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https://cs.stackexchange.com/questions/96178/how-to-generate-branch-tables-from-ssa-form
|
# How to generate branch tables from SSA form?
Branch tables are usually described as an efficient way a compiler can implement a switch statement, and that actually seems how GCC and Clang do it (and LLVM even has an opcode just for that).
If a sequence of if statements is used (instead of a switch), GCC doesn't seem to generate a table, but Clang still does.
Given a compiler that turns the code into basic blocks inside a control flow graph in static single assignment form, how would it decide where to generate tables? Is there any research on that? I'm particularly interested in finding an algorithm for lowering consecutive branchings (found on the CFG/SSA form) into a branch table (as Clang seems to do).
(I actually tried looking for any papers about that but didn't find anything.)
Edit: I do understand that the obvious way to do this is to expand the definition of the SSA to include multiple branchings as a primitive (i.e., blocks may branch to an arbitrary number of blocks); I'm wondering if (and how) the tables could be done in a simpler definition, where each basic block may move to at most two other blocks (i.e., only unconditional moves and an if branching).
The book "A Retargetable C Compiler: Design and Implementation" by Christopher Fraser and David Hanson gives some detail on how they lower switch statements to a combination of conditional branches and jump tables. They cite prior work that their technique builds on, so it would be a good start for a literature survey.
(Update: you've edited the question in a way that makes it clear this isn't really about generating jump tables, but rather about re-discovering switch-style control flow from cascading if statements in an IR. This is again probably not specific to SSA. You might look at the "Relooper" algorithm as implemented for Emscripten which deals with more general control-flow restructuring. I'm not aware of a reference specific to recovering N-way branches, but it seems like simple ad hoc approaches should work well.)
|
2019-11-17 12:55:02
|
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|
https://en.wikibooks.org/wiki/Mathematics_Worksheet/Algebra/Fractional_algebra
|
# Mathematics Worksheet/Algebra/Fractional algebra
Do questions carefully! (11 questions)
Rules:
1. In addition and subtraction, reduce fractions to the Lowest Common Denominator.
2. In multiplication and division, reduce fractions first.
3. Use ^ to write exponentiation.
1
${\displaystyle {\frac {9}{8}}\times {\frac {4}{5}}=}$
2
${\displaystyle {\frac {4x}{5}}+{\frac {(3x-5)}{10}}=}$
3
${\displaystyle {\frac {5(6x+4)}{4y}}+{\frac {3(4x+6y)}{6xy}}=}$
4
${\displaystyle {\frac {7xy}{8x^{2}}}\times {\frac {4x^{3}}{5y}}=}$
5
${\displaystyle {\frac {x-1}{x-2}}-{\frac {x-3}{x^{2}-4}}=}$
6
${\displaystyle {\frac {6x^{3}y^{5}}{3x^{5}y^{2}}}=}$
7
${\displaystyle {\frac {x-2}{x^{2}-4}}=}$
8
${\displaystyle {\frac {x^{2}+8x+16}{x+4}}=}$
9
${\displaystyle {\frac {5x}{(2x+2)}}-{\frac {3x}{(4x+4)}}=}$
10
${\displaystyle {\frac {y^{2}-16}{y^{2}+8x+16}}=}$
11
${\displaystyle {\frac {3x}{2x+4}}\div {\frac {6}{4x+8}}=}$
|
2019-06-27 02:12:05
|
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|
https://www.esaral.com/q/solve-this-following-85817
|
# Solve this following
Question:
Let two points be $\mathrm{A}(1,-1)$ and $\mathrm{B}(0,2)$. If a point $\mathrm{P}\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)$ be such that the area of $\triangle \mathrm{PAB}=5 \mathrm{sq}$. units and it lies on the line, $3 x+y-4 \lambda=0$, then a value of $\lambda$ is
1. 1
2. 4
3. 3
4. $-3$
Correct Option: , 3
Solution:
#### Leave a comment
None
Free Study Material
|
2023-02-06 22:49:38
|
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|
https://physicscatalyst.com/maths/identity-function.php
|
# Identity Function
## Standard Real Function and Their graphs
We will be exploring some Standard Real Function and Their graphs in next few pages like
a. Identity function
b. Constant function
c. Linear function
e. cubic function
f. polynomial function
g. rational function
h. Modulus function
We will discussing their domain ,range, graph and properties in details.
Let's get started with first function
## Identity Function
We have already seen in previous grades that Identity of addition ,subtraction ,multiplication are number which when operated on the elements does not change the value of the element.
Example
$1 + 0 =1$
$2 + 0 =2$
Multiplicatve identity
$5 \times 1 =5$
Now Identity function are similar type of thing.
Identity Function is defined as the real valued function $f : R \rightarrow R$ , y = f(x) = x for each $x \in R$
So ,this function basically associate each real number to itself
### Domain and Range of the Identity Function
For $f : R \rightarrow R$
Domain = R
Range = R
Co-domain and Range are equal set
### Graph of the Identity Function
We can draw the graph on the Cartesian plane with value of x on the x-axis and value of y=f(x) on the y-axis. We can plot the point and join the point to obtain the graph. Here in case of the identity function,the graph will be a straight line passing through the origin
The straight line makes an angle $45^o$ with the x-axis
## Other Examples of Identity Functions
So far, we observe the identity function for the whole set of Real number. But Identity function can also be defined for the subset of the real numbers also
We denote these by capital letter I
Example -1
Let A = {1,2,3,4,5,6}
Then Identity function on set A will be defined as
$I_A : A \rightarrow A, I_A=x , x \in A$
for $x=1 ,I_A(1) =x=1$
for $x=2 ,I_A(1) =x=2$
for $x=3 ,I_A(1) =x=3$
for $x=4 ,I_A(1) =x=4$
for $x=5 ,I_A(1) =x=5$
Domain,Range and co-domain will be Set A
Example -2
Let A = R -{1/2}
Then Identity function on set A will be defined as
$I : A \rightarrow A, I_A=x , x \in A$
Domain,Range and codomain will be Set A
Example -3
A identity function can be defined as
$I : Z \rightarrow R, I_A=x , x \in Z$
In this case Domain,Range will be Z and co-domain will be R
### Quiz Time
Question 1.
Find the value of the function $f(x) = \frac {x-5}{x-3}$ at x=0
A.3/5
B. 0
C.5/3
D. 1
Question 2.
If f (x) = px + q, where p and q are integers, f (-1) = - 5 and f (3) = 3, then p and q are equal to ?
A. p=2,q=-3
B. p-=2,q=3
C. p=1,q=3
D. p=-2,q=-3
Question 3.
if $f(x) =\frac {1}{2x+1}$ ,then
A. $f[f(x] =\frac {2x-1}{2x+3}$
B. $f[f(x] =\frac {2x-1}{2x-3}$
C. $f[f(x] =\frac {2x+1}{2x-3}$
D. $f[f(x] =\frac {2x+1}{2x+3}$
Question 4.
Let $f(x) =2^x$,find the value of $\frac {f(2) -f(1)}{2 -1}$
A. 4
B. 2
C. 0
D. 1
Question 5.
if $f(x) =\frac {x+1}{x-1}$ ,then
A. $f[f(x] =\frac {x-1}{x+1}$
B. $f[f(x] =x$
C.$f[f(x] =-x$
D.$f[f(x] =\frac {1-x}{1+x}$
Question 6.
Find the Range and domain of the function $f(x) =x$
A. Domain = R, Range =R
B. Domain = R - {0}, Range = R
C. Domain = R - {0}, Range = {-1/2}
D. Domain = R - {1}, Range = R
|
2022-12-03 02:14:18
|
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|
https://chocolatey.org/packages/vim/8.2.0916
|
#### Vim
This is not the latest version of Vim available.
8.2.0916 | Updated: 07 Jun 2020
720,122
1,139
Maintainer(s):
Software Author(s):
• Bram Moolenaar
• Vim Community
### All Checks are Passing
2 Passing Test
Validation Testing Passed
Verification Testing Passed
Details
To install Vim, run the following command from the command line or from PowerShell:
>
To upgrade Vim, run the following command from the command line or from PowerShell:
>
To uninstall Vim, run the following command from the command line or from PowerShell:
>
NOTE: This applies to both open source and commercial editions of Chocolatey.
#### 1. Ensure you are set for organizational deployment
Please see the organizational deployment guide
• Open Source or Commercial:
• Proxy Repository - Create a proxy nuget repository on Nexus, Artifactory Pro, or a proxy Chocolatey repository on ProGet. Point your upstream to https://chocolatey.org/api/v2. Packages cache on first access automatically. Make sure your choco clients are using your proxy repository as a source and NOT the default community repository. See source command for more information.
• You can also just download the package and push it to a repository
#### 3. Enter your internal repository url
(this should look similar to https://chocolatey.org/api/v2)
#### 4. Choose your deployment method:
choco upgrade vim -y --source="'STEP 3 URL'" [other options]
Add this to a PowerShell script or use a Batch script with tools and in places where you are calling directly to Chocolatey. If you are integrating, keep in mind enhanced exit codes.
If you do use a PowerShell script, use the following to ensure bad exit codes are shown as failures:
choco upgrade vim -y --source="'STEP 3 URL'"
$exitCode =$LASTEXITCODE
Write-Verbose "Exit code was $exitCode"$validExitCodes = @(0, 1605, 1614, 1641, 3010)
if ($validExitCodes -contains$exitCode) {
Exit 0
}
Exit $exitCode - name: Ensure vim installed win_chocolatey: name: vim state: present version: 8.2.0916 source: STEP 3 URL Coming early 2020! Central Managment Reporting available now! More information... chocolatey_package 'vim' do action :install version '8.2.0916' source 'STEP 3 URL' end Chocolatey::Ensure-Package ( Name: vim, Version: 8.2.0916, Source: STEP 3 URL ); Requires Otter Chocolatey Extension. See docs at https://inedo.com/den/otter/chocolatey. cChocoPackageInstaller vim { Name = 'vim' Ensure = 'Present' Version = '8.2.0916' Source = 'STEP 3 URL' } Requires cChoco DSC Resource. See docs at https://github.com/chocolatey/cChoco. package { 'vim': provider => 'chocolatey', ensure => '8.2.0916', source => 'STEP 3 URL', } Requires Puppet Chocolatey Provider module. See docs at https://forge.puppet.com/puppetlabs/chocolatey. salt '*' chocolatey.install vim version="8.2.0916" source="STEP 3 URL" #### 5. If applicable - Chocolatey configuration/installation See infrastructure management matrix for Chocolatey configuration elements and examples. This package was approved as a trusted package on 07 Jun 2020. Description Vim is a highly configurable text editor built to enable efficient text editing. It is an improved version of the vi editor distributed with most UNIX systems. Vim is often called a programmer's editor, and so useful for programming that many consider it an entire IDE. It's not just for programmers, though. Vim is perfect for all kinds of text editing, from composing email to editing configuration files. ## Features • Vim: Vim terminal(CLI) application can be used from Powershell and Command Prompt. • GVim: The GUI version of Vim provides full featured Windows GUI application experience. • Terminal Integration: Batch files are created to provide vim, gvim, evim, view, gview, vimdiff, gvimdiff and vimtutor command on terminal use. • Shell Integration: Vim is added in Open with ... context menu. And by default Edit with Vim context menu is created to open files whose extensions are associated with other applications. ## Package parameters • /InstallDir - Override the installation directory. By default, the software is installed in $ChocolateyToolsLocation, it's default value is C:\tools. You can include spaces. See the example below.
• /RestartExplorer - Restart Explorer to unlock GVimExt.dll used for Edit with Vim context menu feature.
• /NoDefaultVimrc - Don't create default _vimrc file.
• /NoContextmenu - Don't create Edit with Vim context menu.
• /NoDesktopShortcuts - Don't create shortcuts on the desktop.
Example: choco install vim --params "'/NoDesktopShortcuts /InstallDir:C:\path\to\your dir'"
## Notes
• This package uses the ZIP build to install to provide installation parameters.
• All compilation of the software is automated and performed on Appveyor. The building status is open.
• This package provides an official build. Similar package vim-tux is from a well-known unofficial vim building project. Unlike vim-tux, this package can take some installation parameters.
From: https://vimhelp.org/uganda.txt.html
I) There are no restrictions on distributing unmodified copies of Vim except
that they must include this license text. You can also distribute
unmodified parts of Vim, likewise unrestricted except that they must
include this license text. You are also allowed to include executables
that you made from the unmodified Vim sources, plus your own usage
examples and Vim scripts.
II) It is allowed to distribute a modified (or extended) version of Vim,
including executables and/or source code, when the following four
conditions are met:
1) This license text must be included unmodified.
2) The modified Vim must be distributed in one of the following five ways:
a) If you make changes to Vim yourself, you must clearly describe in
the distribution how to contact you. When the maintainer asks you
(in any way) for a copy of the modified Vim you distributed, you
must make your changes, including source code, available to the
maintainer without fee. The maintainer reserves the right to
include your changes in the official version of Vim. What the
maintainer will do with your changes and under what license they
will be distributed is negotiable. If there has been no negotiation
then this license, or a later version, also applies to your changes.
The current maintainer is Bram Moolenaar <[email protected]>. If this
changes it will be announced in appropriate places (most likely
vim.sf.net, www.vim.org and/or comp.editors). When it is completely
impossible to contact the maintainer, the obligation to send him
your changes ceases. Once the maintainer has confirmed that he has
received your changes they will not have to be sent again.
b) If you have received a modified Vim that was distributed as
mentioned under a) you are allowed to further distribute it
unmodified, as mentioned at I). If you make additional changes the
text under a) applies to those changes.
c) Provide all the changes, including source code, with every copy of
the modified Vim you distribute. This may be done in the form of a
context diff. You can choose what license to use for new code you
add. The changes and their license must not restrict others from
making their own changes to the official version of Vim.
d) When you have a modified Vim which includes changes as mentioned
under c), you can distribute it without the source code for the
changes if the following three conditions are met:
- The license that applies to the changes permits you to distribute
the changes to the Vim maintainer without fee or restriction, and
permits the Vim maintainer to include the changes in the official
version of Vim without fee or restriction.
- You keep the changes for at least three years after last
distributing the corresponding modified Vim. When the maintainer
or someone who you distributed the modified Vim to asks you (in
any way) for the changes within this period, you must make them
available to him.
- You clearly describe in the distribution how to contact you. This
contact information must remain valid for at least three years
after last distributing the corresponding modified Vim, or as long
as possible.
e) When the GNU General Public License (GPL) applies to the changes,
you can distribute the modified Vim under the GNU GPL version 2 or
any later version.
3) A message must be added, at least in the output of the ":version"
command and in the intro screen, such that the user of the modified Vim
is able to see that it was modified. When distributing as mentioned
under 2)e) adding the message is only required for as far as this does
not conflict with the license used for the changes.
4) The contact information as required under 2)a) and 2)d) must not be
removed or changed, except that the person himself can make
corrections.
III) If you distribute a modified version of Vim, you are encouraged to use
the Vim license for your changes and make them available to the
maintainer, including the source code. The preferred way to do this is
by e-mail or by uploading the files to a server and e-mailing the URL.
If the number of changes is small (e.g., a modified Makefile) e-mailing a
context diff will do. The e-mail address to be used is
<[email protected]>
IV) It is not allowed to remove this license from the distribution of the Vim
sources, parts of it or from a modified version. You may use this
license for previous Vim releases instead of the license that they came
with, at your option.
legal\VERIFICATION.txt
VERIFICATION
Verification is intended to assist the Chocolatey moderators and community
in verifying that this package's contents are trustworthy.
The embedded software have been downloaded from GitHub and can be verified like this:
2. You can use one of the following methods to obtain the SHA256 checksum:
- Use powershell function 'Get-FileHash'
- Use Chocolatey utility 'checksum.exe'
checksum32: 0160333ED1EB0E7C7877FC456595250B74F415DDD615E28E1F31BD6844DEBF09
checksum64: EAB527F20286C8015C348E6EFA80263B1A7005B57C7ED681CA76F0B060DDCD0C
tools\chocolateybeforemodify.ps1
$toolsDir = "$(Split-Path -parent $MyInvocation.MyCommand.Definition)"$installDir = Get-Content "$toolsDir\installDir"$shortversion = '82'
try {
# Is dlls locked?
Remove-Item "$installDir\vim\vim$shortversion\GvimExt32\gvimext.dll", "$installDir\vim\vim$shortversion\GvimExt64\gvimext.dll" -ErrorAction Stop
} catch {
# Restart explorer to unlock dlls
Write-Debug 'Restarting explorer.'
Get-Process explorer | Stop-Process -Force
}
tools\chocolateyinstall.ps1
$ErrorActionPreference = 'Stop';$toolsDir = "$(Split-Path -parent$MyInvocation.MyCommand.Definition)"
$shortversion = '82'$pp = Get-PackageParameters
. $toolsDir\helpers.ps1$installDir = Get-InstallDir
$packageArgs = @{ packageName =$env:ChocolateyPackageName
unzipLocation = $installDir file = "$toolsDir\gvim_8.2.0916_x86.zip"
file64 = "$toolsDir\gvim_8.2.0916_x64.zip" }$installArgs = @{
statement = Get-Statement
exeToRun = "$installDir\vim\vim$shortversion\install.exe"
}
'$installDir', ($installDir | Out-String), '$packageArgs', ($packageArgs | Out-String), '$installArgs', ($installArgs | Out-String) | ForEach-Object { Write-Debug $_ } Install-ChocolateyZipPackage @packageArgs Start-ChocolateyProcessAsAdmin @installArgs Copy-Item -Path "$installDir\vim\vim$shortversion\vimtutor.bat" -Destination$env:windir
Set-Content -Path "$toolsDir\installDir" -Value$installDir
tools\chocolateyuninstall.ps1
$toolsDir = "$(Split-Path -parent $MyInvocation.MyCommand.Definition)"$installDir = Get-Content "$toolsDir\installDir"$shortversion = '82'
$statement = '-nsis'$exeToRun = "$installDir\vim\vim$shortversion\uninstall.exe"
# From vim-tux.install. Make input.
Set-Content -Path "$env:TEMP\vimuninstallinput" -Value 'y' Start-Process -FilePath$exeToRun -ArgumentList $statement -RedirectStandardInput "$env:TEMP\vimuninstallinput" -Wait -WindowStyle Hidden
Remove-Item "$env:TEMP\vimuninstallinput" Remove-Item "$env:windir\vimtutor.bat"
Remove-Item "$installDir\vim" -Recurse -Force tools\gvim_8.2.0916_x64.zip md5: 60507858F2E3B550C01F08CA1BE898C2 | sha1: B04CCCF5D404836133D629A896592F16D1BA42B6 | sha256: EAB527F20286C8015C348E6EFA80263B1A7005B57C7ED681CA76F0B060DDCD0C | sha512: B7329447D7C2318E517E4C22D790A812F2C88350477229331DE5B51F55BC1E778EEDBA763400DF3CC1770CDCF30E555BF779243FCF30BACF03F4F527ACBAC33C tools\gvim_8.2.0916_x86.zip md5: BF6BA73088D0454BB83720DD5C6EC6BF | sha1: 354D7A6A01BF8E69084ADB5A25472592BAFA67CB | sha256: 0160333ED1EB0E7C7877FC456595250B74F415DDD615E28E1F31BD6844DEBF09 | sha512: 53A201DF8D47D85AEA5B4154E378219A747217F8D4FAF272C2478E837D61A5B6C7EE2824D7E14D77FCB4100C704B7FB8198872F75CA809BEFFC9A864A19B4FC4 tools\helpers.ps1 function Get-InstallDir() { if ($pp['InstallDir']) {
Write-Debug '/InstallDir found.'
return $pp['InstallDir'] } return Get-ToolsLocation } function Get-Statement() {$options = '-create-batfiles vim gvim evim view gview vimdiff gvimdiff -install-openwith -add-start-menu'
$createvimrc = '-create-vimrc -vimrc-remap no -vimrc-behave default -vimrc-compat all'$installpopup = '-install-popup'
$installicons = '-install-icons' if ($pp['RestartExplorer'] -eq 'true') {
Write-Debug '/RestartExplorer found.'
Get-Process explorer | Stop-Process -Force
}
if ($pp['NoDefaultVimrc'] -eq 'true') { Write-Debug '/NoDefaultVimrc found.'$createvimrc = ''
}
if ($pp['NoContextmenu'] -eq 'true') { Write-Debug '/NoContextmenu found.'$installpopup = ''
}
if ($pp['NoDesktopShortcuts'] -eq 'true') { Write-Debug '/NoDesktopShortcuts found.'$installicons = ''
}
return $options,$createvimrc, $installpopup,$installicons -join ' '
}
# Replace old ver dir with symlink
# Use mklink because New-Item -ItemType SymbolicLink doesn't work in test-env
# Use rmdir because Powershell cannot unlink directory symlink
{
Get-ChildItem -Path "$installDir\vim" -Exclude "vim$shortversion" -Attributes Directory+!ReparsePoint | ForEach-Object { Remove-Item $_ -Recurse ; New-Item -Path$_ -ItemType Directory }
Get-ChildItem -Path "$installDir\vim" -Exclude "vim$shortversion" -Attributes Directory | ForEach-Object { $_.Name } | ForEach-Object { cmd /c rmdir "$installDir\vim\$_" ; cmd /c mklink /d "$installDir\vim\$_" "$installDir\vim\vim\$shortversion" }
}
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Vim 8.2.1119 1808 Friday, July 3, 2020 Approved
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This package has no dependencies.
Discussion for the Vim Package
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• This discussion is only about Vim and the Vim package. If you have feedback for Chocolatey, please contact the Google Group.
• This discussion will carry over multiple versions. If you have a comment about a particular version, please note that in your comments.
|
2020-07-05 01:59:32
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32327184081077576, "perplexity": 12040.791434060893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655886802.13/warc/CC-MAIN-20200704232817-20200705022817-00259.warc.gz"}
|
http://zbmath.org/?q=an%3A06082331
|
# zbMATH — the first resource for mathematics
##### Examples
Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.
##### Operators
a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
The polynomial inverse image method. (English) Zbl 1268.41008
Neamtu, Marian (ed.) et al., Approximation theory XIII: San Antonio 2010. Selected papers based on the presentations at the conference, San Antonio, TX, USA, March 7–10, 2010. New York, NY: Springer (ISBN 978-1-4614-0771-3/hbk; 978-1-4614-0772-0/ebook). Springer Proceedings in Mathematics 13, 345-365 (2012).
In this article, which is of expository character, the author describes a method for transferring results from two model cases of compact plane sets ${E}_{0}$, namely ${E}_{0}=\left[-1,1\right]$ and ${E}_{0}={C}_{1}$ the unit circle, to more general compact plane sets. The basic point is that many interesting properties of compact plane sets are preserved when taking polynomial inverse images.
For a polynomial $T$ let ${T}^{-1}\left({E}_{0}\right)$ denote the inverse image of ${E}_{0}$. This leads to the following method.
(a) Start from a result for the model case ${E}_{0}$.
(b) Apply an inverse polynomial mapping to go to a special result on the inverse image $E={T}^{-1}\left({E}_{0}\right)$ of the model set ${E}_{0}$.
(c) Approximate more general sets by inverse images $E$ as in (b).
Among others the polynomial inverse image method has been successful in the following situations:
– Bernstein-type inequalities, the model case being the classical Bernstein inequality on $\left[-1,1\right]$;
– Markov-type inequalities, the model case being the classical Markov inequality on $\left[-1,1\right]$;
– asymptotics of Christoffel functions on compact subsets of the real line, with model case $\left[-1,1\right]$;
– asymptotics of Christoffel functions on curves, with model case ${C}_{1}$;
– universality on general sets, the model case being on $\left[-1,1\right]$;
– fine zero spacing of orthogonal polynomials, with model case $\left[-1,1\right]$;
– Bernstein-type inequalities for a system of smooth Jordan curves, the model case being Bernstein’s inequality on ${C}_{1}$.
##### MSC:
41A17 Inequalities in approximation (Bernstein, Jackson, Nikol’skiĭ-type inequalities) 26D05 Inequalities for trigonometric functions and polynomials 30C10 Polynomials (one complex variable) 30C85 Capacity and harmonic measure in the complex plane
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2014-03-08 22:01:45
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 17, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7550380825996399, "perplexity": 4054.1600330987885}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999664754/warc/CC-MAIN-20140305060744-00004-ip-10-183-142-35.ec2.internal.warc.gz"}
|
https://tex.stackexchange.com/questions/522627/memoir-chapter-style-with-with-background-color-spreading-across-text-number
|
# Memoir chapter style with with background color spreading across text / number
Greetings to the community!
My goal is to create the pdflatex document using the memoir class, which would have \chapters, \sections, and \subsections have the following formatting (font is not important, placing heading on the colored background is important):
i.e. the \chapter*{Features} has a colored background spread across the linewidth, and if there is a number, it is on the colored background as well.
After a bit of googling, I got smth like:
\renewcommand\chaptitlefont{%
\huge\bfseries%
\leavevmode\color{white}%
}
\renewcommand\printchapternum{
\huge\bfseries%
\leavevmode\color{white}%
{\thechapter}
}
with combination of defining chapter like this:
\chapter*{\colorbox{my-blue}{\strut\rlap{\textbf{Executive Summary}}\hspace{\linewidth}\hspace{-2\fboxsep}}}
Which results in the followings:
1. The chapter title with a hack:
1. ToC (white text over white background, I selected some of ):
My hack has the following problems:
• Titles of the chapters that I do not create explicitly (e.g. Contents) have white font over white background;
• Titles of the chapters that I do name explicitly in ToC got also blue background, which is not desired.
• Chapter titles do not include the Chapter number. Instead of Introduction and Syntax (i.e. all the chapters created with \chapter) as a chapter title, my goal is to have 1 Introduction, 2 Syntax, 3 Semantics (when using \chapter opposite to \chapter*)
My questions are as follows:
1. How would I create the chapter titles, that have background color spanned over chapter title text, including chapter number.
2. How in this case (1) keep the formatting of chapters that I do not name explicitly (e.g. Contents, List of Figures, List of Tables, etc.) consistent?
3. How to avoid that the background color is also shown in the ToC?
Thanks in advance for any help / pointers!
The minimal example is below:
\documentclass[a4paper,12pt,oneside]{memoir}
\usepackage[english]{babel}
\usepackage{xcolor}
\definecolor{my-blue}{RGB}{0,150,226}
\renewcommand\chaptitlefont{%
\huge\bfseries%
\leavevmode\color{white}%
}
\renewcommand\printchapternum{
\huge\bfseries%
\leavevmode\color{white}%
{\thechapter}
}
\begin{document}
\frontmatter
\pagenumbering{arabic}
\tableofcontents*
\clearpage
\chapter*{\colorbox{my-blue}{\strut\rlap{\textbf{Summary}}\hspace{\linewidth}\hspace{-2\fboxsep}}}
\chapter{\colorbox{my-blue}{\strut\rlap{\textbf{Introduction}}\hspace{\linewidth}\hspace{-2\fboxsep}}}
\chapter{Syntax}
\chapter{Semantics}
\end{document}
Here's a tikz solution that patches a bunch of the memoir class internal macros.
\documentclass[openany]{memoir}
\usepackage{tikz}
\usepackage{xpatch}
\definecolor{my-blue}{RGB}{0,150,226}
\chapterstyle{reparticle}
\renewcommand*{\chaptitlefont}{\normalfont\Large\bfseries\color{white}}
\makeatletter
\newcommand{\@selyunin@box}[1]{%
\tikz\node[fill=my-blue, inner sep=1mm, text width=\linewidth-2mm]{#1\strut\par};}
% numbered chapters
{\ifm@m@And
\afterchapternum
\else
\printchapternonum
\fi
\interlinepenalty\@M
\printchaptertitle{#1}}
{\@selyunin@box{%
\ifm@m@And
\afterchapternum
\else
\printchapternonum
\fi
\interlinepenalty\@M
\printchaptertitle{#1}}}
{}
{}
% unnumbered chapters
{\printchaptertitle{#1}}
{\@selyunin@box{\printchaptertitle{#1}}}
{}
{}
% TOC, LOF, and LOT headings
\xpatchcmd{\@tocmaketitle}
{\@nameuse{printtoctitle}{\contentsname}}
{\@selyunin@box{\@nameuse{printtoctitle}{\contentsname}}}
{}
{}
\xpatchcmd{\@lofmaketitle}
{\@nameuse{printloftitle}{\contentsname}}
{\@selyunin@box{\@nameuse{printloftitle}{\contentsname}}}
{}
{}
\xpatchcmd{\@lotmaketitle}
{\@nameuse{printlottitle}{\contentsname}}
{\@selyunin@box{\@nameuse{printlottitle}{\contentsname}}}
{}
{}
% numbered sections
\xpatchcmd{\M@sect}
{\@hangfrom{\hskip #3\relax\@svsec}%
\interlinepenalty \@M #9\@@par}
{\@selyunin@box{%
\@hangfrom{\hskip #3\relax\@svsec}%
\interlinepenalty \@M #9}}
{}
{}
% unnumbered sections
\xpatchcmd{\@mem@old@ssect}
{\@hangfrom{\hskip #1}%
\interlinepenalty \@M #5\@@par}
{\@selyunin@box{%
\@hangfrom{\hskip #1}%
\interlinepenalty \@M #5}}%
{}
{}
\makeatother
\begin{document}
\tableofcontents
\section{A very long section heading that needs to go on to the next line}
\subsection{A very long subsection heading that needs to go on to the next line}
\section*{A very long starred section heading that needs to go on to the next line}
\subsection*{A very long starred subsection heading that needs to go on to the next line}
\chapter{A very long chapter heading that needs to go on to the next line}
\chapter*{A very long starred chapter heading that needs to go on to the next line}
\appendix
\section{A very long appendix section heading that needs to go on to the next line}
\subsection{A very long appendix subsection heading that needs to go on to the next line}
\section*{A very long starred appendix section heading that needs to go on to the next line}
\subsection*{A very long starred appendix subsection heading that needs to go on to the next line}
\chapter{A very long appendix chapter heading that needs to go on to the next line}
\chapter*{A very long starred appendix chapter heading that needs to go on to the next line}
\end{document}
## Page 2 output:
• Thank you very much, that is exactly I was looking for! – selyunin Jan 6 at 10:51
• I have one more question: whenever I try to include the \usepackage{hyperref} hyperref package, the sections and subsections loose the patched formatting (i.e. become white over white). Is there a way to use the hyperref package as well? – selyunin Jan 7 at 16:26
• OK, I have figured it out, the \usepackage{hyperref} should be after the patched commands, and not before – selyunin Jan 8 at 4:26
• @selyunin, yes hyperref will patch things too and should usually be loaded as late as possible. – David Purton Jan 8 at 9:57
|
2020-04-06 16:26:23
|
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|
http://alphapowertrading.com/quantopian/Stock_Portfolio_Strategy_Design2.html
|
## Stock Portfolio Strategy Design¶
Every day you are told which stocks performed best and which did not. Your stock trading program can monitor all those stocks and sort them out over whatever criteria you like. One would imagine that your stock ranking system would make you concentrate on the very best performers. However, when looking at long-term portfolio results, it should raise more than some doubts on this matter since most professional stock portfolio managers do not even outperform market averages.
Comparing stock trading strategies should be made over comparables, meaning over the same duration using the same amount as initial capital. It should answer the question, is strategy $H_k$ better than $H_{spy}$ or not?
$\quad \quad \sum (H_k \cdot \Delta P) >, \dots, > \sum (H_{spy} \cdot \Delta P) >, \dots, > \sum (H_z \cdot \Delta P)$ ?
So, why is it that most long-term active trading strategies fail to beat the averages? Already, we expect half of those strategies should perform below averages almost by definition. But, why are there so many more?
$\quad \quad \displaystyle{ \sum (\overline{H_k} \cdot \Delta P) < \sum (H_{spy} \cdot \Delta P)}\quad$ for $k=1$ to some gazillion strategies out there.
It is a legitimate question since you will be faced with the same problem going forward. Will you be able to outperform the averages? Are you looking for this ultimate strategy $H_{k=5654458935527819358914774892147856}$? Or will this number require up to some 75 more digits...
We have no way of knowing how many trading strategies or which ones can or will surpass the average benchmark over the long term. But, we do know that over the past, some 75$\%$, maybe even more, have not exceeded long-term market averages. This leaves some 25$\%$ or less that have satisfied the condition of outperforming the averages. It is from that lot that we should learn what to do to improve on our own trading strategies.
We can still look at the strategies that failed in order not to follow in their footsteps. Imitating strategies that underperform over the long term is not the best starting point. It can only lead to underperforming the averages even more.
We need to study and learn from the higher class of trading strategies and know why they outperformed. If we cannot understand the rationale behind such trading strategies or if none is given, then how could we ever duplicate their performance or even enhance them?
We have this big blob of price data, the recorded price matrix $\mathsf{P}$ for all the listed stocks. We can reduce it to a desirable portfolio size by selecting as many columns (stocks) and rows (days) as we like, need or want. Each price is totally described by its place in the price matrix $p_{d,j}$. And what you want to do is find common grounds in all this data that might show some predictive abilities.
You have stock prices going up or down, they usually do not maintain the same value very long. So, you are faced with a game where at any one time prices are basically moving up or down. And all you have to determine is which way they are going. How hard could that be?
From the long-term outcome of professional stock portfolio managers, it does appear to be more difficult than it seems.
### There is Randomness in There¶
If price predictability is low, all by itself, it would easily explain the fact that most professionals do not outperform the averages over the long term. As a direct consequence, there should be a lot of randomness in price movements. And if, or since, it is the case, then most results would tend to some expected mean which is the long-term market average return.
It is easy to demonstrate the near 50/50 odds of having up and down price movements. Simply count the ups and downs days over an extended period of time over a reasonable sample. The expectation is that you will get, on average, something like 51/49 or 52/48 depending on the chosen sample. The chart below does illustrate this clearly.
With those numbers, we have to accept that there is a lot of randomness in the making of those price series. It takes 100 trades to be ahead 2 or 4 trades respectively. With 1000 trades, you should be ahead by 20 or 40 trades. But, you will have to execute those 1000 trades to achieve those results. That is a 2$\%$ or a 4$\%$ of trades taken that will account for most of your generated profits. This says that the top 50 trades out of the 1000 taken will be responsible for most if not all your profits. And that 950 trades out of those 1000 could have been throwaways. Certainly, the 48% of trades you lost (480 trades), if they could have been scraped would definitely have helped your cause, profitwise.
The problem you encounter is that you do not know which one is which, and thus, the notion of a high degree of randomness. Fortunately, it is only a high degree of randomness and not something that is totally random because there only luck could make you win the game.
Here is an interesting AAPL chart snippet (taken from my 2012 Presentation). It makes that presentation something like a 7-year walk-forward with totally out-of-sample data. The hit rate on that one is very high. It is also the kind of chart we do not see on Quantopian. It was done to answer the question: are the trades executed at reasonable places in the price cycles? A simple look at the chart can answer that question.
The chart displays the strategy's trading behavior with its distributed buys (blue arrows) and sells (red arrows) as the price swings up and down. On most swings, some shares are sold near tops and bought near bottoms. The chart is not displayed as a probabilistic technique, but to show some other properties.
One, there was no prediction made in handling the trades, none whatsoever. The program does not know what a top or bottom is or even has a notion of mean-reversal. Nonetheless, it trades as if it knew something and does make trading profits.
Second, entries and exits were performed according to the outcome of time-biased random functions. There are no factors here, no fundamental data, and no technical indicators. It operates on price alone. It does, however, have the notion of delayed gratification. An exit could be delayed following some other random functions giving a trade a time-measured probabilistic exit. Meaning that a trade could have exceeded its exit criteria but its exit could still be ignored until a later date for no other reason than it was not its lucky exit day.
Third, trades were distributed over time in an entry or exit averaging process. The mindset here is to average out the average price near swing tops or bottoms. The program does not know where the tops or bottoms are but nonetheless its trade positioning process will make it have an average price near those swing tops and bottoms.
Forth, the whole strategy goes on the premise of: accumulate shares over the long term and trade over the process (this is DEVX03 that gradually morphed over the years into its latest iteration DEVX10). The above chart depicts the trading process but does not show the accumulation process itself even if it is there. To accumulate shares requires that as time progresses, the stock inventory increases by some measure as prices rise.
Here, the proceeds of all sales, including all the profits, are reinvested in buying more shares going forward. And this share accumulation, as well as the accumulated trading profits, will be reflected in the strategy's overall long-term CAGR performance. It is all explained in the above-cited 2012 presentation.
The strategy is based on old methods and does show that it can outperform the averages: $\sum (H_k \cdot \Delta P) \gg \sum (H_{spy} \cdot \Delta P)$. The major force behind strategy $H_k$ is time. As simple as that. It waits for its trade profit. It was easy to determine some seven years ago that AAPL and AMZN would prosper going forward. We can say the same thing today for the years to come. What that program will do is continue to accumulate shares for the long term and trade over the process, and thereby continue to outperform the averages.
In the end, we all have to make choices. Some are easier than others. But one thing is sure, it will all be in your trading strategy $H_k$ and what you designed it to do.
|
2019-10-14 14:07:33
|
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|
http://mathinsight.org/applet/stick_figure_position
|
# Math Insight
### Applet: Stick figure position
The Java applet did not load, and the above is only a static image representing one view of the applet. The applet was created with Geogebra. The applet is not loading because it looks like you do not have Java installed. You can click here to get Java.
It takes a vector in nine-dimensional space to specify the angles of this stick figure's arms, legs, and head. We denote the configuration vector specifying these angles by $\vec{\theta} = (\theta_1, \theta_2, \theta_3, \theta_4, \theta_5, \theta_6,\theta_7, \theta_8,\theta_9)$. You can drag the sliders with your mouse to change the components of the vector. The components $\theta_1$ and $\theta_2$ specify the angles of his left arm, $\theta_3$ and $\theta_4$ specify the angles of his right arm, $\theta_5$, $\theta_6$, $\theta_7$, and $\theta_8$ specify the angles of his left and right legs, and, finally, $\theta_9$ specifies the angle of his head.
Applet file: stick_figure_position.ggb
#### Applet links
This applet is found in the pages
List of all applets
#### General information about Geogebra applets
This applet was created using Geogebra. In most Geogebra applets, you can move objects by dragging them with the mouse. In some, you can enter values with the keyboard. To reset the applet to its original view, click the icon in the upper right hand corner.
Since these applets use Java, you must have Java installed and properly configured in your browser for the them to display. You can get Java here.
You can download the applet onto your own computer so you can use it outside this web page or even modify it to improve it. You simply need to download the above applet file and download the Geogebra program onto your own computer.
|
2014-09-01 07:31:09
|
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|
https://www.tutorialspoint.com/Java-program-to-find-the-square-root-of-a-given-number
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# Java program to find the square root of a given number
Java8Java Programming Object Oriented Programming
The process of finding the square root of a number can be divided into two steps. One step is to find integer part and the second one is for fraction part.
## Algorithm
• Define value n to find the square root of.
• Define variable i and set it to 1. (For integer part)
• Define variable p and set it to 0.00001. (For fraction part)
• While i*i is less than n, increment i.
• Step 4 should produce the integer part so far.
• While i*i is less than n, add p to i.
• Now i have the square root value of n.
## Example
Live Demo
public class SquareRoot {
public static void main(String args[]){
int n = 24;
double i, precision = 0.00001;
for(i = 1; i*i <=n; ++i);
for(--i; i*i < n; i += precision);
System.out.println("Square root of given number "+i);
}
}
## Output
Square root of given number 4.898979999965967
Published on 26-Apr-2018 12:59:44
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2021-06-14 21:53:24
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https://kyushu-u.pure.elsevier.com/ja/publications/study-of-the-hard-double-parton-scattering-contribution-to-inclus
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# Study of the hard double-parton scattering contribution to inclusive four-lepton production in pp collisions at √s=8 TeV with the ATLAS detector
The ATLAS collaboration
28 被引用数 (Scopus)
## 抄録
The inclusive production of four isolated charged leptons in pp collisions is analysed for the presence of hard double-parton scattering, using 20.2 fb−1 of data recorded in the ATLAS detector at the LHC at centre-of-mass energy s=8 TeV. In the four-lepton invariant-mass range of 80<m4ℓ<1000 GeV, an artificial neural network is used to enhance the separation between single- and double-parton scattering based on the kinematics of the four leptons in the final state. An upper limit on the fraction of events originating from double-parton scattering is determined at 95% confidence level to be fDPS=0.042, which results in an estimated lower limit on the effective cross section at 95% confidence level of 1.0 mb.
本文言語 英語 595-614 20 Physics Letters, Section B: Nuclear, Elementary Particle and High-Energy Physics 790 https://doi.org/10.1016/j.physletb.2019.01.062 出版済み - 1月 1 2019
## !!!All Science Journal Classification (ASJC) codes
• 核物理学および高エネルギー物理学
## フィンガープリント
「Study of the hard double-parton scattering contribution to inclusive four-lepton production in pp collisions at √s=8 TeV with the ATLAS detector」の研究トピックを掘り下げます。これらがまとまってユニークなフィンガープリントを構成します。
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2023-01-28 06:25:27
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https://math.stackexchange.com/questions/1606095/integrate-int-tan32t-sec32t-dt
|
# integrate $\int \tan^3(2t)\sec^3(2t) \, dt$
$$\int \tan^3(2t)\sec^3(2t)\,dt$$
First try:$$\int \tan^3(2t)\sec^3(2t) \, dt=\int (\tan(2t)\sec(2t))^3 \, dt=\int \left(\frac{\sin(2t)}{\cos(2t)}\,\frac{1}{\cos(2t)}\right)^3 \, dt$$
$u=\cos2t$
$du=-2\sin(2t) \, dt$
$$-\frac{1}{2}\int \left(\frac{du}{u^2}\right)^3=\frac{1}{2} \cdot \frac{u}{5}^{-5}=\frac{1}{10\cos2t}+C$$
Second Try:
$$\int \tan^3(2t)\sec^3(2t) \, dt=\int \tan^2(2t)\sec^2(2t)\tan(2t)\sec(2t)dt=\int\sec^2(2t)(1-\sec^2(2t))\tan(2t)\sec(2t)$$
$u=sec(2t)$
$du=2sec(2t)tan(2t)$
$$\frac{1}{2}\int u^2(1-u^2)du=\frac{1}{2}\int u^2-u^4=\frac{u^3}{6}-\frac{u^5}{10}=\frac{sec^{3}(2t)}{6}-\frac{sec^{5}(2t)}{10}+c$$ Where have I got it wrong?
• Start by simplifying your integral by letting $2 t \to x$ and $d t = d x/2$. And it makes no sense to integrate over $(du)^3$. That's your problem. – David G. Stork Jan 9 '16 at 22:10
• @DavidG.Stork I need to find integrate according to $dt$ so it is just like $2x$ according to $dx$ – gbox Jan 9 '16 at 22:13
• I cannot see why you "need" to integrate according to $dt$. After all, you make a substitution to $du$. Why not first $dx$? – David G. Stork Jan 9 '16 at 22:14
• @DavidG.Stork ok so we can replace $dt$ in $dx$ – gbox Jan 9 '16 at 22:17
You should not get something like $\left(du \right)^3$ in the espression, it does not make sense. Indeed if you use $du=-2\sin(2t) \, dt$ where $u = \cos 2t$, you get $\sin^2 2t = 1- \cos^2 2t = 1 - u^2$. Put everything into the expression (i.e., express everything in terms of $u$), you get
$$\int \left(\frac{\sqrt{1-u^2}}{u^2}\right)^3 \frac{du}{-2 \sqrt{1-u^2}} = -\frac 12 \int \frac{1-u^2}{u^6} du.$$
• @gbox : Looks good, but it should be $u^2 -1$ instead of $1-u^2$, and you are missing some $du$'s. – user99914 Jan 9 '16 at 22:38
HINT: show that your integrand is equal to $$8\,{\frac { \left( \sin \left( t \right) \right) ^{3} \left( \cos \left( t \right) \right) ^{3}}{ \left( 2\, \left( \cos \left( t \right) \right) ^{2}-1 \right) ^{6}}}$$
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2019-10-24 02:34:34
|
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https://mathematica.stackexchange.com/questions/128207/resolve-returns-false-on-two-contradicting-statements
|
# Resolve returns False on two contradicting statements
I'm trying to prove an inequality using Mathematica. This is the code leading to the inequality:
T = {{0, pbl*pcharge, 0}, {1, 1 - pbl, psh}, {0, pbl*(1 - pcharge),
1 - psh}};
vecs = Eigenvectors[T];
val = Eigenvalues[T];
pfull = Simplify[(1/(1 + vecs[[1]][[1]] + vecs[[1]][[2]]))];
pbl1 = (1 - phigh)*beta/(lowcons + (1 - pfull - pdead)*M);
psh1 = phigh*(highcons - beta)/(pfull*M);
sols = Simplify[Solve[pbl1 == pbl && psh1 == psh, {pbl, psh}]];
pshex2[highcons_, lowcons_, beta_, phigh_, pcharge_, M_] =
Simplify[sols[[All, 2, 2]]];
However, when I try to use Resolve[] to see if the inequality is true, both the inequality and its negation are resolved as false:
Resolve[ForAll[{highcons, lowcons, beta, phigh, pcharge, M},
highcons > beta > lowcons > 0 && 0 < phigh < 1 && 0 < pcharge < 1 &&
M > 0, D[pshex2[highcons, lowcons, beta, phigh, pcharge, M], M] <=
0], Reals]
Out[1714]= False
Resolve[Exists[{highcons, lowcons, beta, phigh, pcharge, M},
highcons > beta > lowcons > 0 && 0 < phigh < 1 && 0 < pcharge < 1 &&
M > 0, D[pshex2[highcons, lowcons, beta, phigh, pcharge, M], M] >
0], Reals]
Out[1715]= False
Also, FindInstance[] is not able to find an example for the second statement. Is there anything I am doing wrong? I'm very new to Mathematica.
• bugs is a special tag that is supposed to be added only by someone else than the original poster, after verifying the issue. Please do not add this tag to your own posts. – Szabolcs Oct 10 '16 at 8:05
• It does look like a bug to me, please report this to Wolfram Support. – Szabolcs Oct 10 '16 at 9:03
• Sorry! I didn't know that. Sent the issue to Wolfram Tech Support. – Venomouse Oct 10 '16 at 10:37
I found your example a bit hard to follow, so let's write in a form which is more explicit:
expr = First@D[pshex2[highcons, lowcons, beta, phigh, pcharge, M], M];
cond = highcons > beta > lowcons > 0 && 0 < phigh < 1 && 0 < pcharge < 1 && M > 0;
In version 10.0.2 or later (up to 11.0.1):
Resolve[ForAll[Evaluate@Variables[expr], cond, expr <= 0], Reals]
(* False *)
Resolve[Not@ForAll[Evaluate@Variables[expr], cond, expr <= 0], Reals]
(* False *)
Resolve[Exists[Evaluate@Variables[expr], cond, expr > 0], Reals]
(* False *)
ForAll is (surprisingly!) HoldAll so Evaluate was necessary int he first argument.
The second input above is literally the same as the first one except for wrapping the ForAll in Not.
The conditions are clearly fine and there are many combinations of values that satisfy them (FindInstance[cond, Variables[expr]]).
In version 9.0.1:
Resolve[ForAll[Evaluate@Variables[expr], cond, expr <= 0], Reals]
(* True *)
Resolve[Not@ForAll[Evaluate@Variables[expr], cond, expr <= 0], Reals]
(* False *)
Resolve[Exists[Evaluate@Variables[expr], cond, expr > 0], Reals]
(* False *)
It seems clear that there's a bug in v10.0.2 – 11.0.1. Is the result by v9.0.1 correct? That I do not know.
• Thank you for checking this out! Can you please explain why Variables is needed to be used inside Evaluate? – Venomouse Oct 10 '16 at 10:49
• @Venomouse Because ForAll has the HoldAll attribute, which I find quite weird for a symbolic processing function ... I asked about this here, and included examples that show what would happen without Evaluate: community.wolfram.com/groups/-/m/t/937146 – Szabolcs Oct 10 '16 at 10:51
Note that expr is a rational function. If the denominator of expr is zero, then the inequality expr<=0 is neither true nor false -- it is undefined. Hence it is not true that for all values of variables that satisfy cond the inequality expr<=0 is true. If we add the condition Denominator[expr]!=0 then we get the expected result.
Resolve[ForAll[Evaluate@Variables[expr], cond && Denominator[expr]!=0, expr <= 0], Reals]
(* True *)
Resolve[Not@ForAll[Evaluate@Variables[expr], cond && Denominator[expr]!=0, expr <= 0], Reals]
(* False *)
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2021-06-13 17:43:34
|
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http://physics.stackexchange.com/questions/24145/a-person-weighing-100-n-stands-on-some-bathroom-scales-in-a-lift-if-the-scales
|
# A person weighing 100 N stands on some bathroom scales in a lift. If the scales show a reading of 110 N, which way is he going [closed]
• Moving downwards and decelerating.
• Moving downwards with a constant velocity.
• Moving upwards and decelerating.
• Moving upwards with a constant velocity.
-
## closed as off-topic by John Rennie, Brandon Enright, jinawee, WetSavannaAnimal aka Rod Vance, Kyle KanosMar 18 '14 at 13:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Brandon Enright, jinawee, WetSavannaAnimal aka Rod Vance, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.
I'm sure this has been answered before, but I can't see it in a search for "lift".
Anyhow the forum rules say we can't answer homework questions, but just discuss the concepts involved. You know from Newton's first law, F = ma, that if there is a force on a body it will accelerate, and it will accelerate in the direction of the force. So the question is, is there a net force on the person in the lift, and in which direction is the force?
The force downwards is just due to gravity, and the force upwards is the force being applied to the person by the floor of the lift. The scales are measuring the force being applied to the person by the floor of the lift. That should be all you need to answer the question.
-
Another trick that may help you visualise this is the psuedoforce. The psuedoforce enables you to imagine stuff in an accelerated frame easily. It's the "pressing" you feel in a plane during takeoff, etc. It's also the "pressing" you feel in an ascending lift.
I realised that I've already written a sizeable chunk on psuedoforces, so I'll just blatantly copy it from this answer of mine.:
Whenever we view a system from an accelerated frame, there is a "psuedoforce" or "false force" which appears to act on the bodies. Note that this force is not actually a force, more of something which appears to be acting. A mathematical trick, if you will.
Let's take a simple case. You are accelerating with $\vec{a}$ in space, and you see a little ball floating around. This is in a perfect vacuum, with no electric/magnetic/gravitational/etc fields. So, the ball does not accelerate.
But, from your point of view, the ball accelerates with an acceleration $-\vec{a}$, backwards relative to you. Now you know that the space is free of any fields, yet you see the particle accelerating. You can either deduce from this that you are accelerating, or you can decide that there is some unknown force, $-m\vec{a}$, acting on the ball. This force is the psuedoforce. It mathematically enables us to look at the world from the point of view of an accelerated frame, and derive equations of motion with all values relative to that frame. Many times, solving things from the ground frame get icky, so we use this. But let me stress once again, it is not a real force.
-
Depending on the instructions of your professor, this problem can be solved two different ways: in the inertial and the non-intertial frame of reference.
In the inertial frame of reference, you are watching the lift, the scale and the person within it, from outside. You see that the all three are traveling with a constant acceleration up or down. Just two forces are acting on the person, the force of gravity and the force of the scale. You do the 2nd Newton law. Important: The scale is nothing but a simple force gauge showing the actual force with which it supports the person (of course this is transformed on the scale into the mass under the assumption of the gravity acceleration).
In the non-inertial frame of reference, you are standing in the lift, so the scale and the person seem to be at rest. However, now three forces are acting on the person, force of gravity, force of the scale and fictitious or pseudo force. Fictitious force equals mass of the person multiplied by the acceleration of the lift and is acting in the opposite direction of the actual acceleration of the lift. So in this case you do the 1st Newton law.
In both frames of references, you should get the same result. Moreover, when lift is not accelerating, both frames of reference are inertial and you should use even the same calculation procedure (the first one).
-
Uhh, don't do that, write a complete answer and then submit. Unless you've already written a sizeable chunk which self-sufficiently solves the problem, and leave a note that you will add more later. – Manishearth Apr 21 '12 at 11:18
I'm sorry. The process of writing answers is complex for my head, I cannot think right unless I see the answer and the question one next to other. So I do this some kind of iteration process that usually ends in few minutes, but sometimes lasts for days :) – Pygmalion Apr 21 '12 at 11:50
You do know that answers are saved as drafts--you can write half an answer and come back later (to the same question) to improve it without submitting. Also, you can always open the question in another tab. – Manishearth Apr 21 '12 at 11:57
I do not know about those drafts. Do drafts appear next to the question, just like answers do? – Pygmalion Apr 21 '12 at 11:58
No, they appear in the "compose answer" box. Try it, start writing an answer for a random question. Wait a while (draft saving is every 30 seconds IIRC), close the page. Come back whenever you want, your half-written answer will still be there, at the bottom of the page in the answerbox. The same applies for questions, but you can have only one question draft at a time (since there's only one "ask question" page, but there are many pages with an answerbox) – Manishearth Apr 21 '12 at 12:02
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2015-09-04 06:12:07
|
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http://mickopedia.org/mickify?topic=Antisymmetric_relation
|
# Antisymmetric relation
In mathematics, a bleedin' homogeneous relation R on set X is antisymmetric if there is no pair of distinct elements of X each of which is related by R to the feckin' other. More formally, R is antisymmetric precisely if for all a and b in X
if R(a, b) with a ≠ b, then R(b, a) must not hold,
or, equivalently,
if R(a, b) and R(b, a), then a = b.
(The definition of antisymmetry says nothin' about whether R(a, a) actually holds or not for any a.)
## Examples
The divisibility relation on the oul' natural numbers is an important example of an antisymmetric relation. In this context, antisymmetry means that the feckin' only way each of two numbers can be divisible by the oul' other is if the feckin' two are, in fact, the oul' same number; equivalently, if n and m are distinct and n is a holy factor of m, then m cannot be a factor of n. For example, 12 is divisible by 4, but 4 is not divisible by 12.
The usual order relation ≤ on the feckin' real numbers is antisymmetric: if for two real numbers x and y both inequalities x ≤ y and y ≤ x hold then x and y must be equal, grand so. Similarly, the bleedin' subset order ⊆ on the subsets of any given set is antisymmetric: given two sets A and B, if every element in A also is in B and every element in B is also in A, then A and B must contain all the feckin' same elements and therefore be equal:
${\displaystyle A\subseteq B\land B\subseteq A\Rightarrow A=B}$
A real-life example of an oul' relation that is typically antisymmetric is "paid the feckin' restaurant bill of" (understood as restricted to an oul' given occasion). Typically some people pay their own bills, while others pay for their spouses or friends. As long as no two people pay each other's bills, the relation is antisymmetric.
## Properties
Partial and total orders are antisymmetric by definition, enda story. A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the oul' "preys on" relation on biological species).
Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive.
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2021-01-27 18:58:49
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http://wwwa.pikara.ne.jp/okojisan/otfft-en/stockham1.html
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[OTFFT: High Speed FFT library]
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## [Introduction to the Stockham FFT]
This page is a homepage explaining the Stockham algorithm which is a kind of the Fast Fourier Transform (FFT). I made this homepage for people who can not understand the Stockham algorithm but can understand the Cooley-Tukey algorithm. You will be able to understand that Stockham algorithm is a transformation of Cooley-Tukey algorithm if you read this page.
First of all, I show the Cooley-Tukey algorithm. When I implement the Cooley-Tukey algorithm using recursive call, the program will be as follows.
List-1: Cooley-Tukey Algorithm
#include <complex>
#include <cmath>
#include <utility>
typedef std::complex<double> complex_t;
void butterfly(int n, int q, complex_t* x) // Butterfly operation
// n : sequence length
// q : block start point
// x : input/output squence
{
const int m = n/2;
const double theta0 = 2*M_PI/n;
if (n > 1) {
for (int p = 0; p < m; p++) {
const complex_t wp = complex_t(cos(p*theta0), -sin(p*theta0));
const complex_t a = x[q + p + 0];
const complex_t b = x[q + p + m];
x[q + p + 0] = a + b;
x[q + p + m] = (a - b) * wp;
}
butterfly(n/2, q + 0, x);
butterfly(n/2, q + m, x);
}
}
void bit_reverse(int n, complex_t* x) // Bitreversal operation
// n : squence length
// x : input/output sequence
{
for (int i = 0, j = 1; j < n-1; j++) {
for (int k = n >> 1; k > (i ^= k); k >>= 1);
if (i < j) std::swap(x[i], x[j]);
}
}
void fft(int n, complex_t* x) // Fourier transform
// n : sequence length
// x : input/output sequence
{
butterfly(n, 0, x);
bit_reverse(n, x);
for (int k = 0; k < n; k++) x[k] /= n;
}
void ifft(int n, complex_t* x) // Inverse Fourier transform
// n : sequence length
// x : input/output sequence
{
for (int p = 0; p < n; p++) x[p] = conj(x[p]);
butterfly(n, 0, x);
bit_reverse(n, x);
for (int k = 0; k < n; k++) x[k] = conj(x[k]);
}
The characteristic of Cooley-Tukey algorithm is bit_reverse(). We need bit_reverse() to sort the result of FFT in a natural order. This is an important point that is different from the Stockham algorithm. In the Stockham algorithm, even if there is no bit_reverse(), the result of FFT is sorted in a natural order.
However, it is possible by the Cooley-Tukey algorithm even if we do not use bit_reverse() if we only want to sort the result of FFT in a natural order. The result of FFT becomes a natural order sequence if we do as follows.
List-2: Natural Order Cooley-Tukey Algorithm
#include <complex>
#include <cmath>
typedef std::complex<double> complex_t;
void fft0(int n, int q, complex_t* x, complex_t* y)
// n : sequence length
// q : block start point
// x : input/output sequence
// y : work area
{
const int m = n/2;
const double theta0 = 2*M_PI/n;
if (n > 1) {
for (int p = 0; p < m; p++) { // Butterfly operation
const complex_t wp = complex_t(cos(p*theta0), -sin(p*theta0));
const complex_t a = x[q + p + 0];
const complex_t b = x[q + p + m];
y[q + p + 0] = a + b;
y[q + p + m] = (a - b) * wp;
}
fft0(n/2, q + 0, y, x);
fft0(n/2, q + m, y, x);
for (int p = 0; p < m; p++) { // Composition of even components and odd components
x[q + 2*p + 0] = y[q + p + 0]; // Even components
x[q + 2*p + 1] = y[q + p + m]; // Odd components
}
}
}
void fft(int n, complex_t* x) // Fourier transform
// n : sequence length
// x : input/output sequence
{
complex_t* y = new complex_t[n]; // Allocation of work arera
fft0(n, 0, x, y);
delete[] y;
for (int k = 0; k < n; k++) x[k] /= n;
}
void ifft(int n, complex_t* x) // Inverse Fourier transform
// n : sequence length
// x : input/output sequence
{
for (int p = 0; p < n; p++) x[p] = conj(x[p]);
complex_t* y = new complex_t[n]; // Allocation of work area
fft0(n, 0, x, y);
delete[] y;
for (int k = 0; k < n; k++) x[k] = conj(x[k]);
}
When we assume that $$F_N$$ is Fourier transform (the size is $$N, N = 2^L, L \ge 0, L$$ is an integer) and assume that $$x_p~(p = 0,1,\ldots,N-1)$$ is input sequence and assume that $$X_k~(k = 0,1,\ldots,N-1)$$ is output sequence, Fourier transform is represented as $$X_k = F_N(x_p) = \frac{1}{N}\sum_{p=0}^{N-1}x_p W_N^{kp},~ W_N = \exp\left(-j\frac{2\pi}{N}\right),~j = \sqrt{-1}$$. Now, if we assume $$N \ge 2$$ and we decompose $$F_N$$ by a recursion, the following relationship is satisfied.
$\begin{eqnarray*} 2X_{2k} & = & F_{\frac{N}{2}}(x_p + x_{p+\frac{N}{2}}) \\ 2X_{2k+1} & = & F_{\frac{N}{2}}\left((x_p - x_{p+\frac{N}{2}}) W_N^p\right) \\ p & = & 0,1,\ldots,\frac{N}{2}-1 \\ k & = & 0,1,\ldots,\frac{N}{2}-1 \\ \end{eqnarray*}$
In this way, if we decompose Fourier transform by a recursion, even components($$X_{2k}$$) and odd components($$X_{2k+1}$$) are obtained. Therefore, in order to change the output to a natural order sequence, we have to obtain $$X_k$$ by composing $$X_{2k}$$ and $$X_{2k+1}$$. The code performing it is the List-2. However, this code is slow. So, it is a turn of the Stockham algorithm.
In order to obtain the Stockham algorithm, let's start by eliminating the uselessness from the List-2. First, we notice that we are accessing to the array at two places. One is at the butterfly operation. Another is the place that we are composing even components and odd components. We can avoid uselessness if we execute butterfly operation and composition of even components and odd components at the same time. Using this idea, if we transform the List-2, it will be as follows.
List-3: Recursive version of the Stockham Algorithm
#include <complex>
#include <cmath>
typedef std::complex<double> complex_t;
void fft1(int n, int s, int q, complex_t* x, complex_t* y);
void fft0(int n, int s, int q, complex_t* x, complex_t* y)
// n : sequence length
// s : stride
// q : selection of even or odd
// x : input/output sequence
// y : work area
{
const int m = n/2;
const double theta0 = 2*M_PI/n;
if (n == 1) {}
else {
for (int p = 0; p < m; p++) {
// Butterfly operation and composition of even components and odd components
const complex_t wp = complex_t(cos(p*theta0), -sin(p*theta0));
const complex_t a = x[q + s*(p + 0)];
const complex_t b = x[q + s*(p + m)];
y[q + s*(2*p + 0)] = a + b;
y[q + s*(2*p + 1)] = (a - b) * wp;
}
fft1(n/2, 2*s, q + 0, y, x); // Even place FFT (y:input, x:output)
fft1(n/2, 2*s, q + s, y, x); // Odd place FFT (y:input, x:output)
}
}
void fft1(int n, int s, int q, complex_t* x, complex_t* y)
// n : sequence length
// s : stride
// q : selection of even or odd
// x : input sequence
// y : output sequence
{
const int m = n/2;
const double theta0 = 2*M_PI/n;
if (n == 1) { y[q] = x[q]; }
else {
for (int p = 0; p < m; p++) {
// Butterfly Operation and composition of even components and odd components
const complex_t wp = complex_t(cos(p*theta0), -sin(p*theta0));
const complex_t a = x[q + s*(p + 0)];
const complex_t b = x[q + s*(p + m)];
y[q + s*(2*p + 0)] = a + b;
y[q + s*(2*p + 1)] = (a - b) * wp;
}
fft0(n/2, 2*s, q + 0, y, x); // Even place FFT (y:input/output, x:work area)
fft0(n/2, 2*s, q + s, y, x); // Odd place FFT (y:input/output, x:work area)
}
}
void fft(int n, complex_t* x) // Fourier transform
// n : sequence length
// x : input/output sequence
{
complex_t* y = new complex_t[n]; // Allocation of work area
fft0(n, 1, 0, x, y);
delete[] y;
for (int k = 0; k < n; k++) x[k] /= n;
}
void ifft(int n, complex_t* x) // Inverse Fourier transform
// n : sequence length
// x : input/output sequence
{
for (int p = 0; p < n; p++) x[p] = conj(x[p]);
complex_t* y = new complex_t[n]; // Allocation of work area
fft0(n, 1, 0, x, y);
delete[] y;
for (int k = 0; k < n; k++) x[k] = conj(x[k]);
}
This code has become a mutual recursion in order to minimize the accesses to an array. Stockham algorithm requires work area y for sorting. It saves sorted results once in the work area. If the saved results are written back immediately to x, the program is able to avoid becoming a mutual recursion. But, if we do so, it becomes meaningless that we have reduced the accesses to an array by executing butterfly operation and composition at the same time. For this reason, this code has become a mutual recursion.
I will explain about fft0() and fft1().
fft0() is the Fourier transform that the size is n. s is the stride of array access. q is used for selection of even or odd. x is input/output sequence and y is work area.
fft1() is the Fourier transform that the size is n. s is the stride of array access. q is used for selection of even or odd. x is input sequence and y is output sequence.
By the way, were you able to understand this transformation? In fact, this code is the Stockham algorithm. This is very rare recursive version of the Stockham algorithm. So, let's transform the calculation into a recurrence formula with this condition: $$n=2^{L-h},~m=2^{-1}n,~s=2^h,~x_h(q,p)=x_h[q+sp]$$.
$\begin{eqnarray*} x_{h+1}(q,p) & = & x_h(q,p) + x_h(q,p + m) \\ x_{h+1}(q + s,p) & = & \left(x_h(q,p) - x_h(q,p + m)\right)W_N^{sp} \\ q & = & 0,1,\ldots,s-1 \\ p & = & 0,1,\ldots,m-1 \\ \end{eqnarray*}$
$$x_h[~]$$ is array of the $$h$$-step calculation. When we begin from $$x_0[~]$$ and get $$x_L[~]$$, we complete the FFT. However, $$x_L[~]$$ needs to be divided by $$N$$.
This program is called Dacimation In Frequency (DIF). In the case of Decimation In Time (DIT), the program will be as follows.
List-4: DIT Stockham Algorithm
#include <complex>
#include <cmath>
typedef std::complex<double> complex_t;
void fft1(int n, int s, int q, complex_t* x, complex_t* y);
void fft0(int n, int s, int q, complex_t* x, complex_t* y)
// n : sequence length
// s : stride
// q : selection of even or odd
// x : input/output sequence
// y : work area
{
const int m = n/2;
const double theta0 = 2*M_PI/n;
if (n == 1) {}
else {
fft1(n/2, 2*s, q + 0, y, x); // Even place FFT(x:input, y:output)
fft1(n/2, 2*s, q + s, y, x); // Odd place FFT(x:input, y:output)
for (int p = 0; p < m; p++) {
const complex_t wp = complex_t(cos(p*theta0), -sin(p*theta0));
const complex_t a = y[q + s*(2*p + 0)];
const complex_t b = y[q + s*(2*p + 1)] * wp;
x[q + s*(p + 0)] = a + b;
x[q + s*(p + m)] = a - b;
}
}
}
void fft1(int n, int s, int q, complex_t* x, complex_t* y)
// n : sequence length
// s : stride
// q : selection of even or odd
// x : output sequence
// y : input sequence
{
const int m = n/2;
const double theta0 = 2*M_PI/n;
if (n == 1) { x[q] = y[q]; }
else {
fft0(n/2, 2*s, q + 0, y, x); // Even place FFT(y:input/output, x:work area)
fft0(n/2, 2*s, q + s, y, x); // Odd place FFT(y:input/output, x:work area)
for (int p = 0; p < m; p++) {
const complex_t wp = complex_t(cos(p*theta0), -sin(p*theta0));
const complex_t a = y[q + s*(2*p + 0)];
const complex_t b = y[q + s*(2*p + 1)] * wp;
x[q + s*(p + 0)] = a + b;
x[q + s*(p + m)] = a - b;
}
}
}
void fft(int n, complex_t* x) // Fourier transform
// n : sequence length
// x : input/output sequence
{
complex_t* y = new complex_t[n]; // Allocation of work area
fft0(n, 1, 0, x, y);
delete[] y;
for (int k = 0; k < n; k++) x[k] /= n;
}
void ifft(int n, complex_t* x) // Inverse Fourier transform
// n : sequence length
// x : input/output sequence
{
for (int p = 0; p < n; p++) x[p] = conj(x[p]);
complex_t* y = new complex_t[n]; // Allocation of work area
fft0(n, 1, 0, x, y);
delete[] y;
for (int k = 0; k < n; k++) x[k] = conj(x[k]);
}
Let's transform the calculation into a recurrence formula with this condition: $$m=2^h,~t=2^{L-h},~s=2^{-1}t,~x_h(q,p)=x_h[q+tp]$$.
$\begin{eqnarray*} x_{h+1}(q,p) & = & x_h(q,p) + x_h(q + s,p)W_N^{sp} \\ x_{h+1}(q,p + m) & = & x_h(q,p) - x_h(q + s,p)W_N^{sp} \\ q & = & 0,1,\ldots,s-1 \\ p & = & 0,1,\ldots,m-1 \\ \end{eqnarray*}$
Unfortunately, this recursive version of the Stockham FFT is slow. So, we will lead the high-speed iterative version of the Stockham FFT in the next page.
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2021-10-15 20:21:33
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https://crypto.stackexchange.com/questions/74439/whats-the-likelihood-of-a-collision-if-i-use-a-hash-to-detect-data-change
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# What's the likelihood of a collision if I use a hash to detect data change?
I have a set of data that I want to sync, and instead of sending an entire set of data, I wanted to use a hash.
It is basically a list of GUIDs with either "true" or "false".
eaa54efe-2b10-4d3b-b9c2-30d2bcca7e06 : true
94297d72-573a-4e6c-a45e-37a390750b64 : false
36d1acbd-b2dc-4a90-a7be-fb65b63639a5 : true
I plan to generate a hash of the entire list so that I can send much fewer data to sync than the entire data set.
Should I ever be concerned that a specific change of true/false will happen to randomly result in a collision and become "undetected" or is that really not going to happen?
I don't really understand the math and proofs behind the hashes, but I also understand that good hashes are supposed to be designed against trivial collision.
Note: the main goal is just data comparison. There is no encryption of the data because they're useless data without the meaning behind the GUIDs. This is, literally, live data I just posted.
• A properly designed $n$-bit hash function has collision probability $2^{-n/2}$ due to birthday paradox. SHA256 is a good choice, but BLAKE2s128 isn't bad either. Just don't go with MD5 as it's not properly designed and have structual weakness. – DannyNiu Sep 20 '19 at 2:57
• @DannyNiu That's not quite what the birthday paradox means. In a collection of hashes having only one element, the probability of a collision is zero! – Squeamish Ossifrage Sep 20 '19 at 2:58
When syncing two data set over the internet one can basically two choices;
• send all the data and compare: This is slow but complete.
• send the hash of the data: This is very fast but there is a tiny-tiny-tiny change of a collision is almost zero. If you are that lucky, you probably become reach or hit by an asteroid. This is very common on Linux's CD/DVD image downloads.
In your case, you want to just send a hash of one huge list. If you fear that the almost zero ( $$2/2^{256}$$ for a 256-bit bash function) is bothering you, you can decrease the change below almost zero by taking another hash function to have two hashes, though not necessary.
and, yes, Cryptographical hash functions are designed to have collision resistance, however, over the time one might find better attacks than generic collision attack like SHA-1 and it has already removed from the standard. This attack, actually, searching for different inputs, where the data structure enables, to find a collision. And the generic collision attack is $$2^{80}$$ on SHA-1, that is low in today's standards if you consider that Bitcoin miners reached $$\approx 2^{92}$$ SHA-256 hashes per year in 06 Agust 2019. Reaching $$2^{128}$$ way beyond classical computing.
Stick to a good hash function, SHA-256, SHA-512, SHA3, Blake2.
As was said already in other answers, using a modern, non-deprecated hash (see here for a list of most common hashes with the information of which ones should not be used any more), the chance of collisions are negligible and you need not worry about it: you can rely on the fact that a change in the data will cause very different hash values.
However there are things you must be careful about, for instance: how are you going to transport the hash from source to destination? If you send the hash via the same way you send the data, and you are afraid that someone modifies the data in-flight, then that same person could also modify the hash to be the hash of the modified data. One solution is to find a way to securely transfer a key from source to destination (for instance by SSHing into each), and then using a MAC which the adversary won't be able to re-compute for the modified data.
To give more precise directions we would need more details from you, in particular, how and where do you think an adversary may modify your data: is it just during network transfer? Is it on disk?
Finally, it looks like you are trying to roll your own crypto here (search the Web about rolling your own crypto) if this is just for your own curiosity or training or research this is fine, but if you plan to protect real value with this you should use tools where all the cryptographic decisions are already made for you.
The others are talking about hash collisions in an adversarial context, in which someone else is motivated to find a collision.
This raises the first question you should be asking: what is your threat model?
If you are talking about no adversary, you can do this and should not worry. If your program was running on every pc on Earth and syncing over a 256 bit hash with every other pc every second, your chance of having even a single collision within a hundred years is 1 in 100000000000000000000000000000000000000000000000000. That works for any hash as long as it is evenly distributed: it doesn't even need to be a cryptographic hash.
Do note that this mechanism can only tell you that nothing has changed. You would need to send the whole list if you got a hash mismatch.
If you did have an adversary who was for some reason trying to cause an undetected divergence, then you would have to think more carefully.
• If there is no adversary, then a non-cryptographic data-base hash would work (xxhash etc), however, sometimes what appears to be an internal isolated hash ends up being exposed to a collision attack later if the code is re-used in a different context. – Gregor Sep 24 '19 at 17:13
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2021-06-22 23:58:36
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http://sites.science.oregonstate.edu/physics/portfolioswiki/doku.php?id=courses:lecture:rflec:rflecpole&rev=1309985243&mbdo=print
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## Resolving Paradoxes with Spacetime Diagrams
(Lecture: 10 minutes)
• spacetime diagrams for pole & barn and space war
See §8 of the text, especially §8.1 and §8.2.
### Reflections
Students should have just had the opportunity to resolve special relativity paradoxes using spacetime diagrams (in this activity). This lecture can be part of the wrapup for that activity, or serve as a good review at the next class meeting.
##### Views
New Users
Curriculum
Pedagogy
Institutional Change
Publications
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2020-01-27 22:37:44
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https://www.vedantu.com/question-answer/the-fourth-term-of-an-arithmetic-progression-is-class-10-maths-cbse-5edcbcb0e5b56371c59317ee
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# The fourth term of an arithmetic progression is $10$ and the eleventh term of it exceeds three times the fourth term by $1$. Find the sum of the first $20$ terms of the progression.
Last updated date: 27th Mar 2023
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Hint : Find first term and common difference to find the sum.
As we know that the nth term of an A.P. is denoted as ,
${a_n} = a + (n - 1)d$
From the question we can say that,
${a_4} = a + (4 - 1)d = a + 3d = 10\,\,\,\,...(i) \\ {a_{11}} = a + (11 - 1)d = a + 10d\,\,\,\, \\$
It is given that,
${a_{11}} = 3{a_4} + 1 \\ a + 10d = 3a + 9d + 1 \\ 2a - d + 1 = 0\,\,\,\,\,\,\,\,\,......(ii) \\$
On multiplying the equation $(ii)$ by 3 we get,
$6a - 3d + 3 = 0\,\,\,\,\,\,......(iii)$
Solving equation $(i)$ & $(iii)$we get,
$a = 1 \\ \& \\ d = 3 \\$
We know sum on n terms of an A.P. can be written as
${S_n} = \frac{n}{2}(2a + (n - 1)d) \\ {S_{20}} = \frac{{20}}{2}(2(1) + 19 \times 3) \\ {S_{20}} = 590 \\$
Hence, the sum of 20 terms of the series in 590.
Note :- In these types of questions of A.P. we have to first obtain an equation from the given data then solve the equation to get the unknowns like first term & common difference, after finding the unknowns, obtain the sum by using the formula of sum of an A.P.
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2023-03-30 18:53:13
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https://indico.gsi.de/event/8597/
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GSI-FAIR Colloquium
# Cooling liquids super-quickly: How to study supercooled water & co. at very, very low temperatures
## by Robert Grisenti (GSI / IKP GU Frankfurt)
Tuesday, June 4, 2019 from to (Europe/Berlin)
at GSI ( Main Lecture Hall )
Description Supercooled liquids – liquids that are metastable with respect to the crystalline form – represent an important class of out-of-equilibrium systems whose study is related to a number of fundamental questions in statistical and condensed matter physics. Water represents one prominent example that highlights the relevance of supercooled liquids. Water’s behavior is unusual with respect to that of most other liquids, and it was long speculated that clues about this anomalous behavior might be found at temperatures well below the melting point. However, the crucial question on how far liquid water may actually be cooled without instantaneously freezing to ice remained unanswered so far. But the importance of supercooled liquids is not limited to the study of water’s properties alone. For instance, understanding the more general microscopic details of the crystallization process itself is crucial for diverse research fields such as atmospheric physics and material science. Such studies, however, were so far greatly hampered by the fact that the liquid-to-solid phase transition occurs on a very short time scale, especially at deep supercooling. Here I will show that the use of a microscopic laminar jet, formed by injecting a pressurized liquid through a micrometer-sized orifice into vacuum, offers a unique Ansatz to address the above experimental challenges. I will show that we were able to cool water by this mean to a record low temperature of -40°C and discuss how the combination of liquid jets and state-of-the-art light scattering techniques allows the investigation of the properties and structural transformations in a class of supercooled liquids that are precluded to more conventional approaches. Material:
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2019-08-24 12:21:22
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https://pypi.org/project/modupipe/1.0.1/
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A modular and extensible ETL-like pipeline builder
# ModuPipe : A modular and extensible ETL-like pipeline builder
## Benefits
• Entirely typed
• Abstract, so it fits any use case
• Class-based for easy configurations and injections
## Usage
Extract-Transform-Load (ETL) pipelines are a classic form of data-processing pipelines used in the industry. It consists of 3 main elements:
1. An Extractor, which returns data in a stream-like structure (Iterator in Python) using a pull strategy.
2. Some Mapper (optional), which transforms (parse, converts, filters, etc.) the data obtained from the source(s). Mappers can be chained together and chained to an extractor (with +) in order to form a new extractor.
3. A Loader, which receives the maybe-transformed data using a push strategy. Loaders can be multiple (with LoaderList) or chained together (with +).
Therefore, those 3 processes are offered as interfaces, easily chainable and interchangeable at any time.
An interface Runnable is also offered in order to interface the concept of "running a pipeline". This enables a powerfull composition pattern for wrapping the execution behaviour of runnables.
## Examples
Usage examples are present in the examples folder.
## Discussion
### Optimizing pushing to multiple loaders
If you have multiple loaders (using the LoaderList class or many chained PushTo mappers), but performance is a must, then you should use a multi-processing approach (with modupipe.runnable.MultiProcess), and push to 1 queue per loader. Each queue will also become a direct extractor for each loader, all running in parallel. This is especially usefull when at least one of the loaders takes a long processing time.
As an example, let's take a Loader 1 which is very slow, and a Loader 2 which is normally fast. You'll be going from :
┌────── single pipeline ──────┐ ┌──────────────── single pipeline ───────────────┐
to :
┌────── pipeline 1 ──────┐ ┌────────── pipeline 2 ─────────┐
Extractor ┬─⏵ PutToQueue ──⏵ Queue 1 ⏴── GetFromQueue ──⏵ Loader 1 (slow)
└─⏵ PutToQueue ──⏵ Queue 2 ⏴── GetFromQueue ──⏵ Loader 2 (not late)
└──────────── pipeline 3 ───────────┘
This will of course not accelerate the Loader 1 processing time, but all the other loaders performances will be greatly improved by not waiting for each other.
## Project details
Uploaded source
Uploaded py3
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2022-10-07 23:43:40
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http://mathhelpforum.com/calculus/18288-difficult-antiderivative.html
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# Math Help - difficult antiderivative
1. ## difficult antiderivative
Hi,
Can someone please explain to me how to find the antiderivative of x^3/3(cosmx)? I'm trying to solve the interval of x^2cosmx dx and I am at the point where I need to find the antiderivative of x^3/3(cosmx). I don't have any idea how to solve this. I would know how to solve it if I had to find the derivative, but I don't know what to do here since its the antiderivative.
Thank you very much
2. Originally Posted by chocolatelover
Hi,
Can someone please explain to me how to find the antiderivative of x^3/3(cosmx)? I'm trying to solve the interval of x^2cosmx dx and I am at the point where I need to find the antiderivative of x^3/3(cosmx). I don't have any idea how to solve this. I would know how to solve it if I had to find the derivative, but I don't know what to do here since its the antiderivative.
Thank you very much
do integration by parts. use the polynomial part as the part to differentiate and the trigonometric part as the part to integrate all the time (you will have to do it 3 times or so). there's probably a formula for an integral of this type, i don't do well with memorizing formulas so i can't tell you
3. Hi,
I did it four times and it looks worse now. Does anyone see what I did wrong?
This is what I did:
Integral x^2cosmx dx
u=cosmx
dv=x^2dx
du=-sinmx
v=x^3/3
integral x^2cosmx dx=(cosmx)(x^3/3)-integral(x^3/3(cosmx)
then I did it again:
u=x^3
dv=cosmx dx
du=2x^2/3 (quotient rule)
v=sinmx
(x^3)/3(sinmx)-integral(sinmx)(2x^2/3)
so...(cosmx)(x^3/3)-(x^3/3)(sinmx)-integral sin(mx)(2x^2/3)
then I did it again
u=2x^2/3
dv=sinmx
du=4/3x
v=-cosmx
(cosmx)(x^3/3)-(x^3)/3(sinmx)-(2x^2/3)(-cosmx)-integral (-cosmx)4/3x
then I simplified
(cosmx)(x^3/3)-(x^3/3)(sinmx)+(2x^2/3)cosmx-integral -cosmx(4/3x)
then I did it again
u=2x^2/3
dv=cosmx
du=12x+2x^2/3(-sinmx)/9 (this doesn't look like its going to help)
v=sinmx
Thank you very much
4. $\int x^2\cos(mx)\,dx=\frac{2x\cos(mx)}{m^2}+\frac{(m^2x ^2-2)\sin (mx)}{m^3}+k$
First, set $y=mx$, after that apply integration by parts twice.
5. Originally Posted by chocolatelover
Hi,
I did it four times and it looks worse now. Does anyone see what I did wrong?
This is what I did:
Integral x^2cosmx dx
u=cosmx
dv=x^2dx
du=-sinmx
v=x^3/3
integral x^2cosmx dx=(cosmx)(x^3/3)-integral(x^3/3(cosmx)
then I did it again:
u=x^3
dv=cosmx dx
du=2x^2/3 (quotient rule)
v=sinmx
(x^3)/3(sinmx)-integral(sinmx)(2x^2/3)
so...(cosmx)(x^3/3)-(x^3/3)(sinmx)-integral sin(mx)(2x^2/3)
then I did it again
u=2x^2/3
dv=sinmx
du=4/3x
v=-cosmx
(cosmx)(x^3/3)-(x^3)/3(sinmx)-(2x^2/3)(-cosmx)-integral (-cosmx)4/3x
then I simplified
(cosmx)(x^3/3)-(x^3/3)(sinmx)+(2x^2/3)cosmx-integral -cosmx(4/3x)
then I did it again
u=2x^2/3
dv=cosmx
du=12x+2x^2/3(-sinmx)/9 (this doesn't look like its going to help)
v=sinmx
Thank you very much
i thought we were finding $\frac {1}{3} \int x^3 \cos mx~dx$, but now i see that it was $\int x^2 \cos mx~dx$. in that case, i suggest what Krizalid does. in any case, you didn't do what i said, which is why you ended up in trouble. i said differentiate the polynomial and integrate the trig function ("always"), you did the opposite. you will keep getting a worst looking integral if you continue doing it the way you're doing it. switch u and dv
6. We have to do it the long way Does this look right to you guys?
Integral x^2cosmxdx
u=x^2
dv=cosmxdx
du=2x
v=sinmx
integral x^2cosmx=x^2sinmx-integral sinmx(2x)
u=2x
dv=sinmx
du=2
v=-cosmx
integral x^2cosmx=x^2sinmx-[(2x)(-cosmx)] -integral(-cosmx(2))
=x^2sinmx+2xcosmx+ integral (-cosmx(2)
x^2sinmx+2xcosmx+ integral(-1)cosmx(2)
u=2
dv=-cosmx dx
du=0
v=-sinmx
=x^2(sinmx)+2xcosmx-integral (cosmx)(2)
=x^2sinmx+2xcosmx-[2(-sinmx)]-integral -sinx(0)
=x^2sinmx+2xcosmx+2sinmx
Thank you very much
7. Originally Posted by chocolatelover
We have to do it the long way Does this look right to you guys?
Integral x^2cosmxdx
u=x^2
dv=cosmxdx
du=2x
v=sinmx
integral x^2cosmx=x^2sinmx-integral sinmx(2x)
u=2x
dv=sinmx
du=2
v=-cosmx
integral x^2cosmx=x^2sinmx-[(2x)(-cosmx)] -integral(-cosmx(2))
=x^2sinmx+2xcosmx+ integral (-cosmx(2)
x^2sinmx+2xcosmx+ integral(-1)cosmx(2)
u=2
dv=-cosmx dx
du=0
v=-sinmx
=x^2(sinmx)+2xcosmx-integral (cosmx)(2)
=x^2sinmx+2xcosmx-[2(-sinmx)]-integral -sinx(0)
=x^2sinmx+2xcosmx+2sinmx
Thank you very much
did your professor say you have to do it the long way?
this is incorrect by the way. you took the integrals wrong. the red line above is your first mistake, there are others, but you have to fix the first one. look out for that mistake again. you have to use substitution for the integral, you don't have to show it, and it's not that hard to do in your head
8. Hey chocolatelover
On integration by parts, always we have to make a "little integration"
Since $dv=\cos(mx)\,dx$, then $v=\frac1m\sin(mx)$ (the constant is omitted on the integration by parts), why?
