) Paid for advertisement ₹ 30,000 by cheque.
Assume CGST @6% and SGST @6%.
To Cheque A/c
(Purchased goods and paid by cheque)
Devki Nandan & Sons A/c
(Sold goods toDevki Nandan & Sons)
(Purchased goods on credit)
Input CSGT (Balance) = Rs 7,140
Input SGST (Balance) = Rs 7,140
Note: The answer doesn’t match with the book. If the last transaction is ignored, then the answer shall match. It seems like there is some printing error in the textbook.
Record the following transactions in the books of Sahdev & Sons assuming all transactions have been entered within the state of Bihar, Charging CGST and SGST @ 9% each.
(i) Bought goods from Nanak Bros. for ₹ 4,00,000 at 10% trade discount and 3% cash discount on purchase price. 25% of the amount paid at the time of purchase.
(ii) Sold goods to Kumar & Sons. for ₹ 2,00,000 at 20% trade discount and 5% cash discount on sale price. 60% of the amount received by Cheque.
(iii) Received from Gopi Chand ₹ 38,000 by Cheque after deducting 5% cash discount.
(iv) Paid ₹ 20,000 for rent by Cheque.
(v) Paid ₹ 50,000 for salaries by Cheque.
(vi) Goods worth ₹ 10,000 distributed as free samples.
(vii) ₹ 5,000 due from Chanderkant are bad-debts.
(viii) Sold household furniture for ₹ 15,000 and the proceeds were invested into business.
To Nanak Bros. A/c
(Purchased goods)
Kumar & Sons A/c
(Sold goods)
To Gopi Chand A/c
(Received from Gopi Chand in full settlement)
(Paid salaries)
To Chanderkant A/c
(Debtor proved bad)
(Invested money into business)
Pass entries in the books of Mr. Roopani of Gujarat assuming CGST @9% and SGST@9%.
(i) Purchased goods for ₹ 2,00,000 from Suryakant of Jaipur (Rajasthan) on Credit.
(ii) Sold goods for ₹ 1,50,000 to Mr. Pawar of Mumbai (Maharashtra) and the cheque received was sent to bank.
(iii) Sold goods for ₹ 2,50,000 within the state on credit.
(iv) Paid insurance premium of 20,000 by cheque.
(v) Purchased furniture for office for ₹ 60,000 by cheque.
To Suryakant’s A/c
(Sold goods and received cheque)
(Paid for insurance premium)
(Input Tax credit up to Rs.27,000 availed and balance to be adjusted against Output CGST)
(Input tax credit availed and balance paid)
GST Common Set Off Procedure:
Pass entries in the books of Sh. Jagdish Mishra of Lucknow (U.P.) assuming CGST @ 6% and SGST @ 6%:
March 5 Purchased goods for ₹ 2,50,000 from Virender Yadav of Patna (Bihar).
March 12 Sold goods costing ₹ 60,000 at 50% profit to Partap Sinha of Ranchi (Jharkhand).
March 14 Purchased goods for ₹ 70,000 from Ram Nath of Kanpur (U.P.) against cheque.
March 18 Sold goods at Varanasi (U.P.) Costing ₹ 2,25,000 at
33\frac{1}{3}%
profit less trade discount 10% against cheque which was deposited into bank.
March 20 Paid rent ₹ 25,000 by cheque.
To Virender Yadav A/c
Partap Sinha A/c
(Sold goods on credit)
(Purchased goods against cheque)
(Sold goods against cheque)
(Paid rent by cheque)
Pass entries in the books of all parties in the following cases assuming CGST @ 6% and SGST @ 6%:
March 1 Mahesh Chandra of Bihar purchased goods for ₹ 1,00,000 from Sunil Soren of Jharkhand and sold the same to Deepak Patnaik of Odisha for ₹ 1,50,000.
March 5 Deepak Patnaik sold goods to Suresh Yadav of Odisha for ₹ 1,80,000.
March 10 Suresh Yadav sold goods to Ravi Chakravarti of West Bengal for ₹ 2,50,000.
March 14 Ravi Chakravarti sold goods costing ₹ 2,50,000 to Sanjay Diwedi of West Bengal at a profit of 40% on cost.
Journal of Mahesh Chand, Bihar
To Sunil Soren
(Purchased goods plus 12% IGST)
(Sold goods plus 12% IGST)
Journal of Sunil Soren, Jharkhand
Journal of Deepak Patnaik, Odisha
To Mahesh Chand
(Sold goods plus 6% CGST and SGST each)
Journal of Suresh Yadav, Odisha
To Deepak Patnaik
Journal of Ravi Chakravarti, West Bengal
To Suresh Yadav
(Sold goods costing ₹. 2,50,000 at 40% Profit plus 6% CGST and SGST each)
Pass entries in the books of Krishnan of Bengaluru (Karnataka) in the following cases:
I Purchased goods from Karunakaran of Chennai for ₹ 1,00,000.
(IGST @18%)
II Sold goods to Ganeshan of Bengaluru for ₹ 1,50,000.
(CGST @6% and SGST @6%)
III Sold goods to S. Nair of Kerala for ₹ 2,60,000.
IV Purchased a Motor-bike for ₹ 80,000 from Bajaj Ltd. against cheque.
V Paid rent ₹ 30,000 by cheque.
VI Purchased goods from Ram Mohan Rai of Bengaluru for ₹ 2,00,000.
VII Paid insurance premium ₹ 10,000 by cheque.
VIII Received commission ₹ 20,000 by cheque which is deposited into bank.
IX Payment made of balance amount of GST.
To Karunakaran A/c
Ganeshan A/c
S. Nair A/c
(Purchased Motor bike against cheque)
To Ram Mohan Rai A/c
(Paid insurance premium by cheque)
(Received commission and deposited in bank)
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Calculus/Integral Test for Convergence - Wikibooks, open books for an open world
Calculus/Integral Test for Convergence
1 Integral Test
Integral TestEdit
The next test for convergence for infinite series is the integral test. The integral test utilizes the fact that an integral is essentially an Riemann Sum—which is itself an infinite sum—over an infinite interval which is useful because integration is relatively straight forward and familiar.
Integral Test for Convergence and Divergence
{\displaystyle S=\sum _{n=j}^{\infty }{s_{n}}}
, and if the discrete function
{\displaystyle s(n)}
is continuous and decreasing on the interval
{\displaystyle [j,\infty )}
{\displaystyle \int _{j}^{\infty }{s(n)dn}}
is convergent, so is
{\displaystyle S}
{\displaystyle \int _{j}^{\infty }{s(n)dn}}
is divergent, so is
{\displaystyle S}
Let us look at why this this test works. As an example, we'll use the harmonic series
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}}
. The harmonic series is well known series that actually happens to be divergent. This series actually passes the limit test, however it does not pass the integral test. If we want to approximate the integral, we can use rectangles like in a Riemann Sum:
Note that this right-handed method will always under approximate the integral (assuming that the function is decreasing on our selected interval). This implies that if the right-handed sum is equal to the actual infinite series, then the integral itself must be greater than the sum. This can help show convergence, because if the integral from the starting point to infinity is convergent, then by the comparison test the original function on that interval must also be convergent. And so we can see that the integral test is actually a "special case" of the comparison test.
But what about divergence? This case is also satisfied—if we use a left-handed approximation instead of a right handed one, we see that we again attain the original series, however there is an important difference:
The key difference, in this case, is that the integral becomes an under approximation for the series, and we can use the new "series" of the integral to show divergence with the comparison test.
This test is useful but is unfortunately only useful on functions that can be integrated and are decreasing in size. The latter may seem like a trivial and unnecessary addition, however consider how this test works; it relies on the fact that the integral of a function decreasing on an interval will always yield an under/over approximation of a series; if the function is not decreasing everywhere on the interval, the integral will not necessarily yield an under/over approximation every time.
Use the integral test to determine if the following series is convergent or divergent:
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}
The improper integral from
{\displaystyle 1}
{\displaystyle \lim _{x\rightarrow \infty }{\frac {-1}{n}}-{\frac {-1}{(1)}}}
. The limit of the first term is zero, and so because the integral converges the series does too.
Determine if the series
{\displaystyle \sum _{x=1}^{\infty }{\frac {x^{2}+1}{x}}}
This series does not pass the first requirement that the series is decreasing over the desired interval;
{\displaystyle \lim _{x\rightarrow \infty }{\frac {x^{2}+1}{x}}=\infty }
. Applying the integral test would still show the series' convergence.
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(n-3)^{2}+1}}}
However, this series is not decreasing everywhere on the interval. However, it has a relative maximum at
{\displaystyle n=1}
, after which it does decrease forever. And so we can write this series as
{\displaystyle \sum _{n=1}^{2}{\frac {1}{(n-3)^{2}+1}}+\sum _{n=3}^{\infty }{\frac {1}{(n-3)^{2}+1}}}
. Integrating the function yields the improper integral
{\displaystyle \tan ^{-1}(n-3)}
{\displaystyle 3}
to infinity, which converges, and so the series converges too.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Calculus/Integral_Test_for_Convergence&oldid=3679950"
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Measurements in Separated and Transitional Boundary Layers Under Low-Pressure Turbine Airfoil Conditions | J. Turbomach. | ASME Digital Collection
Ralph J. Volino,
United States Naval Academy, Department of Mechanical Engineering, Annapolis, MD 21402
National Aeronautics and Space Administration, Glenn Research Center at Lewis Field, Cleveland, OH 44135
Volino, R. J., and Hultgren, L. S. (February 1, 2000). "Measurements in Separated and Transitional Boundary Layers Under Low-Pressure Turbine Airfoil Conditions ." ASME. J. Turbomach. April 2001; 123(2): 189–197. https://doi.org/10.1115/1.1350408
Detailed velocity measurements were made along a flat plate subject to the same dimensionless pressure gradient as the suction side of a modern low-pressure turbine airfoil. Reynolds numbers based on wetted plate length and nominal exit velocity were varied from 50,000 to 300,000, covering cruise to takeoff conditions. Low and high inlet free-stream turbulence intensities (0.2 and 7 percent) were set using passive grids. The location of boundary-layer separation does not depend strongly on the free-stream turbulence level or Reynolds number, as long as the boundary layer remains nonturbulent prior to separation. Strong acceleration prevents transition on the upstream part of the plate in all cases. Both free-stream turbulence and Reynolds number have strong effects on transition in the adverse pressure gradient region. Under low free-stream turbulence conditions, transition is induced by instability waves in the shear layer of the separation bubble. Reattachment generally occurs at the transition start. At
Re=50,000
the separation bubble does not close before the trailing edge of the modeled airfoil. At higher Re, transition moves upstream, and the boundary layer reattaches. With high free-stream turbulence levels, transition appears to occur in a bypass mode, similar to that in attached boundary layers. Transition moves upstream, resulting in shorter separation regions. At Re above 200,000, transition begins before separation. Mean velocity, turbulence, and intermittency profiles are presented.
gas turbines, boundary layer turbulence, velocity measurement, laminar to turbulent transitions, bubbles, pressure control, flow separation
Airfoils, Boundary layers, Bubbles, Pressure, Separation (Technology), Turbines, Turbulence, Flow (Dynamics), Reynolds number, Pressure gradient, Shear (Mechanics)
Hourmouziadis, J., 1989, “Aerodynamic Design of Low Pressure Turbines,” AGARD Lecture Series 167.
The Role of Laminar–Turbulent Transition in Gas Turbine Engines
Sharma, O. P., Ni, R. H., and Tanrikut, S., 1994, “Unsteady Flow in Turbines,” AGARD-LS-195, Paper No. 5.
Morkovin, M. V., 1978, “Instability, Transition to Turbulence and Predictability,” NATO AGARDograph No. 236.
Free-Stream Turbulence and Concave Curvature Effects on Heated Transitional Boundary Layers
Boundary Layer Transition Under High Free-Stream Turbulence and Strong Acceleration Conditions: Part 1: Mean Flow Results; Part 2: Turbulent Transport Results
Boundary-Layer Transition in Accelerating Flow With Intense Freestream Turbulence: Part 1: Disturbances Upstream of Transition Onset,” ASME
Transition in a Separation Bubble
Boundary Layer Development in Axial Compressors and Turbines: Part 3 of 4—LP Turbines
Qiu, S., and Simon, T. W., 1997, “An Experimental Investigation of Transition as Applied to Low Pressure Turbine Suction Surface Flows,” ASME Paper No. 97-GT-455.
Murawski, C. G., Sondergaard, R., Rivir, R. B., Simon, T. W., Vafai, K., and Volino, R. J., 1997, “Experimental Study of the Unsteady Aerodynamics in a Linear Cascade With Low Reynolds Number Low Pressure Turbine Blades,” ASME Paper No. 97-GT-95.
Experimental Investigation of Boundary Layer Behavior in a Simulated Low Pressure Turbine
Dorney, D. J., Ashpis, D. E., Halstead, D. E., and Wisler, D. C., 1999, “Study of Boundary Layer Development in a Two-Stage Low-Pressure Turbine,” AIAA Paper No. 99-0742; also NASA TM-1999-208913.
Chernobrovkin, A., and Lakshminarayana, B., 1999, “Turbulence Modeling and Computation of Viscous Transitional Flow for Low Pressure Turbines,” Proc. 4th International Symposium on Engineering Turbulence Modeling and Measurements, Corsica, France.
Huang, P. G., and Xiong, G., 1998, “Transition and Turbulence Modeling of Low Pressure Turbine Flows,” AIAA Paper No. 98-0039.
Sohn, K. H., and Reshotko, E., 1991, “Experimental Study of Boundary Layer Transition With Elevated Freestream Turbulence on a Heated Flat Plate,” NASA CR 187068.
Hultgren, L. S., and Volino, R. J., 2000, “Measurements in Separated and Transitional Boundary Layers Under Low-Pressure Turbine Airfoil Conditions,” NASA TM to be published.
Turbulent/Non-Turbulent Decisions in an Intermittent Flow
Kestoras
Fluid Mechanics and Heat Transfer Measurements in Transitional Boundary Layers Conditionally Sampled on Intermittency
Volino, R. J., 1998, “Wavelet Analysis of Transitional Flow Data Under High Free-Stream Turbulence Conditions,” ASME Paper No. 98-GT-289.
Velocity and Temperature Profiles in Turbulent Boundary Layers Experiencing Streamwise Pressure Gradients
Narasimha, R., 1984, “Subtransitions in the Transition Zone,” Proc. 2nd IUTAM Symposium on Laminar–Turbulent Transition, Novosibirsk, pp. 141–151.
Some Properties of Boundary Layer Flow During the Transition From Laminar to Turbulent Motion
Narasimha, R., 1998, “Post-Workshop Summary,” Minnowbrook II—1997 Workshop on Boundary Layer Transition in Turbomachines, LaGraff, J. E., and Ashpis, D. E., eds., NASA CP 1998-206958, pp. 485–495.
Approximate Calculations of the Laminar Boundary Layer
Davis, R. L., Carter, J. E., and Reshotko, E., 1985, “Analysis of Transitional Separation Bubbles on Infinite Swept Wings,” AIAA Paper No. 85-1685.
, 1936, “Air Flow in the Boundary Layer Near a Plate,” NACA Report 562.
An Experimental Investigation of Transition as Applied to Low Pressure Turbine Suction Surface Flows
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{\displaystyle {\text{LevelAdvantage}}={\text{AlliedShipLevel}}-{\text{EnemyShipLevel}}}
{\displaystyle {\text{LevelAdvantage}}={\text{AlliedShipLevel}}-{\text{EnemyShipLevel}}+{\text{MaximumDangerLevel}}}
{\displaystyle -25\leq {\text{LevelAdvantage}}\leq 25}
{\displaystyle {\text{DamageReduction}}=1-{\dfrac {100-2\times {\text{LevelAdvantage}}}{100}}\times {\dfrac {100-2\times ({\text{MaximumDangerLevel}}-{\text{DangerLevel}})}{100}}}
{\displaystyle {\text{DamageBonus}}={\dfrac {2\times {\text{LevelAdvantage}}}{100}}}
{\displaystyle {\begin{aligned}{\text{ACV}}&={\frac {\text{Aviation}}{100}}\times \left({\dfrac {\text{ShipLevel}}{100}}+0.8\right)\\&\times ({\text{Fighters}}\times 10+{\text{DiveBombers}}\times 6+{\text{TorpedoBombers}}\times 5+{\text{Seaplanes}}\times 4)+{\text{Extra}}\end{aligned}}}
{\displaystyle {\text{EffectiveEnemyACV}}={\text{InitialEnemyACV}}\times {\dfrac {8000}{{\text{TotalAlliedAA}}+8000}}}
{\displaystyle {\dfrac {\text{AllyACV}}{\text{EffectiveEnemyACV}}}={\begin{cases}{\text{Air Supremacy}}&X>2\\{\text{Air Superiority}}&1.3<X\leq 2\\{\text{Air Parity}}&0.75<X\leq 1.3\\{\text{Air Denial}}&0.5<X\leq 0.75\\{\text{Air Incapability}}&0<X\leq 0.5\end{cases}}}
{\displaystyle {\text{FleetMovement}}(X)=({\text{AverageFleetSpeed}})\times \left(1-{\dfrac {{\text{NumberofShips}}-1}{50}}\right)}
{\displaystyle {\text{TileMovement}}={\begin{cases}2&X\leq 25\\3&25<X\leq 36\\4&36<X\end{cases}}}
{\displaystyle {\text{ReconValue}}={\sqrt[{3}]{({\text{TotalEvasion}}+{\text{TotalAviation}})^{2}}}=({\text{TotalEvasion}}+{\text{TotalAviation}})^{2/3}}
{\displaystyle {\text{AmbushAvoidanceRate}}={\dfrac {({\text{TotalEvasion}})^{2/3}}{({\text{TotalEvasion}})^{2/3}+{\text{MapAvoidanceValue}}}}+{\text{EquipmentRate}}}
{\displaystyle {\text{EncounterRate}}={\dfrac {({\text{MapSurveyValue}})\times ({\text{Steps}}-1)}{4\times ({\text{MapSurveyValue}}+{\text{ReconValue}})}}-{\text{EquipmentRate}}+{\text{BaseAmbushRate}}\times (0.05)}
{\displaystyle {\begin{aligned}{\text{DamagePerTick}}&=\left({\text{WeaponBaseDamage}}\times {\text{WeaponCoefficient}}\times {\text{SlotEfficiency}}\times {\dfrac {100+{\text{Firepower}}\times (1+{\text{FormationBonus}}+{\text{SkillBonus}})}{100}}\times {\text{BurnCoefficient}}+{\text{BurnDamage}}\right)\\&\times (1+\mathrm {SkillBonus} )\end{aligned}}}
{\displaystyle {\begin{aligned}{\text{DamagePerTick}}&={\text{WeaponBaseDamage}}\times {\text{WeaponCoefficient}}\times {\text{SlotEfficiency}}\\&\times {\dfrac {100+{\text{CorrespondingStat}}\times (1+{\text{FormationBonus}}+{\text{SkillBonus}})}{100}}\times {\text{FloodCoefficient}}+{\text{FloodDamage}}\end{aligned}}}
These are damage values for single shot/torpedo/bomb of the weapon. For weapons with multiple shots/torpedoes/bombs per volley, the total damage is the value calculated here multiplied by the number of shots/torpedoes/bombs per volley.
{\displaystyle {\text{CriticalBit}}}
{\displaystyle 1}
if the damage is a critical hit, and
{\displaystyle 0}
{\displaystyle {\begin{aligned}{\text{FinalDamage}}&=\left[{\text{WeaponBaseDamage}}\times {\text{WeaponCoefficient}}\times {\text{SlotEfficiency}}\times \left(1+{\frac {\text{Firepower}}{100}}\times (1+{\text{FormationBonus}}+{\text{FPSkillBonus}})\right)+{\text{random}}(\{0,1,2\})\right]\\&\times {\text{ArmorModifier}}\times {\text{LevelAdvantage}}\times (1+{\text{AmmoBuff}}+{\text{DamageBonus}})\\&\times (1+{\text{AmmoTypeModifier}})\times (1+{\text{EnemyDebuff}})\times (1+{\text{HunterSkill}})\times (1+{\text{ManualBit}}\times (0.2+{\text{ManualModifier}}))\\&\times \left(1+{\text{CriticalBit}}\times (0.5+{\text{CriticalModifier}})\right)\end{aligned}}}
{\displaystyle {\begin{aligned}{\text{FinalDamage}}&=\left[{\text{WeaponBaseDamage}}\times {\text{WeaponCoefficient}}\times {\text{SlotEfficiency}}\times \left(1+{\frac {\text{Torpedo}}{100}}\times (1+{\text{FormationBonus}}+{\text{TRPSkillBonus}})\right)+{\text{random}}(\{0,1,2\})\right]\\&\times {\text{ArmorModifier}}\times {\text{LevelAdvantage}}\times (1+{\text{AmmoBuff}}+{\text{DamageBonus}})\\&\times (1+{\text{AmmoTypeModifier}})\times (1+{\text{EnemyDebuff}})\times (1+{\text{HunterSkill}})\\&\times \left(1+{\text{CriticalBit}}\times (0.5+{\text{CriticalModifier}})\right)\end{aligned}}}
{\displaystyle {\begin{aligned}{\text{FinalDamage}}&=\left[{\text{LoadDamage}}\times {\text{SlotEfficiency}}\times {\frac {100+{\text{AirPower}}\times (1+{\text{StatBonus}})}{100}}+{\text{random}}(\{0,1,2\})\right]\\&\times {\text{ArmorModifier}}\times {\text{LevelAdvantage}}\times (1+{\text{AmmoBuff}}+{\text{DamageBonus}})\times (1+{\text{ACVBonus}})\\&\times (1-{\text{EnemyAirDamageResistance}})\times (1+{\text{EnemyDebuff}})\\&\times \left(1+{\text{CriticalBit}}\times (0.5+{\text{CriticalModifier}})\right)\end{aligned}}}
{\displaystyle {\begin{aligned}{\text{FinalDamage}}&=\left[{\text{LoadDamage}}\times {\text{SlotEfficiency}}\times {\frac {100+{\text{AirPower}}\times 0.8\times (1+{\text{StatBonus}})}{100}}+{\text{random}}(\{0,1,2\})\right]\\&\times {\text{ArmorModifier}}\times {\text{LevelAdvantage}}\times (1+{\text{AmmoBuff}}+{\text{DamageBonus}})\times (1+{\text{ACVBonus}})\\&\times (1-{\text{EnemyAirDamageResistance}})\times (1+{\text{EnemyDebuff}})\\&\times \left(1+{\text{CriticalBit}}\times (0.5+{\text{CriticalModifier}})\right)\end{aligned}}}
{\displaystyle {\begin{aligned}{\text{FinalDamage}}&=\left({\text{CrashDamage}}\times {\frac {100+{\text{AirPower}}}{100}}+{\frac {\text{ShipLevel}}{2}}\right)\times ({\text{CurrentPlaneHP}}\%\times 0.7+0.3)\\&\times {\text{LevelAdvantage}}\times (1+{\text{EnemyDebuff}})\times (1+{\text{DamageBonus}})\\&\times (1-{\text{EnemyDefenseBuff}})\times {\text{EnemyAirDamageResistance}}\end{aligned}}}
{\displaystyle {\begin{aligned}{\text{CriticalRate}}&=0.05+\left({\dfrac {\text{AttackerHit}}{{\text{AttackerHit}}+{\text{DefenderEvasion}}+2000}}\right)\\&+\left({\dfrac {{\text{AttackerLuck}}-{\text{DefenderLuck}}+{\text{LevelDifference}}}{5000}}\right)+{\text{AttackerSkillAndEquipment}}\end{aligned}}}
{\displaystyle {\text{ReloadTime}}={\text{WeaponReloadTime}}\times {\sqrt {\dfrac {200}{{\text{Reload}}\times (1+{\text{StatBonus}})+100}}}}
{\displaystyle {\text{LaunchCooldown}}={\dfrac {\sum _{i}^{N}\left({\text{cooldown}}({\text{Plane}}_{i})\times {\text{numberof}}({\text{Plane}}_{i})\right)}{\sum _{i}^{N}{\text{numberof}}({\text{Plane}}_{i})}}\times 2.2\times {\sqrt {\dfrac {200}{{\text{Reload}}\times (1+{\text{StatBonus}})+100}}}+0.1}
{\displaystyle {\begin{aligned}{\text{FinalDamage}}&=\sum _{\mathrm {for\ each\ ship} }\left(\sum _{\mathrm {for\ each\ gun} }{\left({\text{EquipmentDamage}}\times {\text{SlotEfficiency}}\right)}\times {\dfrac {100+{\text{AA}}(1+{\text{FormationBonus}}+{\text{StatBonus}})}{100}}\right)\end{aligned}}}
{\displaystyle {\text{ReloadTime}}={\dfrac {\sum _{i}^{N}{\text{reloadtime}}({\text{Equipment}}_{i})}{\text{NumberOfAAGuns}}}+0.5}
{\displaystyle {\text{Radius}}={\dfrac {\sum _{i}^{N}{\text{range}}({\text{AAGun}}_{i})}{\text{NumberOfAAGuns}}}}
{\displaystyle \min \left(0.05\times {\text{OwnShipMaxHP}},0.025\times {\sqrt {{\text{OwnShipMaxHP}}\times {\text{OtherShipMaxHP}}}}\right)\times \left(1+{\text{OtherShipRamAttackBonus}}\right)\times \left(1-{\text{OwnShipRamDefenseBonus}}\right)}
{\displaystyle {\text{AviationDamageReceived}}={\dfrac {1}{1+{\frac {\text{AA}}{150}}\cdot \left(1+{\text{SkillBonus}}\right)}}+\left({\text{CloakBit}}*0.1\right)}
{\displaystyle {\text{FinalDamageReduction}}=1-\displaystyle \prod _{i}^{N}(1-{\text{DamageReduction}}_{i})}
{\displaystyle {\begin{aligned}{\text{Accuracy}}=0.1&+{\dfrac {\text{AttackerHit}}{{\text{AttackerHit}}+{\text{DefenderEvasion}}+2}}\\[2ex]&+{\dfrac {{\text{AttackerLuck}}-{\text{DefenderLuck}}+{\text{LevelDifference}}}{1000}}\\[2ex]&+{\text{AccuracyVsShipTypeSkill}}-{\text{EvasionRateSkill}}\end{aligned}}}
{\displaystyle {\text{eHP}}={\dfrac {\text{Base HP}}{\text{Accuracy}}}}
{\displaystyle {\text{Accuracy}}={\dfrac {\text{AttackerHit}}{{\text{AttackerHit}}+{\text{DefenderEvasion}}}}}
{\displaystyle T=\mathrm {Oxygen} /10}
{\displaystyle {\mbox{Node HP Reduction Percentage}}=0.15\times {\sqrt {\text{Submarine Fleet Power}}}+0.25\times ({\text{Average Submarine Fleet Level}}-{\text{Enemy Level}})}
{\displaystyle {\begin{aligned}{\text{ShipPower}}&={\dfrac {\text{Health}}{5}}+{\text{Firepower}}+{\text{Torpedo}}+{\text{Airpower}}+{\text{AA}}+{\text{Reload}}+{\text{ASW}}+{\text{Speed}}\\&+2\times ({\text{Evasion}}+{\text{Hit}})+\sum {\text{EquipmentRarity}}+{\text{Modernization}}\end{aligned}}}
{\displaystyle {\text{StatBonus}}}
: Ship Skills and Equipment Skills. In
{\displaystyle {\text{Stat}}(1+{\text{StatBonus}})}
it refers to skills that buff the ship stats, like Queen Elizabeth's "Queen's Orders" or Little Beaver Squadron Tag's skill.
{\displaystyle {\text{FormationBonus}}}
: Refers to the bonus from fleet Formation
{\displaystyle {\text{EquipmentDamage}}}
: Damage of a single shot of the weapon, i.e. the number you see without multiplying. The damage of the planes are dependent on the type of the bombs/torpedoes they carry.
{\displaystyle {\text{ArmorModifier}}}
: A modifier based on the damage receiver's armor type and the calibre/type/load of the guns/torpedoes/bombs.
{\displaystyle {\text{LevelAdvantage}}}
: A value based on the difference in level in the shooter and the one being hit. It is calculated with these formulas:
{\displaystyle {\text{LevelAdvantage}}=1+({\text{AttackerLevel}}-{\text{DefenderLevel}}+{\text{MaximumDangerLevel}})\times 0.02}
{\displaystyle {\text{LevelAdvantage}}=1+({\text{AttackerLevel}}-{\text{DefenderLevel}})\times 0.02}
{\displaystyle {\text{AmmoBuff}}}
: Refers to Fleet Ammo bonus.
{\displaystyle {\text{DamageBonus}}}
: Skills that increase or decrease damage, like AmmoBuff above, Focused Assault, or Tirpitz's "Lone Queen of the North".
{\displaystyle {\text{AmmoTypeModifier}}}
: Skills that directly buff(or nerf) a type of ammo, like Belfast's "Burn Order" or Kinu's "Demon's Wish".
{\displaystyle {\text{EnemyDebuff}}}
: Skills that debuff the enemy, like Helena's "Radar Scan".
{\displaystyle {\text{HunterSkill}}}
: Skills that increases the damage dealt to specific types of enemies, like Deutschland's "Pocket Battleship".
{\displaystyle {\text{EquipmentReloadTime}}}
: The time you see on the equipment when it is not equipped on anyone.
{\displaystyle {\text{DelayBeforeVolley}}}
{\displaystyle {\text{DelayAfterVolley}}}
: The delay before volley is dependent on the type of ship. It is 0.16 seconds for DD, 0.18 for CL, 0.2 for CA. The delay after volley is 0.1 for all ships.
{\displaystyle {\text{VolleyDuration}}}
: The animation time, dependent on the equipment.
Retrieved from ‘https://azurlane.koumakan.jp/w/index.php?title=Combat&oldid=219820’
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Lanchester's laws - Wikipedia
Lanchester's laws are mathematical formulae for calculating the relative strengths of military forces. The Lanchester equations are differential equations describing the time dependence of two armies' strengths A and B as a function of time, with the function depending only on A and B.[1][2]
In 1915 and 1916, during World War I, M. Osipov and Frederick Lanchester independently devised a series of differential equations to demonstrate the power relationships between opposing forces.[3]: vii–viii Among these are what is known as Lanchester's linear law (for ancient combat) and Lanchester's square law (for modern combat with long-range weapons such as firearms).
Zoologists have found chimpanzees intuitively follow Lanchester's square law before engaging another troop of chimpanzees. A group of chimpanzees will not attack another group unless the numerical advantage is at least a factor of 1.5.[4]
1 Lanchester's linear law
2 Lanchester's square law
2.2 Example equations
2.3 Relation to the salvo combat model
3 Lanchester's law in use
Lanchester's linear law[edit]
Lanchester's square law[edit]
Example equations[edit]
Suppose that two armies, Red and Blue, are engaging each other in combat. Red is shooting a continuous stream of bullets at Blue. Meanwhile, Blue is shooting a continuous stream of bullets at Red.
