url stringlengths 14 2.42k | text stringlengths 100 1.02M | date stringlengths 19 19 | metadata stringlengths 1.06k 1.1k |
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https://www.loudwallpapers.com/power-plant-wallpaper/ | # Power Plant Wallpapers
Power plant came into existence since 18th century. We have collection of quality power plant wallpapers of thermal power plant, nuclear power plant, hydro power plant, etc. with highest resolutions.
Did you know?
Power plants generate electricity which can be used in various industries. Power plants are also called as ‘Power Station’ or ‘Power Generator’.
Download power plant wallpapers HD resolution for free. Also, if you have any royalty free images then do send us.
Source for few wallpapers:
https://www.jpl.nasa.gov/news/news.php?feature=5866
https://en.wikipedia.org/wiki/List_of_largest_power_stations
https://commons.wikimedia.org/wiki/File:Bellefonte_Nuclear_Power_Plant.jpg
https://en.wikipedia.org/wiki/Power_station | 2021-05-18 19:59:37 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8355575203895569, "perplexity": 6376.239879014077}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "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-2021-21/segments/1620243991514.63/warc/CC-MAIN-20210518191530-20210518221530-00009.warc.gz"} |
https://zenodo.org/record/1326265/export/dcite4 | Presentation Closed Access
# Metamodel-based uncertainty quantification for braided composites with manufacturing considerations
Georgios Balokas; Benedikt Kriegesmann; Steffen Czichon; Raimund Rolfes
### DataCite XML Export
<?xml version='1.0' encoding='utf-8'?>
<identifier identifierType="DOI">10.5281/zenodo.1326265</identifier>
<creators>
<creator>
<creatorName>Georgios Balokas</creatorName>
<affiliation>ELAN-AUSY GmbH</affiliation>
</creator>
<creator>
<creatorName>Benedikt Kriegesmann</creatorName>
<affiliation>TUHH</affiliation>
</creator>
<creator>
<creatorName>Steffen Czichon</creatorName>
<affiliation>Fraunhofer IWES</affiliation>
</creator>
<creator>
<creatorName>Raimund Rolfes</creatorName>
<affiliation>Leibniz Universität Hannover</affiliation>
</creator>
</creators>
<titles>
<title>Metamodel-based uncertainty quantification for braided composites with manufacturing considerations</title>
</titles>
<publisher>Zenodo</publisher>
<publicationYear>2018</publicationYear>
<subjects>
<subject>Braided composites</subject>
<subject>Multiscale analysis</subject>
<subject>Monte Carlo simulation</subject>
<subject>Stiffness and strength</subject>
<subject>Neural networks</subject>
<subject>Polynomial chaos expansions</subject>
<subject>Kriging modeling</subject>
</subjects>
<dates>
<date dateType="Issued">2018-07-11</date>
</dates>
<resourceType resourceTypeGeneral="Text">Presentation</resourceType>
<alternateIdentifiers>
<alternateIdentifier alternateIdentifierType="url">https://zenodo.org/record/1326265</alternateIdentifier>
</alternateIdentifiers>
<relatedIdentifiers>
<relatedIdentifier relatedIdentifierType="DOI" relationType="IsVersionOf">10.5281/zenodo.1326264</relatedIdentifier>
</relatedIdentifiers>
<rightsList>
<rights rightsURI="info:eu-repo/semantics/closedAccess">Closed Access</rights>
</rightsList>
<descriptions>
<description descriptionType="Abstract"><p>Stochastic analysis in engineering sciences takes into account the uncertainties that may exist and affect a certain physical system in an a priori unknown manner. As the design of structures gets increasingly complex over the years, the impact of those uncertainties onto the system response has to be studied in order to implement numerical procedures for virtual testing platforms.<br>
Braided composites are of special interest for the aerospace and the automotive industry, due to their excellent performance in terms of stiffness/strength-to-weight ratio, delamination resistance, impact properties etc. The complexity of such materials sets a computational challenge when it comes to robust and reliable simulations. Efficiency also plays an important role for probabilistic assessments since the response variability needs repetitive procedures in order to be calculated (e.g. Monte Carlo simulations). Hence, the aim of this work is to present an uncertainty quantification framework for braided composites simulation, dealing with the stochastic stiffness and strength prediction via numerical multiscale analysis. The numerical burden of Monte Carlo analysis is bypassed with various metamodeling techniques, such as Neural Networks, Polynomial Chaos expansion and Kriging modeling. Uncertainties accounting for material properties randomness, geometric randomness but also for random spatial variations caused by manufacturing aspects (e.g. fabric compaction during molding,&nbsp;jamming actions during braiding), are propagating through the scales to the final scatter of the mechanical properties of the macroscale.<br>
Results offer a perspective on the variability influence of the random parameters, an overview of the performance of several surrogate models and also highlight the importance of realistic uncertainty quantification. Furthermore, this work provides a useful guidance for uncertainty propagation assessment with advanced non-intrusive metamodeling techniques.</p></description>
</descriptions>
<fundingReferences>
<fundingReference>
<funderName>European Commission</funderName>
<funderIdentifier funderIdentifierType="Crossref Funder ID">10.13039/100010661</funderIdentifier>
<awardNumber awardURI="info:eu-repo/grantAgreement/EC/H2020/642121/">642121</awardNumber>
<awardTitle>FULLY INTEGRATED ANALYSIS, DESIGN, MANUFACTURING AND HEALTH-MONITORING OF COMPOSITE STRUCTURES</awardTitle>
</fundingReference>
</fundingReferences>
</resource>
37
0
views | 2023-02-02 14:04:10 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2519701421260834, "perplexity": 4593.266227662769}, "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-2023-06/segments/1674764500028.12/warc/CC-MAIN-20230202133541-20230202163541-00580.warc.gz"} |
http://mathhelpforum.com/differential-geometry/85807-function-limits-sequence-limits.html | ## Function limits and sequence limits
Hi, I am really stuck on both parts of this analysis question could somebody please help me? I don't really know how to start, I have the definition for part a but nothing else.
1.(a). For a function (f) from the reals to the reals define what is meant by f(x) tends to infinity as x tends to negative infinity and prove it holds if and only if whenever a real sequence (x_n) tends to negative infinity f(x_n) tends to infinity.
b). If a function f from the reals to the reals is continuous and tends to infinity as x tends to negative infinity show that there exists a sequence x_n which tends to negative infinity as x tends to infinity such that f(x_n)=n for all but a finite number of n. | 2017-01-19 05:44:51 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9752268195152283, "perplexity": 103.08411144121558}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2017-04/segments/1484560280483.83/warc/CC-MAIN-20170116095120-00334-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.luogu.org/problemnew/show/CF119A | # CF119A Epic Game
• 229通过
• 426提交
• 题目来源
• 评测方式 RemoteJudge
• 标签 最大公约数,gcd 模拟
• 难度 入门难度
• 时空限制 2000ms / 256MB
• 提示：收藏到任务计划后，可在首页查看。
# 题目描述
Simon和Antisimon在玩石子游戏。
共有n颗石子，Simon先拿。
Simon能拿当前n和a的最大公约数，Antisimon能拿当前n和b的最大公约数。
当有一个人不能拿时（n=0）那个人就输了。
求谁赢了。
# 输入输出格式
输入格式
一行，a，b，n（1<=a,b,n<=100）
输出格式
一行，如果Simon赢了，输出0；Antisimon赢了，输出1.
# 说明
gcd(0,x)=gcd(x,0)=x;
对于样例1：
Simon拿gcd(3,9)=3颗
Antisimon拿gcd(5,6)=1颗
Simon拿gcd(3,5)=1颗
Antisimon拿gcd(5,4)=1颗
Simon拿gcd(3,3)=3颗
Antisimon输了
感谢@引领天下 提供的翻译
## 题目描述
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number $a$ and Antisimon receives number $b$ . They also have a heap of $n$ stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given $a$ , $b$ and $n$ who wins the game.