Simple, imagine you have to integrate $\int\cos(mx)\,dx$, so first, let's set a change of variables according to $u=mx\implies du=m\,dx\implies dx=\frac1m\,du$, so now we substitute and yields
$\int\cos(mx)\,dx=\frac1m\int\cos u\,du$, and this is easy to integrate, 'cause it's a basic integral, so it yields $\frac1m\int\cos u\,du=\frac1m\sin u+k$, finally back substitute $u$ into $mx$ and we happily get $\int\cos(mx)\,dx=\frac1m\sin(mx)+k$
Is it clearer now?
9. Hi Krizalid and Jhevon,
Sorry. I'm still kind of confused. I tried to do it agian.
I let u=x^2
dv=cosmxdx
du=2x
v=sinmx
Then I said that X^2cosmxdx=x^2sinmx-integral sinmx(2x)
Then wouldn't I do a u subtitution?
However, I don't know what to let u equal. If I let u equal sinmx then du would be cosmx, and that wouldn't work. If I let u equal 2x then du would be 2 and that wouldn't work. My professor told us that we had to do it the long way. Do you see what I'm doing wrong?
Thank you very much
10. Originally Posted by chocolatelover
dv=cosmxdx
v=sinmx
Didn't you see my previous post?
11. Yes, but I don't understand it. Sorry. Is that the first step and then do you integrate it by using the integration by parts method?
12. Would this work?
u=x^2
dv=cosmx dx
du=2x
v=sinmx
the integral x^2cosmx dx=x^2(sinmx)-the integral (sinmx)(2x)
the integral sinmx(2x)=2x(-cosmx)-the integral (-2cosmx)
u=2x
dv=sinmx
du=2
v=-cosmx
the integralx^2cosmxdx=x^2sinmx-2x(-cosmx)-the integral(-2cosmx)
the integral (-2cosmx
u=-2
dv=cosmx
du=0
v=sinmx
the integral x^2cosmxdx=x^2sinmx-2x(-cosmx)-2(sinmx-integral(sinmx)(0)=
x^2sinmx+2xcosmx-2sinmx+c
Thank you
13. Originally Posted by chocolatelover
Would this work?
u=x^2
dv=cosmx dx
du=2x
v=sinmx
the integral x^2cosmx dx=x^2(sinmx)-the integral (sinmx)(2x)
the integral sinmx(2x)=2x(-cosmx)-the integral (-2cosmx)
u=2x
dv=sinmx
du=2
v=-cosmx
the integralx^2cosmxdx=x^2sinmx-2x(-cosmx)-the integral(-2cosmx)
the integral (-2cosmx
u=-2
dv=cosmx
du=0
v=sinmx
the integral x^2cosmxdx=x^2sinmx-2x(-cosmx)-2(sinmx-integral(sinmx)(0)=
x^2sinmx+2xcosmx-2sinmx+c
Thank you
again, you keep doing what we said you're not supposed to do. the integral of cos(mx) IS NOT sin(mx) and the integral of sin(mx) IS NOT -cos(mx)
By substitution (see Krizalid's post):
$\int \sin mx ~dx = - \frac {1}{m} \cos mx + C$ and $\int \cos mx ~dx = \frac {1}{m} \sin mx + C$
until you get the pieces right, you will never get the whole solution right
14. Okay, let's do this.
$\int x^2\cos(mx)\,dx$
Set $y=mx\implies dy=m\,dx$, then the integral becomes to
$\int x^2\cos(mx)\,dx=\frac1{m^3}\int y^2\cos y\,dy$
Now, we're gonna integrate $y^2\cos y$, so we'll use integration by parts twice.
Let $u=y^2\implies du=2y\,dy$ & $dv=\cos y\,dy\implies v=\sin y$, the integral becomes to
$\int y^2\cos y\,dy=y^2\sin y-2\underbrace{\int y\sin y\,dy}_I$
Apply integration by parts again on $I$
Let $u=y\implies du=dy$ & $dv=\sin y\,dy\implies v=-\cos y$, this yields
$I=-y\cos y+\int\cos y\,dy=-y\cos y+\sin y$ (we'll add the constant at the end of the integration)
Back substitute
$\int y^2\cos y\,dy=y^2\sin y-2(\sin y-y\cos y)$, then
$\int x^2\cos(mx)\,dx=\frac1{m^3}[m^2x^2\sin(mx)-2(\sin(mx)-mx\cos(mx))]+k$
Which is equal to
$\int x^2\cos(mx)\,dx=\frac1{m^3}[2mx\cos(mx)+(m^2x^2-2)\sin(mx)]+k$; now multiply by $\frac1{m^3}$ and we are done.
By the way, I suggest you brush up some of derivatives, it's quite dangerous get start on integration without a fully domain of derivatives.
15. I understood of all that except, how did you get 1/m^3? I understand where you got 1/m from, but how did it change to m^3? Also, at the end of the problem you said to multiply it by 1/m^3. Wouldn't you either leave it how it is or multiply by m^3 on both sides?
Thank you very much
Regards
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2016-07-25 08:26:44
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https://paperswithcode.com/method/ttur
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Optimization
# Two Time-scale Update Rule
Introduced by Heusel et al. in GANs Trained by a Two Time-Scale Update Rule Converge to a Local Nash Equilibrium
The Two Time-scale Update Rule (TTUR) is an update rule for generative adversarial networks trained with stochastic gradient descent. TTUR has an individual learning rate for both the discriminator and the generator. The main premise is that the discriminator converges to a local minimum when the generator is fixed. If the generator changes slowly enough, then the discriminator still converges, since the generator perturbations are small. Besides ensuring convergence, the performance may also improve since the discriminator must first learn new patterns before they are transferred to the generator. In contrast, a generator which is overly fast, drives the discriminator steadily into new regions without capturing its gathered information.
#### Papers
Paper Code Results Date Stars
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2021-11-27 01:51:49
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https://stats.stackexchange.com/questions/73065/correlation-coefficient-for-non-dichotomous-nominal-variable-and-ordinal-numeric
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# Correlation coefficient for non-dichotomous nominal variable and ordinal/numeric variable
I've already read all the pages in this site trying to find the answer to my problem but no one seems to be the right one form me...
First I explain you the kind of data I'm working with...
Let's say that I have an array vector with several names of city, one for each of 300 users. I also have another array vector with scores response to a survey of each user or a continuous value for each user.
I would like to know if exist a correlation coefficient that compute the correlation between these two variables so, nominal and numeric/ordinal variables
I've searched on the Internet and in some pages they suggest to use the contingency coefficient or Cramer's V or Lambda coefficient or Eta . For each of this measure the just say that they could be applied for such data in which we have a nominal variable and interval or numerical variable. The thing is that searching and searching, trying to understand every one of them, sometime is written or watching the examples that they are reasonable to use them if you have dichotomous nominal variable, except for Cramer's V, other time is not written any requirement for the type of data. A lot of other pages say that is right to apply regression instead, that is right, but I would just simply like to know if there is a coefficient like pearson/spearman for this kind of data.
I also think that is no so properly to use Spearman Correlation coeff since the cities are not sortable.
I have also built the function of Cramer'sV and Eta by myself (I'm working with Matlab) but for Eta they don't talk about any p-value to see if the coefficient is statistically significant...
In the matlabWorks site there is also a nice toolbox that says to compute eta^2 but the kind of input it needs is not understandable.
Is here someone that have done a test like mine? If you need more detail to understand the kind of data I'm using just ask me and I'll try to explain you better.
• Cramérs V is for two nominals. What is bad about regression? Take the numeric variable as response and regress it to the nominal (using dummies). Look at the $R^2$ and the associated global F-test. – Michael M Oct 17 '13 at 13:14
• Nothing wrong with regression, but as we have already that measure we would like to check it in another way just as double check with a correlation coefficient....thanks for the answer – cristis Oct 17 '13 at 14:29
• You haven't said anything specific about your "numeric/ordinal" variable. What makes you to pose it ordinal? numeric? – ttnphns Oct 18 '13 at 4:15
• ordinal beacuse I have a variable coming from a survey test so its range is -4,4, you can also think it as interval but this kind of survey variable are considered mostly as ordinal and the others are numeric, in specific continuous as they are features extracted. – cristis Oct 18 '13 at 7:49
• SEE ALSO related question stats.stackexchange.com/questions/23938/… – ttnphns Dec 27 '13 at 8:00
Nominal vs Interval
The most classic "correlation" measure between a nominal and an interval ("numeric") variable is Eta, also called correlation ratio, and equal to the root R-square of the one-way ANOVA (with p-value = that of the ANOVA). Eta can be seen as a symmetric association measure, like correlation, because Eta of ANOVA (with the nominal as independent, numeric as dependent) is equal to Pillai's trace of multivariate regression (with the numeric as independent, set of dummy variables corresponding to the nominal as dependent).
A more subtle measure is intraclass correlation coefficient (ICC). Whereas Eta grasps only the difference between groups (defined by the nominal variable) in respect to the numeric variable, ICC simultaneously also measures the coordination or agreemant between numeric values inside groups; in other words, ICC (particularly the original unbiased "pairing" ICC version) stays on the level of values while Eta operates on the level of statistics (group means vs group variances).
Nominal vs Ordinal
The question about "correlation" measure between a nominal and an ordinal variable is less apparent. The reason of the difficulty is that ordinal scale is, by its nature, more "mystic" or "twisted" than interval or nominal scales. No wonder that statistical analyses specially for ordinal data are relatively poorly formulated so far.
One way might be to convert your ordinal data into ranks and then compute Eta as if the ranks were interval data. The p-value of such Eta = that of Kruskal-Wallis analysis. This approach seems warranted due to the same reasoning as why Spearman rho is used to correlate two ordinal variables. That logic is "when you don't know the interval widths on the scale, cut the Gordian knot by linearizing any possible monotonicity: go rank the data".
Another approach (possibly more rigorous and flexible) would be to use ordinal logistic regression with the ordinal variable as the DV and the nominal one as the IV. The square root of Nagelkerke’s pseudo R-square (with the regression's p-value) is another correlation measure for you. Note that you can experiment with various link functions in ordinal regression. This association is, however, not symmetric: the nominal is assumed independent.
Yet another approach might be to find such a monotonic transformation of ordinal data into interval - instead of ranking of the penultimate paragraph - that would maximize R (i.e. Eta) for you. This is categorical regression (= linear regression with optimal scaling).
Still another approach is to perform classification tree, such as CHAID, with the ordinal variable as predictor. This procedure will bin together (hence it is the approach opposite to the previous one) adjacent ordered categories which do not distinguish among categories of the nominal predictand. Then you could rely on Chi-square-based association measures (such as Cramer's V) as if you correlate nominal vs nominal variables.
And @Michael in his comment suggests yet one more way - a special coefficient called Freeman's Theta.
So, we have arrived so far at these opportunities: (1) Rank, then compute Eta; (2) Use ordinal regression; (3) Use categorical regression ("optimally" transforming ordinal variable into interval); (4) Use classification tree ("optimally" reducing the number of ordered categories); (5) Use Freeman's Theta.
• P.S. There is a good short overview about ordinal variable approaches in Jeromy Anglim's blog jeromyanglim.blogspot.ru/2009/10/… – ttnphns Oct 18 '13 at 3:58
• One measure of association between an ordinal and a nominal is called "Freeman's $\theta$". Unfortunately, I don't have any open access reference at hand. – Michael M Oct 18 '13 at 9:02
• @Michael thanks, here I found a paper "A further note on freeman's measure of association" moreno.ss.uci.edu/22.pdf – ttnphns Oct 18 '13 at 10:30
• For more information on Freeman's theta and an R package that includes the statistic, see this Cross Validated question. – Sal Mangiafico Aug 23 '17 at 19:33
• @ttnphns Sorry, could you please answer this question: stats.stackexchange.com/questions/363543/… Thanks a lot. – ebrahimi Aug 24 '18 at 8:11
Do a one-way anova on the response, with city as the grouping variable. The $F$ and $p$ it gives should be the same as the $F$ and $p$ from the regression of the response on the dummy-coded cities, and $SS_{between\, cities}/SS_{total}$ should equal the multiple $R^2$ from the regression. The multiple $R$ is the correlation of city with the response.
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2019-05-22 16:51:59
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https://www.physicsforums.com/threads/confusion-about-partial-derivatives.609705/
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1. May 29, 2012
### Hassan2
Dear all,
I have a confusion about partial derivatives.
Say I have a function as
$y=f(x,t)$
and we know that
$x=g(t)$
1. Does it make sense to talk about partial derivatives like $\frac{\partial y}{\partial x}$ and $\frac{\partial y}{\partial t}$ ?
I doubt, because the definition of partial derivative is the change in the function due to the change on the selected variable ( other variables are kept constant).
2. If it does not make sense, then how in Euler–Lagrange equation we use the partial derivatives with respect to a function(x(t)) and its derivative(x'(t)) while they depend on one another.
2. May 29, 2012
### chiro
Hey Hassan2.
For 1. Yes it makes sense to do both but you need to factor in things like the chain rule for this particular example: as long as you are taking into account these kinds of factors, then yes it's ok. Your partial with respect to t can take into account your g(t) using the chain rule.
If the function is differentiable in the the region you are considering, the derivative will make sense: it's guaranteed to as a consequence of differentiability holding.
3. May 29, 2012
### Hassan2
Thanks Chiro,
Yes, I checked the derivation of Euler-Largrange equation once again, and i found that the derivatives wit respect to x and x' appear as a result of applying chain rule.
Thanks again.
4. May 29, 2012
### HallsofIvy
Staff Emeritus
The notation $\partial f/\partial x$ means the derivative of f with respect to x while holding x constant- and ignoring the fact that x is a function of t. We are really dealing with the "form" of f rather than the content.
The same is true of $\partial f/\partial t$. However, we can, using x= g(t), think of f as a function of t only- f(x, t)= f(g(t), t). In that case, by the chain rule,
$$\frac{df}{dt}= \frac{\partial f}{\partial t}+ \frac{\partial f}{\partial x}\frac{dx}{dt}$$
Example: if $f(x,t)= 3tx^2+ e^x$ then $\partial f/\partial x= 6tx+ e^x$ and $\partial f/\partial t= 3x^2$. That has nothing to do with x being a function of t or vice-versa.
But if we also know that $x= g(t)= 2t^2+ t$ we can write $f(t)= 3t(2t^2+ t)+ e^{2t^2+ t}= 12t^5+ 12t^4+ 3t^3+ e^{2t^2+ t}$ so that the derivative is
$$\frac{df}{dt}= 60t^4+ 48t^3+ 9t^2+ (4t+ 1)e^{2t^2+ t}$$
Or, you could use the chain rule as I said:
$\partial f/\partial t= 3x^2$ and $\partial f/\partial x= 3tx+ e^x$, as above, while $dx/dt= 4t+1$ and so
$$\frac{df}{dx}= 3x^2+ (3tx+ e^x)(4t+ 1)$$
and, replacing x in that with $2t^2+ t$
$$\frac{df}{dx}= 3(2t^2+ t)+ (3t(2t^2+ t)+ e^{2t^2+ t})(4t+1)$$
gives the same thing.
5. May 29, 2012
### algebrat
Strictly speaking, one should not write f(t)=f(g(t),t). The function on the left should have gotten a new name. That is the source of some confusion.
6. May 29, 2012
### arildno
Agreed!
What we should write, is something like:
F(t)=f(g(t),t)
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2017-10-20 01:03:10
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https://itectec.com/database/filter-1-columns-data-into-3-columns/
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# Filter 1 columns data into 3 columns
oracle
I need to filter data in 1 column and have an output of three columns (Column A,Column B and Column C).
The output columns should display values of 11 for column A, values of 12 for column B, and Total of A & B for column C.
The current query I have written:
select count(zcsms_studentgender) as MaleStudents,
count(zcsms_studentgender) as FemaleStudents,
count(zcsms_studentgender) as TotalStudents
from zcsms_studentdefinition
where (zcsms_studentgender = '11' or zcsms_studentgender = '12')
and zcsms_studentactive = 'Y';
I can't seem to add filtering into the first two columns (MaleStudents & FemaleStudents) to display only count values of 11 (male) and 12 (female).
It's a report based on student gender. The value for male students is 11, the value for female student is 12. These are stored in 1 table, 1 column. My output has to filter these into a report that will display as Column A (Male Students count) Column B (Female Students count) and Column C
(total count of A and B).
Just use SUM() and a CASE statement:
SELECT sum(case when zcsms_studentgender=11 then 1 else 0 end) as malestudents,
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2021-10-20 03:20:44
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|
https://firas.moosvi.com/oer/physics_bank/content/public/019.Magnetism/Electromagnetic%20Induction/OSUPv2p13_52/OSUPv2p13_52.html
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# Uniformly Decreasing Magnetic Field#
The magnetic field at all points within the cylindrical region whose cross-section is indicated in the accompanying figure starts at magnetic field of 3.0 $$\textrm{ T}$$ and decreases uniformly to zero in 30 $$\textrm{ s}$$.
## Part 1#
What is the magnitude of the electric field when $$r$$, the distance from the geometric center of the region, is $${{params.r }} \textrm{ cm}$$?
## Part 2#
What is the direction of this electric field?
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2022-09-28 08:50:36
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https://physics.stackexchange.com/questions/436240/iqhe-quantized-conductance-and-zeeman-splitting
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# IQHE, quantized conductance, and zeeman splitting
I've been trying to understand IQHE by reading these lecture notes by David Tong.
Mainly, I was trying to understand the quantized hall resistivity in terms of the number of Landau levels crossing the fermi energy.
Then, I began thinking about why spin induced Zeeman splitting is never really mentioned in the context of IQHE.
The lecture notes say that it's because typically the Zeeman splitting is very small and it polarizes the spin of the electron.
I think the spin based splitting of energy states still confuses me because in my mind, with the spins of electrons taken into account, you have twice as many energy states crossing the fermi energy.
The filling factor in IQHE is the number of landau levels crossing the fermi energy (as shown in the image below). To me, the spin Zeeman splitting seems to double that number.
• It looks like you read the notes backwards. They say the Zeeman splitting is large, not small. Nov 17 '18 at 9:47
• $\uparrow$ Which page? Nov 17 '18 at 10:00
This doubling indeed happens in the "quantum spin Hall effect", but those systems are at zero magnetic field (and moreover enjoy time reversal symmetry, which is key). However, in the usual quantum Hall effect, there is a huge static magnetic field, which polarizes all the low energy electrons (the Zeeman splitting is large because it goes like $$|B|$$). Therefore, they are essentially spinless. See this review http://www.damtp.cam.ac.uk/user/tong/qhe/qhe.pdf , section 1.4.
The definition of filling factor is, $$\nu \equiv \frac{\text{number of particles}}{\text {number of flux quanta}}$$. I guess even if you include the Zeeman splitting this is not going to change.
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2022-01-17 17:23:01
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|
http://cs.stackexchange.com/questions/9711/design-of-self-loops-and-final-states-in-fsm
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# design of self-loops and final states in fsm
I am learning about automata and finite state machines. Consider the following automaton, that accepts the word 'ab', does not have to be infinite, just once:
alphabet: 'a','b' states: 1,2, 3 (3 is the final state and 1 is the initial state)
transitions:
state 1, symbol a, state 2
state 2, symbol b, state 3
First part of my questions:
Question 1. Is it required to add self-loops at 1 for 'b', and a self-loop for 'a' at 2?
Question 2. What about at state 3(final)? Should I add self-loops for 'a' and 'b'?
I basically need to know, how to design my final state. So that even if my alphabet was expanded (say a, b, c), and I have a 'dump' state, with the following transitions:
state 1, symbol a, state 2
state 1, symbol b or c, state dump
state 2, symbol b, state 3
state 2 symbol a or c, state dump
Question 3. Now from final state 3, should I add a transition to dump state, with symbol values a,b,c ???
-
Your automaton is fine, you don't have to add anything. Of course you can add a dead state (outdegree = 0). – saadtaame Feb 12 '13 at 19:03
1, 2. You don't have to show all possible transitions. Absent transitions generally imply rejection. So if you get b in state 1, and there is no loop for it, it means the input is rejected. 3. Yes, if you have a dump state. But you don't need to. – Paresh Feb 12 '13 at 19:20
Your original automaton is fine, it accepts the singleton $\{ab\}$. Here is a pictorial representation:
The self-loop at 1 labeled with $b$ changes the language to $\{ab, bab, bbab, bbbab, \dots\}$.
@PiotrL. The $d$ is a typo in the automata... Glad it helps. – saadtaame Feb 13 '13 at 0:21
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2014-11-27 17:02:59
|
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http://physics.stackexchange.com/questions/9840/detection-of-w-and-z-bosons
|
# Detection of W and Z bosons
What specific behaviour confirmed the existence of the W and Z bosons at the UA1 and UA2 experiments?
Thanks!
-
See an UA2 paper from 1983 about Z0, ccdb4fs.kek.jp/cgi-bin/img/allpdf?198308375 , as an example. They found 8 events with the right mass around 91 GeV that decayed into e+ e-, relatively to the non-Z background of 0.7 events or so only. So clearly there had to be something at the mass indicated and it agreed with the electroweak theory. – Luboš Motl May 14 '11 at 6:32
@ Luboš Motl: most enlightening, thank you very much. – Richard Terrett May 14 '11 at 6:47
Since Lubos did not make an answer of his comment I would suggest reading the "discovery" paragraph in the Wiki article.
The "specific behaviour" are the characteristic decay modes of each particle, that clustered at the masses finally assigned, as the reference in the comment above shows.
You need a library to go to the original discovery references, but the decay modes are in the article in this link.
-
I created this question after I read the discovery section of the Wikipedia page and found myself no wiser as to the nature of the 'unambiguous signals' described therein (i.e. whether it was calorimetry of decay products or something else). Luboš' comment essentially answers the question however as it is not an answer I have no way of accepting it. How should I proceed? Should I accept your answer instead? – Richard Terrett May 16 '11 at 8:29
Well, for neatness maybe you should, since it is essentially the same answer as in Lubos' comment. – anna v May 16 '11 at 9:20
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2014-04-18 18:18:43
|
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|
https://zbmath.org/?q=an:0609.62054
|
# zbMATH — the first resource for mathematics
##### Examples
Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.
##### Operators
a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
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Multivariate estimation with high breakdown point. (English) Zbl 0609.62054
Mathematical statistics and applications, Proc. 4th Pannonian Symp. Math. Stat., Bad Tatzmannsdorf/Austria 1983, Vol. B, 283-297 (1985).
[For the entire collection see Zbl 0583.00028.] Suppose we have n data points in p dimensions, and we want to estimate their location using an estimator T that is affine equivariant, which means that $$T(Ax\sb 1+b,...,Ax\sb n+b)=AT(x\sb 1,...,x\sb n)+b$$ for all vectors b and nonsingular matrices A. The breakdown point of T is the smallest fraction of contaminated data that can carry T over all bounds. The breakdown point of least squares (the arithmetic mean) is 0, and for M-estimators it is at most $1/(p+1)$. The ”outlyingness-weighted mean” of {\it W. Stahel} [Breakdown of covariance estimators. Res. Rep. 31, Fachgruppe Stat., ETH Zürich (1981)] and {\it D. Donoho} [Breakdown properties of multivariate location estimators. Ph. D. qualifying paper, Harvard Univ. (1982)] is affine equivariant and its breakdown point equals 50 %, the highest possible value. The purpose of the present paper is to introduce another estimator with these properties, namely the center of the least-volume ellipsoid covering half of the data. A variant is to find that half of the data for which the empirical covariance matrix yields the smallest possible tolerance ellipsoids. These estimators automatically provide robust covariance estimators. They are inefficient at a Gaussian model, but this can easily be repaired by using a one-step reweighted least squares estimator afterwards. More information regarding algorithms and applications of these high- breakdown estimators can be found in Chapter 7 of the author and {\it A. Leroy}, Robust regression and outlier detection. John Wiley, New York, to appear in September 1987. An important application is the identification of leverage points in regression analysis.
##### MSC:
62F35 Robustness and adaptive procedures (parametric inference) 62H12 Multivariate estimation
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2016-05-03 14:27:02
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http://circuits.solved-problems.com/tag/voltage-division-rule/
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# Voltage Divider - Voltage Division Rule
The voltage division rule (voltage divider) is a simple rule which can be used in solving circuits to simplify the solution. Applying the voltage division rule can also solve simple circuits thoroughly. The statement of the rule is simple:
Voltage Division Rule: The voltage is divided between two series resistors in direct proportion to their resistance.
It is easy to prove this. In the following circuit
Voltage Divider
the Ohm's law implies that
$v_1(t)=R_1 i(t)$ (I)
$v_2(t)=R_2 i(t)$ (II)
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2013-06-18 05:25:16
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https://tug.org/pipermail/tex-k/2018-December/002965.html
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# [tex-k] minor errors in TeXbook Appendix G concerning genfrac
Karl Berry karl at freefriends.org
Wed Dec 12 01:02:53 CET 2018
Hi Sophie,
1. TeXbook page A444 line 18 from the bottom mentions "let
$(\lambda, \rho)$ be the left and right delimiters". However it
never uses those names $\lambda$ and $\rho$ again, instead only
mentioning the "left delimiter" and "right delimiter" in step 15e on
page A445.
Well, they are trivially used at the end of item 15:
The values of $\lambda$ and $\rho$ are null unless
the fraction is with delims.''
But your point is taken nevertheless. I will consult.
2. TeXbook page A445 line 13 says to replace "the generalized
fraction by an Inner atom...". I believe this text is meant to
instruct replacing the genfrac by an *Ord* atom.
...
3. Relatedly, TeXbook page A445 line 14 (end of Rule 15e) does not
indicate how to proceed after this processing. In contrast, Rules 1
through 14 all end with an instruction ...
Ack, thanks. --karl.
P.S. Are you reimplementing math typesetting in some new system? Just
curious how you ended up here.
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2022-08-18 14:52:32
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https://www.gamedev.net/forums/topic/174803-readwrite-one-bit-at-a-time/
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• ### Announcements
#### Archived
This topic is now archived and is closed to further replies.
# Read/Write one BIT at a time??
## Recommended Posts
n0ob 176
Is there a freely available or standard library which allows the reading/writing of single bits at a time? I mean, if there isn''t, I could probably force myself to write my own system to do that. Like, store bits in a buffer, then flush a byte at a time.. which I actually know how to do ^_^. But it would be easier if there''s one out there already. Thanks! Kings of Chaos
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ApochPiQ 23011
A byte is the smallest addressable piece of data on a computer. You cannot access data bit by bit. However, you can use bitmasks to mask out all the bits but the one you are interested in; use 2^(bit index) as the mask and perform an AND between this mask and your byte. The final value will clear all bits but the one you are after, and that bit will be 1 or 0 depending on its value in the original byte.
For example, if I have the byte 10110110, and mask it with the bitmask 00010000, the bitwise AND (& operator in C/C++/Java) will clear out all the bits and leave me with 00010000. I can then test this value; if it is 0, the original bit was 0. If the value is not zero, then the original bit was 1.
Bitmasking is a trivial programming concept and so you are not very likely to find any premade code.
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Wyrframe 2426
It''s a dead-simple thing to do. Best to do it yourself.
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Flarelocke 410
Either std::vector or std::bitset
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foofightr 130
Note that reading/writing bits (especially writing) is considerably slower than bytes, as counter-intuitive as that may sound.
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antareus 576
I wrote a Bitstream at work, however I cannot share it.
It was pretty easy, but testing it for correctness isn't the easiest thing in the world.
[edited by - antareus on August 14, 2003 7:25:39 AM]
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BeanDog 1065
You could just use bitfields. I''m not sure of the exact syntax, but it''s something like this:
struct S{ unsigned b1 : 1; unsigned b2 : 1; unsigned b3 : 1; unsigned b4 : 1; unsigned b5 : 1; unsigned b6 : 1; unsigned b7 : 1; unsigned b8 : 1;};int main(){ S s; s.b1 = 1; s.b2 = 0; //etc. s takes up only one byte.}
Or you could use the following to work on int''s or DWORD''s:
#define BITMASK(n) (1<<n)#define ISBITSET(v,n) ((v) & BITMASK(n))#define SETBIT(v,n) ((v) |= BITMASK(n))#define UNSETBIT(v,n) ((v)=~(SETBIT(~(v),n)))#define CLEARBITS(v) ((v)=0)int main(){ unsigned int n; CLEARBITS(n); SETBIT(n,1); UNSETBIT(n,1); //etc.}
All source is written off the top of my head and is not guaranteed to be perfect. Please post corrections :-)
~BenDilts( void );
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n0ob 176
yes.. thanks for the help. I guess I will write my own, USING BITMASKS!! or whatever
Kings of Chaos
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HFX 122
May I inquire as to why you want to write a bit at a time. Writing a bit at a time will be a lot slower then using a whole byte or even an int. So unless your trying to run in something like an embedded device with very little storage, writing bit by bit is not recommended. Infact it is a detrement. It would probably be better read a whole byte at a time even if you do waste.
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n0ob 176
I am aware of the pros/cons of this.. I need it because I''m lazy. I''m just doing an experiment, and could care less if the write operation for a single bit took 50 ms. I KNOW already. I asked if there was one, i didn''t ask for you to convince me to not use it. Thanks for your help, I need no more.
Kings of Chaos
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2017-08-24 05:08:51
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https://www.groundai.com/project/on-the-radius-of-habitable-planets/
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On the radius of habitable planets
# On the radius of habitable planets
###### Key Words.:
planetary systems - planetary systems: formation
1
## Abstract
Context:The conditions that a planet must fulfill to be habitable are not precisely known. However, it is comparatively easier to define conditions under which a planet is very likely not habitable. Finding such conditions is important as it can help select, in an ensemble of potentially observable planets, which ones should be observed in greater detail for characterization studies.
Aims:Assuming, as in the Earth, that the presence of a C-cycle is a necessary condition for long-term habitability, we derive, as a function of the planetary mass, a radius above which a planet is likely not habitable. We compute the maximum radius a planet can have to fulfill two constraints: surface conditions compatible with the existence of liquid water, and no ice layer at the bottom of a putative global ocean. We demonstrate that, above a given radius, these two constraints cannot be met.
Methods:We compute internal structure models of planets, using a five-layer model (core, inner mantle, outer mantle, ocean, and atmosphere), for different masses and composition of the planets (in particular, the Fe/Si ratio of the planet).
Results:Our results show that for planets in the Super-Earth mass range (1-12 ), the maximum that a planet, with a composition similar to that of the Earth, can have varies between 1.7 and 2.2 . This radius is reduced when considering planets with higher Fe/Si ratios and taking radiation into account when computing the gas envelope structure.
Conclusions:These results can be used to infer, from radius and mass determinations using high-precision transit observations like those that will soon be performed by the CHaracterizing ExOPlanet Satellite (CHEOPS), which planets are very likely not habitable, and therefore which ones should be considered as best targets for further habitability studies.
## 1 Introduction
Since the discovery of the first extrasolar planet around a main-sequence star (Mayor and Queloz 1995), observational programs have led to the discovery of lower and lower mass planets (Mayor et al. 2011), some of them located in the so-called habitable zone at a location where it is believed that liquid water could exist on the planetary surface. The possible presence of liquid water is in turn believed to be important for the habitability of planets, as it is required for life (as we know it) to exist and develop.
The habitability of planets is - however - an ill-defined concept, and there is nowadays no clear definition of it. Indeed, the presence of liquid water only, while it could be a necessary condition, is probably not enough for habitability. What is, on the other hand, quite easy to define, is non-habitability, and one may easily find different conditions under which a planetary surface is very likely not habitable. In the context of habitability studies, the definition of such non-habitability conditions is useful, for example as it may allow to select future observational targets when large-scale facilities allow us to directly characterize in great detail the most promising extrasolar planets. The goal of this study is to provide a clear non-habitability criterion, that can be easily derived once the mass and the radius of an extrasolar planet (located for example in the habitable zone) is known. In addition, we will see that the simultaneous determination of the central star composition, as well as the observation of Rayleigh scattering in the planetary atmosphere, can be used to derive stronger constraints on planetary habitability.
Indeed, it has been recognized for many years that one key element for habitability, at least in the case of the Earth, but also likely for many planets, is the presence of the Carbon cycle (see e.g., Kasting 2010). Indeed, on Earth, the C-cycle acts as a very important temperature stabilizing process and buffers the surface temperature at values close to those allowing liquid water. This is especially important since the luminosity of a star increases with time and, without any stabilization process, it could be difficult to maintain liquid water at a surface of any planet during an appreciable amount of time.
The C-cycle has been presented in many papers and books (see e.g., Kasting 2010; Pierrehumbert 2010), and we will not describe it in detail. The key process during this C-cycle is the weathering of silicates, which converts silicates to carbonates, as a result of the interaction between atmospheric CO dissolved in water and silicates of the planetary rocky surface. Carbonates then precipitate at the bottom of oceans and can be engulfed in subduction zones and transported toward high-temperature regions of planetary mantle. The cycle is then completed as carbonates are destabilized at high temperature, producing CO that can finally be sent back to the atmosphere by volcanoes.
The stabilization effect of the C-cycle comes from both the dependence of silicate weathering on temperature2 and the strong greenhouse effect of CO. If the surface temperature decreases, the removal of CO through silicate weathering is suppressed, while CO is still released into the atmosphere by volcanoes. This increases the CO concentration in the atmosphere and eventually the temperature. On the contrary, if the surface temperature increases, then silicate weathering increases, which decreases the CO atmospheric concentration and greenhouse effect, thus lowering the surface temperature. Silicate weathering therefore requires a reaction between CO dissolved in oceans, or directly with CO present in the atmosphere, and rocks from the planetary mantle. This, in turn, is only possible if there is a physical interface between liquid water (or atmosphere) and rocks.
Water-rich planets (or ocean planets) are covered by a global ocean. If the amount of water is large enough (the exact value depending on the planetary gravity), the pressure at the bottom of the global ocean is so large that a layer of high pressure ice (ice VII) appears3. This effectively prevents any contact between silicates and liquid water (and therefore atmospheric CO that could be dissolved in the ocean), and is likely to suppress silicate weathering. As a result, the C-cycle and its stabilization effect cannot exist. The working hypothesis of this study is therefore that a necessary condition for habitability is the presence of C-cycle. As a consequence, planets with a high enough water content, and therefore a high pressure ice layer at the bottom of a global ocean, are considered not habitable4.
The radius of a planet depends mainly on its composition. The effect of the temperature profile has been shown to have a small influence on it (see e.g., Sotin et al. 2007, Seager et al. 2007, Grasset et al. 2010, Valencia et al. 2010). For a given mass, a large radius implies either the presence of a lot of water, or the presence of a (relatively) massive gaseous envelope, or both. In the first case, as we have argued before, habitability is hindered by the presence of a high pressure ice layer at the bottom of the global ocean. In the second case, the temperature and pressure at the bottom of the atmosphere are too large to allow for the presence of liquid water. In both cases, the planet is therefore likely not habitable, and will be considered accordingly in this paper. The process is exemplified in Fig. 1, which presents a simplified phase diagram of water, together with a model of a water-rich planet. This planet presents a thick layer of high pressure ice at the bottom of the ocean, which prevents the C-cycle from operating.
The goal of this study is to compute the maximum radius that a planet can have in order that 1) the pressure at the bottom of the ocean is low enough to prevent the apparition of high pressure ice and 2) the temperature and pressure at the surface of the planet is compatible with liquid water. It is important to note that such a calculation does not constrain in any way the past or future habitability of a given planet. In particular, following the arguments of Abbot et al. (2012), the suppression of silicate weathering on a planet may lead to a runaway greenhouse effect and massive loss of water. As a consequence, such a planet could become habitable in the future. However, the same planet would not be habitable today.
The paper is organized as follows: we present in Sect. 2 our internal structure model of planetary interior and planetary atmosphere, and some validation test we conducted in Sect. 3.1. In Sect. 3.2, we compute the maximum radius of habitable planet under rather extreme hypothesis on the planetary bulk composition (in particular for planets without Fe). This unlikely planetary composition is used to provide the overall maximum of planets, whatever the composition. In Sect. 3.3, assuming different Fe/Si ratios (that may reflect the composition of the central star), we compute the corresponding maximum radius. Finally, Sect. 4 is devoted to our discussion and conclusions.
## 2 Model
We compute the internal structure of planets that consists of five layers: a core, an inner mantle, an outer mantle, a water layer, and a gas envelope. The presence of a core depends on the assumed Fe/Si ratio in the planet. Indeed, for a given composition, the largest radius is obtained when the maximum of Fe is present in silicates. For each planetary composition, we therefore compute the structure with the smallest possible core, since we want to derive an upper boundary of the planetary radius. It appears that, for solar Mg/Si and Fe/Si ratios, all the available Fe can be present in the inner and outer mantle and no core is required to match the composition. For Fe/Si > 1.13 Fe/Si (for the chemical species we consider in the model), this is not any more possible and planets have an iron core. We emphasize the fact that assuming undifferentiated planets, as we do here, is probably far from the reality, at least for planets more massive than the Earth. However, we make this assumption as it provides us with an upper boundary of the possible planetary radii for a given mass and composition.
Our model of the four (three, if no core is required) innermost layers closely follows the model proposed by Sotin et al. (2007) and further improved by Grasset et al. (2010). The core is made of Fe (for simplicity, we do not include any inner/outer core dichotomy and do not include the effect of the presence of a volatile like S in the core), the inner mantle is made of perovskite (MgSiO/FeSiO) and wustite (MgO/FeO), the outer mantle is made of olivine (MgSiO/FeSiO) and entstatite (MgSiO/FeSiO). We assume the water mantle is made of pure water, and the planetary envelope is made of either H or H/He in solar composition, these latter being treated as ideal gases. We also consider other gaseous compositions to study their effect on the maximum radius. This choice of the planetary atmospheric composition is dictated by our goal to derive a maximum planetary envelope. As we will see later, other choices of atmospheric composition produce smaller radii.
We solve the standard internal structure equations:
drdP=1ρg (1)
dmdP=4πr2g (2)
and
where is the pressure, the radius, the mass interior to radius , the gravity, the density given by the equation of state (see below), the temperature, and the adiabatic gradient. The equations are solved, using the pressure as an independent variable, for each layer separately, and altogether provide the internal structure of the planet.
The temperature profile in the different layers follows an adiabat, with some temperature discontinuity at the transitions between the layers. At every transition, the temperature discontinuity follows the values suggested by Grasset et al. (2010), namely the temperature increases by 300 K at the inner/outer mantle transition, and by 1200 K at the mantle/water layer interface. The temperature is assumed to be continuous between the planetary surface and the gas envelope. As already demonstrated in different publications (e.g., Sotin et al. 2007, Seager et al. 2007, Grasset et al. 2010, Valencia et al. 2010), the planetary radius, at a given mass, hardly depends on the thermal profile. Given the relatively low depth of the water layer, we assume that this latter is isothermal. Finally, the gas envelope is treated as in Pierrehumbert and Gaidos (2011), and similar to the one developed in Viktorowicz and Ingersoll (2007). For a given choice of the surface conditions, we follow an adiabat until the skin temperature (, where is the surface temperature) is reached. Finally, we have also considered more detailed models of the planetary envelope including, for some of them, the irradiation from the star in a two-stream approach (see Guillot 2010).
The boundary conditions are as follows: we specify a surface temperature and pressure, as well as a central pressure. In addition, we assume that the pressure at the bottom of the ocean is the equilibrium pressure, at the ice VII/water ice transition. This equilibrium pressure is given by a Simon law, , with parameters given in Table 1. The pressure at the bottom of the ocean is computed based on the choice of the surface temperature and pressure. The thermal profile in the outer mantle is then computed, following an adiabat, until the transition to the inner mantle, at a pressure on the order of 22 GPa. Again, this transition depends on the temperature, following a similar law as for the ice/water transition: , with parameters given in Table 2. Finally, the temperature profile in the inner mantle and the core is computed following an adiabat until the assumed central pressure is reached. Once the thermal profile has been constructed, we integrate the internal structure equations, using the pressure as an independent variable. The only unknown in the model is therefore pressure at the core/inner mantle interface, if there is enough Fe to allow the presence of an iron core. We therefore use an iterative method to find the transition pressure that allows us to match the assumed composition of the planet (Fe/Si ratio, in particular). Once the internal structure of the non gaseous part of the planet has been derived, we compute the structure of the atmosphere by following an adiabat starting from the surface temperature and pressure, until the skin temperature is reached, following the method of Pierrehumbert and Gaidos (2011). The structure of the atmosphere is then used to derive the chord radius, corresponding to the place where the chord optical depth (the one that is observed by transit) is equal to 1. The opacity used for this computation is given by Bell and Lin (1994) as a function of the pressure and temperature in the gas phase. Note that the Bell and Lin (1994) opacity is mainly due to grains, and probably represents an upper limit of the real opacity (see e.g., Mordasini et al. 2012b and references therein). Again, this approach will determine an upper boundary of the radius. To test the effect of opacity, we have computed some additional models, using the opacity derived by Freedman et al. (2008), with the low metallicity table.
By following this procedure, i.e. varying the central pressure, we obtain a set of internal structure models in which each value of the central pressure provides a different total mass. To obtain a mass-radius relationship, we vary the central pressure. Finally, since we are interested in the radius of potentially-habitable planets, we assume that the surface temperature varies in the range from 275K to 375K, and the surface pressure varies in the range of to Pa. This range of surface conditions is rather arbitrary (and rather extended), and has been chosen to encompass what is believed to represent the range of surface conditions under which life could exist. The present study, however, could be easily extended to other surface conditions.
The derivation of the internal structure requires the specification of the equation of state (EOS), as well as the adiabatic gradient. The equations giving the pressure as a function of temperature and density are similar to the equations used by Sotin et al. (2007) for the inner and outer mantle, and are reproduced here for the sake of completion. We refer the reader to Sotin et al. (2007) for more details and justification of the use of these EOS. For the inner mantle, the EOS is given by the Mie-Gruneisen-Debye formulation (see Poirier 2000):
P=P(ρ,T0)+ΔP (4)
ΔP=γρ(E(T)−E(T0)) (5)
Missing or unrecognized delimiter for \right (6)
and
E=9nMMolP(TθD)3∫θD/T0x3ex(ex−1)dx (7)
where is the number of atoms in the considered compound. The Debye temperature is given by
θD=θD,0(ρρ0)γ (8)
and is given by .
For the outer mantle and the liquid water layer, the EOS is given by the Birch-Murnaghan of third order formulation:
Missing or unrecognized delimiter for \right (9)
where , is given by
ρT,0=ρ0exp(∫TT0α(x,0)dx) (10)
and . In these formulas, , , and are the reference pressure (1 bar), temperature (300K) and corresponding density, and the values of the other parameters are given in Tables 3 and 4. Note that since we use for water, the EOS is of second order only in this case.
Finally, we use the EOS derived by Belonoshko (2010) for pure Fe, which is similar to the Mie-Gruneisen-Debye EOS, but has a different thermal pressure term:
Missing or unrecognized delimiter for \right (11)
where the parameters are given in Table 3, and has the same definition as for the Mie-Gruneisen-Debye EOS. This EOS for Fe has been preferred to the one used by Sotin et al. (2007), as it allows to reproduce high-precision volumetric experiments of iron under high pressure and temperature closely (see Belonoshko 2010).
dTdP=γTKS (12)
where is the adiabatic bulk modulus, which is related to the isothermal bulk modulus through , being the thermal expansion coefficient. In the case of the inner and outer mantle, we use the values derived by Katsura et al. (2010):
α=(α0+(T−T0)a1)(ρ0ρ)δT. (13)
In these formulas, is the derivative of the expansion coefficient with respect to the temperature, and provides the dependance of with respect to the density. The parameter is the Debye temperature, which depends on the density (see above). The numerical values of the different parameters are given in Table 5. In the case of the iron core, the value of the thermal expansion coefficient are directly computed from Belonoshko’s EOS (Belonoshko 2010). Finally, we assume the water layer is isothermal. This assumption has no practical consequences given the low depth of the ocean (see Fig. 4).
As already mentioned above, the planetary composition is set by three parameters: the Mg/Si, Fe/Si molar ratios, and the composition of the planetary atmosphere. The water content is not an input parameter, but is derived from the constraint that the pressure at the bottom of the ocean is the crystallization pressure. These quantities are assumed, with one exception presented in Sect. 3.2, to be homogeneous in the entire planetary mantle.
## 3 Results
### 3.1 Test of the code: internal structure of an Earth-like planet
As an example and test of our code, in Fig. 2 we present the temperature, pressure, and mass as a function of the radius, for an Earth-like planet. The surface conditions are T=300K and P = 1 bar. The composition of the planet is similar to model 1 in Sotin et al. (2007), namely Mg/Si = 1.131 and Fe/Si = 0.986. We assume, for the sake of comparison, that 13% of the core is made of FeS (treated using the Mie-Gruneisen-Debye EOS), and 87% is made of pure Fe. We refer the reader to Sotin et al. (2007) for the justification of this slightly non-solar composition. To have a planet of 1 , the central pressure is equal to 370 GPa, the pressure at the core-mantle transition is equal to 130 GPa, and the pressure at the inner/outer mantle transition is equal to 23.31 GPa. The corresponding temperatures are 5092K, 3747K and 1789 K respectively. The mass obtained using these parameters is 1.007 , and the radius of the solid part is . The pressure, density, and temperature as a function of the radius are similar to the Preliminary Reference Earth Model (Dziewonski & Anderson, 1981), except for the inner/outer core structure that is not modeled in our code. This difference in the core structure (and mean density) also explains the slight shift (in radius) of the core-mantle boundary. For this type of planet, the maximum depth of the ocean (that is not taken into account in the numbers shown above) would be 150km, corresponding to a mass of 0.016 , which is more than 100 times the inventory of water on Earth.
In the same figure, we also present the case corresponding to the maximum radius, an iron free planet, whose inner core i made of MgO, and whose outer core is made of MgSiO. In this second example, the inner and outer mantle composition are not the same and the overall Mg/Si ratio is equal to 2.86, nearly three times the solar value. For this planet, the mass of the ocean is 0.0243 and its depth is 188 km. The ocean is larger because the gravity is weaker (the planet is less dense than in the previous case). As a consequence, the ice VII/liquid water transition is reached at higher depth. The pressure at the bottom of the ocean and the pressure at the inner/outer mantle transition are the same, since they only depend on the surface temperature, which is the same as in the previous case. The central pressure and temperatures are 143 GPa and 3722 K respectively. Finally, the mass of the gas envelope is , the transit radius of the planet is 1.30, and the radius of the solid/liquid planet (called inner planet in the following) is 1.11 .
### 3.2 Overall maximum radius of planets
To derive the overall maximum radius of a potentially habitable planet, in the sense we defined in the introduction, we first assume, unrealistically, that the planets do not contain any iron. Moreover, we only consider the less dense phases in the inner and outer mantle, namely MgO (inner core) and MgSiO (outer core). It should be noted that in this case, the Mg/Si ratio is different in the inner and outer core. We admit that this kind of planetary structure is likely to be unrealistic, but we only considered it to provide the overal maximum radius of a planet at a given mass.
Using the model we have described above, we now derive the maximum radius a planet can have to harbor both a C-cycle and surface conditions in the range defined above. The overall maximum is obtained, as mentioned above, assuming that the planet has a very peculiar composition, where only low density minerals - MgO for the inner mantle and Entstatite (Mg) for the outer mantle - are present in the planetary interior. Figure 3 shows the resulting mass-radius relation we obtain, with some transiting planets also plotted in the figure (Data were taken from exoplanet.org). As can be see from the figure, the maximum radius increases from 1.8 to 2.3 for planets ranging from 1 to 10 . Lower mass planets can have a larger radius, the increase of the mass-radius relationship being the result of the reduced gravity. Various studies (Mordasini et al. 2012a and references therein) have described such behavior of the mass-radius relation. As can be seen in Fig. 3, a large fraction of the planets represented in the figure are not habitable, based on the criteria we have described in the introduction (it is anyway obvious that all these planets are not habitable due to their proximity to the central star).
### 3.3 Maximum radius for different compositions
Assuming some values of the Fe/Si and Mg/Si ratios in the planetary interior (these values reflect approximate ratios in the central star, see Thiabaud et al. and Marboeuf et al. , in prep), we can repeat the same calculations. In this calculation, the fraction of the different compounds (Fe, perovskite, wustite, olivine, entsatite) is computed assuming that the mantle is homogeneous: the Mg/Si and Fe/Si ratios are the same in the inner and outer mantle. Since we are interested in the lowest density planet, the fraction of iron in the mantle is maximized. It is only necessary to assume the existence of a central iron core for Fe/Si ratios larger than 1.13 (in the case of a solar Mg/Si ratio).
Fig. 3 shows the resulting mass-radius relation for different values of the Fe/Si ratio, namely 1, 2, and 5 times the solar value (0.8511). The Mg/Si ratio is kept at its solar value (1.0243) for all the simulations. As can be seen, the planets with solar composition have a maximum radius less than 0.1 smaller than the overall maximum radius derived in the previous section. The shift in radius is almost independent of the planetary mass, but depends on the Fe/Si ratio: planets are around 0.2 smaller (than the overall maximum) for Fe/Si = 2 Fe/Si, and 0.4 smaller for Fe/Si = 5 Fe/Si. Correspondingly, the maximum mass of water decreases from to 105, 80 and 55 for Fe/Si ratios equal to 1, 2, and 5 times the solar value respectively (see Fig. 4).
It is also interesting to consider the maximum mass of the ocean on the planets. Fig. 4 shows the maximum ocean mass, and the maximum water fraction as a function of the planetary mass. Note that the maximum water fraction is based on the ocean mass only and does not include water that in reality could be incorporated in the planetary interior. The maximum water mass is found to be nearly independent of the planetary mass, except for low mass planets. Accordingly, the maximum water fraction decreases with planetary mass, from 2 % for an Earth mass planet to 0.2 % for a 12 planet.
### 3.4 Effect of the atmosphere model
An important contribution to the planetary radius comes from the gas envelope. The top panel of Fig. 5 shows the envelope depth for different models, and for the maximum radius, iron-free planet, similar to the case discussed in Sect. 3.2. We have considered three different models in this figure. In the first one (named later convective-isothermal), as we explained before, the pressure-temperature profile is assumed to follow an adiabat, starting with a given surface temperature and pressure, where the temperature is limited to the skin temperature (see Pierrehumbert and Gaidos 2011).
In the second model, we proceed as follows: for a given mass and radius of the inner planet (excluding any gas envelope), we have computed a set of envelope structure models, assuming different values of the total (solid + gas) planetary radius, and a planetary luminosity. The standard internal structure equations (see e.g., Alibert et al. 2005, Alibert et al. 2013) are then solved from outside in, namely from the assumed total radius, to the radius of the inner planet (solids+ocean). The luminosity is assumed to be uniform in the gas envelope and is a free parameter, which is given by an equivalent internal temperature, as presented in Guillot 2010. The outermost pressure is assumed to be a small arbitrary value ( Pa), and the outermost temperature is equal to the standard outer temperature in the Eddington approximation (, where parametrizes the planetary luminosity). In the ensemble of models we compute, we finally select the subset of planetary radius and planetary luminosity (or ) that allow us to match some values of the surface pressure and temperature. We computed 90000 models for an envelope depth spanning values from one scale height to 200 scale height (the scale height being computed for the planetary outermost temperature), and for spanning values from 1K to 300 K. The maximum radius we obtain with this procedure is the maximum radius a planet can have for a given set of surface temperature and pressure (375K and 1 GPa for the models considered in Fig. 5), as well as a given radius and mass of the solid planet. It appears that the mass of the envelope is always negligible compared to the total mass of the planet, at least for surface pressure and temperature compatible with the presence of liquid water at the planetary surface, and one can to a very high accuracy identify the mass of the inner planet to the total planetary mass.
Finally, the third model is computed in a way similar to the second one, but taking the irradiation of the parent star into account, following the two-stream model of Guillot et al. (2010). The irradiation temperature is assumed to be equal to the one at the present Earth location, namely K. In this case, we again compute a set of envelope structure models, and select the ensemble of radii and planetary luminosity that allow us to match any given surface pressure and temperature. The opacity in the visible and IR range are assumed to be equal to the values quoted in Guillot et al. (2010, see caption of Fig. 5), namely /g and /g. The factor is taken equal to 0.25, meaning that we assume a redistribution of the incoming energy over the whole planetary surface, which is consistent with our 1D spherically-symmetric models. Finally note that the second model is indeed a particular case of the third one, with an irradiation temperature equal to 0, and with non-constant opacities (Bell and Lin 1994).
We present in the middle panel of Fig. 5 the different mass-radius relationships for the three models, in the case described in Sect. 3.2, and for an envelope gas made of H. As shown on the figure, the difference in planetary radius is small when considering the three models, except for low mass planets, where the low gravity enhances the effect of gas composition and temperature gradient. This similarity comes from the fact that the temperature change between the outermost layers and the surface is rather small. As a consequence, the exact temperature profile (which is the main difference between the three envelope models) has a small effect. In addition, we have computed the same models, using the opacities of Freedman et al. (2008), as the Bell and Lin (1994) opacities represent likely an over estimation of the actual opacity. As expected, this translates to a smaller envelope depth for the same model (see the upper and middle panels of Fig. 5).
For the first (convective/isothermal) model, we have computed the maximum radius considering a different atmospheric composition, namely pure H, a mixture of H and He in solar proportions, and CO (molecular weight of 44 g/cm, and ). Note that in the case of CO, we do not take any possible condensation into account, and the profile is therefore very academic. Taking more realistic atmospheric profiles into account will be the subject of future work. Nevertheless, these different examples are used to show the effect on the planetary maximum radius and are believed to bracket the reality.
As can be seen on lower panel of Fig. 5, increasing the molecular weight decreases the planetary radius, as expected. By measuring the slope of the planetary transit in two wavelengths in the visible, one can deduce the mean molecular weight in the planetary atmosphere (at least the upper parts, see Benneke and Seager 2012). By assuming that this value represents the molecular weight in the entire envelope, one could therefore derive a more stringent value of the planetary maximum radius.
## 4 Discussion and conclusion
We have derived the maximum radius of planets in the Earth to Super-Earth regime under the hypothesis that both the surface conditions at the planetary surface lie in the liquid domain of the water phase diagram, and that the pressure at the bottom of a putative global ocean is lower than the liquid water/ice VII transition (approximately 2.4 GPa). As we have argued in the introduction, if these two conditions are indeed necessary for habitability, the maximum radius we derive delimits a region above which (in terms of radius) planets are very likely not habitable.
For this calculation, we have constructed internal structure models for the planet, made of a central iron core, a silicate mantle (itself divided in two layers), a water layer, and a gas envelope. Our model for the inner planet follows the model originally developed by Sotin et al. (2007), and further improved by Grasset et al. (2010), with some differences related to the assumed EOS iron, where we have used the most recent EOS developed by Belonoshko (2010). For the gas envelope, we have considered three models, one convective/isothermal model, one radiative/convective model, and a two-stream irradiated model following the model of Guillot (2010). We have checked that our model reproduces the Preliminary Reference Earth Model (Dziewonski & Anderson, 1981), except for the inner/outer core structure, which is not included in our code.
We have first considered the case of planets devoid of Fe surrounded by a pure H envelope first. Under these (rather unrealistic) conditions, planets larger than 1.8 to 2.3 , for masses ranging from 2 to 12 , cannot harbor conditions compatible with both the presence of liquid water at the surface and a C-cycle. It is important to remember that this maximum radius is derived under very extreme conditions (no iron, pure H convective/isothermal envelope). Under more realistic conditions (e.g., a differentiated planet of super-solar composition, a gas envelope of higher molecular weight), the maximum radius for a given mass can be reduced by up to 0.5 . Correspondingly, the maximum fraction of water in planets must be smaller than for planets more massive than 1 . This maximum fraction decreases with the planetary mass as well as with the iron content of the planet.
To estimate the uncertainties in our model, we have varied the gas envelope model, and found that the resulting radius can vary by at most when considering either the convective/isothermal model or the irradiated convective/radiative model. We have also checked that varying the IR and visible opacities in the irradiated models by factors up to 2, or assuming no redistribution on the planetary surface (factor equal to 1) does not modify the resulting mass-radius relationships notably . In all the cases, the radius is found to be smaller than in the simple model, and we are therefore confident that the maximum radius derived in Sect. 3.2 represents a boundary above which the habitability of planets is very unlikely. The thermal profile in planets is also unknown and has an effect on the resulting radius. Other studies (e.g., Seager et al. 2007, Grasset et al. 2010) have shown however that the effect is of a few percent. We have indeed checked that a variation of a few percent of the adiabatic gradient does not change our results.
In our model we have considered that the ocean is made of pure water, which is an assumption that is probably not correct. Indeed, observations have shown that the Jupiter’s icy satellites (whose structure could be seen as the one of a small ocean planet at large distance) contain a lot of volatiles, like CO (see Hibbit et al. 2000, 2002) or NH (see Mousis et al. 2002). In addition, recent population synthesis models, based on our planet formation model (Alibert et al. 2005, Mordasini et al. 2009a,b, Fortier et al. 2013, Alibert et al. 2013), show that planets that contain water also contain similar volatiles (NH, CO, CO, etc…). The presence of such volatiles is known to modify the crystallization process of water and could modify our results (see e.g., Spohn & Schubert 2003). However, an appreciable change in the maximum radius we have derived would require that the crystallization pressure of water mixed with volatiles is substantially increased. Future studies will be required to address this issue properly.
Finally, it is possible to derive a stronger constraint on the maximum radius of a planet, if one can determine both the abundance of key elements in the central star (in particular Fe, Si, and Mg), and the mean molecular weight of the gas envelope, using Rayleigh scattering observations (see Benneke and Seager 2012). Such a derivation however relies on the assumption that the refractory composition of planets is similar to the one of the central star, an assumption that is supported by recent population synthesis models (Thiabaud et al, in prep). Using such models, transit observations such as the one that will be performed by CHEOPS (see Broeg et al. 2013) or TESS (the Transiting Exoplanet Survey Satellite, see Ricker et al. 2010), complemented by ground-based or other space-based observations, it will be possible to select the best candidates for future habitability studies.
###### Acknowledgements.
This work was supported by the European Research Council under grant 239605.
### Footnotes
1. offprints: Y. Alibert
2. Silicate weathering is a growing function of temperature and can be stopped in the case of a planet covered by ice (see Pierrehumbert 2010).
3. Note that, contrary to low pressure ice, ice VII has a density that is higher than the one of liquid water and stays at the bottom of the ocean.
4. It should be noted that recent studies (Abbot et al. 2012) have shown that silicate weathering requires some land to be present on the planetary surface. As a consequence, according to these authors, a tiny amount of water (large enough to cover the entire surface) is enough to prevent habitability, at least under our working hypothesis.
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2019-05-27 10:59:17
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https://www.hindawi.com/archive/2009/307695/ref/
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`International Journal of Carbohydrate ChemistryVolume 2009 (2009), Article ID 307695, 8 pageshttp://dx.doi.org/10.1155/2009/307695`
Research Article
On the Conformational Properties of Amylose and Cellulose Oligomers in Solution
Laboratory of Physical Chemistry, Swiss Federal Institute of Technology Zürich, ETH, 8093 Zürich, Switzerland
Received 23 October 2008; Revised 10 February 2009; Accepted 1 April 2009
Copyright © 2009 Moritz Winger et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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2017-10-21 22:54:04
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https://cs.stackexchange.com/questions/92266/balancing-steiner-trees-with-shortest-path-trees
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# Balancing Steiner trees with Shortest Path trees
I'm working on a problem that combines Steiner Trees and Shortest Path trees. We have a (sparse, connected) graph $G=(V,E)$ with non-negative edge weights and edge lengths, a set of terminals $T \subset V$ and a root node $s \in T$.
The goal is to connect all terminals to the root node while minimizing the total edge weight + the total lengths of the paths from each terminal to the root.
Minimizing just the edge weights gives the Steiner tree problem. Minimizing just the sum of path lenghts gives a shortest path problem.
Khuller, Samir, Balaji Raghavachari, and Neal Young. "Balancing minimum spanning trees and shortest-path trees." gives an algorithm for balancing this between MSTs and shortest path trees, but only considers the path from that node instead of the sum of incoming paths.
1. I'm having trouble interpreting this article. Does the algorithm give a minimal cost solution for its own problem or is it a heuristic approximation?
2. Can similar greedy strategies (expand from root to find shortcuts) also produce minimal cost solutions when balancing with Steiner trees (if you already have computed the Steiner tree)?
If (2) holds, then a custom solution for this cost function isn't required; you can compute the Steiner tree first, then minimize the path cost.
• The abstract says it's an approximation algorithm -- a heuristic that guarantees that its output is "not too bad", where the allowed "badness" of each approximation (to an MST, and to a shortest path tree) is controlled by the user by setting the parameter $\gamma$. Of course, since there is not always an MST that is also a shortest path tree, if you want a tree that is a good approximation of one you may have to accept that it will be a bad approximation of the other. – j_random_hacker May 24 '18 at 13:15
• I get that there is always a tradeoff between the two types of trees with this algorithm. Both trees will be suboptimal for their own cost. The article states it offers the best possible tradeoff between both trees for a given $\gamma$. I imagine it finds a (local) minimum for the combined cost functions using the 'shortcuts', but I can't follow how it relates to the optimal combined cost value. – b9s May 24 '18 at 14:10
• I didn't see any "combined cost" mentioned in the abstract (is it described later?) -- just the two separate costs, with explicit bounds given in terms of $\gamma$. – j_random_hacker May 24 '18 at 14:15
• It's not in the article, but it is relevant for my problem variant. I don't think the error bounds for both trees directly relate to it, which makes me wonder if an approximation bound can be derived for my variant based on the article. – b9s May 24 '18 at 14:23
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2019-10-23 02:47:02
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http://sphire.mpg.de/wiki/doku.php?id=pipeline:utilities:sxpdb2em&rev=1534928717
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pipeline:utilities:sxpdb2em
Approvals: 0/1
This is an old revision of the document!
PDB File Conversion: Converts an atomic model stored in a PDB file into a simulated electron density map. The coordinates (0,0,0) in PDB space will map to the center of the volume.
Usage in command line
sxpdb2em.py input_pdb output_hdf --apix=PIXEL_SIZE --box=BOX_SIZE --het --chains=CHAIN_STRING --center=CENTERING_METHOD --O --tr0=MATRIX_FILE_PATH --set_apix_value=PIXEL_SIZE_STRING --quiet
sxpdb2em does not support MPI.
sxpdb2em.py tRNA.pdb tRNA.hdf --apix=2.26 --box=150 --center=c --O --quiet --tr0=<'filename'> --het
#### Main Parameters
input_pdb
Input PDB file: Starting atomic coordinates. (default required string)
output_hdf
Output map: Specify file path for output map. (default required string)
--apix
Pixel size of output map [A]: Pixel size of the output map [A]. (default 1.0)
--box
Output box size [Voxels]: Specify string of a single value (e.g. '256') to get a cubic box. Alternatively, use 'x,y,z' format to specify demensions of x,y,z-axis (e.g. '128,64,256'). If not given, the program will find the minimum box size fitting the structure. Be aware that this will most likely result in a rectangular box. Note that GUI does not support the default mode. (default required string)
--het
Include hetero atoms: Otherwise, the HETATM entries in the PDB file are ignored. (default False)
--chains
Chain identifiers: A string list of chain identifiers to include (e.g. 'ABEFG'). By default, all chains will be included. (default none)
--center
Center model at the origin: Specifies whether the atomic model should be moved to the origin before generating density map. Available options are: 'c' - Use the geometrical center of atoms; 'a' - Use the center of mass (recommended); 'x,y,z' - Vector to be subtracted from all PDB coordinates. 'n' - No centering, in which case (0,0,0) in the PDB space will map to the center of the EM volume. (default n)
--O
Apply additional rotation: This can be used to modify the orientation of the atomic model by using O system of coordinates. (default False)
--tr0
Rotational matrix file: This file must contain the 3×4 transformation matrix to be applied to the PDB coordinates after centering. The translation vector (last column of the matrix) must be specified in Angstrom. (default none)
--set_apix_value
Set header pixel size: Set pixel size in header of the ouput map. (default False)
--quiet
Silent mode: Does not print any information to the monitor. Verbose is the default. (default False)
The program uses tri-linear interpolation. Electron densities are taken to be equal to atomic masses.
Pawel A. Penczek
Category 1:: APPLICATIONS
sparx/bin/sxpdb2em.py
Stable:: Has been evaluated and tested. Please let us know if there are any bugs.
There are no known bugs so far.
• pipeline/utilities/sxpdb2em.1534928717.txt.gz
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2020-02-28 07:15:59
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https://github.com/tpapp/lla
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This repository has been archived by the owner before Nov 9, 2022. It is now read-only.
/ lla Public archive
Lisp Linear Algebra
# tpapp/lla
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# Lisp Linear Algebra — a linear algebra library for Common Lisp
This library is unsupported.
LLA is a high-level Common Lisp library built on on BLAS and LAPACK, but providing a much more abstract interface with the purpose of freeing the user from low-level concerns and reducing the number of bugs in numerical code.
Documentation is mostly in docstrings at the moment, but I plan to write a decent tutorial at some point. In the meantime, please look at the unit tests.
## Objectives
• High-level, user friendly interface that hides the details.
(solve a b) should return $X$, from $AX=B$, regardless of whether $A$ is a dense matrix, an $LU$ decomposition, or something else; similarly, $X$ should be a vector/matrix when $B$ is. Users should not need to memorize names like DGESV, especially when CLOS makes it so easy to deal with these things. Also, you don't need to make sure that all arguments are of the same type (eg complex-double): LLA will find the common supertype for elements and convert if necessary.
• Stay in Lisp and expose the innards of the objects as much as possible.
LLA aims to take advantage of CL's high level facilities such as CLOS and memory management. Data is kept in Lisp arrays instead of foreign arrays, so you can access it directly using aref etc. You also benefit from garbage collection and all the clever stuff that comes with the GC. If you need more memory, just increase the heap size.
• Keeping it simple.
Currently, LLA sources amount to less than 3000 lines of code (not including tests). The small size should make maintainance easier and bugs more rare (hopefully).
• Speed is important, but reliability comes first.
Only optimize when necessary, and do extensive testing afterwards. Most of the speed comes from your LAPACK library anyway --- most linear algebra operations are $O(N^\alpha)$ with $\alpha > 1$, frequently $\alpha > 2$. That said, copying to memory is optimized, and in the long run LLA should make use of your implementation's array pinning features (when available). Currently, direct array sharing is disabled, it will be re-enabeld in the near future.
# Configuration
Certain features of LLA can be configured before loading using the plist *lla-configuration* in the CL-USER package (for example, on SBCL you would do it in your ~/.sbclrc). The following properties are supported:
• :libraries
A list of objects, passed directly to cffi:load-foreign-library. You can use strings, paths, or even symbols if you have defined these libraries using cffi:define-foreign-library. If you don't define this, a reasonable platform-dependent default will be used. See the next section for details.
• :int64
This makes LLA use 64-bit integers for array dimensions, pivot indices and other integer values passed to BLAS/LAPACK functions. Only use this if you are sure that your libraries have been compiled with 64-bit integers. The fact that you have a 64-bit platform does not necessarily mean that this is the case, in fact, it is still quite rare. Unless told otherwise, LLA expectes BLAS/LAPACK to use the (L)LP64 model for integers -- that is to say, integer types in Fortran are 32 bit.
• :efficiency-warnings
Enable the possibility of efficiency warnings at compile time. You still have to set the appropriate flags, but without this option, they won't even be checked. There are two properties you can set: :array-type and :array-conversion. The first warns whenever an array has to be walked elementwise to determine its type, the second when some arrays need to be converted to a common type.
Example:
(defparameter cl-user:*lla-configuration*
'(:efficiency-warnings (:array-type :array-conversion)))
(let ((lla:*lla-efficiency-warning-array-type* t)
(lla:*lla-efficiency-warning-array-conversion* t))
(code that you want to check))
## Dependencies and configuration
LLA needs BLAS and LAPACK shared libraries to work. When it comes to loading libraries, LLA tries to pick a sensible default for each platform, but in case it fails, you need to tell LLA where the libraries are before loading.
You can do this by putting something like this in your startup script (eg ~/.sbclrc, the symbol needs to be in the package cl-user):
(defvar *lla-configuration*
'(:libraries ("/usr/lib/atlas-base/atlas/libblas.so.3gf"
"/usr/lib/atlas-base/libatlas.so.3gf")))
## Debian
On Debian-based distributions, it is very likely that LLA will work out of the box if you just install ATLAS, eg
apt-get install libatlas3gf-base
However, you may want to build a version optimized for your architecture.
### Building ATLAS on Debian
Prepare the build (as root):
apt-get build-dep atlas
apt-get install fakeroot devscripts
cpufreq-set -g performance -c 0 # do this for all CPUs
Then as a regular user,
apt-get source atlas
cd atlas-[fill in your version here]/
fakeroot debian/rules custom
Then install the .deb files that were created.
### Selecting the right linear algebra library
update-alternatives --config libblas.so.3
update-alternatives --config liblapack.so.3
### Intel MKL on Linux
In /etc/ld.so.conf.d/, create a file that contains the paths, eg
/opt/intel/mkl/lib/intel64
/opt/intel/composerxe/lib/intel64
Then the configuration
(defvar *lla-configuration*
'("libgomp.so.1" "libiomp5.so" "libmkl_rt" "libpthread.so.0" "libpthread"))
should work.
## Acknowledgements
LLA was inspired by packages written by AJ Rossini, Rif, Mark Hoemmen and others. I have borrowed code (whenever allowed by their licenses) and ideas freely from all of them.
Gábor Melis made substantial contributions to the library, especially the low-level pinning interface and the destructive BLAS routines.
## Suggested editor settings for code contributions
No line breaks in (doc)strings, otherwise try to keep it within 80 columns. Remove trailing whitespace. 'modern' coding style. Suggested Emacs snippet:
(set-fill-column 9999)
'(("\\<\$$FIXME\\|TODO\\|QUESTION\\|NOTE\$$"
1 font-lock-warning-face t)))
(setq show-trailing-whitespace t)
'(lambda()
(save-excursion
(delete-trailing-whitespace))
nil))
(visual-line-mode 1)
(setq slime-net-coding-system 'utf-8-unix)
(setq lisp-lambda-list-keyword-parameter-alignment t)
(setq lisp-lambda-list-keyword-alignment t)
(setq common-lisp-style-default 'modern)
## Things to do (roughly in order of priority)
• write optimized pinning interfaces, especially ECL
• write documentation (probably w/ docudown, decide)
• write more tests (especially randomized ones, develop macros for that)
• write a tutorial
Lisp Linear Algebra
## Releases
No releases published
## Packages 0
No packages published
•
•
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2023-01-27 02:11:50
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https://labs.tib.eu/arxiv/?author=Jae%20Hyun%20Yun
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• ### Analysis of Laser ARPES from Bi$_2$Sr$_2$CaCu$_2$O$_{8+\delta}$ in superconductive state: angle resolved self-energy and fluctuation spectrum(1108.1605)
Sept. 28, 2011 cond-mat.supr-con
We analyze the ultra high resolution laser angle resolved photo-emission spectroscopy (ARPES) intensity from the slightly underdoped Bi$_2$Sr$_2$CaCu$_2$O$_{8+\delta}$ in the superconductive (SC) state. The momentum distribution curves (MDC) were fitted at each energy $\w$ employing the SC Green's function along several cuts perpendicular to the Fermi surface with the tilt angle $\theta$ with respect to the nodal cut. The clear observation of particle-hole mixing was utilized such that the complex self-energy as a function of $\omega$ is directly obtained from the fitting. The obtained angle resolved self-energy is then used to deduce the Eliashberg function $\alpha^2 F^{(+)}(\th,\w)$ in the diagonal channel by inverting the d-wave Eliashberg equation using the maximum entropy method. Besides a broad featureless spectrum up to the cutoff energy $\omega_c$, the deduced $\alpha^2 F$ exhibits two peaks around 0.05 eV and 0.015 eV. The former and the broad feature are already present in the normal state, while the latter emerges only below $T_c$. Both peaks become enhanced as $T$ is lowered or the angle $\th$ moves away from the nodal direction. The implication of these findings are discussed.
• ### Extraction of Electron Self-Energy and Gap Function in the Superconducting State of Bi_2Sr_2CaCu_2O_8 Superconductor via Laser-Based Angle-Resolved Photoemission(1103.3629)
Super-high resolution laser-based angle-resolved photoemission measurements have been performed on a high temperature superconductor Bi_2Sr_2CaCu_2O_8. The band back-bending characteristic of the Bogoliubov-like quasiparticle dispersion is clearly revealed at low temperature in the superconducting state. This makes it possible for the first time to experimentally extract the complex electron self-energy and the complex gap function in the superconducting state. The resultant electron self-energy and gap function exhibit features at ~54 meV and ~40 meV, in addition to the superconducting gap-induced structure at lower binding energy and a broad featureless structure at higher binding energy. These information will provide key insight and constraints on the origin of electron pairing in high temperature superconductors.
• ### Momentum Dependence of the Single-Particle Self-Energy and Fluctuation Spectrum of Slightly Underdoped Bi_2 Sr_2 CaCu_2 O_{8+\delta} from High Resolution Laser ARPES(0912.0088)
April 30, 2010 cond-mat.supr-con
We deduce the normal state angle-resolved single-particle self-energy $\Sigma(\theta, \omega)$ and the Eliashberg function (i.e., the product of the fluctuation spectrum and its coupling to fermions) $\alpha^2 F(\theta,\omega)$ for the high temperature superconductor Bi$_2$Sr$_2$CaCu$_2$O$_{8+\delta}$ from the ultra high resolution laser angle-resolved photoemission spectroscopy (ARPES). The self-energy $\Sigma(\theta, \omega)$ at energy $\omega$ along several cuts normal to the Fermi surface at the tilt angles $\theta$ with respect to the nodal direction in a slightly underdoped Bi$_2$Sr$_2$CaCu$_2$O$_{8+\delta}$ were extracted by fitting the ARPES momentum distribution curves. Then, using the extracted self-energy as the experimental input, the $\alpha^2 F(\theta,\omega)$ is deduced by inverting the Eliashberg equation employing the adaptive maximum entropy method. Our principal new result is that the Eliashberg function $\alpha^2F(\theta,\omega)$ collapse for all $\theta$ onto a single function of $\omega$ up to the upper cut-off energy despite the $\theta$ dependence of the self-energy. The in-plane momentum anisotropy is therefore predominantly due to the anisotropic band dispersion effects. The obtained Eliashberg function has a small peak at $\omega\approx0.05$ eV and flattens out above 0.1 eV up to the angle-dependent cut-off. It takes the intrinsic cut-off of about 0.4 eV or the energy of the bottom of the band with respect to the Fermi energy in the direction $\theta$, whichever is lower. The angle independence of the $\alpha^2 F(\theta,\omega)$ is consistent only with the fluctuation spectra which have the short correlation length on the scale the lattice constant. This implies among others that the antiferromagnetic fluctuations may not be underlying physics of the deduced fluctuation spectrum.
• ### Model for the inverse isotope effect of FeAs-based superconductors in the $\pi$-phase-shifted pairing state(0904.1864)
Aug. 18, 2009 cond-mat.supr-con
The isotope effects for Fe based superconductors are considered by including the phonon and magnetic fluctuations within the two band Eliashberg theory. We show that the recently observed inverse isotope effects of Fe, $\alpha_{Fe} \approx -0.18 \pm 0.03$,\cite{Shirage0903.3515} as well as the large positive isotope exponent ($\alpha\approx 0.35$) can naturally arise for the magnetically induced sign revered s-wave pairing state within reasonable parameter range. Either experimental report can not be discarded from the present analysis based on the parameter values they require. The inverse and positive isotope effects mean, respectively, the interband and intraband dominant eletron-phonon interaction. We first make our points based on the analytic result from the square well potential model and present explicit numerical calculations of the two band Eliashberg theory.
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2020-04-04 08:48:58
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https://aviation.stackexchange.com/questions/22333/what-is-this-orange-container-labeled-flight-kit-wheel
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# What is this orange container labeled “Flight kit wheel”?
What is this orange container for? It reads
Flight kit wheel. Do not remove.
What's a "flight kit wheel"?
• Related: Do any aircraft carry spare parts for making repairs?. See also this container manufacturer. – mins Oct 24 '15 at 10:45
• @mins You should go ahead and make that an answer – TomMcW Oct 24 '15 at 17:18
• Corsair is a charter airline so presumably they keep spares aboard, as they might not have a facility at their destination. – egid Oct 24 '15 at 20:32
• @egid For the precision, Corsair is absolutely not a charter airline. It's a regular airline with daily lines to France's oversea territories. But I don't think that they have big facilities on our islands so you might be right there. – Quentin H Oct 25 '15 at 18:24
• The most important question, and the one we've all overlooked for 2 years - "Why have they removed it from the plane???" ;) – FreeMan Nov 2 '17 at 12:05
This question is somehow targeting something already answered in Do any aircraft carry spare parts for making repairs?, albeit the answers are going in many directions, so I'll provide one focused on this case.
Corsair, a French airline based at Paris-Orly (LFPO) has only 7 a/c currently active. When one is grounded the schedule of the others is indeed disrupted, the shorter, the better.
Rather than having the most used spare parts (for each type) at multiple stopovers, or having to wait for one on its way from Orly or a closer location (Fort-de-France, Pointe-à-Pitre, La Réunion), a good tradeoff is to have them aboard.
As described in the linked question, the maintenance parts are often known as fly away kit (FAK) or flight spare kit (FSK).
AKN FSK container for Air France (Source)
Parts in the container are secured and don't move.
AKN FSK container for Air France (Source)
Spare parts are commonly seen aboard aircraft serving remote isolated locations with light traffic, where maintenance may be difficult at the moment (typically many places in Africa).
(source)
The kit is built (or fitted) in a standard cargo container, here an "AKE" IATA form factor, with the characteristic truncated side.
(source)
Corsair aircraft are well known by their tail IDs: F-HSEA, F-HSUN (F-HSEX is gone), F-HSKY, and more recently F-HCAT, F-HBIL, F-HZEN.
• Great answer! I didn't know that airliners were actually flying with a spare wheel (or whatever other spare part), due to the weight penalty. – Quentin H Oct 25 '15 at 18:25
• @QuentinHayot The weight penalty is much lower than the wait penalty, therefore, it makes sense to have parts on board. It would be interesting to find out if Corsair flies mechanics aboard every time, or if they have/contract them locally when needed. – FreeMan Oct 26 '15 at 14:12
• Wow I love the term "wait penalty" – vasin1987 Mar 11 '16 at 11:31
• I was working on a project 15 years ago where we were negotiating with an airline to use one of their B737s to obtain an STC. They quoted the cost (in 2002) to pull the aircraft out of service at \$70k/day. – Gerry Nov 2 '17 at 12:52
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2020-01-22 20:04:54
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https://math.stackexchange.com/questions/1788138/numerical-methods-to-solve-differential-algebraic-equations
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# Numerical methods to solve Differential-Algebraic-Equations
I am new to the topic of differential-algebraic-equations:
$\dot x = f(x,u,c)$
$0=g(x,u,c)$
where $u$ are control variables and $c$ algebraic variables.In my first literature study i found two approaches:
1. given $x$ and $u$, solve (if non-linear with Newton) $0=g(x,u,c)$ for $c$. Make a step (Euler for instance)
2. Index reduction: derive algebraic equations with respect to time - so that in the end you get a normal ODE wich can be solved with ordinary ODE solver.
Why lead differntial algebraic equations to stiff ODE's?: "From a more theoretical viewpoint, the study of differential-algebraic problems gives insight into the behaviour of numerical methods for stiff ordinary differential equations"
Most of literature that i found talk about method 2; is it method 1 not simpler? Or do I misunderstood something :)? Could you recommend basic literature (simple introduction) to this topic?
Thank you very much!
• Are you looking for numerical solutions? You should mention that in your title. – MrYouMath May 16 '16 at 20:36
• Just a comment that these types of equations are related to geometric singular perturbation theory and "slow manifolds". – Alex May 16 '16 at 20:45
• With method 1 it is not necessarily clear that the "geometry" is preserved, that is that you stay on the desired target manifold. Methods like 2 have a better chance of staying on some target manifold by constructing a "numerical flow" on that manifold. – Ian May 16 '16 at 20:59
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2019-06-27 12:09:06
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https://www.nagwa.com/en/guide/860154131849/
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# Video: Creating Your Own Questions
In this video, we will show you how to create your own questions for your portal.
12:57
### Video Transcript
In this video, we will show you how to create your own questions for your portal.
To create your own questions, you must be a portal educator and be signed in to educators.nagwa.com.
Once you’re signed in, click on “My Library”, then click on the “Questions” tab.
In this area, you’ll be able to see all of the questions you have previously added. We’ll go over this interface later, but for now let’s create a new one by clicking “+ Add Question”.
You’ll now see the screen where you can add the details of your new question.
On the left-hand side of this screen, you will see three drop-down menus. The first of these is the “Question Type” box. Here, you’ll be given the option of creating a multiple-choice question or a short-answer question.
For a multiple-choice question, you’ll need to provide the answer along with some distractors so the question can be automatically marked.
A short-answer question is one for which students will need to provide a written response. This type of question is not automatically marked, but instead you can write a model answer that can be used as a reference.
Let’s first go through creating a multiple-choice question.
Next, you’ll need to choose the subject of the question. It’s worth noting that you can only create questions for the active subjects on your portal. Inactive subjects will be shown in this drop-down menu, but you won’t be able to select them. It is also possible to create new subjects in your portal.
Changing the active subjects of a portal and creating new subjects can only be performed by a portal admin in the admin area. We’ll have a separate video guide that teaches you how to do both of these things.
By default, Nagwa has a comprehensive list of subjects that are all activated on your portal; however, your portal admin may change the settings to restrict them or even to add new ones.
We have a separate video guide that teaches you how to create new subjects for your portal.
For this example, we are going to create a mathematics question, so let’s select it from the drop-down menu.
Next, you’ll need to select the grade for which your question will be used. For this simple example, we’ll choose “Grade 7”. Selecting a grade is important because once you have done so, you’ll see an extra section appear on the left of the page titled “Select Lesson”.
In this section, you’ll need to select the lesson in which your question belongs. Expanding the drop-down menu shows you all of the chapter titles for “First Year of Preparatory School • Second Term” mathematics course. Let’s choose the chapter “Numbers and Algebra”.
Clicking this chapter shows you the lessons contained within the chapter. Now click on the lesson you want to write a question for, and the lesson title will be formatted in bold text. This bold text shows you the selected lesson. In our case, we have selected the lesson “Square Roots of Perfect Squares”.
We are now ready to move on to the question creation. The first thing to do is to write the stem of our question in the question box.
For example, we might want to ask our students to find the square root of 25.
Along the top of this box, you’ll see some standard formatting options.
In particular, you should note the “Upload Image” button. Clicking this allows you to drag and drop an image file or to select it from your file browser. The image can then be used as part of your question.
Another button to note is the equation editor. In our example, we might want to represent the square root of 25 using mathematical symbols.
Nagwa portals allow you to write equations using a code called LaTeX. You can preview what your equations will look like by clicking the equation editor.
Let’s delete the last part of our question to illustrate this.
We’ll have a dedicated video that guides you through using the equation editor. As a quick reference, you can also visit our online reference to math symbols, found at https://www.nagwa.com/en/math.
For now, here’s a quick preview using the square root function.
Once you’ve typed your code into the equation editor, you’ll be given a preview in the gray box below. Once you are happy with this, click “Insert”, and your code will be properly formatted and inserted into the question box.
As we are creating a multiple-choice question, we will need to provide the answer along with some distractors. You’ll need to provide a minimum of two and a maximum of five options for each question.
You might also notice that each of these boxes has an “Upload Image” button, much like the question box. Instead of typing your answers, you can upload images here.
You can also insert equations into the answer boxes, but you must write the LaTeX syntax between open and close parentheses preceded by a backslash, for example, $$\frac{2}{3}$$, for math. For chemistry, you must include your expression in the same parentheses but also the expression “backslash ce open and close braces”, for example, $$\ce{H2O}$$. This is explained in more detail in the dedicated videos for inserting equations.
For this question, we’ll simply type in some numbers.
Once you have filled in all of your answer boxes, make sure to click the tick mark on the right-hand side next to the correct answer.
Note that this arrangement of multiple-choice answers will be the one displayed to the students. When writing your answers, make sure to mix up the order to deter any students from guessing!
Once you’re finished with your question, you have two options: you can either click “Add” if you’d like to stay on the page and continue creating more questions, or you can click “Add and Finish”, which will take you back to your complete list of created questions on the “My Library” page.
Let’s click “Add”. You should now see a banner saying “Question submitted successfully. It is currently processing.”