Let symbol A represent the number of soldiers in the Red force. Each one has offensive firepower α, which is the number of enemy soldiers it can incapacitate (e.g., kill or injure) per unit time. Likewise, Blue has B soldiers, each with offensive firepower β.
Lanchester's square law calculates the number of soldiers lost on each side using the following pair of equations.[5] Here, dA/dt represents the rate at which the number of Red soldiers is changing at a particular instant. A negative value indicates the loss of soldiers. Similarly, dB/dt represents the rate of change of the number of Blue soldiers.
{\displaystyle {\frac {\mathrm {d} A}{\mathrm {d} t}}=-\beta B}
{\displaystyle {\frac {\mathrm {d} B}{\mathrm {d} t}}=-\alpha A}
The solution to these equations shows that:
If α=β, i.e. the two sides have equal firepower, the side with more soldiers at the beginning of the battle will win;
If A=B, i.e. the two sides have equal numbers of soldiers, the side with greater firepower will win;
If A>B and α>β, then Red will win, while if A<B and α<β, Blue will win;
If A>B but α<β, or A<B but α>β, the winning side will depend on whether the ratio of β/α is greater or less than the square of the ratio of A/B. Thus, if numbers and firepower are unequal in opposite directions, a superiority in firepower equal to the square of the inferiority in numbers is required for victory; or, to put it another way, the effectiveness of the army rises proportionate to the square of the number of people in it, but only linearly with their fighting ability.
The first three of these conclusions are obvious. The final one is the origin of the name "square law".
Relation to the salvo combat model[edit]
Lanchester's equations are related to the more recent salvo combat model equations, with two main differences.
Lanchester's law in use[edit]
In modern warfare, to take into account that to some extent both linear and the square apply often, an exponent of 1.5 is used.[8][9][10][3]: 7-5–7-8
Attempts have been made to apply Lanchester's laws to conflicts between animal groups.[11] Examples include tests with chimpanzees [4] and fire ants.[12] The chimpanzee application was relatively successful; the fire ant application did not confirm that the square law applied.
Lotka–Volterra equations similar mathematical model for predator-prey dynamics
Petrie multiplier similar mathematical model for sexism
Dupuy, Col T N (1979). Numbers, Predictions and War. Macdonald and Jane's.
Lanchester, Frederick W. (1916). Aircraft in Warfare.
^ Lanchester F.W., Mathematics in Warfare in The World of Mathematics, Vol. 4 (1956) Ed. Newman, J.R., Simon and Schuster, 2138–2157; anthologised from Aircraft in Warfare (1916)
^ "Lanchester Equations and Scoring Systems - RAND".
^ a b Osipov, M. (1991) [1915]. Translated by Helmbold, Robert; Rehm, Allan. "The Influence of the Numerical Strength of Engaged Forces on Their Casualties" Влияние Уисленности Сражающихся Сторонъ На Ихъ Потери (PDF). Tsarist Russian Journal Military Collection Военный Сборник. US Army Concepts Analysis Agency. Archived (PDF) from the original on 4 November 2021. Retrieved 23 January 2022.
^ a b Wilson, M. L., Britton, N. F., & Franks, N. R. (2002). Chimpanzees and the mathematics of battle. Proceedings of the Royal Society B: Biological Sciences, 269, 1107-1112. doi:10.1098/rspb.2001.1926
^ Taylor JG. 1983. Lanchester Models of Warfare, volumes I & II. Operations Research Society of America.
^ Armstrong MJ, Sodergren SE, 2015, Refighting Pickett's Charge: mathematical modeling of the Civil War battlefield, Social Science Quarterly.
^ MacKay N, Price C, 2011, Safety in Numbers: Ideas of concentration in Royal Air Force fighter defence from Lanchester to the Battle of Britain, History 96, 304–325.
^ Race to the Swift: Thoughts on Twenty-First Century Warfare by Richard E. Simpkin
^ "Lanchester's Laws and Attrition Modeling, Part II". 9 July 2010.
^ "Asymmetric Warfare: A Primer".
^ Clifton, E. (2020). A Brief Review on the Application of Lanchester's Models of Combat in Nonhuman Animals. Ecological Psychology, 32, 181-191. doi:10.1080/10407413.2020.1846456
^ Plowes, N. J. R., & Adams, E. S. (2005). An empirical test of Lanchester's square law: mortality during battles of the fire ant Solenopsis invicta. Proceedings of the Royal Society B: Biological Sciences, 272, 1809-1814. doi:10.1098/rspb.2005.3162
Retrieved from "https://en.wikipedia.org/w/index.php?title=Lanchester%27s_laws&oldid=1083739488"
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Fast Input & Output · USACO Guide
HomeGeneralFast Input & Output
Example TaskStandard I/OSlowFastcin/coutscanf/printfFile I/OSlowFastEven Faster MethodsAdditional Notesios::sync_with_stdio(false)cin.tie(nullptr)
The USACO Instructions Page briefly mentions some ways of speeding up I/O, but how much of a difference do these actually make? We'll use the following task to benchmark I/O speed:
The input consists of two integers
M
0\le M\le 1
N
1\le N\le 10^6
), followed by a sequence of
N
non-negative integers each less than
10^9+7
M=0
, output the sum of the input sequence modulo
10^9+7
M=1
, output the sum of each prefix of the input sequence modulo
10^9+7
Randomly generating test data results in input and output files each ~10MB large. It is possible to see input files this large (the 11th input file for Robotic Cow Herd is ~10.3MB large), though not output files (the largest we know of is due to Minimum Cost Paths, which has an output file ~2.8MB large).
Some simple methods of I/O don't come close to running under the time limit:
cin/cout + endl (5.8s)
Scanner + System.out.println (16.7s)
input + print (18.9s)
cin/cout
If using cin and cout, include the following two lines.
If you include ios::sync_with_stdio(false), then mixing C (scanf, printf) and C++ (cin, cout) style I/O may produce unexpected results. The upside is that both cin / cout become faster.
Including cin.tie(nullptr) will reduce the runtime if you are interleaving cin and cout (as is the case in the task at hand).
You can find more information about these lines at the end of this module.
cin/cout + unsync + \n (0.41s)
Using endl instead of "\n" will flush the output buffer and cause the above method to be quite slow:
cin/cout + unsync + endl (5.0s)
Though for interactive problems, you need to flush the output buffer every time you use cout. Any one of the following will have the same effect:
Not including cin.tie(nullptr)
Writing cout << endl instead of cout << "\n"
Writing cout << "\n" << flush instead of cout << "\n"
scanf/printf (0.52s)
Use BufferedReader and PrintWriter instead.
BufferedReader + PrintWriter (1.2s)
Use sys.stdin.readline and sys.stdout.write instead.
sys (2.4s)
Pretty similar to standard I/O.
freopen + cin/cout (5.7s)
freopen + cin/cout + unsync + \n (0.42s)
freopen + scanf/printf (0.52s)
ifstream/ofstream (0.43s)
Scanner + PrintWriter (3.4s)
A variant of the above method involves wrapping the BufferedReader with a StreamTokenizer:
StreamTokenizer (1.2s)
readline + write (2.4s)
The input methods described above are easy to type up from scratch and are usually fast enough for USACO contests. But if you're looking for something even faster ...
Using fread and fwrite reduces the runtime even further.
fread/fwrite (0.17s)
Even faster than BufferedReader is a custom-written Fast I/O class that reads bytes directly from an InputStream.
InputStream + PrintWriter (0.84s)
Yet again on C++ I/O
timing various I/O methods
Significance of ios_base::sync_with_stdio(false);
This disables the synchronization between the C and C++ standard streams. By default, all standard streams are synchronized, which in practice allows you to mix C- and C++-style I/O and get sensible and expected results. If you disable the synchronization, then C++ streams are allowed to have their own independent buffers, which makes mixing C- and C++-style I/O an adventure.
cin.tie(nullptr)
Significance of cin.tie(NULL);
This unties cin from cout. Tied streams ensure that one stream is flushed automatically before each I/O operation on the other stream.
By default cin is tied to cout to ensure a sensible user interaction. For example:
Warning: cout.tie(nullptr)
You may see some competitive programmers including this line. This doesn't actually do anything since cout isn't tied to anything. See this post for details.
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Q Find the largest integral x which satisfies the following inequality 4x+19x+5 4x-17x-3 - Maths - Linear Inequalities - 12646083 | Meritnation.com
Q. Find the largest integral x which satisfies the following inequality
\frac{4x+19}{x+5}<\frac{4x-17}{x-3}.
\mathrm{We} \mathrm{have}\phantom{\rule{0ex}{0ex}}\frac{4\mathrm{x}+19}{\mathrm{x}+5}<\frac{4\mathrm{x}-17}{\mathrm{x}-3}\phantom{\rule{0ex}{0ex}}⇒\frac{4\mathrm{x}+19}{\mathrm{x}+5}-\frac{4\mathrm{x}-17}{\mathrm{x}-3}<0\phantom{\rule{0ex}{0ex}}⇒\frac{\left(4\mathrm{x}+19\right)\left(\mathrm{x}-3\right)-\left(4\mathrm{x}-17\right)\left(\mathrm{x}+5\right)}{\left(\mathrm{x}+5\right)\left(\mathrm{x}-3\right)}<0\phantom{\rule{0ex}{0ex}}⇒\frac{4{\mathrm{x}}^{2}-12\mathrm{x}+19\mathrm{x}-57-4{\mathrm{x}}^{2}-20\mathrm{x}+17\mathrm{x}+85}{\left(\mathrm{x}+5\right)\left(\mathrm{x}-3\right)}<0\phantom{\rule{0ex}{0ex}}⇒\frac{4\mathrm{x}+28}{\left(\mathrm{x}+5\right)\left(\mathrm{x}-3\right)}<0\phantom{\rule{0ex}{0ex}}⇒\frac{4\left(\mathrm{x}+7\right)}{\left(\mathrm{x}+5\right)\left(\mathrm{x}-3\right)}<0\phantom{\rule{0ex}{0ex}}⇒\frac{\left(\mathrm{x}+7\right)}{\left(\mathrm{x}+5\right)\left(\mathrm{x}-3\right)}<\frac{0}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{\left(\mathrm{x}+7\right)}{\left(\mathrm{x}+5\right)\left(\mathrm{x}-3\right)}<0\phantom{\rule{0ex}{0ex}}\mathrm{Now} \mathrm{we} \mathrm{use} \mathrm{method} \mathrm{of} \mathrm{interval}\phantom{\rule{0ex}{0ex}}\left(-\infty \right)-----\left(-7\right)+++++\left(-5\right)-----\left(3\right)++++\left(\infty \right)\phantom{\rule{0ex}{0ex}}\mathrm{We} \mathrm{need} <0, \phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{\in }\left(\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{-}\mathbf{7}\right)\mathbf{\cup }\left(\mathbf{-}\mathbf{5}\mathbf{,}\mathbf{3}\right)\mathbf{ }\left[\mathbf{Answer}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}
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01 November 05 On multilinear oscillatory integrals, nonsingular and singular
Michael Christ, Xiaochun Li, Terence Tao, Christoph Thiele
Michael Christ,1 Xiaochun Li,2 Terence Tao,1 Christoph Thiele1
2Department of Mathematics, University of Illinois at Urbana-Champaign Urbana
Duke Math. J. 130(2): 321-351 (01 November 05). DOI: 10.1215/00127094-8229909
Basic questions concerning nonsingular multilinear operators with oscillatory factors are posed and partially answered.
{L}^{p}
norm inequalities are established for multilinear integral operators of Calderón-Zygmund type which incorporate oscillatory factors
{e}^{iP}
P
is a real-valued polynomial
Michael Christ. Xiaochun Li. Terence Tao. Christoph Thiele. "On multilinear oscillatory integrals, nonsingular and singular." Duke Math. J. 130 (2) 321 - 351, 01 November 05. https://doi.org/10.1215/00127094-8229909
Vol.130 • No. 2 • 01 November 05
Michael Christ, Xiaochun Li, Terence Tao, Christoph Thiele "On multilinear oscillatory integrals, nonsingular and singular," Duke Mathematical Journal, Duke Math. J. 130(2), 321-351, (01 November 05)
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Maximum dip to migrate - SEG Wiki
During migration, we can specify the maximum dip we want migrated in the section. This may be useful when we need to suppress the steeply dipping coherent noise. Figure 4.2-8 shows migrations of the dipping events with four different maximum allowable dips. For a 4 ms/trace dip limit, events with dips greater than this value are suppressed. Similarly, for an 8 ms/trace dip value, events with dips greater than this value are suppressed. When the dip value is 12 ms/trace, no suppression occurs, since all events in the input section have dips less than this value. Limiting the dip parameter is a way to reduce computational cost, since it is related to aperture width (equation 1 ), which determines the cost.
{\displaystyle d_{x}={\frac {v^{2}t}{4}}{\frac {\Delta t}{\Delta x}},}
From Figure 4.2-1, note that the Kirchhoff migration impulse response can be limited to various maximum dips. The smaller the maximum allowable dip, the smaller the aperture. This combination of maximum aperture width and maximum dip limit determines the actual effective aperture width used in migration. In particular, diffraction hyperbolas along which summation is done are truncated beyond the specified maximum dip limit.
A field data example of testing the maximum dip parameter is shown in Figure 4.2-9. Some steep dips are lost on the section that corresponds to the 2 ms/trace maximum allowable dip. The 8 ms/trace dip appears to be optimum. The maximum dip parameter must be chosen carefully so that the steep dips of interest in the input section are preserved. Finally, dip value can be changed spatially and in time; however, practical implementation can be cumbersome.
Figure 4.2-1 Migration can be confined to a range of dips present on a seismic section. The impulse response for the dip-limited migration operator is a truncated semicircle. Dip angle θ is measured from the vertical axis.
Figure 4.2-8 Tests for maximum dip to migrate in Kirchhoff migration: (a) a zero-offset section that contains a diffraction hyperbola with 2500-m/s velocity, (b) desired migration; Kirchhoff migration using (c) 4-ms/trace, (d) 8-ms/trace, (e) 12-ms/trace, and (f) 24-ms/trace maximum dip.
Figure 4.2-9 Tests for maximum dip to migrate in Kirchhoff migration: A low value for maximum dip to migrate can be hazardous. All dips of interest must be preserved during migration.
The phase-shift method of migration (migration principles and Section D.7) allows vertical variations in velocity and is accurate for up to dips of 90 degrees. Figure 4.5-1 shows the impulse response of the phase-shift algorithm. Clearly, for a constant-velocity medium, this response is equivalent to that of the Stolt migration. The impulse response shown in Figure 4.5-1 is considered to be the desired impulse response for 2-D zero-offset migration, and as such, responses of all migration algorithms discussed in this chapter are benchmarked against it.
Figure 4.5-2 The impulse response of the f − k migration operator has a truncated semicircular shape when a maximum dip limit is imposed. For comparison, the desired response shape has been superimposed on the f − k responses.
As with the Kirchhoff summation method, migration with the phase-shift method can be limited to smaller dips by truncating the semicircular wavefront (Figure 4.5-2). This dip filtering capability is useful in rejecting coherent noise from the stacked section while migrating the data. If migration is constrained to small dip values, then the steeply dipping reflectors may be filtered out unintentionally. Edge effects also are pronounced when a very narrow range of dips is passed. Note the linear streaks on the impulse response with a dip limit of 2 ms/trace (Figure 4.5-2).
The dip-filtering action caused by imposing a dip limit on the impulse response also is visible on the results shown in Figure 4.5-3. Note that steep dips greater than the specified maximum dip to migrate have been annihilated. On the field data example shown in Figure 4.5-4, severe dip filtering action of the 2 ms/trace maximum dip has caused smearing and eliminated virtually all of the signal contained in the section.
Figure 4.5-3 Tests for maximum dip to migrate in phase-shift migration: (a) a zero-offset section that contains dipping events with 3500-m/s velocity, (b) desired migration; phase-shift migrations using (c) 2-ms/trace, (d) 4-ms/trace, (e) 8-ms/trace, and (f) 16-ms/trace maximum dip limit.
Figure 4.5-4 Tests for maximum dip to migrate in phase-shift migration: A low value for maximum dip to migrate can be hazardous. All dips of interest must be preserved during migration.
Retrieved from "https://wiki.seg.org/index.php?title=Maximum_dip_to_migrate&oldid=19767"
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Frank Cce Everyday Science for Class 8 Science Chapter 13 - Sound
Frank Cce Everyday Science Solutions for Class 8 Science Chapter 13 Sound are provided here with simple step-by-step explanations. These solutions for Sound are extremely popular among Class 8 students for Science Sound Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Frank Cce Everyday Science Book of Class 8 Science Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Frank Cce Everyday Science Solutions. All Frank Cce Everyday Science Solutions for class Class 8 Science are prepared by experts and are 100% accurate.
4 - infrasonic
In hide and seek game, the blind folded person guesses the person closest to him by recognising the direction of loud sounds.
I would suggest my parents to buy the house which is three lanes away from the roadside, because there we will be less noise pollution. The house in front of the road may have noises due to vehicles and hustle and bustle; and noise pollution can cause severe physiological and psychological problems.
Sound needs a material medium to propagate. It cannot travel through vacuum.
(c) boy's voice
A boy's voice has the highest frequency because of his thin and small vocal chords, which can vibrate with a greater frequency.
(b) unbearable and shrill
Frequency of sound above 80 dB is unbearable and shrill.
A flute is a wind instrument, while the others are stringed instruments.
Dholak is not a wind instrument.
1.The unit of frequency is hertz.
2. Frequencies less then 20 Hz are called infrasonic.
3. Only vibrating bodies produce sound.
4. The unit of loudness of sound is decibel.
1. Unit of frequency (c) Hertz
2. Instrument producing sound of single frequency. (e) Turning fork
3. Maximum displacement of an oscillating object. (b) Amplitude
4. Sound with frequency more then 20000 Hz (d) Ultrasonics
5. Low frequency sound which we cannot hear. (a) Infrasonics
3. False. Humans can hear sound frequencies between 20 Hz to 20000 Hz.
4. False. Material medium is necessary for the propagation of sound.
5. False. The louder the sound, the larger will be the amplitude of the vibrating body.
Plants help in reducing noise pollution by absorbing the sounds from the surroundings.
The vibrations from the surroundings are collected by the ear pinna, which send these vibrations to ear drum. The ear drum vibrates as it receives these collected sound and transmits these vibrations to the internal ear. From the internal ear, these vibrations are transmitted to the brain in the form of electro-chemical signals. In the absence of ear drum, the vibrations of the sound will not reach the brain and we will not be able to hear.
Sound requires a material medium for its propagation. We cannot hear any sound on the moon, because it does not have an atmosphere and sound cannot propagate through vacuum.
The voices of men, women and children differ from each other due to the following factors:
(1) The pitch is one of the most essential factors to determine the voice of an individual. For example, men have lower-pitched voices, while women have higher-pitched voices.
(2) The second factor which causes the difference in voices is the length and thickness of vocal cords. Men have broader vocal chords as compared with women and children. In children, the vocal chords are the narrowest and they have very high pitched voices.
Hertz (Hz) is standard international unit of frequency.
A sound which is unpleasant to our ear is called noise.
Galton's whistle is used to give instructions while training dogs.
If a body makes 1 oscillation in a second, then the frequency of the body is said to be 1 Hertz. Hertz (Hz) is the unit of frequency.
Decibel (dB) is the unit of loudness.
Vibration is the back and forth movement of a body about its mean position.
Sound is a form of energy which is produced by vibrating objects.
Ultrasonic sounds are those sounds that have frequencies more than 20,000 Hz.
Infrasonic sounds are those sounds that have frequencies less than 20 Hz.
The following are the two uses of ultrasonic sound:
(a) They are used in industries to detect the cracks in metals.
(b) They are also used to measure the thickness of a material.
Shrillness is the effect of sound on the ear. The difference in various musical notes is recognisable due to the difference in their pitches or shrillness.
Two characteristics of sound are as follows:
(a) Pitch, which depends upon the frequency of vibration.
(b) Loudness, which depends upon the amplitude of vibration.
Galton's whistle is an instrument which generates ultrasonic vibrations. It is used to train dogs.
Air from the lungs makes the vocal cords vibrate.
Pitch Loudness
1. Pitch determines the shrillness of a sound. 1. Loudness of sound determines the degree of sensation of a sound.
2. It depends on the frequency of the vibrating body. 2. It depends on the amplitude of vibration.
Frequency: Frequency of a vibrating body may be defined as the number of vibration made per second.
Time period: Time period of a vibrating body may be defined as the time taken by the body to complete one vibration.
Amplitude: Amplitude is the maximum distance travelled by a vibrating body from its mean position to either side of the extreme positions.
Sound pollution is the disturbance produced in the environment by the excess of noise.
Two reasons of sound pollution are:
1. Industries: They use many heavy machines which are a major source of noise pollution.
2. Vehicles: In cities, large numbers of vehicles run at the same time, creating a lot of noises through horns and engines.
The following are three ill effects created by noise pollution:
1. Chronic exposure to noise pollution can result in loss of hearing.
2. Noise pollution can cause anger, tension and may also affect the sleep pattern of a person.
3. Noise pollution reduces concentration and results in loss of work efficiency.
(a) To prove that sound can move in water, take a bucket full of water and a squeaking toy in a polythene bag. Now, hold the toy inside the water and squeeze it. When we hold our ear against the side of the bucket, we can hear the squeak. This proves that sound can travel through liquid.
(b) To prove that sound can move in solid, take a wooden stick and hold one end of it near your ear. Now, ask any one to gently knock at the other end of the stick. We will see that we can hear the sound of knocking clearly. This proves that sound can travels through solid.
(c) To prove that sound cannot travel through vacuum, suspend an electric bell in a bell jar and connect it to a battery. As the bell rings, we can hear its sound. Now pump out the air from the jar through an outlet in the jar using a vacuum pump. On ringing the bell, we cannot hear anything once the air has been pumped out. This shows that sound cannot travel through vacuum.
To measure the speed of sound in air, we will need two wooden blocks, a measuring tape, a scale and a friend to help with the experiment.
Find a large open area and choose two spots on the opposite sides of the area. Give the wooden blocks to your friend. Make yourself stand at one spot and your friend at the other. Measure the distance between the spots using the measuring tape. Signal your friend to hit the blocks against each other. Start your stopwatch as you see the blocks hit each other. Note the reading on the stopwatch the moment you hear the sound from the blocks.
We can calculate the speed using the formula speed=
\frac{\mathrm{distance}}{\mathrm{time}}
Two reasons of sound pollution are as follows:
1. Industries: They use many heavy machines which is a major source of noise pollution.
2. Vehicles: In cities, large number of vehicles run at the same time creating a lot of noises through engines and horns.
Infrasonic sounds Ultrasonic sounds
1. Infrasonic sounds are the sounds that have frequencies less than 20 Hz. 1. Ultrasonic sounds are the sounds that have frequencies more than 20,000 Hz.
2. Whales and snakes uses infrasonic sound to communicate with each other. 2. Bats produces ultrasonic to communicate with each other.
3. They are utilised for monitoring earthquakes, charting rock and petroleum formations below the earth. 3. They are utilised for detecting cracks in metals, to find the depth of the sea, etc.
Pitch of a sound depends upon the following:
1. Frequency of the vibrating body
The pitch of sound increases with the increase in frequency.
2. Relative motion between the source and the listener
When either the source or the listener is approaching, the pitch of the sound increases and vice versa.
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Average Cost: Definition, Formula & Examples | StudySmarter
Businesses produce and sell a variety of products in different market structures at different price levels. To maximize their profit in the market, they have to take the costs of the production into account as well. To understand how the firms calculate the cost functions and derive their production plan, we should have a close look at two main cost types: marginal cost and average cost. In this article, we will learn all about the average cost, its equation, and what the average cost function looks like with various examples. Ready to deep dive, let’s go!
Average Cost, also called average total cost (ATC), is the cost per output unit. We can calculate the average cost by dividing the total cost by the total output quantity.
Average Cost equals the per-unit cost of production which is calculated by dividing the total cost by the total output.
Total cost means the sum of all costs, including the fixed and variable costs. Therefore, Average Cost is also often called the total cost per unit or the average total cost.
The average cost is important for firms since it shows them how much each unit of output cost them.
Remember, marginal cost shows how much an additional unit of output costs the firm to produce.
We can calculate the average cost using the following equation, where TC stands for the total cost and Q means the total quantity.
The average cost equation is:
ATC=\frac{TotalCost}{Quantityofoutput}=\frac{TC}{Q}
How can we calculate the average cost using the average cost equation?
Let's say the Willy Wonka chocolate firm produces chocolate bars. Their total costs and different levels of quantity are given in the following table. Using the average cost formula, we divide the total cost by the corresponding quantity for each level of quantity in the third column:
Table 1: Calculating Average Cost
As we see in this example, we should divide the total cost by the quantity of output to find the average cost. For instance, for a total cost of $3500, we can produce 1500 chocolate bars. Therefore, the average cost for the production of 1500 chocolate bars is $2.33. This demonstrates average cost decreasing as the fixed costs are spread between more output.
Components of the Average Cost equation
Average total cost breaks into two components: average fixed cost, and average variable cost.
Average fixed cost (AFC) shows us the total fixed cost for each unit. To calculate the AFC, we should divide the total fixed cost by the total quantity:
AFC=\frac{Fixed\mathrm{cos}t}{Quantityofoutput}=\frac{FC}{Q}
Fixed costs are not connected to the quantity of produced output. Fixed costs the firms have to pay, even at a production level of 0. Let's say a firm has to spend $2000 a month for rent and it does not matter whether the firm is active that month or not. Thus, $2000, in this case, is a fixed cost.
Average variable cost (AVC) equals the total variable cost per unit of produced quantity. Similarly, to calculate the AVC, we should divide the total variable cost by the total quantity:
AVC=\frac{Variable\mathrm{co}st}{Quantityofoutput}=\frac{VC}{Q}
Variable costs are production costs that differ depending on the total output of production.
A firm decides to produce 200 units. If raw materials cost $300 and labor to refine them costs $500.
$300+$500=$800 variable cost.
$800/200(units) =$4 Average Variable Cost.
The average cost is the sum of the fixed cost and average cost. Thus, if we add the average fixed cost and average variable cost, we should find the average total cost.
TotalAverageC\mathrm{os}t=Averagevariable\mathrm{cos}t\left(AVC\right)+Averagefixed\mathrm{cos}t\left(AFC\right)
The Average Fixed Cost and the Spreading Effect
The average fixed cost decreases with increasing produced quantity because the fixed cost is a fixed amount. This means it does not change with the produced amount of units.
You can think of the fixed cost as the amount of money you need to open a bakery. This includes, for instance, necessary machines, stands, and tables. In other words, fixed costs equal the required investment you need to make to start producing.
Since the total fixed cost is fixed, the more you produce, the average fixed cost per unit will decrease further. This is the reason why we have a falling average fixed cost curve in Figure 1 above.
This effect is called the spreading effect since the fixed cost is spread over the produced quantity. Given a certain amount of fixed cost, the average fixed cost decreases as the output increases.
The Average Variable Cost and the Diminishing Returns Effect
On the other hand, we see a rising average variable cost. Each unit of output that the firm produced additionally adds more to the variable cost since a rising amount of variable input would be necessary to produce the additional unit. This effect is also known as diminishing returns to the variable input
This effect is called the diminishing returns effect. Since a greater amount of variable input would be necessary as the output increases, we have higher average variable costs for higher levels of produced outputs.
The U-shaped Average Total Cost Curve
How do the spreading effect and diminishing returns effect cause the U-shape of the Average Cost Function? The relationship between these two affects the shape of the Average Cost Function.
For lower levels of output, the spreading effect dominates the diminishing returns effect, and for higher levels of output, the contrary holds. At low levels of output, small increases in output cause large changes in average fixed cost.
Assume a firm has a fixed cost of 200 in the beginning. For the first 2 units of production, we would have a $100 average fixed cost. After the firm produces 4 units, the fixed cost decreases by half: $50. Therefore, the spreading effect has a strong influence on the lower levels of quantity.
At high levels of output, the average fixed cost is already spread over the produced quantity and has a very small influence on the average total cost. Therefore, we don't observe a strong spreading effect anymore. On the other hand, diminishing returns generally increase as quantity rises. Therefore, the diminishing returns effect dominates the spreading effect for a large number of quantities.
It is very important to understand how to calculate the Average Cost using the total fixed cost and average variable cost. Let's practice calculating the Average Cost and have a closer look at the example of the Willy Wonka chocolate firm. After all, we all like chocolate, right?
In the below table, we have columns for the produced quantity, the total cost as well as the average variable cost, average fixed cost, and average total cost.
Average fixed cost ($)
Average variable cost ($)
Average total cost ($)
Table 2. Average Cost Example
As the Willy Wonka chocolate firm produces more chocolate bars, the total costs are increasing as expected. Similarly, we can see that the variable cost of 1 unit is $6, and the average variable cost increases with each additional unit of chocolate bar. The fixed cost equals $54 for the 1 unit of chocolate, the average fixed cost is $54. As we learn, the average fixed costs decrease as the total quantity increase.
At a quantity level of 8, we see that fixed costs have spread out across the total output($13.5). While the average variable cost is increasing($12), it increases less than the average fixed cost decreases. This results in a lower average total cost($18.75). This is the most efficient quantity to produce, as the average total cost is minimized.
Similarly, at a quantity level of 10, we can observe that despite the average fixed cost ($5.4) being minimized, the variable cost ($14) has increased as a result of diminishing returns. This results in a higher average total cost($19.4), which shows that the efficient production quantity is lower than 10.
The surprising aspect is the average total cost, which is first decreasing and then increasing as the quantity rises. It is important to distinguish between the total cost and the average total cost since the former always increases with additional quantity. However, the average total cost function has a U-shape and first falls and then rises as the quantity increases.
The average total cost function has a U-shape, which means it is decreasing for low levels of output and increases for larger output quantities.
In Figure 1, we will analyze the Average Cost Function of the Bakery ABC. Figure 1 illustrates how the average cost changes with different levels of quantity. The quantity is shown on the x-axis, whereas the cost in dollars is given on the y-axis.
Figure 1. Average Cost Function - StudySmarter
On the first look, we can see that the Average Total Cost Function has a U-shape and decreases up to a quantity (Q) and increases after this quantity (Q). The average fixed cost decreases with the increasing quantity and the average variable cost has an increasing path in general.
The U-shape structure of the Average Cost Function is formed by two effects: the spreading effect and the diminishing returns effect. The average fixed cost and average variable cost are responsible for these effects.
Average Cost and Cost Minizimation
At the point Q where the diminishing returns effect and the spreading effect balance each other, the average total cost is at its minimum level.