## 输入输出格式
输入格式：
The only string contains space-separated integers $a$ , $b$ and $n$ ( $1<=a,b,n<=100$ ) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
输出格式：
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
## 输入输出样例
输入样例#1： 复制
3 5 9
输出样例#1： 复制
0
输入样例#2： 复制
1 1 100
输出样例#2： 复制
1
## 说明
The greatest common divisor of two non-negative integers $a$ and $b$ is such maximum positive integer $k$ , that $a$ is divisible by $k$ without remainder and similarly, $b$ is divisible by $k$ without remainder. Let $gcd(a,b)$ represent the operation of calculating the greatest common divisor of numbers $a$ and $b$ . Specifically, $gcd(x,0)=gcd(0,x)=x$ .
In the first sample the game will go like that:
• Simon should take $gcd(3,9)=3$ stones from the heap. After his move the heap has $6$ stones left.
• Antisimon should take $gcd(5,6)=1$ stone from the heap. After his move the heap has $5$ stones left.
• Simon should take $gcd(3,5)=1$ stone from the heap. After his move the heap has $4$ stones left.
• Antisimon should take $gcd(5,4)=1$ stone from the heap. After his move the heap has $3$ stones left.
• Simon should take $gcd(3,3)=3$ stones from the heap. After his move the heap has $0$ stones left.
• Antisimon should take $gcd(5,0)=5$ stones from the heap. As $0<5$ , it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As $n$ is even, Antisimon takes the last stone and Simon can't make a move after that.
提示
标程仅供做题后或实在无思路时参考。
请自觉、自律地使用该功能并请对自己的学习负责。
如果发现恶意抄袭标程，将按照I类违反进行处理。 | 2019-04-20 19:07:46 | {"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": 0.42307430505752563, "perplexity": 1788.0154261740327}, "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-2019-18/segments/1555578529962.12/warc/CC-MAIN-20190420180854-20190420202854-00235.warc.gz"} |
https://www.nag.com/numeric/mb/nagdoc_mb/manual_25_1/html/f07/f07cpf.html | Integer type: int32 int64 nag_int show int32 show int32 show int64 show int64 show nag_int show nag_int
Chapter Contents
Chapter Introduction
NAG Toolbox
# NAG Toolbox: nag_lapack_zgtsvx (f07cp)
## Purpose
nag_lapack_zgtsvx (f07cp) uses the $LU$ factorization to compute the solution to a complex system of linear equations
$AX=B , ATX=B or AHX=B ,$
where $A$ is a tridiagonal matrix of order $n$ and $X$ and $B$ are $n$ by $r$ matrices. Error bounds on the solution and a condition estimate are also provided.
## Syntax
[dlf, df, duf, du2, ipiv, x, rcond, ferr, berr, info] = f07cp(fact, trans, dl, d, du, dlf, df, duf, du2, ipiv, b, 'n', n, 'nrhs_p', nrhs_p)
[dlf, df, duf, du2, ipiv, x, rcond, ferr, berr, info] = nag_lapack_zgtsvx(fact, trans, dl, d, du, dlf, df, duf, du2, ipiv, b, 'n', n, 'nrhs_p', nrhs_p)
## Description
nag_lapack_zgtsvx (f07cp) performs the following steps:
1 If ${\mathbf{fact}}=\text{'N'}$, the $LU$ decomposition is used to factor the matrix $A$ as $A=LU$, where $L$ is a product of permutation and unit lower bidiagonal matrices and $U$ is upper triangular with nonzeros in only the main diagonal and first two superdiagonals. 2 If some ${u}_{ii}=0$, so that $U$ is exactly singular, then the function returns with ${\mathbf{info}}=i$. Otherwise, the factored form of $A$ is used to estimate the condition number of the matrix $A$. If the reciprocal of the condition number is less than machine precision, ${\mathbf{info}}\ge {\mathbf{n}}+1$ is returned as a warning, but the function still goes on to solve for $X$ and compute error bounds as described below. 3 The system of equations is solved for $X$ using the factored form of $A$. 4 Iterative refinement is applied to improve the computed solution matrix and to calculate error bounds and backward error estimates for it.
## References
Anderson E, Bai Z, Bischof C, Blackford S, Demmel J, Dongarra J J, Du Croz J J, Greenbaum A, Hammarling S, McKenney A and Sorensen D (1999) LAPACK Users' Guide (3rd Edition) SIAM, Philadelphia http://www.netlib.org/lapack/lug
Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore
Higham N J (2002) Accuracy and Stability of Numerical Algorithms (2nd Edition) SIAM, Philadelphia
## Parameters
### Compulsory Input Parameters
1: $\mathrm{fact}$ – string (length ≥ 1)
Specifies whether or not the factorized form of the matrix $A$ has been supplied.
${\mathbf{fact}}=\text{'F'}$
dlf, df, duf, du2 and ipiv contain the factorized form of the matrix $A$. dlf, df, duf, du2 and ipiv will not be modified.
${\mathbf{fact}}=\text{'N'}$
The matrix $A$ will be copied to dlf, df and duf and factorized.
Constraint: ${\mathbf{fact}}=\text{'F'}$ or $\text{'N'}$.
2: $\mathrm{trans}$ – string (length ≥ 1)
Specifies the form of the system of equations.
${\mathbf{trans}}=\text{'N'}$
$AX=B$ (No transpose).
${\mathbf{trans}}=\text{'T'}$
${A}^{\mathrm{T}}X=B$ (Transpose).
${\mathbf{trans}}=\text{'C'}$
${A}^{\mathrm{H}}X=B$ (Conjugate transpose).
Constraint: ${\mathbf{trans}}=\text{'N'}$, $\text{'T'}$ or $\text{'C'}$.
3: $\mathrm{dl}\left(:\right)$ – complex array
The dimension of the array dl must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$
The $\left(n-1\right)$ subdiagonal elements of $A$.
4: $\mathrm{d}\left(:\right)$ – complex array
The dimension of the array d must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The $n$ diagonal elements of $A$.
5: $\mathrm{du}\left(:\right)$ – complex array
The dimension of the array du must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$
The $\left(n-1\right)$ superdiagonal elements of $A$.
6: $\mathrm{dlf}\left(:\right)$ – complex array
The dimension of the array dlf must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$
If ${\mathbf{fact}}=\text{'F'}$, dlf contains the $\left(n-1\right)$ multipliers that define the matrix $L$ from the $LU$ factorization of $A$.
7: $\mathrm{df}\left(:\right)$ – complex array
The dimension of the array df must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
If ${\mathbf{fact}}=\text{'F'}$, df contains the $n$ diagonal elements of the upper triangular matrix $U$ from the $LU$ factorization of $A$.
8: $\mathrm{duf}\left(:\right)$ – complex array
The dimension of the array duf must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$
If ${\mathbf{fact}}=\text{'F'}$, duf contains the $\left(n-1\right)$ elements of the first superdiagonal of $U$.
9: $\mathrm{du2}\left(:\right)$ – complex array
The dimension of the array du2 must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-2\right)$
If ${\mathbf{fact}}=\text{'F'}$, du2 contains the ($n-2$) elements of the second superdiagonal of $U$.
10: $\mathrm{ipiv}\left(:\right)$int64int32nag_int array
The dimension of the array ipiv must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
If ${\mathbf{fact}}=\text{'F'}$, ipiv contains the pivot indices from the $LU$ factorization of $A$.
11: $\mathrm{b}\left(\mathit{ldb},:\right)$ – complex array
The first dimension of the array b must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
The second dimension of the array b must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{nrhs_p}}\right)$.
The $n$ by $r$ right-hand side matrix $B$.
### Optional Input Parameters
1: $\mathrm{n}$int64int32nag_int scalar
Default: the first dimension of the array b and the dimension of the arrays d, df, ipiv.