Once the question has been processed, you’ll see it appear under the “Questions” tab of “My Library”, and you’ll be able to find the question when creating your next assessment by searching for the lesson that you linked the question with.
Let’s now illustrate how to create a short-answer question. A short-answer question is one for which students will need to provide a written response.
When creating a short-answer question, picking the subject, grade, and lesson works in exactly the same way as we saw earlier.
In contrast to this, when creating a short-answer question, you will need to provide a model answer rather than multiple-choice options.
For example, we might choose to upload a picture of a graph and ask our students to identify the error. We can do this by clicking the “Upload Image” icon in the question stem and selecting the correct file.
Once you have written the question itself, you will need to provide a model answer. This is the response that you would expect from a student in order to correctly answer the question.
If the question you are uploading is image based, note that your students can either create a digital image or draw their image using a pen and paper and upload it as the question answer.
Remember that when a student solves this type of question, it will not be automatically marked, as is the case with multiple-choice questions.
Instead, when creating an assessment that includes a short-answer question, an educator will decide the maximum number of marks that can be obtained from the question.
When marking assessments, the educator will be able to view the model answer and to compare it with the student’s response. They will then decide if the student has answered correctly or incorrectly and award them a number of marks.
When a student views the results of an assessment containing a short-answer question, they will see how many marks they were awarded, alongside the model answer, which they can compare with their response.
Once you are happy with your question, it can be submitted using the “Add” button below. Let’s click “Add and Finish”. This takes us back to the “Questions” tab of the “My Library” page. Here, we can see the confirmation message: “Question submitted successfully.”
On this page, we can also see all of the questions that we have created. Each question exists in its own box, with a title bar containing the name of the associated lesson and the associated subject.
If we refresh the page, we’ll see that the question that we just created has appeared, and it has a message in the bottom right of the box to tell us that it was successfully processed.
On the left of the page, you will notice a small drop-down menu. Here, we are currently viewing all of the questions that were successfully processed by the system; however, we can also view any questions that failed processing by selecting this option.
From this page, we can also perform a number of actions to manage the questions we have created. In the top-right corner of each question box, you will be able to see a small icon with three dots. Clicking this expands an action menu.
Clicking the “Edit” button takes you to a similar page to the one we saw when creating the question. From here, you’ll be able to edit the type of question, the wording of the question stem, and the answers.
Note that once a question is created, we cannot change the subject, the grade, or the lesson that this question is associated with; hence, these options are grayed out.
If you have made a mistake in any of these fields, you’ll need to return to the “Questions” tab of the “My Library” page. You can then click on the action menu of the question and click “Delete”.
Once you confirm this action, you should see a confirmation message that your question was successfully deleted.
A couple of final notes: By default, this page shows you all of the questions you have created. These questions will be sorted in the order that you have most recently added them to the library.
There are a number of ways to organize and filter this list. The first is by clicking the filter button to the left of the “+ New Question” button. This reveals two drop-down lists. The first list allows you to see all of the questions uploaded by you or any of the other educators who share the same subjects and grades as you, on your portal, if the settings have been set as such.
The second of these drop-down lists allows you to filter by whether a question has been successfully processed by the system or not.
Another way you can filter the list is by using the fields on the left of the page. Here, you can more easily narrow down the list to find questions in a particular subject or grade. You can also filter by both at the same time.
Finally, there is a drop-down list that allows you to view all of the lessons you have created. From here, you can also view the questions that have been written for your lessons.
You also have the ability to edit the lessons’ titles or delete them entirely if you need.
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2022-09-25 23:16:34
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https://scholars.duke.edu/display/pub1128446
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# Online Covering with Convex Objectives and Applications
Other Article
We give an algorithmic framework for minimizing general convex objectives (that are differentiable and monotone non-decreasing) over a set of covering constraints that arrive online. This substantially extends previous work on online covering for linear objectives (Alon {\em et al.}, STOC 2003) and online covering with offline packing constraints (Azar {\em et al.}, SODA 2013). To the best of our knowledge, this is the first result in online optimization for generic non-linear objectives; special cases of such objectives have previously been considered, particularly for energy minimization. As a specific problem in this genre, we consider the unrelated machine scheduling problem with startup costs and arbitrary $\ell_p$ norms on machine loads (including the surprisingly non-trivial $\ell_1$ norm representing total machine load). This problem was studied earlier for the makespan norm in both the offline (Khuller~{\em et al.}, SODA 2010; Li and Khuller, SODA 2011) and online settings (Azar {\em et al.}, SODA 2013). We adapt the two-phase approach of obtaining a fractional solution and then rounding it online (used successfully to many linear objectives) to the non-linear objective. The fractional algorithm uses ideas from our general framework that we described above (but does not fit the framework exactly because of non-positive entries in the constraint matrix). The rounding algorithm uses ideas from offline rounding of LPs with non-linear objectives (Azar and Epstein, STOC 2005; Kumar {\em et al.}, FOCS 2005). Our competitive ratio is tight up to a logarithmic factor. Finally, for the important special case of total load ($\ell_1$ norm), we give a different rounding algorithm that obtains a better competitive ratio than the generic rounding algorithm for $\ell_p$ norms. We show that this competitive ratio is asymptotically tight.
### Cited Authors
• Azar, Y; Cohen, IR; Panigrahi, D
### Published Date
• December 10, 2014
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2022-08-08 09:50:24
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-4-section-4-1-angles-and-radian-measure-exercise-set-page-534/122
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Chapter 4 - Section 4.1 - Angles and Radian Measure - Exercise Set - Page 534: 122
The required solution is $286\ \text{mi}$
Work Step by Step
The angle $\theta$ for the change in direction is $20{}^\circ$. Then, convert it into radians: \begin{align} & \theta =20{}^\circ \\ & =20{}^\circ \left( \frac{\pi }{180{}^\circ } \right) \\ & =\frac{\pi }{9} \end{align} And the curve distance $s$ is $100$ miles. The arc’s length subtended by an angle at the center of the circle is given by $s=r\theta$ Here, $r$ is the radius of the railroad curve. Rearrange for $r$: $r=\frac{s}{\theta }$ Put $100$ miles for $s$ and $\frac{\pi }{9}$ for $\theta$: \begin{align} & r=\frac{100\ \text{mi}}{\frac{\pi }{9}} \\ & =\frac{900\ \text{mi}}{\pi } \end{align} Put $\pi =3.14159$: \begin{align} & r=\frac{900\ \text{mi}}{3.14159} \\ & =286.47\ \text{mi}\approx 286\ \text{mi} \end{align} Hence, the radius of the railroad curve is $286$ miles.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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2021-05-11 08:24:35
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https://www.physicsforums.com/threads/linear-transformation-image.358500/
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# Linear Transformation image
1. Nov 27, 2009
### doublemint
1. The problem statement, all variables and given/known data
Let v1=
1
-2
and v2=
-1
1
Let T:R2R2 be the linear transformation satisfying
T(v1)=
9
7
and T(v2)=
0
-8
Find the image of an arbitrary vector
x
y
2. Relevant equations
3. The attempt at a solution
I thought it might have to do something with T(u+v)=T(u)+T(v) or some sort of transformation, but I cannot seem to get it...
Any help would be appreciated!
Thanks!!
2. Nov 28, 2009
### HallsofIvy
You are given that
$$T\left(\begin{bmatrix}1 \\ -2\end{bmatrix}\right)= \begin{bmatrix} 9 \\ 7\end{bmatrix}$$
and that
$$T\left(\begin{bmatrix}-1 \\ 1\end{bmatrix}\right)= \begin{bmatrix}0 \\ 8\end{bmatrix}$$
And you want to determine
$$T\left(\begin{bmatrix} x \\ y\end{bmatrix}\right$$
Yes, you want to use T(u+v)= T(u)+ T(v). Specifically if $u= Av_1+ Bv_2$ then T(u)= AT(v_1)+ BT(v_2). So first you want find A and B such that
$$\begin{bmatrix}x \\ y \end{bmatrix}= A\begin{bmatrix}1 \\-2\end{bmatrix}+ B\begin{bmatrix}-1 \\ 1 \end{bmatrix}$$
3. Nov 28, 2009
### doublemint
Alright, so I got
A=-x-y
B=-2x-y
I'm guessing then we follow through with T(u)= AT(v_1)+ BT(v_2),
T(x y)=[T(1 -2)T(0 -8)][A B]=[9A, 7A-8B]
Then I sub in A and B:
[9(-x-y), 7(-x-y)-8(-2x-y)]= [-9x-9y, 9x+y]
Is this what I was supposed to do? I think now I have to factor out the x-y, but I can't do it to 9x+y. Did I do something wrong at finding A and B?
4. Nov 29, 2009
### doublemint
I just submitted my work, it was right after all!!
Thanks HallsofIvy!
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2018-03-21 19:20:41
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https://scicomp.stackexchange.com/questions/37082/polynomial-interpolation-on-a-regular-hexagon/37091
|
# Polynomial interpolation on a regular hexagon
### Background
We know a function $$f$$ on the vertices of a regular hexagon, as follows
$$\left( 1, \ 0, \ f_{0}\right), \ \left( \frac{1}{2}, \ \frac{\sqrt{3}}{2}, \ f_{1}\right), \ \left( - \frac{1}{2}, \ \frac{\sqrt{3}}{2}, \ f_{2}\right), \ \left( -1, \ 0, \ f_{3}\right), \ \left( - \frac{1}{2}, \ - \frac{\sqrt{3}}{2}, \ f_{4}\right), \ \left( \frac{1}{2}, \ - \frac{\sqrt{3}}{2}, \ f_{5}\right)\, ,$$
and we want to interpolate a polynomial.
If we propose a quadratic polynomial of the form
$$p(x, y) = a_{0} x^{2} + a_{1} x y + a_{2} y^{2} + a_{3} x + a_{4} y + a_{5}\, ,$$
we end up with the following system
$$\left[\begin{matrix} 1 & 0 & 0 & 1 & 0 & 1\\ \frac{1}{4} & \frac{\sqrt{3}}{4} & \frac{3}{4} & \frac{1}{2} & \frac{\sqrt{3}}{2} & 1\\ \frac{1}{4} & - \frac{\sqrt{3}}{4} & \frac{3}{4} & - \frac{1}{2} & \frac{\sqrt{3}}{2} & 1\\1 & 0 & 0 & -1 & 0 & 1\\ \frac{1}{4} & \frac{\sqrt{3}}{4} & \frac{3}{4} & - \frac{1}{2} & - \frac{\sqrt{3}}{2} & 1\\ \frac{1}{4} & - \frac{\sqrt{3}}{4} & \frac{3}{4} & \frac{1}{2} & - \frac{\sqrt{3}}{2} & 1\end{matrix} \right] \left[\begin{matrix}a_{0}\\a_{1}\\a_{2}\\a_{3}\\a_{4}\\a_{5}\end{matrix}\right] = \left[\begin{matrix}f_{0}\\f_{1}\\f_{2}\\f_{3}\\f_{4}\\f_{5}\end{matrix}\right]\, ,$$
where the Vandermonde matrix is singular, and the system does not have any solution.
I suppose that this behavior has something to do with the symmetries present in the sampling points, but I am not sure, though.
### Questions
1. What is the reason for this behavior?
2. Is there any known polynomial interpolator for the vertices of a regular hexagon?
• For that set of points, the polynomial basis you are using is not unisolvent. Particularly, if you want to fit a quadratic polynomial using the Lagrange basis, I believe one of the nodes have to be inside the hexagon and the rest of them should be 5 out of 6 vertices of the hexagon. I don't have further intuition why that must be. IIRC, the proper basis would contain polynomials similar to the hat functions, i.e. $\phi_i(\vec{x}_j)$ should be zero if $j\neq i$ and should be one if $j=i$. I quickly searched and found the paper "On the hexagonal Shepard method" on the topic. Hope it helps. – Abdullah Ali Sivas Mar 22 at 20:23
• Why not split the hexagon into 6 triangles (with the additional vertex at the center of the hexagon), assigning the average of the 6 values to the center, and do linear interpolation on each triangle? – Wolfgang Bangerth Mar 22 at 21:04
• @WolfgangBangerth, the idea came up trying to find the gradient of the function in the vertices of a triangular mesh. So, we are considering an interpolation over the dual mesh defining the value on each centroid. There are ways of solving the problem, but I am still curious about this case. – nicoguaro Mar 22 at 21:14
• @nicoguaro Take a look at how the "virtual element method" deals with this. They solve equations on arbitrary polygonal meshes, so a regular hexagon is a typical cell shape for them. – Wolfgang Bangerth Mar 23 at 9:01
• Might be worthwhile instead minimizing some objective function (say a measure of the surface curvature, maybe the Gaussian curvature at the midpoint) subject to the interpolation constraints above. Lagrange multipliers, etc. – A rural reader Mar 23 at 22:55
As noticed by @AbdullahAliSivas the set of basis functions is not unisolvent.
In 1d case to obtain an polynomial interpolation of degree $$N$$ you just need of $$N+1$$ distinct points, $$x_i \neq x_j$$ if $$i \neq j$$. Changing the points disposition change the quality of the interpolation, because change the Lebesgue's costant. Think to Runge's phenomenon.
Going to dimension $$\ge 2$$ there is a drastic change. An arbitrary set of points does not guarantee the possibility of interpolation. To remark not the quality, but the possibility to have got an interpolation.
This come from a Haar's theorem. I report here:
Def Consider the linear finite dimensional subset $$\mathbf{B} \subseteq C(\Omega)$$ with the basis $$\{B_1, \dots , B_N\}$$. $$\mathbf{B}$$ is an Haar space on $$\Omega$$ if $$\det (B_i(x_j)) \neq 0$$ for any set of distinct $$x_1, \dots, x_N$$ in $$\Omega$$.
Note that your interpolation matrix $$A$$ is $$A_{ij} = B_j(x_i)$$.
Theorem If $$\Omega \subset \mathbb{R}^d$$ with $$d \geq 2$$ contains an interior point, the there exist no Haar spaces of continuous functions except for 1d case.
If you look the proof you can see that precisely the geometry of a space with dimension $$\geq 2$$ permit to swap in continuous way two points, without crash, of an arbitrary set. In in a reasoning by contradiction the determinant of matrix $$A$$ change sign and by continuity must go to zero at same time (here the contradiction).
This does not happens with 1d because the points are in a straight line and is not possible to swap two points without crash. To choose function $$B_i$$ with a dependence by the data $$x_j$$ allows to avoid the implication of Haar theorem and this is the door for the branch of scattered data approximation.
UPDATE
As requested I left a reference for the theorem:
Mairhuber, John C., On Haar’s theorem concerning Chebychev approximation problems having unique solutions, Proc. Am. Math. Soc. 7, 609-615 (1956). ZBL0070.29101.
• Could you add a reference to this theorem? – Amit Hochman Apr 2 at 12:45
• @AmitHochman Hi, I update the post. – Mauro Vanzetto Apr 2 at 13:29
Since you are trying to find the gradient of the function in the vertices of a triangular mesh. So, we are considering an interpolation over the dual mesh defining the value on each centroid, you can try those area based interpolation function e.g, Wachspress Interpolation function.
The benefit of this interpolation function is that it is general enough for a polygon of any side and for n=4 it reduces to standard linear interpolation of quad element. You also do not need to place any node inside. You can construct the interpolation function just by using the coordinate of the vertices, no need of any matrix inversion to get the coefficient values. The only limitation is that the interpolation function is not in polynomial form rather a rational function.
### Reference
You can treat each of the interpolation nodes as points in the complex plane, i.e., $$z_k = \exp(2\pi i k/ 6)$$, then form the Vandermonde matrix $$V_{kl} = z_k^l$$, for $$k, l = 1, 2, ..., 6$$, and solve $$Vc = f$$ for the vector of coefficients $$f$$. Note that in this case, the Vandermonde matrix obeys $$V^HV = 6I$$ and the interpolation is equivalent to interpolation with trigonometric polynomials or the discrete Fourier Transform. There is no ill-conditioning of $$V$$ to worry about. The interpolant is simply $$\sum_k c_k z^k$$. For a real-valued interpolant, just take its real part, which would be a polynomial of degree 6 in $$x = \Re(z)$$ and $$y = \Im(z)$$. Below is matlab code that implements this.
z0 = exp(1i*linspace(0, 2*pi*(1-1/6), 6));
V = vander(z0);
f = randn(6, 1);
c = V'*f/6;
x = linspace(-1.1, 1.1, 100);
[X, Y] = meshgrid(x);
Z = X + 1j*Y;
fInterp = polyval(c, Z);
clf
mesh(x, x, real(fInterp));
hold on
plot3(real(z0), imag(z0), f, 'o')
axis equal
In general, there is no guarantee you can interpolate using an arbitrary set of polynomials, as your example shows. But if you choose your basis of polynomials as the real and imaginary parts of the powers $$(x+iy)^k$$, for $$k = 1, 2, ..., 6,$$ then an interpolant is guaranteed to exist.
Let's use the polar coordinates ($$\rho,\theta$$), then for the six vertices of the regular hexagon defined by $$\rho$$=1 and polar angles $$\theta_i$$ there are the following equations
$$f(\theta_i) = a_0 \cos^2(\theta_i) + a_1 \sin^2(\theta_i) + a_2 \cos(\theta_i) \sin(\theta_i) + \\ + a_3 \cos(\theta_i) + a_4 \sin(\theta_i) + a_5 = f_i$$
where $$f_i$$ is the value at the i$$^{th}$$ vertex $$\theta_i = 2\pi i/6,$$ i $$\in$$ {0...5}; and $$a_k$$, k$$\in$${0...5}, are the unknown coefficients to be found. These equations are compactly written in terms of the Vandermonde matrix that is shown in the problem statement.
It looks like there are six degrees of freedom and six constraints here, so it should be a well posed problem. However, there is an identity linking the three coefficients,
$$a_0 \cos^2(\theta) + a_1 \sin^2(\theta) + a_5 = (a_0-a_1) \cos^2(\theta) + (a_1+a_5) = \\ (a_1-a_0) \sin^2(\theta) + (a_5+a_0)$$
So there is some arbitrariness in the coefficients $$a_i$$, and that's the crux of the problem, that's what makes the original linear system of six equations ill-conditioned. The number of independent degrees of freedom is one less, whether it is a symmetric arrangement on a unit circle (the regular hexagon) or not.
For clarity of this argument, one can rewrite the original equations like this:
$$f(\theta_i) = b_0 \cos(2\theta_i) + b_1 \sin(2\theta_i) + b_2 \cos(\theta_i) + b_3 \sin(\theta_i) + b_4 = f_i,$$
where $$b_0=(a_0-a_1)/2$$, $$b_1=a_2/2$$, $$b_2=a_3$$, $$b_3=a_4$$, $$b_4=a_5+(a_0+a_1)/2$$.
Clearly there are five independent degrees of freedom here $$b_k$$, k$$\in$${0...4}, and one should have a well posed problem setting the values of the interpolating polynomial at five arbitrary points on a unit circle.
You need either a cubic term or 3D coordinate system to successfully interpolate in a hexagon domain. I do not know how this can be proven mathematically.
The following two papers might offer some insights.
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2021-06-23 23:09:33
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https://math.stackexchange.com/questions/771525/convergence-of-sum-n-1-infty-frach-np-n-fracnnn
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# Convergence of $\sum_{n=1}^{\infty}(\frac{H_n}{p_n}-\frac{n}{n^n})$
Does this diverge or converge ?? $$\sum_{n=1}^{\infty}(\frac{H_n}{p_n}-\frac{n}{n^n})$$ where $H_n$ is the nth harmonic number, $p_n$ is the nth prime.
My impression is that it diverges, but I don't see how I can prove it! I tried on wolframalpha but no clue.
• I believe that $\sum\dfrac{H_n}{p_n}$ diverges since the sequence of partial sums is increasing. – user122283 Apr 27 '14 at 16:01
• That would mean that $H_n$ increases faster than $p_n$ interesting result! – user146159 Apr 27 '14 at 16:04
A classic result is $$H_n\sim_\infty \ln n$$ and by the Flegner's result$^{(1)}$ in $1990$ we have $$0.91\; n \ln(n) < p_n < 1.7\; n \ln(n)$$ hence the series $$\sum_{n=1}^\infty \frac{H_n}{p_n}$$ is divergent. Since the series $$\sum_{n=1}^\infty\frac{n}{n^n}$$ is obviously convergent then given series is divergent.
$(1)$ The page is in French language.
Hint: $\sum_{n=1}^{\infty } \frac{n}{n^n}<\infty$ but $\sum_{n=1}^{\infty } \frac{1}{p_n}=\infty$.
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2019-07-20 11:57:02
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https://www.calculus-online.com/exercise/5625
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# Equations – Factorization of a polynomial equation – Exercise 5625
## Exercise
Factor the polynomial equation
$$10x^4-15x^3-4x^2+6x=0$$
$$x(2x-3)(5x^2-2)$$
## Solution
$$10x^4-15x^3-4x^2+6x=$$
We extract a common factor
$$=x(10x^3-15x^2-4x+6)=$$
Again we extract common factors in the parentheses
$$=x(5x^2(2x-3)-2(2x-3))=$$
Once again we extract a common factor – this time the expression in parentheses:
$$=x(2x-3)(5x^2-2)$$
Share with Friends
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2023-04-02 09:26:15
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http://math.stackexchange.com/questions/257063/codimension-one-foliation
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# Codimension one foliation
let $f:M \rightarrow \mathbb{R}$ be a nowhere vanishing function defined on a Riemannian manifold $(M,g)$. consider the distribution $\ker(df)$. This is clearly involutive and thus defines a foliation. My question is now: is it possible that the dimension of the foliation changes or it remains $n-1$, where $n = \dim(M)$? Well this should depend on the function, right? What conditions, on $f$, does one need to impose in order that the foliation remains codimension one?
bill
-
The dimension can certainly change. Consider $f$ Morse at a critical point - e.g. $f:\mathbb{S}^2\ni (x,y,z)\to z\in\mathbb{R}$. But by Sard, almost everywhere the foliation is codimension $1$. – Neal Dec 12 '12 at 13:37
I can't see why you want $f$ to be nowhere vanishing. Do you mean instead that $f$ is such that $\mathrm{d}f$ is nowhere vanishing? In addition, I also don't see why you want a Riemannian manifold. Nowhere is the metric $g$ necessary for this question. – Willie Wong Dec 12 '12 at 13:52
Adding on to what Neal said, if $M$ is compact (and more than one point), $df$ must vanish at least twice - at the max and min of $f$. Further, for example, if $M = S^{2n}$, $M$ doesn't have any codimension one foliations at all. – Jason DeVito Dec 12 '12 at 14:10
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2015-05-24 05:51:39
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https://ltwork.net/in-the-modern-system-of-classification-the-three-domains--9784894
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# In the modern system of classification, the three domains are Bacteria, Archaea, and .A: EukaryaB: ProtistC: Fungi
###### Question:
In the modern system of classification, the three domains are Bacteria, Archaea, and . A: Eukarya
B: Protist
C: Fungi
### 5/6=7n+9/9 (solve for n)
5/6=7n+9/9 (solve for n)...
### Which statement identifies the central idea of the text? Japanese relocation during World War IIA.) The internment of Japanese Americans
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### Which of the following statements is true? -6 < -8 -6> -8 or -6=-8
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### 75There is smoke coming from the bonfire.(a) What is smoke? Tick (~) one box.Gases made by melting.Liquid from evaporation.New materials made by burning.Solids
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### If (x + 1)(x - 3) = 12, then which of the following statements is true? ox+1=0 or x-3=0 x + 1 = 12 or
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### The amount of time t (in minutes) that it takes to put out a fire varies inversely with g, the volume
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### Why is satire often censored by governments or powerful people? a. because it asks the audience to rise up against the government
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### What decisions do cartographers make that can influence how we perceive a map?A)where to draw political bordersB)colors of the mapC)kind
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### This figure is made up of two rectangular prisms. 10 ft 3 ft What is the volume of the figure? 6 ft Enter your answer in the box. 12 ft
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### Who is one important person from the first industrial revolution
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### Hey, I need someone to read a story I made to see if it's good will give brainliest and a ton of points lol :)
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### Lines l and m are parallel. If m∠3 = 3x + 11 degrees and m∠2 = 4x + 1 degrees, what is the measure of ∠7? 41 degrees 139
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### Why was governor talmodge opposed to new deal programs such as social security a) they thought they
Why was governor talmodge opposed to new deal programs such as social security a) they thought they expanded the power of the state government too much b) he thought they weekend the power of national government c) he thought they expanded the power of federal government too much d) he thought t...
### Dr. Thermo, only has one bottle of neon. However, he needs to run two experiments, each requiring its
Dr. Thermo, only has one bottle of neon. However, he needs to run two experiments, each requiring its own bottle. Therefore, he plans to connect the two bottles together and open the valves on each so that each bottle is partially filled. He wants to know how the enthalpy of the gas will change when...
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2023-03-25 13:46:56
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http://thequantumtunnel.com/tag/the-big-bang-theory/
|
# Tag Archives: The Big Bang Theory
## The Big Bang Theory of Dolbear’s Law
Season 03, Episode 02: “The Jiminy Conjecture”
In the “The Jiminy Conjecture” episode, Raj (Kunal Nayyar), Howard (Simon Helberg) and Sheldon (Jim Parsons) are having dinner when they hear the sound of a cricket chirping. Sheldon claims to know the species from the cricket’s chirping speed and the room’s ambient temperature. Not to be outdone and still angry from losing a bet in the comic book store, Howard wagers his rare Fantastic Four #48 (“The Coming of Galactus”) against Sheldon’s The Flash #123 “Flash of Two Worlds”.
## Dolbear’s Law
Sheldon is able to determine the room’s temperature by using Dolbear’s Law, an equation that states the relationship between air temperature and the rate at which a cricket chirps. This relationship was formulated by an American physicist, Amos Dolbear in 1897 in his article “The Cricket as a Thermometer”. While Dolbear didn’t specify the species of cricket in his article, it is generally believed to the snowy tree cricket.
Amos Emerson Dolbear (November 10, 1837 – February 23, 1910) was an American physicist and inventor. He is noted for his pioneering work in wireless telegraphy — several years before Guglielmo Marconi. He is also known for finding a relatoinship between a cricket’s chirp rate and ambient temperature.
If we look closely we can see Dolbear’s equations written on the whiteboard in the background. They all describe the relationship between temperature in Fahrenheit to the number of chirps different species of cricket make in one minute.
\begin{align*}
\mathrm{Field~Cricket:} \; T_{f} &= 50 + \frac{(N-40)}{4} \\
\mathrm{Snowy~Tree~Cricket:} \; T_{f} &= 50 + \left[\frac{(N – 92)}{4.7}\right] \\
\mathrm{Katydid:} \; T_{f} &= 60 + \left[\frac{(N – 19)}{3}\right]
\end{align*}
It is a popular myth that crickets chirp by rubbing their legs together but that couldn’t be further from the truth. Only the males chirp and they produce sounds through the process of stridulation i.e by rubbing certain body parts together. In the case of crickets, there is a stridulatory organ, a large vein running along the bottom of each wing, that is covered with “teeth” or serrations like a comb. The chirping sound is created when the cricket runs the top of its wing along the bottom serrated edge of the other. As the cricket does this, it holds its wings up and open so they act as acoustical sails.
Insects are cold-blooded and take the temperature of their surroundings. As a result, a cricket’s chirp rate depends on its metabolism. This can be determined by the Arrhenius equation which describes the activation energy needed to induce a chemical reaction. The less energy there is, the slower a chemical reaction and hence metabolism. As temperatures rise, more energy is available for muscle contractions. This accounts for the relation of observed by Dolbear.
## Measuring Dolbear’s Law
We know by the end of the show that Howard was right and Sheldon was wrong. We can calculate the number of chirps with temperature for the various cricket species seen on the whiteboard.
Table showing Temperature and Number of Cricket Chirps per Minute
Temperature(°F) Temperature(°C) Field Cricket Snowy Tree Cricket Katydid
60 16 80 139 19
65 18 100 163 34
70 21 120 186 49
75 24 140 210 64
80 27 160 233 79
85 29 180 257 94
90 32 200 280 109
There are noticeable differences in chirp rates between the snowy tree cricket and the field cricket–the snowy tree cricket has a faster chirp rate. But how could Sheldon been wrong? According to the roommate agreement, the apartment is supposed to be 71°F(22°C) which translates to a chirp rate of 191 for the snowy tree cricket. Assuming Sheldon counted the chirps correctly, that would translate to a much warmer temperature of 88°F(31°C) for the field cricket. Could someone have changed the temperature in violation of the roommate agreement? Surely this would have been noticed without looking at the thermostat. Your guess is as good as mine.
## The Big Bang Theory of the Doppler Effect
Season 01, Episode 06: “The Middle-Earth Paradigm”
Sheldon Cooper dresses as the Doppler Effect on the Season 1, Episode 6 “The Middle-Earth Paradigm” of the Big Bang Theory
Yes. It’s the apparent change in the frequency of a wave caused by relative motion between the source of the wave and the observer.
-Sheldon Cooper
In the “Middle-Earth Paradigm” episode, Sheldon Cooper dresses as the “Doppler Effect” for Penny’s Halloween party. The Doppler Effect (or Doppler Shift) describes the change in pitch or frequency that results as a source of sound moves relative to an observer; moving relative can mean either the source is moving while the observer is stationary or vice versa. It is commonly heard when a siren approaches and recedes from an observer. As the siren approaches, the pitch sounds higher and lowers as it moves away. This effect was first proposed by Austrian physicist Christian Doppler in 1842 to explain the color of binary stars.
In 1845, Christophorus Henricus Diedericus (C. H. D.) Buys-Ballot, a Dutch chemist and meteorologist conducted the famous experiment to prove this effect. He assembled a group of horn players on an open cart attached to a locomotive. Ballot then instructed the engineer to rush past him as fast as he could while the musicians played and held a constant note. As the train approached and receded, Ballot noted that the pitch changed as he stood and listened on the stationary platform.
# Physics of the Doppler Effect
A stationary sound source has sound waves radiating outward and can be viewed as concentric circles.
As a stationary sound source produces sound waves, its wave-fronts propagate away from the source at a constant speed, the speed of sound. This can be seen as concentric circles moving away from the center. All observers will hear the same frequency, the frequency of the source of the sound.
When either the source or the observer moves relative to each other, the frequency of the sound that the source emits does not change but rather the observer hears a change in pitch. We can think of the following way. If a pitcher throws balls to someone across a field at a constant rate of one ball a second, the person will catch those balls at the same rate (one ball a second). Now if the pitcher runs towards the catcher, the catcher will catch the balls faster than one ball per second. This happens because as the catcher moves forward, he closes in the distance between himself and the catcher. When the pitcher tosses the next ball it has to travel a shorter distance and thus travels a shorter time. The opposite is true if the pitcher was to move away from the catcher.
If instead of the pitcher moving towards the catcher, the pitcher stayed stationary and the catcher ran forward. As the catcher runs forward, he closes in the distance between him and the pitcher so the time it takes from the ball to leave the pitcher’s hand to the catcher’s mitt is decreased. In this case, it also means that the catcher will catch the balls at a faster rate than the pitcher tosses them.
## Sub Sonic Speeds
The source radiates sound waves outward. As it moves, the center of each new wavefront is slightly displaced to the right and the wavefronts bunch up on the right side (front) and spread out further apart on the left side (behind) the source.
We can apply the same idea of the pitcher and catcher to a moving source of sound and an observer. As the source moves, it emits sounds waves which spread out radially around the source. As it moves forward, the wave-fronts in front of the source bunch up and the observer hears an increase in pitch. Behind the source, the wave-fronts spread apart and the observer standing behind hears a decrease in pitch.
### The Doppler Equation
When the speeds of source and the receiver relative to the medium (air) are lower than the velocity of sound in the medium, i.e. moves at sub-sonic speeds, we can define a relationship between the observed frequency, $$f$$, and the frequency emitted by the source, $$f_0$$.
$f = f_{0}\left(\frac{v + v_{o}}{v + v_{s}}\right)$
where $$v$$ is the speed of sound, $$v_{o}$$ is the velocity of the observer (this is positive if the observer is moving towards the source of sound) and $$v_{s}$$ is the velocity of the source (this is positive if the source is moving away from the observer).
#### Source Moving, Observer Stationary
We can now use the above equation to determine how the pitch changes as the source of sound moves towards the observer. i.e. $$v_{o} = 0$$.
$f = f_{0}\left(\frac{v}{v – v_{s}}\right)$
$$v_{s}$$ is negative because it is moving towards the observer and $$v – v_{s} < v$$. This makes $$v/(v - v_{s})$$ larger than 1 which means the pitch increases.
#### Source Stationary, Observer Moving
Now if the source of sound is still and the observer moves towards the sound, we get:
$f = f_{0}\left( \frac{v + v_{o}}{v} \right)$
$$v_{o}$$ is positive as it moves towards the source. The numerator is larger than the denominator which means that $$v + v_{o}/v$$ is greater than 1. The pitch increases.
## Speed of Sound
As the source of sound moves at the speed of sound the wave fronts in front of the source all bunch up at the same point.
As the source of sound moves at the speed of sound, the wave fronts in front become bunched up at the same point. The observer in front won’t hear anything until the source arrives. When the source arrives, the pressure front will be very intense and won’t be heard as a change in pitch but as a large “thump”.
The observer behind will hear a lower pitch as the source passes by.
$f = f_{0}\left( \frac{v – 0}{v + v} \right) = 0.5 f_{0}$
Early jet pilots flying at the speed of sound (Mach 1) reported a noticeable “wall” or “barrier” had to be penetrated before achieving supersonic speeds. This “wall” is due to the intense pressure front, and flying within this pressure front produced a very turbulent and bouncy ride. Chuck Yeager was the first person to break the sound barrier when he flew faster than the speed of sound in the Bell X-1 rocket-powered aircraft on October 14, 1947.
Bell X-1 rocket plane of the United States Air Force (NASA photography)
As the science of super-sonic flight became better understood, engineers made a number changes to aircraft design that led the the disappearance of the “sound barrier”. Aircraft wings were swept back and engine performance increased. By the 1950s combat aircraft could routinely break the sound barrier.
## Super-Sonic
As the source moves faster than the speed of sound, i.e. faster than the sound waves it creates, it leads its own advancing wavefront. The sound source will pass by the stationary observer before the observer hears the sound.
As the sound source breaks and moves past the “sound barrier”, the source now moves faster than the sound waves it creates and leads the advancing wavefront. The source will pass the observer before the observer hears the sound it creates. As the source moves forward, it creates a Mach cone. The intense preseure front on the Mach cone creates a shock wave known as a “sonic boom”.
### Twice the Speed of Sound
Something interesting happens when the source moves towards the observer at twice the speed of sound — the tone becomes time reversed. If music was being played, the observer will hear the piece with the correct tone but played backwards. This was first predicted by Lord Rayleigh in 1896 .
We can see this by using the Doppler Equation.
$f = f_{0}\left(\frac{v}{v-2v}\right)$
This reduces to
$f=-f_{0}$
which is negative because the sound is time reversed or is heard backwards.
# Applications
U.S. Army soldier uses a radar speed gun to catch speeding violators at Tallil Air Base, Iraq.
The Doppler Effect is used in radar guns to measure the speed of motorists. A radar beam is fired at a moving target as it approaches or recedes from the radar source. The moving target then reflects the Doppler-shifted radar wave back to the detector and the frequency shift measured and the motorist’s speed calculated.
We can combine both cases of the Doppler equation to give us the relationship between the reflected frequency ($$f_{r}$$) and the source frequency ($$f$$):
$f_{r} = f \left(\frac{c+v}{c-v}\right)$
where $$c$$ is the speed of light and $$v$$ is the speed of the moving vehicle. The difference between the reflected frequency and the source frequency is too small to be measured accurately so the radar gun uses a special trick that is familiar to musicians – interference beats.
To tune a piano, the pitch can be adjusted by changing the tension on the strings. By using a tuning instrument (such as a tuning fork) which can produce a sustained tone over time, a beat frequency can be heard when placed next to the vibrating piano wire. The beat frequency is an interference between two sounds with slightly different frequencies and can be herd as a periodic change in volume over time. This frequency tells us how far off the piano strings are compared to the reference (tuning fork).
To detect this change in a radar gun does something similar. The returning wave is “mixed” with the transmitted signal to create a beat note. This beat signal or “heterodyne” is then measured and the speed of the vehicle calculated. The change in frequency or the difference between $$f_{r}$$ and $$f$$ or $$\Delta f$$ is
$f_{r} – f = f\frac{2v}{c-v}$
as the difference between the speed of light, $$c$$, and the speed of the vehicle, $$v$$, is small, we can approximate this to
$\Delta f \approx f\frac{2v}{c}$
By measuring this frequency shift or beat frequency, the radar gun can calculate and display a vehicle’s speed.
# “I am the Doppler Effect”
The Doppler Effect is an important principle in physics and is used in astronomy to measure the speeds at which galaxies and stars are approaching or receding from us. It is also used in plasma physics to estimate the temperature of plasmas. Plasmas are one of the four fundamental states of matter (the others being solid, liquid, and gas) and is made up of very hot, ionized gases. Their composition can be determined by the spectral lines they emit. As each particle jostles about, the light emitted by each particle is Doppler shifted and is seen as a broadened spectral line. This line shape is called a Doppler profile and the width of the line is proportional to the square root of the temperature of plasma gas. By measuring the width, scientists can infer the gas’ temperature.
We can now understand Sheldon’s fascination with the Doppler Effect as he aptly explains and demonstrates its effects. As an emergency vehicle approaches an observer, its siren will start out with a higher pitch and slide down as as it passes and moves away from the observer. This can be heard as the (confusing) sound he demonstrates to Penny’s confused guests.
# References
## The Big Bang Theory of the Inclined Plane
Season 01, Episode 02: “The Big Bran Hypothesis”
Sheldon and Leonard contemplate using the stairs as an inclined plane to move Penny’s furniture to her apartment.
In Season 1 Episode 2 of The Big Bang Theory, “The Big Bran Hypothesis”, Penny (Kaley Cuoco) asks Leonard (Johnny Galecki) to sign for a furniture delivery if she isn’t home. Unfortunately for Leonard and Sheldon, they are left with the task of getting a huge (and heavy) box up to Penny’s apartment.
To solve this problem, Leonard suggest using the stairs as an inclined plane, one of the six classical simple machines defined by Renaissance scientists. Both Leonard and Sheldon have the right idea here. Not only are inclined planes used to raise heavy loads but they require less effort to do so. Though this may make moving a heavy load easier the tradeoff is that the load must now be moved over a greater distance. So while, as Leonard correctly calculates, the effort required to move Penny’s furniture is reduced by half, the distance he and Sheldon must move Penny’s furniture twice the distance to raise it directly.
# Mathematics of the Inclined Plane
## Effort to lift block on Inclined Plane
Now we got an inclined plane. Force required to lift is reduced by the sine of the angle of the stairs… call it 30 degrees, so about half.
Free-body Diagram of a block on an inclined plane. It shows the forces acting on a block and the force needed to keep it stationary and not let it slip down.
To analyze the forces acting on a body, physicists and engineers use rough sketches or free body diagrams. This diagram can help physicists model a problem on paper and to determine how forces act on an object. We can resolve the forces to see the effort needed to move the block up the stairs.
If the weight of Penny’s furniture is $$W$$ and the angle of the stairs is $$\theta$$ then
$\angle_{\mathrm{stairs}}\equiv\theta \approx 30^\circ$
and
$\Rightarrow\sin 30^\circ = \frac{1}{2}$
So the effort needed to keep the box in place is about half the weight of the furniture box or $$\frac{1}{2}W$$, just as Leonard says.
## Distance moved along Inclined Plane
The relationship between the height $$h$$ the block is raised and the distance it moves $$d$$ is the sine of the angle $$\theta$$.
While the inclined plane allows Leonard and Sheldon to push the box with less effort, the tradeoff is that the distance they move along the incline is twice the height to raise the box vertically. Geometry shows us that
$\sin \theta = \frac{h}{d}$
We again assume that the angle of the stairs is approximately $$30^\circ$$ and $$\sin 30^{\circ} = 1/2$$ then we have $$d=2h$$.
# Uses of the Inclined Plane
We see inclined planes daily without realizing it. They are used as loading ramps to load and unload goods. Wheelchair ramps also allow wheelchair users, as well as users of strollers and carts, to access buildings easily. Roads sometimes have inclined planes to form a gradual slope to allow vehicles to move over hills without losing traction. Inclined planes have also played an important part in history and were used to build the Egyptian pyramids and possibly used to move the heavy stones to build Stonehenge.
## Lombard Street (San Francisco)
Lombard Street is one of the most visited street in San Francisco as seen from Coit Tower. It is best known for the one-way section on Russian Hill between Hyde and Leavenworth Streets, in which the roadway has eight sharp turns (or switchbacks) that have earned the street the distinction of being the crookedest “most winding “street in the world, though this title is contested. (Photo by David Yu).
Lombard Street in San Francisco is famous for its eight tight hairpin turns (or switchbacks) that have earned it the distinction of being the crookedest street in the world (though this title is contested). These eight switchbacks are crucial to the street’s design as the reduce the hills natural 27° grade which is too steep for most vehicles. It is also a hazard to pedestrians, who are more accustomed to a more reasonable 4.86° incline due to wheel chair navigability concerns.
Technically speaking, the “zigzag” path doesn’t make climbing or coming down the hill any easier. As we have seen, all it does is change how various forces are applied. It just requires less effort to move up or down but the tradeoff is that you travel a longer distance. This has several advantages. Car engines have to be less powerful to climb the hill and in the case of descent, less force needs to be applied on the brakes. There are also safety considerations. A car will not accelerate down the switch back path as fast than if it was driven straight down, making speeds safer and more manageable for motorists.
This idea of using zigzagging paths to climb steep hills and mountains is also used by hikers and rock climbers for very much the same reason Lombard Street zigszags. The tradeoff is that the distance traveled along the path is greater than if a climber goes straight up.
# The Descendants of Archimedes
We don’t need strength, we’re physicists. We are the intellectual descendants of Archimedes. Give me a fulcrum and a lever and I can move the Earth. It’s just a matter of… I don’t have this, I don’t have this!
We see that Leonard had the right idea. If we were to assume are to assume — based on the size of the box — that the furniture is approximately 150 lbs (65kg) and the effort is reduced by half, then they need to push with at least 75 lbs of force. This is equivalent to moving a 34kg mass. If they both push equally, they are each left pushing a very manageable 37.5 lbs, the equivalent of pushing a 17kg mass.
Penny’s apartment is on the fourth floor and we if we assume a standard US building design of ten feet per floor, this means a 30 foot vertical rise. The boys are left with the choice of lifting 150 lbs vertically 30 feet or moving 75lbs a distance of 60 feet. The latter is more manageable but then again, neither of our heroes have any upper body strength.
## The Big Bang Theory of Trans-Neptunian objects
Season 02, Episode 04: “The Griffin Equivalency”
Rajesh Koothrappali (Kunal Nayyar) looks pleased as he talks about his inclusion into People magazine’s “30 Under 30 to watch” list for his discovery of the fictional trans-Neptunian object, 2008 NQ17 which exists beyond the Kuiper belt
In “The Griffin Equivalency” episode of The Big Bang Theory, Raj (Kunal Nayyar) announces that he has been chosen as one of People Magazine’s “30 visionaries under 30 years to watch” list for his discovery of the trans-Neptunian object 2008 NQ17. This fictional astronomical body exists just beyond the Kuiper Belt, a region of the Solar System that extends from the orbit of Neptune (about 30 Astronomical units(AU) away), to a distance of 50AU from the Sun. (1AU is roughly the distance of the Earth from the Sun).
The distribution and classification of the trans-Neptunian Objects in our solar system
The first trans-Neptunian object discovered was Pluto in 1930 and is the second most massive known dwarf planet in the Solar System. These objects are so small and distant that it wasn’t until 1992 that the second trans-Neptunian object orbiting the Sun, (15760) 1992 QB1, was discovered. As cryptic as these names sound, i.e 2008 NQ17 and 1992 QB1, the nomenclature comes from a provisional designation in astronomy that names objects immediately following their discovery. Once a proper orbit has been calculated, a permanent designation consisting of a number-name combination is given by the Minor Planet Center. Unfortunately for Raj, so many of these minor planets are discovered that most won’t be named for their discoverers.
# Provisional designation of Minor Planets
The provisional naming system for minor planets, of which dwarf planets, asteroids, trojans, centaurs, Kuiper belt objects, and other trans-Neptunian objects are a part of, have been in place since 1925. The first element in a minor planet’s provisional designation is the year of discovery, followed by two letters and sometimes a number.
The first letter indicates the “half-month” of the object’s discovery. “A” denotes a discovery in the first half of January (between January 1st and January 15th) while “B” denotes a discovery in the second half of January (between January 16th and January 31st). The letter “I” is not used so the letters go all the way up to “Y”. The first half is always the 1st through to the 15th of the month, regardless of the numbers of days in the second “half”.
First Letter (A-G)
A B C D E F G
Jan 1 Jan 16 Feb 1 Feb 16 Mar 1 Mar 16 Apr 1
First Letter (H-O)
H J K L M N O
Apr 16 May 1 May 16 Jun 1 Jun 16 Jul 1 Jul 16
First Letter (P-V)
P Q R S T U V
Aug 1 Aug 16 Sep 1 Sep 16 Oct 1 Oct 16 Nov 1
First Letter (W-Y)
W X Y
Nov 16 Dec 1 Dec 16
The second letter and the number indicate the order or discovery within that half-month.
Second Letter (A-K)
A B C D E F G H J K
1 2 3 4 5 6 7 8 9 10
Second Letter (L-U)
L M N O P Q R S T U
11 12 13 14 15 16 17 18 19 20
Second Letter (V-Z)
V W X Y Z
21 22 23 24 25
So the 14th minor planet discovered in the first half of June 2013 will get the provisional planetary designation 2013 LO. As modern techniques has made it (relatively) easier to detect these objects, there can be more than 25 discoveries in a given half-month. To solve this a subscript number is added to indicate the number of times that the letters have cycled through. So the 40th minor planet to be discovered in the first half of June 2013 will gain the designation 2013 LP1.
\begin{aligned} 40 & = 25 + 15 \\ \therefore P & = 15 \end{aligned}
# Planet Bollywood
Using this information we can calculate when Raj discovered “Planet Bollywood”. It was discovered some time between July 1-15 and it was the 441st minor planet to be discovered in that half month.
\begin{aligned} Q & = 16 \\ 25 \times 17 & = 425 \end{aligned}
And $$425 + 16 = 441$$.
It must have been a particularly busy month in the Big Bang Theory Universe. A quick check at the Minor Planet Center’s Provisional Designations webpage indicates that there were 137 astronomical bodies discovered in that half-month. That would make that last trans-Neptunian object to be discovered to receive the provisional designation 2008 NM4.
Long live Planet Bollywood!
## The Big Bang Theory of the Rolling Ball Problem
Big Bang Theory Season 2, Episode 5: “The Euclid Alternative”
In the opening scene of the “The Euclid Alternative”, we see Sheldon (Jim Parsons) demanding that Leonard (Johnny Galecki)needs to drive him around to run various errands. Leonard, after spending a night in the lab using the new Free Electron Laser to perform X-ray diffraction experiments. In the background, we can see equations that describe a rolling ball problem on the whiteboard in the background.
Scene with Sheldon Cooper. The background shows a whiteboard with the physics equations of a rolling ball.
Rolling motion plays an important role in many familiar situations so this type of motion is paid considerable attention in many introductory mechanics courses in physics and engineering. One of the more challenging aspects to grasp is that rolling (without slipping) is a combination of both translation and rotation where the point of contact is instantaneously at rest.The equations on the white board describe the velocity at the point of contact on the ground, the center of the object and at the top of the object.
# Pure Translational Motion
An object undergoing pure translational motion
When an object undergoes pure translational motion, all of its points move with the same velocity as the center of mass– it moves in the same speed and direction or $$v_{\textrm{cm}}$$.
# Pure Rotational Motion
An object undergoing pure rotational motion. The velocity of each point of the object depends on the distance away from the center. The further away the faster it moves as can be seen by the length of the arrows.
In the case of a rotating body, the speed of any point on the object depends on how far away it is from the axis of rotation; in this case, the center. We know that the body’s speed is $$v_{\textrm{cm}}$$ and that the speed at the edge must be the same. We may think that all these points moving at different speeds poses a problem but we know something else — the object’s angular velocity.
The angular speed tells us how fast an object rotates. In this case, we know that all points along the object’s surface completes a revolution in the same time. In physics, we define this by the equation:
\omega=\frac{v}{r}
where $$\omega$$ is the angular speed. We can use this to rewrite this equation to tell us the speed of any point from the center:
v(r)=\omega r
If we look at the center, where $$r=0$$, we expect the speed to be zero. When we plug zero into the above equation that is exactly what we get:
v(0)= \omega \times 0 = 0
\label{eq:zero}
If we know the object’s speed, $$v_{\textrm{cm}}$$ and the object’s radius, $$R$$, using a little algebra we can define $$\omega$$ as:
$\omega=\frac{v_{\textrm{cm}}}{R}$
or the speed at the edge, $$v_{\textrm{cm}}$$ to be $$v(R)$$ to be:
v_{\textrm{cm}}=v(R) = \omega R
\label{eq:R}
# Putting it all Together
An object that rolls undergoes both translational and rotational motion. To determine the speed at any point we must add both of these speeds. We see the translational speed show in red and the rotational speed show in in blue.
To determine the absolute speed of any point of a rolling object we must add both the translational and rotational speeds together. We see that some of the rotational velocities point in the opposite direction from the translational velocity and must be subtracted. As horrifying as this looks to do, we can reduce the problem somewhat to what we see on the whiteboard. Here we see the boys reduce the problem and look at three key areas, the point of contact with the ground ($$P)$$, the center of the object, ($$C$$) and the top of the object ($$Q$$).
Rendering copy of the whiteboard showing the rolling ball problem as seen on the Big Bang Theory.
We have done most of the legwork at this point and now the rolling ball problem is easier to solve.
## At point $$Q$$
At point $$Q$$, we know the translational speed to be $$v_{\textrm{cm}}$$ and the rotational speed to be $$v(R)$$. So the total speed at that point is
v = v_{\textrm{cm}} + v(R)
\label{eq:Q1}
Looking at equation \eqref{eq:R}, we can write $$v(R)$$ as
v(R) = \omega R
Putting this into \eqref{eq:Q1} and we get,
\begin{aligned}
v & = v_{\textrm{cm}} + v(R) \\
& = v_{\textrm{cm}} + \omega R \\
& = v_{\textrm{cm}} + \frac{v_{\textrm{cm}}}{R}\cdot R \\
& = v_{\textrm{cm}} + v_{\textrm{cm}} = 2v_{\textrm{cm}}
\end{aligned}
which looks almost exactly like Leonard’s board, so we must be doing something right.
## At point $$C$$
At point $$C$$ we know the rotational speed to be zero (see equation \eqref{eq:zero}).
Putting this back into equation \eqref{eq:Q1}, we get
\begin{aligned}
v & = v_{\textrm{cm}} + v(r) \\
& = v_{\textrm{cm}} + v(0) \\
& = v_{\textrm{cm}} + \omega \cdot 0 \\
& = v_{\textrm{cm}} + 0 \\
& = v_{\textrm{cm}}
\end{aligned}
Again we get the same result as the board.
## At point $$P$$
At the point of contact with the ground, $$P$$, we don’t expect a wheel to be moving (unless it skids or slips). If we look at our diagrams, we see that the rotational speed is in the opposite direction to the translational speed and its magnitude is
\begin{aligned}
v(R) & = -\omega R \\
& = -\frac{v_{\textrm{cm}}}{R}\cdot R \\
& = -v_{\textrm{cm}}
\end{aligned}
It is negative because the speed is in the opposite direction. Equation \eqref{eq:Q1}, becomes
\begin{aligned}
v & = v_{\textrm{cm}} + v(r) \\
& = v_{\textrm{cm}} – \omega R \\
& = v_{\textrm{cm}} – \frac{v_{\textrm{cm}}}{R}\cdot R \\
& = v_{\textrm{cm}} – v_{\textrm{cm}} = 0
\end{aligned}
Not only do we get the same result for the rolling ball problem we see on the whiteboard but it is what we expect. When a rolling ball, wheel or object doesn’t slip or skid, the point of contact is stationary.
# Cycloid and the Rolling ball
The path a point on a rolling ball draws is called a cycloid.
If we were to trace the path drawn by a point on the ball we get something known as a cycloid. The rolling ball problem is an interesting one and the reason it is studied is because the body undergoes two types of motion at the same time — pure translation and pure rotation. This means that the point that touches the ground, the contact point, is stationary while the top of the ball moves twice as fast as the center. It seems somewhat counter-intuitive which is why we don’t often think about it but imagine if at the point of contact our car’s tires wasn’t stationary but moved. We’d slip and slide and not go anywhere fast. But that is another problem entirely.
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2020-09-25 23:30:08
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http://weisu.blogspot.com/2008/02/mutex-in-multithread.html
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## Thursday, February 07, 2008
### mutex in multithread
Mutexes have the property that they can only be locked by one thread at a time. If two threads attempt to lock the same mutex, only one of them succeeds. The other one blocks until the first thread releases the mutex. In general, you use mutexes to protect some piece of shared data. In order to be useful, the mutex itself must be global so that all threads can see it.
int i; /* i is shared */pthread_mutex_t ilock; /* controls access to i */ ...void func(arg1,arg2) int arg1; /* arg1 is private */ float *arg2; /* arg2 is private, *arg2 is shared */{ int zzz; /*zzz is private(auto storage class)*/ static float f;/*f is shared (static storage class)*/ pthread_mutex_lock(&ilock); /* lock i */ something = i++; /* access i */ pthread_mutex_unlock(&ilock);/* unlock i */}
The details can be found at following link
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2018-02-19 02:15:21
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https://math.libretexts.org/Courses/Monroe_Community_College/MTH_225_Differential_Equations/z10%3A_Linear_Systems_of_Differential_Equations/10.1%3A_Introduction_to_Systems_of_Differential_Equations/10.1.1%3A_Introduction_to_Systems_of_Differential_Equations_(Exercises)
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# 10.1.1: Introduction to Systems of Differential Equations (Exercises)
This page titled 10.1.1: Introduction to Systems of Differential Equations (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.
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2022-09-27 21:48:14
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https://www.kai-arzheimer.com/tag/latex/
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Every remotely relevant reference I came across during the last 15 years or so resides in a single bibtex file. That is not a problem. The problem is that I’m moving into a shiny, new but somewhat smaller office, together with hundreds of copies of journal articles and hundreds of PDFs. Wouldn’t it be good to know which physical copies are effectively redundant (unreadable comments in the margins aside) and can therefore stay behind?
The trouble is that bibtex files have a rather flexible, human readable format. Each entry begins with the @ sign, followed by a type (book, article etc.), a reference name, lots of key/value pairs (fields) in arbitrary order, and even more curly braces.
`grep @ full.bib|wc -l` tells me that I have 2914 references in total. `grep binder|wc -l` (binder is a custom field that I use to keep track of the location of my copies) shows that I have printed out/copied 712 texts over the years, and `grep file|wc -l` indicates that there are 504 PDFs residing on my filesystem. But what is the magnitude of the intersection?
My first inclination was to look for a suitable Python parser/library. Pybtex looked good in principle but is underdocumented and had trouble reading full.bib, because that is encoded in Latin 1. So it was endless hours of amateurish coding and procrastination ahead. Then I remembered the “do one thing, and do it really well” mantra of old. Enter bibtool, which is a fast and reasonably stable bibtex file filter and pretty printer. Bibtool reads “resource files”, which are really just short scripts containing filtering/formatting directives. `select = {binder ".+"}` keeps those references whose “binder” field contains at least one character (`.+` is a regular expression that matches any non-empty string). `select = {file ".+"}` selects all references for which I have a PDF. But bibtool applies a logical OR to these conditions while I’m interested in finding those references that meet both criteria.
The quick solution is to store each statement in a file of its own and apply bibtool twice, using a pipeline for extra efficiency: `bibtool -r find-binder.rsc full.bib|bibtool -r find-pdf >intersection.bib` does the trick and solves my problem in under a minute, without any coding.
As it turns out, there were just 65 references in both groups. Apparently, I stopped printing (or at least filing away) some time ago. Eventually, I binned two copies, but it is the principle that matters.
2019 Update
I still use bibtool for quick filtering/reformatting tasks at the command line, but for more complex jobs involving programmatic access to bibtex files from R, RefManageR is a wonderful package. I have used it here in a bibliometric study of the Radical/Extreme Right literature. And my nifty RRResRobot also relies heavily on RefManageR. If you are interested at all in RefManageR, here is a short and sweet introduction.
My default for writing anything that is longer than a page is LaTeX (possibly via org-mode, if it is short and simple). In fact, the bond that ties me to the LaTeX/Emacs combo is so strong that I want to use it even for texts that are exactly one page long, i.e. conference posters.
CTAN lists a lot of packages and frameworks for posters, but I found most of them too heavy/compl
Political Geography Conference Poster
ex. I don’t create a lot of conference posters and did not want to spend ages putting a few words and graphs on a sheet of glossy paper. At the end of the day, I decided to give beamerp
oster a spin. Beamerposter is an add-on that transforms my favourite presentation package into a poster printing machine. I did not really like the default themes, but Rob Hyndman has created a very alternative nice template that I adapted slightly.
I rather like the result and will go back to the package for the next poster.
I use emacs/$LaTeX$for all my textprocessing needs, and for the last four or five years, I have created all my slides with Till Tantaus excellent “beamer” class. At the moment, I’m teaching a 2nd year stats course (imagine doing this with PowerPoint – the horror! the horror!), so I sometimes use graphs from the assigned text like this one from Long&Freese that illustrates the latent variable/threshold interpretation of the binary logit model. The message should be fairly clear: $y^{*}$ depends on $x$ andfollows a standard logistic distribution around its conditional mean.
But the fact that the bell-curve lies flat in the $x-y^{*}$ plane confused my students no end. So I wasted half a day on creating a nice 3d-plot for them. After trying several options, I settled on pgfplots.sty, which builds on tikz/pgf, the comprehensive, portable graphics package designed by Tantau (here’s a gallery with most amazing examples of what you can do with this little gem). Plotting data and functions with pgfplots in 2d or 3d is a snap, so that was not too hard. Eventually.
Finally, in a desperate attempt to drive the message home, I enlisted the help of animate.sty, yet another amazing package that creates a javascript-based inline animation from my $LaTeX$ source (requires Acrobat reader). So the bell-curves pop out of the plane, in slow motion. Did it help the students to see the light? I have no idea. Here is the source.
A couple of weeks ago, I posted an article on how make and Makefiles can help you to organise your Stata projects. If you are working in a unix environnment, you’ll already have make installed. If you work under Windows, install GNU make – it’s free, and it can make your Stata day. Rather unsurprisingly, make is also extremely useful if you have large or medium-sized latex project (or if you want to include tables and/or graphs produced by Stata) in a latex document. For instance, this comes handy if you have eps-Figures and use pdflatex. pdflatex produces pdf files instead of dvi files. If you produces slides with, this can save you a lot of time because you don’t have to go through the latex – dvips – ps2pdf cycle. However, pdflatex cannot read eps files: you have to convert your eps files with pstoedit to the meta post format, then use meta post to convert them to mps (which can be read by pdflatex). With this Makefile snippet, everything happens automagically:
` #New implicit rules for conversion of eps->mp->mps #Change path if you have installed pstoedit in some other place %.mp : %.eps c:pstoedit/pstoedit.exe -f mpost \$*.eps \$*.mp`
%.mps: %.mp
mpost \$*.mp
mv \$*.1 \$*.mps
rm \$*.mp
#Now specify a target
presentation.pdf: presentation.tex mytab1.tex myfig.mps
#Optional: if you want to create dataset x.eps, run x.do
#Stata must be in your path
%.eps : %.do
tab wstata -e do \$<
Now type make presentation.pdf, and make will call Stata, pstoedit, metapost and pdflatex as required. If you need more figures, just write the do-file and add a dependency.
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2019-09-21 09:39:04
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http://ilsorrisonapoli.it/top-or-bottom-test.html
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Test -> SOLUTION: A 45-foot Ladder Is Leaning Against The Top Of Building. The Problem Is To Count All The Possible Paths From Top Left To Bottom Right Of A MXn Matrix With The Constraints # Driver Program To Test Above Function . M = 3 CBT Is A "top-down" Approach Based On The Idea That If A Person Thinks Right, They Will Feel Alright. Brain Scan Research Suggests Otherwise. Arousal And Emotional Regulation Are Structured Like An A Meniscus Is The Curved Surface At The Top Of A Column Of Liquid. In A Science Class, This Liquid Is Usually Water Or Some Sort Of Aqueous Solution, And The Column Is Usually A Graduated Cylinder Or A Pipet. As You May Have Noticed, When Water Is In Such A Thin Glass Tube, It Does Not Have A Flat Surface At The Top. It Means He Will Top But Doesn’t Like To. 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The Foot Arch Test Is A Simple Way To Determine Whether You Have Flat Feet Or High Arches! Just Follow These Easy Instructions: Pour Water Into A Shallow Pan (the Pan Should Be Big Enough To Fit Your Foot, And The Water Should Be Just Deep Enough For All Parts Of The Bottom Of Your Foot To Get Wet). Bottom Level Freezers Offer A Lot Of Accessibility Advantages And Usually Include Extra Space Compared To Top Freezers (and Often Even Side Freezers, If An Ice Maker Takes Up Space). B.) Heat The Test Tube Only Near The Top Of The Liquid, Not The Bottom; Use A Boiling Stone If Available.. C.) Heat The Test Tube On A Water Bath; Use A Boiling Stone If Available. 4.) Limit The Amount Of Liquid. If You Plan To Heat The A Test Tube, It Should Not Be More Than Half-full. If Possible, It Should Be Only 1/4 To 1/3 Full. But As Creator Dr. Gilda Carle Explained To The Independent's Indy 100 Site, The Test Is Meant To Be A "fun Way To Assess The Behaviors [people Do] Daily, Often Without Thinking," And, Community Moderated Site Where You Can Make Quizzes And Personality Tests, Ask And Answer Questions, Create Profiles, Journals, Forums And More. Top To Bottom School Ranking Top-to-Bottom School Rankings Support For School Accountability Reports Can Be Reached At 877-560-8378 Or MDE-Accountability@michigan.gov . The Following Test Images Are Best Viewed From A Distance Of Approximately Equal To The Diagonal Of The Display. Viewing Angle And Gamma The Word 'lagom' On This Image Should Blend In With The Background Everywhere; Otherwise The Gamma Curve Of Your Monitor Is Dependent On The Viewing Angle, Which Is The Case With Most Displays Based On TN Place A 10 Cm Piece Of Tape Labeled "top" Onto The First. Peel Up Both Together. 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Test In Sitting CSS Paged Media Module Level 3 Editor’s Draft, 22 July 2019 @page { @bottom-left, @bottom-center, @bottom-right { Border-top: 1px Solid Gray; } @bottom-center { Content: "at " Counter(pag Wi-Fi 6 Adapters Like This One Are Available Online For$40 Or Less. Rivet Networks We Can Still Test Each Router's Top Transfer Speeds By Measuring Its Ability To Move Files Around Locally, Though. Get The Latest News And Analysis In The Stock Market Today, Including National And World Stock Market News, Business News, Financial News And More Perform A Top-heavy Test Each Year. The Top-heavy Rules Generally Ensure That The Lower Paid Employees Receive A Minimum Benefit If The Plan Is Top-heavy. A Plan Is Top-heavy When, As Of The Last Day Of The Prior Plan Year, The Total Value Of The Plan Accounts Of Key Employees Is More Than 60% Of The Total Value Of The Plan Assets. 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Is That A Correct Reading Or Do I Have A Thermostat Problem Or An Element Problem. The Bottom Element Is New. It Has To Determine The Potential Of A Producing Formation, The Operator May Order A Drill Stem Test (DST). The DST Crew Makes Up The Test Tool On The Bottom Of The Drill Stem, Then Lowers It To The Bottom Of The Hole. Weight Is Applied To The Tool To Expand A Hard Rubber Sealer Called A Packer. Directions. In A Small Bowl, Beat The Cream Cheese, Sugar, Egg And Salt Until Smooth. Stir In Chips; Set Aside. For Cupcakes, In A Large Bowl, Beat The Sugar, Water, Oil, Egg, Vinegar And Vanilla Until Well Blended. Combine The Flour, Cocoa, Baking Soda And Salt; This Test Jar Is Large Enough To Hold Any Of The Hydrometers Midwest Sells. The Longer Size Of This Test Jar Works Well For Those Who Make Higher Alcohol Beers Or For Wine Makers.The Screw-off Base Allows For Easy Cleaning.Not For Use With Liquids That Need The Proof And Tralle Hydrometer. Earlier National Test Data From NAEP Showed That Top Students Were Improving More Than Bottom Students. There Were Many Examples Of The Bottom Students Slipping. Related Stories: Top US Students Fare Poorly In International PISA Test Scores, Shanghai Tops The World, Finland Slips The Right Way To Wipe . After Comfortably Passing A Stool, Always Remember To Wipe From Front To Back, Avoiding Any Skin-to-skin Contact With Stool.Simply Reach Behind Your Back And Between Your Legs, Using Plenty Of Crumpled Or Folded Toilet Tissue, And Wipe Backward From The Perineum (the Space Between The Genitals And Anus) Toward And Past The Anus. Angiography: This Test Looks For Problems With Your Arteries In Your Hands, Arms, Feet, And Legs. Before The X-ray, A Dye Is Put Into A Thin Tube Through A Small Cut In Your Groin. The Dye Helps The Arteries Show Up Better On These X-ray Pictures. You Might Also Be Offered This Appointment If You Have A Sore Or A Lump Inside Your Bottom. The Appointment Might Be To Check For Cancer Of The Colon Or Rectum Or Of The Anus, Depending On Where Inside Your Bottom The Symptoms Are. For More Information On Tests For Blood In Your Faeces (stools) Or A CA125 Test See Test Results. The Ralston NPAK-TRS0-2MS0-2FSA Liquid Trap With 1/4" Female NPT Top Port, 1/4" Male NPT Bottom Port, Eliminates Unwanted Fluid Or Moisture That Can Contaminate Any Gas System. It Features A Quick-release Lid, Which Allows For Easy Cleaning Of Dirt Or Liquid Without Tools. It Entraps Fluid And Debris Out Of System To Prevent Blocking Or Plugging. TEST REQUESTED . 10 Routine Well (Total Coliform And E.coli)** Non-Drinking Dairy Water COLLECTION LOCATION COUNTY. DATE COLLECTED (YYYY/MM/DD) COLLECTOR LAST NAME, FIRST NAME. COLLECTION POINT (ex: Sink, Outside Spigot) COLLECTION LOCATION STREET ADDRESS. CITY STATE. ZIP CODE SUBMITTER TELEPHONE NUMBER/EXT. 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The Perimeter Of Each Of The Four Areas (content, Padding, Border, And Margin) Is Called An "edge", So Each Box Has Four Edges: Content Edge Or Inner Edge Mit Top-down Werden Herstellungsverfahren Bezeichnet, Deren Ursprung Und Methodik Eher Den Ansätzen Aus Mikrosystemtechnik, Wie Z. B. Lithografie, Entsprechen, Während Bottom-up-Methoden Die Physikalisch-chemischen Prinzipien Der Molekularen/atomaren Selbstassemblierung Und Selbstorganisation Ausnutzen, Z. B. Bei DNA-Origami Und DNA-Maschinen. {dedeSolution / Factory AutomationeZhejiang HuaRay Technology {dedeSolution / TransporteZhejiang HuaRay Technology Co., Ltd Ø Vacuum Testing Is Conveniently Performed By Means Of A Metal Testing Box, 6 Inches Wide X 30 Inches Long, With A Glass Window In The Top. The Open Bottom Is Sealed Against The Tank Surface By A Sponge-rubber Gasket. Suitable Connections, Valves, And Gauges Should Be Provided. It's So Contagious #b-navbar {height:0px;visibility:hidden 9 Common Digestive Conditions From Top To Bottom. Experts Caution That More Research Is Needed To Test The Efficacy Of This Alternative Approach To Care. By Linda Thrasybule May 20, 2020. Document.write(' '); } //--> = 11) Document.write ZDNet's Technology Experts Deliver The Best Tech News And Analysis On The Latest Issues And Events In IT For Business Technology Professionals, IT Managers And Tech-savvy Business People. * The Change For The Output For The File / Export /Transform Dialog Box Remembers The File Name If Present. * Web Page With Image Now Has References In The Test E.g. "1.2 Helper". Space Has Been Added At The End Of The Page So Clicking On An Argument In The Image Displayes The Appropriate Element Text At The Top Of The Page. Questo, Comunque, Ritarda La Fase Di Test Delle Ultime Unità Funzionali Di Un Sistema Finché Una Parte Significativa Della Progettazione Non è Stata Completata. L'approccio Bottom-up Enfatizza La Codifica E La Fase Di Test Precoce, Che Può Iniziare Appena Il Primo Modulo è Stato Specificato. Number, Ranking And Time Sequence Test Questions & Answers For AIEEE,Bank Exams,CAT, Bank Clerk,Bank PO : Sanjeev Ranks Seventh From The Top And Twenty Eighth From The Bottom In A Class. Xcode Post Build Script" /> <meta http-equiv="-UA-Compatible" content="ie=edge" /> <title>Top Or Bottom Test</keyword> <text> Are You A Top, A Switch Or A Bottom? You Know You Want To Know! I'd Even Venture To Say You NEED To! So, Just How Kinky Are You? Take This Top Or Bottom Quiz Now And Put An End To Your Wonderings! Have Fun, And Please Share With Friends! It's All For Fun, Of Course.:-) Are You A Bottom, Top Or A Switch? 14 Questions - Developed By: Salem - Developed On: 2020-04-24 - 17,784 Taken - 15 People Like It A Series Of Scenarios Where I Will Define What Are You (pls Don’t Take It Seriously) Search, Watch, And Cook Every Single Tasty Recipe And Video Ever - All In One Place! Gradually, The Word Top Is Referred To As The Stronger Half And Bottom The Weaker Half In A Relationship. And In Japanese Culture, Seme Is Used To Describe Top And Uke To Describe Bottom. Do You Want To Know Whether You Are A Seme, An Uke Or Neither Of Them? Come And Take This Relaxing Quiz. See If You Are A Bottom Or A Top. Warning: These Results Are Most Likely Not Accurate And This Quiz, Nor It's Questions Are Meant To Be Taken Seriously. You Shouldn't Trust Us To Determine Or Label Your Sex Life. Maybe You're A Top, Because You Love Taking Control And Getting Exactly What You Want. Or Maybe You're A Bottom, Because You Like Being Dominated In The Bedroom? Fun. This Test Is Not Based On Any Scientific Study Whatsoever. It Is Intended For Fun Only So Do Not Treat The Result Too Seriously :) But Don't Fret, Just Answer These 15 Questions To Reveal Your Top And Bottom Percentages. When You Answer One Question, The Next Will Appear. Good Luck, And May The Odds Be Ever In Your Topping Or Are You A Top Or A Bottom? It's Time To Get Some Insight. Answer These Questions About Your Sex Life, And Personality, And We'll Tell You If You're A Top Or A Bottom. Are You A Top, Bottom, Or Switch? Lily Green. 1. 6. What Is Your Ideal Date Night Like? I Pick Up My Date And We Both Go To See A New Movie. Are You A Top Or Bottom In Your Relationships? Have A Wonderful Day Follow Me On Insta @y3urika @Relatablenotrly: 678,903 People Diagnosed “Hare Tugged Off The Roots At The Bottom And The Tassels At The Top And Put Them In A Pile For Bear.” The BDSM Test Is A Fun And Educational Test To Determine What Kind Of Kinkster You Are. It Was Founded In 2014 With As Its Main Mission To Make A Simple, Accessible Test To Help Beginning Kinksters Determine Which Labels Are Or Aren't Suitable For Them; And To Be A Fun Experience For Everyone Taking It, Beginners And Experts Alike. SEe If You Are On Top Or Bottom And What Type Of Screwer You Are! Get Started With These Questions 1. Do You Like To Cry? *laughing Ass Off* If It Gets Me What I Want. Is It The Top Or Bottom Number? In The Past, It Was Commonly Thought That The Diastolic Pressure Or Bottom Number Was The Most Important Blood Pressure Number. This Mistaken Idea Dates Back To One Of The Early Studies Of The Treatment Of Blood Pressure, Where Only Patients With Diastolic Blood Pressure Over 90 Mm Were Allowed Into The Study. Scroll To Top. #Women. QUIZ: Spot The Top, Bottom, Or Switch! How Good Is Your Bedroom Gaydar? By Arielle Scarcella. October 31 2014 2:14 PM EDT. Do You Think You Know A Top, Bottom, Or Switch I Learned In These Conversations That The Formula By Which Masculinity Is Determined In Relationships Resembles One Of My Ninth-grade Math Teacher's Indecipherable Equations: If Fashionable + Emotional X Beyoncé = Bottom, Then Football + Boot-cut Jeans X Dirty Finger Nails = Top. Did You Think You Were A Top, But Actually You're A Switch - Almost A Bottom? Its Time To Know Once And For All What Type Of Lesbian Are You? Are You Really A Butch, Or Maybe Your A Femme Trapped In Butches Body. SEe If You Are On Top Or Bottom And What Type Of Screwer You Are! Get Started With These Questions 1. Do You Like To Cry? *laughing Ass Off* If It Gets Me What I Want. You Ever Wondered How Will Be Your Sex Life Marriage. Many Of Us Think About It, But We Get Confused So Easily. Now, No Need To Get Confused. The Below Quiz Will Help You Out To Find The Answer. Sure, People Have A Preference, But Now Could Be The Perfect Time To Escape The Top Or Bottom Prison You Live In. So, With The Help Of Some Experts, Let’s Take A Moment To Dismantle What You Teenage Is A Crazy Time – Your Hormones Are Raging, You Have Your Eye On A Guy.. Or A Girl, You Want To Look Good And Also Feel Comfy, You Develop A Sense Of Style, An Affinity For Certain Sports, You Are Unsure Of What You Want To Do In Your Life, And Much More. Seme Means The One Who Is On Top, And Uke Being The One On The Bottom During Sex. So, Given A Yaoi Scenario, Which One Would You Most Likely Be Between Uke And Seme? Take The Quiz And Get To Find Out. Blood Pressure Measurements Are Given As A "top" And "bottom" Number. The First Reflects Systolic Blood Pressure, The Amount Of Pressure In The Arteries As The Heart Contracts. I Wanna Get On Top And Make Them Scream My Name! I Wanna Make Out-of-this-world Love To My Partner When He/she Comes Home Tonight. I Want To Be Taken And Caged In My Partner`s Amazing Arms. That Is, A Switch Who Prefers To Evenly Top And Bottom Will Do Best With Someone The Same; A Switch Who Prefers To Top Will Do Best With One Who Prefers To Bottom; And, Likewise, One Who Prefers To Bottom Will Do Best With One Who Prefers To Top. Just Like With Many Other Aspects Of Human Sexuality, It’s A Spectrum. About Quizz Creator. With Free Quiz Creator Tool/Software, The Developing Of New Quiz About Any Subject Is A Lot Easier And Time Saving. Just Select The Subject About Which You Want To Create A New Quiz And Go Ahead With Your Questions And Their Options. Terms Used For Sex Between Two Women Or Two Men. The Top Is The Pleasure Giver And The One On Bottom Is Receiving. They Are Not The Same As Dom And Sub Because Even Though Many Bottoms Are Subs And Tops Are Doms, There Are Variations Such As Service Tops And Power Bottoms. UQuiz.com Is A Free Online Quiz Making Tool. Make Quizzes, Send Them Viral. Generate Leads, Increase Sales And Drive Traffic To Your Blog Or Website. Intermediate Numbers Indicate Intermediate Levels Of Agreement. These Statements Test Your Tendency To Use Top Brain Vs. Bottom Brain Processing, And Will Assess Your Dominant Thinking Mode. Take The Quiz You And Your Partner Are Having An Ice Cream Is Related To Are You A Top Or A Bottom?. Here You Can Create Your Own Quiz And Questions Like You And Your Partner Are Having An Ice Cream Also And Share With Your Friends. These Questions Will Build Your Knowledge And Your Own Create Quiz Will Build Yours And Others People Knowledge. Start Studying Tops And Bottoms Comprehension Review. Learn Vocabulary, Terms, And More With Flashcards, Games, And Other Study Tools. ♡ SUBSCRIBE TO MY CHANNEL » Http://bit.ly/SubscribeJCharlesBeauty For New Videos Every Week!Hi Guys! I've Been Doing So Many Tutorials Lately So I Wanted To Top Or Bottom?! Find Out The Truth!!! And Then Don't Forget To Subscribe. ;) You Can Take These Quizzes Here: Https://www.buzzfeed.com/tomphillips/how-gay-ar How Kinky Are You? How Far Are You Willing To Go, With Your Lover? This Fan Favorite Kinky Quiz Will Determine That For You! We Will Ask You A Series Of Questions, PLEASE BE HONEST! For Accurate Results, Honesty Is Your Best Friend! We Created An Amazing List Of Questions To Test Your Inner Kinky Side. Top Or Bottom. Jndhjny. @piroreos: 40,169 People Diagnosed 0 Love Tweets Daily Results Result Patterns 3: Enter Your Name For Diagnosis Top Vs Bottom. I Love Astrology And Everything Even Remotely Related To It. For Me, Everything Is Written In The Stars, And Zodiac Signs Are One Of The Best Ways To Understand And Know A Person. Christmas Carol Personality Test! Core Beliefs: Your Location On The Political Spectrum Disney's Sleeping Beauty Quiz Hogwarts House Sorter Sleepy Hollow Selector What JoJo Character Are You Most Like? What Type Of Communist Are You? What Wild Animal Are You? What Would You Look Like As Your Anime Character? Which Gilmore Girls Guy Is Right For Rags To Riches: Answer Questions In A Quest For Fame And Fortune. Tops And Bottoms. A Great Reading Practice! Tools Ready To See Whether You're More Dominant Or Submissive - Or Somewhere In Between? This Very Brief Quiz Will Give You An Idea! We Promise It Will At Least Get You Thinking! Special Note: These Questions Are Posed As If All Activities Are Consensual, Occurring Between Loving Adults. The Terms Top, Bottom And Versatile Do Not Necessarily Refer Literally To Physical Position During Sex. For Example, If The Inserting Partner Lies On His Back And The Receptive Partner Straddles Him, The Inserting Partner Is Still Considered The Top , And The Receptive Partner The Bottom , Despite Their Reverse Physical Arrangement. Independent Samples Test Box . This Is The Next Box You Will Look At. At First Glance, You Can See A Lot Of Information And That Might Feel Intimidating. But Don’t Worry, You Actually Only Have To Look At Half Of The Information In This Box, Either The Top Row Or The Bottom Row. Levene’s Test For Equality Of Variances You Should First Think Of The Test As Part Of The Loop Code. If The Test Logically Belongs At The Start Of The Loop Processing, Then It's A Top-of-the-loop Test. If The Test Logically Belongs At The End Of The Loop (i.e. It Decides If The Loop Should Continue To Run), Then It's Probably A Bottom-of-the-loop Test. Which Type Of Lesbian Are You? My Results: The Undefinable You Just Like Ladies, And That Does Nothing To Affect Your Lifestyle Or Appearance. You Are You, And You Are Unique! Using Modified Code From Lazybear's CVI 0.17% Indicator And Some Logic, This Script Can Help Identify Value Buy And Sell Opportunities At A Very Early Phase. Best Used As A Confirmation Tool. This Is A Beta, So Use It As Such. Enjoy. Sometimes "top" Indications In An Uptrend Mean To Look For An Opportunity To Add To Position While "bottom" Indications During A Downtrend Can Represent The Same I Am On Top Irl And A Bottom Feeder On The Internet. Oh, Sex? Ummm Depends On What I Hook And How Rough She Is. Most Women Surprising Only Want Bottom But I Have Found A Few That Need To Get Off On Top. Me, I Don't Give A Sht. I Guess That Would Be A Flap-jack. Start Studying Reading Test-Top Or Bottom. Learn Vocabulary, Terms, And More With Flashcards, Games, And Other Study Tools. Top Or Bottom Quizzes Tags » Top Or Bottom. Name Creator Views Rating; Test His Love By Answering Some Simple Questions And Find Out If He Loves You! Quizrocket.com. Bottom-up Testing Is A Specific Type Of Integration Testing That Tests The Lowest Components Of A Code Base First. More Generally, It Refers To A Middle Phase In Software Testing That Involves Taking Integrated Code Units And Testing Them Together, Before Testing An Entire System Or Code Base. IT Professionals Refer To Design Assemblies, Or Top/bottom 3D Can Be Largely Understood The Same As SBS 3D, Except That The Entire Frame For Each Eye Is Respectively Scaled Down In A Vertical Way To Fit The Top-half And Bottom-half Of The Frame. Sometimes When Your Display Device Has The Specific Requirement Of The Input As Top/bottom 3D, This Option Is Ready For You. Systolic ; Diastolic: The Top Number (systolic) Is The Pressure Reading When The Heart Pumps And Pushed Blood Through The Arteries. The Bottom Number (diastolic) Is The Pressure When The Heart Rests Between Beats. Answered On Dec 5, 2017 1 About This Page: This Page Was Last Updated On Friday, January 29, 2021 At 03:00 AM EST.. Cases, Deaths, And Testing In All 50 States U.S.: Are We Testing Enough? This Graph Shows The Total Number Of Cases, Deaths, And Tests Performed In Each State Per 100,000 People. A Good Double Top/Bottom Should Take Out A Recent Top/Bottom Followed By A Fast Rejection. The Fast Rejection Shows The Strong Momentum Of The Move. The Strong Momentum Is A Sign That The Smart Money Has Most Probably Finished Its Accumulation Phase And Starts Its Profit Release Phase (the Big Move). Top Blade (or Simply Blade) Steak Is A Small Shoulder Cut. It Is An All-purpose Steak. While It Is Very Tender And Richly Flavored, A Line Of Gristle That Runs Through The Center Of The Meat Makes It A Poor Option For Serving Whole. Remove The Gristle And Slice The Steak Thinly For Stir-fries Or Cut Into Cubes For Kebabs Or Stews. Tenderness: *** To Check That Your Yeast Is Healthy, Run The Following Test: In A Small Bowl, Mix 1 Teaspoon Of Yeast (active Dry Or Instant) With ½ Teaspoon Of Sugar And 1 Tablespoon Of Room-temperature Water. The Mixture Should Look Bubbly Within 10 Minutes; If It Doesn’t, It’s Time To Buy A Fresh Supply Of Yeast. The Bottom Upper Is Not Only A Bottom Up Thinker, But A Bottom Up Doer, Too. He Strives, To The Best Of His Abilities, To Be Economically Productive And Independent. He Favours The Economic Free Market, Which He Sees As The Best Way To Achieve The Common Good; That Is, The Good Of Every Individual Who Is Willing To Put In The Effort To Be First Of All Gay Is A Male Who Like To Have Sex With Males. A TOP Gay Likes To Be On The Top In The Position Of Having Sex Whereas In The Case Of BOTTOM Gay Its The Other Way Round. Top Shelf Vs. Bottom Shelf Rum – What Makes The Better Rum And Coke? Posted By Terry Boram July 21, 2020 RUM Tags: American Pride Bacardi Rum And Coke Rum Review - Taste Test Of The Finest Rums Of The World 4 Min Read Use 40-pound Test And A ½-ounce Weight For A Bottom Rig, And Titanium Or Cable With No Weight For Kings. 9) Offshore, Slow-Trolling: 30-Second Kingfish Rig (BRT Pg 145) The Top-down Integration Testing Works Through Big To Small Components While The Bottom-up Approach Is Inverse Of It. Conclusion In Both Of The Approaches, Top-down And Bottom-up Integration Testing The Top-down Produces More Redundant Results And Lead To Additional Efforts In The Form Of Overheads. An Oral Bottom Is The Exclusively Receptive Partner In Oral Sex, Providing The Penetrative Partner, Or Oral Top, With Unreciprocated Fellatio And Irrumatio. Versatile Edit Versatile Refers To A Person Who Enjoys Both Topping And Bottoming, Or Being Dominant And Submissive, And May Alternate Between The Two In Sexual Situations. Experimented By: Molly Unger Observations That's How The Ice Cracks!! Melting Part This Is What I Started Out With. These Are Two Layers That I Did. By: Molly Unger Thanks !!!! Answers I Noticed That There Were Bubbles Coming Up From The Bottom. The Purple At The Top Was Losing Test Your Reaction Time A Fun Test Of Your Finger Clicking Reaction Time. Tap Reaction Time — Another Online Reaction Time Test, In Which You Respond To A Change In Screen Color By Tapping The Screen. Light Board Reaction Timer — Description Of A Reaction And Coordination Test Used In Testing Boxers. A Quiz And An Attached Worksheet Will Estimate Your Cognizance Of Top-down And Bottom-up Processing And Attention. You Will Need To Understand The Essentials Of The Lesson, Like Multitasking And In A Constant Quest To Evaluate Anchor-holding Power For Its Own products And The Competition, Fortress Marine Anchors Set Up A Rigorous Test Of 12 Comparably Sized, Premium-brand Anchors In Typical Mud/clay Bottom Of The Chesapeake Bay At Solomons, Maryland. Brian Sheehan And Several Other Fortress Executives Invited A Dozen Boating Bottom, Top, Versatile. Dominant, Submissive, Switch. Safe, Bare. For Gay And Bi Men, Much Of Our Lives Are Built On These Words. Some Of Us Embrace Them, Others Reject Them. The Bottom Number (diastolic) Is The Pressure Measured Between Heartbeats. Here's A Look At Blood Pressure Categories And What They Mean. If Your Top And Bottom Numbers Fall Into Two Different Ranges, Your Correct Blood Pressure Category Is The Higher One. Bottom-up Definition: 1. Considering The Smaller Or Less Important Parts Or Details Of A Plan, Organization, Etc. First…. Learn More. Identify The Heating Elements In Your Oven At The Top And Bottom. The Heating Elements Are The Big Coils At The Top And Bottom Of Your Oven. Open Your Oven Door And Remove The Metal Racks. Then, Look At The Very Bottom Of The Oven And Look For A 0.5–1 In (1.3–2.5 Cm) Thick Metal Coil That Loops Around The Bottom Of The Oven. While There Are Many Methodologies For Preparing A Financial Forecast, Two Of The Most Common Are Top-down And Bottom-up Analyses. A Top-down Analysis Starts With A Business Assessing The Market As A Whole. First You Determine The Current Market Size Available For Your Business And Factor In Relevant Sales Trends. Top-down Estimating Is Carried Out By Senior Management Based On The General Information Available About The Project, While Bottom-up Estimating Is Carried Out By Subject-matter Experts Based On Cold Water Enters And Remains At The Bottom Of The Water Heater. So Our Hot Water Is Plenty Hot But We Run Out In Just A Few Minutes. Short Hot Shower - Bad Bottom Water Heater Element. As We're About To Show In Detail, One Can Test A Water Heater Heating Element By A Simple Procedure Using A Continuity Tester Or A Volt-ohm Meter- A VOM. The Top May Sometimes Even Be The Partner Who Is Following Instructions, I.e., They Top When, And In The Manner, Requested By The Bottom. Contrast This With The Pure Dominant, Who Might Give Orders To A Submissive, Or Otherwise Employ Physical Or Psychological Techniques Of Control, But Might Instruct The Submissive To Perform The Act On Them. This Top/bottom Division Matters Because These Two Parts Of The Brain Have Different Cognitive Functions, A Fact First Discovered In The Context Of Visual Perception And Supported In A Milestone Report In 1982 By National Medal Of Science Winner Mortimer Mishkin And Leslie G. Ungerleider, Of The National Institute Of Mental Health. You Are A Switch. You Can Start Off As A Bottom And Feel Like A Top Later. This Is Super Normal And Can Be A Lot Of Fun. "The Tables Have Turned" Doesn't Have A Negative Connotation Here. On A Traffic Signal Arranged Vertically, Red Is Always On Top And Green On The Bottom. When Arranged Horizontally, Red Is Always On The Left And Green On The Right. Looking For The Abbreviation Of Bottom? Find Out What Is The Most Common Shorthand Of Bottom On Abbreviations.com! The Web's Largest And Most Authoritative Acronyms And Abbreviations Resource. There Are Also Two Approaches: Top-down Estimating And Bottom-up Estimating. In The Top-down Approach You Will Estimate The Duration Of Deliverables And/or Major Deliverables. In Bottom-up Estimating You Provide Detailed Estimates For Each Individual Task Making Up Your Deliverables. So, Bottom-up Processing Is Data-driven, And Your Perception Of What It Is That You're Looking At Directs Your Cognitive Awareness Of The Object. So, In Contrast, Top-down Processing Basically Uses Your Background Knowledge, So Uses Your Background Knowledge To Influence Perception. So, Let's Look At This Example Over Here. BP: The Systolic (top) Number Is The Pressure Generated By The Force Of Your Left Ventricle During Contraction (systole). The Diastolic (bottom) Number Is The Residual Pressure In The Aorta During The Resting Phase (diastole) & Reflects The Elastic Properties Of The Arterial System. Figure 2 Neural Tuning For Three-dimensional Configuration Of Surface Fragments. (a) Top 50 Stimuli Across Eight Generations (400 Stimuli) For A Single IT Neuron Recorded From The Ventral Bank Of The Superior Temporal Sulcus (17.5 Mm Anterior To The Interaural Line). (b) Bottom 50 Stimuli For The Same Cell. The Two Chambers On The Top Of The Heart, Called Atria, Collect Blood From The Body And The Lungs. The Two Chambers On The Bottom Of The Heart, Called Ventricles, Pump It To The Lungs And The Body. This Animation Shows How The Different Functions Work Together To Keep Blood Flowing. Loops With The Test At The Top Of The Loop (while And For Loops) Are Called Top-driven Loops. Loops Implemented With The Test At The Bottom (a Do Loop) Are Called Bottom-driven Loops. This Location For The Test Has Some Odd (and Sometimes Undesirable) Effects. Examine The Following: The Bottom Row Has A Lot Of The Less Often Used Letters, So Get Ready For Some Weird Words! Easy Top Row Words. 4. Typing Test. Typing Games. Digital Literacy. When The Frictional Losses Of The Top Performing Ceramic Sample And Top Performing Steel Samples Are Compared (Gold Race At 0.29 Watts And Hawk Racing At 0.32 Watts), The Difference Is 0.03 Watts. This Illustrates That Certain Steelbearing Bottom Brackets Perform As Efficiently As The Best Ceramic-bearing Bearing Bottom Brackets (within 3/100 In A Top-down Organized Firm The Decision-making Is At The Top And The Knowledge Is At The Bottom. This Top-down Approach Is The Dominating Management Style. Managers Or Supervisors Tell Employees What They Have To Do And How They Have To Do It. They Do Not Get Much Feedback From Those In Lower Hierarchical Positions. 15 Top Marketing Trends To Keep An Eye On In 2021. Pressure Test Your Marcom Program To Improve Morale And The Bottom Line. While I Most Definitely Enjoyed Janet Stevens' Tops & Bottoms (and Do Understand Why And How It Won A Caldecott Honour Designation), On A Purely Personal Level, I Actually Find The Illustrations A Bit Too Brash, Bold And In-your-face For My Own Tastes (and Having To Turn The Book On Its Side, Really Rather Majorly Annoyed Me At First). The Bottom Of The Woman. Conveniently, Many Modern Over-under Shotgun Barrel Selectors Are Marked “U” So You Know The “under” Barrel Is Selected, Or “O” For The “over” Barrel. Then There Are Those Shotguns With The Dot Codes And I Could Never Remember Whether The Single Or The Double Dot Was The Over Or The Under Barrel. Remington Sa 700 Detachable Mag Bottom Metal Stealth And Mossberg 590 Shockwave Gun Test Range365 Reviews : You Finding Where To Buy Remington Sa 700 Detachable Mag Bottom Metal Stealth And Mossberg 590 Shockwave Gun Test Range365 For Cheap Best Price. A Doctor Has Uploaded A Video Purportedly Showing The Inefficency Of Facemasks, By Blowing Clouds Of Vape Smoke Through The Sides Of A Mask. Dr Ted Noel, An Anesthesiologist With 36 Years Of Experience, Recently Uploaded A Video Contrasting The Mainstream Media Narrative That Face Masks Are The Ultimate Solution To Protecting Oneself From Spreading Or Catching The Coronavirus. Ernie Ball Skinny Top Heavy Bottom Slinky Strings Are The Perfect Hybrid Set For Those Who Like Thick Bottom Strings Without Sacrificing The Ability To Solo On Smaller Strings. These Strings Are Precision Manufactured To The Highest Standards And Most Exacting Specifications To Ensure Consistency, Optimum Performance, And Long Life. Definition And Usage. The Bottom Property Affects The Vertical Position Of A Positioned Element. This Property Has No Effect On Non-positioned Elements. If Position: Absolute; Or Position: Fixed; - The Bottom Property Sets The Bottom Edge Of An Element To A Unit Above/below The Bottom Edge Of Its Nearest Positioned Ancestor. After Spending 14 Years At The Pentagon In The Business Of Buying Weapons, He Concludes That It Is "a Corrupt Business -- Ethically And Morally Corrupt From Top To Bottom." The Reform Movement He Test Your Typing Speed With Bottom Row Letters. Keep Your Fingers On The Home Row And Move/reach Down To The Bottom Row Keyboard Letter Keys! LEFT Hand Fingers Reach To BVCXZ Keys And RIGHT Hand Fingers Reach To NM,./ Keys. Practice Long And Hard Then Take The Test. You Can Always Come Back And Practice More If You Need To. Free 2-day Shipping. Buy Hurricane TBR30-2 Size 2 Top Bottom Rig W/ 30lb Test X 32" Monofilament Leader At Walmart.com Top Down And; Bottom Up Approach; As The Name Suggests, Big Bang Form Of Testing All The Modules Are Combined To Form A Complete System And Then Tested For Bugs. Integration Testing. Top Down Is Systematic Approach Where The Top Level Modules Are First Tested And Then One By One The Sub Modules Are Added And Tested. An Electrocardiogram Is A Simple, Painless Test That Measures Your Heart’s Electrical Activity. It’s Also Known As An ECG Or EKG. Every Heartbeat Is Triggered By An Electrical Signal That Starts At To Test It, We Consider The Case Of The Mean Being Significantly Greater AND Significantly Less Than X. With A 5% Significance Level, The Sample Mean Is Considered Significantly Different From X When The Test Statistic Falls Into The Critical Region — The Bottom AND Top 2.5% Of The Test Statistic’s Probability Distribution. The Test Tracks Your Body Response -- Specifically, If Your Body Makes A Molecule Called Immunoglobulin E (IgE). A High Level Of IgE Can Mean You Have An Allergy. Skin Prick Test: The Skin Prick BREASTS-IN-THE-MIDDLE TEST When Your Breasts Are In Their Cups And Hoisted To Where They Should Be, Your Bra Cup Apex Should Be Halfway Between The Top Of Your Shoulder And Your Elbow. If Lower Than This, Your Band Is Probably Too Large And Not Giving Your Breasts Enough Support. Multiply The Top Numbers: 1 3 × 9 16 = 1 × 9 = 9 . Step 2. Multiply The Bottom Numbers: 1 3 × 9 16 = 1 × 9 3 × 16 = 9 48. Step 3. Simplify The Fraction: 9 48 = 3 16 (This Time We Simplified By Dividing Both Top And Bottom By 3) 1. Do You See These 4 Boxes? In The Top Row The Pictures Go Together In A Certain Way. Now Look At The Bottom Row. Do You See The Empty Box? Which Of The 4 Pictures On The Side Goes With The Picture In The Bottom Box The Same Way The 2 Pictures In The Top Row Go Together? Shop A Large Selection Of Disposable Plastic Test Tubes Products And Learn More About Globe Scientific 12 X 75mm Round Bottom Polystyrene Test Tubes:Test Tubes, 1000/Bag, The Disadvantages Of Top-Down And Bottom-Up Strategic Management Although Many Companies Choose The Top-down Approach, There Are Several Drawbacks, Including The Fact That Executives Can Ignore Or Fail To Seek The Input Of Talented And Creative Middle Managers And Employees. Editor's Notes. October 15, 2019: Choosing A Bottom Paint For Your Boat Is Dependent On The Ways In Which You Use And Store Your Vessel. It's A Task That Needs To Be Addressed Yearly, Even If You've Used A Multi-season Formula, It's Smart To Double Check If It Needs Any Touch Ups. Q6: Describe The Interaction Between Your Bottom And Top Tapes In Terms Of Conservation Of Charge. As The Tape Is Pulled Apart The Original Number Of Movable Charged Particles (presumably Electrons) Are Unevenly Distributed Between The Pieces Of Tape. No Net Charge Is Created In The Process. The Bottom Of The Ladder Is 27 Feet Away From The Building. How Tall Is The Building? Algebra -> Test -> SOLUTION: A 45-foot Ladder Is Leaning Against The Top Of Building. The Problem Is To Count All The Possible Paths From Top Left To Bottom Right Of A MXn Matrix With The Constraints # Driver Program To Test Above Function . M = 3 CBT Is A "top-down" Approach Based On The Idea That If A Person Thinks Right, They Will Feel Alright. Brain Scan Research Suggests Otherwise. Arousal And Emotional Regulation Are Structured Like An A Meniscus Is The Curved Surface At The Top Of A Column Of Liquid. In A Science Class, This Liquid Is Usually Water Or Some Sort Of Aqueous Solution, And The Column Is Usually A Graduated Cylinder Or A Pipet. As You May Have Noticed, When Water Is In Such A Thin Glass Tube, It Does Not Have A Flat Surface At The Top. It Means He Will Top But Doesn’t Like To. I’m A Bottom, And My Profile Says Versatile.” Jason Will Top On Occasion, When He Loses The Game Known As “Race To The Bottom,” But He Doesn’t Bottom Up Marketing Is The Process Of Developing A Marketing Strategy Within An Organization By Finding A Workable Tactic And Then Building On The Tactic To Create A Powerful Strategy. Bottom Up Marketing Is Advocated By The Authors As An Alternative To Traditional Top Down Marketing The Two Numbers That Measure Your Blood Pressure Are Written Like A Fraction: One Number On Top And One On The Bottom. For Example, What Many People Consider Normal Blood Pressure Is Read As 120/80. The Number On Top Is The Systolic Blood Pressure. It Measures The Pressure Inside Your Blood Vessels At The Moment Your Heart Beats. At Least 2,641 New Coronavirus Deaths And 133,914 New Cases Were Reported In The United States On Jan. 30. Over The Past Week, There Has Been An Average Of 151,038 Cases Per Day, A Decrease Of 33 A Modified Procedure Is Presented In This Study To Evaluate The Equivalent Top-down Load-displacement Curve In A Bottom-up Pile Load Test Considering Elastic Shortening. On The Basis Of The Results Practise Ranking Test Questions And Answers For Free At Smartkeeda. Ranking Test Reasoning Questions Are Asked In Almost All The Bank PO Pre And Clerk Pre Exams. Number And Ranking Questions Are Very Important And Bank Exam Aspirants Must Be Aware Of The Shortcuts To Solve Ranking Test Questions. Improve Your Math Knowledge With Free Questions In "Top And Bottom" And Thousands Of Other Math Skills. America's Test Kitchen Will Not Sell, Rent, Or Disclose Your Email Address To Third Parties Unless Otherwise Notified. Your Email Address Is Required To Identify You For Free Access To Content On The Site. You Will Also Receive Free Newsletters And Notification Of America's Test Kitchen Specials. A More Practical Method Of Establishing A Density Gradient Is By Placing Layer After Layer Of Sucrose Solutions Of Different Concentrations, Thus, Densities, In A Test Tube, With The Heaviest Layer At The Bottom And The Lightest Layer At The Top. The Cell Fraction To Be Separated Is Placed On Top Of The Layer. Top Down. This Is An Approach To Integration Testing Where Top-level Units Are Tested First And Lower Level Units Are Tested Step By Step After That. This Approach Is Taken When Top-down Development Approach Is Followed. Test Stubs Are Needed To Simulate Lower Level Units Which May Not Be Available During The Initial Phases. Bottom Up 4. Test The Valve Itself To Make Sure It's Functioning Properly. Stand Clear Of The Opening If The Valve Isn't Connected To An Overflow Tube. 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Bottom-Up Design: Any Design Method In Which The Most Primitive Operations Are Specified First And The Combined Later Into Progressively Larger Units Until The Whole Problem Can Be Solved: The Converse Of TOP-DOWN DESIGN. For Example, A Communications Program Might Be Built By First Writing A Routine To Fetch A Single Byte From The Communications Port And Working Up From That. View Test Results. As You Run, Write, And Rerun Your Tests, Test Explorer Displays The Results In Groups Of Failed Tests, Passed Tests, Skipped Tests And Not Run Tests.The Details Pane At The Bottom Or Side Of The Test Explorer Displays A Summary Of The Test Run. Falcon™ Round-Bottom Polypropylene Test Tubes With Cap Manufactured From Polypropylene For Excellent Chemical And Thermal Resistance. Corning™ Falcon™Round-Bottom Polypropylene Tubes Are Tested To Withstand 3000 X G Centrifugation. TRANSMISSION EXCHANGE CO Is Located At 1803 NE MLKing Blvd., Portland, OR 97212. 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The Schedule You Create Is Based On Direct Input From Experts Who Will Be Implementing The Project; It’s Also A Useful Technique To Build Teamwork. 5.1 The Bonding Strength Test For The Top And Bottom Layers Of The Geosynthetic Clay Liner Is Intended To Be An Index Test. It Is Anticipated That The Results Of The Test Will Be Used To Evaluate The Quality Of The Bonding Process. Before Removing The Needle From The Tube, Also Streak The Surface Of The Agar With The Needle. Incubate The Tube At 37 Degrees Celsius Overnight. After Incubation, Observe The Color Of Both The Top And The Bottom Of The Agar Slant. If The Entire Slant Is Yellow, The Bacteria Were Able To Ferment Lactose, Sucrose, Or Both. Top Products. IPhone 12 Pro Max With Facetime 256GB Pacific Blue 5G – Middle East Version; Alcatel 3T 10 (8088X) – 10-inch Single SIM Tablet – Midnight Blue; Apple IPad 10.2 128GB Space Gray; Apple IPhone 12 Pro 128GB 6 GB RAM, Graphite Monitor Test Patterns The Patterns On The Right Represent An Ultimate Test Of Monitor Quality And Calibration. If Your Monitor Is Functioning Properly And Calibrated To Gamma = 2.2 Or 1.8, The Corresponding Pattern Will Appear Smooth Neutral Gray When Viewed From A Distance. Test Case1: Test A, X, Y, And Z Individually – Where Test A Comes Under Top Layer Test And Test X, Y And Z Comes Under Bottom Layer Tests Test Case2: Test A, G, H And I Test Case3: Test G, X, And Y Test Case4: Test Hand Z Test Case5: Test A, G, H, I, X, Y, And Z. Merits Of Sandwich Testing Methodology. It Is Very Beneficial For A Big Project Do You Think The Ice Cube In Salt Water Is Melting From The Top Down Or From The Bottom Up? . Why Do You Think So? . What Test Would You Do To Verify Your Idea? . How Would You Explain What You See In The Test? . How Does The Density Of An Object Or A fl Uid Aff Ect Its fl Oating Or Sinking Behavior In Another fl Uid? The Way Project Decisions Are Made And How The Project Is Has Also Evolved. While Top-down Planning Is Still Prevalent In Many Organizations, The Bottom-up Planning Method Is Also Widely Used. Planning Projects Top-down. Top-down Is Still A Widely Used Approach To Project Planning. The Foot Arch Test Is A Simple Way To Determine Whether You Have Flat Feet Or High Arches! Just Follow These Easy Instructions: Pour Water Into A Shallow Pan (the Pan Should Be Big Enough To Fit Your Foot, And The Water Should Be Just Deep Enough For All Parts Of The Bottom Of Your Foot To Get Wet). Bottom Level Freezers Offer A Lot Of Accessibility Advantages And Usually Include Extra Space Compared To Top Freezers (and Often Even Side Freezers, If An Ice Maker Takes Up Space). B.) 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Viewing Angle And Gamma The Word 'lagom' On This Image Should Blend In With The Background Everywhere; Otherwise The Gamma Curve Of Your Monitor Is Dependent On The Viewing Angle, Which Is The Case With Most Displays Based On TN Place A 10 Cm Piece Of Tape Labeled "top" Onto The First. Peel Up Both Together. While Still Together, Stroke The Two Pieces Of Tape With Your Thumb And Index Finger To Discharge Them. Finally, Carefully Peel The Two Apart. The Perfect Top/Bottom Dynamic. If You Are A Gay Man, You May Have A Position Preference, But Chances Are You Will Change It Up Every Now And Then. Whether You Are A Guy Who Prefers To Top Or Just A Big ‘ole Bottom, Make Sure You Know The Ins And Outs Of Both Roles So That You Can Be Pleasing To Your Partner While Protecting Yourself. M-Lab Conducts The Test And Publicly Publishes All Test Results To Promote Internet Research. Published Information Includes Your IP Address And Test Results, But Doesn't Include Any Other Information About You As An Internet User. Learn More About How The Test Works. To Determine What Your Numbers Are, Most Eye Doctors Use The Snellen Chart. It Has Lines Of Letters On It That Start Out Large At The Top And End Up Small At The Bottom. The 20/20 Line Is Near The Bottom Of The Chart. In A Doctor's Office, It Is Usually Placed 20 Feet Away From You. The Pascal-A Test Occupies A Significant Place In The History Of Nuclear Testing Since It Was The First Test To Be That Could Be Called A Contained Underground Test. Pascal-A (originally Named Galileo-A) Was A One-point Safety Test, An Attempt To Verify A Primary Design That Would Have A Small Maximum Energy Release If Accidentally Detonated. Locate The Water Valve At The Bottom Of The Water Heater. Attach A Garden Hose To The Emptying Valve And Turn The Valve Using A Wrench. 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Top-down Combines, Tests, And Debugs Top-level Routines That Become The Test 'harness' Or 'scaffolding' For Lower-level Units. Bottom-up Combines And Tests Low-level Units Into Progressively Larger Modules And Subsystems. 'Big Bang' Testing Is Give The Bottom Shelf A Little More Love. J.P. Wiser’s Triple Barrel Rye Type: Canadian Whiskey Price: $19.99 The Alluring Bottle Draws You In, Its Price Point Keeps Your Attention, And The The Top Layer Is Clear And The Bottom Layer Is Also Mostly Clear With A Slight Hint Of Green/yellow. Cl 2 + I − Yes. Top Layer Is Deep Fuchsia, There Is A Middle Layer Of Yellow And The Bottom Is Clear. Cl 2 + Br − Yes. Top Is Light Yellow Which Fades To Line Of Deeper Yellow And Bottom Is Clear. Br 2 + I − Yes To Test The Defrost Heater You’ll Need To Use A @multimeter To Do An Ohm Test. Please Note That A Continuity Test Will Not Work On Defrost Heater Because The Ohm Value Is Too High To Register On A Continuity Test. If You Have An @autoranging Multimeter, Then Turn It To The Home Setting. Practice Makes Perfect And This Typing Practice Will Help You Learn The Top Row Keys On The Keyboard. Created Just Like The Typing Practice: Home Row Keys And Typing Practice: Bottom Row Keys This Keyboarding Practice Can Help You Type Faster. Top 12 Natural Remedies For Eczema. Written By Jennifer Berry 3 Studies Cited . Natural Ways To Cleanse Your Lungs. Written By Jamie Eske 5 Studies Cited . Fruits For People With Diabetes. No Matter What Your Gender, This Quiz Will Tell You If You Have More Male Or Female Characteristics. Reading Your Soil Texture Jar Test. Your Mason Jar Soil Test Will Be Easy To Decipher. The Heaviest Material, Including Gravel Or Coarse Sand, Will Sink To The Very Bottom, With Smaller Sand On Top Of That. Above The Sand You’ll See Silt Particles, With Clay At The Very Top Of The Jar. 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2021-05-06 21:31:12
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https://www.fishtanklearning.org/curriculum/math/8th-grade/solving-one-variable-equations/lesson-7/
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Match Fishtank is now Fishtank Learning!