The relationship between the average total cost curve and marginal cost curve is illustrated in Figure 2 below.
Figure 2. Average Cost and Cost Minizimation - StudySmarter
The corresponding quantity where the average total cost is minimized is called the minimum-cost output, which equals Q in Figure 2. Further, we see that the bottom of the U-shaped average total cost curve is also the point where the marginal cost curve intersects the average total cost curve. This is in fact not a coincidence but a general rule in the economy: the average total cost equals marginal cost at the minimum-cost output.
Average Cost - Key takeaways
Average fixed cost (AFC) shows us the total fixed cost for each unit and Average variable cost (AVC) equals the total variable cost per unit of produced quantity.
The average cost is the sum of the fixed cost and average variable cost. Thus, if we add the average fixed cost and average variable cost, we should find the average total cost.
The U-shape structure of the Average Cost Function is formed by two effects: the spreading effect and the diminishing returns effect.
For lower levels of output, the spreading effect dominates the diminishing returns effect, and for higher levels of output, the contrary holds.
Average Cost equals the cost of production per output unit.
How to calculate the average cost?
Average Cost is calculated by dividing the total cost by the total output.
What is the average cost function?
Why is the long-run average cost curve U-shaped?
What is an example of average cost?
The total cost of $20,000, we can produce 5000 chocolate bars. Therefore, the average cost for the production of 5000 chocolate bars is $4.
Final Average Cost Quiz
How can we calculate the average cost?
Which one is the definition of Average variable cost (AVC)?
The average variable cost equals the total variable cost per unit of produced quantity.
Which one is the definition of Average fixed cost (AFC)?
The average fixed cost shows us the total fixed cost for each unit.
How does the average fixed cost change with an additional unit of production?
The average fixed cost decreases with increasing produced quantity
Why does the average fixed cost decrease with increasing produced quantity?
Since the total fixed cost is fixed, the more you produce, the average fixed cost per unit will decrease further.
What is the spreading effect?
Since the total fixed cost is fixed, the more you produce, the average fixed cost per unit will decrease further. This is the reason why we have a falling average fixed cost curve.
Since the fixed cost is spread over the produced quantity, given a certain amount of fixed cost, the average fixed cost decreases as the output increases.
What is the diminishing returns effect?
Since a greater amount of variable input would be necessary as the output increases, there are higher average variable costs for higher levels of produced outputs.
How do the spreading effect and diminishing returns effect cause the U-shape of the Average Cost Function?
What is the minimum-cost output?
The corresponding quantity where the average total cost is at its minimum level.
The average total cost function has a U-shape, which means it increases for low levels of output and decreases for larger output quantities.
If we add the average fixed cost and average variable cost, we should find the average total cost.
If a firm has an average variable cost of $20 and an average fixed cost of $10, what is the average total cost?
If a firm has an average total cost of $20 and an average fixed cost of $10, what is the average variable cost?
of the users don't pass the Average Cost quiz! Will you pass the quiz?
Income Inequality Learn
Markets for Factors of Production Learn
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Intimal Hyperplasia and Wall Shear in Arterial Bypass Graft Distal Anastomoses: An In Vivo Model Study | J. Biomech Eng. | ASME Digital Collection
Department of Mechanical Engineering, University of Louisville, Louisville, KY 40292
Mary M. Evancho,
Mary M. Evancho
Division of Surgical Research, Summa Health System, Akron, OH 44309
Rick L. Sims,
Rick L. Sims
Nancy V. Rodway,
Nancy V. Rodway
Department of Pathology, VA Medical Center, Canton, OH 44322
Andrea Gobin,
Department of Biomedical Engineering, Rice University, Houston, TX 77251-1892
Division of Surgical Research, Summa Health System, Akron, OH 44309; Department of Biomedical Engineering, The University of Akron, Akron, OH 44235-0302
Contributed by the Bioengineering Division for publication in the JOURNAL OF BIOMECHANICAL ENGINEERING. Manuscript received by the Bioengineering Division June 2, 2000; revised manuscript received May 16, 2001. Associate Editor: A. P. Yoganathan.
Keynton, R. S., Evancho , M. M., Sims, R. L., Rodway, N. V., Gobin, A., and Rittgers, S. E. (May 16, 2001). "Intimal Hyperplasia and Wall Shear in Arterial Bypass Graft Distal Anastomoses: An In Vivo Model Study ." ASME. J Biomech Eng. October 2001; 123(5): 464–473. https://doi.org/10.1115/1.1389461
The observation of intimal hyperplasia at bypass graft anastomoses has suggested a potential interaction between local hemodynamics and vascular wall response. Wall shear has been particularly implicated because of its known effects upon the endothelium of normal vessels and, thus, was examined as to its possible role in the development of intimal hyperplasia in arterial bypass graft distal anastomoses. Tapered (4–7 mm I.D.) e-PTFE synthetic grafts 6 cm long were placed as bilateral carotid artery bypasses in six adult, mongrel dogs weighing between 25 and 30 kg with distal anastomotic graft-to-artery diameter ratios (DR) of either 1.0 or 1.5. Immediately following implantation, simultaneous axial velocity measurements were made in the toe and artery floor regions in the plane of the anastomosis at radial increments of 0.35 mm, 0.70 mm, and 1.05 mm using a specially designed 20 MHz triple crystal ultrasonic wall shear rate transducer. Mean, peak, and pulse amplitude wall shear rates (WSRs), their absolute values, the spatial and temporal wall shear stress gradients (WSSG), and the oscillatory shear index (OSI) were computed from these velocity measurements. All grafts were harvested after 12 weeks implantation and measurements of the degree of intimal hyperplasia (IH) were made along the toe region and the artery floor of the host artery in 1 mm increments. While some IH occurred along the toe region (8.35±23.1 μm) and was significantly different between DR groups
p<0.003,
the greatest amount occurred along the artery floor (81.6±106.5 μm, mean±S.D.)
p<0.001
although no significant differences were found between DR groups. Linear regressions were performed on the paired IH and mean, peak, and pulse amplitude WSR data as well as the absolute mean, peak, and pulse amplitude WSR data from all grafts. The mean and absolute mean WSRs showed a modest correlation with IH
(r=−0.406
and −0.370, respectively) with further improvements seen
(r=−0.482
and −0.445, respectively) when using an exponential relationship. The overall best correlation was seen against an exponential function of the OSI
r=0.600.
Although these correlation coefficients were not high, they were found to be statistically significant as evidenced by the large F-statistic obtained. Finally, it was observed that over 75 percent of the IH occurred at or below a mean WSR value of 100 s−1 while approximately 92 percent of the IH occurred at or below a mean WSR equal to one-half that of the native artery. Therefore, while not being the only factor involved, wall shear (and in particular, oscillatory wall shear) appears to provide a stimulus for the development of anastomotic intimal hyperplasia.
surgery, blood vessels, haemorheology
Shear (Mechanics), Shear rate, Surgery, Vessels, Flow (Dynamics), Hemodynamics
Secondary Femoro-Popliteal Reconstruction
Biologic Fate of Autogenous Vein Implants as Arterial Substitutes: Clinical, Angiographic and Histopathologic Observations in Femoro-popliteal Operations for Atherosclerosis
A Review of the Histologic Changes in Vein-to-Artery Grafts, With Particular Reference to Intimal Hyperplasia
DeWeese, J. A., 1978, “Anastomotic Neointimal Hyperplasia,” in: Vascular Grafts, P. N. Sawyer and M. J. Kaplitt, eds., New York: Apple-Century-Crofts, pp. 291–307.
Anastomotic Hyperplasia
Mechanisms of Arterial Graft Failure: I. Role of Cellular Proliferation in Early Healing of PTFE Prostheses
Management of Early and Late Thrombosis of Expanded Polytetrafluoroethylene (PTFE) Femoropopliteal Bypass Grafts: Favorable Prognosis with Appropriate Reoperation
Mechanism of Arterial Graft Failure. II. Chronic Endothelial and Smooth Muscle Cell Proliferation in Healing PTFE Prostheses
Hemodynamics of Vein Graft Stenosis
The Influence of Hypertension on Injury-Induced Myointimal Thickening
Y. P. T.
Acute Reductions in Blood Flow and Shear Stress Induce Platelet-Derived Growth Factor-A Expression in Baboon Prosthetic Grafts
Pompeselli
Conformational Stress Anastomotic Hyperplasia
Kinetics of Vein Graft Hyperplasia: Association With Tangential Stress
Velocity Distribution and Intimal Proliferation in Autologous Vein Grafts in Dogs
Quantitative Inverse Correlation of Wall Shear Stress With Experimental Intimal Thickening
Surg. Forum: Cong. Amer. Coll. Surg.
SG & O
Factors Affecting Patency of Femoro-popliteal Bypass Grafts
Distal Arteriovenous Fistula as an Adjunct to Maintaining Arterial and Graft Patency for Limb Salvage
Bharadvaj, B. K., Daddario, D. M., et al., 1982, “Flow Studies at Arterial Anastomoses,” Proc. 35th ACEMB, Philadelphia.
Optimal Graft Diameter: Effect of Wall Shear Stress on Vascular Healing
Effect of Wall Shear Stress on Intimal Thickening of Arterially Transplanted Autogenous Veins in Dogs
Hemodynamic Patterns in Two Flow Models of End-to-Side Vascular Graft Anastomoses: Effects of Pulsatility, Flow Division, Reynolds Number, and Hood Length
Effects of Fluid Shear Stress on Gene Regulation of Vascular Cells
Laminar Shear Stress: Mechanisms by Which Endothelial Cells Transduce an Atheroprotective Force
Intimal Changes in Arterio-venous Bypass Grafts
Flow Disturbances at the Distal End-to-Side Anastomosis
Models of Side-to-end Anastomoses: Effects of Angle and Flow Split
Callow, A. D., 1986, “History of Vascular Graft Development,” in: Vascular Graft Update: Safety and Performance, 1st ed., Philadelphia: American Society for Testing and Materials, pp. 11–15.
The Significance of Graft Diameter
Design, Fabrication and in Vitro Evaluation of an in Vivo Ultrasonic Doppler Wall Shear Rate Measuring Device
Wall Shear Stress Rather Than Shear Rate Regulates Cytoplasmic Ca++ Responses to Flow in Vascular Endothelial Cells
Fluid Flow Decreases Preproendothelin mRNA Levels and Suppresses Endothelin-1 Peptide Release in Cultured Human Endothelial Cells
New. Engl. J. Med.
Flow Input Waveform Effects on the Temporal and Spatial Shear Stress Gradients in a Femoral Graft-Artery Connector
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Linearizability - Wikipedia
(Redirected from Atomic operation)
This article duplicates the scope of other articles, specifically serializability and atomicity (database systems). Please discuss this issue on the talk page and edit it to conform with Wikipedia's Manual of Style. (November 2018)
In grey a linear sub-history, processes beginning in b do not have a linearizable history because b0 or b1 may complete in either order before b2 occurs.
In concurrent programming, an operation (or set of operations) is linearizable if it consists of an ordered list of invocation and response events (callbacks), that may be extended by adding response events such that:
The extended list can be re-expressed as a sequential history (is serializable).
That sequential history is a subset of the original unextended list.
Informally, this means that the unmodified list of events is linearizable if and only if its invocations were serializable, but some of the responses of the serial schedule have yet to return.[1]
In a concurrent system, processes can access a shared object at the same time. Because multiple processes are accessing a single object, there may arise a situation in which while one process is accessing the object, another process changes its contents. Making a system linearizable is one solution to this problem. In a linearizable system, although operations overlap on a shared object, each operation appears to take place instantaneously. Linearizability is a strong correctness condition, which constrains what outputs are possible when an object is accessed by multiple processes concurrently. It is a safety property which ensures that operations do not complete in an unexpected or unpredictable manner. If a system is linearizable it allows a programmer to reason about the system.[2]
1 History of linearizability
2 Definition of linearizability
2.1 Linearizability versus serializability
2.2 Linearization points
3 Primitive atomic instructions
4 High-level atomic operations
5 Examples of linearizability
5.1.1 Non-atomic
5.3 Fetch-and-increment
History of linearizability[edit]
Definition of linearizability[edit]
A history is a sequence of invocations and responses made of an object by a set of threads or processes. An invocation can be thought of as the start of an operation, and the response being the signaled end of that operation. Each invocation of a function will have a subsequent response. This can be used to model any use of an object. Suppose, for example, that two threads, A and B, both attempt to grab a lock, backing off if it's already taken. This would be modeled as both threads invoking the lock operation, then both threads receiving a response, one successful, one not.
A sequential history is one in which all invocations have immediate responses; that is the invocation and response are considered to take place instantaneously. A sequential history should be trivial to reason about, as it has no real concurrency; the previous example was not sequential, and thus is hard to reason about. This is where linearizability comes in.
A history is linearizable if there is a linear order
{\displaystyle \sigma }
of the completed operations such that:
For every completed operation in
{\displaystyle \sigma }
, the operation returns the same result in the execution as the operation would return if every operation was completed one by one in order
{\displaystyle \sigma }
If an operation op1 completes (gets a response) before op2 begins (invokes), then op1 precedes op2 in
{\displaystyle \sigma }
Linearizability versus serializability[edit]
This reordering is sensible provided there is no alternative means of communicating between A and B. Linearizability is better when considering individual objects separately, as the reordering restrictions ensure that multiple linearizable objects are, considered as a whole, still linearizable.
Linearization points[edit]
In the examples below, the linearization point of the counter built on compare-and-swap is the linearization point of the first (and only) successful compare-and-swap update. The counter built using locking can be considered to linearize at any moment while the locks are held, since any potentially conflicting operations are excluded from running during that period.
Primitive atomic instructions[edit]
Most[citation needed] processors include store operations that are not atomic with respect to memory. These include multiple-word stores and string operations. Should a high priority interrupt occur when a portion of the store is complete, the operation must be completed when the interrupt level is returned. The routine that processes the interrupt must not modify the memory being changed. It is important to take this into account when writing interrupt routines.
When there are multiple instructions which must be completed without interruption, a CPU instruction which temporarily disables interrupts is used. This must be kept to only a few instructions and the interrupts must be re-enabled to avoid unacceptable response time to interrupts or even losing interrupts. This mechanism is not sufficient in a multi-processor environment since each CPU can interfere with the process regardless of whether interrupts occur or not. Further, in the presence of an instruction pipeline, uninterruptible operations present a security risk, as they can potentially be chained in an infinite loop to create a denial of service attack, as in the Cyrix coma bug.
The C standard and SUSv3 provide sig_atomic_t for simple atomic reads and writes; incrementing or decrementing is not guaranteed to be atomic.[3] More complex atomic operations are available in C11, which provides stdatomic.h. Compilers use the hardware features or more complex methods to implement the operations; an example is libatomic of GCC.
The ARM instruction set provides LDREX and STREX instructions which can be used to implement atomic memory access by using exclusive monitors implemented in the processor to track memory accesses for a specific address.[4] However, if a context switch occurs between calls to LDREX and STREX, the documentation notes that STREX will fail, indicating the operation should be retried.
High-level atomic operations[edit]
Examples of linearizability[edit]
Increment - adds 1 to the value stored in the counter, return acknowledgement
Read - returns the current value stored in the counter without changing it.
Non-atomic[edit]
The first process reads the value in the register as 0.
The first process adds one to the value, the counter's value should be 1, but before it has finished writing the new value back to the register it may become suspended, meanwhile the second process is running:
Read value in register Ri.
Add one to the value.
Read registers R1, R2, ... Rn.
Return the sum of all registers.
This implementation solves the problem with our original implementation. In this system the increment operations are linearized at the write step. The linearization point of an increment operation is when that operation writes the new value in its register Ri. The read operations are linearized to a point in the system when the value returned by the read is equal to the sum of all the values stored in each register Ri.
Compare-and-swap[edit]
Fetch-and-increment[edit]
^ a b c d Herlihy, Maurice P.; Wing, Jeannette M. (1990). "Linearizability: A Correctness Condition for Concurrent Objects". ACM Transactions on Programming Languages and Systems. 12 (3): 463–492. CiteSeerX 10.1.1.142.5315. doi:10.1145/78969.78972. S2CID 228785.
^ Shavit, Nir and Taubenfel,Gadi (2016). "The Computability of Relaxed Data Structures: Queues and Stacks as Examples" (PDF). Distributed Computing. 29 (5): 396–407. doi:10.1007/s00446-016-0272-0. S2CID 16192696. {{cite journal}}: CS1 maint: multiple names: authors list (link)
^ Kerrisk, Michael (7 September 2018). The Linux Programming Interface. No Starch Press. ISBN 9781593272203 – via Google Books.
^ "ARM Synchronization Primitives Development Article".
^ a b Fich, Faith; Hendler, Danny; Shavit, Nir (2004). "On the inherent weakness of conditional synchronization primitives". Proceedings of the twenty-third annual ACM symposium on Principles of distributed computing – PODC '04. New York, NY: ACM. pp. 80–87. doi:10.1145/1011767.1011780. ISBN 978-1-58113-802-3. S2CID 9313205.
Herlihy, Maurice P.; Wing, Jeannette M. (1987). Axioms for Concurrent Objects. Proceedings of the 14th ACM SIGACT-SIGPLAN Symposium on Principles of Programming Languages, POPL '87. p. 13. doi:10.1145/41625.41627. ISBN 978-0-89791-215-0. S2CID 16017451.
Herlihy, Maurice P. (1990). A Methodology for Implementing Highly Concurrent Data Structures. ACM SIGPLAN Notices. Vol. 25. pp. 197–206. CiteSeerX 10.1.1.186.6400. doi:10.1145/99164.99185. ISBN 978-0-89791-350-8.
Herlihy, Maurice P.; Wing, Jeannette M. (1990). "Linearizability: A Correctness Condition for Concurrent Objects". ACM Transactions on Programming Languages and Systems. 12 (3): 463–492. CiteSeerX 10.1.1.142.5315. doi:10.1145/78969.78972. S2CID 228785.
Aphyr. "Strong Consistency Models". aphyr.com. Aphyr. Retrieved 13 April 2018.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Linearizability&oldid=1087435303"
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Suction_cup Knowpia
A suction cup, also known as a sucker, is a device or object that uses the negative fluid pressure of air or water to adhere to nonporous surfaces, creating a partial vacuum.[1]
A transparent suction cup
The pressure on a suction cup as exerted by collisions of gas molecules holds the suction cup in contact with the surface.
One cup suction lifter.
Suction cups are peripheral traits of some animals such as octopuses and squids, and have been reproduced artificially for numerous purposes.[2]
The working face of the suction cup is made of elastic, flexible material and has a curved surface.[3] When the center of the suction cup is pressed against a flat, non-porous surface, the volume of the space between the suction cup and the flat surface is reduced, which causes the air or water between the cup and the surface to be expelled past the rim of the circular cup. The cavity which develops between the cup and the flat surface has little to no air or water in it because most of the fluid has already been forced out of the inside of the cup, causing a lack of pressure. The pressure difference between the atmosphere on the outside of the cup and the low-pressure cavity on the inside of the cup keeps the cup adhered to the surface.
Suction cup pressed on a window
When the user ceases to apply physical pressure to the outside of the cup, the elastic substance of which the cup is made tends to resume its original, curved shape. The length of time for which the suction effect can be maintained depends mainly on how long it takes for air or water to leak back into the cavity between the cup and the surface, equalizing the pressure with the surrounding atmosphere. This depends on the porosity and flatness of the surface and the properties of the cup's rim. A small amount of mineral oil or vegetable oil is often employed to help maintain the seal.
The force required to detach an ideal suction cup by pulling it directly away from the surface is given by the formula:
{\displaystyle F=AP}
A is the area of the surface covered by the cup,
P is the pressure outside the cup (typically atmospheric pressure)
This is derived from the definition of pressure, which is:
{\displaystyle P=F/A}
For example, a suction cup of radius 2.0 cm has an area of
{\displaystyle \pi }
(0.020 m)2 = 0.0013 square meters. Using the force formula (F = AP), the result is F = (0.0013 m2)(100,000 Pa) = about 130 newtons.
The above formula relies on several assumptions:
The outer diameter of the cup does not change when the cup is pulled.
No air leaks into the gap between the cup and the surface.
The pulling force is applied perpendicular to the surface so that the cup does not slide sideways or peel off.
Artificial useEdit
SatNav devices often ship with suction cup holders for mounting on windscreens.
GoPro camera attached to car with suction cup
Artificial suction cups are believed to have first been used in the third century, B.C., and were made out of gourds. They were used to suction "bad blood" from internal organs to the surface. Hippocrates is believed to have invented this procedure.[citation needed]
The first modern suction cup patents were issued by the United States Patent and Trademark Office during the 1860s. TC Roche was awarded U.S. Patent No. 52,748 in 1866 for a "Photographic Developer Dipping Stick"; the patent discloses a primitive suction cup means for handling photographic plates during developing procedures. In 1868, Orwell Needham patented a more refined suction cup design, U.S. Patent No. 82,629, calling his invention an "Atmospheric Knob" purposed for general use as a handle and drawer opening means.[4][5]
Suction cups have a number of commercial and industrial applications:
To attach an object to a flat, nonporous surface, such as a refrigerator door or a tile on a wall. This is also used for mooring ships[6][7]
To move an object, such as a pane of glass or a raised floor tile, by attaching the suction cup to a flat, nonporous part of the object and then sliding or lifting the object.
In some toys, such as Nerf darts.
A toilet plungers[8]
To climb up almost or completely vertically up or down a flat, nonporous surface, such as the sides of some buildings. This is part of buildering, which is also known as urban climbering. [9]
To hold an object still while it is worked on, such as holding a piece of glass while performing edge grinding application.
On May 25, 1981, Dan Goodwin, a.k.a. SpiderDan, scaled Sears Tower, the former world's tallest building, with a pair of suction cups. He went on to scale the Renaissance Center in Dallas, the Bonaventure Hotel in Los Angeles, the World Trade Center in New York City, Parque Central Tower in Caracas, the Nippon TV station in Tokyo, and the Millennium Tower in San Francisco.[10][11][12]
Wikimedia Commons has media related to Suction cups.
Self-sealing suction cup
^ ""Suction Cup" m-w.com". Merriam Webster: An Encyclopædia Britannica Company. Retrieved 2012-06-01.
^ "Well-Armed Design: 8 Octopus-Inspired Technologies". livescience.com. 29 September 2014. Retrieved July 30, 2015.
^ ""Suction Cup" google.com". Google Patents. Retrieved 2012-06-01.
^ "United States Patent 52,748".
^ "First inland vacuum-based mooring system installed on St. Lawrence Seaway locks". Professional Mariner. September 2015. Retrieved 11 March 2017.
^ Hands Free Mooring on YouTube
^ . 2006-04-24 https://web.archive.org/web/20060424182733/http://www.suctioncupmuseum.com/html/history.html. Archived from the original on April 24, 2006. Retrieved 2012-01-27. {{cite web}}: Missing or empty |title= (help)CS1 maint: unfit URL (link)
^ "Man climbs skyscraper with suction cups". BBC News. 2010-09-07. Retrieved 2012-01-27.
^ Spider-man aka SpiderDan Goodwin scales the Sears Tower V2 - YouTube
^ Spider-man aka SpiderDan Goodwin the Skyscraperman scales the Millennium Tower in San Francisco - YouTube
^ "αποφραξεις τιμες (Greece)". Ventouza. 25 February 2018.
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Example: Visualizing Custom Data on a Choropleth Map - Maple Help
Home : Support : Online Help : Statistics and Data Analysis : DataSets Package : Example: Visualizing Custom Data on a Choropleth Map
Visualizing Custom Data on a Choropleth Map
This application generates a choropleth map visualizing the average number of children born to a woman in a number of European countries for the calendar year 2015.
Data source: Eurostat - http://ec.europa.eu/eurostat/tgm/table.do?tab=table&init=1&language=en&pcode=tsdde220 (accessed 9 April 2017)
Data Description: The mean number of children that would be born alive to a woman during her lifetime if she were to survive and pass through her childbearing years conforming to the fertility rates by age of a given year.
Import fertility rate data
First a DataFrame is created to contain the data set. Note that this data is for the year 2015.
FertilityData := Import("datasets/fertility_rates.csv", base = datadir);
\textcolor[rgb]{0,0,1}{\mathrm{FertilityData}}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{}& \textcolor[rgb]{0,0,1}{\mathrm{Fertility Rate - 2015}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Belgium}}& \textcolor[rgb]{0,0,1}{1.7}\\ \textcolor[rgb]{0,0,1}{\mathrm{Bulgaria}}& \textcolor[rgb]{0,0,1}{1.53}\\ \textcolor[rgb]{0,0,1}{\mathrm{Czech Republic}}& \textcolor[rgb]{0,0,1}{1.57}\\ \textcolor[rgb]{0,0,1}{\mathrm{Denmark}}& \textcolor[rgb]{0,0,1}{1.71}\\ \textcolor[rgb]{0,0,1}{\mathrm{Germany}}& \textcolor[rgb]{0,0,1}{1.5}\\ \textcolor[rgb]{0,0,1}{\mathrm{Estonia}}& \textcolor[rgb]{0,0,1}{1.58}\\ \textcolor[rgb]{0,0,1}{\mathrm{Ireland}}& \textcolor[rgb]{0,0,1}{1.92}\\ \textcolor[rgb]{0,0,1}{\mathrm{Greece}}& \textcolor[rgb]{0,0,1}{1.33}\\ \textcolor[rgb]{0,0,1}{\mathrm{...}}& \textcolor[rgb]{0,0,1}{\mathrm{...}}\end{array}]
Generate a world map and restrict the view to Europe
europeMap := DataSets:-Builtin:-WorldMap():
The SetView command restricts the view frame for the World map to a given area.
SetView(europeMap,-11,36,32,87):
Define the colors used to shade the map
In the following section, we define two colors to be used to generate the gradient in the ChoroplethMap.
Lowest fertility value:
c1 := ColorTools:-Color("RGB",[0/255,55/255,0/255]);
\textcolor[rgb]{0,0,1}{\mathrm{c1}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{〈}\colorbox[rgb]{0,0.215686274509804,0}{$\textcolor[rgb]{1,1,1}{RGB : 0 0.216 0}$}\textcolor[rgb]{0,0,1}{〉}
Highest fertility value:
c2 := ColorTools:-Color("RGB",[230/255,230/255,150/255]);
\textcolor[rgb]{0,0,1}{\mathrm{c2}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{〈}\colorbox[rgb]{0.901960784313726,0.901960784313726,0.588235294117647}{$RGB : 0.902 0.902 0.588$}\textcolor[rgb]{0,0,1}{〉}
Generate a choropleth
The ChoroplethMap command generates a map that is shaded in proportion to the measurements for the displayed variable, the fertility rate. The plots:-display
FertilityMap := ChoroplethMap(europeMap, FertilityData, [c1, c2], mapdata=fine, watercolor="White"):
Additional plot options can be added using the plots:-display command:
plots:-display(FertilityMap, axes = none, size = [800, 800], title = "Number of Children Per Female (2015)", titlefont = [Arial, Bold, 18]);
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Home : Support : Online Help : Programming : Grid Package : Wait
wait for parallel computation to finish
Wait(node1, node2, ...)
The Wait command stops execution until one or more remote compute nodes are finished processing.
When node is given as a parameter, the current process will wait until that specified node is finished. If the given node was finished prior to calling Wait (or never started a job), Wait will return immediately.
When no parameters are given, the call to Wait will block until all compute nodes are finished processing.
The Wait command is intended to be called by the main Maple session, not the compute nodes. See the Barrier command for information about syncing up remote compute nodes.
The Wait command is only available in local Grid mode.
In this example we start 1 job and wait for it to finish
\mathrm{Grid}:-\mathrm{Setup}\left(\mathrm{numnodes}=4\right):
\mathrm{Grid}:-\mathrm{Run}\left(1,"for i from 1 to 10^6 do od:"\right)
\mathrm{Grid}:-\mathrm{Wait}\left(1\right)
In this example we start multiple jobs and wait for them to finish
\mathrm{Grid}:-\mathrm{Run}\left(0,"for i from 1 to 10^5 do od: i;"\right)
\mathrm{Grid}:-\mathrm{Run}\left(1,"for i from 1 to 10^6 do od: i;"\right)
\mathrm{Grid}:-\mathrm{Run}\left(2,"for i from 1 to 10^7 do od: i;"\right)
\mathrm{Grid}:-\mathrm{Wait}\left(\right)
\mathrm{Grid}:-\mathrm{GetLastResult}\left(0\right)
\textcolor[rgb]{0,0,1}{100001}
\mathrm{Grid}:-\mathrm{GetLastResult}\left(1\right)
\textcolor[rgb]{0,0,1}{1000001}
\mathrm{Grid}:-\mathrm{GetLastResult}\left(2\right)
\textcolor[rgb]{0,0,1}{10000001}
The Grid[Wait] command was introduced in Maple 2015.
Grid:-Barrier
Grid:-WaitForFirst
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Multi-position Dimensional Synthesis of a Spatial 3-RPS Parallel Manipulator | J. Mech. Des. | ASME Digital Collection
Multi-position Dimensional Synthesis of a Spatial
3-RPS
Parallel Manipulator
Nalluri Mohan Rao,
Nalluri Mohan Rao
J.N.T.University
, Kakinada - 533 003, A.P., India
K. Mallikarjuna Rao
Mohan Rao, N., and Mallikarjuna Rao, K. (August 15, 2005). "Multi-position Dimensional Synthesis of a Spatial
3-RPS
Parallel Manipulator." ASME. J. Mech. Des. July 2006; 128(4): 815–819. https://doi.org/10.1115/1.2205872
This paper presents dimensional synthesis of a 3 degrees of freedom (DOF) spatial 3-revolute-prismatic-spherical (RPS) parallel manipulator. Tsai and Kim ((2003) ASME J. Mech. Des., 125, pp. 92–97) have shown that the dimensional synthesis can be carried out for at the most six prescribed positions and orientations of the moving platform. The method of synthesis is modified (least-square technique) to make it possible to synthesize the 3-RPS manipulator for any number of positions and orientations of the moving platform. The effectiveness of the modified method of synthesis is demonstrated by considering an example for ten-position synthesis. The modified method of synthesis is an approximation method.
manipulators, least squares approximations
Manipulators, Reactor protection systems
Universal Tyre Test Machine
Proc. Of the 9th international congress of F.I.S.I.T.A.
A Platform with Six Degrees of Freedom
Proc. Inst. Mech. Engg.