$n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}\ge 0$.
2: $\mathrm{nrhs_p}$int64int32nag_int scalar
Default: the second dimension of the array b.
$r$, the number of right-hand sides, i.e., the number of columns of the matrix $B$.
Constraint: ${\mathbf{nrhs_p}}\ge 0$.
### Output Parameters
1: $\mathrm{dlf}\left(:\right)$ – complex array
The dimension of the array dlf will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$
If ${\mathbf{fact}}=\text{'N'}$, dlf contains the $\left(n-1\right)$ multipliers that define the matrix $L$ from the $LU$ factorization of $A$.
2: $\mathrm{df}\left(:\right)$ – complex array
The dimension of the array df will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
If ${\mathbf{fact}}=\text{'N'}$, df contains the $n$ diagonal elements of the upper triangular matrix $U$ from the $LU$ factorization of $A$.
3: $\mathrm{duf}\left(:\right)$ – complex array
The dimension of the array duf will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$
If ${\mathbf{fact}}=\text{'N'}$, duf contains the $\left(n-1\right)$ elements of the first superdiagonal of $U$.
4: $\mathrm{du2}\left(:\right)$ – complex array
The dimension of the array du2 will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-2\right)$
If ${\mathbf{fact}}=\text{'N'}$, du2 contains the ($n-2$) elements of the second superdiagonal of $U$.
5: $\mathrm{ipiv}\left(:\right)$int64int32nag_int array
The dimension of the array ipiv will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
If ${\mathbf{fact}}=\text{'N'}$, ipiv contains the pivot indices from the $LU$ factorization of $A$; row $i$ of the matrix was interchanged with row ${\mathbf{ipiv}}\left(i\right)$. ${\mathbf{ipiv}}\left(i\right)$ will always be either $i$ or $i+1$; ${\mathbf{ipiv}}\left(i\right)=i$ indicates a row interchange was not required.
6: $\mathrm{x}\left(\mathit{ldx},:\right)$ – complex array
The first dimension of the array x will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
The second dimension of the array x will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{nrhs_p}}\right)$.
If ${\mathbf{info}}={\mathbf{0}}$ or $\mathbf{n}+{\mathbf{1}}$, the $n$ by $r$ solution matrix $X$.
7: $\mathrm{rcond}$ – double scalar
The estimate of the reciprocal condition number of the matrix $A$. If ${\mathbf{rcond}}=0.0$, the matrix may be exactly singular. This condition is indicated by ${\mathbf{info}}>{\mathbf{0}} \text{and} {\mathbf{info}}\le \mathbf{n}$. Otherwise, if rcond is less than the machine precision, the matrix is singular to working precision. This condition is indicated by ${\mathbf{info}}\ge {\mathbf{n}}+1$.
8: $\mathrm{ferr}\left({\mathbf{nrhs_p}}\right)$ – double array
If ${\mathbf{info}}={\mathbf{0}}$ or $\mathbf{n}+{\mathbf{1}}$, an estimate of the forward error bound for each computed solution vector, such that ${‖{\stackrel{^}{x}}_{j}-{x}_{j}‖}_{\infty }/{‖{x}_{j}‖}_{\infty }\le {\mathbf{ferr}}\left(j\right)$ where ${\stackrel{^}{x}}_{j}$ is the $j$th column of the computed solution returned in the array x and ${x}_{j}$ is the corresponding column of the exact solution $X$. The estimate is as reliable as the estimate for rcond, and is almost always a slight overestimate of the true error.
9: $\mathrm{berr}\left({\mathbf{nrhs_p}}\right)$ – double array
If ${\mathbf{info}}={\mathbf{0}}$ or $\mathbf{n}+{\mathbf{1}}$, an estimate of the component-wise relative backward error of each computed solution vector ${\stackrel{^}{x}}_{j}$ (i.e., the smallest relative change in any element of $A$ or $B$ that makes ${\stackrel{^}{x}}_{j}$ an exact solution).
10: $\mathrm{info}$int64int32nag_int scalar
${\mathbf{info}}=0$ unless the function detects an error (see Error Indicators and Warnings).
## Error Indicators and Warnings
Cases prefixed with W are classified as warnings and do not generate an error of type NAG:error_n. See nag_issue_warnings.
${\mathbf{info}}<0$
If ${\mathbf{info}}=-i$, argument $i$ had an illegal value. An explanatory message is output, and execution of the program is terminated.
W ${\mathbf{info}}>0 \text{and} {\mathbf{info}}<{\mathbf{n}}$
Element $_$ of the diagonal is exactly zero. The factorization has not been completed, but the factor $U$ is exactly singular, so the solution and error bounds could not be computed. ${\mathbf{rcond}}=0.0$ is returned.
W ${\mathbf{info}}>0 \text{and} {\mathbf{info}}={\mathbf{n}}$
Element $_$ of the diagonal is exactly zero. The factorization has been completed, but the factor $U$ is exactly singular, so the solution and error bounds could not be computed. ${\mathbf{rcond}}=0.0$ is returned.
W ${\mathbf{info}}={\mathbf{n}}+1$
$U$ is nonsingular, but rcond is less than machine precision, meaning that the matrix is singular to working precision. Nevertheless, the solution and error bounds are computed because there are a number of situations where the computed solution can be more accurate than the value of rcond would suggest.
## Accuracy
For each right-hand side vector $b$, the computed solution $\stackrel{^}{x}$ is the exact solution of a perturbed system of equations $\left(A+E\right)\stackrel{^}{x}=b$, where
$E ≤ c n ε L U ,$
$c\left(n\right)$ is a modest linear function of $n$, and $\epsilon$ is the machine precision. See Section 9.3 of Higham (2002) for further details.
If $x$ is the true solution, then the computed solution $\stackrel{^}{x}$ satisfies a forward error bound of the form
$x-x^∞ x^∞ ≤ wc condA,x^,b$
where $\mathrm{cond}\left(A,\stackrel{^}{x},b\right)={‖\left|{A}^{-1}\right|\left(\left|A\right|\left|\stackrel{^}{x}\right|+\left|b\right|\right)‖}_{\infty }/{‖\stackrel{^}{x}‖}_{\infty }\le \mathrm{cond}\left(A\right)={‖\left|{A}^{-1}\right|\left|A\right|‖}_{\infty }\le {\kappa }_{\infty }\left(A\right)$. If $\stackrel{^}{x}$ is the $j$th column of $X$, then ${w}_{c}$ is returned in ${\mathbf{berr}}\left(j\right)$ and a bound on ${‖x-\stackrel{^}{x}‖}_{\infty }/{‖\stackrel{^}{x}‖}_{\infty }$ is returned in ${\mathbf{ferr}}\left(j\right)$. See Section 4.4 of Anderson et al. (1999) for further details.
The total number of floating-point operations required to solve the equations $AX=B$ is proportional to $nr$.
The condition number estimation typically requires between four and five solves and never more than eleven solves, following the factorization. The solution is then refined, and the errors estimated, using iterative refinement.
In practice the condition number estimator is very reliable, but it can underestimate the true condition number; see Section 15.3 of Higham (2002) for further details.
The real analogue of this function is nag_lapack_dgtsvx (f07cb).
## Example
This example solves the equations
$AX=B ,$
where $A$ is the tridiagonal matrix
$A = -1.3+1.3i 2.0-1.0i 0.0i+0.0 0.0i+0.0 0.0i+0.0 1.0-2.0i -1.3+1.3i 2.0+1.0i 0.0i+0.0 0.0i+0.0 0.0i+0.0 1.0+1.0i -1.3+3.3i -1.0+1.0i 0.0i+0.0 0.0i+0.0 0.0i+0.0 2.0-3.0i -0.3+4.3i 1.0-1.0i 0.0i+0.0 0.0i+0.0 0.0i+0.0 1.0+1.0i -3.3+1.3i$
and
$B = 2.4-05.0i 2.7+06.9i 3.4+18.2i -6.9-05.3i -14.7+09.7i -6.0-00.6i 31.9-07.7i -3.9+09.3i -1.0+01.6i -3.0+12.2i .$
Estimates for the backward errors, forward errors and condition number are also output.