# Solving One-Variable Equations
## Objective
Write and solve multi-step equations to represent situations, including variables on both sides of the equation.
## Common Core Standards
### Core Standards
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• 8.EE.C.7 — Solve linear equations in one variable.
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• 7.EE.B.4
## Criteria for Success
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1. Use expressions and equations to model and solve real-world situations (MP.4).
2. De-contextualize a situation to represent it algebraically, and re-contextualize to interpret the solution in context of the problem (MP.2).
3. Build fluency in solving equations with rational numbers and equations on one or both sides of the equation.
## Tips for Teachers
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The following tools may be useful for this lesson: calculators.
#### Remote Learning Guidance
If you need to adapt or shorten this lesson for remote learning, we suggest prioritizing Anchor Problem 1 or 2 (benefit from worked examples) and Anchor Problem 3 (can be done independently). Find more guidance on adapting our math curriculum for remote learning here.
#### Fishtank Plus
Subscribe to Fishtank Plus to unlock access to additional resources for this lesson, including:
• Problem Set
• Student Handout Editor
• Google Classrom Integration
• Vocabulary Package
## Anchor Problems
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### Problem 1
You have a coupon worth $18 off the purchase of a scientific calculator. At the same time, the calculator is offered with a discount of 15%, but no further discounts may be applied. For what tag price on the calculator do you pay the same amount for each discount? #### Guiding Questions Create a free account or sign in to access the Guiding Questions for this Anchor Problem. #### References Illustrative Mathematics Coupon versus Discount Coupon versus Discount, accessed on Aug. 31, 2017, 2:12 p.m., is licensed by Illustrative Mathematics under either the CC BY 4.0 or CC BY-NC-SA 4.0. For further information, contact Illustrative Mathematics. ### Problem 2 You and your brother go to the same store to buy fruit. You purchase a large watermelon for$7.71 and 2.4 pounds of red grapes. Your brother buys a platter of pre-cut fruit for 9.95 and 1.6 pounds of red grapes. The two of you spend the same amount of money. How much do the grapes cost per pound? #### Guiding Questions Create a free account or sign in to access the Guiding Questions for this Anchor Problem. ### Problem 3 Solve the equations for the variable. a. ${-4.2x+6-8.3x={1\over2}(-9x+4)}$ b. ${{{5-x}\over8} = {{{1\over4}x-5}\over3}}$ #### Guiding Questions Create a free account or sign in to access the Guiding Questions for this Anchor Problem. ## Problem Set ? With Fishtank Plus, you can download a complete problem set and answer key for this lesson. Download Sample The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set. • Start each student with a multi-step equation; each student completes one line in the solution of the equation and then passes to the next student; the next student checks the first line and then completes the next line in the solution; this continues until a solution is found and then the next student checks the answer using substitution. • Include additional application problems similar to the Anchor Problems and Target Task. ## Target Task ? Melanie is looking for a summer job. After a few interviews, she ends up with two job offers. • Blue Street Café pays11.75 per hour plus $33 from tips each week. • Fashion Icon Factory Store pays$14.50 per hour with no tips.
1. If Melanie plans to work 10 hours per week, which job offer should she take to maximize her earnings?
2. What if Melanie works 20 hours per week?
3. How many hours would Melanie need to work in order for the pay at each job to be the same? Write and solve an equation.
### Mastery Response
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Create a free account or sign in to view Mastery Response
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2021-01-27 15:43:49
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http://www.control.tu-berlin.de/Trajectory_Planning_and_Control_of_Preferential_Crystallization_Processes
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# Trajectory Planning and Control of Preferential Crystallization Processes
## Contents
### Abstract
In this project, we investigate trajectory planning and feedback control of preferential crystallization processes for the separation of enantiomers. Our current research builds on prior work describing the use of orbital flatness properties for the control of certain classes of crystallization models.
### Description
This project aims at using preferential crystallization for the separation of enantiomers. Enantiomers are chiral substances, which share many physical and chemical properties, but can differ, e.g., with respect to their metabolic effects. Control and optimization of preferential crystallization has recently attracted increased attention. In previous joint work with the PCF group at MPI Magdeburg (e.g. [1]), two different operating modes were suggested. In cyclic operating mode, batch crystallization steps for the two substances to be separated alternate. In each step, seed crystals for one of the two enantiomer species are added to an oversaturated solution. For conglomerate forming enantiomers, these crystals will grow and stay pure even for approximately racemic solutions. If operated in the so called meta-stable area, secondary nucleation dominates primary nucleation. Hence, within each step, crystals of the seeded species will nucleate (and grow), but nucleation of the other species will be negligible. However, this will decrease the concentration of the seeded (preferred) enantiomer in the liquid phase, while the corresponding concentration of the other (counter) enantiomer will remain approximately constant. This, in turn, implies that eventually nucleation (and also growth) of the counter enantiomer will become significant, and, to maintain purity constraints, the process therefore has to be stopped.
Enantioseparation in two coupled vessels.
In coupled mode, crystal-free liquid is continuously exchanged between two crystallizer vessels, and sufficiently pure fractions of both enantiomer species can be obtained simultaneously by adding the respective seed crystals to the two vessels. The main advantage of the coupled mode is the approximately symmetric decrease in concentrations of both enantiomer species in the liquid phase. This allows for a control scheme where, by suitably manipulating the vessel temperatures, supersaturation of the preferred enantiomer in the respective vessel can be kept constant without causing the corresponding supersaturation of the counter enantiomer to steadily increase (as would be the case in a batch step in the cyclic operating mode).
Our research builds on prior work on the use of orbital flatness concepts for trajectory planning and control of single-substance crystallization. Although moment models describing preferential crystallization typically do not exhibit the orbital flatness property, we have developed an inversion procedure which, for a single vessel batch process, can be briefly summarized as follows:
(i) The model contains two population balance equations (PBEs), one for each enantiomer species. One species can be selected, and, as in the single substance case, scaling time with the growth rate of its crystal population,
Time scaling turns characteristic curves into straight lines.
$d\tau_{E_1}=G_{E_1}(t)\cdot dt,$
turns characteristic curves of the respective PBE into straight lines. Then, a specified final crystal size distribution (CSD) for that population can be easily mapped into a desired temporal evolution of the boundary condition, i.e., a desired temporal evolution of the (scaled) nucleation rate:
$\frac{B_{E_1}(t)}{G_{E_1}(t)}=f_{E_1}(0,t).$
(ii) Based on this information, a closed moment model for both species can be dynamically inverted in scaled time. In contrast to the single-substance case, this moment model exhibits nontrivial zero dynamics, even in transformed time. Therefore, differential equations have to be solved to finally obtain the desired temperature profile.
Depending on the application, a number of idealizing assumptions can be justified. For example, neglection of nucleation of the counter enantiomer may be appropriate if the focus lies on properties of the final CSD of the preferred enantiomer, and the assumption can be justified a posteriori. Dynamic inversion is then greatly simplified. Furthermore, optimal control solutions developed recently for single-substance crystallization [2] can be adopted.
The investigation of crystallization processes with size-dependent growth rate has profited from a cooperation with the PSE-group at MPI Magdeburg. This issue has been addressed in [3] [4] [5]. The latter suggests a finite ODE model which can approximate the underlying infinite-dimensional system with a required accuracy. This scheme is suited for use in control problems, such as dynamic inversion and optimal control. For instance, in [6] an efficient solution to optimal control of a single-substance batch crystallization process with size-dependent growth rate is found.
Feedback control becomes necessary in the face of uncertainties and disturbances:
• To reliably estimate purity from the available sensor data (temperature and concentration measurements), a nucleation observer was developed [7], which, for certain parametric model uncertainties and measurement uncertainty, provides a worst-case estimate of product purity.
• A controller for keeping concentrations of the two enantiomer species symmetric in the coupled mode was designed and experimentally validated [8] in cooperation with the PCF-group at MPI Magdeburg. Although, because of the poor quality of sensor data, symmetry could only be achieved in a very approximate manner, this turned out to be not a fundamental problem: investigations by the PCF-group seem to indicate that moderate asymmetries in the concentrations do not generally have a negative effect on purity [9].
• We also investigate the adaption of a number of robust optimization and control schemes developed for single-substance crystallization [10].
### Publications
1. Angelov, I., Raisch, J., Elsner, M. P., Seidel-Morgenstern, A.. Optimal operation of enantioseparation by batch-wise preferential crystallization. Chemical Engineering Science, 63 (5):1282–1292, 2008.
2. Hofmann, S., Raisch, J.. Application of Optimal Control Theory to a Batch Crystallizer using Orbital Flatness. In 16th Nordic Process Control Workshop, Lund, Sweden, 25-27th of August 2010; Nordic Working Group on Process Control, pages 61–67, 2010.
3. Bajcinca, N.. Forward and inverse integration of population balance equations with size-dependent growth rate. DYCOPS 2010, 9th International Symposium on Dynamics and Control of Process Systems, 2010, Leuven Belgium, pages 401–406, 2010.
4. Bajcinca, N., Qamar, S., Flockerzi, D.. Methods for integration and dynamic inversion of population balance equations with size-dependent growth rate. In 4th International Conference on Population Balance Modelling, Berlin, pages 227–243, 2010.
5. Bajcinca, N., Qamar, S., Flockerzi, D., Sundmacher, K.. Integration and dynamic inversion of population balance equations with size-dependent growth rate. Chemical Engineering Science, 66 (17):3711–3720, 2011.
6. Bajcinca, N., Hofmann, S.. Optimal control for batch crystallization with size-dependent growth kinetics. In American Control Conference 2011, San Francisco, USA, 2011.
7. Hofmann, S., Eicke, M., Elsner, M. P., Seidel-Morgenstern, A., Raisch, J.. A Worst-Case Observer for Impurities in Enantioseparation by Preferential Crystallization. In Proc. of ESCAPE-21, Chalkidiki, Greece, pages 860–864, 2011.
8. Hofmann, S., Eicke, M., Elsner, M. P., Raisch, J.. Innovative Control Strategies for Coupled Preferential Crystallization. In Proc. of WCPT6, April 2010.
9. Eicke, M., Hofmann, S., Raisch, J., Elsner, M. P., Seidel-Morgenstern, A.. Separation of Enantiomers by Coupled Preferential Crystallization - Impact of Initial Conditions on Process Performance. In Proc. of the 2010 AiChE Annual Meeting, 2010.
10. Bajcinca, N., Hofmann, S., Raisch, J., Sundmacher, K.. Robust and optimal control scenarios for batch crystallization processes. In Proc. of ESCAPE-20, Ischia, Naples, Italy, pages 1605–1610, 2010.
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2017-09-22 02:38:34
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https://cp4space.hatsya.com/2014/10/31/menage-primes/?replytocom=2061
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# Ménage primes
The Online Encyclopedia of Integer Sequences has recently celebrated its fiftieth anniversary, amassed over 250 000 sequences of varying importance, and heralded the 75th birthday of its creator Neil Sloane. It is somewhat surprising, therefore, that no-one appears to have investigated the intersection of two particularly natural integer sequences (catalogued as A000040 and A000179 in the OEIS).
In this article, I shall attempt to persuade you that investigating this new sequence of ménage primes is a good idea for several reasons. But until then, let us consider the problems that motivated this line of research.
### Vortices and knots
Before the structure of the atom was properly understood, Lord Kelvin conjectured that atoms were vortex rings in a hypothetical ubiquitous fluid known as the luminiferous aether. The shape of the vortex ring writhes and undulates, but up to homotopy it remains constant throughout its lifetime. Begun was the idea that chemical elements correspond to homotopy classes of embeddings of the circle (more commonly known as knots), and classifying them would yield a mathematical explanation for the origin of the periodic table*.
* In reality, the periodic table stems from solving the Schrödinger equation for the potential effected by a point charge (the nucleus).
Realising that knots exhibit ‘unique prime factorisation’, Tait et al began enumerating the prime knots according to the minimal number of crossings when the knot is drawn as a diagram in the plane:
Shown above are all distinct prime knots with eight or fewer crossings. With the exception of the last three, these are all alternating knots, meaning that the string passes alternately over and under the crossings.
A knot diagram can be considered to be a 4-regular plane graph, where each vertex is decorated to specify which path goes over and which goes under. We can add further structure by 2-colouring the regions and assigning a direction to the knot. For example, the second knot in the third row yields this diagram:
Each edge can be labelled from 1 to 2n, where n is the number of crossings, according to the order in which they are traversed. We define the parity of an edge to be whether the blue region is to the left or the right of the arrow on the edge. Observe that the parity alternates between successive edges in the traversal, so this corresponds exactly to the parity of the label.
At each vertex, an odd edge and an even edge converge. This yields a bijection between the odd and even edges of the diagram. Moreover, note that in a reduced diagram (with no trivial redundant crossings), consecutive edges never point to the same vertex.
Now draw a complete bipartite graph K(n, n), where each vertex corresponds to an edge in the diagram. We have a natural matching (the bijection between the odd and even edges of the graph described above) and a natural Hamiltonian cycle (the order of traversal). These share no edges. There are two dual ways to view this as a graph-theoretic problem:
• Counting Hamiltonian cycles of a complete bipartite graph with a perfect matching removed.
• Counting perfect matchings of a complete bipartite graph with a Hamiltonian cycle removed.
This problem was later rediscovered by Edouard Lucas in an entirely different setting.
### Formal combinatorics
At High Table, n Fellows and n guests (one for each Fellow) are seated around the perimeter of a topological disc. To maximise social interaction, the following rules are observed:
• No two Fellows may be seated adjacent to each other.
• No Fellow may be seated adjacent to his or her guest.
In how many ways is this possible? This is known as the ménage problem, due to an outmoded heteronormative statement of the problem involving married couples instead of ordered pairs of the form (Fellow, guest). One such arrangement for n = 7 is illustrated below and obtained from the knot diagram:
The crimson lines biject each Fellow (odd number) with his or her guest (even number); these correspond to edges converging to the same vertex in the knot diagram.
We can eliminate much redundancy in our enumeration by exploiting the vast amount of symmetry in the situation:
• Any permutation of the n ordered pairs yields another valid arrangement (order n!).
• Interchanging each Fellow simultaneously with his or her guest results in another valid arrangement (order 2).
Removing the symmetry is equivalent to imposing that the Fellows are initially seated in a particular arrangement. Hence, our question reduces to:
Given an initial arrangement of n Fellows occupying alternate seats, in how many ways can we seat their guests in the remaining n seats such that no Fellow is adjacent to his or her guest?
The number M(n) is called the nth ménage number. The first few ménage numbers are:
• M(3) = 1 (the unique arrangement involving the Fellows being diametrically opposite their respective guests)
• M(4) = 2 (two chiral solutions)
• M(5) = 13
• M(6) = 80
• M(7) = 579
An explicit formula is given in the OEIS (sequence A000179).
### Which ménage numbers are prime?
Observe that M(4) and M(5) are both prime, which confirms that we have successfully factored out all of the redundant symmetries. Continuing further, one discovers that the next ménage prime is M(13) = 775596313, followed a while afterwards by M(53) = 567525075138663383127158192994765939404930668817780601409606090331861.
I decided to do an overnight Mathematica run for values of n up to circa 10000. In the process, it found a spectacular 30281-digit ménage prime, M(8645). To put this into perspective, this is larger than any prime known before the discovery of 2^132049 − 1 in 1983. This suggests the following questions:
• Are there infinitely many ménage primes?
• If so, what is their asymptotic growth rate?
The first of these is extremely difficult, and the second is at least as hard (although we provide some heuristics in the next section). A more achievable goal is the following:
Find another ménage prime beyond the five already discovered.
If my heuristics are correct, the expected number of ménage primes between M(10^4) and M(10^5) is around 0.25, which is both low enough for me to be willing to offer the following prize and high enough that I’m slightly worried someone may win it.
Specifically, the first person to discover a new ménage prime can choose either of the following prizes:
• A vibrant pink wig (RRP 11.00) so you can self-identify as Minaj Prime (pun definitely intended!).
• Alternatively, the publication of a video of me performing a Minaj song of your choice whilst wearing the aforementioned vibrant pink wig.
[Update: it’s been over three years without any more primes being discovered, and I’ve long since lost the wig, so unfortunately I have redacted the prize.]
### Heuristics
Although the ménage numbers are not exactly equidistributed modulo p for all p (for example, only 2/9 of them are divisible by 3), they do seem to asymptotically be prime as often as one would expect from a randomly-chosen number of a similar size. That is to say, M(n) appears to be prime with probability 1/log(M(n)).
Now we need an asymptotic formula for the ménage numbers. Since M(n) counts the number of permutations such that both σ(i) and σ(i) + 1 are derangements, we expect there to be roughly n!/e² arrangements. It can be shown that this estimate is asymptotically correct (the ratio tends to 1 as n tends to infinity), which by Stirling’s formula is in turn asymptotically equal to:
$\sqrt{2 \pi n} n^n e^{-(n+2)}$
The sum of the reciprocals of the logarithms of these numbers diverges, which (by Borel-Cantelli) suggests that there are infinitely many ménage primes. However, it diverges sufficiently slowly that we only expect around log(log(m)) such primes for n < m. This means that in the long run, the number of ménage primes below a given value N is expected to be around log(log(invgamma(N))), where invgamma is the inverse Gamma function.
With these heuristic estimates, the expected number of ménage primes M(n) with n between 10^4 and 10^5 is roughly 0.25. It is entirely plausible that people may attempt to search further than 10^5; increasing this upper bound to 1.5 × 10^7 (corresponding to primes of over 100 million digits, therefore eligible for the 150 000 dollar prize offered by the Electronic Frontier Foundation) gives an expected value closer to 0.64.
Anyway, good night for now — I have a couple of sequences to add to the OEIS (due to appear as A249510 and A249394).
This entry was posted in Uncategorized. Bookmark the permalink.
### 6 Responses to Ménage primes
1. “To put this into perspective, this is larger than any prime known before the discovery of 2^132049 − 1 in 1983.” And yet the number doesn’t rate an entry in Henri & Renaud Lifchitz’s current top 10000 probable prime list! A word of caution to anyone using PrimeQ in Mathematica 10: It’s not as robust as Mathematica 9 in its small-factors subroutine. I’d say it’s a bug.
2. apgoucher says:
Indeed, computing power has increased massively! #mooreslaw
I used PrimeQ in Mathematica 8, although I was generating the ménage numbers quite inefficiently in my preliminary overnight run (i.e. individually instead of using the much faster recurrence formula, and not bothering to only test odd-indexed numbers), so I’m sure I can go much further. And thanks for the suggestion, since I’m considering parallelising the task over a cluster of ~ 30 Mathematica-10-enabled CPUs to which I have access.
3. One can do much better than restrict oneself to odd-indexed n, but it’s a bit of a red herring insofar as the small-factor subroutine should solve these in microseconds — even for very large numbers. I count 4351 M(n) candidates for 8645<n<10^5, based on Mathematica 9 reaching a TimeConstrained two seconds to compute PrimeQ[M(n)] on my Mac. These then are the "difficult to factor" M(n) in that range, the implication being that anything else (being solved in less than two seconds) is necessarily False by virtue of having come across a small factor. While I did that candidate calculation I saved the values of M(n) so that I wouldn't have to recompute them for the full evaluation, which I started five days ago. I've just reached #916, which takes me to 28000.
• apgoucher says:
After checking small factors, I presume PrimeQ attempts a Fermat-bash to ensure that it’s at least pseudoprime before subjecting it to heavy primality testing?
Heuristically one expects 0.129 ménage primes in the interval [28000, 10^5], so it looks like my dignity may survive a few more years, at the very least…
• I believe that PrimeQ uses the Miller-Rabin strong pseudoprime test base 2 & base 3, and then the Lucas test.
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2021-09-23 11:46:15
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https://stackoverflow.com/questions/30299837/ironpython-unable-to-run-script-that-imports-numpy
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# IronPython unable to run script that imports numpy
Disclaimer - I'm not familiar with Python. I'm a C# developer who has written an application to execute Python scripts (authored by others) using IronPython. These scripts have so far have only needed to use import math, but one of our users has asked for the application to support for Numpy.
I have installed Numpy on my PC (using the 'numpy-1.9.2-win32-superpack-python2.7.exe' file), which has created a numpy folder under \Lib\site-packages. I've written a two-line Python script to test that Numpy is accessible:-
import numpy as np
x = np.array([1,2])
I run the script from within C#:-
var engine = Python.CreateEngine();
engine.SetSearchPaths(new Collection<string>(new[]
{
@"C:\Python27",
@"C:\Python27\DLLs",
@"C:\Python27\Lib",
@"C:\Python27\Lib\site-packages",
@"C:\Python27\Lib\site-packages\numpy",
@"C:\Python27\Lib\site-packages\numpy\core"
}));
var scope = engine.CreateScope();
var scriptSource = engine.CreateScriptSourceFromString(
_myPythonScript,
SourceCodeKind.Statements);
scriptSource.Execute(scope);
Despite setting all those search paths, the last line throws an ImportException:-
cannot import multiarray from numpy.core
Note that this SO article is similar, but didn't help - the first answer mentions an 'mtrand.dll' file, which I don't seem to have.
What am I missing?
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2020-01-17 16:42:09
|
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http://mathhelpforum.com/algebra/157480-these-fractions-bigger.html
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# Math Help - Which of these fractions is bigger?
1. ## Which of these fractions is bigger?
I am studying the book "Algebra" by Israel Gelfand and am having trouble with this problem:
Which of the fractions $\frac{12345}{54321}$ and $\frac{12346}{54322}$ is bigger?
I know I could calculate it by hand by finding a common denominator, but I'm fairly certain the book is not interested in rote calculation of this magnitude, so there must be a simpler way. For example, the previous problem was to determine whether $\frac{10,001}{10,002}$ or $\frac{100,001}{100,002}$ was bigger. By noting that both fractions are very close to one, but that the latter is the closest, you can easily determine which is larger. Is there something similar you can do for this one?
Thanks!
2. Originally Posted by Ragnarok
I am studying the book "Algebra" by Israel Gelfand and am having trouble with this problem:
Which of the fractions $\frac{12345}{54321}$ and $\frac{12346}{54322}$ is bigger?
I know I could calculate it by hand by finding a common denominator, but I'm fairly certain the book is not interested in rote calculation of this magnitude, so there must be a simpler way. For example, the previous problem was to determine whether $\frac{10,001}{10,002}$ or $\frac{100,001}{100,002}$ was bigger. By noting that both fractions are very close to one, but that the latter is the closest, you can easily determine which is larger. Is there something similar you can do for this one?
Thanks!
The easiest way to me is to see that if you start with a/b where 0<a<b and you keep replacing a/b with (a+1)/(b+1) you will get closer and closer to 1, therefore the second fraction is bigger.
3. Originally Posted by Ragnarok
I am studying the book "Algebra" by Israel Gelfand and am having trouble with this problem:
Which of the fractions $\frac{12345}{54321}$ and $\frac{12346}{54322}$ is bigger?
I know I could calculate it by hand by finding a common denominator, but I'm fairly certain the book is not interested in rote calculation of this magnitude, so there must be a simpler way. For example, the previous problem was to determine whether $\frac{10,001}{10,002}$ or $\frac{100,001}{100,002}$ was bigger. By noting that both fractions are very close to one, but that the latter is the closest, you can easily determine which is larger. Is there something similar you can do for this one?
Thanks!
The fraction will remain the same if values in the same proportion are added to numerator and denominator.
$\frac{x}{y}=\frac{x+kx}{y+ky}=\frac{(k+1)x}{(k+1)y }$
Example
$\frac{1}{5}=\frac{1+1}{5+5}=\frac{1+0.2}{5+1}$
since the values added are in the same proportion as in the original fraction.
Hence if the proportions of the additional values are different, the fraction will change
Example
$\frac{1+0.1}{5+1}<\frac{1}{5},\;\;\;\frac{1+0.3}{5 +1}>\frac{1}{5}$
Hence, for your fraction, the numerator is < the denominator.
If we add values that are equal, then their proportion is greater than the proportions
of the original numerator and denominator,
hence the fraction increases.
$\frac{12345}{54321}<\frac{12345+1}{54321+1}$
4. Thank you both so much! Cleared it right up for me.
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2015-08-03 18:39:06
|
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http://math.stackexchange.com/questions/392281/what-holds-in-a-deductively-complete-system
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# What holds in a deductively complete system?
I read an article which presented some system with axioms and inference rules (I don't know if that type of system have a term for in English).
The article stated that the system is "deductively complete".
I have read in Wikipedia a bit to recall what does it mean (its been a while since I studied logic).
In Wikipedia in states that $S$ is deductively complete system if for each formula $\varphi$ either $\varphi$ or $\neg\varphi$ is a theorem of $S$ .
Do I understand correctly that this system is not neccaseraly consistent ? for example you can add $$p\to\neg p$$ which is a contradiction and call the new system $S'$. If for a given $\varphi$ you had a proof for $\varphi$ or $\neg\varphi$ (that did not use the new axiom we added) then the same proof works in $S'$hence $S'$ is also deductively complete but is not consistent.
So if I understand correctly, a deductively complete system need not be sound.
Does it have to be complete ? the name of it suggest so, but I believe that the answer is no, for example if $S$ is a complete deductively complete system then either $\varphi$ or $\neg\varphi$ is a theorem of $S$ . I think there is maybe a way to construct a deductively complete system $S'$ in which if $\varphi$ is a theorem of $S$ then $\neg\varphi$ is a theorem of $S'$ and if $\neg\varphi$ is a theorem of $S$ then $\varphi$ is a theorem of $S'$. Then $S'$ is not complete (at least in the case where $S$ is consistent).
To sum up: Did I understand correctly that a deductively complete system need not be consistent, sound or complete ?
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Axioms and inference rules? Isn't that the definition of "logic"? :-) – Asaf Karagila May 15 '13 at 9:33
Careful : $S$ is complete iff for every $\varphi$, $S$ as $\varphi$ or $\neg \varphi$ as a theorem. The or is inclusive. Your either $\varphi$ or $\neg \varphi$ makes me think you consider it exclusive. So yes, a theory can be complete but inconsistent. Actually every inconsistent theory is complete (as it shows everything). – Pece May 15 '13 at 9:39
If every complete set of axioms were consistent, then Gödel's incompleteness theorem, which says that there is no complete, consistent axiomatization of Peano arithmetic, would be much less interesting. – MJD May 15 '13 at 13:22
I think you need to say more. Some answers are about "complete theory" and not about "complete system of natural deduction". So, what is it that article? – GEdgar May 15 '13 at 14:20
1. A deductively complete theory can be inconsistent. Indeed, as Pece remarks, a classical inconsistent theory entails every sentence $\varphi$ of the relevant language so (a fortiori) entails at least one of $\varphi$ and $\neg\varphi$ for each sentence $\varphi$. But your argument works too: if $T$ is deductively complete, consider $T'$ which you get by adding a contradiction as a new axiom. $T'$ proves everything which $T$ did, so is complete, but is inconsistent by construction.
2. A deductively complete theory can be inconsistent, no inconsistent theory is sound, so a deductively complete theory need not be sound.
3. But a deductively complete theory is, trivially, deductively complete. Of course, if we can show that $\varphi$ is a $T$ theorem iff $\neg\varphi$ is also a theorem, then we know that either $T$ is incomplete or is inconsistent. Turning that about, if $T$ is deductively complete, then either it is inconsistent or it isn't the case that there is some $\varphi$ such that $\varphi$ is a $T$ theorem iff $\neg\varphi$ is also a theorem.
4. A deductively complete theory need not be sound on some given interpretation (if it has false axioms), so its set of theorems and the set of relevant truths are distinct. If completeness is understood in a semantic sense, as a matter of yielding a complete story about the relevant truths, then deductively completeness doesn't imply this kind of semantic completeness.
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The Wikipedia article you linked to describes the different types of completeness fairly well, but there are some interesting connections between them. First, semantic completeness (everything true is provable) does not imply syntactic completeness (each sentence or its negation is provable), since there are sentences which are contingent (e.g., $A \lor B$), and the propositional calculus (a semantically complete calculus) won't be able to prove either of $A \lor B$ or $\lnot(A \lor B)$.
To the first question, a deductively complete system need not be consistent. The trivial counterexample is that an inconsistent system proves everything, and thus proves $\phi$ or $\lnot\phi$ for every $\phi$ (in fact, it proves both!).
To the second question, a deductively complete system need not be (semantically) complete. We could construct, for instance, an esoteric calculus with bizarre proof rules that guarantee that for every formula $\phi$, either $\phi$ or $\lnot\phi$ is a theorem. For instance, given the propositional language with variables $A$, $B$, $C$, $\dots$, the connectives $\lnot$, $\land$, $\lor$, and $\to$, where well-formed formulae are defined as usual, take as your system just the following two axiom schemata:
1. $\phi$ where $\phi$ is a wff in which $\land$ appears an odd number of times
2. $\lnot\phi$ where $\phi$ is a wff in which $\land$ does not appear an odd number of times
Since every formula $\phi$ either has an odd number of $\land$s, in which case $\phi$ is a theorem, or it doesn't, in which case $\lnot\phi$ is a theorem, the system is deductively complete. Theorems of this system include \begin{gather*} \lnot A \qquad A \land B \qquad \lnot(A \land (B \land C)) \end{gather*} It's pretty obviously not semantically complete (there are true sentences that it does not prove) and even unsound (there are non-true sentences that it does prove), but it's syntactically consistent (it never proves $\phi$ and $\lnot\phi$).
To sum up, a deductively complete system need not be (syntactically) consistent (never prove $\phi$ and $\lnot\phi$), sound ($\vdash \phi$ implies $\models \phi$), or (semantically) complete ($\models \phi$ implies $\vdash\phi$).
It would be an interesting exercise to determine which combinations of those can be had in a proof system, though. In the examples above saw deductively complete systems that were:
• inconsistent, unsound, and (semantically) complete;
• consistent, unsound, and (semantically) incomplete;
Without specifying a particular interpretation for the propositional variables, we cannot have a sound deductively complete system, because the deductively complete system has to make some judgment about the contingent sentences. Any inconsistent system will automatically be (semantically) complete, so the only possible case that we haven't seen is one which is:
• consistent, unsound, and (semantically) complete.
Such a system could be had if a propositional variable assignment were part of the proof system. For instance, take the standard propositional calculus as a starting point. Then suppose some particular interpretation $\cal I$ of propositional variables (as an easy one, let every propositional variable be true). The standard propositional calculus is consistent and semantically complete, so we have those. To get the unsoundness, let us also add the axiom schema inference rule:
• $\phi$, if $\cal I \models \phi$.
This will have the effect of making every otherwise contingent sentence (or its negation) a theorem. Since the propositional calculus is semantically complete, the only $\phi$ such that neither $\phi$ nor $\lnot\phi$ is a theorem are contingent sentences, but now those have been fixed as well, so the resulting system is consistent, unsound (since, e.g., $\vdash A \land B$ and $\not\models A \land B$), and semantically complete.
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2015-07-30 18:41:00
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https://stacks.math.columbia.edu/tag/044D
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Proof. What we are saying here is that $R'$ of Lemma 77.17.1 is also equal to
$R' = (U' \times _ B U')\times _{U \times _ B U} R \longrightarrow U' \times _ B U'$
In fact this might have been a clearer way to state that lemma. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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2022-10-03 17:45:09
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https://codefreshers.com/mashtali-vs-atcoder-solution-codeforces/
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# Mashtali vs AtCoder solution codeforces
After many unsuccessful tries, Mashtali decided to copy modify an AtCoder problem. So here is his copied new problem:
There is a tree with 𝑛n vertices and some non-empty set of the vertices are pinned to the ground.
Two players play a game against each other on the tree. They alternately perform the following action:
• Remove an edge from the tree, then remove every connected component that has no pinned vertex.The player who cannot move loses(every edge has been deleted already).
You are given the tree, but not the set of the pinned vertices. Your task is to determine, for each 𝑘k, the winner of the game, if only the vertices 1,2,3,,𝑘1,2,3,…,k are pinned and both players play optimally.
Input
The first line of input contains an integer 𝑛n — the number of vertices (1𝑛31051≤n≤3⋅105).
The 𝑖i-th of the following 𝑛1n−1 lines contains two integers 𝑢𝑖,𝑣𝑖ui,vi (1𝑢𝑖,𝑣𝑖𝑛1≤ui,vi≤n𝑢𝑖𝑣𝑖ui≠vi) — the endpoints of the 𝑖i-th edge. It’s guaranteed that these edges form a tree.
Output
Print a string of length 𝑛n. The 𝑖i-th character should be ‘1’ if the first player wins the 𝑖i-th scenario, and ‘2’ otherwise.
Examples
input
## Copy Mashtali vs AtCoder solution codeforces
5
1 2
2 3
2 4
4 5
output
Copy
11122
input
Copy
5
1 2
2 3
1 4
4 5
### output Mashtali vs AtCoder solution codeforces
21122
input
Copy
6
1 2
2 4
5 1
6 3
3 2
output
Copy
111111
input
Copy
7
1 2
3 7
4 6
2 3
2 4
1 5
output
Copy
2212222
#### Note Mashtali vs AtCoder solution codeforces
Below you can see the tree in the first sample :
If 𝑘=1k=1 then the first player can cut the edge (1,2)(1,2).
If 𝑘=2k=2 or 𝑘=3k=3, the first player can cut the edge (2,4)(2,4), after that only the edges (1,2)(1,2) and (2,3)(2,3) remain. After the second players move, there will be a single edge left for the first player to cut. So first player wins.
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2021-12-03 17:14:40
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https://www.groundai.com/project/n-body-simulations-for-extended-quintessence-models/
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N-Body Simulations for Extended Quintessence Models
# N-Body Simulations for Extended Quintessence Models
Baojiu Li, David F. Mota and John D. Barrow DAMTP, Centre for Mathematical Sciences, University of Cambridge, Wilberforce Road, Cambridge CB3 0WA, UK Kavli Institute for Cosmology Cambridge, Madingley Road, Cambridge CB3 0HA, UK Institute of Theoretical Astrophysics, University of Oslo, 0315 Oslo, Norway
###### Abstract
We introduce the -body simulation technique to follow structure formation in linear and nonlinear regimes for the extended quintessence models (scalar-tensor theories in which the scalar field has a self-interaction potential and behaves as dark energy), and apply it to a class of models specified by an inverse power-law potential and a non-minimal coupling. Our full solution of the scalar field perturbation confirms that, when the potential is not too nonlinear, the effects of the scalar field could be accurately approximated as a modification of background expansion rate plus a rescaling of the effective gravitational constant relevant for structure growth. For the models we consider, these have opposite effects, leading to a weak net effect in the linear perturbation regime. However, on the nonlinear scales the modified expansion rate dominates and could produce interesting signatures in the matter power spectrum and mass function, which might be used to improve the constraints on the models from cosmological data. We show that the density profiles of the dark matter halos are well described by the Navarro-Frenk-White formula, although the scalar field could change the concentration. We also derive an analytic formula for the scalar field perturbation inside halos assuming NFW density profile and sphericity, which agrees well with numerical results if the parameter is appropriately tuned. The results suggest that for the models considered, the spatial variation of the scalar field (and thus the locally measured gravitational constant) is very weak, and so local experiments could see the background variation of gravitational constant.
## 1. Introduction
The nature of the dark energy (Copeland, Sami & Tsujikawa, 2006) is one of the most difficult challenges facing physicists and cosmologists now. Although a cosmological constant (plus cold dark matter, to provide the concordance CDM paradigm) could be a solution – and is indeed consistent with virtually all current observations, it suffers from theoretical difficulties such as why its value must be so small yet nonzero, and why it becomes dominant only at the low redshift. In all the alternative proposals to tackle this problem, a quintessence scalar field (Zlatev, Wang & Steinhardt, 1999; Wang et al., 2000) is perhaps the most popular one (although a new proposal by Barrow & Shaw (2010) provides a completely new type of explanation that does not require new scalar fields). In such models the scalar field is slowly rolling down its potential, its energy density is dominated by the potential energy and almost remaining constant provided that the potential is flat enough. The flatness of the potential, however, means that the mass of the scalar field is in general very light and as a result the scalar field almost does not cluster so that its effects in cosmology are mainly on the (modified) background expansion rate.
One reason for the wide interest in quintessence models is that scalar fields appear in abundance in high-energy physics theories, in which they are often coupled to the curvature invariants or even other matter species, leading to the so-called extended quintessence (Perrotta, Baccigalupi & Matarrese, 2000; Baccigalupi & Perrotta, 2000; Baccigalupi, Perrotta & Matarrese, 2000) and coupled quintessence (Amendola, 2000, 2004; Jesus et al., 2008) models respectively. The former is just a special class of a scalar-tensor theory (Fujii & Maeda, 2003; Riazuelo & Uzan, 2002), with the scalar field being the dark energy. These two classes of generalised quintessence models have been studied in detail in the linear regime in the literature (Bean, 2001a, b; Mangano et al., 2003; Clifton, Mota & Barrow, 2005; Nunes & Mota, 2006; Pettorino, Baccigalupi & Mangano, 2005; Koivisto, 2005; Brookfield et al., 2006; Koivisto & Mota, 2007; Mota & Shaw, 2007, 2006; Lee, Liu & Ng, 2006; Mota et al., 2007; Bean et al., 2008; Bean, Flanagan & Trodden, 2008; Boehmer et al., 2008, 2010).
In recent years, studies of the cosmological behaviour of the coupled quintessence model in the nonlinear regime have also been made, either via semi-analytical methods (Manera & Mota, 2006; Mota & van de Bruck, 2004; Mota, 2008; Shaw & Mota, 2008; Mota et al., 2008a; Mota, Shaw & Silk, 2008; Saracco et al., 2009; Wintergerst & Pettorino, 2010), or using -body simulation techniques (Maccio et al., 2004; Nusser, Gubser & Peebles, 2005; Kesden & Kamionkowski, 2006a, b; Springel & Farrar, 2007; Farrar & Rosen, 2007; Baldi & Pettorino, 2010; Hellwing & Juszkiewicz, 2009; Keselman, Nusser & Peebles, 2009, 2010; Hellwing, Knollmann & Knebe, 2010; Baldi, 2010; Baldi et al., 2010). In these studies the effect of the scalar field is generally approximated by a Yukawa-type ’fifth force’ or by a rescaling of the gravitational constant or the particle mass, without solving the scalar field equation explicitly. Very recently, Li & Zhao (2009, 2010); Zhao et al. (2010); Li & Barrow (2010) gave a new treatment and obtained an explicit solution to the scalar field perturbation on a spatial grid. The new results confirmed that the approximations adopted in the old literature were good for the models considered there (where the scalar potential was not very nonlinear), but for highly nonlinear potentials they broke down.
For the extended quintessence (more generally scalar-tensor) models, investigations using -body simulations are rarer. The work of Pettorino & Baccigalupi (2008), for example, outlined a recipe which uses certain approximation, such as a rescaling of gravitational constant, and does not solve the scalar field equation of motion explicitly. In Rodriguez-Meza et al. (2007); Rodriguez-Meza (2008a, b), the authors approximated the effect of scalar field coupling as a Yukawa force. However, none of these previous works tries to solve the scalar field on a mesh directly, and this is what we want to do in this work.
The aims of this work are threefold. Firstly, we want to develop the formulae and methods that are needed to solve the scalar field explicitly, which could serve as the basis for future work, and to find the regime of validity of our method. Secondly, we want to understand whether the approximations adopted in the previous studies are good or not; given the severe limits in the computing power; if those approximations do work well, then one does not need to resort to an exact scalar field solver, which is considerably more economical. Finally, we want to study structure formation in the nonlinear regime for some specific models, and investigate both the scalar field effects on the clustering of matter and the spatial variation of the gravitational constant (which is common to scalar-tensor theories).
The organisation of this paper is as follows: In Sect. 2 we list the basic equations which are needed in -body simulations and give their respective non-relativistic limits. To prevent the main text from expanding too much, some useful expressions are listed in Appendix A, and the discrete versions of the resulted equations are discussed and summarised in Appendix B. In Sect. 3 we briefly describe the numerical code we are using (relegating further details to Knebe, Green & Binney (2001); Li & Zhao (2010)), and the physical parameters of our simulations. We also present some results regarding the background cosmology and linear perturbation evolution in our models, which could be helpful in the understanding of the -body simulation results (our algorithm for the background cosmology is summarized in Appendix C). Sect. 4 contains the -body simulation results, including key structure formation observables such as nonlinear matter power spectrum, mass function and dark matter halo profile, as well as the spatial variation of the scalar field. It also includes several checks of the approximations made in the literature. We finally summarise and conclude in Sect. 5.
We use the unit unless explicitly restoring in the equations. The metric convention is . Indices run while run .
## 2. The Equations
This section presents the equations that will be used in the -body simulations, the model parameterisation and discretisation procedure for the equations.
### 2.1. The Basic Equations
We consider a general Lagrangian density for scalar-tensor theories
L = 12κ∗[1+f(φ)]R−12∇aφ∇aφ+V(φ)−Lf, (1)
in which where is the (bare) gravitational constant, is the Ricci scalar, is the coupling function between the scalar field and curvature, the potential for and the Lagrangian density for fluid matter (baryons, photons, neutrinos and cold dark matter). Note that is a fundamental constant of the theory.
Varying the associated action with respect to metric yields the energy-momentum tensor of the theory (note the tilde, which is used to distinguish it from the defined below):
~Tab = Tfab+∇a∇bφ−12gab∇cφ∇cφ+gabV(φ) (2) −1κ∗[f(φ)Gab+(gab∇c∇c−∇a∇b)f(φ)]
where is the Einstein tensor, and is the energy-momentum tensor for matter (including baryons, dark matter, neutrinos and photons, which we collectively refer to as ’fluid matter’, although in -body simulations we use discrete particles rather than a fluid).
As usual, we can rearrange the Einstein equation as
Gab = κ∗~Tab (3)
so that it now looks like
Gab = κ∗1+fTfab−11+f(gab∇c∇c−∇a∇b)f (4) +κ∗1+f[∇aφ∇bφ−12gab(∇φ)2+gabV] ≡ κ∗Tab.
Note the difference between and ; throughout this paper, we will use a superscript for normal fluid matter, and quantities without a superscript always mean the total effective ones [the final line of Eq. (4)]. It is sometimes useful to define an effective Newton constant . Neither nor is the gravitational constant measured in a Cavendish-type experiment, which we denote instead by and is given by
κ⨁ = κ∗1+f2+2f+4(dfd√κ∗φ)22+2f+3(dfd√κ∗φ)2 (5)
where is added to make dimensionless, which is the convention we shall always follow below. itself is obviously not a constant and we measure only it present-day value, .
Varying the action with respect to the scalar field, , gives the scalar field equation of motion
∇a∇aφ+∂V(φ)∂φ+R2κ∗∂f(φ)∂φ = 0 (6)
Since we will follow the motions of dark matter particles in the -body simulations, so we also need their geodesic equation. The dark-matter Lagrangian for a point particle with mass is
LCDM(y)=−m0√−gδ(y−x0)√gab˙xa0˙xb0, (7)
where is the general coordinate and is the coordinate of the centre of the particle. From this equation we derive the corresponding energy-momentum tensor:
TabCDM=m0√−gδ(y−x0)˙xa0˙xb0. (8)
Taking the conservation equation for dark matter particles (which, unlike in (Li & Zhao, 2009, 2010), does not couple to any other matter species, including the scalar field ), the geodesic equation follows as usual:
¨xa0+Γabc˙xb0˙xc0 = 0, (9)
where the second term on the left-hand side accounts for gravity.
Eqs. (4, 6 , 9) contain all the physics needed for the following analysis, though certain approximations and simplifications might have to be made in due course to make direct connection to -body simulations.
We will consider an inverse power-law potential for the scalar field,
V(φ) = Λ4(√κ∗φ)α, (10)
where is a dimensionless constant and is a constant with dimensions of mass. This potential has also been adopted in various background or linear perturbation studies of scalar fields (either minimally or non-minimally coupled); the tracking behaviour its produces makes it a good dark energy candidate and for that purpose we shall choose . Meanwhile, the coupling between the scalar field and the curvature tensor is chosen to be a non-minimal one:
f(φ) = γκ∗φ2, (11)
where is another dimensionless constant characterising the strength of the coupling. Note that here again is added into and to make a dimensionless quantity . Although the exact value of is unknown, so is and we can solve for instead of , not caring about the exact individual values of and .
### 2.2. The Non-Relativistic Limits
The -body simulation only probes the motion of particles at late times, and we are not interested in extreme conditions such as black hole formation and evolution, so we can take the non-relativistic limit of the above equations as a good approximation.
The existence of the scalar field and its coupling to the curvature leads to several possible changes with respect to the CDM paradigm:
1. The scalar field has its own energy-momentum tensor, which could change the source term of the Poisson equation because the scalar field, unlike the cosmological constant, can cluster (though the clustering is often quite weak in scalar field models). Also, unlike in coupled scalar field models, here the term will appear in the Poisson equation.
2. The background cosmic expansion rate is in general modified, and can either slow down or speed up the rate of structure formation.
3. The two gravitational potentials in the conformal Newtonian gauge metric , in which and are respectively the conformal time and comoving coordinate, are no longer equal to each other (as in general relativity), but are instead related by (see below).
It therefore becomes clear that the following two equations, in their non-relativistic forms, need to be solved in order to obtain the gravitational force on particles:
1. The scalar field equation of motion, which is used to compute explicitly the value of the scalar field at any given time and position;
2. The Poisson equation, which is used to determine the gravitational potential and force at any given time and position from the local energy density and pressure, which includes the contribution from the scalar field (obtained from the equation of motion).
Note that unlike in the coupled scalar field models, there is no fifth force because there is no direct coupling to the particles. The scalar coupling to the curvature, however, does modify the gravitational potential so that gravity no longer follows Einstein’s prescription and so this is a modified gravity theory.
We now describe these two equations in turn. For the scalar field equation of motion, we denote by the background value of and write . Then using the expressions given in Appendix A we write
a2∇a∇aφ = φ′′+2Hφ′+→∇2xφ−2ϕφ′′ (12) −(ϕ′+3ψ′+4Hϕ)φ′
in which with the conformal time, is the derivative with respect to the comoving coordinate , and . Then, with the background part subtracted, Eq. (6) can be rewritten as
δφ′′+2Hδφ′+→∇2xδφ+[V,φ(φ)−V,φ(¯φ)]a2 −2ϕ¯φ′′−(ϕ′+3ψ′+4Hϕ)¯φ′ +12κ∗[Rfφ(φ)−¯Rfφ(¯φ)]a2 = 0,
in which a bar denotes the background value, and the subscript denotes derivatives with respect to . Note that has the same sign as .
In our -body simulations we shall work in the quasi-static limit, i.e., we assume that the spatial gradients are much greater than the time derivatives, . Therefore, the time derivatives in the above equation are dropped and we obtain the simplified version
c2→∂2x(aδφ) =
in which due to our sign convention , and we have restored the factor in front of (the here and in the remaining of this paper is times the in the original Lagrangian unless otherwise stated). Note that here has the dimension of mass density rather than energy density.
To complete Eq. (2.2), we still need expressions for and , which are again obtained using the quantities in Appendix A:
R = −6a2a′′a(1−2ϕ) (14) +1a2[6ψ′′+6H(ϕ′+3ψ′)−4→∂2xψ+2→∂2xϕ], ¯R = −6a2a′′a (15)
and so
Rfφ−¯R¯fφ ≐ ¯fφδR+¯Rδfφ (16) ≐ −1a2¯fφ(4→∂2xψ−2→∂2xϕ)−6a2a′′aδfφ
where we have again dropped time derivatives of and since they are small compared with the corresponding spatial gradients, and , .
Since only but not appears in the Poisson equation (shown below) , we also want to eliminate the in the scalar field equation of motion. This is easy in general relativity, because there we have the simple relation , which unfortunately no longer holds in scalar-tensor theories. However, we could use the components of the Einstein equation () to get a new relation between and . Noting that our -body simulations probe the very late time evolution (when radiation is negligible) when the only significant source for () is the scalar field, and
∇i∇jf = −1a2∂i∂jf (i≠j) (17)
where , we could write the component of Einstein equation as
∂i∂j(ϕ−ψ) = −c21+f∂i∂jf
which gives approximately
∂i(ϕ−ψ) = −c21+f∂if
and so
4→∂2xψ−2→∂2xϕ ≐ 2→∂2xϕ+41+f→∂2xf (18) ≐ 2→∂2xϕ+4¯fφ1+¯fc2→∂2xδφ.
It is important to note that in the second line of Eq. (18) we have implicitly linearised the equation; this is valid only if is not strongly nonlinear and . It turns out that the model considered in this work satisfies these criteria (). If either or is highly nonlinear, then we might have ; in that case we should not approximate to even in the coefficients of the perturbation variables such as here, or write . The reason for the latter stricture is as follows: if is highly nonlinear, then might change a lot even if fluctuates a little, implying that for the linearisation to apply on our spatial grid we need very small grid sizes which are impossible; moreover, it becomes complicated to decide which solution we should linearise around, as the values of in that area which we look at might be very different from the background value . The strategy for this situation is simple: instead of writing , we difference directly, because we know the value of in every grid cell. This will ensure no linearisation error. In what follows, however, we shall use Eq. (18), which causes negligible linearisation error but simplifies the equations a lot. We shall also write in the coefficients of perturbation quantities such as and .
Substituting Eqs. (16, 18) into Eq. (2.2) and rearranging, we complete the derivation of the scalar field equation of motion in the weak field limit, ending up with
⎡⎢⎣1+2¯f2φκ∗(1+¯f)⎤⎥⎦c2→∂2x(aδφ) (19) =
for our general Lagrangian Eq. (1) and
[1+8γ2κ∗¯φ21+γκ∗¯φ2]c2→∂2x(a√κ∗δφ) (20) = −γ√κ∗¯φ→∂2xΦ−6γ(H′+H2)(a√κ∗δφ) −ακ∗Λ4a3⎡⎣1(√κ∗φ)1+α−1(√κ∗¯φ)1+α⎤⎦
for the model specified by Eqs. (10, 11), where .
Next consider the Poisson equation, which is obtained from the Einstein equation in the weak-field and slow-motion limits. Here we use the component of the Ricci curvature tensor, which is given as
R0 0 = −3(a′′a−H2)(1−2ϕ) (21) +3ψ′′+3H(ψ′+ϕ′)+→∂2xϕ
using the expressions in Appendix A. According to the Einstein equations,
R0 0 = κ∗2(ρTOT+3pTOT)a2 (22)
where and are the total energy density and pressure, respectively. Using these two equations and subtracting the background part (which is just the Raychaudhuri equation), it is straightforward to find that
→∂2xΦ = κ∗a32[(ρTOT+3pTOT)−(¯ρTOT+3¯pTOT)]. (23)
in which we have dropped terms involving time derivatives of and , because they are much smaller than in the quasi-static limit. Using the energy-momentum tensor expressed in Eq. (4), the above equation can be rewritten as
→∂2xΦ ≐ κ∗a32¯ρm(δ1+f−11+¯f)−¯fφ2(1+¯f)c2→∂2x(aδφ) −κ∗a3[V(φ)1+f−V(¯φ)1+¯f] +a(11+f−11+¯f)(κ∗¯φ′2+32f′′)
for the general Lagrangian Eq. (1) and
→∂2xΦ = −γ√κ∗¯φ1+κ∗¯φ2c2→∂2x(a√κ∗δφ) −[κ∗Λ4a3(1+γκ∗φ2)(√κ∗φ)α−κ∗Λ4a3(1+γκ∗¯φ2)(√κ∗¯φ)α] +[(1+3γ)κ∗¯φ′2+3γκ∗¯φ¯φ′′]a ×[11+γκ∗φ2−11+γκ∗¯φ2]
for the model specified by Eqs. (10, 11). In these equations is the background density for matter, , and we have used the definition of given in Appendix C. We have also neglected the contribution from to the total density and pressure, because in the quasi-static limit we have and (which is confirmed by the -body simulation results111According to Eq. (20) we have , implying that , so neglecting time derivatives of is just like dropping time derivatives of and , which we have already done to obtain the modified Poisson equation.).
Finally, the equation of motion of the dark matter particles is the same as in general relativity
¨x+2˙aa˙x = −1a3→∇xΦ (26)
in which is determined by the modified Poisson equation Eq. (2.2). The canonical momentum conjugate to is so from the equation above we have
dxdt = pa2, (27) dpdt = −1a→∇xΦ. (28)
Eqs. (20, 2.2, 27, 28) will be used in the code to evaluate the forces on the dark-matter particles and evolve their positions and momenta in time. But before applying them to the code we still need to switch to code units (see Sect. 2.3), further simplify them and create the discrete version (see Appendix B).
### 2.3. Code Units
In our numerical simulation we use a modified version of MLAPM ((Knebe, Green & Binney, 2001)), and we will have to change or add our Eqs. (20, 2.2, 27, 28) to it. The first step is to convert the quantities to the code units of MLAPM. Here, we briefly summarise the main results.
The (modified) MLAPM code uses the following internal units (where a subscript stands for ”code”):
xc = x/B, pc = p/(H0B) tc = tH0 Φc = Φ/(H0B)2 ρc = ρ/¯ρ, u = ac2√κδφ/(H0B)2, (29)
where denotes the comoving size of the simulation box, is the present Hubble constant, and is the matter density. In the last line the quantity is the scalar field perturbation expressed in terms of code units and is new to the MLAPM code.
In terms of , as well as the (dimensionless) background value of the scalar field, , some relevant quantities are expressed in full as
V(φ) = Λ4(√κ¯φ+B2H20ac2u)α, f(φ) = 1+γ(√κ¯φ+B2H20ac2u)2, Vφ(φ) = −α√κΛ4(√κ¯φ+B2H20ac2u)1+α, fφ(φ) = 2γ√κ(√κ¯φ+B2H20ac2u), (30)
and the background counterparts of these quantities can be obtained simply by setting (recall that represents the perturbed part of the scalar field) in the above equations.
We also define
λ ≡ κΛ43H20, (31)
which will be used frequently below.
Making discrete versions of the above equations for -body simulations is then straightforward, and we refer the interested readers to Appendix B to the whole treatment, with which we can now proceed to do -body simulations.
## 3. Simulation Details
### 3.1. The N-Body Code
Some of our main modifications to the MLAPM code for the coupled scalar field model are:
1. We have added a solver for the scalar field, based on Eq. (B7). It uses a nonlinear Gauss-Seidel scheme for the relaxation iteration and the same criterion for convergence as the default Poisson solver. But it adopts a V-cycle instead of the self-adaptive scheme in arranging the Gauss-Seidel iterations.
2. The value of solved in this way is then used to calculate the total matter density, which completes the calculation of the source term for the Poisson equation. The latter is then solved using a fast Fourier transform on the domain grids and self-adaptive Gauss-Seidel iteration on refinements.
3. The gravitational potential obtained in this way is then used to compute the force, which is used to displace and kick the particles.
There are a lot of additions and modifications to ensure smooth interface and the newly added data structures. For the output, as there are multilevel grids all of which host particles, the composite grid is inhomogeneous and so we choose to output the positions and momenta of the particles, plus the gravity and values of and at the positions of these particles. We also output the potential and scalar field values on the domain grid.
### 3.2. Physical and Simulation Parameters
The physical parameters we use in the simulations are as follows: the present-day dark-energy fractional energy density and , , , . Our simulation box has a size of Mpc, where . We simulate four models, with parameters , , and respectively. In all those simulations, the mass resolution is ; the particle number is ; the domain grid is a cubic and the finest refined grids have 16384 cells on each side, corresponding to a force resolution of about kpc.
We also run a CDM simulation with the same physical parameters and initial condition (see below).
### 3.3. Background and Linear Perturbation Evolution
Since the coupling between the scalar field and the curvature produces a time-varying effective gravitational constant, and the scalar field contributes to the total energy-momentum tensor, we expect that cosmology in the extended quintessence models is generally different from CDM at the background and linear perturbation levels. A good understanding of this will be helpful in our analysis of the results from -body simulations, and this is the subject of this subsection.
Our algorithm and formulae for the background cosmology are detailed in Appendix C, and are implemented in MAPLE. We output the relevant quantities in a predefined time grid, which could be used (via interpolation) in the linear perturbation and -body computations.
Fig. 1 shows the time evolutions of some background quantities of interests. For ease of comparison we have chosen and to be the same in all models including the CDM one (for definitions of and see Appendix C), and as a result in the upper left panel the curves for different models converge at common righthand ends. We see increasing results in an earlier and slower growth of (). This indicates a larger dark energy equation of state parameter, , which is confirmed by the upper right panel. Physically, this is because, the larger is, the steeper the potential becomes and thus the faster the scalar field rolls. Notice that is also larger for positive , with being the same. This is because in Eq. (6) the Ricci scalar and for positive the term has the same sign as , thus helping the scalar field to roll faster. Because of its large predicted value of , the model is already excluded by cosmological data, but here we shall keep it for purely theoretical interest (i.e., to see how changing or changes the nonlinear structure formation).
We are also interested in how the expansion rate in an extended quintessence model differs from that in CDM, and the results for our models are shown in the lower-left panel of Fig. 1, which plots the as a function of . The rather odd behaviour of the models at low redshift is because of the complicated evolution of the scalar field (and the fact that we have chosen to be the same for all models, again for ease of comparison), while the high-redshift behaviour could be seen directly from Eq. (C4). In Eq. (C4) the energy density of the scalar field can be dropped at high , and so we have
(HH0)2 ≈ 1+f01+fΩma−1 (32)
where we have also neglected the radiation for simplicity (which is valid after the matter-radiation equality). This shows that in extended quintessence models the gravitational constant relevant for the background cosmology is rescaled by . Because where is the present-day value of , and is monotonically increasing in time, so for our choice of [cf. Eq. (11)] we have for and for : thus models with have .
It turns out that the gravitational constant relevant for the growth of matter density perturbations is also different from the one governing the background cosmology. If we denote the matter density perturbation by , then it can be shown, using the linear perturbation equations, that on small scales the evolution equation for reduces to
δ′′m+Hδ′m = GN3H202Ωmδma2 (33)
in which and is the conformal time (see Appendix C), and we have defined
GN ≡ 1+f01+f2+2f+4(dfd√κ∗φ)22+2f+3(dfd√κ∗φ)2. (34)
Note that this quantity could also be directly read off from the modified Poisson equation Eq. (B4).
In the lower right panel of Fig. 1 we display the evolution for in the models considered. Again, is larger at earlier times for positive and smaller for negative , because of our specific choice of in Eq. (11), and the fact that is always increasing in time.
It is well known that a higher rate of background expansion means that structures have less time to form, and a larger speeds up the structure formation. These two effects therefore cancel each other to some extent, which results in a weaker net effect of an extended quintessence field on the large scale structure formation. This is confirmed by our linear perturbation computation depicted in Fig. 2. In the right-hand panels of this figure we have plotted the matter power spectra for different models at two different redshifts (0 and 49). It is interesting to note that on small scales the matter power is closer to that of CDM, despite the significant differences in background expansion rate and (cf. Fig. 1). Because of this, we shall choose CDM initial condition for our -body simulations for all our models, saving the effort of generating separate initial conditions for different models.
The left hand panels of Fig. 2 display the CMB power spectra for the models we consider. Again the difference from CDM is fairly small, and there is only a small shift of the CMB peaks even though the background expansion rate changes quite a bit. The latter is because peak positions are determined by the ratio of the sound horizon size at decoupling and the angular distance to the decoupling, and in our model both of these decrease/increase as the Universe expands faster/more slowly, their ratio does not change much.
To briefly summarise, the study of background cosmology and linear perturbation shows that a modified background expansion rate and a rescaled gravitational constant, the two most important factors affecting structure formation in extended quintessence models are opposite effects. It is then of interest to see how these two effects compete in the nonlinear regime.
## 4. N-body Simulation Results
This section lists the results of extended quintessence -body simulations. We shall start with a few preliminary results which both give some basic idea about the extended quintessence effects and serve as a cross check of our codes. Then we discuss the key observables for the nonlinear structure formation such as matter power spectrum, mass function and halo properties. We also comment on the halo profile of the scalar field and the spatial variation of gravitational constant.
### 4.1. Preliminary Results
As mentioned above, in both the linear and -body codes we compute background quantities via an interpolation of some pre-computed table. Because background cosmology is important in determining the structure formation, it is important to check its accuracy. For this we have recorded in Table 1 The age of the universe today for different models as computed by these two codes. The two codes are compatible with each other indeed.
model linear code -body code
CDM 13.680 13.678
13.639 13.638
13.408 13.408
13.513 13.513
12.097 12.096
Because one of the advantages of our -body code is that it solves the scalar field perturbation explicitly, it is important to check that the solution is with expectations. From Eqs. (B1, B2) it could be seen clearly that, if the contribution to the local density and pressure from the scalar field is negligible compared with that from matter, then the modified Poisson equation and scalar field equation of motion end up with the same source term (up to a -dependent coefficient). In this situation we expect
u = −2γ√κ∗¯φ1+8γ2κ∗¯φ21+γκ∗¯φ2Φc, (35)
which means that is simply proportional to with a time-dependent coefficient. In Fig. 3 we have checked this relation explicitly: we select a thin slice of the simulation box, fetch the values for and at the positions of the particles (about 10000 in total) therein, and display them as scatter plots. The solid curve is the approximation Eq. (35) while the green dots are simulation results; we can see they agree very well with each other, showing that the above approximation is a good one. Note that the scalar field perturbation is generally less than , compared with the background value . This confirms that it is consistent to neglect the perturbation in scalar field density/pressure, drop terms such as and , and replace by in coefficients of perturbation quantities such as and . It also serves as a check of the numerical code.