Kinematic Analysis of a Three Degrees of Freedom In-Parallel Actuated Manipulator
Proc. IEEE International Conf. On Robotics and Automation
Dynamic Analysis of a Three Degrees of Freedom In-Parallel Actuated Manipulator
Kinematics of Three a Degrees-Of- Freedom Motion Platform for a Low Cost Driving Simulator
Parenti-Casltelli
Direct Singular Positions of 3RPS Parallel Manipulator
Kinematic Synthesis of a Spatial 3-RPS Parallel Manipulator
A Characterization of the Workspace Boundary of Three-Revolute Manipulators
Adaptive Fuzzy Compensation Control for Manipulator in Reduction Cell Monitor
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Problem 1d correction, hint - Murray Wiki
For part (d) of problem 1 of Hw#2, every instance of
{\displaystyle t}
{\displaystyle \tau }
, so the wording should be changed to read: "Consider the case where
{\displaystyle \zeta =0}
{\displaystyle v(\tau )=\sin \omega \tau ,\omega >1}
{\displaystyle z(\tau )}
, the normalized output of the oscillator, with initial conditions
{\displaystyle z_{1}(0)=z_{2}(0)=0}
If you've already solved it using
{\displaystyle t}
{\displaystyle \tau }
you will get equal credit (it is just a little bit more complex).
To solve this problem, you can use the "method of undetermined coefficients" (see, for example, http://www.efunda.com/math/ode/linearode_undeterminedcoeff.cfm) to solve for the steady-state frequency response solution. Then you can add to it a homogeneous solution that cancels the initial condition from the steady state so that the given initial conditions are satisfied. i.e., find
{\displaystyle z_{homog.}}
{\displaystyle z=z_{homog.}+z_{partic.}}
|
The first paper, named "On a Heuristic Viewpoint Concerning the Production and Transformation of Light", ("Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt") was specifically cited for his Nobel Prize. In this paper, Einstein extends Planck's hypothesis (
{\displaystyle E=h\nu }
) of discrete energy elements to his own hypothesis that electromagnetic energy is absorbed or emitted by matter in quanta of
{\displaystyle h\nu }
(where h is Planck's constant and
{\displaystyle \nu }
is the frequency of the light), proposing a new law
{\displaystyle E_{\max }=h\nu -P\,}
to account for the photoelectric effect, as well as other properties of photoluminescence and photoionization. In later papers, Einstein used this law to describe the Volta effect (1906), the production of secondary cathode rays (1909) and the high-frequency limit of Bremsstrahlung (1911). Einstein's key contribution is his assertion that energy quantization is a general, intrinsic property of light, rather than a particular constraint of the interaction between matter and light, as Planck believed. Another, often overlooked result of this paper was Einstein's excellent estimate (6.17
{\displaystyle \times }
1023) of Avogadro's number (6.02
{\displaystyle \times }
1023). However, Einstein does not propose that light is a particle in this paper; the "photon" concept was not proposed until 1909 (see below).
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Contest – HBO-i APC
The contest will be held over 5 hours. In these 5 hours you will try and solve as many of the approximately 10 so called problems. You will do this in a team of three people on one computer.
The problems consist of a descriptive text and an example. Every single problem expects an input file which you process and give the correct output file. These problems will be mathematical or algorithmic.
Example: Stand on Zanzibar
Source: BAPC 2015
Picture by: Mysid
As soon as Zanzibar has
1\phantom{\rule{thinmathspace}{0ex}}000\phantom{\rule{thinmathspace}{0ex}}000
1\, 000\, 000
turtles, the island is totally covered with turtles, and both reproduction and import come to a halt. Please help Zanzi! Write a program that computes the lower bound of import turtles, given a sequence, as described above.
The input starts with a line containing an integer
T
T
1\le T\le 13
1 \leq T \leq 13
), the number of test cases. Then for each test case:
One line containing a sequence of space-separated, positive integers (
\le 1\phantom{\rule{thinmathspace}{0ex}}000\phantom{\rule{thinmathspace}{0ex}}000
\leq 1\, 000\, 000
), non-decreasing, starting with one or more ones. For convenience, a single space and a 0 are appended to the end of the sequence.
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SymmetryPlot - Maple Help
Home : Support : Online Help : Statistics and Data Analysis : Statistics Package : Visualization : SymmetryPlot
SymmetryPlot(X, plotoptions)
The plotoptions argument can contain one or more of the options accepted by the plots[display] command. See plot[options] for details.
The SymmetryPlot command generates a symmetry plot for the specified data. The elements of X are sorted and plotted in such a way to show which way the data is skewed. If the data is indexed as X[i], med is the median of the data and n is the total number of observations, then a plot of X[n+1-i] - med versus med-X[i] for all i is produced together with the y=x line.
If the data set is symmetric with respect to the median, then the plot produced will have points lying along the line y=x. Any departure from this symmetry will be visible as points lying off of this line. If the data is skewed to the right, the points will lie above the line. If the data is skewed to the left, the points will lie below the line.
The parameter X is the data to be plotted. It is either a single data sample - given as e.g a Vector - or a list of data samples and does not need to be one dimensional, though it will be treated as though it were. There must be at least two points present for this function to create a plot.
This function is part of the Statistics package, so it can be used in the short form SymmetryPlot(..) only after executing the command with(Statistics). However, it can always be accessed through the long form of the command by using Statistics[SymmetryPlot](..).
\mathrm{with}\left(\mathrm{Statistics}\right):
This data is skewed to the right, and so the points lie above y=x.
\mathrm{data1}≔[1.3,3.2,2.1,4.76,7.33,2,0.91,5.5,1,2.4]:
\mathrm{SymmetryPlot}\left(\mathrm{data1}\right)
This data is skewed to the left, and so the points lie below y=x.
\mathrm{data2}≔\mathrm{Array}\left([[1,4,5,13],[15,12,14,16]]\right):
\mathrm{SymmetryPlot}\left(\mathrm{data2}\right)
This data is symmetric, and so the points fall on the line y=x.
\mathrm{data3}≔\mathrm{Vector}\left([1,2,3,4,5,6,7,8,9,10]\right):
\mathrm{SymmetryPlot}\left(\mathrm{data3}\right)
Added options can customize the plot:
\mathrm{SymmetryPlot}\left(\mathrm{data3},\mathrm{symbol}=\mathrm{circle},\mathrm{symbolsize}=15,\mathrm{thickness}=2,\mathrm{axis}=[\mathrm{gridlines}=[\mathrm{linestyle}=\mathrm{dot}]]\right)
\mathrm{data4}≔\mathrm{Array}\left([[1,4,5,13],[15,12,14,16],[6,2,14,13,12],[2,9,16,12,10],[3,8,2,10,14]]\right):
\mathrm{SymmetryPlot}\left(\mathrm{data4}\right)
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E.S. Earth in the Universe (Topic 3) Quiz - Quizizz
E.S. Earth in the Universe (Topic 3)DRAFT
dbor_06804
by dbor_06804
From your location, what is the best definition of a celestial object?
any object in the universe above Earth's atmosphere
any object in the universe outside our solar system
any object in the universe outside our home galaxy
The age of the universe is measured in 10 to 20
almost all galaxies appear to be moving away from Earth at tremendous velocities
few galaxies other than our own exist
all galaxies are approximately the same size
all galaxies are spiral in shape
Background radiation detected in space is believed to be evidence that
the universe began with an explosion
all universe is contracting
all matter in the universe is stationary
galaxies are evenly spaced throughout the the universe
Which statement best describes how most galaxies generally move
galaxies move toward one another
galaxies move away from one another
galaxies move randomly
galaxies do not move
Billions of stars in the same region of the universe are called
In which group are the parts listed in order from oldest to youngest?
solar system, Milky Way Galaxy, universe
Milky Way Galaxy, solar system, universe
universe, solar system, Milky Way Galaxy
universe, Milky Way Galaxy, solar system
The great system of 200 billion stars to which the sun and our solar system belong to is the
A star differs from a planet in that a star
has a fixed orbit
is self-luminous
revolves about the sun
shines by reflected light
The sun's energy is most likely the result of of the
fusion of hydrogen atoms
transformation of the sun's gravitational potential energy to heat energy
radioactive decay of uranium and thorium atoms
Nuclear fusion can only occur in areas of
As star color changes from blue to red, the surface temperature of the star
A luminosity and temperature of stars diagram classifies a star of high temperature and low luminosity as a
Base your answer to this question on the Characteristics of Stars diagram in the Earth Science Reference Tables.
A main sequence star is 1000 times more luminous than the sun. The temperature is likely to be most nearly
A white dwarf star has a temperature of 13,000 K. What is the probable luminosity of the star?
A giant star has a luminosity of 300. Its color is most likely to be
An orange star has a temperature of 4000 K and is 500,000 times as luminous as the sun. To which group does it belong?
The nearest star to the sun is Alpha Centauri. How does this star compare to the sun?
it is much hotter
it has a different color
it has a higher luminosity
it is much smaller in diameter
The sun is best described as a
medium-sized star
The sun is brighter than any star in the group of
Which type of star is associated with the last stage in the evolutionary development of most stars?
Comets are considered to belong to the solar system if they
glow by reflected light
revolve about the sun
have elliptical orbits
have uniform periods of revolution
A person observes that a bright object streaks across the nighttime sky in a few seconds. What is this object most likely to be
A belt of asteroids is located an average of distance of 503 million kilometers from the sun. Between which two planets is this belt located?
Which is not included in our solar system?
in the last billion years when meteorites, asteroids, or comets collided with Earth, which has occurred?
Craters have formed
The whole Earth's surface has melted
All life has been destroyed
The oceans have largely evaporated
Today it is most commonly believed that our solar system formed
by gravitational collapse of a gas-dust "cloud"
from material exploded from the sun
at the time of the Big Bang that formed the present universe
by fusion of matter between the sun and a passing star
The Jovian planets have more gravitational pull than the terrestrial planets. Therefore they have
a shorter year
higher average density
more low-density gases in their atmosphere
Which graph best illustrates the relationship between the diameter of a planet versus the distance of the planet from the sun?
Which graph best illustrates the relationship between the time it takes a planet to make one revolution around the sun versus the distance of the planet from the sun?
Use your Earth Science Reference Tables for the following question
The planet that has the shortest day is
Which member of the solar system has an equatorial diameter of
3.48\times10^3
The planets known as "gas giants," or Jovians, include Jupiter, Uranus and
Which three planets are known as terrestrial planets because of their high density and rocky composition.
Astronomers have observed a reddish spot on the surface of Jupiter. From observations of this spot, it is possible to estimate the
period of Jupiter's rotation
period of Jupiter's revolution
pressure of Jupiter's atmoshpere
temperature of Jupiter's surface
Which planet takes longer for one spin on its axis than for one orbit around the sun.
Which planet revolves fastest in its orbit?
Which diagram best approximates the shape of the path of Earth as it travels around the sun?
Use your Earth Science Reference Table to answer the following question
Which planet's orbit is most nearly circular?
As the distance between two objects in the universe increases, the gravitational attraction between these two objects
The diagram shows Earth (E) in orbit about the sun. If gravitational force were suddenly eliminated, toward which position would Earth then move?
masses are small and the objects are close together
masses are small and the objects are far apart
masses are large and the objects are close together
masses are large and the objects are far apart
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Simultaneous confidence intervals for ranks using the partitioning principle
2021 Simultaneous confidence intervals for ranks using the partitioning principle
Diaa Al Mohamad, Erik van Zwet, Aldo Solari, Jelle Goeman
Diaa Al Mohamad,1 Erik van Zwet,1 Aldo Solari,2 Jelle Goeman1
1Leiden University Medical Center, Einthovenweg 20. 2333 ZC Leiden, The Nethlerlands
2University of Milano-Bicocca, 1 Piazza dell’Ateneo Nuovo. 20126 Milano, Italy
We consider the problem of constructing simultaneous confidence intervals (CIs) for the ranks of n means based on their estimates together with the (known) standard errors of those estimates. We present a generic method based on the partitioning principle in which the parameter space is partitioned into disjoint subsets and then each one of them is tested at level α. The resulting CIs have then a simultaneous coverage of
1-\mathit{\alpha }
. We show that any procedure which produces simultaneous CIs for ranks can be written as a partitioning procedure. We present a first example where we test the partitions using the likelihood ratio (LR) test. Then, in a second example we show that a recently proposed method for simultaneous CIs for ranks using Tukey’s honest significant difference test has an equivalent procedure based on the partitioning principle. By embedding these two methods inside our generic partitioning procedure, we obtain improved variants. We illustrate the performance of these methods through simulations and real data analysis on hotel ratings. While the novel method that uses the LR test and its variant produce shorter CIs when the number of means is small, the Tukey-based method and its variant produce shorter CIs when the number of means is high.
This research is supported by VIDI grant number 639.072.412.
Diaa Al Mohamad. Erik van Zwet. Aldo Solari. Jelle Goeman. "Simultaneous confidence intervals for ranks using the partitioning principle." Electron. J. Statist. 15 (1) 2608 - 2646, 2021. https://doi.org/10.1214/21-EJS1847
Keywords: likelihood ratio test , rankings , simple order , Tukey’s HSD
Diaa Al Mohamad, Erik van Zwet, Aldo Solari, Jelle Goeman "Simultaneous confidence intervals for ranks using the partitioning principle," Electronic Journal of Statistics, Electron. J. Statist. 15(1), 2608-2646, (2021)
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Normal Criterion Concerning Shared Values
2012 Normal Criterion Concerning Shared Values
Wei Chen, Yingying Zhang, Jiwen Zeng, Honggen Tian
We study normal criterion of meromorphic functions shared values, we obtain the following. Let
F
be a family of meromorphic functions in a domain
D
, such that function
f\in F
has zeros of multiplicity at least 2, there exists nonzero complex numbers
{b}_{f},{c}_{f}
f
\left(\text{i}\right){b}_{f}/{c}_{f}
is a constant;
\left(\text{i}\text{i}\right)\mathrm{min} \left\{\sigma \left(0,{b}_{f}\right),\sigma \left(0,{c}_{f}\right),\sigma \left({b}_{f},{c}_{f}\right)\ge m\right\}
m>0
\left(\text{i}\text{i}\text{i}\right)\left(1/{c}_{f}^{k-1}\right)\left({f}^{\prime }{\right)}^{k}\left(z\right)+f\left(z\right)\ne {b}_{f}^{k}/{c}_{f}^{k-1}
\left(1/{c}_{f}^{k-1}\right)\left({f}^{\prime }{\right)}^{k}\left(z\right)+f\left(z\right)={b}_{f}^{k}/{c}_{f}^{k-1}⇒f\left(z\right)={b}_{f}
F
is normal. These results improve some earlier previous results.
Wei Chen. Yingying Zhang. Jiwen Zeng. Honggen Tian. "Normal Criterion Concerning Shared Values." J. Appl. Math. 2012 1 - 7, 2012. https://doi.org/10.1155/2012/312324
Wei Chen, Yingying Zhang, Jiwen Zeng, Honggen Tian "Normal Criterion Concerning Shared Values," Journal of Applied Mathematics, J. Appl. Math. 2012(none), 1-7, (2012)
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Fourier transform - MATLAB fourier - MathWorks Italia
Fourier Transform of Common Inputs
Fourier Transforms Involving Dirac and Heaviside Functions
Specify Fourier Transform Parameters
Fourier Transform of Array Inputs
If Fourier Transform Cannot Be Found
fourier(f) returns the Fourier Transform of f. By default, the function symvar determines the independent variable, and w is the transformation variable.
fourier(f,transVar) uses the transformation variable transVar instead of w.
fourier(f,var,transVar) uses the independent variable var and the transformation variable transVar instead of symvar and w, respectively.
Compute the Fourier transform of common inputs. By default, the transform is in terms of w.
Unit impulse (Dirac delta)
Step (Heaviside)
Right-sided exponential
Also calculate transform with condition a > 0. Clear assumptions.
Double-sided exponential
Assume a > 0. Clear assumptions.
Assume b and c are real. Simplify result and clear assumptions.
Bessel of first kind with nu = 1
Compute the Fourier transform of exp(-t^2-x^2). By default, symvar determines the independent variable, and w is the transformation variable. Here, symvar chooses x.
Specify the transformation variable as y. If you specify only one variable, that variable is the transformation variable. symvar still determines the independent variable.
Specify both the independent and transformation variables as t and y in the second and third arguments, respectively.
Compute the following Fourier transforms. The results are in terms of the Dirac and Heaviside functions.
Specify parameters of the Fourier transform.
Compute the Fourier transform of f using the default values of the Fourier parameters c = 1, s = -1. For details, see Fourier Transform.
Change the Fourier parameters to c = 1, s = 1 by using sympref, and compute the transform again. The result changes.
Change the Fourier parameters to c = 1/(2*pi), s = 1. The result changes.
Preferences set by sympref persist through your current and future MATLAB® sessions. Restore the default values of c and s by setting FourierParameters to 'default'.
Find the Fourier transform of the matrix M. Specify the independent and transformation variables for each matrix entry by using matrices of the same size. When the arguments are nonscalars, fourier acts on them element-wise.
If fourier is called with both scalar and nonscalar arguments, then it expands the scalars to match the nonscalars by using scalar expansion. Nonscalar arguments must be the same size.
If fourier cannot transform the input then it returns an unevaluated call.
Return the original expression by using ifourier.
x (default) | symbolic variable
Independent variable, specified as a symbolic variable. This variable is often called the "time variable" or the "space variable." If you do not specify the variable, then fourier uses the function symvar to determine the independent variable.
w (default) | v | symbolic variable | symbolic expression | symbolic vector | symbolic matrix
Transformation variable, specified as a symbolic variable, expression, vector, or matrix. This variable is often called the "frequency variable." By default, fourier uses w. If w is the independent variable of f, then fourier uses v.
The Fourier transform of the expression f = f(x) with respect to the variable x at the point w is
F\left(w\right)=c\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right){e}^{iswx}dx.
c and s are parameters of the Fourier transform. The fourier function uses c = 1, s = –1.
If any argument is an array, then fourier acts element-wise on all elements of the array.
To compute the inverse Fourier transform, use ifourier.
fourier does not transform piecewise. Instead, try to rewrite piecewise by using the functions heaviside, rectangularPulse, or triangularPulse.
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Data Types · USACO Guide
HomeGeneralData Types
sizes + ranges
1.3 - Working with numbers
Integers, Modular arithmetic, Floating point numbers
2.3 - Variables & Types
C++: Common Fundamental Data Types
Note: These numbers may vary depending on your machine and/or compiler. For more fundamental data types, check the the first resource in the table above.
Description 32-bit integer 64-bit integer Double-precision float True/False value 8-bit character
Size (bytes) 4 8 8 1 1
-2^{31}
2^{31}-1
-2^{63}
2^{63}-1
-1.7E+308 to +1.7E+308
0
1
(true or false)
-2^7
2^7-1
Java: Common Primitive Data Types
For more primitive data types, check the the first resource in the table above.
Description 32-bit integer 64-bit integer Double-precision float True/False value 16-bit Unicode character
Size (bytes) 4 8 8 1 bit(*) 2
-2^{31}
2^{31}-1
-2^{63}
2^{63}-1
-1.7E+308 to +1.7E+308 true/false \u0000 to \uffff (
0-65535
*Note: It's unlikely that booleans will actually use only 1 bit of memory, as in most cases data types must be aligned to bytes. However, only one bit of information can be stored in them.
Description Arbitrary-size integer Double-precision (64 bit) IEEE 754 float True/False value String
Values Any integer -1.7E+308 to +1.7E+308 true/false Any length of text
There are several main data types that are used in contests: integers, floating point numbers, booleans, characters, and strings. Assuming that you are familiar with the language you are using, this should be mostly review.
The normal 32-bit integer data type (int in C++ and Java) supports values between
-2\,147\,483\,648
2\,147\,483\,647
, which is roughly equal to
\pm 2 \cdot 10^9
Some problems require you to use 64-bit integers (long long in C++ and long in Java) instead of 32-bit integers (int). 64-bit integers are less likely to have overflow issues, since they can store any number between
-9\,223\,372\,036\,854\,775\,808
9\,223\,372\,036\,854\,775\,807
which is roughly equal to
\pm 9 \times 10^{18}
. In Python, ints have unlimited size.
Sometimes (but not always) a USACO problem statement (ex. Haircut) will contain a warning such as the following:
Contest problems are usually set such that the 64-bit integer is sufficient, so for lower divisions it might be a good idea to use 64-bit integers in place of 32-bit integers everywhere. Of course, you shouldn't do this when time and/or memory limits are tight, which may be the case in higher divisions of USACO. Also note that in Java, you will need to cast long back to int when accessing array indices.
Additionally, there exist 16-bit integers (short in C++ and Java). However, these are generally not useful as the extra memory saved by using them is usually negligible. Unsigned integers (unsigned int, unsigned long long, etc.) also exist. They aren't used as frequently, though the 2-fold increase in size is sometimes the difference between overflowing and not overflowing.
Floating point numbers are used to store decimal values. It is important to know that floating point numbers are not exact, because the binary architecture of computers can only store decimals to a certain precision. Hence, we should always expect that floating point numbers are slightly off, so it's generally a bad idea to compare two floating-point numbers for exact equality (==).
Contest problems will usually accommodate the inaccuracy of floating point numbers by checking if the absolute or relative difference between your output and the answer is less than some small constant like
\epsilon=10^{-9}
If your output is
x
y
, the absolute difference is
|x-y|
x
y
, the relative difference is
\frac{|x-y|}{|y|}
This is not the case for USACO, where problems generally have a unique correct output. So when floating point is necessary, the output format will be something along the lines of "Print
10^6
times the maximum probability of receiving exactly one accepted invitation, rounded down to the nearest integer." (ex. Cow Dating).
Boolean variables have two possible states: true and false. We'll usually use booleans to mark whether a certain process is done, and arrays of booleans to mark which components of an algorithm have finished. Booleans require 1 byte (8 bits) of storage, not 1 bit, wasting the other 7 bits of storage. To use less memory, one can use bitsets (std::bitset in C++ / BitSet in Java). Unfortunately, bitsets are not available in Python.
Character variables represent a single character. They are returned when you access the character at a certain index within a string. Characters are represented using the ASCII standard, which assigns each character to a corresponding integer. This allows us to do arithmetic with them; for example, both cout << ('f' - 'a'); in C++ and System.out.print('f' - 'a'); in Java will print 5. In Java, characters are 16 bits, while in C/C++, characters are 8 bits.
Strings are effectively arrays of characters. You can easily access the character at a certain index and take substrings of the string (charAt() and substring() in Java).
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Does Labor-Saving R&D Hurt Labors?
Department of Economics, National Central University, Taiwan
\pi =\left[P-c-w\left(1-x\right)\right]Q-{x}^{2}/2
\varphi =w\left(\text{1}-x\right)Q
Q=\frac{a-w}{2-{w}^{2}}
x=\frac{w\left(a-w\right)}{2-{w}^{2}}
\frac{\partial Q}{\partial w}=-\frac{2-w\left(2a-w\right)}{{\left(2-{w}^{2}\right)}^{2}}\gtrless 0
a\gtrless {a}_{1}
\frac{\partial x}{\partial w}=\frac{a\left(2+{w}^{2}\right)-4w}{{\left(2-{w}^{2}\right)}^{2}}\gtrless 0
a\gtrless {a}_{\text{2}}
{w}^{*}=\left\{\begin{array}{l}a-\sqrt{{a}^{2}-2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a>\sqrt{2},\\ \frac{2-\sqrt{4-2{a}^{2}}}{a},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a\le \sqrt{2}.\end{array}
{x}^{*}=\left\{\begin{array}{l}\frac{1}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{for}\text{\hspace{0.17em}}a>\sqrt{2},\\ \frac{2-\sqrt{4-2{a}^{2}}}{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a\le \sqrt{2},\end{array}
{Q}^{*}=\left\{\begin{array}{l}\frac{a+\sqrt{{a}^{2}-2}}{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a>\sqrt{2},\\ \frac{a}{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a\le \sqrt{2},\end{array}
{\pi }^{*}=\left\{\begin{array}{l}\frac{{a}^{2}-2+a\sqrt{{a}^{2}-2}}{8},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{for}\text{\hspace{0.17em}}a>\sqrt{2},\\ \frac{{a}^{2}-2+\sqrt{{a}^{2}-2}}{8},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a\le \sqrt{2}.\end{array}
{\varphi }^{*}=\left\{\begin{array}{l}\frac{1}{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a>\sqrt{2},\\ \frac{{a}^{2}}{8},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a\le \sqrt{2}.\end{array}
{w}^{\ast }-{w}^{\text{o}}=\left\{\begin{array}{l}\frac{a-2\sqrt{{a}^{2}-2}}{2}\gtrless 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{for}\text{\hspace{0.17em}}a\lessgtr 2\sqrt{\frac{2}{3}},\\ \frac{4-{a}^{2}-2\sqrt{4-2{a}^{2}}}{2a}>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a\le \sqrt{2},\end{array}
{Q}^{\ast }-{Q}^{\text{o}}=\left\{\begin{array}{l}\frac{\sqrt{{a}^{2}-2}}{4}>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a>\sqrt{2},\\ 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{for}\text{\hspace{0.17em}}a\le \sqrt{2},\end{array}
{\pi }^{\ast }-{\pi }^{\text{o}}=\left\{\begin{array}{l}\frac{{a}^{2}-4+2a\sqrt{{a}^{2}-2}}{16}\gtrless 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{for}\text{\hspace{0.17em}}a\gtrless \frac{2\left({3}^{3/4}\right)}{3},\\ \frac{{a}^{2}-4+2\sqrt{4-{a}^{2}}}{16}<0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }\text{for}\text{\hspace{0.17em}}a\le \sqrt{2}.\end{array}
{\varphi }^{\ast }-{\varphi }^{\text{o}}=\left\{\begin{array}{l}0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a\le \sqrt{2},\\ \frac{2-{a}^{2}}{8}<0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}a>\sqrt{2},\end{array}
a<2\sqrt{6}/3
a>2\sqrt{6}/3
Wang, Y.-D. (2019) Does Labor-Saving R&D Hurt Labors? Theoretical Economics Letters, 9, 1737-1743. https://doi.org/10.4236/tel.2019.96111
1. International Federation of Robotics (2018) Executive Summary World Robotics 2018 Industrial Robots.
2. Menezes-Filho, N. and Van Reenen, J. (2003) Unions and Innovation: A Survey of the Theory and Empirical Evidence. In: Addison, J.T. and Schnabel, C., Eds., International Handbook of Trade Unions, Edward Elgar Publishing Ltd., Cheltenham, 293-334.
3. Tauman, Y. and Weiss, Y. (1987) Labor Unions and the Adoption of New Technology. Journal of Labor Economics, 5, 477-501. https://doi.org/10.1086/298158
4. Dowrick, S. and Spencer, S.J. (1994) Union Attitudes to Labor-Saving Innovation: When are Unions Luddites. Journal of Labor Economics, 12, 316-344. https://doi.org/10.1086/298359
5. Ulph, A. and Ulph, D. (1989) Labour Markets and Innovation. Journal of Japanese and International Economics, 3, 403-423. https://doi.org/10.1016/0889-1583(89)90011-7
6. Ulph, A. and Ulph, D. (1998) Labour Markets, Bargaining and Innovation. European Economic Review, 42, 931-939. https://doi.org/10.1016/S0014-2921(97)00147-5
7. Ulph, A. and Ulph, D. (2001) Strategic Innovation with Complete and Incomplete Labour Market Contracts. Scandinavian Journal of Economics, 103, 265-282. https://doi.org/10.1111/1467-9442.00244
8. Haucap, J. and Wey, C. (2004) Unionisation Structures and Innovation Incentives. Economic Journal, 114, 149-165. https://doi.org/10.1111/j.0013-0133.2004.00203.x
9. Mukherjee, A., Broll, U. and Mukherjee, S. (2007) Unionized Labor Market and Licensing by a Monopolist. Journal of Economics, 93, 59-79. https://doi.org/10.1007/s00712-007-0293-z
10. Manasakis, C. and Petrakis, E. (2009) Union Structure and Firms Incentives for Cooperative R&D Investments. Canadian Journal of Economics, 42, 656-693. https://doi.org/10.1111/j.1540-5982.2009.01523.x
11. Mukherjee, A. and Pennings, E. (2011) Unionization Structure, Licensing and Innovation. International Journal of Industrial Organization, 29, 232-241. https://doi.org/10.1016/j.ijindorg.2010.06.001
12. Kesavayuth, K. and Zikos, V. (2013) International and National R&D Networks in Unionized Oligopoly. Labour, 27, 18-37. https://doi.org/10.1111/labr.12006
13. Basak, D. and Mukherjee, A. (2017) Labour Unionisation Structure and Product Innovation. International Review of Economics and Finance, 55, 98-110. https://doi.org/10.1016/j.iref.2017.12.013
14. Beladi, H. and Mukherjee, A. (2017) Union Bargaining Power, Subcontracting and Innovation. Journal of Economic Behavior & Organization, 137, 90-104. https://doi.org/10.1016/j.jebo.2017.02.013
a=\left(2+{w}^{2}\right)/2w\equiv {a}_{1}
a=4w/\left(2+{w}^{2}\right)\equiv {a}_{2}
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Model shunt RLC network - Simulink - MathWorks France
Model shunt RLC network
The Shunt RLC block models the shunt RLC network described in the block dialog box, in terms of its frequency-dependent S-parameters.
For the given resistance, inductance, and capacitance, the block first calculates the ABCD-parameters at each frequency contained in the vector of modeling frequencies, and then converts the ABCD-parameters to S-parameters using the RF Toolbox™ abcd2s function. See the Output Port block reference page for information about determining the modeling frequencies.
For this circuit, A = 1, B = 0, C = Y, and D = 1, where
Y=\frac{-LC{\omega }^{2}+j\left(L/R\right)\omega +1}{jL\omega }
\omega =2\pi f
The shunt RLC object is a two-port network as shown in the following circuit diagram.
Resistance (Ohms) — Resistance of shunt RLC network
Scalar value for the resistance. The value must be non-negative.
Inductance (H) — Inductance of shunt RLC network
Capacitance (F) — Capacitance of shunt RLC network
General Passive Network | LC Bandpass Pi | LC Bandpass Tee | LC Bandstop Pi | LC Bandstop Tee | LC Highpass Pi | LC Highpass Tee | LC Lowpass Pi | LC Lowpass Tee | Series C | Series L | Series R | Series RLC | Shunt C | Shunt L | Shunt R
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Induction [Isabelle/HOL Support Wiki]
Trace: • Induction
reference:induct
Applying Induction Rules By Hand
Induction methods can be used with induction rules to prove theorems over inductive datatypes or sets as well as recursive functions.
Note that induction typically creates multiple new subgoals: one for each base case and one for the inductive step.
Performs an induction over a free variable. Consider this example:
lemma "a#l = [a]@l"
induct tries to find a suitable induction rule. See below for details how to have it use a specific rule.
Performs an induction over a free or meta-quantified variable that should not occur among the assumptions.
lemma "⋀a l. a#l = [a]@l"
induct_tac tries to find a suitable induction rule. See below for details how to have it use a specific rule.
Theorems about recursive functions are proved by induction.
Generalise goals for induction by
replacing constants with variables.
universally quantifying all free variables (except the induction variable itself).