```function f07cp_example
fprintf('f07cp example results\n\n');
% Tridiagonal matrix stored by diagonals
du = [ 2 - 1i 2 + 1i -1 + 1i 1 - 1i ];
d = [-1.3 + 1.3i -1.3 + 1.3i -1.3 + 3.3i -0.3 + 4.3i -3.3 + 1.3i];
dl = [ 1 - 2i 1 + 1i 2 - 3i 1 + 1i ];
n = numel(d);
% Rhs B
b = [ 2.4 - 5.0i 2.7 + 6.9i;
3.4 + 18.2i -6.9 - 5.3i;
-14.7 + 9.7i -6.0 - 0.6i;
31.9 - 7.7i -3.9 + 9.3i;
-1 + 1.6i -3.0 + 12.2i];
% Input parameters
fact = 'No factors';
trans = 'No transpose';
dlf = dl;
df = d;
duf = du;
du2 = complex(zeros(n-2,1));
ipiv = zeros(n,1,'int64');
% Solve
[dlf, df, duf, du2, ipiv, x, rcond, ferr, berr, info] = ...
f07cp( ...
fact, trans, dl, d, du, dlf, df, duf, du2, ipiv, b);
disp('Solution(s)');
disp(x);
disp('Backward errors (machine-dependent)');
fprintf('%10.1e',berr);
fprintf('\n');
disp('Estimated forward error bounds (machine-dependent)');
fprintf('%10.1e',ferr);
fprintf('\n\n');
disp('Estimate of reciprocal condition number');
fprintf('%10.1e\n',rcond);
```
```f07cp example results
Solution(s)
1.0000 + 1.0000i 2.0000 - 1.0000i
3.0000 - 1.0000i 1.0000 + 2.0000i
4.0000 + 5.0000i -1.0000 + 1.0000i
-1.0000 - 2.0000i 2.0000 + 1.0000i
1.0000 - 1.0000i 2.0000 - 2.0000i
Backward errors (machine-dependent)
3.6e-17 1.0e-16
Estimated forward error bounds (machine-dependent)
5.5e-14 7.7e-14
Estimate of reciprocal condition number
5.4e-03
``` | 2021-07-26 16:46:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 198, "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": 0.9973999857902527, "perplexity": 1784.1343174555525}, "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-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00272.warc.gz"} |
https://hal.archives-ouvertes.fr/hal-03424086 | # Regular braneworlds with bulk fluids
Abstract : We review studies on the singularity structure and asymptotic analysis of a 3-brane (flat or curved) embedded in a five-dimensional bulk filled with a perfect fluid' with an equation of state with the pressure' and the `density' of the fluid depending on the fifth space coordinate. Regular solutions satisfying positive energy conditions in the bulk exist only in the cases of a flat brane with an EoS parameter equal to -1, or of AdS branes for EoS parameter values in suitable intervals. More cases can be found by gluing two regular branches of solutions at the position of the brane. However, only the case of a flat brane with an EoS parameter equal to -1 leads to finite Planck mass on the brane and thus localises gravity. In a more recent work, we showed that a way to rectify the previous findings and obtain a solution for a flat brane in a finite range of the EoS parameter, which is both free from finite-distance singularities and compatible with the physical conditions of energy and finiteness of four-dimensional Planck mass, is by introducing a bulk fluid component that satisfies a nonlinear equation of state.
Keywords :
Document type :
Preprints, Working Papers, ...
Domain :
https://hal.archives-ouvertes.fr/hal-03424086
Contributor : Inspire Hep Connect in order to contact the contributor
Submitted on : Wednesday, November 10, 2021 - 12:08:07 PM
Last modification on : Tuesday, November 16, 2021 - 5:25:19 AM
### Citation
Ignatios Antoniadis, Spiros Cotsakis, Ifigeneia Klaoudatou. Regular braneworlds with bulk fluids. 2021. ⟨hal-03424086⟩
Record views | 2021-11-28 21:38:30 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6144787669181824, "perplexity": 1620.4349134265303}, "config": {"markdown_headings": true, "markdown_code": false, "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-2021-49/segments/1637964358591.95/warc/CC-MAIN-20211128194436-20211128224436-00544.warc.gz"} |
https://people.maths.bris.ac.uk/~matyd/GroupNames/128/D4.3SD16.html | Copied to
clipboard
G = D4.3SD16order 128 = 27
3rd non-split extension by D4 of SD16 acting via SD16/C8=C2
p-group, metabelian, nilpotent (class 3), monomial
Series: Derived Chief Lower central Upper central Jennings
Derived series C1 — C42 — D4.3SD16
Chief series C1 — C2 — C22 — C2×C4 — C42 — C4×D4 — C8×D4 — D4.3SD16
Lower central C1 — C22 — C42 — D4.3SD16
Upper central C1 — C22 — C42 — D4.3SD16
Jennings C1 — C22 — C22 — C42 — D4.3SD16
Generators and relations for D4.3SD16
G = < a,b,c,d | a4=b2=c8=1, d2=a2, bab=dad-1=a-1, ac=ca, bc=cb, dbd-1=ab, dcd-1=a2c3 >
Subgroups: 184 in 83 conjugacy classes, 34 normal (32 characteristic)
C1, C2, C2, C4, C4, C22, C22, C8, C2×C4, C2×C4, D4, D4, Q8, C23, C42, C22⋊C4, C4⋊C4, C4⋊C4, C2×C8, C2×C8, SD16, C22×C4, C2×D4, C2×Q8, C4×C8, C22⋊C8, D4⋊C4, Q8⋊C4, C4⋊C8, C2.D8, C4×D4, C4⋊Q8, C22×C8, C2×SD16, C4.10D8, C82C8, C8×D4, D4.D4, D4⋊Q8, C4.SD16, D4.3SD16
Quotients: C1, C2, C22, D4, C23, SD16, C2×D4, C4○D4, C4⋊D4, C2×SD16, C4○D8, C8⋊C22, D4.2D4, C88D4, D4.5D4, D4.3SD16
Character table of D4.3SD16