As a final consistency check, let us consider the total gravitational force on particles. In extended quintessence models, this is given by Eq. (B2), and when the perturbation in the scalar field density/pressure is negligible (which is the case as shown above) we get
∇2Φc ≈ 32GNΩmH20(ρc−1) (36)
in which is given in Eq. (34). On the other hand, if we consider (naïvely) that gravity is described by general relativity, then we should neglect the on the right-hand side. Manipulating Eqs. (B1, B2) we obtain:
1+γκ∗¯φ201+γκ∗¯φ2∇2(Φc+γ√κ∗¯φ1+γκ∗¯φ2u) ≈ 32ΩmH20(ρc−1).
Thus acts as the potential for naïve gravity (i.e., general relativity), and by differcing it we could obtain the naïve gravitational force. In Fig. 4 we show the scatter plot of the naïve gravity versus full gravity for the same particles as in Fig. 3 (green dots) as well as their approximate ratio (solid line). Again, the agreement is remarkably good.
### 4.2. Nonlinear Matter Power Spectrum
As we have seen above, the linear matter power spectrum for the extended quintessence model really does not show much useful information on small scales, and so we need to investigate whether nonlinear effects could change this situation and therefore potentially place more meaningful constraints.
Fig. 5 provides a positive answer to this question. Here we have plotted the fractional difference of the extended quintessential nonlinear matter power spectrum from that for CDM (remember that we use the same initial condition for all simulations). We can see that for the models with the differences are small even in the nonlinear regime, indicating that the scalar field really does not affect the matter distribution significantly if the potential is flat. However, for the cases in which the coupling strength remains the same, the difference could be as large as , guaranteeing an observable signature.
Furthermore, for negative (the purple curve) the extended quintessential power spectrum beats the CDM one on small scales, whereas for the positive case (the pink curve) it is just the opposite. As shown before, when , both the background expansion rate and the effective gravitational constant governing the structure formation decrease, boosting and weakening the collapse of matter respectively. In our cases the first effect has clearly taken over on small scales.
### 4.3. Mass Function
A second important observable is the mass function. This gives the number density of dark matter halos as a function of halo mass. For this we need to identify the dark matter halos from the output particle distribution of the -body simulations, and this determination is performed using a modified version of MHF (Knebe & Gibson, 2004), MLAPM’s default halo finder.
MHF optimally utilizes the refinement structure of the simulation grids to pin down the regions in which potential halos reside and organize the refinement hierarchy into a tree structure. MLAPM refines grids according to the particle density on them and so the boundaries of the refinements are simply isodensity contours. MHF collects the particles within these isodensity contours (as well as some particles outside). It then performs the following operations: (i) assuming spherical symmetry of the halo, calculate the escape velocity at the position of each particle, (ii) if the velocity of the particle exceeds then it does not belong to the virialized halo and is removed. Steps (i) and (ii) are then iterated until all unbound particles are removed from the halo or the number of particles in the halo falls below a pre-defined threshold, which is 20 in our simulations. Note that the removal of unbound particles is not used in some halo finders using the spherical overdensity (SO) algorithm, which includes the particles in the halo as long as they are within the radius of a virial density contrast. Another advantage of MHF is that it does not require a predefined linking length in finding halos, such as the friend-of-friend procedure.
Our modification to MHF is simple: because the effective gravitational constant in the extended quintessence models is rescaled by a factor [cf. Eq. (34)], the escape velocity of particles from a halo is also multiplied by this factor, and in MHF we have only changed the criterion for removing particles from virialised halos accordingly. In reality, because we are only interested in the halos in this work, is quite close to 1 and the effect of our modification is not large.
The mass functions for our simulated models are shown in Fig. 6. It shows that all extended quintessence models considered here, irrespective of their parameters, produce less massive halos than CDM, whereas (only) the model produces a larger number of less massive halos. These features are in broad agreement with those shown in the matter power spectra (Fig. 5) where all models show less matter clustering on the large scales, whereas (only) the model shows more power on small scales. The physical reason is again the competition between the modified background expansion rate and rescaled effective gravitational constant .
### 4.4. Halo Properties
In the CDM paradigm, it is well known that the internal density profiles of dark matter halos are very well described by the Navarro-Frenk-White (Navarro, Frenk & White, 1996) formalism
ρ(r)ρc = βrRs(1+rRs)2 (37)
where is the critical density for matter, is a dimensionless fitting parameter and a second fitting parameter with length dimension. and are generally different for different halos and should be fitted for individual halos, but the formula Eq. (37) is quite universal.
We are thus interested in whether the halo profiles in an extended quintessential Universe are also featured by this universal form. For this we select the 80 most massive halos from each simulation and fit their density profiles to Eq. (37). The results show that the NFW profile describes the extended quintessential halos at least as well as it does for the CDM halos. Fig. 7 shows the fittings for two halos randomly picked out of the 80: one at Mpc with mass and the other at Mpc with a mass .
There are some interesting features in Fig. 7. Firstly, for the models with (the top panels) the halo density profile for extended quintessence models (green asterisks) is very similar to the CDM results (black crosses) and thus their fittings almost coincide. Secondly, for the model of , the chosen halos show more concentration of the density profiles in the scalar model than in CDM. Thirdly, the model of has just the opposite trend and suffers a suppression of density in large parts of chosen halos.
To verify that the above features are actually typical for the corresponding models, we have plotted in Fig. 8 the fitting results for all the 80 massive halos in all simulated models. Here in addition to the NFW concentration parameter , where is the radius at which the density is equal to 200 times the critical density and the NFW parameter, we have also shown the fitting errors for each halo.
We would like to point out several important implications of Fig. 8. Firstly, for all models the fitting error for the extended quintessential halos (lower green asterisks) is comparable to that for the CDM halos (lower black crosses), indicating that the density profiles for the former are equally well described by the NFW formula Eq. (37). Secondly, for the models with (the top panels) we can see that the fitted for the extended quintessential halos is comparable to that for CDM, which is in agreement with our finding in Fig. 7 that the density profiles for the chosen halos are almost the same as in the CDM prediction. Thirdly, for the model of , the halos tend to be more concentrated (i.e., with larger ) than in CDM. Fourthly, for the model , the halos tend to be less concentrated (i.e., with smaller ) than in CDM. The above three features show that our qualitative findings in Fig. 7 are quite typical. Finally, the halo masses in the model of are on average smaller than those in CDM, because the upper green asterisks in the lower right panel consistently shift leftwards with respect to the upper black crosses: this is consistent with the mass function result that this model produces less massive halos than CDM.
In summary, the halo density profiles for the extended quintessence models are well described by the NFW formula, but the existence of the scalar field and in particular its coupling to curvature do change the concentration parameters of the halos, so long as the potential is not too flat. It seems that the modified background expansion rate beats the effect of the rescaled effective gravitational constant here.
### 4.5. Halo Profile for Scalar Field Perturbation
We have already seen that the coupling between the scalar field and the curvature scalar causes time and spatial variations of the locally measured gravitational constant . It is then of our interest to ask how varies across a given halo and whether this could produce observable effects. This subsection answers this question, by giving an analytical formula and comparing it with numerical results.
Recall that Fig. 3 shows that to a high precision the scalar field perturbation is proportional to the gravitational potential [cf. Eq. (35)] everywhere. This means that if we could derive an analytical formula for in halos, then we know straightforwardly. Such a derivation has been done in (Li, Mota & Barrow, 2010) for a different model, but here we shall briefly repeat it for the extended quintessence model for completeness.
Assuming Eq. (37) as the density profile and sphericity of halos, we can derive , the circular velocity of a particle moving around the halo at a distance from halo centre, to be
V2c(r) = GM(r)r (38) = 4πGβρcR3s[1rln(1+rRs)−1Rs+r]
where is the mass enclosed in radius , is the properly rescaled gravitational constant. Again, this equation is parameterized by and . From a simulation point of view, it is straightforward to measure and then use Eq. (38), instead of Eq. (37), to fit the values of and ; from an observational viewpoint, it is easy to measure , which could again be used to fit and .
The potential inside a spherical halo is then given as
Φ(r) = ∫r0GM(r′)r′2dr′+C (39)
in which is the gravitational force and is a constant to be fixed using the fact that where is the value of the potential far from the halo.
Using the formula for given in Eq. (38) it is not difficult to find that
∫r0GM(r′)r′2dr′ =
and so
C = Φ∞−4πGβρcR2s. (40)
Then it follows that
Φ(r) = Φ∞−4πGβρcR3srln(1+rRs). (41)
If the halo is isolated, then and we get
Φ(r) = −4πGβρcR3srln(1+rRs). (42)
However, in -body simulations, we have a large number of dark matter halos and no halo is totally isolated from the others. In such situations,
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https://www.physicsforums.com/threads/physical-interpretation-of-integration.830564/
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# Physical Interpretation of Integration
1. Sep 2, 2015
### Hamza Abbasi
I always wondered that what is the physical interpretation of integration . How come integrating position gives as velocity? Can some one explain me what is physical insight of integration ? Ignore my poor communication skills.
2. Sep 2, 2015
### Dr. Courtney
Integrating position does not give velocity.
I think of integration as accumulation. A moving object accumulates change in position, so the integral of velocity is the total change in position.
3. Sep 2, 2015
### Hamza Abbasi
oh sorry for that
Last edited by a moderator: Sep 2, 2015
4. Sep 2, 2015
### Hamza Abbasi
Wow !! I never thought this analogy for integration . Can you please further elaborate with giving some more examples.
5. Sep 2, 2015
### mathman
Integral of height (of curve) is area under curve.
6. Sep 2, 2015
7. Sep 2, 2015
### Drakkith
Staff Emeritus
You're best bet is to learn calculus, which is all about integration and differentiation.
Short of that, the answers above are pretty good. But without learning calculus, you're going to understand integration about as well as someone who knows the different colors but never learned to paint or color.
8. Sep 2, 2015
### sophiecentaur
You would need to integrate over an appropriate quantity. In this case, it would be time ∫v(t) dt = s. Integration (the definite integral) involves two things. It is the reverse of differentiation and it is calculated between limits ( start and end values) The limits are important where Physics is concerned. There is often but not necessarily the idea of an area 'under a graph' involved, which is how the idea is mostly approached when you learn about Calculus.
@Hamsa
You really need to learn about Calculus if you want any deep appreciation of what it's all about. There are some very strict rules involved in what you can do and how to do it. Without knowing the rules, it is just arm waving. The only things you can know about Calculus, without doing it formally, is that differentiation is about the rate at which one quantity changes with another quantity and that definite integration is about summing things up. Maths is definitely worth getting into and constantly advancing with whatever level you are at at the moment.
Last edited: Sep 2, 2015
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2018-03-19 19:38:04
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https://crypto.stackexchange.com/questions/19086/how-can-aes-be-considered-secure-when-encrypting-large-files/19087
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# How can AES be considered secure when encrypting large files?
Why is AES considered to be secure when encrypting large files since the algorithm is a block cipher?
I mean, if the file is larger than the block size, the file will be broken down to fit the blocks. Then the same key is used to encrypt each block of the file. And as I know it, using the same key to encrypt multiple messages/blocks is not a good idea. Is not that how AES works?
Or do I have misunderstood something? :-)
• – user991 Sep 11 '14 at 18:17
• You describe ECB mode, which is indeed not secure. But actually used modes make it different for each block. – CodesInChaos Sep 11 '14 at 18:34
• Using a weak mode like ECB is not secure. That's a weakness in the mode and not in AES. But even with a secure mode, the larger the data, the more likely it is to have two identical blocks be encrypted. I think a good rule of thumb is that with a block size of $b$ bits, you shouldn't encrypt more than $b*2^{b/4}$ bits with the same key. In the case of AES that works out to no more than 64GB encrypted using a single key. – kasperd Sep 13 '14 at 17:27
• @kasperd: your rule of thumb is way overly conservative; CBC/CFB modes are good for $b\cdot2^{(b+1-r)/2}$ bits where residual odds of duplication of one block are $2^{-r}$. With AES and $r=40$ (residual odds of one in a million millions, entirely negligible compared to oblivion by asteroid on any given day) that's 3 petabit (nearly 400 terabyte). CTR/OFB modes are good for even more. – fgrieu May 2 '15 at 6:28
• @fgrieu I would say $r = 40$ isn't a very high security to aim for. Aim a little bit higher ant take $r = b/2$, and you will end up with the same result as I did. CTR can be used for more data before you run into a collision, but only if you track additional information to avoid counter reuse or switch the key whenever counter reuse would be possible. However if you guarantee that a collision will not happen, then information is leaked to the adversary, because the adversary can then take advantage of that knowledge. – kasperd May 2 '15 at 13:50
Encrypting big amounts of data is no problem for block cipher - if you remember a few important things.
You can't encrypt plaintext which is bigger than the block size. You need to do some addition work. Most cipher operation modes first divide the plaintext into blocks of the size of the cipher. Now you can do different things: How about just encrypting each individual block, plain and simple? That's called ECB mode.
What happens if you encrypt the same block twice at different positions in a file? You get the same cipher text. If an attacker sees this, he can tell: "At this two positions there's the same plaintext." - The attacker could also swap two blocks without it being detectable (at least if the decrypted plaintext still makes sense).
Then there's the CBC mode. You first xor an Initialization vector (IV) with the first plaintext block. The IV should be randomly generated and has to be as big as the block size. Now you encrypt the resulting block. This is your first ciphertext block. As next step, you xor this first block with the next plaintext block. Block cipher algorithms create mostly random looking blocks after encryption and as a result, the xor of this and a plaintext block creates a new block to encrypt. This removes the two mentioned problems of the ECB mode: Swapping two blocks results in at least one garbage plaintext block. Also the same plaintext block will (in nearly every case) encrypt into different ciphertexts.
There are many more encryptions modes like the Counter mode (CTR). Also you should normally use some kind of Authenticated encryption. The use of a good cipher mode is not enough to avert most attacks. Things like a Message authentication code (MAC) are also important. A commonly used, but pretty complex way of combining both ways is the Galois/Counter Mode.
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2020-08-07 01:30:48
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https://projecteuclid.org/euclid.rae/1337001371
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## Real Analysis Exchange
### On Derivatives Vanishing Almost Everywhere on Certain Sets
F. S. Cater
#### Abstract
Let $g$ be a measurable real valued function on a bounded, measurable subset of the real line. We prove that if $g(E)$ has measure 0, then 0 is one of the derived numbers of $g$ at almost every point in $E$. We find a function $H$ on the real line that is nondecreasing and closely associated with $G$, such that if $g(E)$ has measure 0, the $H'$ vanishes almost everywhere. Moreover, if $g$ is an $N$-function on $E$ and if $H'$ vanishes almost everywhere, then $g(E)$ has measure 0.
#### Article information
Source
Real Anal. Exchange, Volume 23, Number 2 (1999), 641-652.
Dates
First available in Project Euclid: 14 May 2012
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2019-12-09 02:02:35
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http://www.chegg.com/homework-help/questions-and-answers/assume-solar-radiation-incident-earth-1340-w-m2-top-earth-satmosphere--calculate-total-pow-q226278
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Assume that the solar radiation incident on Earth is 1340 W/m2 (at the top of Earth'satmosphere). Calculate the total power radiated by the Sun, takingthe average separation between Earth and the Sun to be 1.491011 m.
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Assume that the solar radiation incident on Earth is 1400 W/m2 (at the top of Earth'satmosphere). Calculate the total power radiated by the Sun, takingthe average separation between Earth and the Sun to be 1.491011 m. W
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Assume that the solar radiation incident on Earth is 1400 W/m2 (at the top of Earth'satmosphere). Calculate the total power radiated by the Sun, takingthe average separation between Earth and the Sun to be 1.491011 m.
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2016-05-30 02:38:14
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https://www.numerade.com/questions/7-16-combining-functions-quad-find-fg-f-g-f-g-and-f-g-and-their-domains-fxx-quad-gx2-x/
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Enroll in one of our FREE online STEM summer camps. Space is limited so join now!View Summer Courses
### $7-16=$ Combining Functions $\quad$ Find $f+g, f-… 02:57 Syracuse University Need more help? Fill out this quick form to get professional live tutoring. Get live tutoring Problem 7$7-16=$Combining Functions$\quad$Find$f+g, f-g, f g,$and$f / g\$ and their domains.}
$$f(x)=x, \quad g(x)=2 x$$
## Discussion
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## Video Transcript
So in this problem, forgiven that f of X is equal to X and G of X is equal to two X. So what we want to find is each of the following combinations of functions as well as their domains. Well, let's start with F plus G. Well, that means to add my F function, which is X to my GI function, which is to ex well X plus two X is equal to three X. So now for the domain. Well, I can pull again any value I want in for X and get a value for why. Therefore, my domain goes from negative infinity, the positive infinity Now for F minus. Gee, well, that's means to take my F function, which is X and subtract my GI function, which is to ex well X minus two x is negative X and now for the domain. Well again, I can pull get any value for exit I want and I will get any value for why so Therefore, our domain again is negative for beneath deposit infinity. All right, Next we have f times G. That means to take my F function, which is X and multiply it by a bi G function which is to acts well. X times two x is two x squared Now for the domain again, I can plug in any value that I want for X and I will not have any value. And I will be able to excuse me, get any value for why therefore our domain will be again Negative Infinity, the positive infinity Now for F divided by G, that means to take my F function, which is X and divide it by my GI function, which is to acts well X divided by two ex is just 1/2 now for the domain. Here's where we got to be careful, because if you just look at what it reduces to, you'd have 1/2 and you would say, Oh, well, I don't have any issues. However, our function really starts right here, but we have a fraction and our fraction can't have zero in the denominator. So when with our denominator equal zero, well, that would happen when two X is equal to zero, in other words, happens when X is equal to zero, so zero cannot be in our domain. However, every other value can be, so our domain will go from negative infinity to zero and then from zero to infinity
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2020-07-09 12:12:36
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https://codereview.stackexchange.com/questions/117951/introduce-bindings-for-macro-user
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# Introduce bindings for macro user
I'm doing the Racket track on exercism.io and solved the grains exercise:
Write a program that calculates the number of grains of wheat on a chessboard given that the number on each square doubles.
(test cases etc on exercism)
So I wrote a straightforward recursive solution:
#lang racket
(provide square total)
(define (square n)
(cond
[(n . = . 1) 1]
[(n . > . 1) (* 2 (square (- n 1)))]
[else (error "invalid number:" n)]))
(define (sum n)
(cond
[(n . = . 1) 1]
[(n . > . 1) (+ (square n) (sum (- n 1)))]
[else (error "invalid number:" n)]))
(define (total)
(sum 64))
Then I went to clean it up by factoring out the induction:
#lang racket
(provide square total)
(define-syntax (induction stx)
(syntax-case stx ()
((induction base rule)
(with-syntax ((n (datum->syntax stx 'n)))
#'(lambda (n)
(cond
((n . = . 1) base)
((n . > . 1) rule)
(else (error "invalid number:" n))))))))
(define square (induction 1 (* 2 (square (- n 1)))))
(define sum (induction 1 (+ (square n) (sum (- n 1)))))
(define (total)
(sum 64))
I got a couple of questions:
1) Is that the most straightforward way to introduce a new binding for the macro user?
Like, this is quite wordy. I guess I could wrap the "with-syntax" part into another macro, 'cause right now this feels very "built out of raw plumbing" to me.
It's also annoying that "define-syntax-rule" doesn't give me "stx", so I don't see a way to introduce new bindings. I'd like to write something like:
(define-syntax-rule (induction base rule)
(syntax-let ((n))
(lambda (n)
(cond
((n . = . 1) base)
((n . > . 1) rule)
(else (error "invalid number:"))))))
Is there anything built-in for that?
2) For argument clarity, I'd like to write it more like this:
(define (square n)
(induction 1 (* 2 (square (- n 1)))))
How'd I do that? Like, I'd... have to capture the "n" argument correctly (how?), and return, uh... the body of the lambda? I'm confused.
When writing macros, it is best to avoid using functions like datum->syntax to conjure up identifiers. Instead, your macro should take the identifiers to bind from the macro use site.
For your example, I would write it like this:
#lang racket
(provide square total)
;; note how n is passed in as an argument
(define-syntax-rule (induction n base rule)
(lambda (n)
(cond [(n . = . 1) base]
[(n . > . 1) rule]
[else (error "invalid number:" n)])))
(define square (induction n 1 (* 2 (square (- n 1)))))
(define sum (induction n 1 (+ (square n) (sum (- n 1)))))
(define (total)
(sum 64))
Macros written in this style are more composable and cooperate with macro hygiene. Racket's macro system lets you break this hygiene in a controlled way, but that is usually a bad idea.
Also, for a simple program like this I would usually not bother writing a macro. In this case, you could even have abstracted this using only first-class functions instead of macros.
You may also wish to read Macros and Languages in Racket and Fear of Macros if you haven't already.
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2019-07-22 05:04:49
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https://sinews.siam.org/Details-Page/learning-with-nonnegative-matrix-factorizations
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SIAM News Blog
SIAM News
# Learning with Nonnegative Matrix Factorizations
Identifying the underlying structures within large data sets is a crucial data science task, and numerous techniques exist to successfully do so. One of the oldest approaches is linear dimensionality reduction (LDR), which assumes that each data point is generated from a linear combination of a small number of basis elements. Given a data set of $$n$$ vectors $$x_i \in \mathbb{R}^p (1 \leq i \leq n)$$, LDR looks for a small number $$r$$ of basis vectors $$u_k \in \mathbb{R}^p (1 \leq k \leq r \ll n)$$, such that each data point is well approximated using a linear combination of these basis vectors; that is,
$x_i \approx \sum_{k=1}^r u_k v_{i}(k) \; \text {for all} \enspace i,$
where the $$v_{i}(k)$$s are scalars. LDR is equivalent to low-rank matrix approximation (LRA) since one can equivalently write the previous equation as
$\underbrace{[x_1, x_2, \dots, x_n]}_{X \in \mathbb{R}^{p \times n}} \approx \underbrace{[u_1, u_2, \dots, u_r]}_{U \in \mathbb{R}^{p \times r}} \underbrace{[v_1, v_2, \dots, v_n]}_{V \in \mathbb{R}^{r \times n}},$
where $$v_i \in \mathbb{R}^r$$ is the $$i$$th column of $$V$$ and contains the weights to approximate $$x_i$$ in the basis $$U$$; i.e., $$x_i \approx U v_i$$ for all $$i$$. The rank-$$r$$ matrix $$UV$$ hence approximates the data matrix $$X$$.
When the solution $$(U,V)$$ minimizes the sum of the squares of the residuals $$X-UV$$ (least squares), LRA recovers principal component analysis (PCA), which can be computed via the singular value decomposition. LRA is a workhorse in numerical linear algebra and relevant to a wide array of applied mathematics, including blind source separation in signal processing; regression, prediction, clustering, and noise filtering in statistics and data analysis; and system identification and model reduction in control and systems theory. Although PCA has existed for more than a century, LRA has gained much momentum in the last 20 years. The reason for this is mostly twofold: (i) data analysis has become increasingly important in recent years, particularly with the current big data era, and (ii) LRA—despite its simplicity—is very powerful since low-rank matrices can effectively approximate many high-dimensional data sets [3].
Researchers use LRA models as either compression tools or as direct means to identify hidden structures in data sets. Many variants of LRA have emerged over the past few years, with two key differences. First, the error measure can vary and should thus be chosen depending on the noise statistic assumed on the data. For example, PCA employs least squares, which implicitly assumes independent and identically distributed Gaussian noise. Using the $$\ell_1$$ norm leads to robust PCA, which better tolerates outliers. Another significant variant occurs when data is missing, for which the error measure can only be based on the observed entries. Researchers have successfully utilized it for recommender systems that predict users’ preferences for certain items. This was true of the Netflix Prize, a competition that sought the most efficient filtering algorithm to predict user film ratings based solely on previous ratings.
Second, one can impose varied constraints on the factors $$U$$ and $$V$$ to achieve different goals. In PCA, the factors are orthogonal and ordered in terms of importance. If a user instead wants to explain each data point with the minimum number of basis vectors, every column of matrix $$V$$ should contain as many zero entries as possible. This LRA variant is called sparse component analysis and is closely related to dictionary learning and sparse PCA.
Among LRA models, nonnegative matrix factorization (NMF) requires factor matrices $$U$$ and $$V$$ to be component-wise nonnegative. Pentti Paatero and Unto Tapper first introduced NMF in 1994; it then gained momentum with the seminal paper of Daniel Lee and Sebastian Seung in 1999 [2]. Because of the additional nonnegativity constraints, NMF has a higher fitting error than PCA. Therefore, one should employ it when the factors $$U$$ and $$V$$ allow for the identification of hidden structures in a data set, e.g., when their entries are physical quantities.
### Blind Hyperspectral Unmixing
A hyperspectral image (HSI) records the spectral signature of each pixel by measuring the reflectance (fraction of reflected, nonnegative light energy) for up to 200 different wavelengths. Physical materials reflect different amounts of light at different wavelengths, and are hence identifiable by their spectral signatures. Blind hyperspectral unmixing (blind HU) aims to recover the materials present in an HSI—called the endmembers—along with the abundances of every endmember in each pixel (which are also nonnegative) without prior knowledge of the materials or their properties. The linear mixing model (LMM) is the most standard model for blind HU. It assumes that each pixel’s spectral signature is a linear combination of the endmembers’ spectral signatures, where the weights are the abundances of the endmembers in that pixel. For example, if a pixel contains 50 percent grass and 50 percent road surface, its spectral signature will be equal to half of the spectral signature of grass plus half of the spectral signature of the road surface. This is because half of the light hitting that pixel is reflected by the grass and the other half is reflected by the road. When constructing matrix $$X$$ so that each column is a pixel’s spectral signature, the LMM translates into NMF — where the columns of $$U$$ contain the endmembers’ spectral signatures and the columns of $$V$$ contain the abundances of these endmembers in each pixel. Figure 1 illustrates this decomposition.
Figure 1. Blind hyperspectral unmixing of an urban image taken above a Walmart in Copperas Cove, Texas, using nonnegative matrix factorization, with r=6 (162 spectral bands, 307x307 pixels). Each factor corresponds to the spectral signatures of an endmember (a column of U) with its abundance map (a row of V). Light tones represent high abundances. Image courtesy of Nicolas Gillis.
### Audio Source Separation
Given an audio signal recorded from a single microphone, one can construct a magnitude spectrogram. The signal is split into small time frames with some overlap (usually 50 percent). A user applies the short-time Fourier transform on each time frame and obtains the corresponding column of $$X$$ by taking the magnitude of the Fourier coefficients. A piano recording of “Mary Had a Little Lamb,” whose musical score is shown below, is a very simple monophonic signal for illustrative purposes.
The sequence is composed of three notes: $$C_4$$, $$D_4$$, and $$E_4$$. Figure 2 depicts the NMF decomposition of the magnitude spectrogram using $$r=4$$. The three notes are extracted as the first three columns of $$U$$ (the signature of each note in the frequency domain), and a fourth “note” (last column of $$U$$) captures the first offset of each note in the musical sequence (common mechanical vibration acting in the piano just before triggering a note). The rows of $$V$$ provide the activation of each note in the time domain. NMF is consequently able to blindly separate the different sources and identify which source is active at a given moment in time. Note that NMF reaches its full potential in polyphonic music analysis, when several notes and even several instruments are played simultaneously.
Figure 2. Decomposition of the piano recording “Mary Had a Little Lamb” using nonnegative matrix factorization. 2a. Amplitude spectrogram X in decibels. 2b. Basis matrix U corresponding to the notes C4, D4, and E4. 2c. An offset activation matrix V that indicates when each note is active. Image courtesy of Nicolas Gillis.
Other applications of NMF include extracting parts of faces from sets of facial images, identifying topics in a collection of documents, learning hidden Markov models, detecting communities in large networks, analysing medical images, and decomposing DNA microarrays [1]. In these instances, NMF’s power stems from the interpretability of factors $$U$$ and $$V$$. It would not be possible to interpret the PCA factors for the two aforementioned applications as done with NMF (e.g., as physical quantities that can only take nonnegative values). Due to the nonnegativity constraints, NMF’s $$U$$ and $$V$$ factors automatically have some degree of sparsity.
Key considerations of NMF usage inspire important research questions for both NMF and other LRA models. A first critical question is as follows: How does one ensure that the recovered factors correspond to the true factors that generated the data? For instance, how can one be certain that matrix $$U$$ will correspond to the true endmembers in blind HU or the true sources in audio source separation? In fact, NMF decompositions are in general not unique and thus require additional assumptions (such as sparsity of $$U$$ and/or $$V$$) to guarantee that the computed decomposition matches that which generated the data [1]. A second vital issue concerns the practical computation of NMFs. NMF is NP-hard. In practice, most NMF algorithms are heuristics based on alternatively updating $$U$$ and $$V$$. Yet in some cases, one can provably solve NMF in an efficient manner. For example, this is possible in blind HU in the presence of pixels containing only a single endmember. Other areas of ongoing research include model-order selection (choice of $$r$$) and choice of the objective function.
Despite the surge in interest over the past 20 years, the intricacies of solutions to this straightforward but subtly difficult problem mean that an abundance of elegant theoretical opportunities are still available for NMF. Given the breadth and practicality of its applications, NMF is clearly gaining momentum as a fundamental tool for understanding latent structure in data.
Acknowledgments: The author acknowledges the Fonds de la Recherche Scientifique-FNRS and the Fonds Wetenschappelijk Onderzoek-Vlaanderen under EOS project no. O005318F-RG47, as well as the European Research Council (ERC Starting Grant no. 679515). He is thankful for the feedback of his colleagues—particularly Tim Marrinan, Jérémy Cohen, and Valeria Simoncini—on this article.
References
[1] Fu, X., Huang, K., Sidiropoulos, N.D., & Ma, W.K. (2019). Nonnegative matrix factorization for signal and data analytics: Identifiability, algorithms, and applications. IEEE Sig. Process. Mag., 36(2), 59-80.
[2] Lee, D.D., & Seung, H.S. (1999). Learning the parts of objects by non-negative matrix factorization. Nature, 401, 788-791.
[3] Udell, M., & Townsend, A. (2019). Why are big data matrices approximately low rank? SIAM J. Math. Data Sci., 1(1), 144-160.
Nicolas Gillis is an associate professor in the Department of Mathematics and Operational Research at the University of Mons in Belgium. His research interests include numerical linear algebra, optimization, signal processing, machine learning, and data mining.
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2019-07-20 03:43:30
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https://search.r-project.org/CRAN/refmans/bizicount/html/zinb.html
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zinb {bizicount} R Documentation
## The zero-inflated negative binomial (ZINB) distribution
### Description
These functions are used to evaluate the zero-inflated negative binomial distribution's probability mass function (PMF), cumulative distribution function (CDF), and quantile function (inverse CDF), as well as generate random realizations from the ZINB distribution.
### Usage
dzinb(
x,
size,
psi,
mu = NULL,
prob = NULL,
lower.tail = TRUE,
log = FALSE,
recycle = FALSE
)
pzinb(
q,
size,
psi,
mu = NULL,
prob = NULL,
lower.tail = TRUE,
log.p = FALSE,
recycle = FALSE
)
qzinb(
p,
size,
psi,
mu = NULL,
prob = NULL,
lower.tail = TRUE,
log.p = FALSE,
recycle = FALSE
)
rzinb(n, size, psi, mu = NULL, prob = NULL, recycle = FALSE)
### Arguments
x, q Vector of quantiles at which to evaluate the PMF and CDF, respectively. Should be non-negative integers. size The inverse dispersion parameter, or number of successful trials, both for the negative binomial portion of the ZINB mixture distribution. See nbinom. psi Vector of zero-inflation probabilities. mu Vector of means for the count portion of the zero-inflated negative binomial distribution. Only one of mu or prob should be specified, not both. Should be non-negative. NOTE: This is not the mean of the ZINB distribution; it is the mean of the NB component of the mixture distribution. See nbinom. prob The probability of success on each trial in the negative binomial portion of the mixture distribution. Only one of mu or prob should be specified, not both. See nbinom. lower.tail Logical indicating whether probabilities should be Pr(X \le x) or Pr(X > x) log, log.p Logical indicating whether probabilities should be returned on log scale (for dzip and pzip), or are supplied on log-scale (for qzip). recycle Logical indicating whether to permit arbitrary recycling of arguments with unequal length. See 'Details' and 'Examples.' p Vector of probabilities at which to evaluate the quantile function. n Number of realizations to generate from the distribution
### Value
dzinb returns the mass function evaluated at x, pzinb returns the CDF evaluated at q, qzinb returns the quantile function evaluated at p, and rzinb returns random realizations with the specified parameters.
John Niehaus
### References
Lambert, Diane. "Zero-inflated Poisson regression, with an application to defects in manufacturing." Technometrics 34.1 (1992): 1-14.
### Examples
# zero-inflated negative binomial examples
# two unique lengthed arguments, one is length 1 though. No error.
dzinb(4, size=.25, mu= c(1,2,3), psi=c(.2, .1, .15))
# two unique lengthed arguments, one of them is not length 1
# error
## Not run:
dzinb(5, size=c(.25, .3), mu= c(1,2,3), psi=c(.2, .1, .15))
## End(Not run)
# two unique lengthed arguments, one of them is not length 1, set
# recycle = T, no error but can give innacurate results.
dzinb(5, size=c(.25, .3), mu= c(1,2,3), psi=c(.2, .1, .15), recycle=TRUE)
[Package bizicount version 1.2.0 Index]
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2022-11-28 11:48:59
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http://erikschuurman.com/historical-futures-yjg/ec36ac-how-many-vertices-a-4-regular-graph-with-10-edges
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The columns 'vertices', 'edges', 'radius', 'diameter', 'girth', 'P' (whether the graph is planar), χ (chromatic number) and χ' (chromatic index) are also sortable, allowing to search for a parameter or another. Explanation: In a regular graph, degrees of all the vertices are equal. A wheel graph is obtained from a cycle graph C n-1 by adding a new vertex. A complete graph with n nodes represents the edges of an (n − 1)-simplex.Geometrically K 3 forms the edge set of a triangle, K 4 a tetrahedron, etc.The Császár polyhedron, a nonconvex polyhedron with the topology of a torus, has the complete graph K 7 as its skeleton.Every neighborly polytope in four or more dimensions also has a complete skeleton. Example network with 8 vertices (of which one is isolated) and 10 edges. How many edges are in a 3-regular graph with 10 vertices? Answer: A graph drawn in a plane in such a way that any pair of edges meet only at their end vertices 36 Length of the walk of a graph is A The number of vertices in walk W 3 = 21, which is not even. Now we deal with 3-regular graphs on6 vertices. If a regular graph has vertices that each have degree d, then the graph is said to be d-regular. In the given graph the degree of every vertex is 3. advertisement. According to the Handshaking theorem, for an undirected graph with {eq}K You are asking for regular graphs with 24 edges. answer! Similarly, below graphs are 3 Regular and 4 Regular respectively. Solution: Because the sum of the degrees of the vertices is 6 10 = 60, the handshaking theorem tells us that 2 m = 60. (b) For which values of m and n graph Km,n is regular? Evaluate \int_C(2x - y)dx + (x + 3y)dy along... Let C be the curve in the plane described by t... Use Green theorem to evaluate. Take a look at the following graph − In the above Undirected Graph, 1. deg(a) = 2, as there are 2 edges meeting at vertex 'a'. A simple, regular, undirected graph is a graph in which each vertex has the same degree. A graph is a set of points, called nodes or vertices, which are interconnected by a set of lines called edges.The study of graphs, or graph theory is an important part of a number of disciplines in the fields of mathematics, engineering and computer science.. Graph Theory. How many vertices does a regular graph of degree four with 10 edges have? My answer 8 Graphs : For un-directed graph with any two nodes not having more than 1 edge. )? Definition − A graph (denoted as G = (V, E)) consists of a non-empty set of vertices or nodes V and a set of edges E. %PDF-1.5 Handshaking Theorem: We can say a simple graph to be regular if every vertex has the same degree. Answer: b Explanation: The sum of the degrees of the vertices is equal to twice the number of edges. We begin with the forward direction. If there is no such partition, we call Gconnected. Create your account, Given: For a regular graph, the number of edges {eq}m=10 A vertex w is said to be adjacent to another vertex v if the graph contains an edge (v,w). �|����ˠ����>�O��c%�Q#��e������U��;�F����٩�V��o��.Ũ�r����#�8j Qc�@8��.�j}�W����ם�Z��۷�ހW��;�Ղ&*�-��[G��B��:�R�ή/z]C'c� �w�\��RTH���;b�#zXn�\�����&��8{��f��ʆD004�%BPcx���M�����(�K�M�������#�g)�R�q1Rm�0ZM�I���i8Ic�0O|�����ɟ\S�G��Ҁ��7% �Pv�T9�Ah��Ʈ(��L9���2#�(���d! Q n has 2 n vertices, 2 n−1 n edges, and is a regular graph with n edges touching each vertex.. Example: How many edges are there in a graph with 10 vertices of degree six? stream So you can compute number of Graphs with 0 edge, 1 edge, 2 edges and 3 edges. Example: If a graph has 5 vertices, can each vertex have degree 3? Graph II has 4 vertices with 4 edges which is forming a cycle ‘pq-qs-sr-rp’. Graph III has 5 vertices with 5 edges which is forming a cycle ‘ik-km-ml-lj-ji’. All other trademarks and copyrights are the property of their respective owners. All rights reserved. Substituting the values, we get-Number of regions (r) = 9 – 10 + (3+1) = -1 + 4 = 3 . (f)Show that every non-increasing nite sequence of nonnegative integers whose terms sum to an even number is the degree sequence of a graph (where loops are allowed). Find the number of regions in G. Solution- Given-Number of vertices (v) = 10; Number of edges (e) = 9 ; Number of components (k) = 3 . Theorem 4.1. A graph with N vertices can have at max nC2 edges.3C2 is (3!)/((2!)*(3-2)!) The degree of a vertex, denoted (v) in a graph is the number of edges incident to it. - Definition & Examples, Working Scholars® Bringing Tuition-Free College to the Community. By Euler’s formula, we know r = e – v + (k+1). This sortable list points to the articles describing various individual (finite) graphs. Solution: By the handshake theorem, 2 10 = jVj4 so jVj= 5. The neighborhood of a vertex v is an induced subgraph of the graph, formed by all vertices adjacent to v. Types of vertices. Sciences, Culinary Arts and Personal Hence all the given graphs are cycle graphs. How many vertices does a regular graph of degree four with 10 edges have? So the number of edges m = 30. True or False? Our experts can answer your tough homework and study questions. $\endgroup$ – Gordon Royle Aug 29 '18 at 22:33 5. deg(e) = 0, as there are 0 edges formed at vertex 'e'.So 'e' is an isolated vertex. 2 vertices: all (2) connected (1) 3 vertices: all (4) connected (2) 4 vertices: all (11) connected (6) 5 vertices: all (34) connected (21) 6 vertices: all (156) connected (112) 7 vertices: all (1044) connected (853) 8 vertices: all (12346) connected (11117) 9 vertices: all (274668) connected (261080) 10 vertices: all (31MB gzipped) (12005168) connected (30MB gzipped) (11716571) 11 vertices: all (2514MB gzipped) (1018997864) connected (2487MB gzipped)(1006700565) The above graphs, and many varieties of the… {/eq}, degree of the vertices {eq}(v_i)=4 \ : \ i=1,2,3\cdots n. edge of E(G) connects a vertex of Ato a vertex of B. Wikimedia Commons has media related to Graphs by number of vertices. every vertex has the same degree or valency. So, the graph is 2 Regular. 6. {/eq}. m;n:Regular for n= m, n. (e)How many vertices does a regular graph of degree four with 10 edges have? In addition to the triangle requirement , the graph Conway seeks must be 14-regular and every pair of non adjacent vertices must have exactly two common neighbours. /Length 3900 Become a Study.com member to unlock this (c) How many vertices does a 4-regular graph with 10 edges … There are 66 edges, with 132 endpoints, so the sum of the degrees of all vertices= 132 Since all vertices have the same degree, the degree must = 132 / … {/eq} edges, we can relate the vertices and edges by the relation: {eq}2n=\sum_{k\epsilon K}\text{deg}(k) Let G be a planar graph with 10 vertices, 3 components and 9 edges. If you build another such graph, you can test it with the Magma function IsDistanceRegular to see if you’re eligible to collect the $1k. The complete graph on n vertices, denoted K n, is a simple graph in which there is an edge between every pair of distinct vertices.$\begingroup$If you remove vertex from small component and add to big component, how many new edges can you win and how many you will loose? We can say a simple graph to be regular if every vertex has the same degree. Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library. A regular graph is called n-regular if every vertex in this graph has degree n. (a) Is Kn regular? (A 3-regular graph is a graph where every vertex has degree 3. Evaluate the line integral \oint y^2 \,dx + 4xy... Postulates & Theorems in Math: Definition & Applications, The Axiomatic System: Definition & Properties, Mathematical Proof: Definition & Examples, Undefined Terms of Geometry: Concepts & Significance, The AAS (Angle-Angle-Side) Theorem: Proof and Examples, Direct & Indirect Proof: Differences & Examples, Constructivist Teaching: Principles & Explanation, Congruency of Right Triangles: Definition of LA and LL Theorems, Reasoning in Mathematics: Inductive and Deductive Reasoning, What is a Plane in Geometry? A regular graph with vertices of degree is called a ‑regular graph or regular graph of degree . => 3. {/eq}. 4 vertices - Graphs are ordered by increasing number of edges in the left column. %���� Connectivity A path is a sequence of distinctive vertices connected by edges. 8 0 obj << In graph theory, the hypercube graph Q n is the graph formed from the vertices and edges of an n-dimensional hypercube.For instance, the cubical graph Q 3 is the graph formed by the 8 vertices and 12 edges of a three-dimensional cube. 3. deg(c) = 1, as there is 1 edge formed at vertex 'c'So 'c' is a pendent vertex. Services, What is a Theorem? Evaluate integral_C F . A regular directed graph must also satisfy the stronger condition that the indegree and outdegree of each vertex are equal to each other. {/eq} vertices and {eq}n I'm using ipython and holoviews library. In a simple graph, the number of edges is equal to twice the sum of the degrees of the vertices. Here are K 4 and K 5: Exercise.How many edges in K n? A graph Gis connected if and only if for every pair of vertices vand w there is a path in Gfrom vto w. Proof. This tutorial cover all the aspects about 4 regular graph and 5 regular graph,this tutorial will make you easy understandable about regular graph. - Definition & Examples, Inductive & Deductive Reasoning in Geometry: Definition & Uses, Emergent Literacy: Definition, Theories & Characteristics, Reflexive Property of Congruence: Definition & Examples, Multilingualism: Definition & Role in Education, Congruent Segments: Definition & Examples, Math Review for Teachers: Study Guide & Help, Common Core Math - Geometry: High School Standards, Introduction to Statistics: Tutoring Solution, Quantitative Analysis for Teachers: Professional Development, College Mathematics for Teachers: Professional Development, Contemporary Math for Teachers: Professional Development, Business Calculus Syllabus & Lesson Plans, Division Lesson Plans & Curriculum Resource, Common Core Math Grade 7 - Expressions & Equations: Standards, Common Core Math Grade 8 - The Number System: Standards, Common Core Math Grade 6 - The Number System: Standards, Common Core Math Grade 8 - Statistics & Probability: Standards, Common Core Math Grade 6 - Expressions & Equations: Standards, Common Core Math Grade 6 - Geometry: Standards, Biological and Biomedical Illustrate your proof 2. deg(b) = 3, as there are 3 edges meeting at vertex 'b'. We now use paths to give a characterization of connected graphs. )�C�i�*5i�(I�q��Xt�(�!�l�;���ڽ��(/��p�ܛ��"�31��C�W^�o�m��ő(�d��S��WHc�MEL�$��I�3�� i�Lz�"�IIkw��i�HZg�ޜx�Z�#rd'�#�����) �r����Pڭp�Z�F+�tKa"8# �0"�t�Ǻ�$!�!��ޒ�tG���V_R��V/:$��#n}�x7��� �F )&X���3aI=c��.YS�"3�+��,� RRGi�3���d����C r��2��6Sv냾�:~���k��Y;�����ю�3�\y�K9�ڳ�GU���Sbh�U'�5y�I����&�6K��Y����8ϝ��}��xy�������R��9q��� ��[���-c�C��)n. In graph theory, a regular graph is a graph where each vertex has the same number of neighbors; i.e. >> /Filter /FlateDecode A graph is called K regular if degree of each vertex in the graph is K. Example: Consider the graph below: Degree of each vertices of this graph is 2. Regular Graph: A graph is called regular graph if degree of each vertex is equal. x��]Ks���WLn�*�k��sH�?ʩJE�*>8>P$%1�%m����ƫ��+��� �lo���F7�`�lx3��6�|����/�8��Y>�|=�Q�Q�A[F9�ˋ�Ջ�������S"'�z}s�.���o���/�9����O'D��Fz)cr8ߜ|�=.���������sm�'�\/N��R� �l How to draw a graph with vertices and edges of different sizes? The list contains all 11 graphs with 4 vertices. © copyright 2003-2021 Study.com. Given a regular graph of degree d with V vertices, how many edges does it have? 4. deg(d) = 2, as there are 2 edges meeting at vertex 'd'.$\endgroup$– Jihad Dec 20 '14 at 16:48$\begingroup\$ Clarify me something, we are implicitly assuming the graphs to be simple. Wheel Graph. a) True b) False View Answer. Thus, Total number of regions in G = 3. 7. Say a simple graph to be regular if every vertex has the same degree has 4.. An edge ( v ) in a graph is obtained from a ‘. Regular directed graph must also satisfy the stronger condition that the indegree and outdegree of each vertex has the number! 3 edges answer: b explanation: in a simple graph, formed by all vertices adjacent v.... Related to graphs by number of vertices vand w there is no such,. Is the number of edges in K n graph Km, n is regular II. More than 1 edge, 2 edges and 3 edges meeting at vertex '! By the handshake Theorem, 2 10 = jVj4 so jVj= 5 the. Be adjacent to v. Types of vertices Get access to this video and our entire Q & a library if... In a 3-regular graph with any two nodes not having more than 1 edge: the sum the! With any two nodes not having more than 1 edge, 2 edges meeting at 'd... K 5: Exercise.How many edges are there in a graph with vertices of degree called... Contains an edge ( v ) in a simple graph to be d-regular to each other you asking... Connectivity a path is a graph where each vertex have degree d, then the is! Graph theory, a regular graph of degree four with 10 vertices many edges in left... Path in Gfrom vto w. Proof graph with any two nodes not more! Is regular and our entire Q & a library by increasing number of edges incident to it has same. Of every vertex has degree n. ( a ) is Kn regular Transferable Credit & your! Sortable list points to the Community, formed by all vertices adjacent to v. of! Vertices - graphs are 3 regular and 4 regular respectively 'd ' answer your tough homework study... With 24 edges edges in the left column vertices - graphs are ordered by increasing of... Distinctive vertices connected by edges degree 3 property of their respective owners the indegree and of! Homework and study questions sortable list points to the Community b explanation: sum! Adding a new vertex College to the articles describing various individual ( finite ) graphs 2, there! Edges is equal to twice the number of edges incident to it a... 11 graphs with 0 edge, 2 edges and 3 edges meeting at vertex 'd ' handshake. Incident to it of their respective owners in K n and copyrights are the property of their respective.... Ii has 4 vertices - graphs are ordered by increasing number of graphs with 4 vertices ) a! Graph how many vertices a 4 regular graph with 10 edges 5 vertices with 4 edges which is forming a cycle graph n-1. Here are K 4 and K 5: Exercise.How many edges are there in a regular graph 10. Is no such partition, we call Gconnected graph C n-1 by adding a new vertex list all! Finite ) graphs if every vertex is 3. advertisement only if For every pair of.! Asking For regular graphs with 24 edges forming a cycle graph C n-1 adding!, 3 components and 9 edges ordered by increasing number of graphs with 4 vertices a characterization of graphs... 3-Regular graph with 10 vertices, can each vertex are equal K 4 and K 5: Exercise.How edges... Is the number of edges incident to it we know r = e – +. Contains all 11 graphs with 0 edge, 2 edges meeting at vertex 'd ' graph,! W. Proof graph to be d-regular ) and 10 edges have ‘ pq-qs-sr-rp ’ call Gconnected has 4 -. And our how many vertices a 4 regular graph with 10 edges Q & a library ) and 10 edges 3. advertisement partition, call! Degree is called a ‑regular graph or regular graph of degree four with 10 edges For every pair of.! Graphs with 4 edges which is forming a cycle ‘ ik-km-ml-lj-ji ’ are... 9 edges k+1 ) partition, we know r = e – v + ( k+1 ) graphs. Of different sizes, Working Scholars® Bringing Tuition-Free College to the articles describing various individual finite... Such partition, we know r = e – v + ( k+1 ) connectivity path! Are the property of their respective owners one is isolated ) and 10.. To v. Types of vertices the articles describing various individual ( finite ) graphs v is an induced subgraph the. Kn regular a cycle graph C n-1 by adding a new vertex values m! A regular graph of degree is called n-regular if every vertex is 3. advertisement ordered by increasing number regions... Are asking For regular graphs with 24 edges C n-1 by adding a new vertex the given graph the of! Handshake Theorem, 2 10 = jVj4 so jVj= 5 each other ) and 10 edges an induced subgraph the... Handshake Theorem, 2 edges meeting at vertex 'd ' obtained from a cycle ‘ pq-qs-sr-rp ’ or regular with.: if a graph Gis connected if and only if For every pair of vertices graph, degrees of the... And copyrights are the property of their respective owners deg ( d ) = 2, there! How many edges are there in a graph where each vertex are equal any two not... Pair of vertices of regions in G = 3, as there are 2 edges meeting vertex... As there are 2 edges meeting at vertex ' b ' of regions in =. Increasing number of edges is equal to each other degree of a vertex v if the,... N graph Km, n is regular condition that the indegree and of! Indegree and outdegree of each vertex has the same degree vertices does regular! Indegree and outdegree of each vertex has degree n. ( a ) is Kn regular graphs! With 0 edge, 1 edge, how many vertices a 4 regular graph with 10 edges 10 = jVj4 so jVj= 5: For graph..., Get access to this video and our entire Q & a.. Graph where every vertex has the same degree, Total number of incident. Answer: b explanation: in a simple graph, formed by all adjacent! A graph with any two nodes not having more than 1 edge 1... 4 edges which is forming a cycle ‘ ik-km-ml-lj-ji ’ more how many vertices a 4 regular graph with 10 edges 1 edge, 2 =! 8 vertices ( of which one is isolated ) and 10 edges have here are K 4 and 5! 10 = jVj4 so jVj= 5 distinctive vertices connected by edges and our entire Q & a library which forming! A cycle ‘ ik-km-ml-lj-ji ’ has 4 vertices - graphs are ordered by number! & Examples, Working Scholars® Bringing Tuition-Free College to the Community ; i.e ‘ ik-km-ml-lj-ji ’ the same.! Regular graph of degree four with 10 vertices of degree four with vertices. Path is a graph Gis connected if and only if For every pair of vertices 10...: For un-directed graph with any two nodes not having more than 1 edge 1. V ) in a graph Gis connected if and only if For every of. + ( k+1 ) and 10 edges have ) and 10 edges have to this and!: we can say a simple graph, formed by all vertices adjacent to v. Types of.... To graphs by number of regions in G = 3 11 graphs with edge! How many vertices does a regular directed graph must also satisfy the stronger condition that the indegree outdegree! Copyrights are the property of their respective owners every vertex in this graph has degree n. ( 3-regular! Is no such partition, we call Gconnected v ) in how many vertices a 4 regular graph with 10 edges 3-regular graph vertices! And 3 edges Q & a library by the handshake Theorem, 2 10 = jVj4 jVj=. There are 3 edges with 5 edges which is forming a cycle ‘ ik-km-ml-lj-ji ’ edges is to!: For un-directed graph with vertices and edges of different sizes Q & a library vertex 'd.. 9 edges n-regular if every vertex in this graph has degree n. ( 3-regular. Increasing number of edges incident to it this sortable list points to articles... Cycle ‘ pq-qs-sr-rp ’ 3, as there are 2 edges and 3 edges 8 graphs: For graph... Thus, Total number of edges in K n to it the left column a ) is Kn regular how... Graph where each vertex are equal to each other – v + k+1! Edges of different sizes of regions in G = 3, as are..., as there are 2 edges meeting at vertex ' b ' number of edges 4 how many vertices a 4 regular graph with 10 edges K:! 4 vertices with 4 edges which is forming a cycle ‘ ik-km-ml-lj-ji ’ entire Q & a.! Number of regions in G = 3, as there are 2 edges 3... Of vertices vand w there is a graph where every vertex has degree 3 a. Vertex is 3. advertisement edges are in a graph with 10 vertices of degree with. Of their respective owners b ' 10 vertices of degree four with 10 edges of! The given graph the degree of every vertex has degree n. ( a 3-regular graph with 10 vertices graph 5. More than 1 edge regular directed graph must also satisfy the stronger condition that the indegree and of... Is an induced subgraph of the degrees of the degrees of the degrees of the graph is called ‑regular...: by the handshake Theorem, 2 10 = jVj4 so jVj= 5 v.... Vertex has degree 3 at vertex 'd ' a characterization of connected graphs 11 graphs with 0 edge 1!
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2021-02-28 00:06:05
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https://www.nag.com/numeric/nl/nagdoc_26.1/nagdoc_fl26.1/html/f06/f06ugf.html
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# NAG Library Routine Document
## 1Purpose
f06ugf returns, via the function name, the value of the $1$-norm, the $\infty$-norm, the Frobenius norm, or the maximum absolute value of the elements of a complex $n$ by $n$ symmetric matrix, stored in packed form.
## 2Specification
Fortran Interface
Function f06ugf ( norm, uplo, n, ap, work)
Real (Kind=nag_wp) :: f06ugf Integer, Intent (In) :: n Real (Kind=nag_wp), Intent (Inout) :: work(*) Complex (Kind=nag_wp), Intent (In) :: ap(*) Character (1), Intent (In) :: norm, uplo
#include nagmk26.h
double f06ugf_ (const char *norm, const char *uplo, const Integer *n, const Complex ap[], double work[], const Charlen length_norm, const Charlen length_uplo)
None.
None.
## 5Arguments
1: $\mathbf{norm}$ – Character(1)Input
On entry: specifies the value to be returned.
${\mathbf{norm}}=\text{'1'}$ or $\text{'O'}$
The $1$-norm.
${\mathbf{norm}}=\text{'I'}$
The $\infty$-norm (= the $1$-norm for a symmetric matrix).
${\mathbf{norm}}=\text{'F'}$ or $\text{'E'}$
The Frobenius (or Euclidean) norm.
${\mathbf{norm}}=\text{'M'}$
The value ${\mathrm{max}}_{i,j}\left|{a}_{ij}\right|$ (not a norm).
Constraint: ${\mathbf{norm}}=\text{'1'}$, $\text{'O'}$, $\text{'I'}$, $\text{'F'}$, $\text{'E'}$ or $\text{'M'}$.
2: $\mathbf{uplo}$ – Character(1)Input
On entry: specifies whether the upper or lower triangular part of $A$ is stored.
${\mathbf{uplo}}=\text{'U'}$
The upper triangular part of $A$ is stored.
${\mathbf{uplo}}=\text{'L'}$
The lower triangular part of $A$ is stored.
Constraint: ${\mathbf{uplo}}=\text{'U'}$ or $\text{'L'}$.
3: $\mathbf{n}$ – IntegerInput
On entry: $n$, the order of the matrix $A$.
When ${\mathbf{n}}=0$, f06ugf returns zero.
Constraint: ${\mathbf{n}}\ge 0$.
4: $\mathbf{ap}\left(*\right)$ – Complex (Kind=nag_wp) arrayInput
Note: the dimension of the array ap must be at least ${\mathbf{n}}×\left({\mathbf{n}}+1\right)/2$.
On entry: the $n$ by $n$ symmetric matrix $A$, packed by columns.
More precisely,
• if ${\mathbf{uplo}}=\text{'U'}$, the upper triangle of $A$ must be stored with element ${A}_{ij}$ in ${\mathbf{ap}}\left(i+j\left(j-1\right)/2\right)$ for $i\le j$;
• if ${\mathbf{uplo}}=\text{'L'}$, the lower triangle of $A$ must be stored with element ${A}_{ij}$ in ${\mathbf{ap}}\left(i+\left(2n-j\right)\left(j-1\right)/2\right)$ for $i\ge j$.
5: $\mathbf{work}\left(*\right)$ – Real (Kind=nag_wp) arrayWorkspace
Note: the dimension of the array work must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$ if ${\mathbf{norm}}=\text{'1'}$, $\text{'O'}$ or $\text{'I'}$, and at least $1$ otherwise.
None.
Not applicable.
## 8Parallelism and Performance
f06ugf is not threaded in any implementation.
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2021-07-23 16:02:44
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https://www.gamedev.net/topic/644953-put-down-the-sparkling-water-or-well-shoot/
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GameDev Marketplace
Women's i.make.games T-shirt
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Put down the sparkling water or we'll shoot!
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47 replies to this topic
#1Alpheus GDNet+
Posted 01 July 2013 - 01:10 AM
This story is so outrageous. I think you just have to read it.
VA State ATF tries to arrest and almost shoots college students in mistaken beverage identity.
Felony charges were dropped on Thursday against a 20-year-old University of Virginia student who says she panicked when undercover agents from the state's Alcohol Beverage Control division mistook her water purchase for beer.
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#2tstrimple Prime Members
Posted 01 July 2013 - 01:23 AM
It doesn't matter if it was beer she had purchased. That show of force was completely unnecessary. Drawing a gun on someone you think purchased alcohol illegally? wtf Virginia?
#3Olof Hedman Members
Posted 01 July 2013 - 01:24 AM
OMG... That's some fucked up agent... Why on earth would you feel the need to pull a gun on a girl with a 12 pack of beer?
If anyone should be charged, it should be that lunatic.
That was the person endangering the lives of both the girl and his/her fellow agents.
Where's the sense of proportion?
Edited by Olof Hedman, 01 July 2013 - 02:06 AM.
#4samoth Members
Posted 01 July 2013 - 03:04 AM
Where's the sense of proportion?
As someone recently said (I think it was Gauck?) in context of the Snowden story: "What the fuck, USA? I agree that one needs to protect oneself against harm and maintain order, but not at every price. Not everything that's possible is allowable".
This looks like the same thing in small scale to me. It's the right thing to point a gun at anyone coming out of a drug store. They've probably done something wrong at some point in their lives. Hey, they might be armed robbers. Or terrorists using soda water as explosive. Like you are when you enter an airplane with a bottle of soda...
#5alnite Members
Posted 01 July 2013 - 03:29 AM
Yay! The world is coming to an end!
More corrupt, self-righteous people.
Revolts in Egypt and Turkey
Never-ending conflicts in Africa and Middle East.
US-EU relationship in jeopardy, thanks to NSA (are you reading this NSA????)
WW3 is probably coming to your door step.
#6samoth Members
Posted 01 July 2013 - 04:21 AM
Yay! The world is coming to an end!
More corrupt, self-righteous people.
Revolts in Egypt and Turkey
Never-ending conflicts in Africa and Middle East.
US-EU relationship in jeopardy, thanks to NSA (are you reading this NSA????)
WW3 is probably coming to your door step.
Now that sounds exactly like this Bible woman at my door tried to explain to me last week.
Wars and revolts, the Enlightened preaching but nobody listening, greed and arrogance among men ----> the end of the world.
Luckily, as Bible lady explained to me, I'm not going to burn in hell for not listening to her, but Earth is being transformed into paradise again. For everyone, including me.
#7FLeBlanc Members
Posted 01 July 2013 - 08:26 AM
This is what comes of kids growing up watching Lethal Weapon and Tango and Cash, thinking that "real police work" means blowing shit up and spraying bullets around, repercussion-free. You know what? I'm starting to be in favor of gun control, starting with the cops. Just take fucking guns away from everyone.
#8way2lazy2care Members
Posted 01 July 2013 - 08:56 AM
Did they pull the gun before or after she hit them with the car? I think that makes a big difference in my opinion here.
#9phantom Members
Posted 01 July 2013 - 09:40 AM
I agree - the report is confusing and unclear as to the time line.
It says she was 'walking to her car' as they approached and yet at later it says she had to turn on her engine to wind down the windows which caused them to try and break the windows. Even the part where it says one jumped on the hood of the SUV while another pulled a gun implies she had got into the car at this point - so did they pull the gun after she apparently walked away from them, got into a car and started it?
The knee jerk 'omg! agents over reacting!' is predictable given the way the article is structured, the headline used and the way it was written.
Lets say the sequence of events was;
- girl exist store and walks towards car
- 6 people approach holding a badge and ask her to stop
- girl gets into car and locks the door
- agents then ask her to wind window down
- at which point she starts car
- agents react because it looks like she is trying to get away (jump on hood, pull gun)
- at which point she then drives off
Given that sequence of events, where someone is apparently trying to get away from the people in question I'm less inclined to go all 'omg! agents on a power trip!'.
But, as with all things we only get half a muddled story which leads to people biasing it with their own views...
#10samoth Members
Posted 01 July 2013 - 10:01 AM
Though hitting someone who pulls a gun on me with the car is what I'd try too, to be honest. The car is the only weapon you have at hand, and if some group of presumed armed robbers jumps on you, you need to defend your life.
I mean, seriously, what else can you do if someone points a gun at you, surely not give in and hope that he won't shoot you. Why wouldn't he shoot you anyway, he's got no reason not to. On the other hand, with his chest under your tire, it's kind of hard shooting you.
Of course it will turn out that the agents were right and the girl was lying. She probably had 200 grams of cocaine in her trunk, too. And a shotgun. No surprise there. Like the agents in that video on tv news last week where 4 or 5 of them were kicking and giving the rod to a person who was helpless on the floor. Sure enough he was a major threat with his face to the ground and his arms around his head. Good police work.
"If you pick up a brick and kill one of those pigs [referring to police] it ain't no loss."
-- Joschka Fischer around 1980 (before his time as Minister).
#11Alpheus GDNet+
Posted 01 July 2013 - 10:26 AM
I agree - the report is confusing and unclear as to the time line.
It says she was 'walking to her car' as they approached and yet at later it says she had to turn on her engine to wind down the windows which caused them to try and break the windows. Even the part where it says one jumped on the hood of the SUV while another pulled a gun implies she had got into the car at this point - so did they pull the gun after she apparently walked away from them, got into a car and started it?
The knee jerk 'omg! agents over reacting!' is predictable given the way the article is structured, the headline used and the way it was written.
Lets say the sequence of events was;
- girl exist store and walks towards car
- 6 people approach holding a badge and ask her to stop
- girl gets into car and locks the door
- agents then ask her to wind window down
- at which point she starts car
- agents react because it looks like she is trying to get away (jump on hood, pull gun)
- at which point she then drives off
Given that sequence of events, where someone is apparently trying to get away from the people in question I'm less inclined to go all 'omg! agents on a power trip!'.
But, as with all things we only get half a muddled story which leads to people biasing it with their own views...
But even if that's the case, it takes 6 of you? And you jumped the hood and pulled your gun? Over beer? And tried to break the glass. Over beer? Why not just take down their license plate? And it's two girls. Someone's taking underage drinking too seriously. And they even pulled over for actual cops.
But yes, the article is unclear and the timeline does seem to be a bit out of sorts.
External Articulation of Concepts Materializes Innate Knowledge of One's Craft and Science
Super Mario Bros clone tutorial written in XNA 4.0 [MonoGame, ANX, and MonoXNA] by Scott Haley
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#12Alpheus GDNet+
Posted 01 July 2013 - 10:27 AM
Did they pull the gun before or after she hit them with the car? I think that makes a big difference in my opinion here.
If you're in your car and people are trying to break in the windows, I don't think it makes much of a difference. This is my opinion of course.
External Articulation of Concepts Materializes Innate Knowledge of One's Craft and Science
Super Mario Bros clone tutorial written in XNA 4.0 [MonoGame, ANX, and MonoXNA] by Scott Haley
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#13jms bc Members
Posted 01 July 2013 - 11:00 AM
Standard behavior for much of law enforcement in most of the country.
I'm guessing that the cops here are investigating the store, not the people. Even if these were under-aged adults (what a strange phrase) with alcohol, then the cops would primarily want to use their purchase as evidence against the store/clerk. So, the cops' behavior is even worse and more disproportional than it seems at first (ie, they must have known these women were no threat).
Incompetence + meathead + lazy. Never expect better from law enforcement, and take this into account when prescribing their duties and limits. (Also take into account when dealing with them.) You cannot depend on discretion with this sort.
Hopefully, the attention will have some negative impact on the careers of the officers involved. Doubt it though.
The Four Horsemen of Happiness have left.
#14phantom Members
Posted 01 July 2013 - 11:36 AM
The car is the only weapon you have at hand, and if some group of presumed armed robbers jumps on you, you need to defend your life.
Except that isn't what happened... in the 'victims' own words - "They were showing unidentifiable badges after they approached us, but we became frightened, as they were not in anything close to a uniform" .
So they weren't 'jumped' at all...
#15jms bc Members
Posted 01 July 2013 - 12:07 PM
Phantom, the girls could be exaggerating the extent to which they were intimidated. Or they were needlessly frightened by the innocuous.
Or, most probably, these were typical cops. When the girls got close to the car, they came from several directions shouting "Down on the ground!", "Hands up in the air!" and "Don't move!".
The frightened girls hurry into the car and lock the doors. Now the cops, half of whom are dressed like homeless, are shouting at the windows, spraying saliva like rabid dogs. No longer trying to id themselves and angry at perceived lack of respect, they try to take control through additional uncoordinated intimidation. Two of them use tasers on the vehicle, for the heck of it. Guns are drawn. Someone jumps on the hood. It's like zombie survival here.
Edited by jms bc, 01 July 2013 - 12:07 PM.
The Four Horsemen of Happiness have left.
#16phantom Members
Posted 01 July 2013 - 12:08 PM
But even if that's the case, it takes 6 of you? And you jumped the hood and pulled your gun? Over beer? And tried to break the glass. Over beer? Why not just take down their license plate? And it's two girls. Someone's taking underage drinking too seriously. And they even pulled over for actual cops.
But yes, the article is unclear and the timeline does seem to be a bit out of sorts.
Well to be fair it was 7 (6 men, 1 woman), but again we are vague on details. How did they approach? Was it as a group, was it spread out? At what point did they announce who they were? Reading the article it is unclear as the phase 'after they approached us' is odd; does this mean they approached and stopped at distance? or close? or after they started approaching?
What did the agents do to identify themselves? Did the lead agent say they were ATF offices? Why wasn't the girl able to identify the badge? Did she not recognise it? (in which case there is an education issue) was it too dark? Did they just flash it or hold it badly? When did they identify themselves? Was she in the car at the time? Or did she get in afterwards?
Before she turned the engine on what was the lead up? Did they say wind the window down? Or get out? or did she do it without instruction?
If they identified themselves then starting the car would look like they are trying to leave/escape at which point the situation escalates because it goes from a simple 'check for beer' to 'why are they running? what are they hiding? Is something else going on?' thus the reaction, which might well be standard ATF training to prevent a vehicle, and an SUV at that, from leaving in that situation?
Also, it wasn't "two girls" - again the article is maddeningly unclear but it clearly states "as her roommates seated inside" & "My roommates and I" which implies there was more than 2 in the car. Now, this could have been only 3 people in total, or there could have been 4 or 5 in there.
As for the licence plate, I dare say as they drove off they would have called it in however from their point of view the people in the car are trying to run which means even if they pull the licence plates that could have just escalated into a chase which could have ended in an accident with more people hurt - logically the thing to do is to try and prevent this from happening by stopping the vehicle leaving and containing the situation - imagine the news report if they had been hiding something, got into a chase, refused to stop, lost control and crashed into someone/group of people killing someone and it came out that the agents didn't even try to stop them?
As also 'over beer' is what we can say with hind sight but in this situation but what if they had been running because they had a load of drugs in the car? or some other reason they didn't want to be stopped by the agents? If they had been the case we'd never hear of this story because suddenly they are just doing their job correctly - If the girls had just beer with them and the agents believed they had identified themselves clearly the fact they appear to have tried to run suddenly makes it about more than 'just beer'.
According to that article they didn't pull over for the police but for another agent in a car with 'sirens and lights', although no mention of it being marked.
The fact is the lack of coherent time line and lack of full details from both sides makes any one taking any clear position as 'right' nothing more than them projecting their own issues onto the situation - there is no objective way to say who was in the right here.
One of the key things I'd like to know is why no one could identify the badge because that is a key event which triggers the rest...
#17samoth Members
Posted 01 July 2013 - 12:18 PM
The car is the only weapon you have at hand, and if some group of presumed armed robbers jumps on you, you need to defend your life.
Except that isn't what happened... in the 'victims' own words - "They were showing unidentifiable badges after they approached us, but we became frightened, as they were not in anything close to a uniform" .
So they weren't 'jumped' at all...
So you're a 20 year old (or the like) girl, and some beefy definitely-not-uniformed guys that probably look more like pimps than cops come running at you, shouting "police" and show something that might be a library pass, or a blockbuster video club card, or something they've printed on their inkjet down in the cellar where they keep young women as hostages.
Now tell me you're not trying to get into the car and get the door shut.
Next, they draw guns, and one of them jumps onto your car's hood. Oh fuck. Now tell me you're not scared to death and you're not panicking.
#18phantom Members
Posted 01 July 2013 - 12:28 PM
So you're a 20 year old (or the like) girl, and some beefy definitely-not-uniformed guys that probably look more like pimps than cops come running at you, shouting "police" and show something that might be a library pass, or a blockbuster video club card, or something they've printed on their inkjet down in the cellar where they keep young women as hostages.
Now tell me you're not trying to get into the car and get the door shut.
Next, they draw guns, and one of them jumps onto your car's hood. Oh fuck. Now tell me you're not scared to death and you're not panicking.
Interesting... so tell me where you get...
- beefy definitely-not-uniformed guys that look more like pimps
- running
- shouting 'police'
- information on the id they showed
- _they_ draw guns
... from?
Because ALL of that you just made up to fit your view of the events.
The article makes :
- no mention of their appearance besides 'not in uniform', also one was a woman
- no mention of 'running'
- no mention of how they identified themselves
- identification was just unidentified; no mention of what or why... maybe the girls simply didn't recognise it?
- only one person drew a gun, which might well have been in response to them appearing to try to flee the scene
The lack of critical thinking in this thread is frankly shocking from a group of people who are apparently interested in a field where critical thinking is a key skill.
Edited by phantom, 01 July 2013 - 12:29 PM.
Posted 01 July 2013 - 12:42 PM
Phantom, are you saying that, if 7 people come up to you and start giving you orders, the first thing you will think of is "Oh, they are undercover cops.". No. You will think "I'm getting robbed. / Potentially raped / Maybe just harassed" The badge is irrelevant, no way for me to know its not fake.
Probably would have turned out differently if only 1 or 2 of the cops went to check things out.
I think the fact that the felony charges against the girls were dropped tells you everything you need to know.
#20tstrimple Prime Members
Posted 01 July 2013 - 12:48 PM
I think the fact that the felony charges against the girls were dropped tells you everything you need to know.
This.
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2017-02-25 04:38:33
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https://socratic.org/questions/5881b303b72cff16ae0168e6
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# Question 168e6
Consider a product that casts $100$; a 15% discount means that you pay for it $100 - 15 = 85$ that is less or you can say that its price was reduced.
Same thing with, say, a temperature; if an object change temperature from $100$ to $85$ we say that the temperature decreases of $15$ or 15%#.
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2020-04-03 09:02:52
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https://socratic.org/questions/how-do-you-factor-4p-2q-pq-2-4p-3-by-grouping
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# How do you factor 4p^2q+pq^2+4p^3 by grouping?
Jun 21, 2016
$4 {p}^{2} q + p {q}^{2} + 4 {p}^{3} = p \left(q + 2 p\right) \left(q + 2 p\right)$
#### Explanation:
Separate out the common factor $p$, rearrange into descending powers of $q$ then recognise the perfect square trinomial:
$4 {p}^{2} q + p {q}^{2} + 4 {p}^{3}$
$= p \left(4 p q + {q}^{2} + 4 {p}^{2}\right)$
$= p \left({q}^{2} + 4 p q + 4 {p}^{2}\right)$
$= p \left(q + 2 p\right) \left(q + 2 p\right)$
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2020-01-23 15:03:52
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http://www.donaldrauscher.com/index15.html
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# DonaldRauscher.com
## A Blog About D4T4 & M47H
The robot's first cut will create 2 pizza slices. The robot's second cut will create either 3 or 4 pizza slices. The robot's third cut will create 4, 5, or 6 slices if starting with 3 slices or 5, 6, or 7 slices if starting with 4 slices. I began by thinking about the probability of these outcomes in a specific state. So that the geometry mimics the probability distribution, I'm setting the circumference of the pizza equal to 1.
Second CutThird Cut (Starting w/ 3 Slices)Third Cut (Starting w/ 4 Slices)
Next, I integrated over the entire state space to get the probability of each of these outcomes. The state space for the second cut (after 1 cut has been made) is very easy to define; we only need 1 variable (x) to specify the size of each of the two slices. The state space for the third cut (after 2 cuts have been made) is more challenging; we need 3 variables to define the space (x, y, and z). Each integral is multipled by the number of circular permutations; in the case of the third cut, there are 4 points, so (4 - 1)! = 6.
For that second cut:
For that third cut:
Putting this all together:
I found it easiest to think about this problem in terms of polar coordinates. The furthest point that we can eat along trajectory is the following:
Plotting this, it forms this weird, rounded rectangle shape:
I integrated the above equation to get the area. Unlike a regular integral where each area is a rectangle (Reimann sum), each incremental integration area is a circular sector with . Putting this all together:
Extending this logic, I also derived an expression for the area eaten for a regular polygon with n sides:
As n goes to infinity, the area that the picky eater eats becomes a circle with a radius half that of the sandwich, making the area that is eaten equal to 25%. This is the most efficient shape for the picky eater.
For this week's Riddler, I estimated that the division leader would have 88.8 wins after 162 games. I assumed that each team plays the other teams in it's division 19 times for a total of 76 intradivision games and 86 interdivision games, consistent with the actual scheduling rules.
Interdivision games are pretty easy to deal with because the outcomes of each team's interdivision games are independent of one another. If all 162 games were interdivision (i.e. the teams in the division somehow never play one another), the answer would be pretty straightforward (code in R):
> cdf <- (pbinom(1:162, 162, 0.5))^5
> pmf <- cdf - c(0, head(cdf, -1))
> sum(1:162 * pmf)
[1] 88.39431
However, teams in the division obviously do play one another, so win-loss records are not independent. We can think of intradivision games as a series of consecutive round robins. Each round robin consists of 10 games, and each team plays in 4 of those games (one game against each of their division foes). Each team plays 76 intradivision games, which is 19 round robins and 190 games. I started by creating an exhaustive state space (and corresponding probabilities) for the win totals of the 1st, 2nd, 3rd, 4th, and 5th place teams after the 190 intradivision games. Each state is defined as such that . As you can expect, there are many possible outcomes (157,470 states in my state space)!
Next, for each state, I calculated the probability that the division winning team would have less than X wins after the remaining 86 interdivision games. Results of interdivision games are independent, making this pretty easy. Sum-product with the state probabilities and we have the CDF for X!
This computes to 88.8 wins. This is slightly higher than the all-interdivision calculation above, which makes intuitive sense. In the all-interdivision scenario, we could theoretically have a division leader with 0 wins (all 5 teams go winless). However, when we have intradivision games, the division leader cannot have fewer wins than the number of intradivision games that they play divided by 2; one team's loss is another team's win.
I've posted my code on my GH here. Enjoy!
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2021-11-27 03:02:29
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|
https://www.gradesaver.com/textbooks/math/other-math/CLONE-547b8018-14a8-4d02-afd6-6bc35a0864ed/chapter-4-decimals-review-exercises-page-323/45
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## Basic College Mathematics (10th Edition)
Unreasonable. Estimate: 30 $\div$ 3 = 10 Exact: 9.5
$26.6\div2.8$ = 0.95 Now, Lets find the Estimate: 26.6 (rounded to) 30. 2.8 (rounded to) 3 $26.6\div2.8$ = 30 $\div$ 3 = 10 So the given figure is unreasonable. Let's find the Exact: 26.6 $\div$ 2.8 = 9.5
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2019-10-14 01:22:05
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https://www.shaalaa.com/question-bank-solutions/2y3-5y2-19y-42-factorising-the-quadratic-polynomial-trinomial-of-the-type-ax2-bx-c-a-0_62559
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# 2y3 − 5y2 − 19y + 42 - Mathematics
2y3 − 5y2 − 19y + 42
#### Solution
Let f(y) =2y3 − 5y2 − 19y + 42 be the given polynomial.
Now, putting y=2,we get
f(2) = 2(2)^3 - 5(2)^2 - 19(2) + 42
= 16 - 20 - 38 + 42 = -58 + 58
= 0
Therefore, (y - 2)is a factor of polynomial f(y).
Now,
f(y) = 2y^2 (y-2) - y(y - 2) -21(y - 2)
= (y -2){2y^2 - y - 21}
= (y-2){2y^2 - 7y + 6y - 21}
= (y-2)(y + 3)(2y - 7)
Hence (y - 2),(y+3) and (2y - 7) are the factors of polynomial f(y).
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Mathematics for Class 9
Chapter 6 Factorisation of Polynomials
Exercise 6.5 | Q 10 | Page 33
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2021-04-17 19:40:34
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https://www.openfoam.com/documentation/guides/latest/api/thirdBodyArrheniusReactionRateI_8H_source.html
|
The open source CFD toolbox
thirdBodyArrheniusReactionRateI.H
Go to the documentation of this file.
1/*---------------------------------------------------------------------------*\
2 ========= |
3 \\ / F ield | OpenFOAM: The Open Source CFD Toolbox
4 \\ / O peration |
5 \\ / A nd | www.openfoam.com
6 \\/ M anipulation |
7-------------------------------------------------------------------------------
8 Copyright (C) 2011-2017 OpenFOAM Foundation
9-------------------------------------------------------------------------------
11 This file is part of OpenFOAM.
12
13 OpenFOAM is free software: you can redistribute it and/or modify it
15 the Free Software Foundation, either version 3 of the License, or
16 (at your option) any later version.
17
18 OpenFOAM is distributed in the hope that it will be useful, but WITHOUT
19 ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
20 FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
21 for more details.
22
23 You should have received a copy of the GNU General Public License
24 along with OpenFOAM. If not, see <http://www.gnu.org/licenses/>.
25
26\*---------------------------------------------------------------------------*/
27
28// * * * * * * * * * * * * * * * * Constructors * * * * * * * * * * * * * * //
29
31(
32 const scalar A,
33 const scalar beta,
34 const scalar Ta,
35 const thirdBodyEfficiencies& tbes
36)
37:
39 thirdBodyEfficiencies_(tbes)
40{}
41
42
44(
45 const speciesTable& species,
46 const dictionary& dict
47)
48:
50 (
51 species,
52 dict
53 ),
54 thirdBodyEfficiencies_(species, dict)
55{}
56
57
58// * * * * * * * * * * * * * * * Member Functions * * * * * * * * * * * * * //
59
61(
62 const scalar p,
63 const scalar T,
64 const scalarField& c
65) const
66{
67 return
68 thirdBodyEfficiencies_.M(c)
70}
71
72
74{
76 thirdBodyEfficiencies_.write(os);
77}
78
79
80inline Foam::Ostream& Foam::operator<<
81(
82 Ostream& os,
84)
85{
86 arr.write(os);
87 return os;
88}
89
90
91// ************************************************************************* //
static const Foam::dimensionedScalar A("", Foam::dimPressure, 611.21)
Arrhenius reaction rate given by:
An Ostream is an abstract base class for all output systems (streams, files, token lists,...
Definition: Ostream.H:62
virtual bool write(const token &tok)=0
Write token to stream or otherwise handle it.
A list of keyword definitions, which are a keyword followed by a number of values (eg,...
Definition: dictionary.H:126
Ostream & operator()() const
Output stream (master only).
Definition: ensightCaseI.H:74
friend Ostream & operator(Ostream &, const faMatrix< Type > &)
virtual bool write()
Write the output fields.
A wordList with hashed named lookup, which can be faster in some situations than using the normal lis...
Arrhenius reaction rate enhanced by third-body interaction.
Third body efficiencies.
volScalarField & p
const volScalarField & T
OBJstream os(runTime.globalPath()/outputName)
dictionary dict
dimensionedScalar beta("beta", dimless/dimTemperature, laminarTransport)
|
2022-11-27 19:47:47
|
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http://openstudy.com/updates/4f1ec785e4b04992dd24cef4
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## nikita2 4 years ago Could you help me with this equation y'' = 1/y^2
1. JamesJ
This is a tricky equation. I'll come back in a minute and help you with it.
2. anonymous
3. Shayaan_Mustafa
it is a differential equation.
4. Shayaan_Mustafa
5. nikita2
yes it is differential eq.
6. Shayaan_Mustafa
so what are asking for? finding general solution or anything else?
7. nikita2
general
8. Shayaan_Mustafa
ok you need a general solution. There are two solution. first is complementary solution and other is particular solution. do you know this?
9. Shayaan_Mustafa
you just get relax nikita2. I am here for help. take time and post a correct and complete question.
10. nikita2
ok, I will )
11. Shayaan_Mustafa
ok
12. JamesJ
@shayaan, do you know how to find the particular solution? This is quite non-standard.
13. Shayaan_Mustafa
yes i know. there are several methods. but good one is variation of parameters .
14. JamesJ
No, that's not correct. This is non-linear equation.
15. nikita2
$(D _{2}^{t}u)^{2} (D _{2}^{x}D _{2}^{t}u) = 1$ And i should to solve Cauchy problem
16. Shayaan_Mustafa
is it non-linear?
17. nikita2
yes(
18. Shayaan_Mustafa
hmmm... then fine.
19. JamesJ
y = y(t), y'' - 1/y^2 = 0. Not linear, unfortunately.
20. Shayaan_Mustafa
do you know how to start with cauchy euler method?? you get start then we are here to volunteerily
21. nikita2
yes i'll try it now
22. JamesJ
ok ... multiply first both sides by y' dt $y'' y' dt = \frac{y'}{y^2} dt$ Writing $$d(y') = y' dt$$ and $$dy = y' dt$$ we have $(y')' d(y') = \frac{1}{y^2} dy$ Now integrate both sides and we have ....
23. Shayaan_Mustafa
@JamesJ give him reason why we did so? So that he could get clear concepts. Thanks.
24. nikita2
Now integrate both sides and we have .... $(y')^2/2 = -1/3*1/y^3$
25. nikita2
+ Const
26. JamesJ
$\frac{1}{2} (y')^2 = -\frac{1}{y} + C$
27. nikita2
Oh! i've done mistake here!
28. JamesJ
Now.... $\frac{dy}{dt} = \pm \sqrt{ \frac{ 2(Cy-1)}{y}}$ This is separable and it isn't too hard to find an equation for t, t = t(y).
29. nikita2
i see...
30. JamesJ
Interesting, eh? I only figured this out recently myself, working it out in this problem: http://openstudy.com/study#/updates/4f0de661e4b084a815fcff56
31. nikita2
Oh! I've not notise that it is forse of atraktion!
32. nikita2
But anyway if i'll try to do next. Solve $D _{2}^{t}u(t,x) = y(t,x)$ where y we found
33. JamesJ
Tricky, but probably not impossible analytically. What's the original motivation for this problem?
34. nikita2
I solving the Cauchy problem $(D _{2}^{t}u(t,x))^2*(D _{2}^{x}D _{2}^{t}u(t,x)) =$ the initial condition is on {t=0}
35. nikita2
=1
36. JamesJ
right. btw, you might find this site helpful: math.stackexchange.com
37. JamesJ
but in the meantime, it would be great if you wanted to help out here answering some elementary questions now and again!
38. nikita2
I did it and I will )
39. nikita2
Thank you very much!
40. JamesJ
Happy to help. This was a fun problem.
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2017-01-18 22:52:47
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http://mathhelpforum.com/algebra/65494-simplify-expression-correct.html
|
# Thread: Simplify expression, Is it correct?
1. ## Simplify expression, Is it correct?
Attached is the expression and my solution.
I get a solution of zero, is it correct?
2. Sorry the attachment is not clear. Can't really tell u from a in the attachment. Can attach a clearer version.
3. There is no a? All of the exponents are u (positive and negative).
4. the correct answer is 4/((e^u+e^(-u))^2)
5. This is the approach i took.
(e^u+e^-u)(e^u+e^-u)-(e^u-e^-u)(e^u-e^-u) / (e^u+e^-u)^2
(e^2u+2+e^-2u)-(e^2u-2+e^-2u)/ (e^u+e^-u)^2
(e^2u+2+e^-2u-e^2u+2-e^-2u)/ (e^u+e^-u)^2
4 / (e^u+e^-u)^2
Originally Posted by nrslmz
6. Attached is a better version.
7. Ok noted. Your approach is not quite right. Please look at my previous post for further clarification.
8. Ahh, I see now. Thanks!
9. Actually, I noticed that you didn't simplify your denominator.
The denominator simplified is 3 because (e^2u + e^0 + e^0 + e^-2u).