The right-hand side of an equation should (in some sense) be simpler than the left-hand side.
Put all occurrences of the induction variable into the conclusion using
\longrightarrow
The induction methods can not always find a fitting induction rule. In other cases, the one found might be unsuitable to prove the current goal.
!!TODO!!
If you have a specific induction rule you want to apply—for instance one provided by an inductive definition— you can use it as you would any rule. As an example, consider the set of all even natural numbers, defined by:
inductive_set even :: "nat set" where
"0 ∈ even"
| "n ∈ even ⟹ (Suc (Suc n)) ∈ even"
Then, Isabelle automatically generates an induction rule even.induct. If we now want to prove a theorem like
\quad n \in \text{even}\ \Longrightarrow 2\ \text{dvd}\ n
a reasonable approach is to do an induction over all even numbers; this can be done by
apply (erule even.induct)
reference/induct.txt · Last modified: 2011/07/13 12:52 by 131.246.161.187
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Pade approximant - MATLAB pade - MathWorks Italia
Find Padé Approximant for Symbolic Expressions
Specify Expansion Variable
Approximate Value of Function at Particular Point
Increase Accuracy of Padé Approximant
Plot Accuracy of Padé Approximant
pade(f,var)
pade(f,var,a)
pade(___,Name,Value)
pade(f,var) returns the third-order Padé approximant of the expression f at var = 0. For details, see Padé Approximant.
If you do not specify var, then pade uses the default variable determined by symvar(f,1).
pade(f,var,a) returns the third-order Padé approximant of expression f at the point var = a.
pade(___,Name,Value) uses additional options specified by one or more Name,Value pair arguments. You can specify Name,Value after the input arguments in any of the previous syntaxes.
Find the Padé approximant of sin(x). By default, pade returns a third-order Padé approximant.
pade(sin(x))
-(x*(7*x^2 - 60))/(3*(x^2 + 20))
If you do not specify the expansion variable, symvar selects it. Find the Padé approximant of sin(x) + cos(y). The symvar function chooses x as the expansion variable.
pade(sin(x) + cos(y))
(- 7*x^3 + 3*cos(y)*x^2 + 60*x + 60*cos(y))/(3*(x^2 + 20))
Specify the expansion variable as y. The pade function returns the Padé approximant with respect to y.
pade(sin(x) + cos(y),y)
(12*sin(x) + y^2*sin(x) - 5*y^2 + 12)/(y^2 + 12)
Find the value of tan(3*pi/4). Use pade to find the Padé approximant for tan(x) and substitute into it using subs to find tan(3*pi/4).
P = pade(f);
y = subs(P,x,3*pi/4)
(pi*((9*pi^2)/16 - 15))/(4*((9*pi^2)/8 - 5))
Use vpa to convert y into a numeric value.
You can increase the accuracy of the Padé approximant by increasing the order. If the expansion point is a pole or a zero, the accuracy can also be increased by setting OrderMode to relative. The OrderMode option has no effect if the expansion point is not a pole or zero.
Find the Padé approximant of tan(x) using pade with an expansion point of 0 and Order of [1 1]. Find the value of tan(1/5) by substituting into the Padé approximant using subs, and use vpa to convert 1/5 into a numeric value.
p11 = pade(tan(x),x,0,'Order',[1 1])
p11 = subs(p11,x,vpa(1/5))
Find the approximation error by subtracting p11 from the actual value of tan(1/5).
y = tan(vpa(1/5));
error = y - p11
Increase the accuracy of the Padé approximant by increasing the order using Order. Set Order to [2 2], and find the error.
p22 = subs(p22,x,vpa(1/5));
-(3*x)/(x^2 - 3)
The accuracy increases with increasing order.
If the expansion point is a pole or zero, the accuracy of the Padé approximant decreases. Setting the OrderMode option to relative compensates for the decreased accuracy. For details, see Padé Approximant. Because the tan function has a zero at 0, setting OrderMode to relative increases accuracy. This option has no effect if the expansion point is not a pole or zero.
p22Rel = pade(tan(x),x,0,'Order',[2 2],'OrderMode','relative')
p22Rel = subs(p22Rel,x,vpa(1/5));
error = y - p22Rel
p22Rel =
(x*(x^2 - 15))/(3*(2*x^2 - 5))
The accuracy increases if the expansion point is a pole or zero and OrderMode is set to relative.
Plot the difference between exp(x) and its Padé approximants of orders [1 1] through [4 4]. Use axis to focus on the region of interest. The plot shows that accuracy increases with increasing order of the Padé approximant.
expr = exp(x);
fplot(expr - pade(expr,'Order',i))
legend('Order [1,1]','Order [2,2]','Order [3,3]','Order [4,4]',...
title('Difference Between exp(x) and its Pade Approximant')
Input to approximate, specified as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression.
Expansion variable, specified as a symbolic variable. If you do not specify var, then pade uses the default variable determined by symvar(f,1).
number | symbolic number | symbolic variable | symbolic function | symbolic expression
Expansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You also can specify the expansion point as a Name,Value pair argument. If you specify the expansion point both ways, then the Name,Value pair argument takes precedence.
Example: pade(f,'Order',[2 2]) returns the Padé approximant of f of order m = 2 and n = 2.
Expansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You can also specify the expansion point using the input argument a. If you specify the expansion point both ways, then the Name,Value pair argument takes precedence.
Order — Order of Padé approximant
integer | vector of two integers | symbolic integer | symbolic vector of two integers
Order of the Padé approximant, specified as an integer, a vector of two integers, or a symbolic integer, or vector of two integers. If you specify a single integer, then the integer specifies both the numerator order m and denominator order n producing a Padé approximant with m = n. If you specify a vector of two integers, then the first integer specifies m and the second integer specifies n. By default, pade returns a Padé approximant with m = n = 3.
OrderMode — Flag that selects absolute or relative order for Padé approximant
Flag that selects absolute or relative order for Padé approximant, specified as 'absolute' or 'relative'. The default value of 'absolute' uses the standard definition of the Padé approximant. If you set 'OrderMode' to 'relative', it only has an effect when there is a pole or a zero at the expansion point a. In this case, to increase accuracy, pade multiplies the numerator by (var - a)p where p is the multiplicity of the zero or pole at the expansion point. For details, see Padé Approximant.
By default, pade approximates the function f(x) using the standard form of the Padé approximant of order [m, n] around x = x0 which is
\frac{{a}_{0}+{a}_{1}\left(x-{x}_{0}\right)+...+{a}_{m}{\left(x-{x}_{0}\right)}^{m}}{1+{b}_{1}\left(x-{x}_{0}\right)+...+{b}_{n}{\left(x-{x}_{0}\right)}^{n}}.
When OrderMode is relative, and a pole or zero exists at the expansion point x = x0, the pade function uses this form of the Padé approximant
\frac{{\left(x-{x}_{0}\right)}^{p}\left({a}_{0}+{a}_{1}\left(x-{x}_{0}\right)+...+{a}_{m}{\left(x-{x}_{0}\right)}^{m}\right)}{1+{b}_{1}\left(x-{x}_{0}\right)+...+{b}_{n}{\left(x-{x}_{0}\right)}^{n}}.
The parameters p and a0 are given by the leading order term f = a0 (x - x0)p + O((x - x0)p + 1) of the series expansion of f around x = x0. Thus, p is the multiplicity of the pole or zero at x0.
If you use both the third argument a and ExpansionPoint to specify the expansion point, the value specified via ExpansionPoint prevails.
The parameters a1,…,bn are chosen such that the series expansion of the Padé approximant coincides with the series expansion of f to the maximal possible order.
The expansion points ±∞ and ±i∞ are not allowed.
When pade cannot find the Padé approximant, it returns the function call.
For pade to return the Padé approximant, a Taylor or Laurent series expansion of f must exist at the expansion point.
series | taylor
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Verdant Force: Discoveries in Life and Proteomics: Blogger posting with Markdown using StackEdit
I recently discovered StackEdit, a tool for writing and previewing Markdown.
Now, I’ve been using Markdown for quite sometime, it’s useful for everything from electronic lab notebooks to taking notes on informational interviews.
Finally, through StackEdit’s built-in post to Blogger feature, I’m able to abandon the clunky default interface, and write posts the way they were meant to be written!
Additional benefits include easy code snippets…
def stackedit():
print 'I love your product!'
Easy inspirational quotes…
And easy equations (same syntax as LaTeX
, by the way)…
ΔG=12kBTln(Keq)
\Delta G = \frac{1}{2} k_\mathrm{B} T ln(K_\textrm{eq})
Needless to say, I’m excited about the switch, and would highly recommend giving StackEdit a try yourself!
Posted by Verdant Force at Friday, May 16, 2014
Labels: Code, Markdown, Tips
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Enhanced physical downlink control channel (EPDCCH) decoding - MATLAB lteEPDCCHDecode - MathWorks Switzerland
lteEPDCCHDecode
Decode EPDCCH codeword
Decode EPDCCH DCI Message
hestgrid
Syntax Dependent Processing
Enhanced physical downlink control channel (EPDCCH) decoding
[bits,symbols] = lteEPDCCHDecode(enb,chs,sym)
[bits,symbols] = lteEPDCCHDecode(enb,chs,rxsym,hest,noiseest)
[bits,symbols] = lteEPDCCHDecode(enb,chs,rxsym,hest,noiseest,alg)
[bits,symbols] = lteEPDCCHDecode(enb,chs,grid)
[bits,symbols] = lteEPDCCHDecode(enb,chs,rxgrid,hestgrid,noiseest,alg)
[bits,symbols] = lteEPDCCHDecode(enb,chs,sym) returns softbits and received constellation of complex symbols resulting from performing the inverse of enhanced physical downlink control channel (EPDCCH) processing of a single configured EPDCCH candidate given cell-wide settings structure, EPDCCH transmission configuration structure, and EPDCCH symbols. The input symbols are assumed to contain ideal EPDCCH symbols, so no equalization is performed. The output received EPDCCH symbols are QPSK symbol demodulated and descrambled. For more EPDCCH processing information, see lteEPDCCH and TS 36.211[1], Section 6.8A.
When using this syntax, the input structures only require enb.NSubframe and chs.EPDCCHNID.
For more information, see Syntax Dependent Processing.
[bits,symbols] = lteEPDCCHDecode(enb,chs,rxsym,hest,noiseest) performs EPDCCH decoding and equalization for a single configured EPDCCH candidate given cell-wide settings structure, EPDCCH transmission configuration structure, received EPDCCH symbols rxsym, channel estimate hest, and noise estimate noiseest. The output received EPDCCH symbols are equalized, and QPSK symbol demodulated and descrambled.
[bits,symbols] = lteEPDCCHDecode(enb,chs,rxsym,hest,noiseest,alg) performs EPDCCH decoding and equalization for a single configured EPDCCH candidate and provides control over weighting the output soft bits with channel state information (CSI) calculated during the equalization stage using algorithmic configuration structure, alg.
[bits,symbols] = lteEPDCCHDecode(enb,chs,grid) performs EPDCCH decoding for all possible EPDCCH candidate locations given cell-wide settings structure, EPDCCH transmission configuration structure, and the resource element grid across all possible EPDCCH antenna ports. The resource element grid is assumed to contain ideal EPDCCH REs, so no equalization is performed. The decoding consists of extraction of all EPDCCH REs from grid followed by QPSK symbol demodulation. Each EPDCCH candidate is descrambled individually during EPDCCH search. For this syntax, chs.EPDCCHECCE and chs.EPDCCHNID are not required as no candidate-specific resource extraction or descrambling is performed.
[bits,symbols] = lteEPDCCHDecode(enb,chs,rxgrid,hestgrid,noiseest,alg) performs EPDCCH decoding and equalization for all possible EPDCCH candidate locations given cell-wide settings structure, EPDCCH transmission configuration structure, received resource element grid, channel estimate grid, noise estimate, and provides control over weighting the output soft bits with channel state information (CSI) calculated during the equalization stage using algorithmic configuration structure, alg. EPDCCH RE locations extracted from rxgrid and hestgrid are equalized, then QPSK symbol demodulated. Each EPDCCH candidate is descrambled individually during EPDCCH search. For this syntax, chs.EPDCCHECCE and chs.EPDCCHNID are not required as no candidate-specific resource extraction or descrambling is performed.
Modulate and then demodulate EPDCCH symbols for a codeword of random bits.
Initialize the cell-wide settings structure and the EPDCCH transmission channel configuration structure.
Create an input codeword for EPDCCH and generate EPDCCH symbols.
sym = lteEPDCCH(enb,chs,cw);
Decode the symbols and confirm the codeword was successfully recovered.
rxcw = lteEPDCCHDecode(enb,chs,sym);
Perform DCI coding to the capacity of a particular EPDCCH candidate. EPDCCH modulate the coded message and transmit it. Add EPDCCH demodulation reference signals (DMRS) and perform channel estimation. Finally, extract the EPDCCH (and corresponding channel estimate) from the resource grid. Perform EPDCCH demodulation and decode the received DCI message.
Initialize the cell-wide settings structure.
enb.CyclicPrefix = 'Extended';
Initialize the EPDCCH transmission channel configuration structure.
chs.EPDCCHPRBSet = (0:3).';
chs.EPDCCHFormat = 2;
chs.DCIFormat = 'Format2D';
chs.RNTI = 11;
Create a set of random bits representing a DCI message and performing DCI coding to the capacity of a particular EPDCCH candidate.
dciInfo = lteDCIInfo(enb,chs);
dcibits = randi([0 1],dciInfo.(chs.DCIFormat),1);
candidates = lteEPDCCHSpace(enb,chs);
chs.EPDCCHECCE = candidates(1,:);
[ind,info] = lteEPDCCHIndices(enb,chs);
cw = lteDCIEncode(chs,dcibits,info.EPDCCHG);
Generate EPDCCH symbols and resource element grid. Populate the grid.
grid = lteDLResourceGrid(enb,4);
grid(lteEPDCCHDMRSIndices(enb,chs)) = lteEPDCCHDMRS(enb,chs);
Generate channel estimate.
cec.Reference = 'EPDCCHDMRS';
[hestgrid,noiseest] = lteDLChannelEstimate(enb,chs,cec,grid);
[rxsym,hest] = lteExtractResources(ind,grid,hestgrid);
Decode the symbols and DCI message bits. Confirm the DCI message was successfully recovered.
rxcw = lteEPDCCHDecode(enb,chs,rxsym,hest,noiseest);
rxdcibits = lteDCIDecode(dciInfo.(chs.DCIFormat),rxcw);
isequal(dcibits,rxdcibits>0)
{N}_{\text{RB}}^{\text{DL}}
The following parameter is only read when chs.EPDCCHStart is not present.
The following zero power CSI-RS resource parameter is only applicable if one or more of the above zero power subframe configurations are set to any value other than 'Off'.
chs — EPDCCH-specific channel transmission configuration
EPDCCH-specific channel transmission configuration, specified as a structure that can contain the following parameter fields.
1- or 2- element vector specifying the zero-based ECCE index or inclusive [begin, end] ECCE index range according to the aggregation level L, where L = end – begin + 1. The number of ECCEs in the candidate must be a power of 2.
The set of one of several consecutive ECCEs defining the EPDCCH transmission candidate in the overall EPDCCH set.
EPDCCHStart Optional
If this parameter is not present, then the cell-wide CFI parameter is used for the starting symbol.
EPDCCH starting symbol
EPDCCH scrambling sequence initialization
The following parameter applies when EPDCCHType is set to 'Localized'.
sym — EPDCCH modulation symbols
complex-vector
EPDCCH modulation symbols associated with a single EPDCCH transmission in a subframe, specified as a complex vector. This input contains the QPSK symbols.
rxsym — Received EPDCCH symbols
EPDCCHGd-by-NRxAnts matrix
Received EPDCCH symbols, specified as a EPDCCHGd-by-NRxAnts matrix. EPDCCHGd is the EPDCCH symbol capacity, given by the info.EPDCCHGd field of lteEPDCCHIndices. NRxAnts is the number of receive antennas. This matrix contains the elements of the received resource grid in the locations of the EPDCCH REs for the candidate configured via chs.EPDCCHECCE.
EPDCCHGd-by-NRxAnts-by-4 array
Channel estimate, specified as an EPDCCHGd-by-NRxAnts array. EPDCCHGd is the EPDCCH symbol capacity, given by the info.EPDCCHGd field of lteEPDCCHIndices. NRxAnts is the number of receive antennas. The third dimension represents the 4 possible EPDCCH antenna ports (p=107...110). This array contains the elements of the channel estimate array in the locations of the EPDCCH REs for the candidate configured via chs.EPDCCHECCE.
Noise estimate of the noise power spectral density per RE on the received subframe, specified as a numeric scalar. The lteDLChannelEstimate function provides this estimate.
Algorithmic configuration, specified as a structure. The structure must have the field:
Flag provides control over weighting the soft values that are used to determine the output values with the channel state information (CSI) calculated during the equalization process
K-by-L-by-4 array
Resource grid across the four possible EPDCCH ports, specified as a K-by-L-by-4 array. K is the number of subcarriers, L is the number of OFDM symbols in one frame, and 4 is all possible EPDCCH antenna ports (p=107...110).
rxgrid — Received resource elements grid
K-by-L-by-NRxAnts array
Received resource elements grid, specified as a K-by-L-by-NRxAnts array. K is the number of subcarriers, L is the number of OFDM symbols in one frame, and NRxAnts is the number of receive antennas.
hestgrid — Channel estimate grid
K-by-L-by-NRxAnts-by-4 array
Channel estimate grid, specified as a K-by-L-by-NRxAnts-by-4 array. K is the number of subcarriers, L is the number of OFDM symbols in one frame, and NRxAnts is the number of receive antennas. The 4th dimension represents the 4 possible EPDCCH antenna ports (p=107...110).
bits — Decoded bit estimates
column-vector | MTot-by-4 matrix
Decoded bit estimates for the candidate configured via chs.EPDCCHECCE, returned as one of the following:
a column-vector of length EPDCCHG = EDPCCHGd × 2.
MTot-by-4 matrix. MTot is the total number of bits associated with EPDCCHs and 4 is all possible EPDCCH antenna ports (p=107...110). Since the bits output is used as input to lteEPDCCHSearch, where each ECCE candidate has to be descrambled individually, the bits output are not descrambled.
symbols — Received QPSK symbols
column-vector | (MTot / 2)-by-4 matrix
Received QPSK symbols corresponding to bits in bits, specified as one of the following:.
A column-vector of length EPDCCHGd, where EPDCCHGd is the EPDCCH symbol capacity, given by the info.EPDCCHGd field of lteEPDCCHIndices.
(MTot / 2)-by-4 matrix, for all EPDCCH ECCEs and all 4 EPDCCH reference signal ports (p=107...110).
The lteEPDCCHDecode function works with only one EPDCCH-PRB-Set because lteDLChannelEstimate works with only one EPDCCH-PRB-Set. The lteEPDCCHDecode function processing performed depends on which input signals are provided to the function. The figures shown here align available syntaxes with processing performed.
If the symbols for a single configured EPDCCH candidate are input, the function performs symbol demodulation and descrambling. The function assumes the input symbols were already equalized.
If the symbols for a single configured EPDCCH candidate are input along with channel and noise estimates, the function performs MMSE equalization, then symbol demodulation and descrambling. If the optional alg input is provided, CSI weighting is applied to the output bits.
If the resource element grid across all possible EPDCCH antenna ports is input, the function extracts all EPDCCH resource elements and performs EPDCCH decoding for all possible EPDCCH candidate locations. The function assumes the input symbols were already equalized. Each EPDCCH candidate is descrambled individually during EPDCCH search.
If the resource element grid is input, along with channel and noise estimates, the function extracts all EPDCCH resource elements and performs MMSE equalization, then symbol demodulation. If the optional alg input is provided, CSI weighting is applied to the output bits. Each EPDCCH candidate is descrambled individually during EPDCCH search.
lteEPDCCH | lteEPDCCHDMRSIndices | lteEPDCCHIndices | lteEPDCCHSearch | lteEPDCCHSpace | lteEPDCCHPRBS | lteDCIDecode
|
Hilbert transform - MATLAB htrans - MathWorks Italia
Transform Symbolic Expression
Transform Sinc Function
Apply Phase Shifts
Calculate Instantaneous Frequency
H = htrans(f)
H = htrans(f,transVar)
H = htrans(f,var,transVar)
H = htrans(f) returns the Hilbert transform of symbolic function f. By default, the independent variable is t and the transformation variable is x.
H = htrans(f,transVar) uses the transformation variable transVar instead of x.
H = htrans(f,var,transVar) uses the independent variable var and the transformation variable transVar instead of t and x, respectively.
If all input arguments are arrays of the same size, then htrans acts element-wise.
If one input is a scalar and the others are arrays of the same size, then htrans expands the scalar into an array of the same size.
If f is an array of symbolic expressions with different independent variables, then var must be a symbolic array with elements corresponding to the independent variables.
Compute the Hilbert transform of sin(t). By default, the transform returns a function of x.
f = sin(t);
-\mathrm{cos}\left(x\right)
Compute the Hilbert transform of the sinc(x) function, which is equal to sin(pi*x)/(pi*x). Express the result as a function of u.
syms f(x) H(u);
f(x) = sinc(x);
H(u) = htrans(f,u)
-\frac{\frac{\mathrm{cos}\left(\pi u\right)}{u}-\frac{1}{u}}{\pi }
Plot the sinc function and its Hilbert transform.
fplot(f(x),[0 6])
fplot(H(u),[0 6])
legend('sinc(x)','H(u)')
Create a sine wave with a positive frequency in real space.
syms A x t u;
assume([x t],'real')
y = A*sin(2*pi*10*t + 5*x)
A \mathrm{sin}\left(5 x+20 \pi t\right)
Apply a –90-degree phase shift to the positive frequency component using the Hilbert transform. Specify the independent variable as t and the transformation variable as u.
H = htrans(y,t,u)
-A \mathrm{cos}\left(5 x+20 \pi u\right)
Now create a complex signal with negative frequency. Apply a 90-degree phase shift to the negative frequency component using the Hilbert transform.
z = A*exp(-1i*10*t)
A {\mathrm{e}}^{-10 t \mathrm{i}}
H = htrans(z)
A {\mathrm{e}}^{-10 x \mathrm{i}} \mathrm{i}
Create a real-valued signal
\mathit{f}\left(\mathit{t}\right)
with two frequency components, 60 Hz and 90 Hz.
syms t f(t) F(s)
f(t) = sin(2*pi*60*t) + sin(2*pi*90*t)
f(t) =
\mathrm{sin}\left(120 \pi t\right)+\mathrm{sin}\left(180 \pi t\right)
Calculate the corresponding analytic signal
\mathit{F}\left(\mathit{s}\right)
using the Hilbert transform.
F(s) = f(s) + 1i*htrans(f(t),s)
F(s) =
\mathrm{sin}\left(120 \pi s\right)+\mathrm{sin}\left(180 \pi s\right)-\mathrm{cos}\left(120 \pi s\right) \mathrm{i}-\mathrm{cos}\left(180 \pi s\right) \mathrm{i}
Calculate the instantaneous frequency of
\mathit{F}\left(\mathit{s}\right)
{f}_{instant}\left(s\right)=\frac{1}{2\pi }\frac{d\varphi \left(s\right)}{ds},
\varphi \left(s\right)=\mathrm{arg}\left[F\left(s\right)\right]
is the instantaneous phase of the analytic signal.
InstantFreq(s) = diff(angle(F(s)),s)/(2*pi);
assume(s,'real')
simplify(InstantFreq(s))
75
Input, specified as a symbolic expression, symbolic function, symbolic vector, or symbolic matrix.
t (default) | symbolic variable | symbolic vector | symbolic matrix
Independent variable, specified as a symbolic variable, symbolic vector, or symbolic matrix. This variable is usually in the time domain. If you do not specify the variable, then htrans uses t by default. If f does not contain t, then htrans uses the function symvar to determine the independent variable.
x (default) | v | symbolic variable | symbolic vector | symbolic matrix
Transformation variable, specified as a symbolic variable, symbolic vector, or symbolic matrix. This variable is in the same domain as var. If you do not specify the variable, then htrans uses x by default. If x is the independent variable of f, then htrans uses the transformation variable v.
H — Hilbert transform of f
Hilbert transform or harmonic conjugate of the input function f. The output H is a function of the variable specified by transVar.
When htrans cannot transform the input function, it returns an unevaluated call. To return the original expression, apply the inverse Hilbert transform to the output by using ihtrans.
The Hilbert transform H = H(x) of the expression f = f(t) with respect to the variable t at point x is
H\left(x\right)=\frac{1}{\pi }\text{p}\text{.v}\text{.}\underset{-\infty }{\overset{\infty }{\int }}\frac{f\left(t\right)}{x-t}dt.
Here, p.v. represents the Cauchy principal value of the integral. The function f(t) can be complex, but t and x must be real.
To compute the inverse Hilbert transform, use ihtrans. The Hilbert transform of a function is equal to the negative of its inverse Hilbert transform.
For a signal in the time domain, the Hilbert transform applies a –90-degree phase shift to positive frequencies of the corresponding Fourier components. It also applies a 90-degree phase shift to negative frequencies.
For a real-valued signal a, the Hilbert transform b = htrans(a) returns its harmonic conjugate b. The real signal a = real(z) and its Hilbert transform b = imag(z) form the analytic signal z = a + 1i*b.
ihtrans | fourier | ifourier | laplace | ilaplace
|
First Use of Fragile Geologic Features to Set the Design Motions for a Major Existing Engineered Structure | Bulletin of the Seismological Society of America | GeoScienceWorld
Mark W. Stirling *
Department of Geology, University of Otago, Dunedin, New Zealand
Corresponding author: mark.stirliing@otago.ac.nz
Elizabeth R. Abbott;
Elizabeth R. Abbott
Dylan H. Rood;
Royal School of Mines, Imperial College London, London, United Kingdom
Department of Earth and Environmental Science, A.E. Lalonde AMS Laboratory, University of Ottawa, Ottawa, Ontario, Canada
Graeme H. McVerry;
Graeme H. McVerry
Retired, GNS Science, Lower Hutt, New Zealand
Civil and Environmental Engineering, University of California, Berkeley, Berkeley, California, U.S.A.
Rand Huso;
Lisa Luna;
David A. Rhoades;
Peter Silvester;
Contact Energy, Clyde New Zealand
Chris Van Houtte;
Tu Ora/Compass Health, Wellington, New Zealand
Mark W. Stirling, Elizabeth R. Abbott, Dylan H. Rood, Graeme H. McVerry, Norman A. Abrahamson, David J. A. Barrell, Rand Huso, Nicola J. Litchfield, Lisa Luna, David A. Rhoades, Peter Silvester, Russ J. Van Dissen, Chris Van Houtte, Albert Zondervan; First Use of Fragile Geologic Features to Set the Design Motions for a Major Existing Engineered Structure. Bulletin of the Seismological Society of America 2021;; 111 (5): 2673–2695. doi: https://doi.org/10.1785/0120210026
We document the first use of fragile geologic features (FGFs) to set formal design earthquake motions for a major existing engineered structure. The safety evaluation earthquake (SEE) spectrum for the Clyde Dam, New Zealand (the mean 10,000 yr, ka, return period response spectrum) is developed in accordance with official guidelines and utilizes constraints provided by seven precariously balanced rocks (PBRs) located 2 km from the dam site and the local active Dunstan fault. The PBRs are located in the hanging wall of the fault. Deterministic PBR fragilities are estimated from field measurements of rock geometries and are the dynamic peak ground accelerations (PGAs) required for toppling. PBR fragility ages are modeled from
B10e
cosmogenic isotope exposure dating techniques and are in the range of 24–66 ka. The fragility ages are consistent with the PBRs having survived at least two large Dunstan fault earthquakes. We develop a PGA‐based fragility distribution from all of the PBRs, which represents the cumulative toppling probability of a theoretical random PBR as a function of PGA. The fragility distribution is then used to eliminate logic‐tree branches that produce PGA hazard curves that would topple the random PBR with a greater than 95% probability (i.e., less than 5% survival probability) over a time period of 24 ka (youngest PBR fragility age). The mean 10 ka spectrum of the remaining hazard estimates is then recommended as the SEE spectrum for the dam site. This SEE spectrum has a PGA of
0.55g
, which is significantly reduced from the
0.96g
obtained for a preliminary version of the SEE spectrum. The reduction is due to the combined effects of the PBR constraints and a substantial update of the probabilistic seismic hazard model. The study serves as an important proof‐of‐concept for future applications of FGFs in engineering design.
Dunstan Fault
|
Evaluate satisfaction of function matching requirement - MATLAB
dependentVar
indepVar1,...,indepVarN
Match Quadratic Function to Two-Dimensional Variable Data
Class: sdo.requirements.FunctionMatching
Evaluate satisfaction of function matching requirement
evaluation = evalRequirement(requirement,dependentVar)
evaluation = evalRequirement(requirement,dependentVar,indepVar1,...,indepVarN)
evaluation = evalRequirement(requirement,dependentVar) evaluates whether the test data dependentVar matches the function that is specified in the Type property of the requirement object. The software computes the specified function using default independent variable vectors with value [0 1 2 ...]. There is an independent variable vector corresponding to each dimension of dependentVar, and the length of each independent variable vector is the same as the size of dependentVar in the corresponding dimension.
For example, consider a two-dimensional dependentVar of size 3-by-2. To compute a linear function of the form
{a}_{0}+{a}_{1}{X}_{1}+{a}_{2}{X}_{2}
, the software uses the independent variable vectors X1 = [0 1 2] and X2 = [0 1]. The software calculates the fit coefficients a0, a1, and a2 and then calculates the error between the test data and the linear function.
evaluation = evalRequirement(requirement,dependentVar,indepVar1,...,indepVarN) specifies the independent variable vectors to use for computing the function.
requirement — Function matching requirement
sdo.requirements.FunctionMatching object
Function matching requirement, specified as an sdo.requirements.FunctionMatching object. You specify the function to be matched in requirement.Type.
dependentVar — Dependent variable test data to be evaluated
Dependent variable test data to be evaluated, specified as a vector, matrix, or multidimensional array.
indepVar1,...,indepVarN — Independent variable vectors
Independent variable vectors used for computing the function, specified as real, numeric, monotonic vectors. The independent variable vectors must satisfy the following characteristics:
The number of independent variables N must equal the number of dimensions of the test data.
For example, use two independent variables when the test data dependentVar is a matrix, and use three independent variables when the test data is a three-dimensional array.
The first independent variable vector specifies coordinates going down test data rows, and the second independent variable vector specifies coordinates going across test data columns. The Nth independent variable vector specifies coordinates along the Nth dimension of dependentVar.
The number of elements in each independent variable vector must match the size of test data in the corresponding dimension.
The independent variable vectors must be monotonically increasing or decreasing.