class 1 2A 2B 2C 2D 2E 4A 4B 4C 4D 4E 4F 4G 4H 4I 8A 8B 8C 8D 8E 8F 8G 8H 8I 8J 8K 8L 8M 8N size 1 1 1 1 4 4 2 2 2 2 4 4 4 16 16 2 2 2 2 4 4 4 4 4 4 8 8 8 8 ρ1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 trivial ρ2 1 1 1 1 -1 -1 1 1 1 1 -1 1 -1 1 -1 -1 -1 -1 -1 1 1 1 -1 -1 1 1 -1 1 -1 linear of order 2 ρ3 1 1 1 1 -1 -1 1 1 1 1 -1 1 -1 -1 -1 1 1 1 1 -1 -1 -1 1 1 -1 1 1 1 1 linear of order 2 ρ4 1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -1 1 -1 linear of order 2 ρ5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -1 1 linear of order 2 ρ6 1 1 1 1 -1 -1 1 1 1 1 -1 1 -1 1 1 1 1 1 1 -1 -1 -1 1 1 -1 -1 -1 -1 -1 linear of order 2 ρ7 1 1 1 1 -1 -1 1 1 1 1 -1 1 -1 -1 1 -1 -1 -1 -1 1 1 1 -1 -1 1 -1 1 -1 1 linear of order 2 ρ8 1 1 1 1 1 1 1 1 1 1 1 1 1 -1 -1 1 1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 linear of order 2 ρ9 2 2 2 2 -2 -2 -2 2 -2 2 2 -2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 orthogonal lifted from D4 ρ10 2 2 2 2 0 0 2 -2 2 -2 0 -2 0 0 0 2 2 2 2 0 0 0 -2 -2 0 0 0 0 0 orthogonal lifted from D4 ρ11 2 2 2 2 0 0 2 -2 2 -2 0 -2 0 0 0 -2 -2 -2 -2 0 0 0 2 2 0 0 0 0 0 orthogonal lifted from D4 ρ12 2 2 2 2 2 2 -2 2 -2 2 -2 -2 -2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 orthogonal lifted from D4 ρ13 2 2 2 2 0 0 -2 -2 -2 -2 0 2 0 0 0 0 0 0 0 -2i -2i 2i 0 0 2i 0 0 0 0 complex lifted from C4○D4 ρ14 2 2 2 2 0 0 -2 -2 -2 -2 0 2 0 0 0 0 0 0 0 2i 2i -2i 0 0 -2i 0 0 0 0 complex lifted from C4○D4 ρ15 2 2 -2 -2 0 0 0 -2 0 2 -2i 0 2i 0 0 -√-2 √-2 √-2 -√-2 -√2 √2 -√2 -√-2 √-2 √2 0 0 0 0 complex lifted from C4○D8 ρ16 2 2 -2 -2 0 0 0 -2 0 2 2i 0 -2i 0 0 √-2 -√-2 -√-2 √-2 -√2 √2 -√2 √-2 -√-2 √2 0 0 0 0 complex lifted from C4○D8 ρ17 2 2 -2 -2 2 -2 0 2 0 -2 0 0 0 0 0 -√-2 √-2 √-2 -√-2 √-2 -√-2 -√-2 √-2 -√-2 √-2 0 0 0 0 complex lifted from SD16 ρ18 2 -2 2 -2 0 0 -2 0 2 0 0 0 0 0 0 2i -2i 2i -2i 0 0 0 0 0 0 -√-2 -√2 √-2 √2 complex lifted from C4○D8 ρ19 2 -2 2 -2 0 0 -2 0 2 0 0 0 0 0 0 2i -2i 2i -2i 0 0 0 0 0 0 √-2 √2 -√-2 -√2 complex lifted from C4○D8 ρ20 2 2 -2 -2 -2 2 0 2 0 -2 0 0 0 0 0 -√-2 √-2 √-2 -√-2 -√-2 √-2 √-2 √-2 -√-2 -√-2 0 0 0 0 complex lifted from SD16 ρ21 2 -2 2 -2 0 0 -2 0 2 0 0 0 0 0 0 -2i 2i -2i 2i 0 0 0 0 0 0 -√-2 √2 √-2 -√2 complex lifted from C4○D8 ρ22 2 -2 2 -2 0 0 -2 0 2 0 0 0 0 0 0 -2i 2i -2i 2i 0 0 0 0 0 0 √-2 -√2 -√-2 √2 complex lifted from C4○D8 ρ23 2 2 -2 -2 2 -2 0 2 0 -2 0 0 0 0 0 √-2 -√-2 -√-2 √-2 -√-2 √-2 √-2 -√-2 √-2 -√-2 0 0 0 0 complex lifted from SD16 ρ24 2 2 -2 -2 0 0 0 -2 0 2 -2i 0 2i 0 0 √-2 -√-2 -√-2 √-2 √2 -√2 √2 √-2 -√-2 -√2 0 0 0 0 complex lifted from C4○D8 ρ25 2 2 -2 -2 0 0 0 -2 0 2 2i 0 -2i 0 0 -√-2 √-2 √-2 -√-2 √2 -√2 √2 -√-2 √-2 -√2 0 0 0 0 complex lifted from C4○D8 ρ26 2 2 -2 -2 -2 2 0 2 0 -2 0 0 0 0 0 √-2 -√-2 -√-2 √-2 √-2 -√-2 -√-2 -√-2 √-2 √-2 0 0 0 0 complex lifted from SD16 ρ27 4 -4 4 -4 0 0 4 0 -4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 orthogonal lifted from C8⋊C22 ρ28 4 -4 -4 4 0 0 0 0 0 0 0 0 0 0 0 2√2 2√2 -2√2 -2√2 0 0 0 0 0 0 0 0 0 0 symplectic lifted from D4.5D4, Schur index 2 ρ29 4 -4 -4 4 0 0 0 0 0 0 0 0 0 0 0 -2√2 -2√2 2√2 2√2 0 0 0 0 0 0 0 0 0 0 symplectic lifted from D4.5D4, Schur index 2
Smallest permutation representation of D4.3SD16
On 64 points
Generators in S64
(1 45 35 52)(2 46 36 53)(3 47 37 54)(4 48 38 55)(5 41 39 56)(6 42 40 49)(7 43 33 50)(8 44 34 51)(9 62 19 31)(10 63 20 32)(11 64 21 25)(12 57 22 26)(13 58 23 27)(14 59 24 28)(15 60 17 29)(16 61 18 30)
(1 56)(2 49)(3 50)(4 51)(5 52)(6 53)(7 54)(8 55)(9 13)(10 14)(11 15)(12 16)(17 21)(18 22)(19 23)(20 24)(25 60)(26 61)(27 62)(28 63)(29 64)(30 57)(31 58)(32 59)(33 47)(34 48)(35 41)(36 42)(37 43)(38 44)(39 45)(40 46)
(1 2 3 4 5 6 7 8)(9 10 11 12 13 14 15 16)(17 18 19 20 21 22 23 24)(25 26 27 28 29 30 31 32)(33 34 35 36 37 38 39 40)(41 42 43 44 45 46 47 48)(49 50 51 52 53 54 55 56)(57 58 59 60 61 62 63 64)
(1 63 35 32)(2 27 36 58)(3 61 37 30)(4 25 38 64)(5 59 39 28)(6 31 40 62)(7 57 33 26)(8 29 34 60)(9 42 19 49)(10 52 20 45)(11 48 21 55)(12 50 22 43)(13 46 23 53)(14 56 24 41)(15 44 17 51)(16 54 18 47)
G:=sub<Sym(64)| (1,45,35,52)(2,46,36,53)(3,47,37,54)(4,48,38,55)(5,41,39,56)(6,42,40,49)(7,43,33,50)(8,44,34,51)(9,62,19,31)(10,63,20,32)(11,64,21,25)(12,57,22,26)(13,58,23,27)(14,59,24,28)(15,60,17,29)(16,61,18,30), (1,56)(2,49)(3,50)(4,51)(5,52)(6,53)(7,54)(8,55)(9,13)(10,14)(11,15)(12,16)(17,21)(18,22)(19,23)(20,24)(25,60)(26,61)(27,62)(28,63)(29,64)(30,57)(31,58)(32,59)(33,47)(34,48)(35,41)(36,42)(37,43)(38,44)(39,45)(40,46), (1,2,3,4,5,6,7,8)(9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56)(57,58,59,60,61,62,63,64), (1,63,35,32)(2,27,36,58)(3,61,37,30)(4,25,38,64)(5,59,39,28)(6,31,40,62)(7,57,33,26)(8,29,34,60)(9,42,19,49)(10,52,20,45)(11,48,21,55)(12,50,22,43)(13,46,23,53)(14,56,24,41)(15,44,17,51)(16,54,18,47)>;
G:=Group( (1,45,35,52)(2,46,36,53)(3,47,37,54)(4,48,38,55)(5,41,39,56)(6,42,40,49)(7,43,33,50)(8,44,34,51)(9,62,19,31)(10,63,20,32)(11,64,21,25)(12,57,22,26)(13,58,23,27)(14,59,24,28)(15,60,17,29)(16,61,18,30), (1,56)(2,49)(3,50)(4,51)(5,52)(6,53)(7,54)(8,55)(9,13)(10,14)(11,15)(12,16)(17,21)(18,22)(19,23)(20,24)(25,60)(26,61)(27,62)(28,63)(29,64)(30,57)(31,58)(32,59)(33,47)(34,48)(35,41)(36,42)(37,43)(38,44)(39,45)(40,46), (1,2,3,4,5,6,7,8)(9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56)(57,58,59,60,61,62,63,64), (1,63,35,32)(2,27,36,58)(3,61,37,30)(4,25,38,64)(5,59,39,28)(6,31,40,62)(7,57,33,26)(8,29,34,60)(9,42,19,49)(10,52,20,45)(11,48,21,55)(12,50,22,43)(13,46,23,53)(14,56,24,41)(15,44,17,51)(16,54,18,47) );
G=PermutationGroup([[(1,45,35,52),(2,46,36,53),(3,47,37,54),(4,48,38,55),(5,41,39,56),(6,42,40,49),(7,43,33,50),(8,44,34,51),(9,62,19,31),(10,63,20,32),(11,64,21,25),(12,57,22,26),(13,58,23,27),(14,59,24,28),(15,60,17,29),(16,61,18,30)], [(1,56),(2,49),(3,50),(4,51),(5,52),(6,53),(7,54),(8,55),(9,13),(10,14),(11,15),(12,16),(17,21),(18,22),(19,23),(20,24),(25,60),(26,61),(27,62),(28,63),(29,64),(30,57),(31,58),(32,59),(33,47),(34,48),(35,41),(36,42),(37,43),(38,44),(39,45),(40,46)], [(1,2,3,4,5,6,7,8),(9,10,11,12,13,14,15,16),(17,18,19,20,21,22,23,24),(25,26,27,28,29,30,31,32),(33,34,35,36,37,38,39,40),(41,42,43,44,45,46,47,48),(49,50,51,52,53,54,55,56),(57,58,59,60,61,62,63,64)], [(1,63,35,32),(2,27,36,58),(3,61,37,30),(4,25,38,64),(5,59,39,28),(6,31,40,62),(7,57,33,26),(8,29,34,60),(9,42,19,49),(10,52,20,45),(11,48,21,55),(12,50,22,43),(13,46,23,53),(14,56,24,41),(15,44,17,51),(16,54,18,47)]])