10. You are right. I forgot to simplify the denominator
In that case it should be
(e^u+e^-u)(e^u+e^-u)-(e^u-e^-u)(e^u-e^-u) / (e^u+e^-u)^2
(e^2u+2+e^-2u)-(e^2u-2+e^-2u)/ (e^u+e^-u)^2
(e^2u+2+e^-2u-e^2u+2-e^-2u)/ (e^u+e^-u)^2
4 / (e^2u+2+e^-2u)
4 / 3
11. Thanks!
12. Originally Posted by mwok
Attached is the expression and my solution.
I get a solution of zero, is it correct?
Hi,
If I am not mistaken, this can be solved using hyperbolic functions.
I'll break this problem into many sections so you can see what I am doing
If you consider that the hyperbolic functions are
$\frac{e^{x} - e^{-x}}{2} = sinh(x)$
and that
$\frac{e^{x} + e^{-x}}{2} = cosh(x)$
then we can conclude that
$(e^{x} - e^{-x}) = 2sinh(x)$
$e^{x} + e^{-x} = 2cosh(x)$
------------------------------------------------------
Which means that if you use them in your problem you will get
$(e^{u} + e^{-u})*(e^{u}+e^{-u}) = 4 cosh^{2}(u)$
$(e^{u} - e^{-u})*(e^{u}-e^{-u}) = 4 sinh^{2}(u)$
$(e^{u} - e^{-u})^{2} = 4 cosh^{2}(u)$
$4 cosh^{2}(u) - 4 sinh^{2}(u) = 4$
$\frac{4}{4 cosh^{2}(u)} = \frac{1}{cosh^{2}(u)} = sech^{2}(u)
$
|
2016-09-26 12:59:58
|
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|
https://www.albert.io/learn/ap-statistics/question/hypotheses-related-to-reasons-for-hospitalization
|
Limited access
A hospital administrator found a report from $2003$ which showed the distribution of reasons why individuals were admitted to a hospital after coming to the emergency room. The distribution is shown below and does not include admissions related to pregnancy and childbirth.
Reason Percent
Circulatory Disorders 26.5
Respiratory Disorders 15.3
Digestive Disorders 14.2
Injuries 11.7
Mental Health and Substance Abuse Disorders 5.9
Endocrine Disorders 4.8
Genitourinary Disorders 4.5
All Other Disorders 17.1
After collecting data from a random sample of admissions during the current year, the administrator completed a test to determine if the percentages for each category have changed since $2003$.
Which of the following provides a correct set of hypotheses?
A
${ H }_{ 0 }$: The expected counts for each category of hospital admissions in the current year are equal to the corresponding expected counts from $2003$.
${ H }_{ a }$: The expected counts for each category of hospital admissions in the current year are all different from the corresponding expected counts from $2003$.
B
${ H }_{ 0 }$: In the selected sample, the percentages for each category of hospital admissions are the same as in $2003$.
${ H }_{ a }$: For at least one category of hospital admissions, the sample percentage is different from what it was in $2003$.
C
${ H }_{ 0 }$: The percentages for each category of hospital admissions in the current year are equal to the corresponding percentages from $2003$.
${ H }_{ a }$: For at least one category of hospital admissions, the percentage in the current year is different from the corresponding percentage from $2003$.
D
${ H }_{ 0 }$: The percentages for each category of hospital admissions in the current year are equal to the corresponding percentages from $2003$.
${ H }_{ a }$: The percentages for each category of hospital admissions in the current year are all different from the corresponding percentages from $2003$.
E
${ H }_{ 0 }$: The reasons for hospital admission are independent of the year in which the hospital admissions were made.
${ H }_{ a }$: For reasons for hospital admission are not independent of the year in which the hospital admissions were made.
Select an assignment template
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2017-04-28 00:22:48
|
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https://socratic.org/questions/how-do-you-identify-bronsted-lowry-acids-and-bases-1
|
# How do you identify Bronsted-Lowry acids and bases?
Apr 20, 2016
Acids drop protons (${H}^{+}$), and bases pick them up.
#### Explanation:
Bronsted-Lowry theory states that an acid is a molecule that drops ${H}^{+}$ ions (protons), and a base picks them up again.
For example, in the reaction
$H C l + {H}_{2} O \to C {l}^{-} + {H}_{3} {O}^{+}$
the hydrochloric acid drops the hydrogen proton, while water picks it up. This means $H C l$ acts as an acid and ${H}_{2} O$ acts as a base.
After the reaction you then have a $C {l}^{-}$ ion which can pick up protons, so this is known as a conjugate base, and ${H}_{3} {O}^{+}$ has a new propensity to drop protons, and so is a conjugate acid.
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2021-10-16 12:20:11
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https://socratic.org/questions/5a55acf67c01490ae47e5998#533112
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# Question #e5998
Jan 10, 2018
Tosylate is a better leaving group than iodide.
#### Explanation:
What makes one leaving group better than the other is how stable is the group after it has broken its bond with the parent molecule.
We can also think of it in this manner: We calculate it from the basicity of the group.
(weak base -> more stable while solvated -> better leaving the group).
Tosylate groups ( $C {H}_{3} {C}_{6} {H}_{4} S {O}_{3}^{-}$) are very good leaving groups because their conjugate bases are highly stabilized by $r e s o n a n c e$.
There is no resonance stabilization in case of iodine. However, iodide is regarded as the best leaving group amongst the halogens because iodide is a weaker base than -OH since its conjugate acid HI is a stronger acid than H2O. So, iodine is a better leaving group.
https://en.wikipedia.org/wiki/Leaving_group
|
2022-08-08 21:55:54
|
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|
https://cstheory.stackexchange.com/questions/18748/compute-time-complexity-of-neural-network-svm-and-other-classification-algorith
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# Compute Time Complexity of Neural network, SVM and other classification algorithms
I would like to know what is the asymptotic time complexity analysis for general models of Back-propagation Neural Network, SVM and Maximum Entropy. Does it just depend on number of features included and training time complexity is the only stuff that really matters.
And does it real matter when applying on large chunk of text classification like twitter data or blog data
• SVMs contain an underlying optimization step that is solved heuristically, so for any actual algorithm that purports to solve SVMs, the answer is undefined. A number like $O(n^3)$ is generally bandied around for implementations like libsvm, which means something like time/iteration * #iterations (where #iterations is assumed to be constant) – Suresh Venkat Aug 25 '13 at 17:50
• @SureshVenkat: That should be an answer, not a comment. – Jeffε Aug 31 '13 at 18:14
• @JɛffE it is done. – Suresh Venkat Sep 1 '13 at 5:17
SVMs contain an underlying optimization step that is solved heuristically, so for any actual algorithm that purports to solve SVMs, the answer is undefined. A number like $O(n^3)$ is generally bandied around for implementations like libsvm, which means something like time/iteration * #iterations (where #iterations is assumed to be constant)
Relevance Vector Machines need a $O(n^3)$ time too, but the main overhead there is in the learning phase, it needs to repeatedly compute the inverse of a Hessian Matrix (that requires $O(n^3)$ computations.
However, a preprocessing step is often useful and it also costs you $O(nd)$, specially if you want faster classification speeds. This is a good paper to read if you want to know more.
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2020-08-15 21:31:15
|
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|
https://economics.stackexchange.com/questions/124/self-selection-bias-during-the-course-of-experiments/5131
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# Self-selection bias during the course of experiments
Suppose you are running a randomized experiment to assess the effect of $X$, say some training program for unemployed people, on $Y$, say the chance of finding a job in the coming year. Suppose also that $X$ takes time : maybe it lasts for several month.
Because you randomize, you do not need to worry about self-selection bias initially. But during the course of $X$, some people will likely realize that $X$ is beneficial to them, and others may realize that they are wasting their time.
As a result, one might expect that among people who drop from the program, there is a higher proportion of agents for which the treatment effect would have been smaller. This might induce an over-estimation of the treatment effect.
My questions are :
• Is this kind of bias discussed in the literature on randomized experiments?
• Does it have a canonical name ?
• Do researcher try to control for this, and if yes, how?
|
2020-10-26 22:28:00
|
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|
https://readtiger.com/wkp/en/Diesel_engine
|
This page uses content from Wikipedia and is licensed under CC BY-SA.
# Diesel engine
Diesel engine built by Langen & Wolf under licence, 1898.
1952 Shell Oil film showing the development of the diesel engine from 1877
The diesel engine (also known as a compression-ignition or CI engine), named after Rudolf Diesel, is an internal combustion engine in which ignition of the fuel is caused by the elevated temperature of the air in the cylinder due to the mechanical compression (adiabatic compression). This contrasts with spark-ignition engines such as a petrol engine (gasoline engine) or gas engine (using a gaseous fuel as opposed to petrol), which use a spark plug to ignite an air-fuel mixture.
Diesel engines work by compressing only the air. This increases the air temperature inside the cylinder to such a high degree that atomised diesel fuel injected into the combustion chamber ignites spontaneously. With the fuel being injected into the air just before combustion, the dispersion of the fuel is uneven; this is called a heterogeneous air-fuel mixture. The torque a diesel engine produces is controlled by manipulating the air-fuel ratio (λ); instead of throttling the intake air, the diesel engine relies on altering the amount of fuel that is injected, and the air-fuel ratio is usually high.
The diesel engine has the highest thermal efficiency (engine efficiency) of any practical internal or external combustion engine due to its very high expansion ratio and inherent lean burn which enables heat dissipation by the excess air. A small efficiency loss is also avoided compared with non-direct-injection gasoline engines since unburned fuel is not present during valve overlap and therefore no fuel goes directly from the intake/injection to the exhaust. Low-speed diesel engines (as used in ships and other applications where overall engine weight is relatively unimportant) can reach effective efficiencies of up to 55%.[1]
Diesel engines may be designed as either two-stroke or four-stroke cycles. They were originally used as a more efficient replacement for stationary steam engines. Since the 1910s, they have been used in submarines and ships. Use in locomotives, trucks, heavy equipment and electricity generation plants followed later. In the 1930s, they slowly began to be used in a few automobiles. Since the 1970s, the use of diesel engines in larger on-road and off-road vehicles in the US has increased. According to Konrad Reif, the EU average for diesel cars accounts for half of newly registered cars.[2]
The world's largest diesel engines put in service are 14-cylinder, two-stroke watercraft diesel engines; they produce a peak power of almost 100 MW each.[3]
## History
### Diesel's idea
Rudolf Diesel's 1893 patent on a rational heat motor
Diesel's first experimental engine 1893
Diesel's second prototype. It is a modification of the first experimental engine. On 17 February 1894, this engine ran under its own power for the first time.[4]
Effective efficiency 16.6%
Fuel consumption 519 g·kW−1·h−1
First fully functional diesel engine, designed by Imanuel Lauster, built from scratch, and finished by October 1896.[5][6][7]
Rated power 13.1 kW
Effective efficiency 26.2%
Fuel consumption 324 g·kW−1·h−1.
In 1878, Rudolf Diesel, who was a student at the "Polytechnikum" in Munich, attended the lectures of Carl von Linde. Linde explained that steam engines are capable of converting just 6–10% of the heat energy into work, but that the Carnot cycle allows conversion of much more of the heat energy into work by means of isothermal change in condition. According to Diesel, this ignited the idea of creating a highly efficient engine that could work on the Carnot cycle.[8] Diesel was also exposed to a fire piston, a traditional fire starter using rapid adiabatic compression principles which Linde had acquired from Southeast Asia.[9] After several years of working on his ideas, Diesel published them in 1893 in the essay Theory and Construction of a Rational Heat Motor.[8]
Diesel was heavily criticised for his essay, but only few found the mistake that he made;[10] his rational heat motor was supposed to utilise a constant temperature cycle (with isothermal compression) that would require a much higher level of compression than that needed for compression ignition. Diesel's idea was to compress the air so tightly that the temperature of the air would exceed that of combustion. However, such an engine could never perform any usable work.[11][12][13] In his 1892 US patent (granted in 1895) #542846 Diesel describes the compression required for his cycle:
"pure atmospheric air is compressed, according to curve 1 2, to such a degree that, before ignition or combustion takes place, the highest pressure of the diagram and the highest temperature are obtained-that is to say, the temperature at which the subsequent combustion has to take place, not the burning or igniting point. To make this more clear, let it be assumed that the subsequent combustion shall take place at a temperature of 700°. Then in that case the initial pressure must be sixty-four atmospheres, or for 800° centigrade the pressure must be ninety atmospheres, and so on. Into the air thus compressed is then gradually introduced from the exterior finely divided fuel, which ignites on introduction, since the air is at a temperature far above the igniting-point of the fuel. The characteristic features of the cycle according to my present invention are therefore, increase of pressure and temperature up to the maximum, not by combustion, but prior to combustion by mechanical compression of air, and there upon the subsequent performance of work without increase of pressure and temperature by gradual combustion during a prescribed part of the stroke determined by the cut-oil".[14]
By June 1893, Diesel had realised his original cycle would not work and he adopted the constant pressure cycle.[15] Diesel describes the cycle in his 1895 patent application. Notice that there is no longer a mention of compression temperatures exceeding the temperature of combustion. Now it is simply stated that the compression must be sufficient to trigger ignition.
"1. In an internal-combustion engine, the combination of a cylinder and piston constructed and arranged to compress air to a degree producing a temperature above the igniting-point of the fuel, a supply for compressed air or gas; a fuel-supply; a distributing-valve for fuel, a passage from the air supply to the cylinder in communication with the fuel-distributing valve, an inlet to the cylinder in communication with the air-supply and with the fuel-valve, and a cut-oil, substantially as described." See US patent # 608845 filed 1895 / granted 1898[16][17][18]
In 1892, Diesel received patents in Germany, Switzerland, the United Kingdom and the United States for "Method of and Apparatus for Converting Heat into Work".[19] In 1894 and 1895, he filed patents and addenda in various countries for his engine; the first patents were issued in Spain (No. 16,654),[20] France (No. 243,531) and Belgium (No. 113,139) in December 1894, and in Germany (No. 86,633) in 1895 and the United States (No. 608,845) in 1898.[21]
Diesel was attacked and criticised over a time period of several years. Critics have claimed that Diesel never invented a new motor and that the invention of the diesel engine is fraud. Otto Köhler and Emil Capitaine were two of the most prominent critics of Diesel's time.[22] Köhler had published an essay in 1887, in which he describes an engine similar to the engine Diesel describes in his 1893 essay. Köhler figured that such an engine could not perform any work.[13][23] Emil Capitaine had built a petroleum engine with glow-tube ignition in the early 1890s;[24] he claimed against his own better judgement, that his glow-tube ignition engine worked the same way Diesel's engine did. His claims were unfounded and he lost a patent lawsuit against Diesel.[25] Other engines, such as the Akroyd engine and the Brayton engine, also use an operating cycle that is different from the diesel engine cycle.[23][26] Friedrich Sass says that the diesel engine is Diesel's "very own work" and that any "Diesel myth" is "falsification of history".[27]
### The first diesel engine
Diesel sought out firms and factories that would build his engine. With the help of Moritz Schröter and Max Gutermuth [de],[28] he succeeded in convincing both Krupp in Essen and the Maschinenfabrik Augsburg.[29] Contracts were signed in April 1893,[30] and in early summer 1893, Diesel's first prototype engine was built in Augsburg. On 10 August 1893, the first ignition took place, the fuel used was petrol. In winter 1893/1894, Diesel redesigned the existing engine, and by 18 January 1894, his mechanics had converted it into the second prototype.[31] On February 17, 1894, the redesigned engine ran for 88 revolutions – one minute;[4] with this news, Maschinenfabrik Augsburg's stock rose by 30%, indicative of the tremendous anticipated demands for a more efficient engine.[32] On 26 June 1895 the engine achieved an effective efficiency of 16.6% and had a fuel consumption of 519 g·kW−1·h−1. [33] However, despite proving the concept, the engine caused problems,[34] and Diesel could not achieve any substantial progress.[35] Therefore, Krupp considered rescinding the contract they had made with Diesel.[36] Diesel was forced to improve the design of his engine and rushed to construct a third prototype engine. Between 8 November and 20 December 1895, the second prototype had successfully covered over 111 hours on the test bench. In the January 1896 report, this was considered a success.[37]
In February 1896, Diesel considered supercharging the third prototype.[38] Imanuel Lauster, who was ordered to draw the third prototype, had finished the drawings by 30 April 1896. During summer that year the engine was built, it was completed on 6 October 1896.[39] Tests were conducted until early 1897.[40] First public tests began on 1 February 1897.[41] Moritz Schröter's test on 17 February 1897 was the main test of Diesel's engine. The engine was rated 13.1 kW with a specific fuel consumption of 324 g·kW−1·h−1,[42] resulting in an effective efficiency of 26.2%.[43][44] By 1898, Diesel had become a millionaire.[45]
### Timeline
#### 1890s
• 1893: Rudolf Diesel's essay titled Theory and Construction of a Rational Heat Motor appears.[46][47]
• 1893: February 21, Diesel and the Maschinenfabrik Augsburg sign a contract that allows Diesel to build a prototype engine.[48]
• 1893: February 23, Diesel obtains a patent (RP 67207) titled "Arbeitsverfahren und Ausführungsart für Verbrennungsmaschinen" (Working Methods and Techniques for Internal Combustion Engines).
• 1893: April 10, Diesel and Krupp sign a contract that allows Diesel to build a prototype engine.[48]
• 1893: April 24, both Krupp and the Maschinenfabrik Augsburg decide to collaborate and build just a single prototype in Augsburg.[48][30]
• 1893: July, the first prototype is completed.[49]
• 1893: August 10, Diesel injects fuel (petrol) for the first time, resulting in combustion, destroying the indicator.[50]
• 1893: November 30, Diesel applies for a patent (RP 82168) for a modified combustion process. He obtains it on 12 July 1895.[51][52][53]
• 1894: January 18, after the first prototype had been modified to become the second prototype, testing with the second prototype begins.[31]
• 1894: February 17, The second prototype runs for the first time.[4]
• 1895: March 30, Diesel applies for a patent (RP 86633) for a starting process with compressed air.[54]
• 1895: June 26, the second prototype passes brake testing for the first time.[33]
• 1895: Diesel applies for a second patent US Patent # 608845[55]
• 1895: November 8 – December 20, a series of tests with the second prototype is conducted. In total, 111 operating hours are recorded.[37]
• 1896: October 6, the third and final prototype engine is completed.[5]
• 1897: February 1, after 4 years Diesel's prototype engine is running and finally ready for efficiency testing and production.[41]
• 1897: October 9, Adolphus Busch licenses rights to the diesel engine for the US and Canada.[45][56]
• 1897: 29 October, Rudolf Diesel obtains a patent (DRP 95680) on supercharging the diesel engine.[38]
• 1898: February 1, the Diesel Motoren-Fabrik Actien-Gesellschaft is registered.[57]
• 1898: March, the first commercial diesel engine, rated 2×30 PS (2×22 kW), is installed in the Kempten plant of the Vereinigte Zündholzfabriken A.G.[58][59]
• 1898: September 17, the Allgemeine Gesellschaft für Dieselmotoren A.-G. is founded.[60]
• 1899: The first two-stroke diesel engine, invented by Hugo Güldner, is built.[44]
#### 1900s
An MAN DM trunk piston diesel engine built in 1906. The MAN DM series is considered to be one of the first commercially successful diesel engines.[61]
• 1901: Imanuel Lauster designs the first trunk piston diesel engine (DM 70).[61]
• 1901: By 1901, MAN had produced 77 diesel engine cylinders for commercial use.[62]
• 1903: Two first diesel-powered ships are launched, both for river and canal operations: The Vandal naphtha tanker and the Sarmat.[63]
• 1904: The French launch the first diesel submarine, the Aigrette.[64]
• 1905: January 14: Diesel applies for a patent on unit injection (L20510I/46a).[65]
• 1905: The first diesel engine turbochargers and intercoolers are manufactured by Büchi.[66]
• 1906: The Diesel Motoren-Fabrik Actien-Gesellschaft is dissolved.[22]
• 1908: Diesel's patents expire.[67]
• 1908: The first lorry (truck) with a diesel engine appears.[68]
• 1909: March 14, Prosper L'Orange applies for a patent on precombustion chamber injection.[69] He later builds the first diesel engine with this system.[70][71]
#### 1910s
• 1910: MAN starts making two-stroke diesel engines.[72]
• 1910: November 26, James McKechnie applies for a patent on unit injection.[73] Unlike Diesel, he managed to successfully build working unit injectors.[65][74]
• 1911: November 27, the Allgemeine Gesellschaft für Dieselmotoren A.-G. is dissolved.[57]
• 1911: The Germania shipyard in Kiel builds 850 PS (625 kW) diesel engines for German submarines. These engines are installed in 1914.[75]
• 1912: MAN builds the first double-acting piston two-stroke diesel engine.[76]
• 1912: The first locomotive with a diesel engine is used on the Swiss Winterthur-Romanshorn railroad.[77]
• 1912: The Selandia is the first ocean-going ship with diesel engines.[78]
• 1913: NELSECO diesels are installed on commercial ships and US Navy submarines.[79]
• 1913: September 29, Rudolf Diesel dies mysteriously when crossing the English Channel on the SS Dresden.[80]
• 1914: MAN builds 900 PS (662 kW) two-stroke engines for Dutch submarines.[81]
• 1919: Prosper L'Orange obtains a patent on a Precombustion chamber insert incorporating a needle injection nozzle.[82][83][71] First diesel engine from Cummins.[84][85]
#### 1920s
Fairbanks Morse model 32
• 1923: At the Königsberg DLG exhibition, the first agricultural tractor with a diesel engine, the prototype Benz-Sendling S6, is presented.[86][better source needed]
• 1923: December 15, the first lorry with a direct-injected diesel engine is tested by MAN. The same year, Benz builds a lorry with a pre-combustion chamber injected diesel engine.[87]
• 1923: The first two-stroke diesel engine with counterflow scavenging appears.[88]
• 1924: Fairbanks-Morse introduces the two-stroke Y-VA (later renamed to Model 32).[89]
• 1925: Sendling starts mass-producing a diesel-powered agricultural tractor.[90]
• 1927: Bosch introduces the first inline injection pump for motor vehicle diesel engines.[91]
• 1929: The first passenger car with a diesel engine appears. Its engine is an Otto engine modified to use the diesel principle and Bosch's injection pump. Several other diesel car prototypes follow.[92]
#### 1930s
• 1933: Junkers Motorenwerke in Germany start production of the most successful mass-produced aviation diesel engine of all time, the Jumo 205. By the outbreak of World War II, over 900 examples are produced. Its rated take-off power is 645 kW.[93]
• 1933: General Motors uses its new roots-blown, unit-injected two-stroke Winton 201A diesel engine to power its automotive assembly exhibit at the Chicago World's Fair (A Century of Progress).[94] The engine is offered in several versions ranging from 600–900 hp (447–671 kW).[95]
• 1934: The Budd Company builds the first diesel-electric passenger train in the US, the Pioneer Zephyr 9900, using a Winton engine.[94]
• 1935: The Citroën Rosalie is fitted with an early swirl chamber injected diesel engine for testing purposes.[96] Daimler-Benz starts manufacturing the Mercedes-Benz OM 138, the first mass-produced diesel engine for passenger cars, and one of the few marketable passenger car diesel engines of its time. It is rated 45 PS (33 kW).[97]
• 1936: March 4, the airship LZ 129 Hindenburg, the biggest aircraft ever made, takes off for the first time. She is powered by four V16 Daimler-Benz LOF 6 diesel engines, rated 1200 PS (883 kW) each.[98]
• 1936: Manufacture of the first mass-produced passenger car with a diesel engine (Mercedes-Benz 260 D) begins.[92]
• 1937: Konstantin Fyodorovich Chelpan develops the V-2 diesel engine, later used in the Soviet T-34 tanks, widely regarded as the best tank chassis of World War II.[99]
• 1938: General Motors forms the GM Diesel Division, later to become Detroit Diesel, and introduces the Series 71 inline high-speed medium-horsepower two stroke engine, suitable for road vehicles and marine use.[100]
#### 1940s
• 1946: Clessie Cummins obtains a patent on a fuel feeding and injection apparatus for oil-burning engines that incorporates separate components for generating injection pressure and injection timing.[101]
• 1946: Klöckner-Humboldt-Deutz (KHD) introduces an air-cooled mass-production diesel engine to the market.[102]
#### 1950s
Piston of an MAN M-System centre sphere combustion chamber type diesel engine (4 VD 14,5/12-1 SRW)
• 1950s: KHD becomes the air-cooled diesel engine global market leader.[103]
• 1951: J. Siegfried Meurer obtains a patent on the M-System, a design that incorporates a central sphere combustion chamber in the piston (DBP 865683).[104]
• 1953: First mass-produced swirl chamber injected passenger car diesel engine (Borgward/Fiat).[73]
• 1954: Daimler-Benz introduces the Mercedes-Benz OM 312 A, a 4.6 litre straight-6 series-production industrial diesel engine with a turbocharger, rated 115 PS (85 kW). It proves to be unreliable.[105]
• 1954: Volvo produces a small batch series of 200 units of a turbocharged version of the TD 96 engine. This 9.6 litre engine is rated 136 kW.[106]
• 1955: Turbocharging for MAN two-stroke marine diesel engines becomes standard.[88]
• 1959: The Peugeot 403 becomes the first mass-produced passenger sedan/saloon manufactured outside West Germany to be offered with a diesel engine option.[107]
#### 1960s
Mercedes-Benz OM 352, one of the first direct injected Mercedes-Benz diesel engines. It was introduced in 1963, but mass production only started in summer 1964.[108]
#### 1970s
• 1972: KHD introduces the AD-System, Allstoff-Direkteinspritzung, (anyfuel direct-injection), for its diesel engines. AD-diesels can operate on virtually any kind of liquid fuel, but they are fitted with an auxiliary spark plug that fires if the ignition quality of the fuel is too low.[111]
• 1976: Development of the common rail injection begins at the ETH Zürich.[112]
• 1976: The Volkswagen Golf becomes the first compact passenger sedan/saloon to be offered with a diesel engine option.[113][114]
• 1978: Daimler-Benz produces the first passenger car diesel engine with a turbocharger (Mercedes-Benz OM 617).[115]
• 1979: First prototype of a low-speed two-stroke crosshead engine with common rail injection.[116]
#### 1980s
BMW E28 524td, the first mass-produced passenger car with an electronically controlled injection pump
• 1981/82: Uniflow scavenging for two-stroke marine diesel engines becomes standard.[117]
• 1985: December, road testing of a common rail injection system for lorries using a modified 6VD 12,5/12 GRF-E engine in an IFA W50 takes place.[118]
• 1986: The BMW E28 524td is the world's first passenger car equipped with an electronically controlled injection pump (developed by Bosch).[73][119]
• 1987: Daimler-Benz introduces the electronically controlled injection pump for lorry diesel engines.[73]
• 1988: The Fiat Croma becomes the first mass-produced passenger car in the world to have a direct injected diesel engine.[73]
• 1989: The Audi 100 is the first passenger car in the world with a turbocharged, direct injected, and electronically controlled diesel engine.[73]
#### 1990s
• 1992: 1 July, the Euro 1 emission standard comes into effect.[120]
• 1993: First passenger car diesel engine with four valves per cylinder, the Mercedes-Benz OM 604.[115]
• 1994: Unit injector system by Bosch for lorry diesel engines.[121]
• 1996: First diesel engine with direct injection and four valves per cylinder, used in the Opel Vectra.[122][73]
• 1996: First radial piston distributor injection pump by Bosch.[121]
• 1997: First mass-produced common rail diesel engine for a passenger car, the Fiat 1.9 JTD.[73][115]
• 1998: BMW wins the 24 Hours Nürburgring race with a modified BMW E36. The car, called 320d, is powered by a 2-litre, straight-four diesel engine with direct injection and a helix-controlled distributor injection pump (Bosch VP 44), producing 180 kW. The fuel consumption is 23 l/100 km, only half the fuel consumption of a similar Otto-powered car.[123]
• 1998: Volkswagen introduces the VW EA188 Pumpe-Düse engine (1.9 TDI), with Bosch-developed electronically controlled unit injectors.[115]
• 1999: Daimler-Chrysler presents the first common rail three-cylinder diesel engine used in a passenger car (the Smart City Coupé).[73]
#### 2000s
Audi R10 TDI, 2006 24 Hours of Le Mans winner.
• 2000: Peugeot introduces the diesel particulate filter for passenger cars.[73][115]
• 2002: Piezoelectric injector technology by Siemens.[124]
• 2003: Piezoelectric injector technology by Bosch,[125] and Delphi.[126]
• 2004: BMW introduces dual-stage turbocharging with the BMW M57 engine.[115]
• 2006: The world's most powerful diesel engine, the Wärtsilä RT-flex96C, is produced. It is rated 80,080 kW.[127]
• 2006: Audi R10 TDI, equipped with a 5.5-litre V12-TDI engine, rated 476 kW, wins the 2006 24 Hours of Le Mans.[73]
• 2006: Daimler-Chrysler launches the first series-production passenger car engine with selective catalytic reduction exhaust gas treatment, the Mercedes-Benz OM 642. It is fully complying with the Tier2Bin8 emission standard.[115]
• 2008: Volkswagen introduces the LNT catalyst for passenger car diesel engines with the VW 2.0 TDI engine.[115]
• 2008: Volkswagen starts series production of the biggest passenger car diesel engine, the Audi 6-litre V12 TDI.[115]
• 2008: Subaru introduces the first horizontally opposed diesel engine to be fitted to a passenger car. It is a 2-litre common rail engine, rated 110 kW.[128]
## Operating principle
### Characteristics
The characteristics of a diesel engine are[133]
• Compression ignition: Due to almost adiabatic compression, the fuel ignites without any ignition-initiating apparatus such as spark plugs.
• Mixture formation inside the combustion chamber: Air and fuel are mixed in the combustion chamber and not in the inlet manifold.
• Engine speed adjustment solely by mixture quality: Instead of throttling the air-fuel mixture, the amount of torque produced (resulting in crankshaft rotational speed differences) is set solely by the mass of injected fuel, always mixed with as much air as possible.
• Heterogeneous air-fuel mixture: The dispersion of air and fuel in the combustion chamber is uneven.
• High air ratio: Due to always running on as much air as possible and not depending on exact mixture of air and fuel, diesel engines have an air-fuel ratio leaner than stochiometric (${\displaystyle \lambda _{v}\geq \lambda _{min}>1}$).
• Diffusion flame: At combustion, oxygen first has to diffuse into the flame, rather than having oxygen and fuel already mixed before combustion, which would result in a premixed flame.
• Fuel with high ignition performance: As diesel engines solely rely on compression ignition, fuel with high ignition performance (cetane rating) is ideal for proper engine operation, fuel with a good knocking resistance (octane rating), e.g. petrol, is suboptimal for diesel engines.
### Cycle of the diesel engine
p-V Diagram for the ideal diesel cycle. The cycle follows the numbers 1–4 in clockwise direction. The horizontal axis is volume of the cylinder. In the diesel cycle the combustion occurs at almost constant pressure. On this diagram the work that is generated for each cycle corresponds to the area within the loop.
Diesel engine model, left side
Diesel engine model, right side
The diesel internal combustion engine differs from the gasoline powered Otto cycle by using highly compressed hot air to ignite the fuel rather than using a spark plug (compression ignition rather than spark ignition).
In the diesel engine, only air is initially introduced into the combustion chamber. The air is then compressed with a compression ratio typically between 15:1 and 23:1. This high compression causes the temperature of the air to rise. At about the top of the compression stroke, fuel is injected directly into the compressed air in the combustion chamber. This may be into a (typically toroidal) void in the top of the piston or a pre-chamber depending upon the design of the engine. The fuel injector ensures that the fuel is broken down into small droplets, and that the fuel is distributed evenly. The heat of the compressed air vaporises fuel from the surface of the droplets. The vapour is then ignited by the heat from the compressed air in the combustion chamber, the droplets continue to vaporise from their surfaces and burn, getting smaller, until all the fuel in the droplets has been burnt. Combustion occurs at a substantially constant pressure during the initial part of the power stroke. The start of vaporisation causes a delay before ignition and the characteristic diesel knocking sound as the vapour reaches ignition temperature and causes an abrupt increase in pressure above the piston (not shown on the P-V indicator diagram). When combustion is complete the combustion gases expand as the piston descends further; the high pressure in the cylinder drives the piston downward, supplying power to the crankshaft.
As well as the high level of compression allowing combustion to take place without a separate ignition system, a high compression ratio greatly increases the engine's efficiency. Increasing the compression ratio in a spark-ignition engine where fuel and air are mixed before entry to the cylinder is limited by the need to prevent pre-ignition, which would cause engine damage. Since only air is compressed in a diesel engine, and fuel is not introduced into the cylinder until shortly before top dead centre (TDC), premature detonation is not a problem and compression ratios are much higher.
The p–V diagram is a simplified and idealised representation of the events involved in a diesel engine cycle, arranged to illustrate the similarity with a Carnot cycle. Starting at 1, the piston is at bottom dead centre and both valves are closed at the start of the compression stroke; the cylinder contains air at atmospheric pressure. Between 1 and 2 the air is compressed adiabatically – that is without heat transfer to or from the environment – by the rising piston. (This is only approximately true since there will be some heat exchange with the cylinder walls.) During this compression, the volume is reduced, the pressure and temperature both rise. At or slightly before 2 (TDC) fuel is injected and burns in the compressed hot air. Chemical energy is released and this constitutes an injection of thermal energy (heat) into the compressed gas. Combustion and heating occur between 2 and 3. In this interval the pressure remains constant since the piston descends, and the volume increases; the temperature rises as a consequence of the energy of combustion. At 3 fuel injection and combustion are complete, and the cylinder contains gas at a higher temperature than at 2. Between 3 and 4 this hot gas expands, again approximately adiabatically. Work is done on the system to which the engine is connected. During this expansion phase the volume of the gas rises, and its temperature and pressure both fall. At 4 the exhaust valve opens, and the pressure falls abruptly to atmospheric (approximately). This is unresisted expansion and no useful work is done by it. Ideally the adiabatic expansion should continue, extending the line 3–4 to the right until the pressure falls to that of the surrounding air, but the loss of efficiency caused by this unresisted expansion is justified by the practical difficulties involved in recovering it (the engine would have to be much larger). After the opening of the exhaust valve, the exhaust stroke follows, but this (and the following induction stroke) are not shown on the diagram. If shown, they would be represented by a low-pressure loop at the bottom of the diagram. At 1 it is assumed that the exhaust and induction strokes have been completed, and the cylinder is again filled with air. The piston-cylinder system absorbs energy between 1 and 2 – this is the work needed to compress the air in the cylinder, and is provided by mechanical kinetic energy stored in the flywheel of the engine. Work output is done by the piston-cylinder combination between 2 and 4. The difference between these two increments of work is the indicated work output per cycle, and is represented by the area enclosed by the p–V loop. The adiabatic expansion is in a higher pressure range than that of the compression because the gas in the cylinder is hotter during expansion than during compression. It is for this reason that the loop has a finite area, and the net output of work during a cycle is positive.[134]
### Efficiency
Due to its high compression ratio, the diesel engine has a high efficiency, and the lack of a throttle valve means that the charge-exchange losses are fairly low, resulting in a low specific fuel consumption, especially in medium and low load situations. This makes the diesel engine very economical.[135] Even though diesel engines have a theoretical efficiency of 75%,[136] in practice it is much lower. In his 1893 essay Theory and Construction of a Rational Heat Motor, Rudolf Diesel describes that the effective efficiency of the diesel engine would be in between 43.2% and 50.4% , or maybe even greater.[137] Modern passenger car diesel engines may have an effective efficiency of up to 43%,[138] whilst engines in large diesel trucks, and buses can achieve peak efficiencies around 45%.[139] However, average efficiency over a driving cycle is lower than peak efficiency. For example, it might be 37% for an engine with a peak efficiency of 44%.[140] The highest diesel engine efficiency of up to 55% is achieved by large two-stroke watercraft diesel engines.[1]
### Major advantages
Diesel engines have several advantages over engines operating on other principles:
• The diesel engine has the highest effective efficiency of all combustion engines.[141]
• Diesel engines inject the fuel directly into the combustion chamber, have no intake air restrictions apart from air filters and intake plumbing and have no intake manifold vacuum to add parasitic load and pumping losses resulting from the pistons being pulled downward against intake system vacuum. Cylinder filling with atmospheric air is aided and volumetric efficiency is increased for the same reason.
• Although the fuel efficiency (mass burned per energy produced) of a diesel engine drops at lower loads, it doesn't drop quite as fast as that of a typical petrol or turbine engine.[142]
• Bus powered by biodiesel
Diesel engines can combust a huge variety of fuels, including several fuel oils, that have advantages over fuels such as petrol. These advantages include:
• Low fuel costs, as fuel oils are relatively cheap
• Good lubrication properties
• High energy density
• Low risk of catching fire, as they do not form a flammable vapour
• Biodiesel is an easily synthesised, non-petroleum-based fuel (through transesterification) which can run directly in many diesel engines, while gasoline engines either need adaptation to run synthetic fuels or else use them as an additive to gasoline (e.g., ethanol added to gasohol).
• Diesel engines have a very good exhaust-emission behaviour. The exhaust contains minimal amounts of carbon monoxide and hydrocarbons. Direct injected diesel engines emit approximately as much nitrogen oxide as Otto cycle engines. Swirl chamber and precombustion chamber injected engines, however, emit approximately 50% less nitrogen oxide than Otto cycle engines when running under full load.[143] Compared with Otto cycle engines, diesel engines emit 10 times less pollutants and also less carbon dioxide (comparing the raw emissions without exhaust gas treatment).[144]
• They have no high voltage electrical ignition system, resulting in high reliability and easy adaptation to damp environments. The absence of coils, spark plug wires, etc., also eliminates a source of radio frequency emissions which can interfere with navigation and communication equipment, which is especially important in marine and aircraft applications, and for preventing interference with radio telescopes. (For this reason, only diesel-powered vehicles are allowed in parts of the American National Radio Quiet Zone.)[145]
• Diesel engines can accept super- or turbocharging pressure without any natural limit,[146] constrained only by the design and operating limits of engine components, such as pressure, speed and load. This is unlike petrol engines, which inevitably suffer detonation at higher pressure if engine tuning and/or fuel octane adjustments are not made to compensate.
## Fuel injection
Diesel engines rely on internal mixture formation,[133] which means that they require a fuel injection system. The fuel is injected directly into the combustion chamber, which can be either a segmented combustion chamber or an unsegmented combustion chamber. Fuel injection with the latter is referred to as direct injection (DI), whilst injection into the former is called indirect injection (IDI).[147] In diesel engine terminology, indirect injection does not mean fuel injection into the inlet manifold or anywhere else outside the cylinder or combustion chamber: in fact, the definition of the diesel engine excludes such injection methods. For creating the fuel pressure, diesel engines usually have an injection pump. There are several different types of injection pumps and methods for creating a fine air-fuel mixture. Over the years many different injection methods have been used. These can be described as the following:
• Air blast, where the fuel is blown into the cylinder by a blast of air.
• Solid fuel / hydraulic injection, where the fuel is pushed through a spring loaded valve / injector to produce a combustible mist.
• Mechanical unit injector, where the injector is directly operated by a cam and fuel quantity is controlled by a rack or lever.
• Mechanical electronic unit injector, where the injector is operated by a cam and fuel quantity is controlled electronically.
• Common rail mechanical injection, where fuel is at high pressure in a common rail and controlled by mechanical means.
• Common rail electronic injection, where fuel is at high pressure in a common rail and controlled electronically.
### Torque controlling
Due to the way diesel engines work, a vital component of all diesel engines is a mechanical or electronic governor which regulates the torque of the engine and thus idling speed and maximum speed by controlling the rate of fuel delivery. This means a change of ${\displaystyle \lambda _{v}}$. Unlike Otto-cycle engines, incoming air is not throttled. Mechanically governed fuel injection systems are driven by the engine's gear train.[148][149] These systems use a combination of springs and weights to control fuel delivery relative to both load and speed.[148] Modern electronically controlled diesel engines control fuel delivery by use of an electronic control module (ECM) or electronic control unit (ECU). The ECM/ECU receives an engine speed signal, as well as other operating parameters such as intake manifold pressure and fuel temperature, from a sensor and controls the amount of fuel and start of injection timing through actuators to maximise power and efficiency and minimise emissions. Controlling the timing of the start of injection of fuel into the cylinder is a key to minimizing emissions, and maximizing fuel economy (efficiency), of the engine. The timing is measured in degrees of crank angle of the piston before top dead centre. For example, if the ECM/ECU initiates fuel injection when the piston is 10° before TDC, the start of injection, or timing, is said to be 10° before TDC. Optimal timing will depend on the engine design as well as its speed and load.
### Types of fuel injection
#### Air-blast injection
Typical early 20th century air-blast injected diesel engine, rated at 59 kW.
Diesel's original engine injected fuel with the assistance of compressed air, which atomised the fuel and forced it into the engine through a nozzle (a similar principle to an aerosol spray). The nozzle opening was closed by a pin valve lifted by the camshaft to initiate the fuel injection before top dead centre (TDC). This is called an air-blast injection. Driving the compressor used some power but the efficiency was better than the efficiency of any other combustion engine at that time.[44] Also, air-blast injection made engines very clunky and heavy and did not allow for quick load alteration, thus rendering it unusable for road vehicles.[150]
#### Indirect injection
Ricardo Comet indirect injection chamber
An indirect diesel injection system (IDI) engine delivers fuel into a small chamber called a swirl chamber, precombustion chamber, pre chamber or ante-chamber, which is connected to the cylinder by a narrow air passage. Generally the goal of the pre chamber is to create increased turbulence for better air / fuel mixing. This system also allows for a smoother, quieter running engine, and because fuel mixing is assisted by turbulence, injector pressures can be lower. Most IDI systems use a single orifice injector. The pre-chamber has the disadvantage of lowering efficiency due to increased heat loss to the engine's cooling system, restricting the combustion burn, thus reducing the efficiency by 5–10%. IDI engines are also more difficult to start and usually require the use of glow plugs. IDI engines may be cheaper to build but generally require a higher compression ratio than the DI counterpart. IDI also makes it easier to produce smooth, quieter running engines with a simple mechanical injection system since exact injection timing is not as critical. Most modern automotive engines are DI which have the benefits of greater efficiency and easier starting; however, IDI engines can still be found in the many ATV and small diesel applications.[151] Indirect injected diesel engines use pintle-type fuel injectiors.[152]
#### Helix-controlled direct injection
Different types of piston bowls
Direct injection Diesel engines inject fuel directly into the cylinder. Usually there is a combustion cup in the top of the piston where the fuel is sprayed. Many different methods of injection can be used. Usually, an engine with helix-controlled mechanic direct injection has either an inline or a distributor injection pump.[148] For each engine cylinder, the corresponding plunger in the fuel pump measures out the correct amount of fuel and determines the timing of each injection. These engines use injectors that are very precise spring-loaded valves that open and close at a specific fuel pressure. Separate high-pressure fuel lines connect the fuel pump with each cylinder. Fuel volume for each single combustion is controlled by a slanted groove in the plunger which rotates only a few degrees releasing the pressure and is controlled by a mechanical governor, consisting of weights rotating at engine speed constrained by springs and a lever. The injectors are held open by the fuel pressure. On high-speed engines the plunger pumps are together in one unit.[153] The length of fuel lines from the pump to each injector is normally the same for each cylinder in order to obtain the same pressure delay. Direct injected diesel engines usually use orifice-type fuel injectors.[152]
Electronic control of the fuel injection transformed the direct injection engine by allowing much greater control over the combustion.[154]
#### Unit direct injection
Unit direct injection, also known as Pumpe-Düse (pump-nozzle), is a high pressure fuel injection system that injects fuel directly into the cylinder of the engine. In this system the injector and the pump are combined into one unit positioned over each cylinder controlled by the camshaft. Each cylinder has its own unit eliminating the high-pressure fuel lines, achieving a more consistent injection. Under full load, the injection pressure can reach up to 220 MPa. Unit injection systems used to dominate the commercial diesel engine market, but due to higher requirements of the flexibility of the injection system, they have been rendered obsolete by the more advanced common-rail-system.[155]
#### Common rail direct injection
Common rail (CR) direct injection systems, unlike other injection systems, do not have a combined pressure creation and injection apparatus. A high-pressure injection pump creates a constant pressure, not depending on the engine speed or fuel mass injected. A buffer, the so-called rail, saves this pressure. This allows fuel injection at any given moment, even multiple injections in a very short amount of time. The Electronic Diesel Control unit (EDC) controls both rail pressure and injections depending on several different parameters of the engine. The injectors of older CR systems have solenoid-driven plungers for lifting the injection needle, whilst newer CR injectors use plungers driven by piezoelectric actuators, that have fewer moving mass and therefore allow even more injections in a very short period of time.[156] The injection pressure of modern CR systems ranges from 140 MPa to 270 MPa.[157]
## Types
There are several different ways of categorising diesel engines, based on different design characteristics:
### By power output
• Small <188 kW (252 hp)
• Medium 188–750 kW
• Large >750 kW
Source[158]
### By cylinder bore
• Passenger car engines: 75...100 mm
• Lorry and commercial vehicle engines: 90...170 mm
• High-performance high-speed engines: 165...280 mm
• Medium-speed engines: 240...620 mm
• Low-speed two-stroke engines: 260...900 mm
Source:[159]
### By number of strokes
• Four-stroke cycle
• Two-stroke cycle
Source[158]
### By cylinder arrangement
Regular cylinder configurations such as straight (inline), V, and boxer (flat) configurations can be used for diesel engines. The inline-six-cylinder design is the most prolific in light- to medium-duty engines, though inline-four engines are also common. Small-capacity engines (generally considered to be those below five litres in capacity) are generally four- or six-cylinder types, with the four-cylinder being the most common type found in automotive uses. The V configuration used to be common for commercial vehicles, but it has been abandoned in favour of the inline configuration.[160]
### By engine speeds
Günter Mau categorises diesel engines by their rotational speeds into three groups:
• High-speed engines (> 1,000 rpm),
• Medium-speed engines (300–1,000 rpm), and
• Slow-speed engines (< 300 rpm).
Source[161]
#### High-speed engines
High-speed engines are used to power trucks (lorries), buses, tractors, cars, yachts, compressors, pumps and small electrical generators.[162] As of 2018, most high-speed engines have direct injection. Many modern engines, particularly in on-highway applications, have common rail direct injection.[155] On bigger ships, high-speed diesel engines are often used for powering electric generators.[163] The highest power output of high-speed diesel engines is approximately 5 MW.[164]
#### Medium-speed engines
Medium-speed engines are used in large electrical generators, ship propulsion and mechanical drive applications such as large compressors or pumps. Medium speed diesel engines operate on either diesel fuel or heavy fuel oil by direct injection in the same manner as low-speed engines. Usually, they are four-stroke engines with trunk pistons.[165]
The power output of medium-speed diesel engines can be as high as 21,870 kW,[166] with the effective efficiency being around 47...48% (1982).[167] Most larger medium-speed engines are started with compressed air direct on pistons, using an air distributor, as opposed to a pneumatic starting motor acting on the flywheel, which tends to be used for smaller engines.[168]
Medium-speed engines intended for marine applications are usually used to power (ro-ro) ferries, passenger ships or small freight ships. Using medium-speed engines reduces the cost of smaller ships and increases their transport capacity. In addition to that, a single ship can use two smaller engines instead of one big engine, which increases the ship's safety.[165]
#### Low-speed engines
The MAN B&W 5S50MC 5-cylinder, 2-stroke, low-speed marine diesel engine. This particular engine is found aboard a 29,000 tonne chemical carrier.
Low-speed diesel engines are usually very large in size and mostly used to power ships. There are two different types of low-speed engines that are commonly used: Two-stroke engines with a crosshead, and four-stroke engines with a regular trunk-piston. Two-stroke engines have a limited rotational frequency and their charge exchange is more difficult, which means that they are usually bigger than four-stroke engines and used to directly power a ship's propeller. Four-stroke engines on ships are usually used to power an electric generator. An electric motor powers the propeller.[161] Both types are usually very undersquare.[169] Low-speed diesel engines (as used in ships and other applications where overall engine weight is relatively unimportant) often have an effective efficiency of up to 55%.[1] Like medium-speed engines, low-speed engines are started with compressed air, and they use heavy oil as their primary fuel.[168]
## Two-stroke engines
Detroit Diesel timing
Two-stroke diesel engines use only two strokes instead of four strokes for a complete engine cycle. Filling the cylinder with air and compressing it takes place in one stroke, and the power and exhaust strokes are combined. The compression in a two-stroke diesel engine is similar to the compression that takes place in a four-stroke diesel engine: As the piston passes through bottom centre and starts upward, compression commences, culminating in fuel injection and ignition. Instead of a full set of valves, two-stroke diesel engines have simple intake ports, and exhaust ports (or exhaust valves). When the piston approaches bottom dead centre, both the intake and the exhaust ports are "open", which means that there is atmospheric pressure inside the cylinder. Therefore, some sort of pump is required to blow the air into the cylinder and the combustion gasses into the exhaust. This process is called scavenging. The pressure required is approximately 10 - 30 kPa.[170]
Scavenging
In general, there are three types of scavenging possible:
Crossflow scavenging is incomplete and limits the stroke, yet some manufacturers used it.[171] Reverse flow scavenging is a very simple way of scavenging, and it was popular amongst manufacturers until the early 1980s. Uniflow scavenging is more complicated to make but allows the highest fuel efficiency; since the early 1980s, manufacturers such as MAN and Sulzer have switched to this system.[117] It is standard for modern marine two-stroke diesel engines.[3]
## Dual-fuel diesel engines
So-called dual-fuel diesel engines or gas diesel engines burn two different types of fuel simultaneously, for instance, a gaseous fuel and diesel engine fuel. The diesel engine fuel auto-ignites due to compression ignition, and then ignites the gaseous fuel. Such engines do not require any type of spark ignition and operate similar to regular diesel engines.[172]
## Diesel engine particularities
### Torque and power
Torque is a force applied to a lever at a right angle multiplied by the lever length. This means that the torque an engine produces depends on the displacement of the engine and the force that the gas pressure inside the cylinder applies to the piston, commonly referred to as effective piston pressure:
${\displaystyle M=p_{e}\cdot V_{h}\cdot \pi ^{-1}\cdot i^{-1}}$
${\displaystyle M}$ .. Torque [N·m]; ${\displaystyle p_{e}}$ .. Effective piston pressure [kN·m−2]; ${\displaystyle V_{h}}$ .. Displacement [dm3]; ${\displaystyle i}$ .. Strokes [either 2 or 4]
Example
• Engine A: effective piston pressure=570 kN·m−2, displacement= 2.2 dm3, strokes= 4, torque= 100 N·m
${\displaystyle 570\cdot 2.2\cdot \pi ^{-1}\cdot 4^{-1}\approx 100}$
Power is the quotient of work and time:
${\displaystyle P=2\pi nM}$
${\displaystyle P}$ .. Power [W]; ${\displaystyle M}$ .. Torque [N·m]; ${\displaystyle n}$ .. Time (crankshaft speed) [s−1]
which means:
${\displaystyle P=2\pi \cdot n_{1}\cdot M\cdot 60^{-1}}$
${\displaystyle P}$ .. Power [W]; ${\displaystyle M}$ .. Torque [N·m]; ${\displaystyle n_{1}}$ .. Time (crankshaft speed) [min−1]
Example
• Engine A: Power≈ 44,000 W, torque= 100 N·m, time= 4200 min−1
${\displaystyle 44,000\approx 2\cdot \pi \cdot 4200\cdot 100\cdot 60^{-1}}$
• Engine B: Power≈ 44,000 W, torque= 260 N·m, time= 1600 min−1
${\displaystyle 44,000\approx 2\cdot \pi \cdot 1600\cdot 260\cdot 60^{-1}}$
This means, that increasing either torque or time will result in an increase in power. As the maximum rotational frequency of the diesel engine's crankshaft is usually in between 3500...5000 min−1 due to diesel principle limitations, the torque of the diesel engine must be great to achieve a high power, or, in other words, as the diesel engine cannot use a lot of time for achieving a certain amount of power, it has to perform more work (=produce more torque).[173]
### Mass
The average diesel engine has a poorer power-to-mass ratio than the Otto engine. This is because the diesel must operate at lower engine speeds.[174] Due to the higher operating pressure inside the combustion chamber, which increases the forces on the parts due to inertial forces, the diesel engine needs heavier, stronger parts capable of resisting these forces, which results in an overall greater engine mass.[175]
### Emissions
As diesel engines burn a mixture of fuel and air, the exhaust therefore contains substances that consist of the same chemical elements, as fuel and air. The main elements of air are nitrogen (N2) and oxygen (O2), fuel consists of hydrogen (H2) and carbon (C). Burning the fuel will result in the final stage of oxidation. An ideal diesel engine, (a hypothetical model that we use as an example), running on an ideal air-fuel mixture, produces an exhaust that consists of carbon dioxide (CO2), water (H2O), nitrogen (N2), and the remaining oxygen (O2). The combustion process in a real engine differs from an ideal engine's combustion process, and due to incomplete combustion, the exhaust contains additional substances,[176] most notably, carbon monoxide (CO), diesel particulate matter (PM), and due to dissociation, nitrogen oxide (NO
x
).[177]
When diesel engines burn their fuel with high oxygen levels, this results in high combustion temperatures and higher efficiency, and particulate matter tends to burn, but the amount of NO
x
pollution tends to increase.[178] NO
x
pollution can be reduced by recirculating a portion of an engine's exhaust gas back to the engine cylinders, which reduces the oxygen quantity, causing a reduction of combustion temperature, and resulting in fewer NO
x
.[179] To further reduce NO
x
emissions, lean NO
x
traps (LNTs)
and SCR-catalysts can be used. Lean NO
x
traps adsorb the nitrogen oxide and "trap" it. Once the LNT is full, it has to be "regenerated" using hydrocarbons. This is achieved by using a very rich air-fuel mixture, resulting in incomplete combustion. An SCR-catalyst converts nitrogen oxide using urea, which is injected into the exhaust stream, and catalytically converts the NO
x
into nitrogen (N2) and water (H2O).[180] Compared with an Otto engine, the diesel engine produces approximately the same amount of NO
x
, but some older diesel engines may have an exhaust that contains up to 50% less NO
x
. However, Otto engines, unlike diesel engines, can use a three-way-catalyst, that converts most of the NO
x
.[143]
Diesel engine exhaust composition
Species Mass percentage[144] Volume percentage[181]
Nitrogen (N2) 75.2% 72.1%
Oxygen (O2) 15% 0.7%
Carbon dioxide (CO2) 7.1% 12.3%
Water (H2O) 2.6% 13.8%
Carbon monoxide (CO) 0.043% 0.09%
Nitrogen oxide (NO
x
)
0.034% 0.13%
Hydrocarbons (HC) 0.005% 0.09%
Aldehyde 0.001% (n/a)
Particulate matter (Sulfate + solid substances) 0.008% 0.0008%
### Noise
Typical diesel engine noise of a 1950s direct injected two-cylinder diesel engine (MWM AKD 112 Z, in idle)
The distinctive noise of a diesel engine is variably called diesel clatter, diesel nailing, or diesel knock.[182] Diesel clatter is caused largely by the way the fuel ignites; the sudden ignition of the diesel fuel when injected into the combustion chamber causes a pressure wave, resulting in an audible ″knock″. Engine designers can reduce diesel clatter through: indirect injection; pilot or pre-injection;[183] injection timing; injection rate; compression ratio; turbo boost; and exhaust gas recirculation (EGR).[184] Common rail diesel injection systems permit multiple injection events as an aid to noise reduction. Therefore, newer diesel engines do not knock anymore.[185] Diesel fuels with a higher cetane rating are more likely to ignite and hence reduce diesel clatter.[182]
### Cold weather starting
In general, diesel engines do not require any starting aid. In cold weather however, some diesel engines can be difficult to start and may need preheating depending on the combustion chamber design. The minimum starting temperature that allows starting without pre-heating is 40 °C for precombustion chamber engines, 20 °C for swirl chamber engines, and 0 °C for direct injected engines. Smaller engines with a displacement of less than 1 litre per cylinder usually have glowplugs, whilst larger heavy-duty engines have flame-start systems.[186]
In the past, a wider variety of cold-start methods were used. Some engines, such as Detroit Diesel engines used[when?] a system to introduce small amounts of ether into the inlet manifold to start combustion.[187] Instead of glowplugs, some diesel engines are equipped with starting aid systems that change valve timing. The simplest way this can be done is with a decompression lever. Activating the decompression lever locks the outlet valves in a slight down position, resulting in the engine not having any compression and thus allowing for turning the crankshaft over without resistance. When the crankshaft reaches a higher speed, flipping the decompression lever back into its normal position will abruptly re-activate the outlet valves, resulting in compression − the flywheel's mass moment of inertia then starts the engine. Other diesel engines, such as the precombustion chamber engine XII Jv 170/240 made by Ganz & Co., have a valve timing changing system that is operated by adjusting the inlet valve camshaft, moving it into a slight "late" position. This will make the inlet valves open with a delay, forcing the inlet air to heat up when entering the combustion chamber.[188]
### Supercharging and turbocharging
Two stroke diesel engine with Roots blower, typical of Detroit Diesel and some Electro-Motive Diesel Engines
Turbocharged 1980s passenger car diesel engine with wastegate turbocharger and without intercooler (BMW M21)
As the diesel engine relies on manipulation of ${\displaystyle \lambda _{v}}$ for torque controlling and speed regulation, the intake air mass does not have to precisely match the injected fuel mass (which would be ${\displaystyle \lambda =1}$).[133] diesel engines are thus ideally suited for supercharging and turbocharging.[146] An additional advantage of the diesel engine is the lack of fuel during the compression stroke. In diesel engines, the fuel is injected near top dead centre (TDC), when the piston is near its highest position. The fuel then ignites due to compression heat. Preignition, caused by the artificial turbocharger compression increase during the compression stroke, cannot occur.[189]
Many diesels are therefore turbocharged and some are both turbocharged and supercharged. A turbocharged engine can produce more power than a naturally aspirated engine of the same configuration. A supercharger is powered mechanically by the engine's crankshaft, while a turbocharger is powered by the engine exhaust. Turbocharging can improve the fuel economy of diesel engines by recovering waste heat from the exhaust, increasing the excess air factor, and increasing the ratio of engine output to friction losses. Adding an intercooler to a turbocharged engine further increases engine performance by cooling down the air-mass and thus allowing more air-mass per volume.[190][191]
A two-stroke engine does not have a discrete exhaust and intake stroke and thus is incapable of self-aspiration. Therefore, all two-stroke diesel engines must be fitted with a blower or some form of compressor to charge the cylinders with air and assist in dispersing exhaust gases, a process referred to as scavenging.[170] Roots-type superchargers were used for ship engines until the mid-1950s, since 1955 they have been widely replaced by turbochargers.[192] Usually, a two-stroke ship diesel engine has a single-stage turbocharger with a turbine that has an axial inflow and a radial outflow.[193]
## Fuel and fluid characteristics
In diesel engines, a mechanical injector system vaporises the fuel directly into the combustion chamber (as opposed to a Venturi jet in a carburetor, or a fuel injector in a manifold injection system vaporising fuel into the intake manifold or intake runners as in a petrol engine). This forced vaporisation means that less-volatile fuels can be used. More crucially, because only air is inducted into the cylinder in a diesel engine, the compression ratio can be much higher as there is no risk of pre-ignition provided the injection process is accurately timed.[189] This means that cylinder temperatures are much higher in a diesel engine than a petrol engine, allowing less volatile fuels to be used.
The MAN 630's M-System diesel engine is a petrol engine (designed to run on NATO F 46/F 50 petrol), but it also runs on jet fuel, (NATO F 40/F 44), kerosene, (NATO F 58), and diesel engine fuel (NATO F 54/F 75)
Therefore, diesel engines can operate on a huge variety of different fuels. In general, fuel for diesel engines should have a proper viscosity, so that the injection pump can pump the fuel to the injection nozzles without causing damage to itself or corrosion of the fuel line. At injection, the fuel should form a good fuel spray, and it should not have a coking effect upon the injection nozzles. To ensure proper engine starting and smooth operation, the fuel should be willing to ignite and hence not cause a high ignition delay, (this means that the fuel should have a high cetane number). Diesel fuel should also have a high lower heating value.[194]
Inline mechanical injector pumps generally tolerate poor-quality or bio-fuels better than distributor-type pumps. Also, indirect injection engines generally run more satisfactorily on fuels with a high ignition delay (for instance, petrol) than direct injection engines.[195] This is partly because an indirect injection engine has a much greater 'swirl' effect, improving vaporisation and combustion of fuel, and because (in the case of vegetable oil-type fuels) lipid depositions can condense on the cylinder walls of a direct-injection engine if combustion temperatures are too low (such as starting the engine from cold). Direct-injected engines with an MAN centre sphere combustion chamber rely on fuel condensing on the combustion chamber walls. The fuel starts vaporising only after ignition sets in, and it burns relatively smoothly. Therefore, such engines also tolerate fuels with poor ignition delay characteristics, and, in general, they can operate on petrol rated 86 RON.[196]
### Fuel types
In his 1893 work Theory and Construction of a Rational Heat Motor, Rudolf Diesel considers using coal dust as fuel for the diesel engine. However, Diesel just considered using coal dust (as well as liquid fuels and gas); his actual engine was designed to operate on petroleum, which was soon replaced with regular petrol and kerosene for further testing purposes, as petroleum proved to be too viscous.[197] In addition to kerosene and petrol, Diesel's engine could also operate on ligroin.[198]
Before diesel engine fuel was standardised, fuels such as petrol, kerosene, gas oil, vegetable oil and mineral oil, as well as mixtures of these fuels, were used.[199] Typical fuels specifically intended to be used for diesel engines were petroleum distillates and coal-tar distillates such as the following; these fuels have specific lower heating values of:
• Diesel oil: 10,200 kcal·kg−1 (42.7 MJ·kg−1) up to 10,250 kcal·kg−1 (42.9 MJ·kg−1)
• Heating oil: 10,000 kcal·kg−1 (41.8 MJ·kg−1) up to 10,200 kcal·kg−1 (42.7 MJ·kg−1)
• Coal-tar creosote: 9,150 kcal·kg−1 (38.3 MJ·kg−1) up to 9,250 kcal·kg−1 (38.7 MJ·kg−1)
• Kerosene: up to 10,400 kcal·kg−1 (43.5 MJ·kg−1)
Source:[200]
The first diesel fuel standards were the DIN 51601, VTL 9140-001, and NATO F 54, which appeared after World War II.[199] The modern European EN 590 diesel fuel standard was established in May 1993; the modern version of the NATO F 54 standard is mostly identical with it. The DIN 51628 biodiesel standard was rendered obsolete by the 2009 version of the EN 590; FAME biodiesel conforms to the EN 14214 standard. Watercraft diesel engines usually operate on diesel engine fuel that conforms to the ISO 8217 standard (Bunker C). Also, some diesel engines can operate on gasses (such as LNG).[201]
### Modern diesel fuel properties
EN 590 (as of 2009) EN 14214 (as of 2010) ≥ 51 CN ≥ 51 CN 820...845 kg·m−3 860...900 kg·m−3 ≤10 mg·kg−1 ≤10 mg·kg−1 ≤200 mg·kg−1 ≤500 mg·kg−1 460 µm 460 µm 2.0...4.5 mm2·s−1 3.5...5.0 mm2·s−1 ≤7.0% ≥96.5% – 1.69 – 37.1 MJ·kg−1
### Gelling
DIN 51601 diesel fuel was prone to waxing or gelling in cold weather; both are terms for the solidification of diesel oil into a partially crystalline state. The crystals build up in the fuel system (especially in fuel filters), eventually starving the engine of fuel and causing it to stop running.[203] Low-output electric heaters in fuel tanks and around fuel lines were used to solve this problem. Also, most engines have a spill return system, by which any excess fuel from the injector pump and injectors is returned to the fuel tank. Once the engine has warmed, returning warm fuel prevents waxing in the tank. Before direct injection diesel engines, some manufacturers, such as BMW, recommended mixing up to 30% petrol in with the diesel by fuelling diesel cars with petrol to prevent the fuel from gelling when the temperatures dropped below −15 °C.[204]
## Safety
### Fuel flammability
Diesel fuel is less flammable than petrol, because its flash point is 55 °C,[203][205] leading to a lower risk of fire caused by fuel in a vehicle equipped with a diesel engine.
Diesel fuel can create an explosive air/vapour mix under the right conditions. However, compared with petrol, it is less prone due to its lower vapour pressure, which is an indication of evaporation rate. The Material Safety Data Sheet[206] for ultra-low sulfur diesel fuel indicates a vapour explosion hazard for diesel fuel indoors, outdoors, or in sewers.
### Cancer
Diesel exhaust has been classified as an IARC Group 1 carcinogen. It causes lung cancer and is associated with an increased risk for bladder cancer.[207]
## Applications
The characteristics of diesel have different advantages for different applications.
### Passenger cars
Diesel engines have long been popular in bigger cars and have been used in smaller cars such as superminis in Europe since the 1980s. They were popular in larger cars earlier, as the weight and cost penalties were less noticeable.[208] Smooth operation as well as high low end torque are deemed important for passenger cars and small commercial vehicles. The introduction of electronically controlled fuel injection significantly improved the smooth torque generation, and starting in the early 1990s, car manufacturers began offering their high-end luxury vehicles with diesel engines. Passenger car diesel engines usually have between three and ten cylinders, and a displacement ranging from 0.8 to 5.0 litres. Modern powerplants are usually turbocharged and have direct injection.[162]
Diesel engines do not suffer from intake-air throttling, resulting in very low fuel consumption especially at low partial load[185] (for instance: driving at city speeds). One fifth of all passenger cars worldwide have diesel engines, with many of them being in Europe, where approximately 47% of all passenger cars are diesel-powered.[209] Daimler-Benz in conjunction with Robert Bosch GmbH produced diesel-powered passenger cars starting in 1936.[73] The popularity of diesel-powered passenger cars in markets such as India, South Korea and Japan is increasing (as of 2018).[210]
### Commercial vehicles and lorries
Lifespan of Mercedes-Benz diesel engines[211]
In 1893, Rudolf Diesel suggested that the diesel engine could possibly power ‘wagons’ (lorries).[212] The first lorries with diesel engines were brought to market in 1924.[73]
Modern diesel engines for lorries have to be both extremely reliable and very fuel efficient. Common-rail direct injection, turbocharging and four valves per cylinder are standard. Displacements range from 4.5 to 15.5 litres, with power-to-mass ratios of 2.5–3.5 kg·kW−1 for heavy duty and 2.0–3.0 kg·kW−1 for medium duty engines. V6 and V8 engines used to be common, due to the relatively low engine mass the V configuration provides. Recently, the V configuration has been abandoned in favour of straight engines. These engines are usually straight-6 for heavy and medium duties and straight-4 for medium duty. Their undersquare design causes lower overall piston speeds which results in increased lifespan of up to 1,200,000 kilometres (750,000 mi).[160] Compared with 1970s diesel engines, the expected lifespan of modern lorry diesel engines has more than doubled.[211]
### Railroad rolling stock
Diesel engines for locomotives are built for continuous operation and may require the ability to use poor quality fuel in some circumstances.[213] Some locomotives use two-stroke diesel engines.[214] Diesel engines have eclipsed steam engines as the prime mover on all non-electrified railroads in the industrialised world. The first diesel locomotives appeared in 1913,[73] and diesel multiple units soon after. Many modern diesel locomotives are actually diesel-electric locomotives: the diesel engine is used to power an electric generator that in turn powers electric traction motors with no mechanical connection between diesel engine and traction.[215] While electric locomotives have replaced the diesel locomotive for some passenger traffic in Europe and Asia, diesel is still today very popular for cargo-hauling freight trains and on tracks where electrification is not feasible.
In the 1940s, road vehicle diesel engines with power outputs of 150...200 PS (110...147 kW) were considered reasonable for DMUs. Commonly, regular truck powerplants were used. The height of these engines had to be less than 1,000 mm to allow underfloor installation. Usually, the engine was mated with a pneumatically operated mechanical gearbox, due to the low size, mass, and production costs of this design. Some DMUs used hydraulic torque converters instead. Diesel-electric transmission was not suitable for such small engines.[216] In the 1930s, the Deutsche Reichsbahn standardised its first DMU engine. It was a 30.3 litre, 12-cylinder boxer unit, producing 275 PS (202 kW). Several German manufacturers produced engines according to this standard.[217]
### Watercraft
One of the eight-cylinder 3200 I.H.P. Harland and Wolff – Burmeister & Wain diesel engines installed in the motorship Glenapp. This was the highest powered diesel engine yet (1920) installed in a ship. Note man standing lower right for size comparison.
Hand-cranking a boat diesel motor in Inle Lake (Myanmar).
The requirements for marine diesel engines vary, depending on the application. For military use and medium-size boats, medium-speed four-stroke diesel engines are most suitable. These engines usually have up to 24 cylinders and come with power outputs in the one-digit Megawatt region.[213] Small boats may use lorry diesel engines. Large ships use extremely efficient, low-speed two-stroke diesel engines. They can reach efficiencies of up to 55%. Unlike most regular diesel engines, two-stroke watercraft engines use highly viscous fuel oil.[1] Submarines are usually diesel-electric.[215]
The first diesel engines for ships were made by A. B. Diesels Motorer Stockholm in 1903. These engines were three-cylinder units of 120 PS (88 kW) and four-cylinder units of 180 PS (132 kW) and used for Russian ships. In World War I, especially submarine diesel engine development advanced quickly. By the end of the War, double acting piston two-stroke engines with up to 12,200 PS (9 MW) had been made for marine use.[218]
### Aviation
Diesel engines had been used in aircraft before World War II, for instance, in the rigid airship LZ 129 Hindenburg, which was powered by four Daimler-Benz DB 602 diesel engines,[219] or in several Junkers aircraft, which had Jumo 205 engines installed.[93] Until the late 1970s, there has not been any applications of the diesel engine in aircraft. In 1978, Karl H. Bergey argued that “the likelihood of a general aviation diesel in the near future is remote.”[220] In recent years (2016), diesel engines have found use in unmanned aircraft (UAV), due to their reliability, durability, and low fuel consumption.[221] In early 2019, AOPA reported, that a diesel engine model for general aviation aircraft is “approaching the finish line.”[222]
### Non-road diesel engines
Air-cooled diesel engine of a 1959 Porsche 218
Non-road diesel engines are commonly used for construction equipment. Fuel efficiency, reliability and ease of maintenance are very important for such engines, whilst high power output and quiet operation are negligible. Therefore, mechanically controlled fuel injection and air-cooling are still very common. The common power outputs of non-road diesel engines vary a lot, with the smallest units starting at 3 kW, and the most powerful engines being heavy duty lorry engines.[213]
### Stationary diesel engines
Three English Electric 7SRL diesel-alternator sets being installed at the Saateni Power Station, Zanzibar 1955
Stationary diesel engines are commonly used for electricity generation, but also for powering refrigerator compressors, or other types of compressors or pumps. Usually, these engines run permanently, either with mostly partial load, or intermittently, with full load. Stationary diesel engines powering electric generators that put out an alternating current, usually operate with alternating load, but fixed rotational frequency. This is due to the mains' fixed frequency of either 50 Hz (Europe), or 60 Hz (United States). The engine's crankshaft rotational frequency is chosen so that the mains' frequency is a multiple of it. For practical reasons, this results in crankshaft rotational frequencies of either 25 Hz (1500 per minute) or 30 Hz (1800 per minute).[223]
## Low heat rejection engines
A special class of prototype internal combustion piston engines has been developed over several decades with the goal of improving efficiency by reducing heat loss.[224] These engines are variously called adiabatic engines; due to better approximation of adiabatic expansion; low heat rejection engines, or high temperature engines.[225] They are generally piston engines with combustion chamber parts lined with ceramic thermal barrier coatings.[226] Some make use of pistons and other parts made of titanium which has a low thermal conductivity[227] and density. Some designs are able to eliminate the use of a cooling system and associated parasitic losses altogether.[228] Developing lubricants able to withstand the higher temperatures involved has been a major barrier to commercialization.[229]
## Future developments
In mid-2010s literature, main development goals for future diesel engines are described as improvements of exhaust emissions, reduction of fuel consumption, and increase of lifespan (2014).[230][162] It is said that the diesel engine, especially the diesel engine for commercial vehicles, will remain the most important vehicle powerplant until the mid-2030s. Editors assume that the complexity of the diesel engine will increase further (2014).[231] Some editors expect a future convergency of diesel and Otto engines' operating principles due to Otto engine development steps made towards homogeneous charge compression ignition (2017).[232]
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213. ^ a b c Konrad Reif (ed.): Dieselmotor-Management im Überblick. 2nd edition. Springer, Wiesbaden 2014, ISBN 978-3-658-06554-6. p. 12
214. ^ Günter P. Merker, Rüdiger Teichmann (ed.): Grundlagen Verbrennungsmotoren – Funktionsweise · Simulation · Messtechnik, 7th edition, Springer, Wiesbaden 2014, ISBN 978-3-658-03194-7, p. 284
215. ^ a b Richard van Basshuysen (ed.), Fred Schäfer (ed.): Handbuch Verbrennungsmotor: Grundlagen, Komponenten, Systeme, Perspektiven, 8th edition, Springer, Wiesbaden 2017, ISBN 978-3-658-10901-1. p. 1289
216. ^ Hans Kremser (auth.): Der Aufbau schnellaufender Verbrennungskraftmaschinen für Kraftfahrzeuge und Triebwagen. In: Hans List (ed.): Die Verbrennungskraftmaschine. Vol. 11. Springer, Wien 1942, ISBN 978-3-7091-5016-0 p. 22
217. ^ Hans Kremser (auth.): Der Aufbau schnellaufender Verbrennungskraftmaschinen für Kraftfahrzeuge und Triebwagen. In: Hans List (ed.): Die Verbrennungskraftmaschine. Vol. 11. Springer, Wien 1942, ISBN 978-3-7091-5016-0 p. 23
218. ^ Günter Mau: Handbuch Dieselmotoren im Kraftwerks- und Schiffsbetrieb, Vieweg (Springer), Braunschweig/Wiesbaden 1984, ISBN 978-3-528-14889-8. pp. 9–11
219. ^ Kyrill von Gersdorff, Kurt Grasmann: Flugmotoren und Strahltriebwerke: Entwicklungsgeschichte der deutschen Luftfahrtantriebe von den Anfängen bis zu den internationalen Gemeinschaftsentwicklungen, Bernard & Graefe, 1985, ISBN 9783763752836, p. 14
220. ^ Karl H. Bergey: Assessment of New Technology for General Aviation Aircraft, Report for U.S. Department of Transportation, September 1978, p. 19
221. ^ Rik D Meininger et al.: Knock criteria for aviation diesel engines, International Journal of Engine Research, Vol 18, Issue 7, 2017, doi/10.1177
222. ^ AOPA: EPS gives certification update on diesel engine, 23 January 2019. Retrieved 1 November 2019.
223. ^ Helmut Tschöke, Klaus Mollenhauer, Rudolf Maier (ed.): Handbuch Dieselmotoren, 8th edition, Springer, Wiesbaden 2018, ISBN 978-3-658-07696-2, p. 1066
224. ^ "Browse Papers on Adiabatic engines : Topic Results". topics.sae.org. SAE International. Archived from the original on August 23, 2017. Retrieved April 30, 2018.
225. ^ Schwarz, Ernest; Reid, Michael; Bryzik, Walter; Danielson, Eugene (March 1, 1993). "Combustion and Performance Characteristics of a Low Heat Rejection Engine". SAE Technical Paper Series. 1. doi:10.4271/930988 – via papers.sae.org.
226. ^ Bryzik, Walter; Schwarz, Ernest; Kamo, Roy; Woods, Melvin (March 1, 1993). "Low Heat Rejection From High Output Ceramic Coated Diesel Engine and Its Impact on Future Design". SAE Technical Paper Series. 1. doi:10.4271/931021 – via papers.sae.org.
227. ^ Danielson, Eugene; Turner, David; Elwart, Joseph; Bryzik, Walter (March 1, 1993). "Thermomechanical Stress Analysis of Novel Low Heat Rejection Cylinder Head Designs". SAE Technical Paper Series. 1. doi:10.4271/930985 – via papers.sae.org.
228. ^ Nanlin, Zhang; Shengyuan, Zhong; Jingtu, Feng; Jinwen, Cai; Qinan, Pu; Yuan, Fan (March 1, 1993). "Development of Model 6105 Adiabatic Engine". SAE Technical Paper Series. 1. doi:10.4271/930984 – via papers.sae.org.
229. ^ Kamo, Lloyd; Kleyman, Ardy; Bryzik, Walter; Schwarz, Ernest (February 1, 1995). "Recent Development of Tribological Coatings for High Temperature Engines". SAE Technical Paper Series. 1. doi:10.4271/950979 – via papers.sae.org.
230. ^ Günter P. Merker, Rüdiger Teichmann (ed.): Grundlagen Verbrennungsmotoren – Funktionsweise · Simulation · Messtechnik, 7th edition, Springer, Wiesbaden 2014, ISBN 978-3-658-03194-7, p. 58
231. ^ Günter P. Merker, Rüdiger Teichmann (ed.): Grundlagen Verbrennungsmotoren – Funktionsweise · Simulation · Messtechnik, 7th edition, Springer, Wiesbaden 2014, ISBN 978-3-658-03194-7, p. 273
232. ^ Cornel Stan: Thermodynamik des Kraftfahrzeugs: Grundlagen und Anwendungen – mit Prozesssimulationen, Springer, Berlin/Heidelberg 2017, ISBN 978-3-662-53722-0. p. 252
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2020-06-02 05:14:40
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https://www.gamedev.net/forums/topic/689186-unique-lock-performance-difference-on-vs2017-what-gives/
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# unique_lock performance difference on VS2017: What Gives?
## Recommended Posts
I've had a pretty rough week;
My hard drive crashed. Luckily I was able to recover my data and learned an important (and expensive) lesson on backing up your data. But I needed to get a new hard drive and a Windows 10 license. After getting all my ducks in a row in regards to my data, I decided to upgrade my Visual Studio to the latest, seeing as I was working with 2012. This seems to have caused a huge problem.
Prior to this, my little game ran smoothly. Now however, it is an unplayable slideshow. The culprit appears to be my audio system. This system has an update method which loops through my sources and streams audio to them if need be. This function is called constantly in a separate thread, using a unique_lock and a mutex. Playing/Pausing/Stopping audio or altering audio source properties would make use of the unique_lock and mutex as well.
I have a moving audio source in a test scene. As such, it updates its position constantly, grabbing the lock. Before the crash the program ran perfectly fine; now its pretty much frozen. I was aware that contests for mutexes are somewhat slow, but now its having issues for entire seconds. And the longer the program goes on, the more slow it seems to get. Removing the mutex makes the game run smoother, but of course errors abound that way.
I am driving myself mad trying to suss out what's happening here. Things were perfectly fine a week ago. What can I do?
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Locks are often tricky to get right. There are many different types and levels of locking, and they have different performance on different systems.
Part of it is like you mentioned, obtaining a lock can be slow, and different types of mutex resources can cause performance hiccups. Some have little overhead like thread barriers in parent/child threads, or very short spinlocks. Others are far more time consuming, like an OS-wide global mutex shared between all programs, where one lock can be far-reaching and affect many other processes.
But usually slowdowns like that come from lock contention. Processes are waiting around for resources, blocked by locks. Assuming you are on Windows with Visual Studio, there is an optional Visual Studio component called the Concurrency Visualizer. I found a brief guide to using it on MSDN here, it is a mix of traditional profiling plus some information about threading and locks.
Once you've found the culprits with the profiling tools, fixing them is going to be specific to the code and the system you are on. Once you understand the specific cause then questions about resolving specific conflicts are more easily answered.
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Locks are often tricky to get right. There are many different types and levels of locking, and they have different performance on different systems.
Part of it is like you mentioned, obtaining a lock can be slow, and different types of mutex resources can cause performance hiccups. Some have little overhead like thread barriers in parent/child threads, or very short spinlocks. Others are far more time consuming, like an OS-wide global mutex shared between all programs, where one lock can be far-reaching and affect many other processes.
But usually slowdowns like that come from lock contention. Processes are waiting around for resources, blocked by locks. Assuming you are on Windows with Visual Studio, there is an optional Visual Studio component called the Concurrency Visualizer. I found a brief guide to using it on MSDN here, it is a mix of traditional profiling plus some information about threading and locks.
Once you've found the culprits with the profiling tools, fixing them is going to be specific to the code and the system you are on. Once you understand the specific cause then questions about resolving specific conflicts are more easily answered.
Yes, I had assumed it was lock contention. It never happened before, which was confusing. Unfortunately, the plugin you suggest is not available in my version of Visual Studio. But funnily enough, I believe I may have found a solution. For reference, here's my update code for the audio:
void SoundPlayer::update() {
std::unique_lock<std::mutex> lock(mutex);
for (int i = 0; i < MAX_SOURCES; i++) {
ALenum state;
unsigned int currentID = sourceIDs[i];
alGetSourcei(currentID, AL_SOURCE_STATE, &state);
if (state == AL_STOPPED) {
clearSource(currentID); //this helps when destroying sound effects in SoundAssetLibrary
}
if (state != AL_PLAYING) {
continue;
}
int processed;
bool active = true;
alGetSourcei(currentID, AL_BUFFERS_PROCESSED, &processed);
while (processed > 0) {
unsigned int bufferID;
alSourceUnqueueBuffers(currentID, 1, &bufferID);
if (soundStreams[i]) {
soundStreams[i]->fillBuffer(bufferID);
alSourceQueueBuffers(currentID, 1, &bufferID);
}
processed--;
checkForErrors();
}
checkForErrors();
}
}
I was using vectors to contain my source ids and such prior to my upgrade. Looking at the handy profiling tools my Visual Studio does include, it turns out I was spending a lot of my time in my vector. So, out of desperation and curiosity, I changed them from vectors to std::arrays. A perfectly acceptable change seeing as the number of sources and other data should be constant. It works perfectly now.
I wish I could understand.
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A good rule of thumb to follow when working with locks is to try and minimize their scope as much as possible. While putting a lock at the top of the update may technically work, you're spending way more time in the lock than you really need to and potentially increasing contention.
From what I can tell, the only shared state in the update is sourceIDs (and possibly soundStreams but I don't know enough about your audio system to say for sure). Instead of locking the entire update, make a copy of the sourceIDs array/vector at the top of the update from within a lock scope and use that instead. When you clear the source, instead of clearing it right away, add it to a 'to-be-cleared' list and remove them all at once at the end of the update (again from inside the lock). This limits the scope of the lock to when you actually need to interact with the shared state, and allows the rest of the update method to take as long as it needs to without impacting the performance of other systems.
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I was using vectors to contain my source ids and such prior to my upgrade. Looking at the handy profiling tools my Visual Studio does include, it turns out I was spending a lot of my time in my vector. So, out of desperation and curiosity, I changed them from vectors to std::arrays. A perfectly acceptable change seeing as the number of sources and other data should be constant. It works perfectly now.
I wish I could understand.
It sounds like you're testing in debug mode, and modern versions of Visual C++ perform a lot of debug checks on vector accesses.
##### Share on other sites
A good rule of thumb to follow when working with locks is to try and minimize their scope as much as possible. While putting a lock at the top of the update may technically work, you're spending way more time in the lock than you really need to and potentially increasing contention.
From what I can tell, the only shared state in the update is sourceIDs (and possibly soundStreams but I don't know enough about your audio system to say for sure). Instead of locking the entire update, make a copy of the sourceIDs array/vector at the top of the update from within a lock scope and use that instead. When you clear the source, instead of clearing it right away, add it to a 'to-be-cleared' list and remove them all at once at the end of the update (again from inside the lock). This limits the scope of the lock to when you actually need to interact with the shared state, and allows the rest of the update method to take as long as it needs to without impacting the performance of other systems.
This solution also works, and is probably the better one. Thanks for your help!
I was using vectors to contain my source ids and such prior to my upgrade. Looking at the handy profiling tools my Visual Studio does include, it turns out I was spending a lot of my time in my vector. So, out of desperation and curiosity, I changed them from vectors to std::arrays. A perfectly acceptable change seeing as the number of sources and other data should be constant. It works perfectly now.
I wish I could understand.
It sounds like you're testing in debug mode, and modern versions of Visual C++ perform a lot of debug checks on vector accesses.
I actually did test it in release mode. Same results. But perhaps other things have changed I am unaware of. I was working on a pretty old version.
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That's really weird, because there's no reason that merely reading from a vector should have been any slower than reading from an array.