In the requirement object, you can specify centering and scaling of the independent variables using the Centers and Scales properties. The independent variable vectors specified by you are divided by these Scales values after subtracting the Centers values. For more information, see the property descriptions on the sdo.requirements.FunctionMatching reference page.
You can also specify independent variable vectors using a cell array. The number of elements in the cell array must match the number of dimensions in the test data, dependentVar. For example, suppose that dependentVar is two-dimensional, you can use either of the following syntaxes:
evaluation = evalRequirement(requirement,dependentVar,independentVar1,independentVar2);
evaluation = evalRequirement(requirement,dependentVar,{independentVar1,independentVar2});
evaluation — Evaluation of the function matching requirement
Evaluation of the function matching requirement, returned as a scalar, vector, matrix, or array, depending on the value of requirement.Method.
evalRequirement computes an error signal that is the difference between test data and the specified function of the independent variables. The error signal is then processed further to compute evaluation. The value of evaluation depends on the error processing method specified in requirement.Method.
requirement.Method
evaluation is returned as a scalar value equal to the sum of squares of the errors.
A positive value indicates that the requirement is violated, and 0 value indicates that the requirement is satisfied. The closer evaluation is to 0, the better the match between the function and test data.
evaluation is returned as a scalar value equal to the sum of absolute values of the errors.
'Residuals' evaluation is returned as a vector, matrix, or array of the same size as the test data dependentVar. evaluation contains the difference between the test data and the specified function of the independent variables.
Create a requirement object, and specify the function to be matched.
Requirement = sdo.requirements.FunctionMatching('Type','purequadratic');
The object specifies that the variables should match a quadratic function with no cross-terms.
Create 2-dimensional test data for the variable.
[X1,X2] = ndgrid((-1:1),(-4:2:4));
dependentVar = X1.^2 + X2.^2;
Specify independent variable vectors to compute the quadratic function.
The number of independent variable vectors must equal the dimensionality of the test data. In addition, the independent variable vectors must be monotonic and have the same size as the test data in the corresponding dimension.
indepVar1 = (-2:0);
indepVar2 = (-6:2:2);
Evaluate if the test data satisfies the requirement.
evaluation = evalRequirement(Requirement,dependentVar,indepVar1,indepVar2)
The evalRequirement command computes an error signal that is the difference between test data and the function of the independent variable vectors. The error signal is further processed to compute evaluation, based on the error processing method specified in Requirement.Method.
In this example, the processing method has the default value of 'SSE', so evaluation is returned as a scalar value equal to the sum of squares of the errors. evaluation is very close to zero, indicating that the dependentVariable test data almost matches a pure quadratic function.
Create test data with cross-terms.
dependentVariable2 = X1.^2 + X2.^2 + X1.*X2;
Evaluate the requirement for the new test data.
evaluation2 = evalRequirement(Requirement,dependentVariable2,indepVar1,indepVar2)
evaluation2 = 5.3333
The output evaluation2 is greater than evaluation and is substantially different from 0, indicating that dependentVariable2 does not fit a pure-quadratic function as well as dependentVariable fits the function.
|
Analysis | Bean Machine
Inference results are useful not only for learning posterior distributions, but for verifying that inference ran correctly. We'll cover common techniques for analyzing results in this section. As is the case for everything else in this Overview, the code for this section is available as a notebook on GitHub and Colab.
Results of Inference
Bean Machine stores the results of inference in an object of type MonteCarloSamples. Internally, this class uses a dictionary to map random variables to PyTorch tensors of posterior samples. The class can be accessed like a dictionary, and there are additional wrapper methods to make function calls more explicit.
In the Inference section, we obtained results on the disease modeling example by running inference:
<beanmachine.ppl.inference.monte_carlo_samples.MonteCarloSamples>
Extracting Samples for a Specific Variable
In order to perform inference on the random variable reproduction_rate(), we added it to the queries list. We can see that it, and no other random variable, is available in samples:
list(samples.keys())
[RVIdentifier(function=<function reproduction_rate>, arguments=())]
To extract the inference results for reproduction_rate(), we can use get_variable():
samples.get_variable(reproduction_rate())
The result has shape
4 \times 7000
, representing the
7000
samples that we drew in each of the four chains of inference from the posterior distribution.
MonteCarloSamples supports convenient dictionary indexing syntax to obtain the same information:
samples[reproduction_rate()]
Please note that many inference methods require a small number of samples before they start drawing samples that correctly resemble the posterior distribution. The 3000 samples that we specified in num_adaptive_samples were already excluded for us, so nothing needs to be done here. However, if you use no adaptive samples, we recommend you discard at least a few hundred samples before using your inference results.
Extracting Samples for a Specific Chain
We'll see how to make use of chains in Diagnostics; for inspecting the samples themselves, it is often useful to examine each chain individually. The recommended way to access the results of a specific chain is with get_chain():
chain = samples.get_chain(chain=0)
This returns a new MonteCarloSamples object which is limited to the specified chain. Tensors no longer have a dimension representing the chain:
chain[reproduction_rate()]
Visualizing Distributions
Visualizing the results of inference can be a great help in understanding them. Since you now know how to access posterior samples, you're free to use whatever visualization tools you prefer.
reproduction_rate_samples = samples.get_chain(0)[reproduction_rate()]
After running inference it is useful to run diagnostic tools to assess reliability of the inference run. Bean Machine provides two standard types of such diagnostic tools, discussed below.
Bean Machine provides important summary statistics for individual, numerically-valued random variables. Let's take a look at the code to generate them, and then we'll break down the statistics themselves.
Bean Machine's interface to the ArviZ libray makes it easy to generate a Pandas DataFrame presenting these statistics for all queried random variables:
az.rcParams["stats.hdi_prob"] = 0.89
az.summary(samples.to_xarray(), round_to=5)
reproduction_rate() 0.2196 0.0003 0.2192 0.2201 0.0 0.0 19252.377 19175.7875 1.0002
We recommend reading the official ArviZ documentation for a full explanation, but the statistics presented are:
89% highest density interval (HDI).
Markov chain standard error (MCSE).
Effective sample size (ESS)
N_\text{eff}
Convergence diagnostic
\hat{R}
We choose to display the 89% highest density interval (HDI), following recommendations in Statistical Rethinking: A Bayesian Course with Examples in R and Stan (McElreath, 2020). The statistics above provide different insights into the quality of the results of inference. For instance, we can use them in combination with synthetically generated data for which we know ground truth values for parameters and then check to make sure that these values fall within some HDI of our posterior samples. Of course, in doing so it is important to keep in mind that the prior distributions in our model (and not just the data) will have an influence on the posterior distribution. Similarly, we can use the size of the HDI to gain insights into the model: if it is large, this could indicate that either we have too few observations or that the prior is too weak.
\hat{R} \in [1, \infty)
summarizes how effective inference was at converging on the correct posterior distribution for a particular random variable. It uses information from all chains run in order to assess whether inference had a good understanding of the distribution or not. Values very close to
1.0
indicate that all chains discovered similar distributions for a particular random variable. We do not recommend using inference results where
\hat{R} > 1.01
, as inference may not have converged. In that case, you may want to run inference for more samples. However, there are situations in which increasing the number of samples will not improve convergence. In this case, it is possible that the prior is too far from the posterior, or that the particular inference method is unable to reliably explore the posterior distribution.
N_\text{eff} \in [1,
num_samples
]
summarizes how independent posterior samples are from one another. Although inference was run for num_samples iterations, it's possible that those samples were very similar to each other (due to the way inference is implemented), and may not each be representative of the full posterior space. Larger numbers are better here, and if your particular use case calls for a certain number of samples to be considered, you should ensure that
N_\text{eff}
is at least that large. For more information on R-hat and
N_\text{eff}
, see the Diagnostics Section.
In the case of our example model, we have a healthy
\hat{R}
value very close to 1.0, and a healthy relative number of effective samples.
Diagnostic Plots
Bean Machine can also plot diagnostic information to assess health of the inference run. Let's take a look:
{"Reproduction rate": samples[reproduction_rate()]},
az.plot_autocorr({"Reproduction rate": samples[reproduction_rate()]})
The diagnostics output shows two diagnostic plots for individual random variables: trace plots and autocorrelation plots.
Trace plots are simply a time series of values assigned to random variables over each iteration of inference. The concrete values assigned are usually problem-specific. However, it's important that these values are "mixing" well over time. This means that they don't tend to get stuck in one region for large periods of time, and that each of the chains ends up exploring the same space as the other chains throughout the course of inference.
Autocorrelation plots measure how predictive the last several samples are of the current sample. Autocorrelation may vary between -1.0 (deterministically anticorrelated) and 1.0 (deterministically correlated). (We compute autocorrelation approximately, so it may sometimes exceed these bounds.) In an ideal world, the current sample is chosen independently of the previous samples: an autocorrelation of zero. This is not possible in practice, due to stochastic noise and the mechanics of how inference works. The autocorrelation plots here plot how correlated samples from the end of the chain are compared with samples taken from elsewhere within the chain, as indicated by the iteration index on the x axis.
For our example model, we see from the trace plots that each of the chains are healthy: they don't get stuck, and do not explore a chain-specific subset of the space. From the autocorrelation plots, we see the absolute magnitude of autocorrelation to be very small, often well below
0.1
, indicating a healthy exploration of the space.
Congratulations, you've made it through the Overview! If you're looking to get an even deeper understanding of Bean Machine, check out the Framework topics next. Or, if you're looking to get to coding, check out our Tutorials. In either case, happy modeling!
Results of Inference
Extracting Samples for a Specific Variable
Extracting Samples for a Specific Chain
|
Repeated XOR | PicoCTF 2014 Write-ups
1.1. Function Address
1.2. Basic ASM
1.3. Cyborg Secrets
1.4. Towers of Toast
1.5. Police Records
2.1. No Comment
2.2. Internet Inspection
2.4. Injection 1
2.5. Potentially Hidden Password
2.6. secure_page_service
2.7. Make a Face
2.10. Steve's List
3.2. The Valley of Fear
3.3. ZOR
3.5. Repeated XOR
3.8. Low Entropy
3.10. Web Interception
3.11. Revenge of the Bleichenbacher
3.12. RSA Mistakes
4.1. Intercepted Post
4.2. Snapcat
4.3. PNG or Not?
4.4. SSH Backdoor
5.1. Overflow 1
5.5. Execute Me
5.6. OBO
5.7. ROP 1
6.1. Tyrannosaurus Hex
PicoCTF 2014 Write-ups
Repeated XOR - 70 (Cryptography)
Writeup by Gladius Maximus
There's a secret passcode hidden in the robot's "history of cryptography" module. But it's encrypted! Here it is, hex-encoded: encrypted.txt. Can you find the hidden passcode?
Like the title suggests, this is repeating-key XOR. You should try to find the length of the key - it's probably around 10 bytes long, maybe a little less or a little more.
Preform Kasiski elimination to find the key length. For this problem, the keylength is 8. Since the cipher text is encrypted using repeateing xor, you can do a frequency analysis on the 0th, 8th, 16th characters, and then another one on the 1st, 9th, and 17th characters, adn then on the 2nd, 10th, and 18th character.
About repeated XOR
In a repeating XOR cipher, if the key is shorter than the message (it almost always is), the key is duplicated in order to cover the whole message. Then each byte of the plain text is XORed with each according byte of the key. For example, suppose we are trying to encrypt the message 'THIS IS A MESSAGE', with the key 'YOU', we first convert all of the characters to integers using ASCII encoding. (More examples may be found on the Wikipedia page)
T H I S I S A M E S S A G E
Then we duplicate the key and convert it to an integer sequence.
Y O U Y O U Y O U Y O U Y O U Y O
Then we XOR the integers together.
xor --------------------------------------------------------------
13, 7, 28, 10,111, 28, 10,111, 20,121, 2, 16, 10, 28, 20, 30, 10
Do you notice the repeating 28, 10, 111? That is because there was a repeat in the paintext that was a multiple of the keylength apart. In reality, repeats happen quite frequently, but it is just as likely to get a repeat 3 characters apart as it is to get one 4 characters apart. If they are 4 characters apart, a repeat is not shown in the cipher text, because the length of the offset (4) is not a multiple of the key length (3). For example, encrypting 'THIS MESSAGE IS NOT' with the key 'YOU' gives:
xor ----------------------------------------------------------------------
13, 7, 28, 10,111, 24, 28, 28, 6, 24, 8, 16,121, 6, 6,121, 1, 26, 13
XOR has a lot of special properties. First XOR is commutative.
a \oplus b = b \oplus a
\oplus
stands for XOR). It is associative
a \oplus (b \oplus c) = (a \oplus b) \oplus c
. Next, anything XORed with itself is zero:
a \oplus a = 0
, and anything XORed with zero is anything:
a \oplus 0 = a
. Thus we can conclude that
a \oplus b \oplus b = a \oplus (b \oplus b) = a \oplus 0 = a
a \oplus b = c
a \oplus c = a \oplus (a \oplus b) = (a \oplus a) \oplus b = 0 \oplus b = b
If that all went over your head, just know that XOR is its own inverse. That is the most important part of XOR and it will come in handy later.
However, because repeated text shows up frequently in the English language, we can expect that at least some of the time the repeated text will be offset by a multiple of the key. We can count on most of the repeated cipher text being an integer multiple of the key. Thus we can find the key length
First, I made encrypted.py which contains all of the raw data and parsed data. Parsing the data into different forms is usually a good place to start.
from encrypted import enc_numbers
return data[offset:] + data[:offset]
def count_same(a, b):
print ('key lengths')
for key_len in range(1, 33): # try multiple key lengths
freq = count_same(enc_numbers, shift(enc_numbers, key_len))
print ('{0:< 3d} | {1:3d} |'.format(key_len, freq) + '=' * (freq / 4))
# ^ this line does fancy formatting that outputs key_len and then freq and
# then a bar graph
This makes a nice bar graph showing how many characters were the same after shifting the original cipher text over by key_len offset. You can see that 8, 16, and 32 have a much higher amount of same-characters. Thus, we can guess that the key is 8 characters long.
for i in range(0, key_len):
frequency = Counter()
for ch in enc_ascii[i::key_len]:
frequency[ch] += 1
A snipped version of the output is below:
3 | 12 |===
4 | 18 |====
5 | 9 |==
6 | 2 |
7 | 42 |==========
8 | 249 |==============================================================
9 | 64 |================
10 | 2 |
11 | 6 |=
Once we know the key length, each character and the character 8 down are XORed with the same key character. Thus we can do a frequency analysis. Imagin splitting up the message into rows 8 characters long. Each column was encrypted with the same character of the key, so each column should have character frequencies that correspond. Thus we can go through each 0th, 8th, 16th, ... character and pick the most frequently occuring character and then do the same for the 1st, 9th, 17th, etc.
Once we have the most frequently occuring character, we say it decrypts to ' ' (space occurs most frequently in a block of text). Let the most frequently occuring character in the nth column of the cipher text
c_n
. Let the nth character of the key is
k_n
. Let ' ' be
m
m \oplus k_n = c_n
. XOR is its own inverse, so
m \oplus c_n = k_n
. So because XOR is its own inverse, we can find the key by XORing cipher text and known plain text Thus we can find the key if we know the most common character in english and the most common character in the nth column.
We can check this by printing out what other common characters decrypt to. If we have the right answer, we should get other common characters decrypting to t, e, a, i or something like that.
print ('guesses for most common letters')
key_numbers = []
k = ord(frequency.most_common(1)[0][0]) ^ ord(' ')
print ('{k} -> \' \''.format(**locals()))
key_numbers.append(k)
for val, freq in frequency.most_common(10):
others += chr(ord(val) ^ k) + ' '
print ('Other common letters: {others}\n'.format(**locals()))
A snipped version of the output is below. It looks like it worked.
34 -> ' '
e t a o r h n s i
e t i o n a s l r
155 -> ' '
e t o n d i r s a
We are ready to decrypt the whole text.
def decrypt(c_num, k_num):
return ''.join(chr(c ^ k) for c, k in izip(c_num, cycle(k_num)))
print ('decrypting text')
print (decrypt(enc_numbers, key_numbers))
I have attached the complete summarized script containing all of the parts of this writeup here (kasiski.py and its dependency encrypted.py)
89ae5d8b68c8d6323c79e2f8540f0f50728cffb7
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Knapsack DP · USACO Guide
HomeGoldKnapsack DP
TutorialSolution - Dice CombinationsProblemsCSESUSACONT
Authors: Nathan Chen, Michael Cao, Benjamin Qi
Contributor: Neo Wang
Problems that can be modeled as filling a limited-size container with a subset of items.
CSES - Very Easy
7.1, 7.4 - Coin, Knapsack Problems
Solves "Minimizing Coins," 0/1 Knapsack
Videos for common knapsack variations
Knapsack problems generally involve filling a limited container with a subset of items where we want to count or optimize some quantity associated with the items. Almost every time, you can think of each item as having a positive weight, and the total weight of the items we choose must not exceed the capacity of the container, which is some number. Some variations of knapsack-type problems include:
The 0/1 Knapsack problem: Choosing a subset of items such that we maximize their total value and their total weight does not exceed the capacity of the container
Finding all the possible total weights that we can achieve from any subset of items such that their total weight does not exceed the capacity of the container (in the chapter of CPH linked above)
Counting how many sequences of items will fill the container completely, meaning the total weight is exactly the capacity of the container (the order may or may not matter)
The DP solution to knapsack problems usually has the state keeping track of the capacity of the knapsack, and the transitions involve trying to add an item to the knapsack. In competitive programming, you can expect that classical knapsack problems will be given twists, disguises, and extra state information involved.
Solution - Dice Combinations
\mathcal{O}(N)
The problem asks us how many sequences of dice rolls exist such that the sum of the top faces is
N
N \leq 10^6
). To keep up with the knapsack analogy, that means we have infinite numbers of items of weights
1
6
, and we want to count how many sequences of items exist such that if we put items into the container while following the sequence, the container becomes completely full. Note that the order of the items matters in this problem.
For convenience, let
\texttt{dp}[x]
be the number of sequences of dice rolls that add up to
x
. To count how many sequences add up to
N
, or in other words, to find
\texttt{dp}[N]
, let's look at the last dice roll that brings us up to a total sum of
N
If the last roll was a
1
, then we know there are
\texttt{dp}[N-1]
ways to achieve sum
N
when the last roll is
1
. If the last roll was a
2
\texttt{dp}[N-2]
N
2
. Continue this logic for all the dice numbers up to
6
. Considering all those cases together, we have shown that
\texttt{dp}[N] = \texttt{dp}[N-1] + \texttt{dp}[N-2] + \texttt{dp}[N-3] + \texttt{dp}[N-4] + \texttt{dp}[N-5] + \texttt{dp}[N-6].
Apply that same logic we used for
\texttt{dp}[N]
on a general
x
\texttt{dp}[x] = \sum_{i=1}^6\texttt{dp}[x-i].
Start with the base case that
\texttt{dp}[0] = 1
\texttt{dp}[1]
\texttt{dp}[2]
\texttt{dp}[3]
, and so on... can be calculated using the recurrence until we find
\texttt{dp}[N]
. Note in the code that we ignore
\texttt{dp}[x]
x < 0
long long dp[1000001];
int n; n = Integer.parseInt(br.readLine());
long dp[] = new long[n+1];
dp.append(sum(dp[-6:]) % (10 ** 9 + 7))
Minimizing Coins
Very Easy Show Tags Knapsack
Coin Combinations I (Unordered)
Easy Show Tags Knapsack
Coin Combinations II (Ordered)
Hard Show Tags Knapsack
Easy Show Tags DP, Knapsack
Normal Show Tags DP, Knapsack
Hard Show Tags Binary Search, DP, Knapsack
Mooriokart
Very Hard Show Tags Knapsack
Some knapsack problems with number-theoretic twists!
Round Subset
Normal Show Tags Knapsack
Normal Show Tags Exponentiation, Knapsack
Normal Show Tags Knapsack, Prime Factorization
2004 - Maximal
SubmultiplesOfN
Hard Show Tags Knapsack, Sorting
Insane Show Tags Knapsack, Prime Factorization
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Teth - Wikipedia
(Redirected from ط)
This article is about the Semitic letter. For the 5th-century Cornish saint also known as Teth, see Saint Tetha.
Teth, also written as Ṭēth or Tet, is a letter of the Semitic abjads, including Phoenician Ṭēt , Hebrew Tēt ט, Aramaic Ṭēth , Syriac Ṭēṯ ܛ, and Arabic Ṭāʾ ط. It is the 16th letter of the modern Arabic alphabet. The Persian ṭa is pronounced as a hard "t" sound and is the 19th letter in the modern Persian alphabet. The Phoenician letter also gave rise to the Greek theta (Θ), originally an aspirated voiceless dental stop but now used for the voiceless dental fricative. The Arabic letter (ط) is sometimes transliterated as tah in English,[1] for example in Arabic script in Unicode.
2 Arabic ṭāʾ
3 Hebrew Tet
The Phoenician letter name ṭēth may mean "spinning wheel"[2] pictured as (compare Hebrew root ט-ו-י meaning 'spinning' (a thread) which begins with Teth). According to another hypothesis (Brian Colless[citation needed]), the letter possibly continues a Middle Bronze Age glyph named ṭab 'good', Aramaic טַב 'tav', Hebrew טוב 'tov', Syriac ܛܒܐ 'tava', modern Arabic طَيّب 'ṭayyib', all of identical meaning, whose picture is based on the Nefer 'good' hieroglyph common in ancient Egyptian names (e.g. Nefertiti):
Arabic ṭāʾ[edit]
Hebrew Tet[edit]
Similar symbols[edit]
{\displaystyle \otimes }
Retrieved from "https://en.wikipedia.org/w/index.php?title=Teth&oldid=1089371288#Arabic_ṭāʾ"
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Numerical Simulation of the Churning Power Losses in the Automotive Hypoid Gear Reducer
Numerical Simulation of the Churning Power Losses in the Automotive Hypoid Gear Reducer
School of Mechanical and Electronic Engineering, Wuhan University of Technology, Wuhan, China
Improving vehicle transmission efficiency and reducing vehicle fuel consumption is currently one of the main objectives in the automotive field. Reducing gear churning power losses has significant influence on the decreasing vehicle fuel consumption. Based on the two phase flow theory, a 2D two-phase model of the simplified hypoid gear is established to predict the churning losses in different conditions, the VOF method is introduced to track the volume fraction of the free surface, a standard k-ε model is also built to calculate complex turbulence. The oil distributions at the different rotational speed, the different immersion depth and the different viscosity as well as the churning losses of the hypoid gear are obtained and discussed in detail. In general, the churning power losses increase with the increase of the speed, the immersion depth and the viscosity, while the rotational speed shows the greatest influence on the churning losses. It is hoped that this investigation will be helpful in automotive industry applications.
Churning Power Loss, Two Phases Flow, VOF Method, Hypoid Gear
The internal transmission efficiency of the rear axle of a car directly affects the oil consumption of the whole vehicle. The gears and bearings, as the main components of the rear axle, play a key impact on the transmission efficiency of the rear axle. It is critical to figure out the mechanism of its efficiency losses to reduce the oil consumption of the whole vehicle.
In the last years, the churning losses of the gears have been the subject of extensive numerical and experimental investigations. Concli & Gorla computationally and experimentally analyzed the churning power losses in an industrial planetary speed reducers and point out that the VOF method could provide more information on the physical phenomena than any experimental measurement [1] . And then they developed a new approach to predict churning power losses of ordinary gears [2] . Kodela et al. proposed a CFD method to estimate the splash loss for the complete manual transmission gear box at the different operating conditions such as the rotational speed, the temperature and the oil level [3]. Liu et al. numerically investigated the oil distribution and the churning loss of a single-stage gearbox by the finite volume CFD method, and showed that the turbulence of oil increased with the increase of the rotational speed [4]. Wang et al. numerically investigated the churning losses of the spur gear, but found out that there are big differences between simulation values and theoretical calculation [5]. Concli & Gorla further pointed out that the power losses were not directly decreased with the increase of the temperature, and an optimal temperature existed [6].
The above literatures mostly pay attention to investigating the churning losses of the spur gear. However, less investigation on the churning losses of the hypoid gears is discussed. In the present study, the oil distribution of the hypoid gear at the different conditions and the influencing factors of the churning losses such as the speed, the immersion depth and the viscosity are obtained and discussed. It is hoped that this investigation will be helpful for the deepening of our understanding on the oil consumption of the whole vehicle.
2. Governing Equations and Numerical Method
The numerical method stated in this paper depends on the solution of two governing equations that mathematically represent the conservation laws of mass and momentum [7]:
\frac{\partial {u}_{i}}{\partial {x}_{i}}=0
\frac{\partial {u}_{i}}{\partial t}+\frac{\partial \left({u}_{i}{u}_{j}\right)}{\partial {x}_{i}}=-\frac{1}{\rho }\frac{\partial p}{\partial {x}_{i}}+\eta \frac{\partial }{\partial {x}_{i}}\left(\frac{\partial {u}_{j}}{\partial {x}_{i}}+\frac{\partial {u}_{i}}{\partial {x}_{j}}\right)+{f}_{i}
{u}_{i}
is velocity in the
{x}_{i}
direction. Where p is pressure,
\rho
is mixture density,
\eta
is the mixture viscosity,
{u}_{i}
{y}_{i}
{f}_{i}
is acceleration vector in
{x}_{i}
direction. The behavior of a transient incompressible flow is described by these basic equations.
The presence of more than one phase implies the need of additional equations. VOF method has been widely used to track the interface of two phases. Thus, a transportation equation for the volume fraction α is introduced [8]:
\frac{\partial \alpha }{\partial t}+\frac{\partial \left(\alpha {u}_{i}\right)}{\partial {x}_{i}}=0
In the two-phase flow system, the mixture properties of density and viscosity can be given by the volume weighted average of two liquids as:
\rho =\alpha {\rho }_{oil}+\left(1-\alpha \right){\rho }_{air}
\eta =\alpha {\eta }_{oil}+\left(1-\alpha \right){\eta }_{air}
{\rho }_{oil}
is density of oil,
{\rho }_{air}
is density of air,
{\eta }_{oil}
is viscosity of oil,
{\eta }_{air}
is viscosity of air. In order to obtain the solution of equations, the standard
k-\epsilon
turbulence model is applied to calculate the complex turbulent problems with high Reynolds number. The turbulence equation can be written as [9] [10]:
\rho \frac{\partial \left(k{u}_{i}\right)}{\partial {x}_{i}}=\frac{\partial }{\partial {x}_{i}}\left[\left(\eta +\frac{{\eta }_{t}}{{\sigma }_{k}}\right)\frac{\partial k}{\partial {x}_{i}}\right]+\rho \left(S-\epsilon \right)
\rho \frac{\partial \left(\epsilon {u}_{i}\right)}{\partial {x}_{i}}=\frac{\partial }{\partial {x}_{i}}\left[\left(\eta +\frac{{\eta }_{t}}{{\sigma }_{k}}\right)\frac{\partial \epsilon }{\partial {x}_{i}}\right]\text{+}\rho \frac{\epsilon }{k}\left({C}_{1}S-{C}_{2}\epsilon \right)
{\eta }_{t}=\rho {C}_{\eta }\frac{{k}^{2}}{\epsilon }
S=\frac{\eta {}_{t}}{\rho }\left(\frac{\partial {u}_{i}}{\partial {x}_{j}}+\frac{\partial {u}_{j}}{\partial {u}_{i}}\right)\frac{\partial {u}_{i}}{\partial {x}_{j}}
{C}_{\eta }=0.09
{C}_{1}=1.44
{C}_{2}=1.92
{\sigma }_{k}=1.0
{\sigma }_{\epsilon }=1.3
3. Geometrical Model and Boundary Conditions
In order to reduce the number of cells and save time, it’s necessary to simplify the geometrical model of the hypoid gear (Figure 1). The whole model has been discretized with tetrahedral mesh (Figure 2). During the simulation, the meshes of the hypoid gear rotates in the closed domain at the predefined rotational speed.
Table 1 shows the combinations of specific parameters for each simulation. These are respectively simulated at 100 rpm, 400 rpm and 700 rpm, and h represents height of a tooth. Table 2 shows the kinematic viscosity corresponding to each oil type. These are all GL-5 type at 40˚C.
Figure 1. Geometrical model of the hypoid gear.
Table 1. Specific parameters of each simulation at the different ratational speed.
Figure 2. Grid model of the hypoid gear.
Table 2. Viscosity corresponding to oil type.
Figure 3 shows that the power loss increases with the immersion depth. Decreasing the immersion depth from 2.0 h to 1.5 h, the power increases about only 50 watts under the same speed. The relation between the immersion depth and the power loss is relatively weak. Figure 4 shows the effect of the viscosity on the power loss. While the viscosity increases from 59.07 mm2/s to 82.88 m2/s, the power loss increases 8.4 watts at 100 rpm, 21.4 watts at 400 rpm and 991.8 watts at 700 rpm. These results show that the rotational speed has impact on effect extent of the viscosity on churning losses. Furthermore, oil type named GL-5 75W80 have minimum effect on power loss. The diagrams show the impact of the rotational speed on power loss. It can be clearly seen that the power approximately enhances with 90% while the rotational speed is reduced by 300 rpm. Therefore, the rotational speed has the greatest influence on the churning loss. Figure 5 shows contours of volume fraction for the oil distribution at the different rotational speed. The gear churns more oil with high speed, which results in easily generating larger churning torque. This explains the influence of rotational speed on power loss is maximum.
Based on two-phase flow, the VOF method and the standard k-ε model, a new simplified 2D model is established. The oil distributions and the churning power losses of the hypoid gear at the different working conditions with the rotational speed of 100 rpm, 400 rpm and 700 rmp at the immersion depth from 0.5 h to 2 h were numerically investigated.
In general, the churning power losses increased with the increase of the rotational speed and the immersion depth as well as the viscosity, which the rotational speed has the greatest influence on the churning loss. Minimum effect on the power loss is oil type named GL-5 75W80. The difference reaches tens of watts between 100 rpm and 400 rpm but runs up to hundreds of watts at the range of 400 rpm to 700 rpm. The turbulence of oil increases with the rotational speed, which causes higher oil power loss.
Figure 3. The effect of the immersion depth on power loss at three different rotational speed.
Figure 4. The effect of the viscosity on the power loss at three different rotational speed.
Figure 5. (a) Contours of volume fraction for the oil phase at 100 rpm; (b) Contours of volume fraction for the oil phase at 400 rpm; (c) Contours of volume fraction for the oil phase at 700 rpm.
However, the lower speed and the viscosity as well as the immersion depth have influence on gear lubrication and meshing. Thus, the meshing loss of gear transmission system and the impact of lubrication on power loss will be investigated in further studies.
The authors wish to thank the Liuzhou Science and Technology Bureau of Guangxi Province, China, for its support through Liuzhou science and technology research key project (Grant No. 2017AA10102).