Matrix representation of D4.3SD16 in GL4(𝔽17) generated by
1 0 0 0 0 1 0 0 0 0 0 1 0 0 16 0
,
16 0 0 0 0 16 0 0 0 0 0 1 0 0 1 0
,
5 5 0 0 12 5 0 0 0 0 13 0 0 0 0 13
,
16 0 0 0 0 1 0 0 0 0 5 12 0 0 12 12
G:=sub<GL(4,GF(17))| [1,0,0,0,0,1,0,0,0,0,0,16,0,0,1,0],[16,0,0,0,0,16,0,0,0,0,0,1,0,0,1,0],[5,12,0,0,5,5,0,0,0,0,13,0,0,0,0,13],[16,0,0,0,0,1,0,0,0,0,5,12,0,0,12,12] >;
D4.3SD16 in GAP, Magma, Sage, TeX
D_4._3{\rm SD}_{16}
% in TeX
G:=Group("D4.3SD16");
// GroupNames label
G:=SmallGroup(128,411);
// by ID
G=gap.SmallGroup(128,411);
# by ID
G:=PCGroup([7,-2,2,2,-2,2,-2,2,224,141,512,422,1123,136,2804,718,172]);
// Polycyclic
G:=Group<a,b,c,d|a^4=b^2=c^8=1,d^2=a^2,b*a*b=d*a*d^-1=a^-1,a*c=c*a,b*c=c*b,d*b*d^-1=a*b,d*c*d^-1=a^2*c^3>;
// generators/relations
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𝔽 | 2021-12-06 18:33:01 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8835137486457825, "perplexity": 578.9561961577093}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2021-49/segments/1637964363309.86/warc/CC-MAIN-20211206163944-20211206193944-00419.warc.gz"} |
http://aas.org/archives/BAAS/v26n2/aas184/abs/S1503.html | Constraints on the Luminosity Function of Gamma--Ray Bursts Detected by BATSE
Session 15 -- Gamma Ray Bursts
Oral presentation, Monday, 30, 1994, 10:00-11:30
## [15.03] Constraints on the Luminosity Function of Gamma--Ray Bursts Detected by BATSE
J. M. Horack (NASA/MSFC), and A. G. Emslie (UAH)
We have utilized the BATSE--measured gamma--ray burst intensity distribution in conjunction with the general properties satisfied by integral moments to constrain significantly the permissable range of luminosities from which the observed gamma--ray bursts must be drawn. The constraints are independent of the functional form of the burst radial distribution. Given a distance independent luminosity function, the range of luminosity from which 80\$\%\$ of the observed gamma--ray bursts must be drawn may not exceed \$\sim\$6.5, with a 3--sigma upper limit of approximately 12. The finding that a substantial majority of bursts are passable standard--candles has important ramifications for many burst production models. | 2014-08-01 23:11:22 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9644211530685425, "perplexity": 5112.275715836451}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2014-23/segments/1406510275393.46/warc/CC-MAIN-20140728011755-00400-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://openstudy.com/updates/55dfa691e4b03567e10fcae2 | anonymous one year ago **please help!** Using the following equation, find the center and radius of the circle. x^2 - 4x + y^2 + 8y = -4
• This Question is Open
1. anonymous
I would complete the square twice, once for a quadratic in x and once for a quadratic in y.
2. anonymous
can you show me step by step?
3. anonymous
In other words, Add 4 to the 4^2-4x to create a perfect square trinomial. Then add 16 to the y^2 + 8y to make another perfect square trinomial. Then, because 20 is added to the left hand side, 20 must also be added to th right hand side.$x^2-4x+y^2+8y=-4$$x^2-4x+4+y^2+8y+16=-4+20$I combined some stuff. Do you understand what I've done?
4. anonymous
yes sorta.
5. anonymous
Now, consider the LHS as two trinomials. We can group them to make it easier to follow. Plus, we can simplify the RHS$\left( x^2-4x+4 \right)+\left( y^2+8y+16 \right)=16$OK with that?
6. anonymous
yes
7. anonymous
OK. Now, the two trinomials on the LHS are perfect squares. Can you factor them and write them as squares?
8. anonymous
would it be (-4^2−4x+4)+(20^2+8y+16)=16
9. anonymous
Not quite. Let's slow down a bit and take it step by step. The first trinomial on the LHS is$x^2-4x+4$Can you factor this into the product of two binomials?
10. anonymous
x^2-4x+4-4+20? im really bad at math and I'm home schooled so it's even tougher
11. anonymous
Have you multiplied binomials together before? Stuff like$\left( x+2 \right)\left( x-3 \right) = x^2-x-6$
12. anonymous
no :/
13. anonymous
Oh, then we have a problem. Factoring is the reverse of the above. Let me think of there's another way to get at this...
14. anonymous
So you haven't used this technique of 'completing the square' before either?
15. anonymous
nope. I took basics of alegebra then was thrown into geometry
16. anonymous
OK. Do you know what the equation of a circle looks like?
17. anonymous
x^2+y^2=r?
18. anonymous
That's very good, but the radius must also squared. For a circle with center at the origin (0, 0), it's equation would be$x^2+y^2=r^2$
19. anonymous
IN your question, however, the center of the circle is NOT at the origin. So, for a circle with its center at some point (a, b), its equation looks like$\left( x-a \right)^2 + \left( y-b \right)^2 = r^2$Notice that, if the center is at the origin, a=b=0 and we get the equation you proposed earlier. Understand?
20. anonymous
ok
21. anonymous
Alright. So, I've been tying to help you get the equation you were given in the problem$x^2-4x+y^2+8y=-4$to look like$\left( x-a \right)^2+\left( y-b \right)^2 = r^2$
22. anonymous
So far, we've done a couple of things and are at the stage where we have$\left( x^2-4x+4 \right)+\left( y^2+8y+16 \right)=16$Look at the RHS. What number squared is equal to 16?