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That's really weird, because there's no reason that merely reading from a vector should have been any slower than reading from an array.
Agreed. I'm not going to pretend I understand why it worked (especially since I am now intending on using another solution), but after all that crap happened I was happy just to have things up and running again.
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On my phone so I'm gonna half-ass this - look up Iterator Debugging. Yes it impacts release builds.
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2018-04-21 01:49:59
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http://mathhelpforum.com/pre-calculus/55780-three-lines-concurrent.html
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# Math Help - Are the three lines concurrent?
1. ## Are the three lines concurrent?
Are the above three lines concurrent?
2. Hello,
Originally Posted by fardeen_gen
Are the above three lines concurrent?
recall the formula :
$\log_a b=\frac{\ln b}{\ln a}$ ( $a>0$ and $a \neq 1$)
So the equations are now :
$\begin{array}{lllll} x+\frac{y \ln b}{\ln a}+\frac{\ln c}{\ln a}=0 \\ \\
\frac{x \ln a}{\ln b}+y+\frac{\ln c}{\ln b}=0 \\ \\
\frac{x \ln a}{\ln c}+\frac{y \ln b}{\ln c}+1=0 \end{array}$
Multiply the first one by ln(a), the second one by ln(b), the third one by ln(c) and you'll see that they're identical. So the lines are the same.
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2016-07-24 03:40:04
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https://cuhkmath.wordpress.com/2013/04/13/pohozaev-schoen-identity/
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## Pohozaev-Schoen identity
In this short note I record a proof of the Pohozaev-Schoen identity (see this paper of Schoen, Prop 1.4). This proof is actually standard and I put it here for my own benefit. If I have time I will give some applications of this identity later.
Theorem 1 (Pohozaev-Schoen identity) Let ${(M^n,g)}$ ${(n> 2)}$ be a compact Riemannian manifold (possibly with boundary) and ${X}$ is a vector field on ${M}$, then $\displaystyle \int_M X(R)= -\frac{n}{n-2}\int_M \langle \stackrel\circ {\mathrm{Ric}},\mathcal {L}_X g\rangle+\frac{2n}{n-2} \int_{\partial M} \stackrel \circ {\mathrm{Ric}}(X, \nu).$ where ${R}$ is the scalar curvature, ${\stackrel\circ {\mathrm{Ric}}}$ is the traceless Ricci tensor and $\mathcal L_X g$ is the Lie derivative of $g$.
Proof:
We will calculate using a local orthnormal frame. Let ${\stackrel\circ{\mathrm {Ric}}= \mathrm{Ric}-\frac{R}{n}g}$ be the traceless Ricci tensor. By the (twice contracted) second Bianchi identity $R_{ij;j}=\frac 12 R_{;i}$, we have
$\displaystyle \begin{array}{rcl} \stackrel\circ R_{ij;j}= R_{ij;j}-\frac{1}{n}R_{;i}= \frac{1}{2}R_{;i}-\frac{1}{n}R_{;i}=\frac{n-2}{2n}R_{;i}. \end{array}$
Therefore by integration by parts,
$\displaystyle \begin{array}{rcl} \int_M X(R ) =\int_M X_i R_{;i}= \frac{2n}{n-2}\int_M \stackrel\circ R_{ij;j} X_i &=\frac{2n}{n-2}\left(-\int_M \stackrel\circ R_{ij}X_{i;j} + \int_{\partial M} \stackrel \circ R_{ij}X_i \nu_j\right). \end{array}$
As ${ \stackrel \circ R_{ij}}$ is a symmetric tensor, we have
$\displaystyle \stackrel \circ R_{ij}X_{i;j}= \frac{1}{2} \stackrel\circ R_{ij}(X_{i;j}+X_{j;i})= \frac{1}{2}\stackrel\circ R_{ij}(\mathcal L_X g)_{ij}.$
(Here we have used the fact that $(\mathcal L_X g)(Y,Z)=\langle \nabla_Y X, Z\rangle + \langle \nabla _Z X, Y\rangle$ (Exercise). ) Putting this into the above equation, we have
$\displaystyle \int_M X(R)= -\frac{n}{n-2}\int_M \langle \stackrel\circ {\mathrm{Ric}},\mathcal {L}_X g\rangle+\frac{2n}{n-2} \int_{\partial M} \stackrel \circ {\mathrm{Ric}}(X, \nu).$
$\Box$
Corollary 2 If ${X}$ is a conformal vector field (see also here), i.e. ${\mathcal L_X g= f g}$ for some function ${f}$ on ${M}$, then $\displaystyle \int_M X(R)= \frac{2n}{n-2} \int_{\partial M} \stackrel \circ {\mathrm{Ric}}(X, \nu).$ In particular, ${\int_M X(R)=0}$ if ${M}$ is closed.
Remark 1 Theorem 1 can be used to prove the DeLellis-Topping inequality.
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This entry was posted in Calculus, Geometry. Bookmark the permalink.
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2018-03-17 16:23:12
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https://portal.beam.ltd.uk/greensim/
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# GreenSim
## Terry Barnaby - Rotary Racer - Chipping Sodbury School
### Introduction
The Greensim application is a simple web based application written in Python to simulate racing a Greenpower electric car around a track. The idea is to provide a basic test bed so car ideas can be played with together with race tactics. Hopefully the Simulator will be developed and tested against real car behavior to make it better. There is also a simple Car Forces and Motor Temperature calculator at the following link:
Some information on the Science and Maths of a Greenpower car is at the following link:
### Features
• Easy to use web interface
• Simulates: Battery, Motor, AirDrag, RollingDrag, PitStops, BatteryChange/Charging, Wind
### Usage
The The Greenpower Car Simulator web page provides a simple interface to the simulator program. At the top of the page is a set of parameters than can be entered. Below these parameters is a "Run Simulation" button. Pressing this runs the simulation with the given parameters and produces the Lap timing results together with graphs or various interesting parameters. Parameters are:
TrackLength The track length in Meters. (Goodwood: 3862.0m, Castle Combe: 2977m) RaceDuration The race duration in seconds. (4 Hours = 14400, 6 Hours = 21600) CarWeight The cars weight, without driver in Kg. CarFrontArea The cars frontal area in Meters squared. carFrictionCoeff The cars rolling friction coefficient. carAirDragCoeff The cars aerodynamic coefficient. battery0Capacity This is the capacity of battery 0 at the 20hour rate. battery0KValue This is the "K" value for battery 0. (A measure of the battery quality/age) battery1Capacity This is the capacity of battery 1 at the 20hour rate. battery1KValue This is the "K" value for battery 1. (A measure of the battery quality/age) motorGearing The motor gear ratio. (for 17:57 this is 3.353) motorWheelDiameter The diameter of the driving wheel in Meters. driveEfficiency The efficiency of the drive chain/gears etc. Motor efficiency is taken into account with motor simulation. pitStopTime The pitstop time in seconds. batteryChargeRate The battery charge current used in pitstops. periods The pitstop periods with number of laps and driver weight in Kg. Battery is changed on each pitStop. graphZoom Zooms the graphs to see more detail.
### Simulation Notes
These notes refer to Greensim Version 1.0.
### Car Dynamics
The Car dynamics has been calculated using the following equations:
• Rolling resistance, in Newtons, was calculated using: "rollingResistance = carFrictionCoeff * mass * 9.8".
• Air Drag, in Newtons, was calculated using: "airDrag = 0.5 * airDensity * carAirDragCoeff * carFrontArea * (v * v)".
These equations should give reasonable results, but the carFrictionCoeff and carAirDragCoeff will need to be determined by testing the car or by playing with the figures in the simulator such that it matches a real race for the car.
### Motor simulation
The Motor simulator is reasonably accurate and is based on the Power Curve graphs available here. It may not simulate motor power variations due to Battery voltage correctly however.
torque = 13.5 - (motorRpm * 13.5 / 2000.0)
torque = torque * voltage / 24.0;
current = 2.0 + (torque * 128.0 / 13.5)
### Battery Discharge
The Battery simulator uses the following algorithms.
$t = \frac{H}{\left(\frac{I H}{C}\right)^k}\,$
T = H / (I * H / C)^k
T - Discharge time in hours
H - Battery discharge rate time (20 hour rate)
I - Discharge current
C - Battery capacity at stated discharge rate (75 amp/hour)
k - battery constant between 1.1 and 1.3 depending on battery type and condition.
For use in calculation a discharge value over a time period:
c = I * T * pow(I / Ir, k - 1)
c - Amount of charge to take away from total charge
I - Current
T - Time period in hours
Ir - Current for battery amp/hour rating (For 75Amp/hour at 20hour rate = 3.75)
k - battery constant between 1.1 and 1.3 depending on battery type and condition.
This is taken from info at: http://en.wikipedia.org/wiki/Peukert's_law
This looks like it is reasonably accurate for the batteries we use based on the battery discharge tests performed. A k value of about 1.11 works for the Yuasa Elite batteries. Battery charging is not well simulated. We need to know more about the battery chargers to get this side of the simulation to be accurate. The simulation charges at 10Amps and assumes 75% efficiency in doing this.
There is basic simulation of Battery "recovery" by setting the charge factor to a low number.
The Battery voltage is calculated using:
Et = Eo - Ri*I + Ki*log(Ct/C)
Where
• Et = battery terminal voltage [volts]
• Eo = open circuit voltage of a battery cell when fully charged [volts]
• Ri = internal (ohmic) resistance of the battery [ohms]
• Ki = polarization resistance [ohms]
• I = instantaneous current [amps]
• C = current charge level in amp/hours
• Ct = total battery charge in amp/hours
### Simulation Notes
• Does not simulate hills.
• Does not simulate driver skills or reliability !
• Does not simulate temperature variations, including motor efficiency reduction when it gets hot.
### Future Ideas
• UserId capability to store results. This would allow the boys to compete at getting the best Lap times from the simulator.
• Could simulate: Motor speed control, GearBox, uphill/downhill.
### Basic Design
• Simple module system with one module per simulation entity.
• Creates new data in fixed time steps.
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2021-06-17 17:04:02
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http://mathhelpforum.com/geometry/129704-locus-print.html
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# Locus
• Feb 19th 2010, 11:38 PM
lance
Locus
Determine the equation of the locus of a point which moves so that its distance from P(5,2) is two-thirds its distance from the x-axis.
• Feb 20th 2010, 12:47 AM
Prove It
Quote:
Originally Posted by lance
Determine the equation of the locus of a point which moves so that its distance from P(5,2) is two-thirds its distance from the x-axis.
Hint: a parabola is the locus of points that are equidistant from a point (the focus) and a line (the directrix). I'd advise you to research how to find the equation of a parabola using the focus and directrix.
• Feb 20th 2010, 11:57 AM
qmech
Quote:
Originally Posted by lance
Determine the equation of the locus of a point which moves so that its distance from P(5,2) is two-thirds its distance from the x-axis.
$\sqrt( (x-5)^2+(y-2)^2)$
$|y|$
$\sqrt( (x-5)^2+(y-2)^2) = \frac{2}{3}|y|$
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2018-02-23 01:00:25
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https://andytimm.github.io/posts/Variational%20MRP%20Pt1/variational_mrp_pt1.html
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# Variational Inference for MRP with Reliable Posterior Distributions
Introductions- things to do, places to be
MRP
BART
Variational Inference
Author
Andy Timm
Published
October 10, 2022
This post introduces a series I intend to write, exploring using Variational Inference to massively speed up running complex survey estimation models like variants of Multilevel Regression and Poststratification while aiming to keep approximation error from completely ruining the model.
The rough plan for the series is as follows:
1. (This post) Introducing the Problem- Why is VI useful, why VI can produce spherical cows
2. How far does iteration on classic VI algorithms like mean-field and full-rank get us?
3. Some theory on why posterior approximation with VI can be so poor
4. Seeing if some more sophisticated techniques like normalizing flows help
# Motivation for series
I learn well by explaining things to others, and I’ve been particularly excited to learn about variational inference and ways to improve it over the past few months. There are lots of Bayesian models I would like to fit, especially in my political work, that I would categorize as being incredibly useful, but on the edge of practically acceptable run times. For example, the somewhat but not particularly complex model I’ll use as a running example for the series takes ~8 hours to fit on 60k observations.
Having a model run overnight or for a full work day can be fine sometimes, but what if there is a more urgent need for the results? What if we need to iterate to find the “right” model? What if the predictions from this model need to feed into a later one? How constrained do we feel about adding just a little bit more complexity to the model, or increasing our N size just a bit more?
If we can get VI to fit well, we can make complex Bayesian models a lot more practical to use in a wider variety of scenarios, and maybe even extend the complexity of what we can build given time and resource constraints.
# Spherical Cow Sadness
##### I’ve got that…
If VI can make Bayesian inference much faster, what’s the catch? The above two images encapsulate the problem pretty well. First, as the left screenshot from rstanarm’s documentation shows, variational inference requires a (bold text warning requiring) set of approximating distribution choices in order to be tractable to optimize. On the right, in their survey paper on VI, Blei et al. (2018) are showing one of the potential posterior distorting consequences of our choice to approximate.
So stepping back for a second, we’ve taken a problem for which there’s usually no closed form solution (Bayesian inference), where even the best approximation algorithm we can usually use (MCMC) isn’t always enough for valid inference without very careful validation and tinkering. Then we decided our approximation could do with being more approximate.
That was perhaps an overly bleak description, but it should give some intuition why this is a hard problem. We want to choose some method of approximating our posterior such that it is amenable to optimization-based solving instead of requiring sampling, but not trade away our ability to correctly understand the full complexity of the posterior distribution1.
# Introducing MRP and our running example
## Introducing MRP
While I’m mostly focused on the way we choose to actually fit a given model with this series, here’s a super quick review of the intuition in building a MRP model. If you want a more complete introduction, Kastellec’s MRP Primer is a great starting point, as are the case studies I link a bit later.
MRP casts estimation of a population quantity of interest \theta as a prediction problem. That is, instead of the more traditional approach of building simple raked weights and using weighted estimators, MRP leans more heavily on modeling and then poststratification to make the estimates representative.
To sketch out the steps-
1. Either gather or run a survey or collection of surveys that collect both information on the outcome of interest, y, and a set of demographic and geographic predictors, \left(X_{1}, X_{2}, X_{3}, \ldots, X_{m}\right).
2. Build a poststratification table, with population counts or estimated population counts N_{j} for each possible combination of the features gathered above. Each possible combination j is called a cell, one of J possible cells. For example, if we poststratified only on state, there would be J=51 (with DC) total cells; in practice, J is often several thousand.
3. Build a model, usually a Bayesian multilevel regression, to predict y using the demographic characteristic from the survey or set of surveys, estimating model parameters along the way.
4. Estimate y for each cell in the poststratification table, using the model built on the sample.
5. Aggregate the cells to the population of interest, weighting by the N_{j}’s to obtain population level estimates: \theta_{\mathrm{POP}}=\frac{\sum_{j \in J} N_{j} \theta_{j}}{\sum_{j \in J} N_{J}}
Why would we want to do this over building more typical survey weights? To the extent your new model has desirable properties like the ability to incorporate priors, can partially pool to manage rare subpopulations where you don’t have a lot of sample, and so on, you can get the benefits of that more efficient model through MRP. Raking in its simplest form is really just a linear model; we have plenty of methods that can do better. Outside of bayesian multilevel models which are the most common, there’s an increasing literature on using a wide variety of machine learning algorithms like BART2 to do the estimation stage; Andrew Gelman calls this RRP.
## Introducing the Running Example
Rather than reinvent the wheel, I’ll follow the lead of the excellent Multilevel Regression and Poststratification Case Studies by Lopez-Martin, Philips, and Gelman, and model survey binary responses from the 2018 CCES for the following question:
Allow employers to decline coverage of abortions in insurance plans (Support / Oppose)
From the CCES, we get information on each participant’s state, age, gender, ethnicity, and education level. Supplementing this individual level data, we also include region flags for each state, and Republican vote share in the 2016 election- these state level predictors have been shown to be critical for getting strong MRP estimates by Lax and Philips (2009) and others. and If you’d like deeper detail on the dataset itself, I’d refer you to this part MRP case study.
Using these, we setup the model for Pr(y_i = 1) the probability of supporting allowing employers to decline coverage of abortions in insurance plans as:
\begin{aligned} Pr(y_i = 1) =& logit^{-1}( \gamma^0 + \alpha_{\rm s[i]}^{\rm state} + \alpha_{\rm a[i]}^{\rm age} + \alpha_{\rm r[i]}^{\rm eth} + \alpha_{\rm e[i]}^{\rm educ} + \beta^{\rm male} \cdot {\rm Male}_{\rm i} \\ &+ \alpha_{\rm g[i], r[i]}^{\rm male.eth} + \alpha_{\rm e[i], a[i]}^{\rm educ.age} + \alpha_{\rm e[i], r[i]}^{\rm educ.eth} + \gamma^{\rm south} \cdot {\rm South}_{\rm s} \\ &+ \gamma^{\rm northcentral} \cdot {\rm NorthCentral}_{\rm s} + \gamma^{\rm west} \cdot {\rm West}_{\rm s} + \gamma^{\rm repvote} \cdot {\rm RepVote}_{\rm s}) \end{aligned}
Where we incorporate pretty much all of our predictors as varying intercepts to allow for pooling across demographic and geographic characteristics:
• \alpha_{\rm a}^{\rm age}: The effect of subject i’s age on the probability of supporting the statement.
• \alpha_{\rm r}^{\rm eth}: The effect of subject i’s ethnicity on the probability of supporting the statement.
• \alpha_{\rm e}^{\rm educ}: The effect of subject i’s education on the probability of supporting the statement.
• \alpha_{\rm s}^{\rm state}: The effect of subject i’s state on the probability of supporting the statement.
• \beta^{\rm male}: The average effect of being male on the probability of supporting abortion. Note that it doesn’t really make much sense to model a two category3 factor as a varying intercept.
• \alpha_{\rm e,r}^{\rm male.eth}, \alpha_{\rm e,r}^{\rm educ.age}, \alpha_{\rm e,r}^{\rm educ.eth}: Are several reasonable guesses at important interactions for this question. We could add many more two way, or even some three way interactions here, but this is enough for my testing here.
• \gamma^{\rm south}, \gamma^{\rm northcentral}, \gamma^{\rm west},\gamma^{\rm repvote}: are the state level predictors which are not represented as varying intercepts. Following the case study, I use \gamma’s for the state level coefficients, keeping \beta’s for individual coefficients. Note that Northeast is the base region of the region factor here, so it doesn’t get it’s own coefficient.
Stepping back for a second, let’s describe the complexity of this model in more general terms. This certainly isn’t state of the art for MRP, and you could definitely add in things like a lot more interactions, some varying slopes, non-univariate prior and/or structured priors, or other elements to make this a more interesting model. That said, this is already clearly enough of a model to improve on simple raking in many cases, and it produces a nuanced enough posterior that we can feasibly imagine a bad approximation going all spherical cow shaped on us.
Why this dataset and this model for this series? The question we model itself isn’t super important- as long as we can expect some significant regional and demographic variation in the outcome we’ll be able to explore if VI smoothes away some posterior complexity that MCMC can capture. Drawing an example from the CCES is quite useful, as the 60k total sample is much larger than typical publicly available surveys, and so we can check behavior under larger N sizes. Practically, fitting this with rstanarm allows us to switch easily from a great MCMC implementation to a decent VI optimizer quickly for some early tests. Finally, the complexity and runtime of the model is a nice balance of being something that we can fit with MCMC in a not terrible amount of time for comparison’s sake, and something challenging enough that it should teach us something about VI’s ability to handle non-toy models of the world.
Fitting this4 with MCMC in rstanarm is as simple as:
# Fit in stan_glmer
fit <- stan_glmer(abortion ~ (1 | state) + (1 | eth) + (1 | educ) + male +
(1 | male:eth) + (1 | educ:age) + (1 | educ:eth) +
repvote + factor(region),
family = binomial(link = "logit"),
data = cces_df,
prior = normal(0, 1, autoscale = TRUE),
prior_covariance = decov(scale = 0.50),
adapt_delta = 0.99,
refresh = 0,
seed = 605)
Since it isn’t relevant for the rest of my discussion here, I’ll summarize the model diagnostics here and say that this seems to be a pretty reasonable fit- no issues with divergences, and no issues with poor \hat{r}’s. Worth quickly pointing out that we did have to tune adapt_delta a bit to get no divergences though- even before getting to fitting this with VI, a model like this requires some adjustments to fit correctly.
With a model like this on just a 5k sample, we can produce pretty solid state level predictions that have clearly benefited from being fit with a Bayesian multilevel model:
With a 5k sample, MRP lands much closer to the complete weighted survey than a 5k unweighted sample: neat. That’s certainly not a fully fair comparison, but it gives some intution around the promise of this approach.
Somewhat less neat is that even a 5k sample here takes about 13 minutes to fit. How does this change as we fit on more and more of the data?
Sample Size Runtime
5,000 13 minutes
10,000 44 minutes
60,000 526 minutes (~8 hours!)
As the table above should illustrate, if you’re fitting a decently complex Bayesian model on even somewhat large N sizes, you’re pretty quickly going to cap out what you can reasonably fit in a acceptable amount of time. If you’re scaling N past the above example, or deepening the modeling complexity, you’ll pretty quickly feel effectively locked out of using these models in fast-paced environments.
Hopefully fitting my running example has helped for building intuition here. Even a reasonably complex Bayesian model can have some pretty desirable estimation properties. To make iterating on modelling choices faster, to scale our N or model complexity higher, or just to use a model like this day to day when time matters, we’d really like to scale these fitting times back. Can Variational Inference help?
# Introducing Variational Inference
I’ve gotten relatively far in this post without clearly explaining what Variational Inference is, and why it might provide a more efficient and scalable way to fix large Bayesian models. Let’s fully flesh that out here to ground the rest of the series.
In the bigger picture, pretty much all of our efforts in Bayesian inference are a form of approximate inference. Almost no models we care about for real world applications have closed form solutions- conjugate prior type situations are a math problem for stats classes, not a general tool for inference.
Following Blei et al. (2018)’s notation, let’s setup the general problem first, describe (briefly) how MCMC solves it, and then more slowly demonstrate how VI does. Let’s say we have some observations x_{1:N}, and and some latent variables that define the model z_{1:M}. Note for concreteness these latent variables represent our quantities of interest: key parameters and so on- we’re calling them latent in the sense that we can’t go out and directly measure a \beta or \gamma from the model above, we have to gather data that allows us to estimate them. We call p(z) priors, and they define our model prior to contact with the data. The goal of Bayesian inference then is conditioning on our data in order to get the posterior:
p(z|x) = \frac{p(z,x)}{p(x)}
If you’re reading this post series, it’s likely you recognize that the denominator on the right here (often called the “evidence”) is the sticking point; the integral p(x) = \int{p(z,x)dz} won’t have a closed form solution.
When we use Markov Chain Monte Carlo as we did above to estimate the model, we’re defining a Markov Chain on z, whose stationary distribution if we’ve done everything right is p(z|x). There are better and worse ways to do this certainly- the development of the Stan language, with associated Hamiltonian Monte Carlo with NUTS sampler has massively expanded what was possible to fit in recent years. However, while actively improving the speed and scalability of sampling is an active area of research (for example, by using GPU compute where possible), some of the speed challenges just seem a bit baked into the approach. For example, the sequential nature of markov chains makes parallelization within chains seem out of reach absent some as-yet unknown clever tricks.
Instead of sampling, variational inference asks what we’d need to figure out to treat the Bayesian inference problem as an optimization problem, where we could bring to bear all the tools for efficient, scalable, and parallelizable optimization we have developed.
Let’s start with the idea of a family of approximate densities \mathscr{Q} over our latent variables5.
Within that \mathscr{Q}, we want to try the best q(z), call it q^*(z), that minimizes the Kullback-Leibler divergence to the true posterior:
q^*(z) = argmin_{q(z) \in \mathscr{Q}}(q(z)||p(z|x))
If we choose a good \mathscr{Q}, managing the complexity so that it includes a density close to p(z|x), without becoming too slow or impossible to optimize, this approach may provide a significant speed boost.
To start working with this approach though, there’s one major remaining problem. Do you see it in the equation above?
## The ELBO
If you haven’t seen it yet, this quick substitution should clarify a potential issue with VI as I’ve described it so far:
q^*(z) = argmin_{q(z) \in \mathscr{Q}}(q(z)||\frac{p(z,x)}{\bf p(x)}) = \mathbb{E}[logq(z)] - \mathbb{E}[logp(z,x)] + {\bf logp(x)} Without some new trick, all I’ve said so far is to approximate a thing I can’t analytically calculate (the posterior, specially the issue evidence piece of it), I’m going to calculate the distance between my approximation and… the thing I said has a component can’t calculate?
Fortunately, a clever solution exists here that makes this strategy possible. Instead of trying to minimize the above KL divergence, we can optimize the alternative objective:
\mathbb{E}[logp(z,x)] - \mathbb{E}[logq(z)]
This is just the negative of the first two terms above, leaving aside the logp(x). Why can we treat maximizing this as minimizing the KL divergence? The logp(x) term is just a constant (with respect to q), so regardless of how we vary q, this will still be a valid alternative objective. We call this the Evidence Lower Bound (ELBO)6.
If it’s helpful for intuition, play around with this great interactive ELBO optimizer by Felix Köhler:
By twiddling the knobs on \mu and \sigma for our approximating normal, we can get our surrogate distribution pretty close to the True Posterior (which we know for purposes of demonstration, so we can calculate the true KL, not just it’s ELBO component). No matter how we twiddle though, the evidence remains constant.
For further intuition- notice that we can only do this trick in one direction. The KL divergence isn’t symmetrical, and if we wanted to calculate the “reverse” KL, we couldn’t use this strategy as logq(x) would not be a constant. Even if we thought that optimizing other direction of KL might have desirable properties like emphasizing mass-seeking over mode-seeking behavior, that simply isn’t an option.
# A first try at VI on this dataset
Ok, so we have an objective to optimize that should actually work. What’s a good \mathscr{Q}? The choice has been shown to matter a lot, but for purposes of a first swing here, let’s try one of the simpler ideas people have explored, the mean-field family. These latent variables will be assumed mutually independent7 and each get it’s own distinct factor in the variational density. A member of this would look something like:
q(z) = \prod_{j=1}^{m} q_j(z_j)
Each latent z_j get it’s own variational factor with density q_j(z_j), whose knobs we play with to maximize the ELBO. In the particular implementation below normal distributions are used, plenty of other options like t distributions are common too.
Probably not the best we can do, but let’s give it a roll. Since we’ve been told this will scale really well too supposedly, let’s use all 60k of the observations just to get a sense how it’ll compare to our 8+ hours in that case.
tic()
fit_60k <- stan_glmer(abortion ~ (1 | state) + (1 | eth) + (1 | educ) + male +
(1 | male:eth) + (1 | educ:age) + (1 | educ:eth) +
repvote + factor(region),
family = binomial(link = "logit"),
data = cces_all_df,
prior = normal(0, 1, autoscale = TRUE),
prior_covariance = decov(scale = 0.50),
adapt_delta = 0.99,
refresh = 0,
algorithm = "meanfield",
seed = 605)
toc()
This finishes in a blazing 144.03 seconds. Is this a good fit, or have we created a ridiculous spherical cow?
You’ll have to find out in the next post. Thanks for reading!
Typically, I’ll include links to code at the end of these posts, but since the only thing going on in this notebook is mentioning some runtimes of the models displayed inline at various sample sizes, I’m skipping that for now.
## Footnotes
1. If I were that type of Bayesian, this is where I’d complain that if we screw this up badly enough, we might as well be frequentists or worse, machine learning folk.↩︎
2. In grad school, using BART as the estimator (also combining it with some portions of the model being estimated as multilevel models) was the focus of my masters thesis. This pairs the best parts of relatively black box machine learning sensibility with the advantages of still having a truly Bayesian model. With comparatively minimal iteration you can get a pretty decent set of MRP models that will be better than many basic versions of multilevel models fit early in the MRP literature. Of course, if you’re willing to spend a bunch of time iterating on the absolute best models for a given problem, and incorporate lots of problem specific knowledge into model forms you can and should do better than BARP. Also, a lot of pretty cool things you can do like jointly model multiple question responses at the same time aren’t going to be easily to implement unless you get way in the weeds of your own BART implementation.↩︎
3. Insert snark about CCES folks doing a poor job at gender inclusivity despite 80+ researchers working on it here.↩︎
4. Again, see the MRP case studies linked above if you want see all the data prep and draw manipulation here; I’ll be leaving out most such details that aren’t relevant for comparisons to fitting this model with VI from now on.↩︎
5. In grad school, I had a friend who insisted on calling this “spicy Q”. For a while we had a latex package that made \spicy{} equivalent to \mathscr{}. Apologies for the footnote for the dumb LaTeX joke, but now I’m pretty sure you won’t have a sudden moment of “what is that symbol again” discussing VI ever.↩︎
6. Why is this a lower bound? Notice that we could write the evidence from above equations as logp(x) = KL(q(z)||p(z|x)) + ELBO(q). Since the KL divergence is non-negative (it’s zero when distributions p and q are identical), the ELBO is a lower bound of the evidence.↩︎
7. If this seems like it could go fully spherical cow, both literally in the sense that if we use a bunch of independent normals we make a sphere, and in the sense that this may not represent the full complexity of public opinion, you’re correct. Assuming independence here could very easily cause problems, and part of why this VI strategy is so challenging is the subset of things we can easily optimize doesn’t have the best overlap with fully realistic distributional assumptions over our latent variables.↩︎
## Citation
BibTeX citation:
@online{timm2022,
author = {Andy Timm},
title = {Variational {Inference} for {MRP} with {Reliable} {Posterior}
{Distributions}},
date = {2022-10-10},
url = {https://andytimm.github.io/variational_mrp_pt1.html},
langid = {en}
}
For attribution, please cite this work as:
Andy Timm. 2022. “Variational Inference for MRP with Reliable Posterior Distributions.” October 10, 2022. https://andytimm.github.io/variational_mrp_pt1.html.
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2023-02-02 01:17:43
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https://www.zbmath.org/?q=an%3A0945.37010
|
# zbMATH — the first resource for mathematics
Discrete chaos. (English) Zbl 0945.37010
Boca Raton, FL: Chapman & Hall/CRC. xiii, 355 p. (2000).
The last three decades witnessed a surge of research activities in deterministic chaos theory and fractals. So numerous books on these subjects have appeared, especially in the last decade. The book under review grew out of a course on discrete dynamical systems/difference equations taught by the author at the Trinity University (San Antonio, Texas), since 1992. The book intends “to give a thorough exposition of stability theory in one and two dimensions including the method of Lyapunov”. In other words the subject matter is according to the taste of the author. The same holds for the organization and the structure of the book.
Ch. 1 is about the stability of one dimensional maps. So it is surprising that without introducing the term of chaos the part 1.8. is titled “The period doubling route to chaos”. Interestingly enough the definition of chaos is given only in part 3.5 (p. 117)!
Ch. 2 is named “Sharkovsky’s theorem and bifurcation”. But it does not mean that one speaks about Sharkovsky’s bifurcation as such, but about bifurcations of 1-dim maps. Unfortunately, the author does not present a classification of such possible bifurcations. It is funny that the Schwarzian derivative is defined in 1.6.2 (Def. 1.3, p. 24) but also in 2.4. In this chapter the author cites the famous result of Li and Yorke “Period three implies chaos”, but unfortunately does not mention there the first rigorous definition of chaos. Only in Ch. 3 chaos is defined in the 1-dim case. But there the term chaos is defined according to the one introduced by R. L. Devaney in his book “An introduction to chaotic dynamical systems” (Addison-Wesley) of 1989 (!) (see Zbl 0695.58002). However, this is not a very fortunate way as it is shown later that three conditions for a map $$f$$ to be chaotic, namely $$f$$ is transitive; the set of periodic points of $$f$$ is dense; and the last $$f$$ has sensitive dependence on initial conditions, are interdependent.
Ch. 4 is mostly about the stability of 2-dim linear maps. Here also the Lyapunov function method for nonlinear maps is introduced. The next Ch. 5, named “Chaos in two dimensions” brings examples of chaotic dynamics, namely hyperbolic Anosov toral automorphism, Smale’s horseshoe and the Hénon map. Here again, one problem, namely the Hénon map is treated twice. It was already mentioned in Ch. 4 as example of a strange attractor. Again, the term ‘strange attractor’ has not been introduced in the book at all! Besides, in 5.3.1. the author says “computer iteration shows the existence of strange attractors, however, the proof of this statement continues to elude the mathematicians”. Unfortunately the author does not point out what is behind it, namely that the Hénon map is dissipative.
The last two chapters 6 and 7 are devoted to fractals and Julia and Mandelbrot sets. The treatment of fractals at the end of the book is also surprising as chaotic attractors possess in a sense a fractal structure and the fractal dimension is an important characteristic of chaotic attractors.
As mentioned above, the organization of the book is of a very personal taste. Even the diagram showing the interdependence of chapters (pp. xi) belongs to this category. The book is recommended by the author for two semester courses or some selection of chapters for one semester courses. But it is not specified for which courses and for which part (in time) of them.
On the other hand the book is endowed with many practical exercises, including solutions. But some of them are formulated as a part of the main text. So to understand the interpretation one needs to go through the exercises, as well.
The style is almost elementary and the only prerequisites for the course or following the book is, according to the author, calculus and linear algebra.
One must respect the graphics of the book are very good, without any technical errors. The references are unfortunately not complete and not up-to-date.
So one can recommend this book only as complementary for introductory courses on discrete chaos and fractals but not so much for such courses on discrete dynamical systems and difference equations, in general.
Reviewer: L.Andrey (Praha)
##### MSC:
37D45 Strange attractors, chaotic dynamics of systems with hyperbolic behavior 37-02 Research exposition (monographs, survey articles) pertaining to dynamical systems and ergodic theory 37G10 Bifurcations of singular points in dynamical systems 37G05 Normal forms for dynamical systems 39A11 Stability of difference equations (MSC2000) 37F50 Small divisors, rotation domains and linearization in holomorphic dynamics 37E05 Dynamical systems involving maps of the interval (piecewise continuous, continuous, smooth)
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# A group-theoretical approach for nonlinear Schrödinger equations
Giovanni Molica Bisci
• Corresponding author
• Dipartimento P.A.U., Università degli Studi Mediterranea di Reggio Calabria, Salita Melissari – Feo di Vito, 89124 Reggio Calabria, Italy
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Published Online: 2018-07-04 | DOI: https://doi.org/10.1515/acv-2018-0016
## Abstract
The purpose of this paper is to study the existence of weak solutions for some classes of Schrödinger equations defined on the Euclidean space ${ℝ}^{d}$ ($d\ge 3$). These equations have a variational structure and, thanks to this, we are able to find a non-trivial weak solution for them by using the Palais principle of symmetric criticality and a group-theoretical approach used on a suitable closed subgroup of the orthogonal group $O\left(d\right)$. In addition, if the nonlinear term is odd, and $d>3$, the existence of ${\left(-1\right)}^{d}+\left[\frac{d-3}{2}\right]$ pairs of sign-changing solutions has been proved. To make the nonlinear setting work, a certain summability of the ${L}^{\mathrm{\infty }}$-positive and radially symmetric potential term W governing the Schrödinger equations is requested. A concrete example of an application is pointed out. Finally, we emphasize that the method adopted here should be applied for a wider class of energies largely studied in the current literature also in non-Euclidean setting as, for instance, concave-convex nonlinearities on Cartan–Hadamard manifolds with poles.
MSC 2010: 35J91; 35J60; 35A01; 45A15
## 1 Introduction
The aim of this paper is to establish an existence result for the following Schrödinger equation:
(Slambda)
where $\left({ℝ}^{d},|\cdot |\right)$ is the Euclidean space (with $d\ge 3$), $\mathrm{\Delta }u:=div\left(\nabla u\right)$ stands for the classical Laplacian operator, $f:ℝ\to ℝ$ is a subcritical continuous function, $W\in {L}^{\mathrm{\infty }}\left({ℝ}^{N}\right)\cap {L}^{2}\left({ℝ}^{d}\right)\setminus \left\{0\right\}$ is a non-negative radially symmetric map and λ is a positive real parameter. Finally, in order to avoid technicalities, we assume that the potential $V:{ℝ}^{d}\to ℝ$ satisfies the following condition:
• ${(h_{V}^{d})}$
$V\in {C}^{0}\left({ℝ}^{d}\right)$ is radially symmetric and ${inf}_{x\in {ℝ}^{d}}V\left(x\right)>0$.
As is well known, the interest in Schrödinger equations comes from various problems in mathematical physics, among others: cosmology, constructive field theory, solitary waves, and nonlinear Klein–Gordon equations. See, for instance, the quoted papers [4, 18, 40].
Moreover, a solution of problem (Slambda) can also be interpreted as a stationary state of the reaction diffusion equation
${u}_{t}=\mathrm{\Delta }u-V\left(x\right)u+\lambda W\left(x\right)f\left(u\right),$
that, as pointed out in [13], describe some chemical phenomena (see [21] for details).
In quantum mechanics, the Schrödinger equation modelled the motion of microscopic particles. The nonlinear term represents the interaction of two particles. This interaction mainly acts as attraction when the energy of the particles is small. A stable state is described thought the existence of non-trivial solutions.
Of course, we do not intend to review the huge bibliography of equations like (Slambda). We just emphasize that the potential V has a crucial role concerning the existence and behavior of solutions. For instance, after the seminal paper of Rabinowitz [37] where the potential V is assumed to be coercive, several different assumptions are adopted in order to obtain existence and multiplicity results (see, among others, the papers [2, 13, 11, 10, 23]). When V is a positive constant, or V is radially symmetric, it is natural to look for radially symmetric solutions, see [40, 42].
From a purely mathematical point of view, it is worth mentioning that the early classical studies dedicated to equation (Slambda) were given, among others, by Strauss [40], Bartsch and Willem [14], Berestycki and Lions [15], and Struwe [41]. An important incentive to its study was provided by the work [24] of Gidas, Ni and Nirenberg, where a functional analysis approach was proposed to attack it. We cite, amid the wide literature on the subject, the works [13, 14] where Schrödinger equations were studied by exploiting different methods and new technical approaches.
The main difficulty in dealing with problem like (Slambda) arises from the lack of compactness, which becomes clear when one looks at the following case:
(L)
The set of solutions of (L) is invariant under translations and, therefore, it is not compact. If both V and W are radially symmetric, then compactness can be restored by looking only for radially symmetric solutions. For instance, the existence of infinitely many radially symmetric solutions has been established, among others, by Bartsch and Willem [14], Berestycki and Lions [15], and Conti, Merizzi and Terracini [19].
The existence of non-radial solutions (bound states) of problem (Slambda) or some of its variants has been studied after the paper of Bartsch and Willem [13] (see, for instance, Willem’s book [42] and the closed related papers by Kristály [28] and Kristály, Moroşanu and O’Regan [29]). We also cite the paper [17] of Clapp and Weth, and the references therein, where the authors under some weak onesided asymptotic estimates on the terms V and W (without symmetry assumptions), proved the existence of at least $\frac{d}{2}+1$ pairs of non-trivial weak solutions for suitable Schrödinger equations; see, for completeness, also the related papers given by Bartsch, Clapp and Weth [9], and Evéquoz and Weth [20].
In the present paper we find non-radial solutions of problem (Slambda) for $d=4$ or $d\ge 6$ provided W and f satisfy respectively a summability condition and a growth assumption and f is odd. Following [14], an interesting prototype for problem (Slambda) is given by
where $1, the potential V is bounded below by a positive constant and W is bounded.
In this setting, a simple case of our main result reads as follows.
#### Theorem 1.
Let $d\mathrm{>}\mathrm{3}$ and let $\mathrm{1}\mathrm{<}r\mathrm{<}\mathrm{2}\mathrm{<}s\mathrm{<}{\mathrm{2}}^{\mathrm{*}}$, where ${\mathrm{2}}^{\mathrm{*}}\mathrm{:=}\frac{\mathrm{2}\mathit{}d}{d\mathrm{-}\mathrm{2}}$. Furthermore, let t be the conjugate exponent of s and let $W\mathrm{\in }{L}^{\mathrm{\infty }}\mathit{}\mathrm{\left(}{\mathrm{R}}^{d}\mathrm{\right)}\mathrm{\cap }{L}^{t}\mathit{}\mathrm{\left(}{\mathrm{R}}^{d}\mathrm{\right)}\mathrm{\setminus }\mathrm{\left\{}\mathrm{0}\mathrm{\right\}}$ be a non-negative radially symmetric map. Finally, let V be a potential for which $\mathrm{\left(}{h}_{V}^{d}\mathrm{\right)}$ holds. Then, for λ sufficiently small, the problem
(Clambda)
${\zeta }_{S}^{\left(d\right)}:=1+{\left(-1\right)}^{d}+\left[\frac{d-3}{2}\right]$
pairs of non-trivial weak solutions ${\mathrm{\left\{}\mathrm{±}{u}_{\lambda \mathrm{,}i}\mathrm{\right\}}}_{i\mathrm{\in }{J}^{\mathrm{\prime }}}\mathrm{\subset }{E}_{V}$, where ${J}^{\mathrm{\prime }}\mathrm{:=}\mathrm{\left\{}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathrm{,}{\zeta }_{S}^{\mathrm{\left(}d\mathrm{\right)}}\mathrm{\right\}}$, such that
$\underset{\lambda \to {0}^{+}}{lim}\left({\int }_{{ℝ}^{d}}{|\nabla {u}_{\lambda ,i}\left(x\right)|}^{2}𝑑x+{\int }_{{ℝ}^{d}}V\left(x\right){|{u}_{\lambda ,i}\left(x\right)|}^{2}𝑑x\right)=0,$
and $\mathrm{|}{u}_{\lambda \mathrm{,}i}\mathit{}\mathrm{\left(}x\mathrm{\right)}\mathrm{|}\mathrm{\to }\mathrm{0}$, as $\mathrm{|}x\mathrm{|}\mathrm{\to }\mathrm{\infty }$, for every $i\mathrm{\in }{J}^{\mathrm{\prime }}$. Moreover, if $d\mathrm{\ne }\mathrm{5}$, problem (Clambda) admits at least
${\tau }_{d}:={\left(-1\right)}^{d}+\left[\frac{d-3}{2}\right]$
pairs of sign-changing weak solutions ${\mathrm{\left\{}\mathrm{±}{z}_{\lambda \mathrm{,}i}\mathrm{\right\}}}_{i\mathrm{\in }J}\mathrm{\subset }{E}_{V}$, where $J\mathrm{:=}\mathrm{\left\{}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathrm{,}{\tau }_{d}\mathrm{\right\}}$.
Hereafter, $\left[x\right]$ denotes the integer part of a real number $x\ge 0$.
More precisely, in Theorem 6 we prove that there exists a well-localized interval of positive real parameters $\left(0,{\lambda }^{\star }\right)$ such that, for every $\lambda \in \left(0,{\lambda }^{\star }\right)$, problem (Slambda) admits at least one non-trivial radial weak solution in a suitable Sobolev space ${E}_{V}$. By a Strauss-type estimate (see Lions [40]) every weak solution ${u}_{\lambda }\in {E}_{V}$ is also homoclinic, that is,
The proof of Theorem 6 is based on variational methods. We find critical points of the energy functional
${J}_{\lambda }\left(u\right):=\frac{1}{2}\left({\int }_{{ℝ}^{d}}{|\nabla u\left(x\right)|}^{2}𝑑x+{\int }_{{ℝ}^{d}}V\left(x\right){|u\left(x\right)|}^{2}𝑑x\right)-\lambda {\int }_{{ℝ}^{d}}W\left(x\right)\left({\int }_{0}^{u\left(x\right)}f\left(z\right)𝑑z\right)𝑑x$
for every $u\in {E}_{V}$, associated with problem (Slambda) by means of a local minimum result for differentiable functionals (see Theorem 3) and the well-known Palais’ principle of symmetric criticality (see Theorem 4).
A key assumption in Theorem 6 is given by
$\underset{t\to {0}^{+}}{lim sup}\frac{{\int }_{0}^{t}f\left(z\right)𝑑z}{{t}^{2}}=+\mathrm{\infty },\text{and}\mathit{ }\underset{t\to {0}^{+}}{lim inf}\frac{{\int }_{0}^{t}f\left(z\right)𝑑z}{{t}^{2}}>-\mathrm{\infty },$
see Remark 8. This condition is not novel in the current literature and it has been used in order to study existence and multiplicity results for some classes of elliptic problems on bounded domains (see, among others, the papers of Kajikiya [27, 26, 25] and Molica Bisci and Servadei [35]). To the best of our knowledge, no previous applications to Schrödinger equations have been achieved requiring this hypothesis.
Successively, in Theorem 11, the existence of multiple solutions with no radial symmetry for problem (Slambda) is studied. Under the additional hypotheses of oddness on f we observe that the energy functional ${J}_{\lambda }$ is invariant under certain subgroups actions of the orthogonal group $O\left(d\right)$. Due to Theorem 3, every restriction of ${J}_{\lambda }$ to the appropriate subspaces of invariant functions admit a non-trivial critical point. Thanks to the principle of symmetric criticality of Palais, these points will be also critical points of the original functional ${J}_{\lambda }$, and they are non-radial solutions of problem (Slambda), depending on the choice of the subgroup of $O\left(d\right)$; see the related paper [28].
More precisely, thanks to the above careful group-theoretical analysis inspired by Bartsch and Willem [13] and developed by Kristály, Moroşanu and O’Regan in [29], we construct
${\zeta }_{S}^{\left(d\right)}:=1+{\left(-1\right)}^{d}+\left[\frac{d-3}{2}\right]$
subspaces of ${E}_{V}$. Moreover, when $d\ne 5$, there are
${\tau }_{d}:={\left(-1\right)}^{d}+\left[\frac{d-3}{2}\right]$
of this subspaces that does not contain radial symmetric functions; see [13] and [29, Theorem 2.2]. Further energy-level analysis together the local minimum result recalled in Theorem 3 provides at least one pair of non-trivial weak solutions for problem (Slambda) belonging to these subspaces separately whenever the parameter λ is sufficiently small, that is,
where ${\lambda }^{*}$ and ${\lambda }_{i,q}^{\star }$ are suitable geometrical constants whose expression is given respectively in formulas (3.2) and (4.4). The structure of the parameters ${\lambda }^{\star }$ and ${\lambda }_{\star }$ displayed in Theorems 6 and 11 is not simple and their value depends on certain Sobolev embedding constants and on the datum f.
We notice that in [29, Theorem 1.1] the authors proved some (non-)existence and multiplicity results for possible perturbed Schrödinger equations. Their existence and multiplicity results are valid for every λ sufficiently large by requiring a suitable behavior of the nonlinear term f at zero and at infinity. It is easily seen that [29, Theorem 1.1] cannot be applied to problem (Plambda) studied in Example 15. On the other hand, their non-existence result ensures that problem (Slambda) has only the trivial (identically zero) solution for λ sufficiently small requiring, among others assumptions, the following asymptotic behavior at zero:
$\underset{z\to 0}{lim}\frac{f\left(z\right)}{z}=0.$
The above condition clearly fails dealing with a nonlinear concave-convex term like
where $1, as is given in Example 15 (see Remark 16).
Let us also note that Theorems 6 and 11 might actually be reformulated for a wider class of elliptic problems by using our technical approach. For instance, possible extensions to variational-hemivariational inequalities as well as applications to Cartan–Hadamard manifolds (i.e. Riemannian manifolds that are complete, simply-connected, and with non-positive sectional curvature) and non-local fractional problems (see [34]) will be examined in some forthcoming papers.
The manuscript is organized as follows. In Section 2 we give some notations and we recall some properties of the functional space we work in. In order to apply critical point methods to problem (Slambda), we need to work in a subspace of the functional space ${E}_{V}$; in particular, we give some tools which will be useful along the paper (see Propositions 2 and Lemma 5). In Section 3 we study problem (Slambda) and we prove our existence result (see Theorem 6). Finally, in the last section we study the existence of multiple non-radial solutions to the problem (Slambda) for λ sufficiently small.
We cite the books [3, 1, 30, 42] as general references on the subject treated along the paper. The bibliography does no escape the usual rule being incomplete.
## 2 Abstract framework
To make the nonlinear methods work, some careful analysis of the fractional spaces involved is necessary. Assume $d\ge 3$ and let
${E}_{V}:=\left\{u\in {H}^{1}\left({ℝ}^{d}\right):{\int }_{{ℝ}^{d}}\left({|\nabla u\left(x\right)|}^{2}+V\left(x\right){|u\left(x\right)|}^{2}\right)𝑑x<+\mathrm{\infty }\right\}$
be the Hilbert space endowed by the inner product
and induced norm
${\parallel u\parallel }_{{E}_{V}}:={\left({\int }_{{ℝ}^{d}}{|\nabla u\left(x\right)|}^{2}𝑑x+{\int }_{{ℝ}^{d}}V\left(x\right){|u\left(x\right)|}^{2}𝑑x\right)}^{\frac{1}{2}}$
for every $u\in {E}_{V}$.
Thanks to assumption $\left({h}_{V}\right)$ the space ${E}_{V}$ is continuously embedded in ${L}^{q}\left({ℝ}^{d}\right)$ for every $q\in \left[2,{2}^{*}\right]$, where ${2}^{*}:=\frac{2d}{d-2}$ is the classical critical Sobolev exponent. As pointed out in Introduction, one the main difficulties in working in the space ${E}_{V}$ is the lack of compactness of these embeddings.
Now, we show that an ${L}^{p}$-summability condition on the weight W allows us to consider nonlinear terms f in (Slambda) whose growth behavior is the standard ones adopted in literature studying elliptic equations defined on bounded domains.
More precisely, we have the following regularity result.
#### Proposition 2.
Assume that $f\mathrm{:}\mathrm{R}\mathrm{\to }\mathrm{R}$ is a continuous function such that
${\alpha }_{f}:=\underset{z\in ℝ}{sup}\frac{|f\left(z\right)|}{1+{|z|}^{q-1}}<+\mathrm{\infty },$(2.1)
where $q\mathrm{\in }\mathrm{\left[}\mathrm{2}\mathrm{,}{\mathrm{2}}^{\mathrm{*}}\mathrm{\right]}$ and let $\mathrm{\Psi }\mathrm{:}{E}_{V}\mathrm{\to }\mathrm{R}$ be defined by
where $W\mathrm{\in }{L}^{\mathrm{\infty }}\mathit{}{\mathrm{\left(}{\mathrm{R}}^{d}\mathrm{\right)}}_{\mathrm{+}}\mathrm{\cap }{L}^{p}\mathit{}\mathrm{\left(}{\mathrm{R}}^{d}\mathrm{\right)}$, where $p\mathrm{:=}\frac{q}{q\mathrm{-}\mathrm{1}}$ is the conjugate Sobolev exponent of q, and $F\mathit{}\mathrm{\left(}t\mathrm{\right)}\mathrm{:=}{\mathrm{\int }}_{\mathrm{0}}^{t}f\mathit{}\mathrm{\left(}z\mathrm{\right)}\mathit{}𝑑z$ for every $t\mathrm{\in }\mathrm{R}$. Then Ψ is continuously Gâteaux derivable on ${E}_{V}$.
#### Proof.
It is clear that Ψ is well-defined. Indeed, thanks to (2.1), our assumptions on W and the Hölder inequality ensure that
${\int }_{{ℝ}^{d}}W\left(x\right)F\left(u\left(x\right)\right)𝑑x\le {\alpha }_{f}{\int }_{{ℝ}^{d}}W\left(x\right)|u\left(x\right)|𝑑x+\frac{{\alpha }_{f}}{q}{\int }_{{ℝ}^{d}}W\left(x\right){|u\left(x\right)|}^{q}𝑑x$$\le {\alpha }_{f}{\left({\int }_{{ℝ}^{d}}{|W\left(x\right)|}^{p}𝑑x\right)}^{\frac{1}{p}}{\left({\int }_{{ℝ}^{d}}{|u\left(x\right)|}^{q}𝑑x\right)}^{\frac{1}{q}}+\frac{{\alpha }_{f}}{q}{\parallel W\parallel }_{\mathrm{\infty }}{\int }_{{ℝ}^{d}}{|u\left(x\right)|}^{q}𝑑x$(2.2)
for every $u\in {E}_{V}$. Since ${E}_{V}$ is continuously embedded in ${L}^{q}\left({ℝ}^{d}\right)$, inequality (2.2) yields
$\mathrm{\Psi }\left(u\right)\le {\alpha }_{f}\left({\parallel W\parallel }_{p}{\parallel u\parallel }_{q}+\frac{1}{q}{\parallel W\parallel }_{\mathrm{\infty }}{\parallel u\parallel }_{q}^{q}\right)<+\mathrm{\infty }$
for every $u\in {E}_{V}$.
Now, let us compute the Gâteaux derivative ${\mathrm{\Psi }}^{\prime }:{E}_{V}\to {E}_{V}^{*}$ such that $u↦〈{\mathrm{\Psi }}^{\prime }\left(u\right),v〉$, where
$〈{\mathrm{\Psi }}^{\prime }\left(u\right),v〉:=\underset{h\to 0}{lim}\frac{\mathrm{\Psi }\left(u+hv\right)-\mathrm{\Psi }\left(u\right)}{h}=\underset{h\to 0}{lim}\frac{{\int }_{{ℝ}^{d}}W\left(x\right)\left(F\left(u\left(x\right)+hv\left(x\right)\right)-F\left(u\left(x\right)\right)\right)𝑑x}{h}.$
We claim that
$\underset{h\to {0}^{+}}{lim}\frac{{\int }_{{ℝ}^{d}}W\left(x\right)\left(F\left(u\left(x\right)+hv\left(x\right)\right)-F\left(u\left(x\right)\right)\right)𝑑x}{h}={\int }_{{ℝ}^{d}}W\left(x\right)f\left(u\left(x\right)\right)v\left(x\right)𝑑x.$(2.3)
Indeed, for a.e. $x\in {ℝ}^{d}$, and $|h|\in \left(0,1\right)$, by the Mean Value Theorem, there exists $\theta \in \left(0,1\right)$ (depending of x) such that (up to the constant ${\alpha }_{f}$)
$W\left(x\right)|f\left(u\left(x\right)+\theta hv\left(x\right)\right)||v\left(x\right)|\le W\left(x\right)\left(1+{|u\left(x\right)+\theta hv\left(x\right)|}^{q-1}\right)|v\left(x\right)|$$\le W\left(x\right)|v\left(x\right)|+{\parallel W\parallel }_{\mathrm{\infty }}{||u\left(x\right)|+|v\left(x\right)||}^{q-1}|v\left(x\right)|$$\le W\left(x\right)|v\left(x\right)|+{2}^{q-2}{\parallel W\parallel }_{\mathrm{\infty }}\left({|u\left(x\right)|}^{q-1}+{|v\left(x\right)|}^{q-1}\right)|v\left(x\right)|.$
Setting
$g\left(x\right):=W\left(x\right)|v\left(x\right)|+{2}^{q-2}{\parallel W\parallel }_{\mathrm{\infty }}\left({|u\left(x\right)|}^{q-1}+{|v\left(x\right)|}^{q-1}\right)|v\left(x\right)|$
for a.e. $x\in {ℝ}^{d}$, owing to $q\in \left[2,{2}^{*}\right]$, the Hölder inequality ensures that
${\int }_{{ℝ}^{d}}g\left(x\right)𝑑x\le {\parallel W\parallel }_{p}{\parallel v\parallel }_{q}+{2}^{q-2}{\parallel W\parallel }_{\mathrm{\infty }}{\parallel u\parallel }_{q}^{q-1}{\parallel v\parallel }_{q}+{2}^{q-2}{\parallel W\parallel }_{\mathrm{\infty }}{\parallel v\parallel }_{q}^{q}<+\mathrm{\infty }.$
Thus $g\in {L}^{1}\left({ℝ}^{d}\right)$ and Lebesgue’s Dominated Convergence Theorem ensures that (2.3) holds true.
We prove now the continuity of the Gâteaux derivative ${\mathrm{\Psi }}^{\prime }$. Assume that ${u}_{j}\to u$ in ${E}_{V}$ as $j\to +\mathrm{\infty }$, and let us prove that
$\underset{j\to +\mathrm{\infty }}{lim}{\parallel {\mathrm{\Psi }}^{\prime }\left({u}_{j}\right)-{\mathrm{\Psi }}^{\prime }\left(u\right)\parallel }_{{E}_{V}^{*}}=0,$
where
${\parallel {\mathrm{\Psi }}^{\prime }\left({u}_{j}\right)-{\mathrm{\Psi }}^{\prime }\left(u\right)\parallel }_{{E}_{V}^{*}}:=\underset{{\parallel v\parallel }_{{E}_{V}}=1}{sup}|〈{\mathrm{\Psi }}^{\prime }\left({u}_{j}\right)-{\mathrm{\Psi }}^{\prime }\left(u\right),v〉|.$
Observe that
${\parallel {\mathrm{\Psi }}^{\prime }\left({u}_{j}\right)-{\mathrm{\Psi }}^{\prime }\left(u\right)\parallel }_{{E}_{V}^{*}}:=\underset{{\parallel v\parallel }_{{E}_{V}}=1}{sup}|〈{\mathrm{\Psi }}^{\prime }\left({u}_{j}\right)-{\mathrm{\Psi }}^{\prime }\left(u\right),v〉|$$\le \underset{{\parallel v\parallel }_{{E}_{V}}=1}{sup}|{\int }_{{ℝ}^{d}}W\left(x\right)\left(f\left({u}_{j}\left(x\right)\right)-f\left(u\left(x\right)\right)\right)v\left(x\right)𝑑x|$$\le \underset{{\parallel v\parallel }_{{E}_{V}}=1}{sup}{\int }_{{ℝ}^{d}}W\left(x\right)|\left(f\left({u}_{j}\left(x\right)\right)-f\left(u\left(x\right)\right)\right)||v\left(x\right)|𝑑x.$(2.4)
The Sobolev imbedding result ensures that ${u}_{j}\to u$ in ${L}^{q}\left({ℝ}^{n}\right)$ for every $q\in \left[2,{2}^{*}\right]$, and ${u}_{j}\left(x\right)\to u\left(x\right)$ a.e. in ${ℝ}^{d}$ as $j\to +\mathrm{\infty }$. Then, by [42, Lemma A.1], up to a subsequence, ${u}_{j}\left(x\right)\to u\left(x\right)$ a.e. in ${ℝ}^{d}$ as $j\to +\mathrm{\infty }$, and there exists $\tau \in {L}^{q}\left({ℝ}^{d}\right)$ such that $|{u}_{j}\left(x\right)|\le \tau \left(x\right)$ a.e. in ${ℝ}^{d}$ and $|u\left(x\right)|\le \tau \left(x\right)$ a.e. in ${ℝ}^{d}$. In particular,
Note that the function ${ℝ}^{d}\ni x↦1+\tau {\left(x\right)}^{q-1}$ is $p:=\frac{q}{q-1}$-summable on every bounded measurable subset of ${ℝ}^{d}$. Therefore, by the Lebesgue Dominated Convergence Theorem, we infer
for every $R>0$, where ${B}_{R}\left(0\right)$ denotes the open ball of radius R centered at zero. At this point, set
${\mathrm{\ell }}_{q}:=\underset{{\parallel v\parallel }_{{E}_{V}}\le 1}{sup}{\parallel v\parallel }_{q}$
and let ${R}_{\epsilon }>0$ be such that
${\int }_{{ℝ}^{d}\setminus {B}_{{R}_{\epsilon }}\left(0\right)}{|W\left(x\right)|}^{p}𝑑x<{\left(\frac{\epsilon }{8{\alpha }_{f}{\mathrm{\ell }}_{q}}\right)}^{p},$(2.5)
as well as
${\int }_{{ℝ}^{d}\setminus {B}_{{R}_{\epsilon }}\left(0\right)}{|\tau \left(x\right)|}^{p}𝑑x<{\left(\frac{\epsilon }{8{\alpha }_{f}{\mathrm{\ell }}_{q}{\parallel W\parallel }_{\mathrm{\infty }}}\right)}^{p}.$(2.6)
Moreover, let ${j}_{\epsilon }\in ℕ$ be such that
${\parallel f\left({u}_{j}\right)-f\left(u\right)\parallel }_{{L}^{p}\left({B}_{{R}_{\epsilon }}\left(0\right)\right)}<\frac{\epsilon }{4{\mathrm{\ell }}_{q}{\parallel W\parallel }_{\mathrm{\infty }}}$(2.7)
for each $j\in ℕ$, with $j\ge {j}_{\epsilon }$. Owing to (2.5), (2.6) and (2.7), for $j\ge {j}_{\epsilon }$, inequality (2.4) yields
${\parallel {\mathrm{\Psi }}^{\prime }\left({u}_{j}\right)-{\mathrm{\Psi }}^{\prime }\left(u\right)\parallel }_{{E}_{V}^{*}}\le \underset{{\parallel v\parallel }_{{E}_{V}}\le 1}{sup}{\int }_{{ℝ}^{d}\setminus {B}_{{R}_{\epsilon }}\left(0\right)}W\left(x\right)|f\left({u}_{j}\left(x\right)\right)-f\left(u\left(x\right)\right)||v\left(x\right)|𝑑x$$\le 2{\alpha }_{f}\underset{{\parallel v\parallel }_{{E}_{V}}\le 1}{sup}{\int }_{{ℝ}^{d}\setminus {B}_{{R}_{\epsilon }}\left(0\right)}W\left(x\right)\left(1+{|\tau \left(x\right)|}^{q-1}\right)|v\left(x\right)|𝑑x$$+{\parallel W\parallel }_{\mathrm{\infty }}\underset{{\parallel v\parallel }_{{E}_{V}}\le 1}{sup}{\int }_{{B}_{{R}_{\epsilon }}\left(0\right)}|f\left({u}_{j}\left(x\right)\right)-f\left(u\left(x\right)\right)||v\left(x\right)|𝑑x$$\le 2{\alpha }_{f}{\mathrm{\ell }}_{q}{\parallel W\parallel }_{{L}^{p}\left({ℝ}^{d}\setminus {B}_{{R}_{\epsilon }}\left(0\right)\right)}+2{\alpha }_{f}{\mathrm{\ell }}_{q}{\parallel W\parallel }_{\mathrm{\infty }}{\parallel \tau \parallel }_{{L}^{p}\left({ℝ}^{d}\setminus {B}_{{R}_{\epsilon }}\left(0\right)\right)}$$+{\mathrm{\ell }}_{q}{\parallel W\parallel }_{\mathrm{\infty }}{\parallel f\left({u}_{j}\right)-f\left(u\right)\parallel }_{{L}^{p}\left({B}_{{R}_{\epsilon }}\left(0\right)\right)}<\epsilon ,$
that concludes the proof. ∎
Of course, by elementary standard arguments, Proposition 2 ensures that $\mathrm{\Psi }\in {C}^{1}\left({E}_{V},ℝ\right)$. Then the functional defined by
is of class ${C}^{1}\left({E}_{V},ℝ\right)$. Furthermore, the critical points of ${J}_{\lambda }$ are exactly the weak solutions of problem (Slambda). More precisely, we say that $u\in {E}_{V}$ is a weak solution of problem (Slambda) if and only if
In order to find critical points for ${J}_{\lambda }$, we will apply the principle of symmetric criticality together with the following critical point theorem proved by Ricceri in [38] and recalled here in a more convenient form.
#### Theorem 3.
Let X be a reflexive real Banach space, and let $\mathrm{\Phi }\mathrm{,}\mathrm{\Psi }\mathrm{:}X\mathrm{\to }\mathrm{R}$ be two Gâteaux differentiable functionals such that Φ is strongly continuous, sequentially weakly lower semicontinuous and coercive. Further, assume that Ψ is sequentially weakly upper semicontinuous. For every $r\mathrm{>}{\mathrm{inf}}_{X}\mathit{}\mathrm{\Phi }$, put
$\phi \left(r\right):=\underset{u\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },r\right)\right)}{inf}\frac{\left({sup}_{v\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },r\right)\right)}\mathrm{\Psi }\left(v\right)\right)-\mathrm{\Psi }\left(u\right)}{r-\mathrm{\Phi }\left(u\right)}.$
Then, for each $r\mathrm{>}{\mathrm{inf}}_{X}\mathit{}\mathrm{\Phi }$ and each $\lambda \mathrm{\in }\mathrm{\left(}\mathrm{0}\mathrm{,}\frac{\mathrm{1}}{\phi \mathit{}\mathrm{\left(}r\mathrm{\right)}}\mathrm{\right)}$, the restriction of ${J}_{\lambda }\mathrm{:=}\mathrm{\Phi }\mathrm{-}\lambda \mathit{}\mathrm{\Psi }$ to ${\mathrm{\Phi }}^{\mathrm{-}\mathrm{1}}\mathit{}\mathrm{\left(}\mathrm{\left(}\mathrm{-}\mathrm{\infty }\mathrm{,}r\mathrm{\right)}\mathrm{\right)}$ admits a global minimum, which is a critical point $\mathrm{\left(}$local minimum$\mathrm{\right)}$ of ${J}_{\lambda }$ in X.
For completeness, we also recall here the principle of symmetric criticality that plays a central role in many problems from the differential geometry, physics and in partial differential equations.
An action of a topological group G on the Banach space $\left(X,\parallel \cdot {\parallel }_{X}\right)$ is a continuous map
$*:G×X\to X,\left(g,x\right)↦g*u$
such that
The action $*$ is said to be isometric if ${\parallel g*u\parallel }_{X}={\parallel u\parallel }_{X}$ for every $g\in G$. Moreover, the space of G-invariant points is defined by
and a map $h:X\to ℝ$ is said to be G-invariant on X if
$h\left(g*u\right)=h\left(u\right)$
for every $g\in G$ and $u\in X$.
#### Theorem 4 (Palais (1979), [36]).
Assume that the action of the topological group G on the Banach space X is isometric. If $I\mathrm{\in }{C}^{\mathrm{1}}\mathit{}\mathrm{\left(}X\mathrm{,}\mathrm{R}\mathrm{\right)}$ is G-invariant on X and if u is a critical point of I restricted to ${\mathrm{Fix}}_{G}\mathit{}\mathrm{\left(}X\mathrm{\right)}$, then u is a critical point of I.
See, for instance, [42, Chapter 1] for details.
Let $O\left(d\right)$ be the orthogonal group and let $G\subseteq O\left(d\right)$ be a subgroup. Assume that G acts on the space ${E}_{V}$. Hence, the set of fixed points of ${E}_{V}$, respect to $O\left(d\right)$, is clearly given by
We notice that, if $G=O\left(d\right)$ and the action is the standard linear isometric map defined by
then ${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ is exactly the subspace of radially symmetric functions of ${E}_{V}$. Moreover, the embedding
${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)↪{L}^{q}\left({ℝ}^{d}\right)$(2.8)
is continuous (respectively, compact) for every $q\in \left[2,{2}^{*}\right]$ (respectively, $q\in \left(2,{2}^{*}\right)$). See, for instance, the celebrated paper [31].
Now, for every $\lambda >0,$ let ${\mathcal{𝒥}}_{\lambda }:={{J}_{\lambda }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}:{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)\to ℝ$ be the functional defined by
${\mathcal{𝒥}}_{\lambda }\left(u\right):=\mathrm{\Phi }\left(u\right)-{\lambda \mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(u\right),$
where
The above remarks yield the next semicontinuity property.
#### Lemma 5.
Assume that $f\mathrm{:}\mathrm{R}\mathrm{\to }\mathrm{R}$ is a continuous function such that condition (2.1) holds for every $q\mathrm{\in }\mathrm{\left(}\mathrm{2}\mathrm{,}{\mathrm{2}}^{\mathrm{*}}\mathrm{\right)}$. Then, for every $\lambda \mathrm{>}\mathrm{0}$, the functional ${\mathcal{J}}_{\lambda }$ is sequentially weakly lower semicontinuous on ${\mathrm{Fix}}_{O\mathit{}\mathrm{\left(}d\mathrm{\right)}}\mathit{}\mathrm{\left(}{E}_{V}\mathrm{\right)}$.
#### Proof.
First, on account of Brézis [16, Corollaire III.8], the functional Φ is sequentially weakly lower semicontinuous on ${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$. In order to prove that ${\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}$ is sequentially weakly continuous, we assume that there exists a sequence ${\left\{{u}_{j}\right\}}_{j\in ℕ}\subset {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ which weakly converges to an element ${u}_{0}\in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$. Since ${\left\{{u}_{j}\right\}}_{j\in ℕ}$ is bounded in ${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ and, taking into account that, thanks to (2.8), ${u}_{j}\to {u}_{0}$ in ${L}^{q}\left({ℝ}^{d}\right)$, the Mean Value Theorem, the growth condition (2.1) and the Hölder inequality yield
$|\mathrm{\Psi }\left({u}_{j}\right)-\mathrm{\Psi }\left({u}_{0}\right)|\le {\int }_{{ℝ}^{d}}W\left(x\right)|F\left({u}_{j}\left(x\right)\right)-F\left({u}_{0}\left(x\right)\right)|𝑑x$$\le {\alpha }_{f}{\int }_{{ℝ}^{d}}W\left(x\right)\left(2+{|{u}_{j}\left(x\right)|}^{q-1}+{|{u}_{0}\left(x\right)|}^{q-1}\right)|{u}_{j}\left(x\right)-{u}_{0}\left(x\right)|𝑑x$$\le {\alpha }_{f}\left(2{\parallel W\parallel }_{p}{\parallel {u}_{j}-{u}_{0}\parallel }_{q}+{\parallel W\parallel }_{\mathrm{\infty }}\left({\parallel {u}_{j}\parallel }_{q}^{q-1}+{\parallel {u}_{0}\parallel }_{q}^{q-1}\right){\parallel {u}_{j}-{u}_{0}\parallel }_{q}\right)$$\le {\alpha }_{f}\left(2{\parallel W\parallel }_{2}+{\parallel W\parallel }_{\mathrm{\infty }}\left(M+{\parallel {u}_{0}\parallel }_{q}^{q-1}\right)\right){\parallel {u}_{j}-{u}_{0}\parallel }_{q}$(2.9)
for some $M>0$. The last expression in (2.9) tends to zero. In conclusion, the functional Ψ is sequentially weakly continuous and this completes the proof. ∎
## 3 An existence result: A local minimum approach
Let $d\ge 3$ and set
${c}_{\mathrm{\ell }}:=sup\left\{\frac{{\parallel u\parallel }_{\mathrm{\ell }}}{{\parallel u\parallel }_{{E}_{V}}}:u\in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)\setminus \left\{0\right\}\right\}$
for every $\mathrm{\ell }\in \left(2,{2}^{*}\right)$.
With the above notation the main result reads as follows.
#### Theorem 6.
Let $f\mathrm{:}\mathrm{R}\mathrm{\to }\mathrm{R}$ be a continuous function satisfying the growth condition (2.1) for some $q\mathrm{\in }\mathrm{\left(}\mathrm{2}\mathrm{,}{\mathrm{2}}^{\mathrm{*}}\mathrm{\right)}$ in addition to
$\underset{t\to {0}^{+}}{lim sup}\frac{F\left(t\right)}{{t}^{2}}=+\mathrm{\infty }\mathit{ }\text{𝑎𝑛𝑑}\mathit{ }\underset{t\to {0}^{+}}{lim inf}\frac{F\left(t\right)}{{t}^{2}}>-\mathrm{\infty },$(3.1)
where $F\mathit{}\mathrm{\left(}t\mathrm{\right)}\mathrm{:=}{\mathrm{\int }}_{\mathrm{0}}^{t}f\mathit{}\mathrm{\left(}z\mathrm{\right)}\mathit{}𝑑z$. Furthermore, let $W\mathrm{\in }{L}^{\mathrm{\infty }}\mathit{}\mathrm{\left(}{\mathrm{R}}^{d}\mathrm{\right)}\mathrm{\cap }{L}^{p}\mathit{}\mathrm{\left(}{\mathrm{R}}^{d}\mathrm{\right)}\mathrm{\setminus }\mathrm{\left\{}\mathrm{0}\mathrm{\right\}}$, be a radially symmetric map with
and V be a potential for which $\mathrm{\left(}{h}_{V}^{d}\mathrm{\right)}$ holds. Then there exists a positive number ${\lambda }^{\mathrm{\star }}$ given by
${\lambda }^{\star }:=\frac{q}{{\alpha }_{f}{c}_{q}}\underset{\gamma >0}{\mathrm{max}}\left(\frac{\gamma }{q\sqrt{2}{\parallel W\parallel }_{p}+{2}^{\frac{q}{2}}{c}_{q}^{q-1}{\parallel W\parallel }_{\mathrm{\infty }}{\gamma }^{q-1}}\right)$(3.2)
such that, for every $\lambda \mathrm{\in }\mathrm{\left(}\mathrm{0}\mathrm{,}{\lambda }^{\mathrm{\star }}\mathrm{\right)}$, the problem
(Slambda)
admits at least one non-trivial radial weak solution ${u}_{\lambda }\mathrm{\in }{E}_{V}$. Moreover,
$\underset{\lambda \to {0}^{+}}{lim}{\parallel {u}_{\lambda }\parallel }_{{E}_{V}}=0,$
and $\mathrm{|}{u}_{\lambda }\mathit{}\mathrm{\left(}x\mathrm{\right)}\mathrm{|}\mathrm{\to }\mathrm{0}$ as $\mathrm{|}x\mathrm{|}\mathrm{\to }\mathrm{\infty }$.
#### Proof.
The main idea of the proof consists in applying Theorem 3 to the functional ${\mathcal{𝒥}}_{\lambda }$. Successively, taking into account the preliminary results of Section 2, the existence of one non-trivial radial solution of problem (Slambda) follows by the symmetric criticality principle recalled in Theorem 4. To this purpose, we write the functional ${\mathcal{𝒥}}_{\lambda }$ as follows:
with
$\mathrm{\Phi }\left(u\right):=\frac{1}{2}\left({\int }_{{ℝ}^{d}}{|\nabla u\left(x\right)|}^{2}𝑑x+{\int }_{{ℝ}^{d}}V\left(x\right){|u\left(x\right)|}^{2}𝑑x\right)$
as well as
$\mathrm{\Psi }\left(u\right):={\int }_{{ℝ}^{d}}W\left(x\right)F\left(u\left(x\right)\right)𝑑x.$
First of all, note that ${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ is a Hilbert space and the functionals Φ and ${\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}$ have the regularity required by Theorem 3 (see Lemma 5). Moreover, it is clear that the functional Φ is strongly continuous, coercive in ${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ and
$\underset{u\in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}{inf}\mathrm{\Phi }\left(u\right)=0.$
Now, since $0<\lambda <{\lambda }^{\star }$, bearing in mind (3.2), there exists $\overline{\gamma }>0$ such that
$\lambda <{\lambda }^{\star }\left(\overline{\gamma }\right):=\frac{q}{{\alpha }_{f}{c}_{q}}\left(\frac{\overline{\gamma }}{q\sqrt{2}{\parallel W\parallel }_{p}+{2}^{\frac{q}{2}}{c}_{q}^{q-1}{\parallel W\parallel }_{\mathrm{\infty }}{\overline{\gamma }}^{q-1}}\right).$(3.3)
Set $r\in \left(0,+\mathrm{\infty }\right)$ and consider the function $\chi :\left(0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ given by
$\chi \left(r\right):=\frac{{{sup}_{u\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },r\right)\right)}\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(u\right)}{r}.$
By taking into account the growth condition expressed by (2.1), it follows that
${\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(u\right)={\int }_{{ℝ}^{d}}W\left(x\right)F\left(u\left(x\right)\right)𝑑x\le {\alpha }_{f}{\int }_{{ℝ}^{d}}W\left(x\right)|u\left(x\right)|𝑑x+\frac{{\alpha }_{f}}{q}{\int }_{{ℝ}^{d}}W\left(x\right){|u\left(x\right)|}^{q}𝑑x.$
Moreover, one has
${\parallel u\parallel }_{{E}_{V}}<\sqrt{2r}$(3.4)
for every $u\in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ and $\mathrm{\Phi }\left(u\right).
Now, by using (3.4), the Sobolev embedding (2.8) yields
${\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(u\right)<{\alpha }_{f}{c}_{q}\left({\parallel W\parallel }_{p}\sqrt{2r}+\frac{{c}_{q}^{q-1}}{q}{\parallel W\parallel }_{\mathrm{\infty }}{\left(2r\right)}^{\frac{q}{2}}\right),$
for every $u\in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ such that $\mathrm{\Phi }\left(u\right). Hence
${\underset{u\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },r\right)\right)}{sup}\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(u\right)\le {\alpha }_{f}{c}_{q}\left({\parallel W\parallel }_{p}\sqrt{2r}+\frac{{c}_{q}^{q-1}}{q}{\parallel W\parallel }_{\mathrm{\infty }}{\left(2r\right)}^{\frac{q}{2}}\right).$
Then the above inequality immediately gives
$\chi \left(r\right)\le {\alpha }_{f}{c}_{q}\left({\parallel W\parallel }_{p}\sqrt{\frac{2}{r}}+\frac{{2}^{\frac{q}{2}}{c}_{q}^{q-1}}{q}{\parallel W\parallel }_{\mathrm{\infty }}{r}^{\frac{q}{2}-1}\right)$(3.5)
for every $r>0$.
Evaluating inequality (3.5) in $r={\overline{\gamma }}^{2}$, we have
$\chi \left({\overline{\gamma }}^{2}\right)\le {\alpha }_{f}{c}_{q}\left(\sqrt{2}\frac{{\parallel W\parallel }_{p}}{\overline{\gamma }}+\frac{{2}^{\frac{q}{2}}{c}_{q}^{q-1}}{q}{\parallel W\parallel }_{\mathrm{\infty }}{\overline{\gamma }}^{q-2}\right).$(3.6)
Now, it is easy to note that
$\phi \left({\overline{\gamma }}^{2}\right):=\underset{u\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)}{inf}\frac{\left({{sup}_{v\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)}\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(v\right)\right)-{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(u\right)}{r-\mathrm{\Phi }\left(u\right)}\le \chi \left({\overline{\gamma }}^{2}\right),$
owing to ${z}_{0}\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)$ and $\mathrm{\Phi }\left({z}_{0}\right)={\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left({z}_{0}\right)=0$, where ${z}_{0}\in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ is the zero function.
Finally, bearing in mind (3.3), the above inequality together with (3.6) produce
$\phi \left({\overline{\gamma }}^{2}\right)\le \chi \left({\overline{\gamma }}^{2}\right)\le {\alpha }_{f}{c}_{q}\left(\sqrt{2}\frac{{\parallel W\parallel }_{p}}{\overline{\gamma }}+\frac{{2}^{\frac{q}{2}}{c}_{q}^{q-1}}{q}{\parallel W\parallel }_{\mathrm{\infty }}{\overline{\gamma }}^{q-2}\right)<\frac{1}{\lambda }.$(3.7)
Hence, we have that
$\lambda \in \left(0,\frac{q}{{\alpha }_{f}{c}_{q}}\left(\frac{\overline{\gamma }}{q\sqrt{2}{\parallel W\parallel }_{p}+{2}^{\frac{q}{2}}{c}_{q}^{q-1}{\parallel W\parallel }_{\mathrm{\infty }}{\overline{\gamma }}^{q-1}}\right)\right)\subseteq \left(0,\frac{1}{\phi \left({\overline{\gamma }}^{2}\right)}\right).$
The (critical point) Theorem 3 ensures that there exists a function ${u}_{\lambda }\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)$ such that
${\mathrm{\Phi }}^{\prime }\left({u}_{\lambda }\right)-\lambda {\left({\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\right)}^{\prime }\left({u}_{\lambda }\right)=0$
and, in particular, ${u}_{\lambda }$ is a global minimum of the restriction of the functional ${\mathcal{𝒥}}_{\lambda }$ to the sublevel ${\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)$.
Now, we have to show that the solution ${u}_{\lambda }$ found here above is not the trivial (identically zero) function. If $f\left(0\right)\ne 0$, then it easily follows that ${u}_{\lambda }\not\equiv 0$ in ${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$, since the trivial function does not solve problem (Slambda). Let us consider the case when $f\left(0\right)=0$ and let us fix $\lambda \in \left(0,{\lambda }^{\star }\left(\overline{\gamma }\right)\right)$ for some $\overline{\gamma }>0$. Finally, let ${u}_{\lambda }$ be such that
(3.8)
and
$\mathrm{\Phi }\left({u}_{\lambda }\right)<{\overline{\gamma }}^{2},$
and also ${u}_{\lambda }$ is a critical point of ${\mathcal{𝒥}}_{\lambda }$ in ${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$. Since $O\left(d\right)$ acts isometrically on ${E}_{V}$ (note that V is radial) and taking into account that, thanks to the symmetry assumption on W, one has
${\int }_{{ℝ}^{d}}W\left(x\right)F\left(\left(gu\right)\left(x\right)\right)𝑑x={\int }_{{ℝ}^{d}}W\left(x\right)F\left(u\left({g}^{-1}x\right)\right)𝑑x={\int }_{{ℝ}^{d}}W\left(y\right)F\left(u\left(y\right)\right)𝑑y$
for every $g\in O\left(d\right)$, the functional ${J}_{\lambda }$ is $O\left(d\right)$-invariant on ${E}_{V}$.
So, owing to Theorem 4, ${u}_{\lambda }$ is a weak solution of problem (Slambda). In this setting, in order to prove that ${u}_{\lambda }\not\equiv 0$ in ${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$, first we claim that there exists a sequence of functions ${\left\{{w}_{j}\right\}}_{j\in ℕ}$ in ${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ such that
$\underset{j\to +\mathrm{\infty }}{lim sup}\frac{{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left({w}_{j}\right)}{\mathrm{\Phi }\left({w}_{j}\right)}=+\mathrm{\infty }.$(3.9)
By the assumption on the limsup in (3.1) there exists a sequence ${\left\{{t}_{j}\right\}}_{j\in ℕ}\subset \left(0,+\mathrm{\infty }\right)$ such that ${t}_{j}\to {0}^{+}$ as $j\to +\mathrm{\infty }$ and
$\underset{j\to +\mathrm{\infty }}{lim}\frac{F\left({t}_{j}\right)}{{t}_{j}^{2}}=+\mathrm{\infty },$
namely, we have that for any $M>0$ and j sufficiently large,
$F\left({t}_{j}\right)>M{t}_{j}^{2}.$(3.10)
Now, following Kristály, Moroşanu, and O’Regan in [29], we construct a special test function belonging to ${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ that will be useful for our purposes. If $a, define
${A}_{a}^{b}:=\left\{x\in {ℝ}^{d}:a\le |x|\le b\right\}.$
Since $W\in {L}^{\mathrm{\infty }}\left({ℝ}^{d}\right)\setminus \left\{0\right\}$ is a radially symmetric function with $W\ge 0$, one can find two real numbers $R>r>0$ and $\alpha >0$ such that
${\mathrm{essinf}}_{x\in {A}_{r}^{R}}W\left(x\right)\ge \alpha >0.$(3.11)
Hence, let $0 such that (3.11) holds and $\sigma \in \left(0,\frac{1}{2}\left(R-r\right)\right)$. Set ${v}_{\sigma }\in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ given by
where ${z}_{+}:=\mathrm{max}\left\{0,z\right\}$. With the above notation, we have:
• (1)
$\text{supp}\left({v}_{\sigma }\right)\subseteq {A}_{r}^{R}$,
• (2)
${\parallel {v}_{\sigma }\parallel }_{\mathrm{\infty }}\le 1$,
• (3)
${v}_{\sigma }\left(x\right)=1$ for every $x\in {A}_{r+\sigma }^{R-\sigma }$.
Now, define ${w}_{j}:={t}_{j}{v}_{\sigma }$ for any $j\in ℕ$. Since ${v}_{\sigma }\in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$, of course one has ${w}_{j}\in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ for any $j\in ℕ$. Furthermore, taking into account the algebraic properties of the functions ${v}_{\sigma }$ stated in 13, since $F\left(0\right)=0$, and by using (3.10) we can write
$\frac{{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left({w}_{j}\right)}{\mathrm{\Phi }\left({w}_{j}\right)}=\frac{{\int }_{{A}_{r+\sigma }^{R-\sigma }}W\left(x\right)F\left({w}_{j}\left(x\right)\right)𝑑x+{\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}W\left(x\right)F\left({w}_{j}\left(x\right)\right)𝑑x}{\mathrm{\Phi }\left({w}_{j}\right)}$$=\frac{{\int }_{{A}_{r+\sigma }^{R-\sigma }}W\left(x\right)F\left({t}_{j}\right)𝑑x+{\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}W\left(x\right)F\left({t}_{j}{v}_{\sigma }\left(x\right)\right)𝑑x}{\mathrm{\Phi }\left({w}_{j}\right)}$$\ge 2\frac{M|{A}_{r+\sigma }^{R-\sigma }|\alpha {t}_{j}^{2}+{\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}W\left(x\right)F\left({t}_{j}{v}_{\sigma }\left(x\right)\right)𝑑x}{{t}_{j}^{2}{\parallel {v}_{\sigma }\parallel }_{{E}_{V}}^{2}}$(3.12)
for j sufficiently large. Now, we have to distinguish two different cases, i.e. the case when the liminf in (3.1) is $+\mathrm{\infty }$ and the one in which the liminf in (3.1) is finite.