Zou, L., Du, M.Y., Jia, B., Xu, J.L. and Ren, L.S. (2018) Numerical Simulation of the Churning Power Losses in the Automotive Hypoid Gear Reducer. Journal of Applied Mathematics and Physics, 6, 1951-1956. https://doi.org/10.4236/jamp.2018.69166
1. Concli, F. and Gorla, C. (2011) Computational and Experimental Analysis of the Churning Power Losses in an Industrial Planetary Speed Reducers. Proceedings of the 9th International Conference on Advances in Fluid Mechanics, Split, 26-28 June 2012, 287-298.
2. Concli, F., Gorla, C., Della Torre A. and Montenegro, G. (2015) Churning Power Losses of Ordi-nary Gears: A New Approach Based on the Internal Fluid Dynamics Simulations. Lubrication Science, 27, 313-326. https://doi.org/10.1002/ls.1280
3. Kodela, C., Kraetschmer, M. and Basa, S. (2015) Churning Loss Estimation for Manual Transmission Gear Box Using CFD. SAE International Journal of Passenger Cars-Mechanical Systems, 8.
4. Liu, H., Jurkschat, T., Lohner, T. and Stahl, K. (2017) Determination of Oil Distribution and Churning Power Loss of Gearboxes by Finite Volume CFD Method. Tribology International, 109, 346-354. https://doi.org/10.1016/j.triboint.2016.12.042
5. Wang, B., Liu, L.J. and Wang, P. (2014) Simulation of Churning Losses in Geared Box. Applied Mechanics and Materials, 703, 241-244. https://doi.org/10.4028/https://www.scientific.net/AMM.703.241
6. Concli, F. and Gorla, C. (2013) Influence of Lubricant Temperature, Lubricant Level and Rotational Speed on the Churning Power Loss in an Industrial Planetary Speed Reducer: Computational and Experimental Study. International Journal of Computational Methods and Experimental Measurements, 1, 353-366. https://doi.org/10.2495/CMEM-V1-N4-353-366
7. Gao, D., Morley, N.B. and Dhir, V. (2003) Numerical Simulation of Wavy Falling Film Flow Using VOF Method. Journal of Computational Physics, 192, 624-642. https://doi.org/10.1016/j.jcp.2003.07.013
8. Theodorakakos, A. and Bergeles, G. (2004) Simulation of Sharp Gas-Liquid Interface Using VOF Method and Adaptive Grid Local Refinement around the Interface. International Journal for Numerical Methods in Fluids, 45, 421-439. https://doi.org/10.1002/fld.706
9. Hou, Q. and Zou, Z. (2006) Comparison between Standard and Renormalization Group k-.EPSILON. Models in Numerical Simulation of Swirling Flow Tundish. Isij International, 45, 325-330. https://doi.org/10.2355/isijinternational.45.325
10. Sinha, K. and Balasridhar, S.J. (2013) Comparison between Standard and Renormalization Group k-.EPSILON. Models in Numerical Simulation of Swirling Flow Tundish. AIAA Journal, 51, 1872-1882. https://doi.org/10.2514/1.J052289
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Compute positive-sequence active and reactive powers - Simulink - MathWorks Nordic
Power (Positive-Sequence)
Compute positive-sequence active and reactive powers
The Power (Positive-Sequence) block computes the positive-sequence active power P (in watts) and reactive power Q (in vars) of a periodic set of three-phase voltages and currents. To perform this computation, the block first computes the positive sequence of the input voltages and currents with a running window over one cycle of the specified fundamental frequency. These formulas are then evaluated:
\begin{array}{c}P=3×\frac{|{V}_{1}|}{\sqrt{2}}×\frac{|{I}_{1}|}{\sqrt{2}}×\mathrm{cos}\left(\phi \right)\\ Q=3×\frac{|{V}_{1}|}{\sqrt{2}}×\frac{|{I}_{1}|}{\sqrt{2}}×\mathrm{sin}\left(\phi \right)\\ \phi =\angle {V}_{1}-\angle {I}_{1}\end{array}
V1 is the positive-sequence component of input Vabc. I1 is the positive-sequence component of input Iabc.
A current flowing into an RL circuit produces a positive P and a positive Q.
As this block uses a running average window, one cycle of simulation must complete before the outputs give the correct value. For the first cycle of simulation, the output is held constant using the values specified by the Voltage initial input and Current initial input parameters.
Specify the voltage initial magnitude and phase used by the block to compute the constant output for the first cycle of simulation. Default is [0, 0].
Specify the current initial magnitude and phase used by the block to compute the constant output for the first cycle of simulation. Default is [0, 0].
The three-phase voltage signal.
The three-phase current signal.
The positive-sequence active power, in watts.
The positive-sequence reactive power, in vars.
The power_ThreePhasePower model compares the outputs of the Power (Positive-Sequence) block with the Power (3ph, Instantaneous) and the Power (dq0, Instantaneous) block. The Power (Positive-Sequence) block gives ripple-free accurate results even with the presence of zero and negative sequences in the voltage supply.
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Integration by substitution - MATLAB changeIntegrationVariable - MathWorks Italia
changeIntegrationVariable
Find Closed-Form Solution
Numerical Integration with High Precision
G = changeIntegrationVariable(F,old,new)
G = changeIntegrationVariable(F,old,new) applies integration by substitution to the integrals in F, in which old is replaced by new. old must depend on the previous integration variable of the integrals in F and new must depend on the new integration variable. For more information, see Integration by Substitution.
When specifying the integrals in F, you can return the unevaluated form of the integrals by using the int function with the 'Hold' option set to true. You can then use changeIntegrationVariable to show the steps of integration by substitution.
Apply a change of variable to the definite integral
{\int }_{\mathit{a}}^{\mathit{b}}\mathit{f}\left(\mathit{x}+\mathit{c}\right)\text{\hspace{0.17em}}\mathit{dx}
Define the integral.
syms f(x) y a b c
F = int(f(x+c),a,b)
{\int }_{a}^{b}f\left(c+x\right)\mathrm{d}x
Change the variable
\mathit{x}+\mathit{c}
in the integral to
\mathit{y}
G = changeIntegrationVariable(F,x+c,y)
{\int }_{a+c}^{b+c}f\left(y\right)\mathrm{d}y
\int \mathrm{cos}\left(\mathrm{log}\left(\mathit{x}\right)\right)\mathit{dx}
using integration by substitution.
F = int(cos(log(x)),'Hold',true)
\int \mathrm{cos}\left(\mathrm{log}\left(x\right)\right)\mathrm{d}x
Substitute the expression log(x) with t.
G = changeIntegrationVariable(F,log(x),t)
\int {\mathrm{e}}^{t} \mathrm{cos}\left(t\right)\mathrm{d}t
H = release(G)
\frac{{\mathrm{e}}^{t} \left(\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right)\right)}{2}
Restore log(x) in place of t.
H = simplify(subs(H,t,log(x)))
\frac{\sqrt{2} x \mathrm{sin}\left(\frac{\pi }{4}+\mathrm{log}\left(x\right)\right)}{2}
Compare the result to the integration result returned by int without setting the 'Hold' option to true.
Fcalc = int(cos(log(x)))
\frac{\sqrt{2} x \mathrm{sin}\left(\frac{\pi }{4}+\mathrm{log}\left(x\right)\right)}{2}
Find the closed-form solution of the integral
\int \mathit{x}\text{\hspace{0.17em}}\mathrm{tan}\left(\mathrm{log}\left(\mathit{x}\right)\right)\mathit{dx}
Define the integral using the int function.
F = int(x*tan(log(x)),x)
\int x \mathrm{tan}\left(\mathrm{log}\left(x\right)\right)\mathrm{d}x
The int function cannot find the closed-form solution of the integral.
Substitute the expression log(x) with t. Apply integration by substitution.
\frac{{\mathrm{e}}^{2 t} {}_{2}{\mathrm{F}\text{hypergeom}}_{1}\left(1,-\mathrm{i};\mathrm{ }1-\mathrm{i};\mathrm{ }-{\mathrm{e}}^{2 t \mathrm{i}}\right) \mathrm{i}}{2}+{\mathrm{e}}^{t \left(2+2 \mathrm{i}\right)} {}_{2}{\mathrm{F}\text{hypergeom}}_{1}\left(1,1-\mathrm{i};\mathrm{ }2-\mathrm{i};\mathrm{ }-{\mathrm{e}}^{2 t \mathrm{i}}\right) \left(-\frac{1}{4}-\frac{1}{4} \mathrm{i}\right)
The closed-form solution is expressed in terms of hypergeometric functions. For more details on hypergeometric functions, see hypergeom.
Compute the integral
{\int }_{0}^{1}{\mathit{e}}^{\sqrt{\mathrm{sin}\left(\mathit{x}\right)}}\mathit{dx}
numerically with high precision.
Define the integral
{\int }_{0}^{1}{\mathit{e}}^{\sqrt{\mathrm{sin}\left(\mathit{x}\right)}}\mathit{dx}
. A closed-form solution to the integral does not exist.
F = int(exp(sqrt(sin(x))),x,0,1)
{\int }_{0}^{1}{\mathrm{e}}^{\sqrt{\mathrm{sin}\left(x\right)}}\mathrm{d}x
You can use vpa to compute the integral numerically to 10 significant digits.
F10 = vpa(F,10)
F10 =
1.944268879
Alternatively, you can use the vpaintegral function and specify the relative error tolerance.
Fvpa = vpaintegral(exp(sqrt(sin(x))),x,0,1,'RelTol',1e-10)
1.944268879
The vpa function cannot find the numerical integration to 70 significant digits, and it returns the unevaluated integral in the form of vpaintegral.
1.944268879138581167466225761060083173280747314051712224507065962575967
To find the numerical integration with high precision, you can perform a change of variable. Substitute the expression
\sqrt{\mathrm{sin}\left(\mathit{x}\right)}
\mathit{t}
. Compute the integral numerically to 70 significant digits.
G = changeIntegrationVariable(F,sqrt(sin(x)),t)
{\int }_{0}^{\sqrt{\mathrm{sin}\left(1\right)}}\frac{2 t {\mathrm{e}}^{t}}{\sqrt{1-{t}^{4}}}\mathrm{d}t
G70 = vpa(G,70)
G70 =
1.944268879138581167466225761060083173280747314051712224507065962575967
F — Expression containing integrals
Expression containing integrals, specified as a symbolic expression, function, vector, or matrix.
old — Subexpression to be substituted
symbolic scalar variable | symbolic expression | symbolic function
Subexpression to be substituted, specified as a symbolic scalar variable, expression, or function. old must depend on the previous integration variable of the integrals in F.
new — New subexpression
New subexpression, specified as a symbolic scalar variable, expression, or function. new must depend on the new integration variable.
Mathematically, the substitution rule is formally defined for indefinite integrals as
\int f\left(g\left(x\right)\right)\text{\hspace{0.17em}}g\text{'}\left(x\right)\text{\hspace{0.17em}}dx={\left(\int f\left(t\right)\text{\hspace{0.17em}}dt\right)|}_{t=g\left(x\right)}
and for definite integrals as
\underset{a}{\overset{b}{\int }}f\left(g\left(x\right)\right)\text{\hspace{0.17em}}g\text{'}\left(x\right)\text{\hspace{0.17em}}dx=\underset{g\left(a\right)}{\overset{g\left(b\right)}{\int }}f\left(t\right)\text{\hspace{0.17em}}dt.
integrateByParts | release | int | diff | vpaintegral
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Let f:r rarr r,f(x)={[|x-[x]|","[x]],[|x-[x+1]|","[x]]:} is o-Turito
Let denotes greatest integer function, then is equal to
Answer:The correct answer is: 5Since these two lines are intersecting so shortest distance between the lines will be 0.
{\int }_{0}^{1} |\mathrm{sin} 2\pi x|\mid dx
Given plane y + z + 1 = 0 is parallel to x-axis as 0.1 + 1.0 + 1.0 = 0 but normal to the plane will be perpendicular to x-axis. Hence (c) is the correct answer.
{\int }_{0}^{1} |\mathrm{sin} 2\pi x|\mid dx
{\int }_{-\pi /4}^{\pi /4} \frac{{e}^{x}\left(x\mathrm{sin}x\right)}{{e}^{2x}-1}dx
Let direction cosines of straight line be l, m, n \ 4l + m + n = 0 l – 2m + n = 0 Þ
\frac{l}{3}=\frac{m}{-3}=\frac{n}{-9}
\frac{l}{-1}=\frac{m}{+1}=\frac{n}{3}
\ Equation of straight line is
\frac{x-2}{-1}=\frac{y+1}{1}=\frac{z+1}{3}
. Hence (c) is the correct choice.
{\int }_{-\pi /4}^{\pi /4} \frac{{e}^{x}\left(x\mathrm{sin}x\right)}{{e}^{2x}-1}dx
\frac{l}{3}=\frac{m}{-3}=\frac{n}{-9}
\frac{l}{-1}=\frac{m}{+1}=\frac{n}{3}
\frac{x-2}{-1}=\frac{y+1}{1}=\frac{z+1}{3}
{\int }_{0}^{100} \left\{\sqrt{x}\right\}dx
(where {x} is the fractional part of x) is
{\int }_{0}^{100} \left\{\sqrt{x}\right\}dx
(where a, b are integers) =
The centre of the sphere is (1, 2, –3) so if other extremity of diameter is (x1, y1, z1), then
= 1, = 2, = –3
\ Required point is (0, 5, 7).
The foot of the perpendicular on the line drown from the origin is C if the line cuts the x-axis and y-axis at A and B respectively then BC : CA is
Let direction ratios of the required line be <a, b, c>
Therefore a - 2 b - 2 c = 0
And 2 b + c = 0
Þ c = - 2 b
a - 2 b + 4b = 0 Þ a = - 2 b
Therefore direction ratios of the required line are <- 2b, b, - 2b> = <2, - 1, 2>
direction cosines of the required line
The shortest distance between the two straight line
\frac{x-4/3}{2}=\frac{y+6/5}{3}=\frac{z-3/2}{4}
\frac{5y+6}{8}=\frac{2z-3}{9}=\frac{3x-4}{5}
\frac{x-4/3}{2}=\frac{y+6/5}{3}=\frac{z-3/2}{4}
\frac{5y+6}{8}=\frac{2z-3}{9}=\frac{3x-4}{5}
A bob of mass M is suspended by a massless string of length L. The horizontal velocity
v
at position A is just sufficient to make it reach the point B. The angle
\theta
at which the speed of the bob is half of that at A, satisfies
Velocity of the bob at the point A
v=\sqrt{5gL}
{\left(\frac{v}{2}\right)}^{2}={v}^{2}-2gh\left(ii\right)
h=L\left(1-\mathrm{cos}\theta \right)\left(iii\right)
Solving Eqs.\left(i\right), \left(ii\right)and \left(iii\right), we get
\mathrm{cos}\theta =-\frac{7}{8}
or \theta ={cos}^{-1}\left(-\frac{7}{8}\right)=151°
v
\theta
v=\sqrt{5gL}
{\left(\frac{v}{2}\right)}^{2}={v}^{2}-2gh\left(ii\right)
h=L\left(1-\mathrm{cos}\theta \right)\left(iii\right)
Solving Eqs.\left(i\right), \left(ii\right)and \left(iii\right), we get
\mathrm{cos}\theta =-\frac{7}{8}
or \theta ={cos}^{-1}\left(-\frac{7}{8}\right)=151°
A bob of mass is suspended by a massless string of length . The horizontal velocity at position is just sufficient to make it reach the point . The angle at which the speed of the bob is half of that at , satisfies
A piece of wire is bent in the shape of a parabola
y=k{x}^{2} \left(y
-axis vertical) with a bead of mass
m
on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the
x
-axis with a constant acceleration
a
. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the
y
ma\mathrm{cos}\theta =mg\mathrm{cos}\left(90-\theta \right)
⇒\frac{a}{g}=\mathrm{tan}\theta ⇒\frac{a}{g}=\frac{dy}{dx}
⇒\frac{d}{dx}{\left(kx\right)}^{2}=\frac{a}{g}⇒x=\frac{a}{2gk}
y=k{x}^{2} \left(y
m
x
a
y
ma\mathrm{cos}\theta =mg\mathrm{cos}\left(90-\theta \right)
⇒\frac{a}{g}=\mathrm{tan}\theta ⇒\frac{a}{g}=\frac{dy}{dx}
⇒\frac{d}{dx}{\left(kx\right)}^{2}=\frac{a}{g}⇒x=\frac{a}{2gk}
A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out length where is in metre and t is in second. The radius of the path is 20 m. The acceleration of P when t =2s is nearly
A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and material are now added to P as shown in the figure. The ray will suffer
As the prisms Q and R are of the same material and have identical shape they combine to form a slab with parallel faces. Such a slab does not cause any deviation.
In the given figure, what is the angle of prism
Angle of prism is the angle between incident and emergent surfaces.
A ray of light is incident on an equilateral glass prism placed on a horizontal table. For minimum deviation which of the following is true
In minimum deviation position refracted ray inside the prism is parallel to the base of the prism
The equation of the plane containing the line where al + bm + cn is equal to
Since these two lines are intersecting so shortest distance between the lines will be 0.
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Compute active and reactive powers of voltage-current pair at fundamental frequency - Simulink - MathWorks Nordic
Compute active and reactive powers of voltage-current pair at fundamental frequency
The Power block computes the active power (P), in watts, and the reactive power (Q), in vars, of a voltage-current pair at fundamental frequency. To perform this computation, the block first determines the fundamental values (magnitude and phase) of the two input signals V & I. The following P and Q are then calculated:
\begin{array}{l}P=\frac{|V|}{\sqrt{2}}×\frac{|I|}{\sqrt{2}}×\mathrm{cos}\left(\phi \right)\\ Q=\frac{|V|}{\sqrt{2}}×\frac{|I|}{\sqrt{2}}×\mathrm{sin}\left(\phi \right)\\ \phi =\angle V-\angle I\end{array}
As this block uses a running average window, one cycle of simulation must complete before the outputs give the correct value. For the first cycle of simulation, the outputs are held constant using the values specified by the Voltage initial input and Current initial input parameters.
Specify the fundamental frequency, in hertz, of the input signals. Default is 60.
Voltage initial input [ Mag, Phase (degrees) ]
Specify the voltage initial magnitude and phase to set a constant output for the first cycle of simulation. Default is [1, 0].
Current initial input [ Mag, Phase (degrees) ]
Specify the current initial magnitude and phase to set a constant output for the first cycle of simulation. Default is [1, 0].
The voltage signal.
The current signal.
The active power of voltage-current pair signal, in watts.
The reactive power of voltage-current pair signal, in vars.
The power_Power model shows the accuracy of the Power block in evaluating the active and reactive powers of a voltage-current set.
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I am sorry to trouble you, but will you tell me how I can best send back the large Box.2 Is it not too heavy for Parcels Delivery? Secondly how address it? How to the Library?— You must let me know what the carriage will come to.— I will return all immediately I get your answer, except, perhaps 2d, & except 3d & 4th vols. of Ledebour,3 if you can spare them, but I fear these will be just the Books you are likely most to want. But if you can spare them I will return each vol. as soon as finished.
I shd. like also to keep Grisebach.4 also Cybele;5 this latter, probably, can best be spared. Grisebach I have prepared for classifying, but do not care much about.— So please give me orders, which of the above Books I may keep: Ledebour is by far most useful on account of ranges: I think
\frac{1}{2}
Ledebour will be done by time Box goes.—
I shall be very glad to do 2 or 3 vols. of Decandolle;6 if you will think which wd. be best:— I shd. think, but will be guided entirely by you, one with some one huge nat. Family, & one with several small ones.—
Also I shd. especially be thankful for any Flora of Holland with varieties marked.— These can be sent in about 3 weeks time, as by enclosed address.
Thank you much for enquiring after our children: Etty is better but far from strong:7 Lenny much better, but has attacks of intermittent pulse.8
How busy you appear to be!
Pray give my very kind remembrances to Mrs. Hooker; How long it is since I have seen her: I hope she is thoroughily well.
I have sent little notice to Gardeners’ Chron. on fertilisation of Kidney Beans:9 I have not yet looked to your case of hybrid kidney Beans.—
You talked of coming here for a Sunday: is there any hope of it? I shd. most thoroughily enjoy it & I have a frightful lot of questions to torture you with you unfortunate wretch.—
Cd. you lay your hand on paper in which a Fucus is mentioned not capable of crossing reciprocally?10 it is not worth a hunt.—
If you cannot come here you must let me pay you a call of an hour or two, & do some heavy questioning.—
I have just looked at my list of queries, but it is not so tremendous, as I had fancied.—
The year is given by CD’s reference to his notice on the fertilisation of kidney beans (letter to Gardeners’ Chronicle, 18 October [1857]).
CD was returning several botanical works lent to him by Hooker (see letter to J. D. Hooker, 30 September [1857]).
Grisebach 1843–4.
Henrietta Darwin was at Moor Park, where she was undergoing hydropathic treatment. She returned to Down after ten weeks of therapy on 31 October 1857 (Emma Darwin’s diary).
Letter to Gardeners’ Chronicle, 18 October [1857].
Thuret 1854–5.
This is CD’s ‘list of queries’ referred to in the letter. It is preserved in DAR 114: 222c.
Adlumia had been mentioned in the letter from Asa Gray, [August 1857], as a plant in which insects could hardly be agents of cross-pollination .
CD refers to Andrew Campbell, a British civil servant in India whom Hooker had met and travelled with during his stay in India (see Correspondence vol. 4). CD was at this time investigating the breeding of yaks (see letter from Robert Schlagintweit, 25 September 1857).
Boreau 1840. CD discussed this query in a note in DAR 47: 192a: ‘Boreau Tom 2. p. 26. Has seen Linum cartharticum with alternate leaves. Is there not division of Linum into 2 sections.’ The note was pasted onto another sheet of paper, and CD wrote: ‘Ch. 7 | Hooker will enquire’. Hooker answered the query in the letter from J. D. Hooker, [6 December 1857]. CD subsequently added in pencil: ‘(not entered)’.
DAR 114: 212, 222c
ALS 6pp CD note AL 2pp inc
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{\displaystyle f:\mathbb {F} _{2^{n}}\rightarrow \mathbb {F} _{2}}
{\displaystyle 0}
{\displaystyle \pm \mu }
{\displaystyle \mu }
called the amplitude o{\displaystyle f}
{\displaystyle F}
{\displaystyle (n,m)}
{\displaystyle F}
{\displaystyle u\cdot F}
{\displaystyle u\neq 0}
{\displaystyle F}
{\displaystyle F}
{\displaystyle a}nd every
{\displaystyle v}
{\displaystyle \{b\in \mathbb {F} _{2}^{n}:D_{a}D_{b}F(x)=v\}}
{\displaystyle x}
{\displaystyle \{b\in \mathbb {F} _{2}^{n}:D_{a}F(b)=D_{a}F(x)+v\}}
{\displaystyle x}
{\displaystyle f_{\phi ,h}}
{\displaystyle f_{\phi ,h}(x,y)=x\cdot \phi (y)+h(y)}
{\displaystyle x\in \mathbb {F} _{2}^{r},y\in \mathbb {F} _{2}^{s}}
{\displaystyle r}
{\displaystyle s}
{\displaystyle n=r+s}
{\displaystyle \phi :\mathbb {F} _{2}^{s}\rightarrow \mathbb {F} _{2}^{r}}
{\displaystyle h:\mathbb {F} _{2}^{s}\rightarrow \mathbb {F} _{2}}
{\displaystyle f_{\phi ,h}}
{\displaystyle W_{f_{\phi ,h}}(a,b)=2^{r}\sum _{y\in \phi ^{-1}(a)}(-1)^{b\cdot y+h(y)}}
{\displaystyle (a,b)}
{\displaystyle \phi }
{\displaystyle f_{\phi ,h}}
{\displaystyle 2^{r}}
{\displaystyle 2^{r+1}}
{\displaystyle f}
{\displaystyle \sum _{a,b\in \mathbb {F} _{2}^{n}}(-1)^{DaDbf(x)}}
{\displaystyle x\in \mathbb {F} _{2}^{n}}
{\displaystyle F}
{\displaystyle (n,m)}
{\displaystyle v\in \mathbb {F} _{2}^{m}}
{\displaystyle \{(a,b)\in (\mathbb {F} _{2}^{n})^{2}:D_{a}D_{b}F(x)=v\}}
{\displaystyle x}
{\displaystyle x}
{\displaystyle v\in \mathbb {F} _{2}^{m}}
{\displaystyle v\neq 0}
{\displaystyle F}
{\displaystyle D_{a}D_{b}F(x)}
{\displaystyle D_{a}F(b)+D_{a}F(x)}
{\displaystyle (a,b)}
{\displaystyle (\mathbb {F} _{2}^{n})^{2}}
{\displaystyle F,G}
{\displaystyle u\cdot F,u\cdot G}
{\displaystyle F(x)=x^{d}}
{\displaystyle \lambda \neq 0}
{\displaystyle |\{(a,b)\in \mathbb {F} _{2^{n}}^{2}:D_{a}F(b)+D_{a}F(x)=v\}|=|\{(a,b)\in \mathbb {F} _{2^{n}}^{2}:D_{a}F(b)+D_{a}F(x/\lambda )=v/\lambda ^{d}\}|.}
{\displaystyle F}
{\displaystyle v\in \mathbb {F} _{2^{n}}}
{\displaystyle |\{(a,b)\in \mathbb {F} _{2^{n}}^{2}:D_{a}F(b)+D_{a}F(1)=v\}|=|\{(a,b)\in \mathbb {F} _{2^{n}}^{2}:D_{a}F(b)+D_{a}F(0)=v\}|;}
{\displaystyle F}
{\displaystyle v\neq 0}
{\displaystyle F}
{\displaystyle (n,m)}
{\displaystyle F}
{\displaystyle v,x\in \mathbb {F} _{2}^{n}}
{\displaystyle |\{(a,b)\in (\mathbb {F} _{2}^{n})^{2}:D_{a}D_{b}F(x)=v\}|=|\{(a,b)\in (\mathbb {F} _{2}^{n})^{2}:F(a)+F(b)=v\}|.}
{\displaystyle F}
{\displaystyle v}
{\displaystyle v\neq 0}
{\displaystyle {\rm {Im}}(D_{a}F)}
{\displaystyle F}
{\displaystyle f}
{\displaystyle {\Delta _{f}}(a)=\sum _{x\in \mathbb {F} _{2}^{n}}(-1)^{f(x)+f(x+a)}}
A{\displaystyle n}
{\displaystyle f}
{\displaystyle x\in \mathbb {F} _{2}^{n}}
{\displaystyle 2^{n}\sum _{a\in \mathbb {F} _{2}^{n}}\Delta _{f}(a)\Delta _{f}(a+x)=\left(\sum _{a\in \mathbb {F} _{2}^{n}}\Delta _{f}^{2}(a)\right)\Delta _{f}(x).}
{\displaystyle (n,m)}
{\displaystyle F}
{\displaystyle x\in \mathbb {F} _{2}^{n},u\in \mathbb {F} _{2}^{m}}
{\displaystyle 2^{n}\sum _{a\in \mathbb {F} _{2}^{n}}\Delta _{u\cdot F}(a)\Delta _{u\cdot F}(a+x)=\left(\sum _{a\in \mathbb {F} _{2}^{n}}\Delta _{u\cdot F}^{2}(a)\right)\Delta _{u\cdot F}(x).}
{\displaystyle F}
{\displaystyle x\in \mathbb {F} _{2}^{n},u\in \mathbb {F} _{2}^{m}}
{\displaystyle \sum _{a\in \mathbb {F} _{2}^{n}}\Delta _{u\cdot F}(a)\Delta _{u\cdot F}(a+x)=\mu ^{2}\Delta _{u\cdot F}(x).}
{\displaystyle F}
{\displaystyle x,v\in \mathbb {F} _{2}^{n}}
{\displaystyle 2^{n}|\{(a,b,c)\in (\mathbb {F} _{2}^{n})^{3}:F(a)+F(b)+F(c)+F(a+b+c+x)=v\}|=|\{(a,b,c,d)\in (\mathbb {F} _{2}^{n})^{4}:F(a)+F(b)+F(c)+F(a+b+c)+F(d)+F(d+x)=v\}|.}
{\displaystyle f:\mathbb {F} _{2^{n}}\rightarrow \mathbb {F} _{2}}
{\displaystyle 0\neq \alpha \in \mathbb {F} _{2}^{n}}
{\displaystyle \sum _{w\in \mathbb {F} _{2}^{n}}W_{f}(w+\alpha )W_{f}^{3}(w)=0.}
{\displaystyle (n,m)}
{\displaystyle F}
{\displaystyle u\in \mathbb {F} _{2}^{m}}
{\displaystyle 0\neq \alpha \in \mathbb {F} _{2}^{n}}
{\displaystyle \sum _{w\in \mathbb {F} _{2}^{n}}W_{F}(w+\alpha ,u)W_{F}^{3}(w,u)=0.}
{\displaystyle F}
{\displaystyle \sum _{w\in \mathbb {F} _{2}^{n}}W_{F}^{4}(w,u)}
{\displaystyle u}
{\displaystyle u\neq 0}
{\displaystyle f:\mathbb {F} _{2^{n}}\rightarrow \mathbb {F} _{2}}
{\displaystyle b\in \mathbb {F} _{2}}
{\displaystyle \sum _{a\in \mathbb {F} _{2}}W_{f}^{4}(a)=2^{n}(-1)^{f(b)}\sum _{a\in \mathbb {F} _{2}^{n}}(-1)^{a\cdot b}W_{f}^{3}(a).}
{\displaystyle (n,m)}
{\displaystyle F}
{\displaystyle b\in \mathbb {F} _{2}^{n}}
{\displaystyle u\in \mathbb {F} _{m}}
{\displaystyle \sum _{a\in \mathbb {F} _{2}^{n}}W_{F}^{4}(a,u)=2^{n}(-1)^{u\cdot F(b)}\sum _{a\in \mathbb {F} _{2}^{n}}(-1)^{a\cdot b}W_{F}^{3}(a,u).}
{\displaystyle F}
{\displaystyle u}
{\displaystyle u\neq 0}
{\displaystyle f}
i{\displaystyle n}
{\displaystyle \left(\sum _{a\in \mathbb {F} _{2}^{n}}W_{f}^{4}(a)\right)^{2}\leq 2^{2n}\left(\sum _{a\in \mathbb {F} _{2}^{n}}W_{f}^{6}(a)\right),}
with equality if and only i{\displaystyle f}
{\displaystyle (n,m)}
{\displaystyle F}
{\displaystyle \sum _{u\in \mathbb {F} _{2}^{m}}\left(\sum _{a\in \mathbb {F} _{2}^{n}}W_{F}^{4}(a,u)\right)^{2}\leq 2^{2n}\sum _{u\in \mathbb {F} _{2}^{m}}\left(\sum _{a\in \mathbb {F} _{2}^{n}}W_{F}^{6}(a,u)\right),}
{\displaystyle F}
{\displaystyle (n,m)}
{\displaystyle \sum _{u\in \mathbb {F} _{2}^{m}}\sum _{a\in \mathbb {F} _{2}^{n}}W_{F}^{4}(a,u)\leq 2^{n}\sum _{u\in \mathbb {F} _{2}^{m}}{\sqrt {\sum _{a\in \mathbb {F} _{2}^{n}}W_{F}^{6}(a,u)}},}
{\displaystyle F}
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IsomorphismClassRepresentatives - Maple Help
Home : Support : Online Help : Mathematics : Algebra : Magma : IsomorphismClassRepresentatives
return a list of isomorphism class representatives for a list of finite magmas
IsomorphismClassRepresentatives( L )
list; list of magmas whose isomorphism class representatives are to be computed
The IsomorphismClassRepresentatives( L ) command returns a list containing one representative from each isomorphism class of the magmas in the input list L. Thus, every magma in L is isomorphic to one of the magmas in the list returned, and no two magmas in the returned list are isomorphic.