23. anonymous
4
24. anonymous
Good. So that's the radius. OK with that much?
25. anonymous
yup
26. anonymous
Good. Now I'm going to cheat a bit and factor those trinomials for you. We have$\left( x^2-4x+x \right)+\left( y^2+8x+16 \right) = 4^2$$\left( x-2 \right)^2+\left( y+4 \right)^2=4^2$Compare that with$\left( x-a \right)^2+\left( y-b \right)^2=r^2$Can you tell what a and b are?
27. anonymous
yes a is 2 and b is 4
28. anonymous
You're right about a, but look carefully at the signs and try b again.
29. anonymous
4^2?
30. anonymous
Nope. If b=4, then you'd have (y - 4)^2. But you want (y + 4)^2. What does b have to be?
31. anonymous
What is y - (-4) ?
32. anonymous
b=-4??
33. anonymous
That's right.$\left( y-\left( -4 \right) \right)^2 = \left( y+4 \right)^2$So b must be -4.
34. anonymous
So now you know a, b, and r. Where's the center of the circle?
35. anonymous
-4x, 8y?
36. anonymous
No. We know that a=2, b=-4 and r=4. The center of the circle is at coordinates (a, b) and the radius is r. So where;s the center of the circle?
37. anonymous
2,-4?
38. anonymous
39. anonymous
4
40. anonymous
41. anonymous
thanks soooooo muchhh!!!
42. anonymous
You're welcome. I would encourage you to get some help learning to multiply and factor polynomials. Those skills will be necessary to move forward. Good luck! | 2017-01-24 23:53:18 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.8072060346603394, "perplexity": 2938.53479155213}, "config": {"markdown_headings": false, "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-2017-04/segments/1484560285315.77/warc/CC-MAIN-20170116095125-00541-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://cracku.in/4-45-part-of-a-tank-is-filled-with-oil-after-taking--x-ssc-cpo-4-july-2017-afternoon-shift | Question 4
# 4/5 part of a tank is filled with oil. After taking out 42 litres of oil the tank is 3/4 part full. What is the capacity (in litres) of the tank?
Solution
Let capacity of tank = $$x$$ litres
According to ques,
=> $$\frac{4x}{5}-42=\frac{3x}{4}$$
=> $$\frac{4x}{5}-\frac{3x}{4}=42$$
=> $$\frac{16x-15x}{20}=42$$
=> $$x=42\times20=840$$
$$\therefore$$ Capacity (in litres) of the tank = 840 litres
=> Ans - (C) | 2022-06-29 15:23:15 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8822414875030518, "perplexity": 2860.207945159739}, "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-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00058.warc.gz"} |
https://theachievablebody.com/04lvvgt2/174734-core-description-sun | {\displaystyle \left\{{\begin{aligned}&&{}^{12}\!\mathrm {C} +{}^{1}\!\mathrm {H} &\rightarrow {}^{13}\!\mathrm {N} +\gamma \\{\text{then}}&&{}^{13}\!\mathrm {N} &\rightarrow {}^{13}\!\mathrm {C} +e^{+}+\nu _{e}\\{\text{then}}&&{}^{13}\!\mathrm {C} +{}^{1}\!\mathrm {H} &\rightarrow {}^{14}\!\mathrm {N} +\gamma \\{\text{then}}&&{}^{14}\!\mathrm {N} +{}^{1}\!\mathrm {H} &\rightarrow {}^{15}\!\mathrm {O} +\gamma \\{\text{then}}&&{}^{15}\!\mathrm {O} &\rightarrow {}^{15}\!\mathrm {N} +e^{+}+\nu _{e}\\{\text{then}}&&{}^{15}\!\mathrm {N} +{}^{1}\!\mathrm {H} &\rightarrow {}^{12}\!\mathrm {C} +{}^{4}\!\mathrm {He} +\gamma \\\end{aligned}}\right.}. Outer core. Not limited to stratigraphic unit outcrop positions 2. + γ Rings. When Core's battle effect is used, a holographic image of a Mars Djinni surrounds itself with a fully-3D special effect resembling a fiery … + The density slowly decreases moving away from the core. Data used to understand depositional environment, reservoir geometry, and reservoir quality. It is not a solid surface, but rather a layer of gasses. Core definition, the central part of a fleshy fruit, containing the seeds. 0 0. + One of 100 billion stars in our galaxy, the Sun provides light and heat to the orbiting planets. What does sun mean? All the energy that comes away from the Sun and all the reaches the Earth started in the core. e Core Description Analysis of conventional core include description of bedding, lithology, sedimentary structures, fossils, and any other macro-features of the rock. + The outer core of the Earth is a liquid layer about 2,260 kilometers thick. C N Normally, protons in atomic nuclei repel each other because they have the same electrical charge. This is the inner most part of the Sun. Data used to understand depositional environment, reservoir geometry, and reservoir quality. + Get the latest updates on NASA missions, watch NASA TV live, and learn about our quest to reveal the unknown and benefit all humankind. [14], The high-energy photons (gamma rays) released in fusion reactions take indirect paths to the Sun's surface. From the inside out, the solar interior consists of: the core (the central region where nuclear reactions consume hydrogen to form helium. At 19% of the solar radius, near the edge of the core, temperatures are about 10 million kelvins and fusion power density is 6.9 W/m3, which is about 2.5% of the maximum value at the solar center. Core. The core of the Sun is considered to extend from the center to about 0.2 to 0.25 of There are two distinct reactions in which four hydrogen nuclei may eventually result in one helium nucleus: the proton-proton chain reaction – which is responsible for most of the Sun's released energy – and the CNO cycle. The Sun’s core is an extremely hot, dense mass of atomic nuclei and electrons. The first reaction in which 4 H nuclei may eventually result in one He nucleus, known as the proton–proton chain reaction, is:[6][11], { [7] Despite its intense temperature, the peak power generating density of the core overall is similar to an active compost heap, and is lower than the power density produced by the metabolism of an adult human. The time that deuterium and helium-3 in the next reactions last, by contrast, are only about 4 seconds and 400 years. Overview. The Sun's radiative zone is the section of the solar interior between the innermost core and the outer convective zone.In the radiative zone, energy generated by nuclear fusion in the core moves outward as electromagnetic radiation.In other words, the energy is conveyed by photons.When the energy reaches the top of the radiative zone, it begins to move in a different … It is the hottest part of the Sun and the Solar system. + The characteristic time for the first reaction is about one billion years even at the high densities and temperatures of the core, due to the necessity for the weak force to cause beta decay before the nucleons can adhere (which rarely happens in the time they tunnel toward each other, to be close enough to do so). Q&A for Work. Anonymous. And the temperature is about 5,500 degrees C. 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Z 11 compass with weather information maven then download the dependency jar and add it to your WEB-INF directory solar. 4, and base Defense by 2 Sun fuses about 600 million tons of hydrogen takes place a! | 2022-01-24 07:31:42 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3192211985588074, "perplexity": 2091.1587532644407}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2022-05/segments/1642320304515.74/warc/CC-MAIN-20220124054039-20220124084039-00313.warc.gz"} |
https://uwaterloo.ca/statistics-and-actuarial-science/events/archive/2018 | # Events - 2018
Thursday, November 29, 2018 — 4:00 PM EST
## Department seminar by Dr. Samuel Drapeau, Shanghai Jiao Tong University
Computational Aspects of Robust Optimized Certainty Equivalent and Option Pricing
We present a robust extension under distribution uncertainty of optimized certainty equivalent that includes the expected shortfall. We show that the infinite dimensional optimization problem can be reduced to a finite one using transport duality methods. Some important cases such as the Expected Shortfall can even be computed explicitly and provide insights about the additional costs from distributional uncertainty.