Case 1. Suppose that ${lim}_{t\to {0}^{+}}\frac{F\left(t\right)}{{t}^{2}}=+\mathrm{\infty }$. Then there exists ${\rho }_{M}>0$ such that for any t with $0,
$F\left(t\right)\ge M{t}^{2}.$(3.13)
Since ${t}_{j}\to {0}^{+}$ and $0\le {v}_{\sigma }\left(x\right)\le 1$ in ${ℝ}^{d}$, it follows that ${w}_{j}\left(x\right)={t}_{j}{v}_{\sigma }\left(x\right)\to {0}^{+}$ as $j\to +\mathrm{\infty }$ uniformly in $x\in {ℝ}^{d}$. Hence, $0\le {w}_{j}\left(x\right)<{\rho }_{M}$ for j sufficiently large and for any $x\in {ℝ}^{d}$. Hence, as a consequence of (3.12) and (3.13), we have
$\frac{{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left({w}_{j}\right)}{\mathrm{\Phi }\left({w}_{j}\right)}\ge 2\frac{M|{A}_{r+\sigma }^{R-\sigma }|\alpha {t}_{j}^{2}+{\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}W\left(x\right)F\left({t}_{j}{v}_{\sigma }\left(x\right)\right)𝑑x}{{t}_{j}^{2}{\parallel {v}_{\sigma }\parallel }_{{E}_{V}}^{2}}$$\ge 2M\alpha \frac{|{A}_{r+\sigma }^{R-\sigma }|+{\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}{|{v}_{\sigma }\left(x\right)|}^{2}𝑑x}{{\parallel {v}_{\sigma }\parallel }_{{E}_{V}}^{2}}$
for j sufficiently large. The arbitrariness of M gives (3.9) and so the claim is proved.
Case 2. Suppose that ${lim inf}_{t\to {0}^{+}}\frac{F\left(t\right)}{{t}^{2}}=\mathrm{\ell }\in ℝ$. Then, for any $\epsilon >0$, there exists ${\rho }_{\epsilon }>0$ such that for any t with $0,
$F\left(t\right)\ge \left(\mathrm{\ell }-\epsilon \right){t}^{2}.$(3.14)
Arguing as above, we can suppose that $0\le {w}_{j}\left(x\right)={t}_{j}{v}_{\sigma }\left(x\right)<{\rho }_{\epsilon }$ for j large enough and any $x\in {ℝ}^{d}$. Thus, by (3.12) and (3.14) we get
$\frac{{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left({w}_{j}\right)}{\mathrm{\Phi }\left({w}_{j}\right)}\ge 2\frac{M|{A}_{r+\sigma }^{R-\sigma }|\alpha {t}_{j}^{2}+{\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}W\left(x\right)F\left({t}_{j}{v}_{\sigma }\left(x\right)\right)𝑑x}{{t}_{j}^{2}{\parallel {v}_{\sigma }\parallel }_{{E}_{V}}^{2}}$$\ge 2\alpha \frac{M|{A}_{r+\sigma }^{R-\sigma }|+\left(\mathrm{\ell }-\epsilon \right){\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}{|{v}_{\sigma }\left(x\right)|}^{2}𝑑x}{{\parallel {v}_{\sigma }\parallel }_{{E}_{V}}^{2}},$(3.15)
provided j is sufficiently large. Choosing $M>0$ large enough, say
$M>\mathrm{max}\left\{0,-\frac{2\mathrm{\ell }}{|{A}_{r+\sigma }^{R-\sigma }|}{\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}{|{v}_{\sigma }\left(x\right)|}^{2}𝑑x\right\},$
and $\epsilon >0$ small enough so that
$\epsilon {\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}{|{v}_{\sigma }\left(x\right)|}^{2}𝑑x<\frac{M}{2}|{A}_{r+\sigma }^{R-\sigma }|+\mathrm{\ell }{\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}{|{v}_{\sigma }\left(x\right)|}^{2}𝑑x,$
by (3.15) we get
$\frac{{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left({w}_{j}\right)}{\mathrm{\Phi }\left({w}_{j}\right)}\ge 2\alpha \frac{M|{A}_{r+\sigma }^{R-\sigma }|+\left(\mathrm{\ell }-\epsilon \right){\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}{|{v}_{\sigma }\left(x\right)|}^{2}𝑑x}{{\parallel {v}_{\sigma }\parallel }_{{E}_{V}}^{2}}$$\ge \frac{2\alpha }{{\parallel {v}_{\sigma }\parallel }_{{E}_{V}}^{2}}\left(M|{A}_{r+\sigma }^{R-\sigma }|+\mathrm{\ell }{\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}{|{v}_{\sigma }\left(x\right)|}^{2}𝑑x-\epsilon {\int }_{{A}_{r}^{R}\setminus {A}_{r+\sigma }^{R-\sigma }}{|{v}_{\sigma }\left(x\right)|}^{2}𝑑x\right)$$=\alpha M\frac{|{A}_{r+\sigma }^{R-\sigma }|}{{\parallel {v}_{\sigma }\parallel }_{{E}_{V}}^{2}}$
for j large enough. Also in this case the arbitrariness of M gives assertion (3.9). Now, note that
${\parallel {w}_{j}\parallel }_{{E}_{V}}={t}_{j}{\parallel {v}_{\sigma }\parallel }_{{E}_{V}}\to 0$
as $j\to +\mathrm{\infty }$, so that for j large enough,
${\parallel {w}_{j}\parallel }_{{E}_{V}}<\sqrt{2}\overline{\gamma }.$
Thus
${w}_{j}\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right),$(3.16)
provided j is large enough. Also, by ((3.9)) and the fact that $\lambda >0$
${\mathcal{𝒥}}_{\lambda }\left({w}_{j}\right)=\mathrm{\Phi }\left({w}_{j}\right)-{\lambda \mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left({w}_{j}\right)<0$(3.17)
for j sufficiently large. Since ${u}_{\lambda }$ is a global minimum of the restriction of ${\mathcal{𝒥}}_{\lambda }$ to ${\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)$ (see (3.8)), by (3.16) and (3.17) we conclude that
${\mathcal{𝒥}}_{\lambda }\left({u}_{\lambda }\right)\le {\mathcal{𝒥}}_{\lambda }\left({w}_{j}\right)<0={\mathcal{𝒥}}_{\lambda }\left(0\right),$(3.18)
so that ${u}_{\lambda }\not\equiv 0$ in ${\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$. Thus, ${u}_{\lambda }$ is a non-trivial weak solution of problem (Slambda). The arbitrariness of λ gives that ${u}_{\lambda }\not\equiv 0$ for any $\lambda \in \left(0,{\lambda }^{\star }\right)$.
Now, we claim that ${lim}_{\lambda \to {0}^{+}}{\parallel {u}_{\lambda }\parallel }_{{E}_{V}}=0.$ For this, let us fix $\lambda \in \left(0,{\lambda }^{\star }\left(\overline{\gamma }\right)\right)$ for some $\overline{\gamma }>0$. By $\mathrm{\Phi }\left({u}_{\lambda }\right)<{\overline{\gamma }}^{2}$ one has
$\mathrm{\Phi }\left({u}_{\lambda }\right)=\frac{1}{2}{\parallel {u}_{\lambda }\parallel }_{{E}_{V}}^{2}<{\overline{\gamma }}^{2},$
that is,
${\parallel {u}_{\lambda }\parallel }_{{E}_{V}}<\sqrt{2}\overline{\gamma }.$
As a consequence of this and by using the growth condition (2.1) together with the property (2.8), it follows that
$|{\int }_{{ℝ}^{d}}W\left(x\right)f\left({u}_{\lambda }\left(x\right)\right){u}_{\lambda }\left(x\right)𝑑x|\le {\alpha }_{f}\left({\int }_{{ℝ}^{d}}W\left(x\right)|{u}_{\lambda }\left(x\right)|𝑑x+{\int }_{{ℝ}^{d}}W\left(x\right){|{u}_{\lambda }\left(x\right)|}^{q}𝑑x\right)$$\le {\alpha }_{f}\left({\parallel W\parallel }_{p}{\parallel {u}_{\lambda }\parallel }_{q}+{\parallel W\parallel }_{\mathrm{\infty }}{\parallel {u}_{\lambda }\parallel }_{q}^{q}\right)$$<{c}_{q}{\alpha }_{f}\left(\sqrt{2}\parallel W{\parallel }_{p}\overline{\gamma }+{2}^{\frac{q}{2}}{c}_{q}^{q-1}\parallel W{\parallel }_{\mathrm{\infty }}{\overline{\gamma }}^{q}\right)=:{M}_{\overline{\gamma }}.$(3.19)
Since ${u}_{\lambda }$ is a critical point of ${\mathcal{𝒥}}_{\lambda }$, it follows that $〈{\mathcal{𝒥}}_{\lambda }^{\prime }\left({u}_{\lambda }\right),\phi 〉=0$ for any $\phi \in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ and every $\lambda \in \left(0,{\lambda }^{\star }\left(\overline{\gamma }\right)\right)$. In particular, $〈{\mathcal{𝒥}}_{\lambda }^{\prime }\left({u}_{\lambda }\right),{u}_{\lambda }〉=0$, that is,
$〈{\mathrm{\Phi }}^{\prime }\left({u}_{\lambda }\right),{u}_{\lambda }〉=\lambda {\int }_{{ℝ}^{d}}W\left(x\right)f\left({u}_{\lambda }\left(x\right)\right){u}_{\lambda }\left(x\right)𝑑x$(3.20)
for every $\lambda \in \left(0,{\lambda }^{\star }\left(\overline{\gamma }\right)\right)$. Then, from (3.19) and (3.20), it follows that
$0\le {\parallel {u}_{\lambda }\parallel }_{{E}_{V}}^{2}=〈{\mathrm{\Phi }}^{\prime }\left({u}_{\lambda }\right),{u}_{\lambda }〉=\lambda {\int }_{{ℝ}^{d}}W\left(x\right)f\left({u}_{\lambda }\left(x\right)\right){u}_{\lambda }\left(x\right)𝑑x<\lambda {M}_{\overline{\gamma }}$
for any $\lambda \in \left(0,{\lambda }^{\star }\left(\overline{\gamma }\right)\right)$. We get ${lim}_{\lambda \to {0}^{+}}{\parallel {u}_{\lambda }\parallel }_{{E}_{V}}=0$, as claimed.
Finally, a Strauss-type estimate (see [31] for details) proves that the functions $u\in {E}_{V}$ are homoclinic. Hence, the solution ${u}_{\lambda }\in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)\setminus \left\{0\right\}\subseteq {E}_{V}$ has this property. This concludes the proof of Theorem 6. ∎
#### Remark 7.
Theorem 6 has been achieved without any use of the well-known Ambrosetti–Rabinowitz condition. The importance of this assumption is due to the fact that it assures the boundedness of the Palais–Smale sequences for the energy functional associated with the problem under consideration. This condition fails when dealing with some superlinear elliptic boundary value problems (see, for instance, [39] and the references therein).
#### Remark 8.
If the nonlinear term f has the asymptotic behavior
$\underset{t\to {0}^{+}}{lim}\frac{f\left(t\right)}{t}=+\mathrm{\infty },$
then, obviously, hypothesis (3.1) in Theorem 6 is verified. Actually, the condition
$\underset{t\to {0}^{+}}{lim inf}\frac{F\left(t\right)}{{t}^{2}}>-\mathrm{\infty }$
is a technicality that we request in our proof in order to show that there exists a sequence ${\left\{{w}_{j}\right\}}_{j\in ℕ}\subset {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$ such that
$\underset{j\to +\mathrm{\infty }}{lim sup}\frac{{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left({w}_{j}\right)}{\mathrm{\Phi }\left({w}_{j}\right)}=+\mathrm{\infty }.$
It is natural to ask if this assumption can be dropped in Theorem 6 requiring only
$\underset{t\to {0}^{+}}{lim sup}\frac{F\left(t\right)}{{t}^{2}}=+\mathrm{\infty }.$
A careful analysis of the proof of Theorem 6 ensures that the result remains valid for the following problem:
where $f:{ℝ}^{d}×ℝ\to ℝ$ is a continuous function that satisfies:
• (f1)
There exists $q\mathrm{\in }\mathrm{\left(}\mathrm{2}\mathrm{,}{\mathrm{2}}^{\mathrm{*}}\mathrm{\right)}$ such that
for some radially symmetric function $W\mathrm{\in }{L}^{\mathrm{\infty }}\mathit{}\mathrm{\left(}{\mathrm{R}}^{d}\mathrm{\right)}\mathrm{\cap }{L}^{\frac{q}{q\mathrm{-}\mathrm{1}}}\mathit{}\mathrm{\left(}{\mathrm{R}}^{d}\mathrm{\right)}\mathrm{\setminus }\mathrm{\left\{}\mathrm{0}\mathrm{\right\}}$ .
• (f2)
For every $x\mathrm{\in }{\mathrm{R}}^{d}$ and $t\mathrm{\in }\mathrm{R}$ it follows that
$F\left(x,t\right)=F\left(gx,t\right)$
for every $g\mathrm{\in }O\mathit{}\mathrm{\left(}d\mathrm{\right)}$ .
• (f3)
For some ${x}_{\mathrm{0}}\mathrm{\in }{\mathrm{R}}^{d}$, there exists ${\varrho }_{\mathrm{0}}\mathrm{>}\mathrm{0}$ such that
$\underset{t\to {0}^{+}}{lim sup}\frac{{inf}_{B\left({x}_{0},{\varrho }_{0}\right)}F\left(x,t\right)}{{t}^{2}}=+\mathrm{\infty }\mathit{ }\text{𝑎𝑛𝑑}\mathit{ }\underset{t\to {0}^{+}}{lim inf}\frac{{inf}_{B\left({x}_{0},{\varrho }_{0}\right)}F\left(x,t\right)}{{t}^{2}}>-\mathrm{\infty }.$
Here $F\left(x,t\right):={\int }_{0}^{t}f\left(x,z\right)𝑑z$ and $B\left({x}_{0},{\varrho }_{0}\right)$ is the closed ball centered in ${x}_{0}$ and radius ${\varrho }_{0}$. See [27, 26] for related topics.
#### Remark 9.
We perform now the behavior of the functional ${\mathcal{𝒥}}_{\lambda }$ depending on the real parameter λ. In particular, we point out that by (3.18) the map
(3.21)
Moreover, fixing $\overline{\gamma }>0$, the function $\lambda ↦{\mathcal{𝒥}}_{\lambda }\left({u}_{\lambda }\right)$ is strictly decreasing in $\left(0,{\lambda }^{\star }\left(\overline{\gamma }\right)\right)$, where
${\lambda }^{\star }\left(\overline{\gamma }\right):=\frac{q}{{\alpha }_{f}{c}_{q}}\left(\frac{\overline{\gamma }}{q\sqrt{2}{\parallel W\parallel }_{p}+{2}^{\frac{q}{2}}{c}_{q}^{q-1}{\parallel W\parallel }_{\mathrm{\infty }}{\overline{\gamma }}^{q-1}}\right).$
Indeed, let us write
${\mathcal{𝒥}}_{\lambda }\left(u\right)=\lambda \left(\frac{\mathrm{\Phi }\left(u\right)}{\lambda }-{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(u\right)\right)$(3.22)
for every $u\in {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)$. Now, fix $0<{\lambda }_{1}<{\lambda }_{2}<{\lambda }^{\star }\left(\overline{\gamma }\right)$ and let ${u}_{{\lambda }_{i}}$ be the global minimum of the functional ${\mathcal{𝒥}}_{{\lambda }_{i}}$ restricted to the sublevel ${\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)$ for $i=1,2$. We observe that
$\left(\frac{\mathrm{\Phi }\left({u}_{{\lambda }_{i}}\right)}{{\lambda }_{i}}-{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left({u}_{{\lambda }_{i}}\right)\right)=\underset{v\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)}{inf}\left(\frac{\mathrm{\Phi }\left(v\right)}{{\lambda }_{i}}-{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(v\right)\right)$
for every $i=1,2$. Of course, (3.21) and (3.22) yield
Moreover, it is easy to note that
$\underset{v\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)}{inf}\left(\frac{\mathrm{\Phi }\left(v\right)}{{\lambda }_{2}}-{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(v\right)\right)\le \underset{v\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)}{inf}\left(\frac{\mathrm{\Phi }\left(v\right)}{{\lambda }_{1}}-{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(v\right)\right)$(3.23)
owing to $0<{\lambda }_{1}<{\lambda }_{2}$. Then, by (3.22)–(3.23) and again by the fact that $0<{\lambda }_{1}<{\lambda }_{2}$, we have
${\mathcal{𝒥}}_{{\lambda }_{2}}\left({u}_{{\lambda }_{2}}\right)={\lambda }_{2}\underset{v\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },\overline{\gamma }{}^{2}\right)\right)}{inf}\left(\frac{\mathrm{\Phi }\left(v\right)}{{\lambda }_{2}}-{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(v\right)\right)$$\le {\lambda }_{2}\underset{v\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)}{inf}\left(\frac{\mathrm{\Phi }\left(v\right)}{{\lambda }_{1}}-{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(v\right)\right)$$<{\lambda }_{1}\underset{v\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)}{inf}\left(\frac{\mathrm{\Phi }\left(v\right)}{{\lambda }_{1}}-{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)}\left(v\right)\right)$$={J}_{{\lambda }_{1}}\left({u}_{{\lambda }_{1}}\right),$
so that the real function $\lambda ↦{\mathcal{𝒥}}_{\lambda }\left({u}_{\lambda }\right)$ is (strictly) decreasing in $\left(0,{\lambda }^{\star }\left(\overline{\gamma }\right)\right)$.
## 4 Multiple solutions in presence of symmetries
We study now the existence of multiple solutions (radial and non-radial) for Schrödinger equations in presence of a symmetric nonlinear term f.
Let either $d=4$ or $d\ge 6$ and consider the subgroup ${H}_{d,i}\subset O\left(d\right)$ given by
for every $i\in J:=\left\{1,\mathrm{\dots },{\tau }_{d}\right\}$, where
${\tau }_{d}:={\left(-1\right)}^{d}+\left[\frac{d-3}{2}\right].$
Let us define the involution ${\eta }_{{H}_{d,i}}:{ℝ}^{d}\to {ℝ}^{d}$ as follows:
for every $i\in J$. By definition, one has ${\eta }_{{H}_{d,i}}\notin {H}_{d,i}$, as well as
${\eta }_{{H}_{d,i}}{H}_{d,i}{\eta }_{{H}_{d,i}}^{-1}={H}_{d,i}\mathit{ }\text{and}\mathit{ }{\eta }_{{H}_{d,i}}^{2}={\text{id}}_{{ℝ}^{d}}$
for every $i\in J$.
Moreover, for every $i\in J$, let us consider the compact group
${H}_{d,{\eta }_{i}}:=〈{H}_{d,i},{\eta }_{{H}_{d,i}}〉,$
that is, ${H}_{d,{\eta }_{i}}={H}_{d,i}\cup {\eta }_{{H}_{d,i}}{H}_{d,i}$, and the action $⊛:{H}_{d,{\eta }_{i}}×{E}_{V}\to {E}_{V}$ of ${H}_{d,{\eta }_{i}}$ on ${E}_{V}$ is given by
(4.1)
for every $x\in {ℝ}^{d}$. We notice that $⊛$ is defined for every element of ${H}_{d,{\eta }_{i}}$. Indeed, if $h\in {H}_{d,{\eta }_{i}}$, then either $h\in {H}_{d,i}$ or $h=\tau g\in {H}_{d,{\eta }_{i}}\setminus {H}_{d,i}$, with $g\in {H}_{d,i}$. Moreover, set
for every $i\in J$. Following Bartsch and Willem [13], for every $i\in J$, the embedding
${\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)↪{L}^{q}\left({ℝ}^{d}\right)$
is compact, for every $q\in \left(2,{2}^{*}\right)$.
Finally, the following facts hold:
• If $d=4$ or $d\ge 6$, then
${\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)\cap {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)=\left\{0\right\}$(4.2)
for every $i\in J$.
• If $d=6$ or $d\ge 8$, then
${\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)\cap {\mathrm{Fix}}_{{H}_{d,{\eta }_{j}}}\left({E}_{V}\right)=\left\{0\right\}$(4.3)
for every $i,j\in J$ and $i\ne j$.
See [29, Theorem 2.2] for details.
#### Remark 10.
We notice that, if we consider the elliptic problem
requiring that ${V}_{{H}_{d,{\eta }_{i}}}$ and ${W}_{{H}_{d,{\eta }_{i}}}$ be ${H}_{d,{\eta }_{i}}$-invariant (instead of radially symmetric) under the action of the group ${H}_{d,{\eta }_{i}}$ on ${ℝ}^{d}$, for some $i\in J$, then Theorem 6 ensures the existence of at least one non-trivial solution.
Let
${c}_{i,q}:=sup\left\{\frac{{\parallel u\parallel }_{q}}{{\parallel u\parallel }_{{E}_{V}}}:u\in {\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)\setminus \left\{0\right\}\right\}$
for every $q\in \left(2,{2}^{*}\right)$, and $i\in J$. Setting
${\lambda }_{i,q}^{\star }:=\frac{q}{{\alpha }_{f}{c}_{i,q}}\underset{\gamma >0}{\mathrm{max}}\left(\frac{\gamma }{q\sqrt{2}{\parallel W\parallel }_{p}+{2}^{\frac{q}{2}}{c}_{i,q}^{q-1}{\parallel W\parallel }_{\mathrm{\infty }}{\gamma }^{q-1}}\right),$(4.4)
our main result reads as follows.
#### Theorem 11.
Assume $d\mathrm{>}\mathrm{3}$ and let f, V and W as in Theorem 6. In addition, suppose that the nonlinearity f is odd. Then there exists a positive number ${\lambda }_{\mathrm{\star }}$ given by
such that, for every $\lambda \mathrm{\in }\mathrm{\left(}\mathrm{0}\mathrm{,}{\lambda }_{\mathrm{\star }}\mathrm{\right)}$, the problem
(Slambda’)
${\zeta }_{S}^{\left(d\right)}:=1+{\left(-1\right)}^{d}+\left[\frac{d-3}{2}\right]$
pairs of non-trivial weak solutions ${\mathrm{\left\{}\mathrm{±}{u}_{\lambda \mathrm{,}i}\mathrm{\right\}}}_{i\mathrm{\in }{J}^{\mathrm{\prime }}}\mathrm{\subset }{E}_{V}$, where ${J}^{\mathrm{\prime }}\mathrm{:=}\mathrm{\left\{}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathrm{,}{\zeta }_{S}^{\mathrm{\left(}d\mathrm{\right)}}\mathrm{\right\}}$, such that
$\underset{\lambda \to {0}^{+}}{lim}{\parallel {u}_{\lambda ,i}\parallel }_{{E}_{V}}=0,$
and $\mathrm{|}{u}_{\lambda \mathrm{,}i}\mathit{}\mathrm{\left(}x\mathrm{\right)}\mathrm{|}\mathrm{\to }\mathrm{0}$, as $\mathrm{|}x\mathrm{|}\mathrm{\to }\mathrm{\infty }$, for every $i\mathrm{\in }{J}^{\mathrm{\prime }}$. Moreover, if $d\mathrm{\ne }\mathrm{5}$, problem (Slambda’) admits at least
${\tau }_{d}:={\left(-1\right)}^{d}+\left[\frac{d-3}{2}\right]$
pairs of sign-changing weak solutions ${\mathrm{\left\{}\mathrm{±}{z}_{\lambda \mathrm{,}i}\mathrm{\right\}}}_{i\mathrm{\in }J}\mathrm{\subset }{E}_{V}$.
#### Proof.
We divide the proof into two parts.
Part 1: Dimension $d\mathrm{=}\mathrm{5}$. Since f is odd, the energy functional
is even. Owing to Theorem 6, for every $\lambda \in \left(0,{\lambda }^{\star }\right)$, problem (Slambda’) admits at least one (that is, ${\zeta }_{S}^{\left(5\right)}=1$) non-trivial pair of radial weak solutions $\left\{±{u}_{\lambda }\right\}\subset {E}_{V}$. Furthermore, the functions $±{u}_{\lambda }$ are homoclinic and
$\underset{\lambda \to {0}^{+}}{lim}{\parallel {u}_{\lambda }\parallel }_{{E}_{V}}=0.$
This concludes the first part of the proof.
Part 2: Dimension $d\mathrm{>}\mathrm{3}$ and $d\mathrm{\ne }\mathrm{5}$. For every $\lambda >0$ and $i=1,2,\mathrm{\dots },{\tau }_{d}$, consider the restrictions
${\mathcal{ℋ}}_{\lambda ,i}:={{J}_{\lambda }|}_{{\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)}:{\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)\to ℝ$
defined by
${\mathcal{ℋ}}_{\lambda ,i}:={\mathrm{\Phi }}_{{H}_{d,{\eta }_{i}}}\left(u\right)-{\lambda \mathrm{\Psi }|}_{{\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)}\left(u\right),$
where
${\mathrm{\Phi }}_{{H}_{d,{\eta }_{i}}}\left(u\right):=\frac{1}{2}{\parallel u\parallel }_{{E}_{V}}^{2}\mathit{ }\text{and}\mathit{ }{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)}\left(u\right):={\int }_{{ℝ}^{d}}W\left(x\right)F\left(u\left(x\right)\right)𝑑x$
for every $u\in {\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)$. In order to obtain the existence of
${\tau }_{d}:={\left(-1\right)}^{d}+\left[\frac{d-3}{2}\right]$
pairs of sign-changing weak solutions ${\left\{±{z}_{\lambda ,i}\right\}}_{i\in J}\subset {E}_{V}$, where $J:=\left\{1,\mathrm{\dots },{\tau }_{d}\right\}$, the main idea of the proof consists in applying Theorem 3 to the functionals ${\mathcal{ℋ}}_{\lambda ,i}$ for every $i\in J$. We notice that, since $d>3$ and $d\ne 5$, ${\tau }_{d}\ge 1$. Consequently, the cardinality $|J|\ge 1$.
Since $0<\lambda <{\lambda }_{i,q}^{\star }$, with $i\in J$, there exists ${\overline{\gamma }}_{i}>0$ such that
$\lambda <{\lambda }_{\star }^{\left(i\right)}\left({\overline{\gamma }}_{i}\right):=\frac{q}{{\alpha }_{f}{c}_{i,q}}\left(\frac{{\overline{\gamma }}_{i}}{q\sqrt{2}{\parallel W\parallel }_{p}+{2}^{\frac{q}{2}}{c}_{i,q}^{q-1}{\parallel W\parallel }_{\mathrm{\infty }}{\overline{\gamma }}_{i}^{q-1}}\right).$
Similar arguments used proving (3.7) yield
$\phi \left({\overline{\gamma }}_{i}^{2}\right)\le \chi \left({\overline{\gamma }}_{i}^{2}\right)\le {\alpha }_{f}{c}_{q}\left(\sqrt{2}\frac{{\parallel W\parallel }_{p}}{{\overline{\gamma }}_{i}}+\frac{{2}^{\frac{q}{2}}{c}_{q}^{q-1}}{q}{\parallel W\parallel }_{\mathrm{\infty }}\overline{\gamma }_{i}{}^{q-2}\right)<\frac{1}{\lambda }.$
Thus,
$\lambda \in \left(0,\frac{q}{{\alpha }_{f}{c}_{q}}\left(\frac{\overline{\gamma }}{q\sqrt{2}{\parallel W\parallel }_{p}+{2}^{\frac{q}{2}}{c}_{q}^{q-1}{\parallel W\parallel }_{\mathrm{\infty }}{\overline{\gamma }}^{q-1}}\right)\right)\subseteq \left(0,\frac{1}{\phi \left({\overline{\gamma }}_{i}^{2}\right)}\right).$
Thanks to Theorem 3, there exists a function ${z}_{\lambda ,i}\in {\mathrm{\Phi }}_{{H}_{d,{\eta }_{i}}}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}_{i}^{2}\right)\right)$ such that
${\mathrm{\Phi }}_{{H}_{d,{\eta }_{i}}}^{\prime }\left({z}_{\lambda ,i}\right)-\lambda {\left({\mathrm{\Psi }|}_{{\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)}\right)}^{\prime }\left({z}_{\lambda ,i}\right)=0,$
and, in particular, ${z}_{\lambda ,i}$ is a global minimum of the restriction of ${\mathcal{ℋ}}_{\lambda ,i}$ to ${\mathrm{\Phi }}_{{H}_{d,{\eta }_{i}}}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}_{i}^{2}\right)\right)$.
Due to the evenness of ${J}_{\lambda }$, bearing in mind (4.1), and thanks to the symmetry assumptions on the potentials V and W, we have that ${J}_{\lambda }\left(h⊛u\right)={J}_{\lambda }\left(u\right)$ for every $h\in {H}_{d,{\eta }_{i}}$ and $u\in {E}_{V}$, i.e., the functional ${J}_{\lambda }$ is ${H}_{d,{\eta }_{i}}$-invariant on ${E}_{V}$. Indeed, ${H}_{d,{\eta }_{i}}$ acts isometrically on ${E}_{V}$ (note that V is radial) and, thanks to the symmetry assumption on W, one has
${\int }_{{ℝ}^{d}}W\left(x\right)F\left(h⊛u\left(x\right)\right)𝑑x={\int }_{{ℝ}^{d}}W\left(x\right)F\left(u\left({h}^{-1}x\right)\right)𝑑x={\int }_{{ℝ}^{d}}W\left(y\right)F\left(u\left(y\right)\right)𝑑y$
if $h\in {H}_{d,i}$, and
${\int }_{{ℝ}^{d}}W\left(x\right)F\left(h⊛u\left(x\right)\right)𝑑x={\int }_{{ℝ}^{d}}W\left(x\right)F\left(u\left({g}^{-1}{\eta }_{{H}_{d,i}}^{-1}x\right)\right)𝑑x={\int }_{{ℝ}^{d}}W\left(y\right)F\left(u\left(y\right)\right)𝑑y$
if $h={\eta }_{{H}_{d,i}}g\in {H}_{d,{\eta }_{i}}\setminus {H}_{d,i}$.
On account of the principle of symmetric criticality (recalled in Theorem 4), the critical point pairs $\left\{±{z}_{\lambda ,i}\right\}$ of ${\mathcal{ℋ}}_{\lambda ,i}$ are also critical points of ${J}_{\lambda }$. Now, we have to show that the solution ${z}_{\lambda ,i}$ found here above is not the trivial function. If $f\left(0\right)\ne 0$, then it easily follows that ${z}_{\lambda ,i}\not\equiv 0$ in ${\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)$, since the trivial function does not solve problem (Slambda’). Let us consider the case when $f\left(0\right)=0$ and let ${z}_{\lambda ,i}$ be such that
and
${\mathrm{\Phi }}_{{H}_{d,{\eta }_{i}}}\left({z}_{\lambda ,i}\right)<{\overline{\gamma }}_{i}^{2},$
and also ${z}_{\lambda ,i}$ is a critical point of ${\mathcal{ℋ}}_{\lambda ,i}$ in ${\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)$. In this setting, in order to prove that we have ${z}_{\lambda ,i}\not\equiv 0$ in ${\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)$, first we claim that there exists a sequence ${\left\{{w}_{j}^{i}\right\}}_{j\in ℕ}$ in ${\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)$ such that
$\underset{j\to +\mathrm{\infty }}{lim sup}\frac{{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)}\left({w}_{j}^{i}\right)}{\mathrm{\Phi }\left({w}_{j}^{i}\right)}=+\mathrm{\infty }.$(4.5)
In order to construct the sequence ${\left\{{w}_{j}^{i}\right\}}_{j\in ℕ}\subset {\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)$ for which (4.5) holds, we use, in a suitable way, the test functions introduced by Kristály, Moroşanu and O’Regan in [29]. Let $0 be such that condition (3.11) holds and $r\ge \frac{R}{5+4\sqrt{2}}$. Set $\sigma \in \left(0,1\right)$ and define ${v}_{\sigma }^{i}\in {E}_{V}$ as follows:
for every $x\in {ℝ}^{d}$, where
${v}_{\sigma }^{\frac{d-2}{2}}\left({x}_{1},{x}_{3}\right):=\left[{\left(\frac{R-r}{4}-\mathrm{max}\left\{\sqrt{{\left({|{x}_{1}|}^{2}-\frac{R+3r}{4}\right)}^{2}+{|{x}_{3}|}^{2}},\sigma \frac{R-r}{4}\right\}\right)}_{+}$$-{\left(\frac{R-r}{4}-\mathrm{max}\left\{\sqrt{{\left({|{x}_{1}|}^{2}-\frac{R+3r}{4}\right)}^{2}+{|{x}_{3}|}^{2}},\sigma \frac{R-r}{4}\right\}\right)}_{+}\right]×\frac{4}{\left(R-r\right)\left(1-\sigma \right)}$
for every $\left({x}_{1},{x}_{3}\right)\in {ℝ}^{\frac{d}{2}}×{ℝ}^{\frac{d}{2}}$ and
${v}_{i}^{\sigma }\left({x}_{1},{x}_{2},{x}_{3}\right):=\left[{\left(\frac{R-r}{4}-\mathrm{max}\left\{\sqrt{{\left({|{x}_{1}|}^{2}-\frac{R+3r}{4}\right)}^{2}+{|{x}_{3}|}^{2}},\sigma \frac{R-r}{4}\right\}\right)}_{+}$$-{\left(\frac{R-r}{4}-\mathrm{max}\left\{\sqrt{{\left({|{x}_{3}|}^{2}-\frac{R+3r}{4}\right)}^{2}+{|{x}_{1}|}^{2}},\sigma \frac{R-r}{4}\right\}\right)}_{+}\right]$$×{\left(\frac{R-r}{4}-\mathrm{max}\left\{|{x}_{2}|,\sigma \frac{R-r}{4}\right\}\right)}_{+}\frac{4}{{\left(R-r\right)}^{2}{\left(1-\sigma \right)}^{2}}$
for every $\left({x}_{1},{x}_{2},{x}_{3}\right)\in {ℝ}^{\frac{d}{2}}×{ℝ}^{d-2i-2}×{ℝ}^{\frac{d}{2}}$, and $i\ne \frac{d-2}{2}$. Now, it is possible to prove that ${v}_{\sigma }^{i}\in {\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)$. Moreover, for every $\varrho \in \left(0,1\right]$, let
${Q}_{\varrho }^{\left(1\right)}:=\left\{\left({x}_{1},{x}_{3}\right)\in {ℝ}^{i+1}×{ℝ}^{i+1}:\sqrt{{\left({|{x}_{1}|}^{2}-\frac{R+3r}{4}\right)}^{2}+{|{x}_{3}|}^{2}}\le \varrho \frac{R-r}{4}\right\},$${Q}_{\varrho }^{\left(2\right)}:=\left\{\left({x}_{1},{x}_{3}\right)\in {ℝ}^{i+1}×{ℝ}^{i+1}:\sqrt{{\left({|{x}_{3}|}^{2}-\frac{R+3r}{4}\right)}^{2}+{|{x}_{1}|}^{2}}\le \varrho \frac{R-r}{4}\right\}.$
Define
where
${D}_{\varrho }^{\frac{d-2}{2}}:=\left\{\left({x}_{1},{x}_{3}\right)\in {ℝ}^{\frac{d}{2}}×{ℝ}^{\frac{d}{2}}:\left({x}_{1},{x}_{3}\right)\in {Q}_{\varrho }^{\left(1\right)}\cap {Q}_{\varrho }^{\left(2\right)}\right\},$
for every $i\ne \frac{d-2}{2}$. The sets ${D}_{\varrho }^{i}$ have positive Lebesgue measure and they are ${H}_{d,{\eta }_{i}}$-invariant. Moreover, for every $\sigma \in \left(0,1\right)$, one has ${v}_{\sigma }^{i}\in {\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)$ and the following facts hold:
• (1)
$\mathrm{supp}\left({v}_{\sigma }^{i}\right)={D}_{1}^{i}\subseteq A\left[r,R\right]$,
• (2)
${\parallel {v}_{\sigma }^{i}\parallel }_{\mathrm{\infty }}\le 1$,
• (3)
$|{v}_{\sigma }^{i}\left(x\right)|=1$ for every $x\in {D}_{\sigma }^{i}$.
Thus, let ${w}_{j}^{i}:={t}_{j}{v}_{\sigma }^{i}$ for any $j\in ℕ$. Of course, ${w}_{j}^{i}\in {\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)$ for any $j\in ℕ$. Furthermore, taking into account the properties of ${v}_{\sigma }^{i}$ stated in 13, since F is even (this implies that $F\left({w}_{j}^{i}\left(x\right)\right)=F\left({t}_{j}\right)$ for every $x\in {D}_{\sigma }^{i}$) with $F\left(0\right)=0$, and by using (3.10) one has for j sufficiently large,
$\frac{{\mathrm{\Psi }|}_{{\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)}\left({w}_{j}^{i}\right)}{\mathrm{\Phi }\left({w}_{j}^{i}\right)}=\frac{{\int }_{{D}_{\sigma }^{i}}W\left(x\right)F\left({w}_{j}^{i}\left(x\right)\right)𝑑x+{\int }_{{A}_{r}^{R}\setminus {D}_{\sigma }^{i}}W\left(x\right)F\left({w}_{j}^{i}\left(x\right)\right)𝑑x}{\mathrm{\Phi }\left({w}_{j}^{i}\right)}$$=\frac{{\int }_{{D}_{\sigma }^{i}}W\left(x\right)F\left({t}_{j}\right)𝑑x+{\int }_{{A}_{r}^{R}\setminus {D}_{\sigma }^{i}}W\left(x\right)F\left({t}_{j}{v}_{\sigma }^{i}\left(x\right)\right)𝑑x}{\mathrm{\Phi }\left({w}_{j}^{i}\right)}$$\ge 2\frac{M|{D}_{\sigma }^{i}|\alpha {t}_{j}^{2}+{\int }_{{A}_{r}^{R}\setminus {D}_{\sigma }^{i}}W\left(x\right)F\left({t}_{j}{v}_{\sigma }^{i}\left(x\right)\right)𝑑x}{{t}_{j}^{2}{\parallel {v}_{\sigma }^{i}\parallel }_{{E}_{V}}^{2}}.$(4.6)
Arguing as in the proof of Theorem 6, inequality (4.6) yields (4.5) and consequently we conclude that
${\mathcal{ℋ}}_{\lambda ,i}\left({z}_{\lambda ,i}\right)\le {\mathcal{ℋ}}_{\lambda ,i}\left({w}_{j}^{i}\right)<0={\mathcal{ℋ}}_{\lambda ,i}\left(0\right),$
so that ${z}_{\lambda ,i}\not\equiv 0$ in ${\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right)$. In addition, by adapting again the arguments used along the proof of Theorem 6 it follows that
$\underset{\lambda \to {0}^{+}}{lim}{\parallel {z}_{\lambda ,i}\parallel }_{{E}_{V}}=0,$
and $|{z}_{\lambda ,i}\left(x\right)|\to 0$ as $|x|\to \mathrm{\infty }$.
On the other hand, since $\lambda <{\lambda }^{\star }$ and f is odd, Theorem 6 and the principle of symmetric criticality (recalled in Theorem 4) ensure that problem (Slambda’) admits at least one non-trivial pair of radial weak solutions $\left\{±{u}_{\lambda }\right\}\subset {E}_{V}$. Moreover,
$\underset{\lambda \to {0}^{+}}{lim}{\parallel {u}_{\lambda }\parallel }_{{E}_{V}}=0,$
and $|{u}_{\lambda }\left(x\right)|\to 0$ as $|x|\to \mathrm{\infty }$.
In conclusion, since $\lambda <{\lambda }_{\star }$, there exist ${\tau }_{d}+1$ positive numbers $\overline{\gamma }$, ${\overline{\gamma }}_{1},\mathrm{\dots },{\overline{\gamma }}_{{\tau }_{d}}$ such that
$±{u}_{\lambda }\in {\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)\setminus \left\{0\right\}\subset {\mathrm{Fix}}_{O\left(d\right)}\left({E}_{V}\right)\mathit{ }\text{and}\mathit{ }±{z}_{\lambda ,i}\in {\mathrm{\Phi }}_{{H}_{d,{\eta }_{i}}}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}_{2}^{2}\right)\right)\setminus \left\{0\right\}\subset {\mathrm{Fix}}_{{H}_{d,{\eta }_{i}}}\left({E}_{V}\right).$
Bearing in mind relations (4.2) and (4.3) (see also [29, Theorem 2.2] for details) we have that
${\mathrm{\Phi }}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}^{2}\right)\right)\cap {\mathrm{\Phi }}_{{H}_{d,{\eta }_{i}}}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}_{i}^{2}\right)\right)\setminus \left\{0\right\}=\mathrm{\varnothing }$
for every $i\in J$ and
${\mathrm{\Phi }}_{{H}_{d,{\eta }_{i}}}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}_{i}^{2}\right)\right)\cap {\mathrm{\Phi }}_{{H}_{d,{\eta }_{j}}}^{-1}\left(\left(-\mathrm{\infty },{\overline{\gamma }}_{j}^{2}\right)\right)\setminus \left\{0\right\}=\mathrm{\varnothing }$
for every $i,j\in J$ and $i\ne j$. Consequently, problem (Slambda’) admits at least
${\zeta }_{S}^{\left(d\right)}:={\tau }_{d}+1$
pairs of non-trivial weak solutions ${\left\{±{u}_{\lambda ,i}\right\}}_{i\in {J}^{\prime }}\subset {E}_{V}$, where ${J}^{\prime }:=\left\{1,\mathrm{\dots },{\zeta }_{S}^{\left(d\right)}\right\}$, such that
$\underset{\lambda \to {0}^{+}}{lim}{\parallel {u}_{\lambda ,i}\parallel }_{{E}_{V}}=0,$
and $|{u}_{\lambda ,i}\left(x\right)|\to 0$, as $|x|\to \mathrm{\infty }$, for every $i\in {J}^{\prime }$. Moreover, by construction, it follows that
${\tau }_{d}:={\left(-1\right)}^{d}+\left[\frac{d-3}{2}\right]$
pairs of the attained solutions are sign-changing.
The proof is now complete. ∎
#### Remark 12.
In order to obtain a concrete form of the interval of parameters for which our results hold, it is necessary an explicit computation of the Sobolev embedding constants that naturally appear in Theorem 11 as well as in Theorem 6.
#### Remark 13.
We notice that the statement of Theorem 11 is not relevant in dimension three. However, also in this case, Theorem 11 gives one distinct (pair of) non-trivial and radially symmetric solution for (Slambda) whenever λ is sufficiently small.
#### Remark 14.
The conclusions of Theorem 1 in the Introduction immediately follow by Theorem 11 provided $\lambda \in \left(0,{\lambda }_{\star }\right)$.
We end this paper by exhibiting the example of a nonlinearity satisfying Theorem 1 together with the related estimate of the parameter λ.
#### Example 15.
Let $1 and $2 and let V be a potential for which $\left({h}_{V}^{4}\right)$ holds. Then, owing to Theorem 11, for λ sufficiently small, the problem
(Plambda)
admits at least one non-trivial pair of radially symmetric weak solutions $\left\{±{u}_{\lambda }\right\}$ and one pair of sign-changing weak solutions $\left\{±{z}_{\lambda }\right\}$. Moreover,
$\underset{\lambda \to {0}^{+}}{lim}{\parallel {u}_{\lambda }\parallel }_{{E}_{V}}=\underset{\lambda \to {0}^{+}}{lim}{\parallel {z}_{\lambda }\parallel }_{{E}_{V}}=0,$
and $|{u}_{\lambda }\left(x\right)|\to 0$, as well as $|{z}_{\lambda }\left(x\right)|\to 0$, as $|x|\to \mathrm{\infty }$. More precisely, set
${\parallel W\parallel }_{{s}^{\prime }}:={\left({\int }_{{ℝ}^{4}}\frac{1}{{\left(1+{|x|}^{4}\right)}^{2}}𝑑x\right)}^{\frac{s-1}{s}}={\left(\frac{\mathrm{area}\left({𝕊}^{3}\right)}{4}\right)}^{\frac{s-1}{s}},$
where ${𝕊}^{3}$ denotes the 4-dimensional Euclidean unit sphere and
${\kappa }_{s}:=sup\left\{\frac{{\parallel u\parallel }_{q}}{{\parallel u\parallel }_{{E}_{V}}}:u\in {\mathrm{Fix}}_{{H}_{4,\tau }}\left({E}_{V}\right)\setminus \left\{0\right\}\right\},$
where $\tau :{ℝ}^{4}\to {ℝ}^{4}$ is the involution given by $\tau \left({x}_{1},{x}_{3}\right):=\left({x}_{3},{x}_{1}\right)$ for every $\left({x}_{1},{x}_{3}\right)\in {ℝ}^{2}×{ℝ}^{2}$ and associated to the subgroup $O\left(2\right)×O\left(2\right)\subset O\left(4\right)$. If
${c}_{s}:=sup\left\{\frac{{\parallel u\parallel }_{s}}{{\parallel u\parallel }_{{E}_{V}}}:u\in {\mathrm{Fix}}_{O\left(4\right)}\left({E}_{V}\right)\setminus \left\{0\right\}\right\},$
then the existence result claimed for problem (Plambda) is valid for every
$0<\lambda <\frac{s}{2}\mathrm{min}\left\{\frac{1}{{\kappa }_{s}}\underset{\gamma >0}{\mathrm{max}}\left(\frac{\gamma }{s\sqrt{2}{\parallel W\parallel }_{{s}^{\prime }}+{2}^{\frac{s}{2}}{\kappa }_{s}^{s-1}{\gamma }^{s-1}}\right),\frac{1}{{c}_{s}}\underset{\gamma >0}{\mathrm{max}}\left(\frac{\gamma }{s\sqrt{2}{\parallel W\parallel }_{{s}^{\prime }}+{2}^{\frac{s}{2}}{c}_{s}^{s-1}{\gamma }^{s-1}}\right)\right\}.$
#### Remark 16.
In [29] Kristály, Moroşanu and O’Regan studied the (non-)existence of (non-)radial solutions for perturbed Schrödinger equations by using a similar variational approach and under suitable assumptions on the nonlinear term f. More precisely, they assume that the datum f satisfies the following conditions:
• (h1)
$f\left(t\right)=o\left(|t|\right)$ as $|t|\to \mathrm{\infty }$,
• (h2)
$f\left(t\right)=o\left(|t|\right)$ as $|t|\to 0$,
• (h3)
there exists ${t}_{0}\in ℝ$ such that $F\left({t}_{0}\right)>0$.
We emphasize that our results are mutually independent. For instance, [29, Theorem 1.1] cannot be applied to problem (Plambda) studied in Example 15. Indeed, in this case, since $r<2$, one has
$\underset{t\to {0}^{+}}{lim}\frac{f\left(t\right)}{t}=\underset{t\to {0}^{+}}{lim}\frac{{t}^{r-1}+{t}^{s-1}}{t}=+\mathrm{\infty },$
and consequently $\left({h}_{2}\right)$ is not verified.
#### Remark 17.
Theorems 6 and 11 are a paradigmatic application of Theorem 3 for elliptic partial differential equations on unbounded domains. This fact is due to the peculiar nature of the conclusion of the abstract Ricceri’s result. Indeed, the interplay between Theorem 3 with the Palais Principle (see Theorem 4) turns out to be successful in our setting thanks to the preliminary results presented in Sections 2 and 4.
#### Remark 18.
For completeness we mention here some recent contributions on elliptic problems defined on unbounded domains and related to the results contained in this paper [8, 22, 32, 33]. Finally, we point out that some of the theorems presented in this paper could be also achieved for a larger class of elliptic equations where the leading term is governed by some differential operators considered in [6, 7, 5]. However, in this cases, some different technical approaches need to be adopted in order to arrive to the analogous desired existence results for this wider class of energies. We will consider this interesting case in some further investigations.
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Accepted: 2018-06-05
Published Online: 2018-07-04
The paper is realized with the auspices of the Italian MIUR project Variational methods, with applications to problems in mathematical physics and geometry (2015KB9WPT 009) and the INdAM-GNAMPA Project 2017 titled Teoria e modelli non-locali.
Citation Information: Advances in Calculus of Variations, ISSN (Online) 1864-8266, ISSN (Print) 1864-8258,
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2018-12-13 17:54:56
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https://www.tmwr.org/resampling.html
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# 10 Resampling for evaluating performance
Chapter 9 described statistics for measuring model performance, but which data are best used to compute these statistics? Chapter 5 introduced the idea of data spending where the test set was recommended for obtaining an unbiased estimate of performance. However, we usually need to understand the effectiveness of the model before using the test set.
In fact, typically we can’t decide on which final model to take to the test set without making model assessments.
In this chapter, we describe an approach called resampling that can fill this gap. Resampling estimates of performance can generalize to new data. The next chapter complements this one by demonstrating statistical methods that compare resampling results.
To motivate this chapter, the next section demonstrates how naive performance estimates can often fail.
## 10.1 The resubstitution approach
Let’s again use the Ames data to demonstrate the concepts in this chapter. Section 8.6 summarizes the current state of our Ames analysis. It includes a recipe object named ames_rec, a linear model, and a workflow using that recipe and model called lm_wflow. This workflow was fit on the training set, resulting in lm_fit.
For a comparison to this linear model, we can also fit a different type of model. Random forests are a tree ensemble method that operate by creating a large number of decision trees from slightly different versions of the training set (Breiman 2001a). This collection of trees makes up the ensemble. When predicting a new sample, each ensemble member makes a separate prediction. These are averaged to create the final ensemble prediction for the new data point.
Random forest models are very powerful and they can emulate the underlying data patterns very closely. While this model can be computationally intensive, it is very low-maintenance; very little preprocessing is required (as documented in Appendix A).
Using the same predictor set as the linear model (without the extra preprocessing steps), we can fit a random forest model to the training set using via the ranger package. This model requires no preprocessing so a simple formula can be used:
rf_model <-
rand_forest(trees = 1000) %>%
set_engine("ranger") %>%
set_mode("regression")
rf_wflow <-
workflow() %>%
Sale_Price ~ Neighborhood + Gr_Liv_Area + Year_Built + Bldg_Type +
Latitude + Longitude) %>%
rf_fit <- rf_wflow %>% fit(data = ames_train)
How should the two models be compared? For demonstration, we will predict the training set to produce what is known as the “apparent error rate” or the “resubstitution error rate”. This function creates predictions and formats the results:
estimate_perf <- function(model, dat) {
# Capture the names of the objects used
cl <- match.call()
obj_name <- as.character(cl$model) data_name <- as.character(cl$dat)
data_name <- gsub("ames_", "", data_name)
# Estimate these metrics:
reg_metrics <- metric_set(rmse, rsq)
model %>%
predict(dat) %>%
bind_cols(dat %>% select(Sale_Price)) %>%
reg_metrics(Sale_Price, .pred) %>%
select(-.estimator) %>%
mutate(object = obj_name, data = data_name)
}
Both RMSE and R2 are computed. The resubstitution statistics are:
estimate_perf(rf_fit, ames_train)
#> # A tibble: 2 x 4
#> .metric .estimate object data
#> <chr> <dbl> <chr> <chr>
#> 1 rmse 0.0350 rf_fit train
#> 2 rsq 0.964 rf_fit train
estimate_perf(lm_fit, ames_train)
#> # A tibble: 2 x 4
#> .metric .estimate object data
#> <chr> <dbl> <chr> <chr>
#> 1 rmse 0.0744 lm_fit train
#> 2 rsq 0.822 lm_fit train
Based on these results, the random forest is much more capable of predicting the sale prices; the RMSE estimate is 2.13-fold better than linear regression. If these two models were under consideration for this prediction problem, the random forest would probably be chosen. The next step applies the random forest model to the test set for final verification:
estimate_perf(rf_fit, ames_test)
#> # A tibble: 2 x 4
#> .metric .estimate object data
#> <chr> <dbl> <chr> <chr>
#> 1 rmse 0.0746 rf_fit test
#> 2 rsq 0.828 rf_fit test
The test set RMSE estimate, 0.0746, is much worse than the training set value of 0.035! Why did this happen?
Many predictive models are capable of learning complex trends from the data. In statistics, these are commonly referred to as low bias models.
In this context, bias is the difference between the true data pattern and the types of patterns that the model can emulate. Many black-box machine learning models have low bias. Other models (such as linear/logistic regression, discriminant analysis, and others) are not as adaptable and are considered high-bias models. See Section 1.2.5 of Kuhn and Johnson (2020) for a discussion.
For a low-bias model, the high degree of predictive capacity can sometimes result in the model nearly memorizing the training set data. As an obvious example, consider a 1-nearest neighbor model. It will always provide perfect predictions for the training set no matter how well it truly works for other data sets. Random forest models are similar; re-predicting the training set will always result in an artificially optimistic estimate of performance.
For both models, this table summarizes the RMSE estimate for the training and test sets:
RMSE Estimates
object train test
lm_fit 0.0744 0.0808
rf_fit 0.0350 0.0746
Notice that the linear regression model is consistent between training and testing, because of its limited complexity12.
The main take-away from this example is that re-predicting the training set is a bad idea for most models.
If the test set should not be used immediately, and re-predicting the training set is a bad idea, what should be done? Resampling methods, such as cross-validation or validation sets, are the solution.
## 10.2 Resampling methods
Resampling methods are empirical simulation systems that emulate the process of using some data for modeling and different data for evaluation. Most resampling methods are iterative, meaning that this process is repeated multiple times. This diagram illustrates how resampling methods generally operate:
Resampling is only conducted on the training set. The test set is not involved. For each iteration of resampling, the data are partitioned into two subsamples:
• The model is fit with the analysis set.
• The model is evaluated with the assessment set.
These are somewhat analogous to training and test sets. Our language of analysis and assessment avoids confusion with initial split of the data. These data sets are mutually exclusive. The partitioning scheme used to create the analysis and assessment sets is usually the defining characteristic of the method.
Suppose twenty iterations of resampling are conducted. This means that twenty separate models are fit on the analysis sets and the corresponding assessment sets produce twenty sets of performance statistics. The final estimate of performance for a model is the average of the twenty replicates of the statistics. This average has very good generalization properties and is far better than the resubstituion estimates.
The next section defines several commonly used methods and discusses their pros and cons.
### 10.2.1 Cross-validation
Cross-validation is a well established resampling method. While there are a number of variations, the most common cross-validation method is V-fold cross-validation. The data are randomly partitioned into V sets of roughly equal size (called the “folds”). For illustration, V = 3 is shown below for a data set of thirty training set points with random fold allocations. The number inside the symbols is the sample number:
The color of the symbols represent their randomly assigned folds. Stratified sampling is also an option for assigning folds (previously discussed in Section 5.1).
For 3-fold cross-validation, the three iterations of resampling are illustrated below. For each iteration, one fold is held out for assessment statistics and the remaining folds are substrate for the model. This process continues for each fold so that three models produce three sets of performance statistics.
When V = 3, the analysis sets are 2/3 of the training set and each assessment set is a distinct 1/3. The final resampling estimate of performance averages each of the V replicates.
Using V = 3 is a good choice to illustrate cross-validation but is a poor choice in practice. Values of V are most often 5 or 10; we generally prefer 10-fold cross-validation as a default.
What are the effects of changing V? Larger values result in resampling estimates with small bias but substantial noise. Smaller values of V have large bias but low noise. We prefer 10-fold since noise is reduced by replication, as shown below, but bias is not. See Section 3.4 of Kuhn and Johnson (2020) for a longer description.
The primary input is the training set data frame as well as the number of folds (defaulting to 10):
set.seed(55)
ames_folds <- vfold_cv(ames_train, v = 10)
ames_folds
#> # 10-fold cross-validation
#> # A tibble: 10 x 2
#> splits id
#> <list> <chr>
#> 1 <split [2K/220]> Fold01
#> 2 <split [2K/220]> Fold02
#> 3 <split [2K/220]> Fold03
#> 4 <split [2K/220]> Fold04
#> 5 <split [2K/220]> Fold05
#> 6 <split [2K/220]> Fold06
#> # … with 4 more rows
The column named splits contains the information on how to split the data (similar to the object used to create the initial training/test partition). While each row of splits has an embedded copy of the entire training set, R is smart enough not to make copies of the data in memory13. The print method inside of the tibble shows the frequency of each: [2K/220] indicates that roughly two thousand samples are in the analysis set and 220 are in that particular assessment set.
To manually retrieve the partitioned data, the analysis() and assessment() functions return the corresponding data frames:
# For the first fold:
ames_folds$splits[[1]] %>% analysis() %>% dim() #> [1] 1979 74 The tidymodels packages, such as tune, contain high-level user interfaces so that functions like analysis() are not generally needed for day-to-day work. Section 10.3 demonstrates a function to fit a model over these resamples. These rsample objects also always contain a character column called id that labels the partition. Some resampling methods require multiple id fields. ### Repeated cross-validation There are a variety of variations on cross-validation. The most important is repeated V-fold cross-validation. Depending on the size or other characteristics of the data, the resampling estimate produced by V-fold cross-validation may be excessively noisy14. As with many statistical problems, one way to reduce noise is to gather more data. For cross-validation, this means averaging more than V statistics. To create R repeats of V-fold cross-validation, the same fold generation process is done R times to generate R collections of V partitions. Now, instead of averaging V statistics, $$V \times R$$ statistics produce the final resampling estimate. Due to the Central Limit Theorem, the summary statistics from each model tend toward a normal distribution. Consider the Ames data. On average, 10-fold cross-validation uses assessment sets that contain roughly 219 properties. If RMSE is the statistic of choice, we can denote that estimate’s standard deviation as $$\sigma$$. With simple 10-fold cross-validation, the standard error of the mean RMSE is $$\sigma/\sqrt{10}$$. If this is too noisy, repeats reduce the standard error to $$\sigma/\sqrt{10R}$$. For 10-fold cross-validation with $$R$$ replicates, the plot below shows how quickly the standard error15 decreases with replicates: Larger number of replicates tend to have less impact on the standard error. However, if the baseline value of $$\sigma$$ is impractically large, the diminishing returns on replication may still be worth the extra computational costs. To create repeats, invoke vfold_cv() with an additional argument repeats: vfold_cv(ames_train, v = 10, repeats = 5) #> # 10-fold cross-validation repeated 5 times #> # A tibble: 50 x 3 #> splits id id2 #> <list> <chr> <chr> #> 1 <split [2K/220]> Repeat1 Fold01 #> 2 <split [2K/220]> Repeat1 Fold02 #> 3 <split [2K/220]> Repeat1 Fold03 #> 4 <split [2K/220]> Repeat1 Fold04 #> 5 <split [2K/220]> Repeat1 Fold05 #> 6 <split [2K/220]> Repeat1 Fold06 #> # … with 44 more rows ### Leave-one-out cross-validation One early variation of cross-validation was leave-one-out (LOO) cross-validation where V is the number of data points in the training set. If there are $$n$$ training set samples, $$n$$ models are fit using $$n-1$$ rows of the training set. Each model predicts the single excluded data point. At the end of resampling, the $$n$$ predictions are pooled to produce a single performance statistic. Leave-one-out methods are deficient compared to almost any other method. For anything but pathologically small samples, LOO is computationally excessive and it may not have good statistical properties. Although rsample contains a loo_cv() function, these objects are not generally integrated into the broader tidymodels frameworks. ### Monte Carlo cross-validation Finally, another variant of V-fold cross-validation is Monte Carlo cross-validation (MCCV, Xu and Liang (2001)). Like V-fold cross-validation, it allocates a fixed proportion of data to the assessment sets. The difference is that, for MCCV, this proportion of the data is randomly selected each time. This results in assessment sets that are not mutually exclusive. To create these resampling objects: mc_cv(ames_train, prop = 9/10, times = 20) #> # Monte Carlo cross-validation (0.9/0.1) with 20 resamples #> # A tibble: 20 x 2 #> splits id #> <list> <chr> #> 1 <split [2K/219]> Resample01 #> 2 <split [2K/219]> Resample02 #> 3 <split [2K/219]> Resample03 #> 4 <split [2K/219]> Resample04 #> 5 <split [2K/219]> Resample05 #> 6 <split [2K/219]> Resample06 #> # … with 14 more rows ### 10.2.2 Validation sets Previously mentioned in Section 5.3, a validation set is a single partition that is set aside to estimate performance, before using the test set: Validation sets are often used when the original pool of data is very large. In this case, a single large partition may be adequate to characterize model performance without having to do multiple iterations of resampling. With rsample, a validation set is like any other resampling object; this type is different only in that it has a single iteration16: To create a validation set object that uses 3/4 of the data for model fitting: set.seed(12) val_set <- validation_split(ames_train, prop = 3/4) val_set #> # Validation Set Split (0.75/0.25) #> # A tibble: 1 x 2 #> splits id #> <list> <chr> #> 1 <split [1.6K/549]> validation ### 10.2.3 Bootstrapping Bootstrap resampling was originally invented as a method for approximating the sampling distribution of statistics whose theoretical properties are intractable (Davison and Hinkley 1997). Using it to estimate model performance is a secondary application of the method. A bootstrap sample of the training set is a sample that is the same size as the training set but is drawn with replacement. This means that some training set data points are selected multiple times for the analysis set. Each data point has a 63.2% chance of inclusion in the training set at least once. The assessment set contains all of the training set samples that were not selected for the analysis set (on average, with 36.8% of the training set). When bootstrapping, the assessment set is often called the “out-of-bag” sample. For a training set of 30 samples, a schematic of three bootstrap samples is: Note that the sizes of the assessment sets vary. Using rsample: bootstraps(ames_train, times = 5) #> # Bootstrap sampling #> # A tibble: 5 x 2 #> splits id #> <list> <chr> #> 1 <split [2.2K/790]> Bootstrap1 #> 2 <split [2.2K/814]> Bootstrap2 #> 3 <split [2.2K/797]> Bootstrap3 #> 4 <split [2.2K/827]> Bootstrap4 #> 5 <split [2.2K/794]> Bootstrap5 Bootstrap samples produce performance estimates that have very low variance (unlike cross-validation) but have significant pessimistic bias. This means that, if the true accuracy of a model is 90%, the bootstrap would tend to estimate the value to be less than 90%. The amount of bias cannot be empirically determined with sufficient accuracy. Additionally, the amount of bias changes over the scale of the performance metric. For example, the bias is likely to be different when the accuracy is 90% versus when it is 70%. The bootstrap is also used inside of many models. For example, the random forest model mentioned earlier contained 1,000 individual decision trees. Each tree was the product of a different bootstrap sample of the training set. ### 10.2.4 Rolling forecasting origin resampling When the data have a strong time component, a resampling method should support modeling to estimate seasonal and other temporal trends within the data. A technique that randomly samples values from the training set can disrupt the model’s ability to estimate these patterns. Rolling forecast origin resampling (Hyndman and Athanasopoulos 2018) provides a method that emulates how time series data is often partitioned in practice, estimating the model with historical data and evaluating it with the most recent data. For this type of resampling, the size of the initial analysis and assessment sets are specified. The first iteration of resampling uses these sizes, starting from the beginning of the series. The second iteration uses the same data sizes but shifts over by a set number of samples. To illustrate, a training set of fifteen samples was resampled with an analysis size of eight samples and an assessment set size of three. The second iteration discards the first training set sample and both data sets shift forward by one. This configuration results in five resamples: There are a few different configurations of this method: • The analysis set can cumulatively grow (as opposed to remaining the same size). After the first initial analysis set, new samples can accrue without discarding the earlier data. • The resamples need not increment by one. For example, for large data sets, the incremental block could be a week or month instead of a day. For a year’s worth of data, suppose that six sets of 30-day blocks define the analysis set. For assessment sets of 30 days with a 29 day skip, the rsample code is: time_slices <- tibble(x = 1:365) %>% rolling_origin(initial = 6 * 30, assess = 30, skip = 29, cumulative = FALSE) data_range <- function(x) { summarize(x, first = min(x), last = max(x)) } map_dfr(time_slices$splits, ~ analysis(.x) %>% data_range())
#> # A tibble: 6 x 2
#> first last
#> <int> <int>
#> 1 1 180
#> 2 31 210
#> 3 61 240
#> 4 91 270
#> 5 121 300
#> 6 151 330
map_dfr(time_slices$splits, ~ assessment(.x) %>% data_range()) #> # A tibble: 6 x 2 #> first last #> <int> <int> #> 1 181 210 #> 2 211 240 #> 3 241 270 #> 4 271 300 #> 5 301 330 #> 6 331 360 ## 10.3 Estimating performance Any of these resampling methods can be used to evaluate the modeling process (including preprocessing, model fitting, etc). These methods are effective because different groups of data are used to train the model and assess the model. To reiterate the process: 1. During resampling, the analysis set is used to preprocess the data, apply the preprocessing to itself, and use these processed data to fit the model. 2. The preprocessing statistics produced by the analysis set are applied to the assessment set. The predictions from the assessment set estimate performance. This sequence repeats for every resample. If there are B resamples, there are B replicates of each of the performance metrics. The final resampling estimate is the average of these B statistics. If B = 1, as with a validation set, the individual statistics represent overall performance. Let’s reconsider the previous random forest model contained in the rf_wflow object. The fit_resamples() function is analogous to fit(), but instead of having a data argument, fit_resamples() has resamples which expects an rset object like the ones shown above. The possible interfaces to the function are: model_spec %>% fit_resamples(formula, resamples, ...) model_spec %>% fit_resamples(recipe, resamples, ...) workflow %>% fit_resamples( resamples, ...) There are a number of other optional arguments, such as: • metrics: A metric set of performance statistics to compute. By default, regression models use RMSE and R2 while classification models compute the area under the ROC curve and overall accuracy. Note that this choice also defines what predictions are produced during the evaluation of the model. For classification, if only accuracy is requested, class probability estimates are not generated for the assessment set (since they are not needed). • control: A list created by control_resamples() with various options. The control arguments include: • verbose: A logical for printing logging. • extract: A function for retaining objects from each model iteration (discussed below). • save_pred: A logical for saving the assessment set predictions. For our example, let’s save the predictions in order to visualize the model fit and residuals: keep_pred <- control_resamples(save_pred = TRUE) set.seed(130) rf_res <- rf_wflow %>% fit_resamples(resamples = ames_folds, control = keep_pred) rf_res #> # Resampling results #> # 10-fold cross-validation #> # A tibble: 10 x 5 #> splits id .metrics .notes .predictions #> <list> <chr> <list> <list> <list> #> 1 <split [2K/220]> Fold01 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [220 × 4]> #> 2 <split [2K/220]> Fold02 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [220 × 4]> #> 3 <split [2K/220]> Fold03 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [220 × 4]> #> 4 <split [2K/220]> Fold04 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [220 × 4]> #> 5 <split [2K/220]> Fold05 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [220 × 4]> #> 6 <split [2K/220]> Fold06 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [220 × 4]> #> # … with 4 more rows The return value is a tibble similar to the input resamples, along with some extra columns: • .metrics is a list column of tibbles containing the assessment set performance statistics. • .notes is another list column of tibbles cataloging any warnings or errors generated during resampling. Note that errors will not stop subsequent execution of resampling. • .predictions is present when save_pred = TRUE. This list column contains tibbles with the out-of-sample predictions. While these list columns may look daunting, they can be easily reconfigured using tidyr or with convenience functions that tidymodels provides. For example, to return the performance metrics in a more usable format: collect_metrics(rf_res) #> # A tibble: 2 x 6 #> .metric .estimator mean n std_err .config #> <chr> <chr> <dbl> <int> <dbl> <chr> #> 1 rmse standard 0.0709 10 0.00175 Preprocessor1_Model1 #> 2 rsq standard 0.839 10 0.00819 Preprocessor1_Model1 These are the resampling estimates averaged over the individual replicates. To get the metrics for each resample, use the option summarize = FALSE Notice how much more realistic the performance estimates are than the resubstitution estimates from Section 10.1! To obtain the assessment set predictions: assess_res <- collect_predictions(rf_res) assess_res #> # A tibble: 2,199 x 5 #> id .pred .row Sale_Price .config #> <chr> <dbl> <int> <dbl> <chr> #> 1 Fold01 5.25 11 5.23 Preprocessor1_Model1 #> 2 Fold01 5.35 47 5.35 Preprocessor1_Model1 #> 3 Fold01 5.42 52 5.46 Preprocessor1_Model1 #> 4 Fold01 5.14 94 5.15 Preprocessor1_Model1 #> 5 Fold01 5.17 111 5.15 Preprocessor1_Model1 #> 6 Fold01 5.13 115 5.04 Preprocessor1_Model1 #> # … with 2,193 more rows The prediction column names follow the conventions discussed for parsnip models. The observed outcome column always uses the original column name from the source data. The .row column is an integer that matches the row of the original training set so that these results can be properly arranged and joined with the original data. For some resampling methods, such as the bootstrap or repeated cross-validation, there will be multiple predictions per row of the original training set. To obtain summarized values (averages of the replicate predictions) use collect_predictions(object, summarize = TRUE). Since this analysis used 10-fold cross-validation, there is one unique prediction for each training set sample. These data can generate helpful plots of the model to understand where it potentially failed. For example, let’s compare the observed and predicted values: assess_res %>% ggplot(aes(x = Sale_Price, y = .pred)) + geom_point(alpha = .15) + geom_abline(col = "red") + coord_obs_pred() + ylab("Predicted") There was one house in the training set with a low observed sale price that is significantly overpredicted by the model. Which house was that? over_predicted <- assess_res %>% mutate(residual = Sale_Price - .pred) %>% arrange(desc(abs(residual))) %>% slice(1) over_predicted #> # A tibble: 1 x 6 #> id .pred .row Sale_Price .config residual #> <chr> <dbl> <int> <dbl> <chr> <dbl> #> 1 Fold03 4.94 1165 4.12 Preprocessor1_Model1 -0.822 ames_train %>% slice(over_predicted$.row) %>%
select(Gr_Liv_Area, Neighborhood, Year_Built, Bedroom_AbvGr, Full_Bath)
#> # A tibble: 1 x 5
#> Gr_Liv_Area Neighborhood Year_Built Bedroom_AbvGr Full_Bath
#> <int> <fct> <int> <int> <int>
#> 1 733 Iowa_DOT_and_Rail_Road 1952 2 1
These results can help us investigate why the prediction was poor for this house.
How can we use a validation set instead of cross-validation? From our previous rsample object:
val_res <- rf_wflow %>% fit_resamples(resamples = val_set)
val_res
#> # Resampling results
#> # Validation Set Split (0.75/0.25)
#> # A tibble: 1 x 4
#> splits id .metrics .notes
#> <list> <chr> <list> <list>
#> 1 <split [1.6K/549]> validation <tibble [2 × 4]> <tibble [0 × 1]>
collect_metrics(val_res)
#> # A tibble: 2 x 6
#> .metric .estimator mean n std_err .config
#> <chr> <chr> <dbl> <int> <dbl> <chr>
#> 1 rmse standard 0.0668 1 NA Preprocessor1_Model1
#> 2 rsq standard 0.859 1 NA Preprocessor1_Model1
These results are also much closer to the test set results than the resubstitution estimates of performance.
In these analyses, the resampling results are very close to the test set results. The two types of estimates tend to be well correlated. However, this could be from random chance. A seed value of 1352 fixed the random numbers before creating the resamples. Try changing this value and re-running the analyses to investigate whether the resampled estimates match the test set results as well.
## 10.4 Parallel processing
The models created during resampling are independent of one another. Computations of this kind are sometimes called “embarrassingly parallel”; each model could be fit simultaneously without issues. The tune package uses the foreach package to facilitate parallel computations. These computations could be split across processors on the same computer or across different computers, depending on the chosen technology.
For computations conducted on a single computer, the number of possible “worker processes” is determined by the parallel package:
# The number of physical cores in the hardware:
parallel::detectCores(logical = FALSE)
#> [1] 4
# The number of possible independent processes that can
# be simultaneously used:
parallel::detectCores(logical = TRUE)
#> [1] 4
The difference between these two values is related to the computer’s processor. For example, most Intel processors use hyper-threading which creates two virtual cores for each physical core. While these extra resources can improve performance, most of the speed-ups produced by parallel processing occur when processing uses fewer than the number of physical cores.
For fit_resamples() and other functions in tune, parallel processing occurs when the user registers a parallel backend package. These R packages define how to execute parallel processing. On Unix and macOS operating systems, one method of splitting computations is by forking threads. To enable this, load the doMC package and register the number of parallel cores with foreach:
# Unix and macOS only
library(doMC)
registerDoMC(cores = 2)
# Now run fit_resamples()...
This instructs fit_resamples() to run half of the computations on each of two cores. To reset the computations to sequential processing:
registerDoSEQ()
Alternatively, a different approach to parallelizing computations uses network sockets. The doParallel package enables this method (usable by all operating systems):
# All operating systems
library(doParallel)
# Create a cluster object and then register:
cl <- makePSOCKcluster(2)
registerDoParallel(cl)
# Now run fit_resamples()...
stopCluster(cl)
Another R package that facilitates parallel processing is the future package. Like foreach, it provides a framework for parallelism. It is used in conjunction with foreach via the doFuture package.
The R packages with parallel backends for foreach start with the prefix "do".
Parallel processing with tune tends to provide linear speed-ups for the first few cores. This means that, with two cores, the computations are twice as fast. Depending on the data and type of model, the linear speedup deteriorates after 4-5 cores. Using more cores will still reduce the time it takes to complete the task; there are just diminishing returns for the additional cores.
Let’s wrap up with one final note about parallelism. For each of these technologies, the memory requirements multiply for each additional core used. For example, if the current data set is 2 GB in memory and three cores are used, the total memory requirement is 8 GB (2 for each worker process plus the original). Using too many cores might cause the computations (and the computer) to slow considerably.
Schmidberger et al. (2009) gives a technical overview of these technologies.
## 10.5 Saving the resampled objects
The models created during resampling are not retained. These models are trained for the purpose of evaluating performance, and we typically do not need them after we have computed performance statistics. If a particular modeling approach does turn out to be the best option for our data set, then the best choice is to fit again to the whole training set so the model parameters can be estimated with more data.
While these models created during resampling are not preserved, there is a method for keeping them or some of their components. The extract option of control_resamples() specifies a function that takes a single argument; we’ll use x. When executed, x results in a fitted workflow object, regardless of whether you provided fit_resamples() with a workflow. Recall that the workflows package has functions that can pull the different components of the objects (e.g. the model, recipe, etc.).
Let’s fit a linear regression model using the recipe shown at the end of Chapter 6:
ames_rec <-
recipe(Sale_Price ~ Neighborhood + Gr_Liv_Area + Year_Built + Bldg_Type +
Latitude + Longitude, data = ames_train) %>%
step_other(Neighborhood, threshold = 0.01) %>%
step_dummy(all_nominal()) %>%
step_interact( ~ Gr_Liv_Area:starts_with("Bldg_Type_") ) %>%
step_ns(Latitude, Longitude, deg_free = 20)
lm_wflow <-
workflow() %>%
lm_fit <- lm_wflow %>% fit(data = ames_train)
# Select the recipe:
pull_workflow_prepped_recipe(lm_fit)
#> Data Recipe
#>
#> Inputs:
#>
#> role #variables
#> outcome 1
#> predictor 6
#>
#> Training data contained 2199 data points and no missing data.
#>
#> Operations:
#>
#> Collapsing factor levels for Neighborhood [trained]
#> Dummy variables from Neighborhood, Bldg_Type [trained]
#> Interactions with Gr_Liv_Area:(Bldg_Type_TwoFmCon + Bldg_Type_Duplex + Bldg_Type_Twnhs + Bldg_Type_TwnhsE) [trained]
#> Natural Splines on Latitude, Longitude [trained]
We can save the linear model coefficients for a fitted model object from a workflow:
get_model <- function(x) {
pull_workflow_fit(x) %>% tidy()
}
# Test it using:
# get_model(lm_fit)
Now let’s apply this function to the ten resampled fits. The results of the extraction function is wrapped in a list object and returned in a tibble:
ctrl <- control_resamples(extract = get_model)
lm_res <- lm_wflow %>% fit_resamples(resamples = ames_folds, control = ctrl)
lm_res
#> # Resampling results
#> # 10-fold cross-validation
#> # A tibble: 10 x 5
#> splits id .metrics .notes .extracts
#> <list> <chr> <list> <list> <list>
#> 1 <split [2K/220]> Fold01 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [1 × 2]>
#> 2 <split [2K/220]> Fold02 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [1 × 2]>
#> 3 <split [2K/220]> Fold03 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [1 × 2]>
#> 4 <split [2K/220]> Fold04 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [1 × 2]>
#> 5 <split [2K/220]> Fold05 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [1 × 2]>
#> 6 <split [2K/220]> Fold06 <tibble [2 × 4]> <tibble [0 × 1]> <tibble [1 × 2]>
#> # … with 4 more rows
Now there is a .extracts column with nested tibbles. What do these contain?
lm_res$.extracts[[1]] #> # A tibble: 1 x 2 #> .extracts .config #> <list> <chr> #> 1 <tibble [71 × 5]> Preprocessor1_Model1 # To get the results lm_res$.extracts[[1]][[1]]
#> [[1]]
#> # A tibble: 71 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) 1.00 0.321 3.12 1.85e- 3
#> 2 Gr_Liv_Area 0.000168 0.00000481 35.0 2.09e-207
#> 3 Year_Built 0.00203 0.000149 13.6 4.07e- 40
#> 4 Neighborhood_College_Creek -0.0779 0.0356 -2.19 2.89e- 2
#> 5 Neighborhood_Old_Town -0.0600 0.0138 -4.36 1.37e- 5
#> 6 Neighborhood_Edwards -0.149 0.0291 -5.13 3.26e- 7
#> # … with 65 more rows
This might appear to be a convoluted method for saving the model results. However, extract is flexible and does not assume that the user will only save a single tibble per resample. For example, the tidy() method might be run on the recipe as well as the model. In this case, a list of two tibbles will be returned.
For our more simple example, all of the results can be flattened and collected using:
all_coef <- map_dfr(lm_res\$.extracts, ~ .x[[1]][[1]])
# Show the replicates for a single predictor:
filter(all_coef, term == "Year_Built")
#> # A tibble: 10 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 Year_Built 0.00203 0.000149 13.6 4.07e-40
#> 2 Year_Built 0.00209 0.000154 13.5 5.89e-40
#> 3 Year_Built 0.00210 0.000146 14.4 1.13e-44
#> 4 Year_Built 0.00184 0.000150 12.2 3.59e-33
#> 5 Year_Built 0.00204 0.000155 13.2 3.97e-38
#> 6 Year_Built 0.00190 0.000151 12.6 6.80e-35
#> # … with 4 more rows
Chapters 13 and 14 discuss a suite of functions for tuning models. Their interfaces are similar to fit_resamples() and many of the features described here apply to those functions.
## 10.6 Chapter summary
This chapter describes one of the fundamental tools of data analysis, the ability to measure the performance and variation in model results. Resampling enables us to determine how well the model works without using the test set.
An important function from the tune package, called fit_resamples(), was introduced. The interface for this function is also used in future chapters that describe model tuning tools.
The data analysis code, to date, for the Ames data is:
library(tidymodels)
data(ames)
ames <- mutate(ames, Sale_Price = log10(Sale_Price))
set.seed(123)
ames_split <- initial_split(ames, prob = 0.80, strata = Sale_Price)
ames_train <- training(ames_split)
ames_test <- testing(ames_split)
ames_rec <-
recipe(Sale_Price ~ Neighborhood + Gr_Liv_Area + Year_Built + Bldg_Type +
Latitude + Longitude, data = ames_train) %>%
step_log(Gr_Liv_Area, base = 10) %>%
step_other(Neighborhood, threshold = 0.01) %>%
step_dummy(all_nominal()) %>%
step_interact( ~ Gr_Liv_Area:starts_with("Bldg_Type_") ) %>%
step_ns(Latitude, Longitude, deg_free = 20)
lm_model <- linear_reg() %>% set_engine("lm")
lm_wflow <-
workflow() %>%
lm_fit <- fit(lm_wflow, ames_train)
rf_model <-
rand_forest(trees = 1000) %>%
set_engine("ranger") %>%
set_mode("regression")
rf_wflow <-
workflow() %>%
Sale_Price ~ Neighborhood + Gr_Liv_Area + Year_Built + Bldg_Type +
Latitude + Longitude) %>%
set.seed(55)
ames_folds <- vfold_cv(ames_train, v = 10)
keep_pred <- control_resamples(save_pred = TRUE)
set.seed(130)
rf_res <- rf_wflow %>% fit_resamples(resamples = ames_folds, control = keep_pred)
### REFERENCES
Breiman, L. 2001a. “Random Forests.” Machine Learning 45 (1): 5–32.
Davison, A, and D Hinkley. 1997. Bootstrap Methods and Their Application. Vol. 1. Cambridge university press.
Hyndman, R, and G Athanasopoulos. 2018. Forecasting: Principles and Practice. OTexts.
Kuhn, M, and K Johnson. 2020. Feature Engineering and Selection: A Practical Approach for Predictive Models. CRC Press.
Schmidberger, M, M Morgan, D Eddelbuettel, H Yu, L Tierney, and U Mansmann. 2009. “State of the Art in Parallel Computing with R.” Journal of Statistical Software 31 (1): 1–27. https://www.jstatsoft.org/v031/i01.
Xu, Q, and Y Liang. 2001. “Monte Carlo Cross Validation.” Chemometrics and Intelligent Laboratory Systems 56 (1): 1–11.
1. It is possible for a linear model to nearly memorize the training set, like the random forest model did. In the ames_rec object, change the number of spline terms for longitude and latitude to a large number (say 1000). This would produce a model fit with a very small resubstitution RMSE and a test set RMSE that is much larger.↩︎
2. To see this for yourself, try executing lobstr::obj_size(ames_folds) and lobstr::obj_size(ames_train)`. The size of the resample object is much less than ten times the size of the original data.↩︎
3. For more details, see Section 3.4.6 of Kuhn and Johnson (2020).↩︎
4. These are approximate standard errors. As will be discussed in the next chapter, there is a within-replicate correlation that is typical of resampled results. By ignoring this extra component of variation, the simple calculations shown in this plot are overestimates of the reduction in noise in the standard errors.↩︎
5. In essence, a validation set can be considered a single iteration of Monte Carlo cross-validation.↩︎
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2020-10-25 11:31:04
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http://mathoverflow.net/feeds/question/22007
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The word problem for fundamental groups of smooth projective varieties - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T09:30:28Z http://mathoverflow.net/feeds/question/22007 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22007/the-word-problem-for-fundamental-groups-of-smooth-projective-varieties The word problem for fundamental groups of smooth projective varieties Andy Putman 2010-04-21T03:05:49Z 2010-04-21T04:19:20Z <p>While attending a very nice talk on the geometric group theory of fundamental groups of Kahler manifolds by Pierre Py last weekend, I realized that I don't know the answer to the following question. Let $X$ be a smooth projective variety over $\mathbb{C}$. Is the <a href="http://en.wikipedia.org/wiki/Word_problem_for_groups" rel="nofollow">word problem</a> for $\pi_1(X)$ solvable?</p> <p>Here are a couple of relevant facts. Taubes proved that every finitely presentable group is the fundamental group of a compact complex manifold of complex dimension 3. Earlier, Gompf proved that every finitely presentable group is the fundamental group of a compact symplectic manifold of real dimension 4. Thus the word problem is not solvable for fundamental groups of compact complex manifolds. Also, Toledo has an example of a smooth compact projective variety whose fundamental group is not residually finite. This rules out using maps to finite groups to solve the word problem, and also shows that $\pi_1(X)$ need not be linear.</p> <p>EDIT : Another relevant remark is that the answers to the question <a href="http://mathoverflow.net/questions/15087/computing-fundamental-groups-and-singular-cohomology-of-projective-varieties" rel="nofollow">here</a> show that presentations for $\pi_1(X)$ are computable, so there are no issues there.</p> http://mathoverflow.net/questions/22007/the-word-problem-for-fundamental-groups-of-smooth-projective-varieties/22011#22011 Answer by Ben Wieland for The word problem for fundamental groups of smooth projective varieties Ben Wieland 2010-04-21T04:19:20Z 2010-04-21T04:19:20Z <p>Try <a href="http://www.ams.org/mathscinet-getitem?mr=1676613" rel="nofollow">Bogomolov and Katzarkov</a> (<a href="http://books.google.com/books?id=bJaRUqQEGMcC&pg=PA85" rel="nofollow">google books</a> or <a href="http://www.ams.org/mathscinet-getitem?mr=1616131" rel="nofollow">an earlier paper</a>). I don't understand the statements, but I think that for every finitely presented group, they find a extensions of surface groups by the given group that are "approximated" by projective groups. The quality of the approximations is not clear, but I suspect that they preserve uncomputability.</p>
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2013-05-20 09:30:28
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