If the magmas in L are already pairwise non-isomorphic, then L itself is returned.
\mathrm{with}\left(\mathrm{Magma}\right):
L≔[\mathrm{seq}]\left(\mathrm{RandomMagma}\left(2\right),i=1..20\right):
C≔\mathrm{IsomorphismClassRepresentatives}\left(L\right)
\textcolor[rgb]{0,0,1}{C}\textcolor[rgb]{0,0,1}{≔}[[\begin{array}{cc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{2}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{2}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{2}\end{array}]]
\mathrm{C2}≔\mathrm{IsomorphismClassRepresentatives}\left(C\right)
\textcolor[rgb]{0,0,1}{\mathrm{C2}}\textcolor[rgb]{0,0,1}{≔}[[\begin{array}{cc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{2}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{2}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{2}\end{array}]]
\mathrm{evalb}\left(C=\mathrm{C2}\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
The Magma[IsomorphismClassRepresentatives] command was introduced in Maple 16.
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Table 4 The ability of 2C loop to buffer TF noise under of parameters in all three models.
Stop Model
Degr. Model
Dual Degr. Model
\stackrel{̄}{\epsilon }
\stackrel{̄}{\epsilon }
\stackrel{̄}{\epsilon }
\begin{array}{c}{h}_{s}=1\\ {h}_{s}=100\\ {h}_{s}=200\\ {h}_{s}=400\end{array}
\begin{array}{c}\hfill -0.71\hfill \\ \hfill -0.49\hfill \\ \hfill -\mathbf{\text{0}}\mathbf{\text{.09}}\hfill \\ \hfill -0.30\hfill \end{array}
\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 34\hfill \\ \hfill 4\hfill \end{array}
\left\{\begin{array}{c}-0.69\hfill \\ -0.57\hfill \\ -0.57\hfill \\ -0.64\hfill \end{array}\right\}
\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}
\begin{array}{c}\hfill -0.70\hfill \\ \hfill -0.63\hfill \\ \hfill \left\{\begin{array}{c}-0.57\hfill \\ -0.59\hfill \end{array}\right\}\hfill \end{array}
\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}
\begin{array}{c}{k}_{q}=0.02\\ {k}_{q}=0.04\\ {k}_{q}=0.08\\ {k}_{q}=0.16\end{array}
\begin{array}{c}\hfill -0.07\hfill \\ \hfill -0.17\hfill \\ \hfill -0.64\hfill \\ \hfill -0.94\hfill \end{array}
\begin{array}{c}\hfill 43\hfill \\ \hfill 26\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}
\begin{array}{c}\hfill -0.38\hfill \\ \hfill -0.57\hfill \\ \hfill -0.75\hfill \\ \hfill -0.94\hfill \end{array}
\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}
\begin{array}{c}\hfill -0.30\hfill \\ \hfill -0.58\hfill \\ \hfill -0.82\hfill \\ \hfill -0.94\hfill \end{array}
\begin{array}{c}\hfill 2\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}
\begin{array}{c}{k}_{s}=0.01\\ {k}_{s}=0.25\\ {k}_{s}=0.50\\ {k}_{s}=0.75\end{array}
\begin{array}{c}\hfill -0.69\hfill \\ \hfill -0.43\hfill \\ \hfill -0.13\hfill \\ \hfill 0.05\hfill \end{array}
\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 30\hfill \\ \hfill 55\hfill \end{array}
\begin{array}{c}\hfill -0.69\hfill \\ \hfill \left\{\begin{array}{c}-0.56\hfill \\ -0.58\hfill \\ -0.61\hfill \end{array}\right\}\hfill \end{array}
\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}
\begin{array}{c}\hfill -0.69\hfill \\ \hfill -0.63\hfill \\ \hfill -0.59\hfill \\ \hfill -0.55\hfill \end{array}
\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}
\begin{array}{c}{h}_{p}\left({h}_{g}\right)=30;{k}_{rs}=1\cdot 1{0}^{-5}\\ {h}_{p}\left({h}_{g}\right)=60;{k}_{rs}=2\cdot 1{0}^{-5}\\ {h}_{p}\left({h}_{g}\right)=120;{k}_{rs}=4\cdot 1{0}^{-5}\\ {h}_{p}\left({h}_{g}\right)=240;{k}_{rs}=8\cdot 1{0}^{-5}\end{array}
\begin{array}{c}\hfill 0.07\hfill \\ \hfill -0.14\hfill \\ \hfill -0.42\hfill \\ \hfill -0.61\hfill \end{array}
\begin{array}{c}\hfill 58\hfill \\ \hfill 26\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}
\begin{array}{c}\hfill -0.64\hfill \\ \hfill \left\{\begin{array}{c}-0.57\hfill \\ -0.56\hfill \end{array}\right\}\hfill \\ \hfill -0.64\hfill \end{array}
\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}
\begin{array}{c}\hfill -0.63\hfill \\ \hfill -0.59\hfill \\ \hfill -0.49\hfill \\ \hfill -0.42\hfill \end{array}
\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}
For each model and for each parameter value we present the average ε value (left column) and the number n of experiments with positive value of this coefficient (right column). The method for calculation of ε is described in section. Negative ε mean that TF noise is dampen in a loop. The extremal ε values achieved at intermediate parameter values are shown in bold. The Wilcoxon Rank-Sum Test was used to test for significance of difference between the values of ε for adjacent parameter values. The differences which are statistically insignificant at the α = 0.05 level are placed in parentheses.
|
DP on Trees - Introduction · USACO Guide
HomeGoldDP on Trees - Introduction
TutorialSolution - Tree MatchingSolution 1Taking No EdgesTaking One EdgeSolution 2 - GreedyProblemsEasierHarder
DP on Trees - Introduction
Authors: Michael Cao, Benjamin Qi
Using subtrees as subproblems.
Silver - (Optional) Introduction to Functional Graphs
DP on Trees and DAGs
bad code format
Don't just dive into trying to figure out a DP state and transitions -- make some observations if you don't see any obvious DP solution! Also, sometimes a greedy strategy suffices in place of DP.
Solution - Tree Matching
In this problem, we're asked to find the maximum matching of a tree, or the largest set of edges such that no two edges share an endpoint. Let's use DP on trees to do this.
Root the tree at node
1
, allowing us to define the subtree of each node.
dp_2[v]
represent the maximum matching of the subtree of
v
such that we don't take any edges leading to some child of
v
dp_1[v]
v
such that we take one edge leading into a child of
v
. Note that we can't take more than one edge leading to a child, because then two edges would share an endpoint.
Taking No Edges
Since we will take no edges to a child of
v
, the children vertices of
v
can all take an edge to some child, or not. Additionally, observe that the children of
v
taking an edge to a child will not prevent other children of
v
from doing the same. In other words, all of the children are independent. So, the transitions are:
dp_2[v] = \sum_{u \in child(v)} \max(dp_1[u], dp_2[u])
Taking One Edge
The case where we take one child edge of
v
is a bit trickier. Let's assume the edge we take is
v \rightarrow u
u \in child(v)
. Then, to calculate
dp_1[v]
for the fixed
u
dp_1[v] = dp_2[u] + 1 + dp_2[v] - \max(dp_2[u], dp_1[u])
In other words, we take the edge
v \rightarrow u
, but we can't take any children of
u
in the matching, so we add
dp_2[u] + 1
. Then, to deal with the other children, we add:
\sum_{w \in child(v), w \neq u} \max(dp_1[w], dp_2[w]).
Fortunately, since we've calculated
dp_2[v]
already, this expression simplifies to:
dp_2[v] - \max(dp_2[u], dp_1[u])
Overall, to calculate the transitions for
dp_1[v]
over all possible children
u
dp_1[v] = \max_{u \in child(v)} (dp_2[u] + 1 + dp_2[v] - \max(dp_2[u], dp_1[u]))
Loop through the children of
v
twice to calculate
dp_1[v]
dp_2[v]
separately! You need to know
dp_2[v]
dp_1[v]
Just keep matching a leaf with the only vertex adjacent to it while possible.
Easy Show Tags DP, Tree
Normal Show Tags DP, Tree
2020 - Village (Minimum)
Normal Show Tags Greedy, Tree
Normal Show Tags Tree
Berland Federalization
Hard Show Tags Greedy, Tree
2004 - Spies
Hard Show Tags Functional Graph
Memory limit for "Spies" is very tight.
These problems are not Gold level. You may wish to return to these once you're in Platinum.
Very Hard Show Tags DP, Tree
Very Hard Show Tags Binary Search, DP, Tree
Insane Show Tags DP, Tree
Insane Show Tags Greedy, Tree
|
On the stability of non-symmetric equilibrium figures of a rotating viscous incompressible liquid | EMS Press
On the stability of non-symmetric equilibrium figures of a rotating viscous incompressible liquid
We consider a classical problem of stability of equilibrium figures of a liquid rotating uniformly as a rigid body about a fixed axis. We connect the problem of stability with the behavior for large
t
of solutions of an evolution problem governing a motion of an isolated liquid mass whose initial data are slight perturbations of the regime of rigid rotation. The main attention is given to the case when the figure is not rotationally symmetric; in this case the regime of a rigid rotation defines a periodic solution of the above-mentioned non-stationary problem. It is proved that the sufficient condition of stability is the positivity of the second variation of the energy functional in an appropriate space of functions.
Vsevolod A. Solonnikov, On the stability of non-symmetric equilibrium figures of a rotating viscous incompressible liquid. Interfaces Free Bound. 6 (2004), no. 4, pp. 461–492
|
Cohomology of mapping class groups and the abelian moduli space | EMS Press
We consider a surface
\Sigma
g \geq 3
, either closed or with exactly one puncture. The mapping class group
\Gamma
\Sigma
acts symplectically on the abelian moduli space
M = \operatorname{Hom}(\pi_1(\Sigma), \operatorname{U}(1)) = \operatorname{Hom}(H_1(\Sigma), \operatorname{U}(1))
, and hence both
L^2(M)
C^\infty(M)
are modules over
\Gamma
. In this paper, we prove that both the cohomology groups
H^1(\Gamma, L^2(M))
H^1(\Gamma, C^\infty(M))
Jørgen Ellegaard Andersen, Rasmus Villemoes, Cohomology of mapping class groups and the abelian moduli space. Quantum Topol. 3 (2012), no. 3, pp. 359–376
|
Conservation Laws, Symmetry Reductions, and New Exact Solutions of the (2 + 1)-Dimensional Kadomtsev-Petviashvili Equation with Time-Dependent Coefficients
2014 Conservation Laws, Symmetry Reductions, and New Exact Solutions of the (2 + 1)-Dimensional Kadomtsev-Petviashvili Equation with Time-Dependent Coefficients
The (2 + 1)-dimensional Kadomtsev-Petviashvili equation with time-dependent coefficients is investigated. By means of the Lie group method, we first obtain several geometric symmetries for the equation in terms of coefficient functions and arbitrary functions of
t
. Based on the obtained symmetries, many nontrivial and time-dependent conservation laws for the equation are obtained with the help of Ibragimov’s new conservation theorem. Applying the characteristic equations of the obtained symmetries, the (2 + 1)-dimensional KP equation is reduced to (1 + 1)-dimensional nonlinear partial differential equations, including a special case of (2 + 1)-dimensional Boussinesq equation and different types of the KdV equation. At the same time, many new exact solutions are derived such as soliton and soliton-like solutions and algebraically explicit analytical solutions.
Li-hua Zhang. "Conservation Laws, Symmetry Reductions, and New Exact Solutions of the (2 + 1)-Dimensional Kadomtsev-Petviashvili Equation with Time-Dependent Coefficients." Abstr. Appl. Anal. 2014 (SI55) 1 - 13, 2014. https://doi.org/10.1155/2014/853578
Li-hua Zhang "Conservation Laws, Symmetry Reductions, and New Exact Solutions of the (2 + 1)-Dimensional Kadomtsev-Petviashvili Equation with Time-Dependent Coefficients," Abstract and Applied Analysis, Abstr. Appl. Anal. 2014(SI55), 1-13, (2014)
|
The figure shows a unifrom conducting structure which carries c-Turito
A particle is projected from the ground with an initial velocity of at an angle of 60 with horizontal. From what height should an another particle be projected horizontally with velocity so that both the particles collide in ground at point if both are projected simultaneously?
Horizontal component of velocity of is 10 or which is equal to the velocity of in horizontal direction. They will collide at if time of flight of the particles are equal or
Ethyl alcohol is also known as:
Let f : R →R and g : R →R be continuous functions, then the value of
{\int }_{-\pi /2}^{\pi \text{/2}}
(f(x) + f(-x)) (g(x) – g(-x)) dx, is equal to
{\int }_{-\pi /2}^{\pi \text{/2}}
|
Differential Privacy - TOИIC
Differential privacy is one of the techniques Tonic uses to ensure the privacy of your data. At a high level, differential privacy limits the influence of a single user’s data on Tonic’s output data, ensuring privacy through a robust notion of plausible deniability. Slightly more precisely, differential privacy is a property of a randomized process on databases which stipulates that altering the database by a single record (or user) can only perturb the output in a limited way.
Differentially private processes have some valuable properties. The first, and perhaps most important, is that no amount of post processing or additional knowledge can break the guarantees of differential privacy. This isn’t true for other data anonymization techniques, e.g. k-anonymity. Additionally, differentially private data can be combined with other differentially private data without losing its protection. In practice this means data protected by a process with differential privacy cannot be reverse engineered, re-identified, or otherwise compromised, no matter the adversary.
Tonic can enable differential privacy for the Categorical and Continuous generators and many of Tonic’s generators enjoy this property for free — any generator that does not use the underlying data at all is considered "data free", which has the property of being differentially private.
Categorical Generator
The categorical generator shuffles the values of a column while preserving the overall frequencies of the values. Differential privacy (enabled by default) will further protect the privacy of your data by:
Adding noise to the frequencies of categories
Suppressing rare categories.
These steps ensure that a single row of source data has limited influence on the output values. By default, the privacy budget for this generator is
\varepsilon = 1,
\delta = \frac{1}{10n}
n
is the number of rows.
Disabling differential privacy is not recommended, but there is one situation where it may be necessary: when the data in each row is unique or nearly unique — as a general rule of thumb, categories represented by fewer than 15 rows are at risk of being suppressed. Tonic will warn you when a column isn’t suitable for differential privacy, and you must disable it in those cases.
Continuous Generator
The continuous generator produces samples which preserve the individual column distributions and correlations between columns. When differential privacy is enabled, noise is added to the individual distributions and the correlation matrix, using the mechanism described in [4]. The default privacy budget for this generator is
\varepsilon = 1,
\delta = \frac{1}{10n}
More details: Mathematical formulation
Differential privacy is a property of a randomized algorithm
\mathcal{M}
which takes as input a database
D
and produces some output
\mathcal{M}(D).
The outputs could be counts or summary statistics or synthetic databases — the specific type is not important for this formulation. For this formulation, we say two databases
D
D'
are neighbors if they differ by a single row. For a given
\varepsilon\geq 0
\mathcal{M}
\varepsilon
differentially private if, for all subset of outputs
\mathcal{O}
\operatorname{P}(\mathcal{M}(D)\in \mathcal{O}) \leq e^\varepsilon \operatorname{P}(\mathcal{M}(D')\in \mathcal{O})
\delta
is non-zero, this is sometimes called approximately differentially private. The parameter
\delta
is often interpreted as an upper bound on the risk a privacy leak, but
\varepsilon
is the privacy budget of the algorithm, and quantifies in a precise sense, an upper bound on how much information an adversary can gain from observing the outputs of the algorithm on an unknown database: Suppose an attacker suspects our secret database
D
is one of two possible neighboring databases
D_1, D_2
, with some fixed odds. If
\mathcal{M}
\varepsilon
differentially private, then observing
\mathcal{M}(D)
updates the the attacker's log odds of
D=D_1
D=D_2
\varepsilon
. The closer
\varepsilon
0
the better the privacy guarantee, as an attacker is more and more limited in what information they can learn from
\mathcal{M}(D)
. Conversely, larger values of
\varepsilon
mean an attacker can possibly learn significant information by observing
\mathcal{M}(D)
A simple example: counting
Suppose we want to count the number of users in a database having some sensitive property — for example, the number of users with a particular medical diagnosis. Dwork, McSherry, Nissim and Smith introduced in [2] the Laplace Mechanism as a way of publishing these counts in a secure way, by adding noise sampled from the Laplace distribution. This noise affords us plausible deniability — if the underlying count changed by
\pm 1
, then the probability of observing the same noisy output does not change by much:
\operatorname{P}( \text{Noised} = k | \text{count} = n \pm 1) \leq \operatorname{exp}(1) \operatorname{P}(\text{Noised} = k | \text{count} = n)
We illustrate this visually, showing the probability density function (pdf) of the observed values given true counts of
n
n + 1
(orange), and
n - 1
The blue shaded region shows that the the possibly noisy count values for
n-1
n+1
lie within a factor of
\exp(1)
of the noisy count values of
n
, so this mechanism is differentially private with
\varepsilon=1
A common relaxation, called approximate differential privacy, allows for flexible privacy analysis with noise drawn from a wider array of distributions than the Laplace distribution. For example, the AnalyzeGauss mechanisms of [4], and differentially private gradient descent of [1], use Gaussian noise as a fundamental ingredient, which requires the following relaxation: For a given
\varepsilon\geq 0
\delta\in[0,1]
\mathcal{M}
(\varepsilon,\delta)
\mathcal{O}
\operatorname{P}(\mathcal{M}(D)\in \mathcal{O}) \leq e^\varepsilon \operatorname{P}(\mathcal{M}(D')\in \mathcal{O}) +\delta
\delta
is often described as the risk of a (possibly catastrophic) privacy violation. While this formal definition does allow, for example, a mechanism revealing a sensitive database with probability
\delta
, in practice this is not a plausible outcome with carefully designed mechanisms. Additionally, taking
\delta
to be small relative to the size of the database ensures the risk of disclosure is low.
Martin Abadi, Andy Chu, Ian Goodfellow, H. Brendan McMahan, Ilya Mironov, Kunal Talwar, and Li Zhang. 2016. Deep Learning with Differential Privacy. In Proceedings of the 2016 ACM SIGSAC Conference on Computer and Communications Security (CCS '16). Association for Computing Machinery, New York, NY, USA, 308–318. DOI:https://doi.org/10.1145/2976749.2978318
Cynthia Dwork, Frank McSherry, Kobbi Nissim and Adam Smith. 2006 Calibrating Noise to Sensitivity in Private Data Analysis. In: Halevi S., Rabin T. (eds) Theory of Cryptography. (TCC '06). Lecture Notes in Computer Science, vol 3876. Springer, Berlin, Heidelberg. DOI:https://doi.org/10.1007/11681878_14
Cynthia Dwork and Aaron Roth. 2014. The Algorithmic Foundations of Differential Privacy. Found. Trends Theor. Comput. Sci. 9, 3–4 (August 2014), 211–407. DOI:https://doi.org/10.1561/0400000042
Cynthia Dwork, Kunal Talwar, Abhradeep Thakurta, and Li Zhang. 2014. Analyze gauss: optimal bounds for privacy-preserving principal component analysis. In Proceedings of the forty-sixth annual ACM symposium on Theory of computing (STOC '14). Association for Computing Machinery, New York, NY, USA, 11–20. DOI:https://doi.org/10.1145/2591796.2591883
|
Axially flexible bar or cable - MATLAB - MathWorks Nordic
Number of flexible elements
Axially flexible bar or cable
The Rod block represents an axially flexible bar or cable in tension or compression.
To represent the bar or cable, the block uses a lumped parameter model. The model is composed of N+1 lumped masses that are connected in series by N sets of parallel-connected spring and damper circuits. The spring represents elasticity. The damper represents material damping.
A single flexible element model exhibits an eigenfrequency that is close to the first eigenfrequency of the distributed parameter model. For more accurate analysis, select 2, 4, 8, or more flexible elements.
The equivalent physical network contains N spring and damper circuits and N+1 mass blocks. The total mass of the rod is distributed evenly over the mass blocks. The stiffness of the spring in each spring and damper circuit is equal to the N times the stiffness of the rod material.
The defining equations for the model are:
\eta =2\zeta \sqrt{\frac{K\cdot m}{2}}
A=\frac{\pi \left({D}^{2}-{d}^{2}\right)}{4}
K=\frac{E\cdot A}{L}
m=A\cdot L\cdot \rho
η is the damping of the rod.
ς is the damping ratio of the rod material.
K is the stiffness of the rod.
A is the cross-sectional area of the rod.
D is the outer diameter of the rod.
d is the inner diameter of the rod, where
d = 0 for a solid rod.
d > 0 for an annular, that is, hollow, rod.
E is Young's modulus, that is, the modulus of elasticity, of the rod material.
L is the length of the rod.
m is the mass of the rod.
ρ is the density of the rod material.
The rod does not buckle, in the case of a bar, or go slack, in the case of a cable, in tension.
The rod has a constant cross-section along its length.
The distributed parameter model is approximated as a finite number of flexible elements, N.
Translational conserving port associated with rod base or input.
Translational conserving port associated with rod follower or output.
The table shows how the visibility of some Rod parameters depends on choices that you make for other Rod parameters. To learn how to read the table, see Parameter Dependencies.
Rod Parameter Dependencies Table
By stiffness and inertia By material and geometry
Stiffness Geometry
Mass Solid Annular
By stiffness and inertia (default) | By material and geometry
Parameterization method.
Each Parameterization option exposes other parameters. For more information, see Rod Parameter Dependencies Table.
Stiffness — Material stiffness
5e8 N/m (default) | non-negative scalar
Material stiffness.
Setting the Parameterization parameter to By stiffness and inertia exposes this parameter. For more information, see Rod Parameter Dependencies Table.
Mass — Rod mass
20 kg (default) | positive scalar
Rod mass.
Geometry — Rod cross-section
Rod cross-sectional geometry.
Setting the Parameterization parameter to By material and geometry exposes this parameter. Each Geometry option exposes other parameters. For more information, see Rod Parameter Dependencies Table.
Length — Rod length
Rod length.
Setting the Parameterization parameter to By material and geometry exposes this parameter. For more information, see Rod Parameter Dependencies Table.
Outer diameter — Rod outer diameter
Rod outer diameter.
Inner diameter — Rod inner diameter
Rod inner diameter. If the rod is solid, specify 0.
Setting the Parameterization parameter to By material and geometry and the Geometry parameter to Annular exposes this parameter. For more information, see Rod Parameter Dependencies Table.
Density — Material density
Young's modulus for the rod material.
Damping ratio — Material damping ratio
0.01 (default) | non-negative scalar
Number of flexible elements — N
Number of flexible elements, N, for the approximation.
A larger number of flexible elements, N, increases the accuracy of the model but reduces simulation performance, that is, simulation speed. The single-element model (N=1) exhibits an eigenfrequency that is close to the first eigenfrequency of the continuous, distributed parameter model.
If accuracy is more important than performance, select 2, 4, 8, or more flexible elements. For example, the four lowest eigenfrequencies are represented with an accuracy of 0.1, 1.9, 1.6, and 5.3 percent, respectively, by a 16-element model.
[0, 0] N/(m/s) (default) | non-negative vector
Viscous friction coefficients at base port, B, and the follower port, F.
Initial deflection — Initial deflection
0 m (default) | non-negative scalar
Deflection of the rod at the start of simulation.
A positive initial deflection results in a positive translation of the base, B, end of the rod relative to the follower, F, end of the rod.
0 m/s (default) | non-negative scalar
Longitudinal velocity of the base, B, end of the rod relative to the follower, F, end of the rod at the start of simulation.
Flexible Shaft | Mass | Translational Damper | Translational Spring
|
Minimum Spanning Trees · USACO Guide
HomeGoldMinimum Spanning Trees
Kruskal'sImplementationSolution - Road ReparationPrim'sComplexitySolution - Road ReparationProblems
Authors: Benjamin Qi, Andrew Wang, Kuan-Hung Chen
A subset of the edges of a connected, undirected, edge-weighted graph that connects all the vertices to each other of minimum total weight, where no cycles are allowed.
Gold - Shortest Paths with Non-Negative Edge Weights
To review a couple of terms:
An undirected edge is an edge that goes both ways
A connected graph is a graph of vertices such that each vertex can reach every other vertex using undirected edges.
A spanning tree is a set of edges that forms a tree and contains every vertex in the original graph
A minimum spanning tree is a spanning tree such that the sum of edge weights are minimized
Road Reparation
15.1 - Kruskal's
Kruskal's with DSU
4.3.2 - Kruskal's
Kruskal's Algorithm finds the MST by greedily adding edges. For all edges not yet in the MST, we can repeatedly add the edge of minimum weight to the MST except when adding edges that would forms a cycle. This can be done by sorting the edges in order of non-decreasing weight. Furthermore, we can easily determine whether adding an edge will create a cycle in constant time using Union Find. Note that since the most expensive operation is sorting the edges, the computational complexity of Kruskal's Algorithm is
\mathcal{O}(E \log E)
Disjoint Set Union + Kruskal
#include "DSU.h"
template<class T> T kruskal(int N, vector<pair<T,pi>> ed) {
sort(all(ed));
T ans = 0; DSU D; D.init(N); // edges that unite are in MST
trav(a,ed) if (D.unite(a.s.f,a.s.s)) ans += a.f;
public static HashMap<Integer, ArrayList<Integer>> MST;
public static PriorityQueue<Edge> pq; //contains all edges
//Assumes that DSU code is included
public static void kruskal() {
if (find(e.start) != find(e.end)) {
MST.get(e.start).add(e.end);
MST.get(e.end).add(e.start);
Solution - Road Reparation
Notice that the road that allows for a "decent route between any two cities," with cost "as small as possible" is the definition of a minimum spanning tree. Thus, we can use our favorite minimum spanning tree algorithm to determine the cost of such a tree by calculating
\sum c
for all edges included in the tree.
However, we must also account for the impossible case, which occurs when any nodes cannot be connected to the tree. Recall that the minimum spanning tree must contain a total of
n-1
edges, so we can use a variable
cnt
that is incremented every time we add an edge to the minimum spanning tree. After running Kruskal's, if
cnt \ne n-1
, then we know that we failed to built the tree properly. Furthermore, since our minimum spanning tree algorithm gurantees no edges are counted twice, we cannot "accidentally" count
n-1
#define trav(a,x) for (auto& a: x)
static int comp;
static int disjoint[];
static int size[];
15.3 - Prim's
4.3.3 - Prim's
Similar to Dijkstra's, Prim's algorithm greedily adds vertices. On each iteration, we add the vertex that is closest to the current MST (instead of closest to the source in Dijkstra's) until all vertices have been added.
The process of finding the closest vertex to the MST can be done efficiently using a priority queue. After removing a vertex, we add all of its neighbors that are not yet in the MST to the priority queue and repeat. To begin the algorithm, we simply add any vertex to the priority queue.
Our implementation has complexity
\mathcal{O}(E \log E)
since in the worst case every edge will be checked and its corresponding vertex will be added to the priority queue.
Alternatively, we may linearly search for the closest vertex instead of using a priority queue. Each linear pass runs in time
\mathcal{O}(V)
, and this must be repeated
V
times. Thus, this version of Prim's algorithm has complexity
\mathcal{O}(V^2)
. As with Dijkstra, this complexity is preferable for dense graphs (in which
E \approx V^2
typedef pair<ll, int> pl;
class prim {
static Map<Integer, ArrayList<Edge>> tree;
static int N, ct;
static long[] dist;
static long max = 10000000000000000 L;
def prim(n, G):
used = [False for _ in range(n)]
pq = [(0, 0)]
weight, node = heapq.heappop(pq)
if used[node]:
Easy Show Tags MST
Normal Show Tags MST, Math
Normal Show Tags MST
Normal Show Tags Binary Search, MST
Moo Network
Hard Show Tags MST
2017 - Sirni
Hard Show Tags MST, NT
2013 - Toll
Insane Show Tags Bitmasks, MST
2011 - Timeismoney
Insane Show Tags Convex, MST
The original problem statement for "Inheritance" is in Japanese. You can find a user-translated version of the problem here.
|
Bent Functions - Boolean Functions
Revision as of 20:03, 10 February 2019 by Nikolay (talk | contribs) (Created page with "= Background and Definition = The covering radius bound states that the nonlinearity <math>nl(F)</math> of any <math>(n,m)</math>-function <math>F</math> satisfies <div>...")
The covering radius bound states that the nonlinearity
{\displaystyle nl(F)}
{\displaystyle (n,m)}
{\displaystyle F}
{\displaystyle nl(F)\leq 2^{n-1}-2^{n/2-1}.}
A function is called bent if it achieves this bound with equality.
Since nonlinearity is a CCZ-invariant, CCZ-equivalence (and therefore also EA-equivalence and affine equivalence) preserves the property of a function of being bent.
The algebraic degree of any bent
{\displaystyle (n,m)}
-function is at most
{\displaystyle n/2}
{\displaystyle (n,m)}
-function is bent if and only if all of its derivatives
{\displaystyle D_{a}F}
{\displaystyle a\neq 0}
are balanced. In this sense, bent functions are also referred to as perfect nonlinear (PN) functions.
Bent (PN)
{\displaystyle (n,m)}
-functions exist only for
{\displaystyle n}
{\displaystyle m\leq n/2}
[1]. Conversely, for any pair of integers
{\displaystyle (n,m)}
satisfying this hypothesis, there exists a bent
{\displaystyle (n,m)}
↑ Nyberg K. Perfect nonlinear S-boxes. InWorkshop on the Theory and Application of of Cryptographic Techniques 1991 Apr 8 (pp. 378-386). Springer, Berlin, Heidelberg.
Retrieved from "https://boolean.h.uib.no/mediawiki/index.php?title=Bent_Functions&oldid=176"
|
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