The general result can be further applied for explicit computation of robust option price where we also provide some explicit formulas in cases of call options. We finally address dual representation of the robust optimized certainty equivalent.
This talk is based on a joint work with Daniel Bartle and Ludovic Tangpi.
Thursday, November 22, 2018 — 4:00 PM EST
## Department seminar by Dr. Bei Jiang, University of Alberta
A Bayesian Approach to Joint Modeling of Matrix-valued Imaging Data and Treatment Outcome with Applications to Depression Studies
In this talk, we discuss a unified Bayesian joint modeling framework for studying association between a binary treatment outcome and a baseline matrix-valued predictor. Specifically, a joint modeling approach relating an outcome to a matrix-valued predictor through a probabilistic formulation of multilinear principal component analysis (MPCA) is developed. This framework establishes a theoretical relationship between the outcome and the matrix-valued predictor although the predictor is not explicitly expressed in the model. Simulation studies are provided showing that the proposed method is superior or competitive to other methods, such as a two-stage approach and a classical principal component regression (PCR) in terms of both prediction accuracy and estimation of association; its advantage is most notable when the sample size is small and the dimensionality in the imaging covariate is large. Finally, our proposed joint modeling approach is shown to be a very promising tool in an application exploring the association between baseline EEG data and a favorable response to treatment in a depression treatment study by achieving a substantial improvement in prediction accuracy in comparison to competing methods.
Wednesday, November 21, 2018 — 4:00 PM EST
## Graduate Student Seminar by Danqiao Guo
Eigen Portfolio Selection: A Robust Approach to Sharpe Ratio Maximization
We show that even when a covariance matrix is poorly estimated, it is still possible to obtain a robust maximum Sharpe ratio portfolio by exploiting the uneven distribution of estimation errors across principal components. This is accomplished by approximating an investor’s view on future asset returns using a few relatively accurate sample principal components. We discuss two approximation methods. The first method leads to a subtle connection to existing approaches in the literature; while the second one is novel and able to address main shortcomings of existing methods.
** Pizza & refreshments will be provided **
Everyone welcome!
Friday, November 16, 2018 — 11:00 AM EST
## Department seminar by Mehdi Molkaraie, ETH Zurich
The Ising model: series expansions and new algorithms
We propose new and simple Monte Carlo methods to estimate the partition function of the Ising model. The methods are based on the well-known series expansion of the partition function from statistical physics. For the Ising model, typical Monte Carlo methods work well at high temperature, but fail in the low-temperature regime. We demonstrate that our proposed Monte Carlo methods work differently: they behave particularly well at low temperature. We also compare the accuracy of our estimators with the state-of-the-art variational methods.
Thursday, November 8, 2018 — 4:00 PM EST
## Department seminar by Dr. Dennis K. J. Lin, Pennsylvania State University
Ghost Data
As natural as the real data, ghost data is everywhere—it is just data that you cannot see. We need to learn how to handle it, how to model with it, and how to put it to work. Some examples of ghost data are (see, Sall, 2017):
(a) Virtual data—it isn’t there until you look at it;
(b) Missing data—there is a slot to hold a value, but the slot is empty;
(c) Pretend data—data that is made up;
(d) Highly Sparse Data—whose absence implies a near zero, and
(e) Simulation data—data to answer “what if.”
For example, absence of evidence/data is not evidence of absence. In fact, it can be evidence of something. More Ghost Data can be extended to other existing areas: Hidden Markov Chain, Two-stage Least Square Estimate, Optimization via Simulation, Partition Model, Topological Data, just to name a few.
Three movies will be discussed in this talk: (1) “The Sixth Sense” (Bruce Wallis)—I can see things that you cannot see; (2) “Sherlock Holmes” (Robert Downey)—absence of expected facts; and (3) “Edge of Tomorrow” (Tom Cruise)—how to speed up your learning (AlphaGo-Zero will also be discussed). It will be helpful, if you watch these movies before coming to my talk. This is an early stage of my research in this area--any feedback from you is deeply appreciated. Much of the basic idea is highly influenced via Mr. John Sall (JMP-SAS).
Thursday, November 1, 2018 — 4:00 PM EDT
## Department seminar by Dr. Bing Li, Pennsylvania State University
Copula Gaussian graphical models for functional data
We consider the problem of constructing statistical graphical models for functional data; that is, the observations on the vertices are random functions. This types of data are common in medical applications such as EEG and fMRI. Recently published functional graphical models rely on the assumption that the random functions are Hilbert-space-valued Gaussian random elements. We relax this assumption by introducing a copula Gaussian random elements Hilbert spaces, leading to what we call the Functional Copula Gaussian Graphical Model (FCGGM). This model removes the marginal Gaussian assumption but retains the simplicity of the Gaussian dependence structure, which is particularly attractive for large data. We develop four estimators, together with their implementation algorithms, for the FCGGM. We establish the consistency and the convergence rates of one of the estimators under different sets of sufficient conditions with varying strengths. We compare our FCGGM with the existing functional Gaussian graphical model by simulation, under both non-Gaussian and Gaussian graphical models, and apply our method to an EEG data set to construct brain networks.
Tuesday, October 30, 2018 — 4:00 PM EDT
## Department seminar by Dr. Jong Soo Hong
Systemic risk and the optimal capital requirements in a model of financial networks and fire sales
I consider an interbank network with fire sales externalities and multiple illiquid assets and study the problem of optimally trading off between capital reserves and systemic risk. I find that the problem of measuring systemic risk and the optimal capital requirements under various liquidation rules can be formulated as a convex and convex mixed integer programming. To solve the convex MIP, I offer an iterative algorithm that converges to the optimal solutions. I show the results of the methodology through numerical examples and provide implications for regulatory policies and related research topics.
Thursday, October 25, 2018 — 4:00 PM EDT
## Department seminar by Dr. Linbo Wang, University of Toronto
Causal Inference with Unmeasured Confounding: an Instrumental Variable Approach
Causal inference is a challenging problem because causation cannot be established from observational data alone. Researchers typically rely on additional sources of information to infer causation from association. Such information may come from powerful designs such as randomization, or background knowledge such as information on all confounders. However, perfect designs or background knowledge required for establishing causality may not always be available in practice. In this talk, I use novel causal identification results to show that the instrumental variable approach can be used to combine the power of design and background knowledge to draw causal conclusions. I also introduce novel estimation tools to construct estimators that are robust, efficient and enjoy good finite sample properties. These methods will be discussed in the context of a randomized encouragement design for a flu vaccine.
Friday, October 19, 2018 — 11:00 AM EDT
## Department seminar by Dr. Donglin Zeng, The University of North Carolina at Chapel Hill
Efficient Estimation, Robust Testing and Design Optimality for Two-Phase Studies
Two-phase designs are cost-effective sampling strategies when some covariates are expensive to be measured on all study subjects. Well-known examples include case-control, case-cohort, nested case-control and extreme tail sampling designs. In this talk, I will discuss three important aspects in two-phase studies: estimation, hypothesis testing and design optimality. First, I will discuss efficient estimation methods we have developed for two-phase studies. We allow expensive covariates to be correlated with inexpensive covariates collected in the first phase. Our proposed estimation is based on maximization of a modified nonparametric likelihood function through a generalization of the expectation-maximization algorithm. The resulting estimators are shown to be consistent, asymptotically normal and asymptotically efficient with easily estimated variances. Second, I will focus on hypothesis testing in two-phase studies. We propose a robust test procedure based on imputation. The proposed procedure guarantees preservation of type I error, allows high-dimensional inexpensive covariates, and yields higher power than alternative imputation approaches. Finally, I will present some recent development on design optimality. We show that for general outcomes, the most efficient design is an extreme-tail sampling design based on certain residuals. This conclusion also explains the high efficiency of extreme tail sampling for continuous outcomes and balanced case-control design for binary outcomes. Throughout the talk, I will present numerical evidences from simulation studies and illustrate our methods using different applications.
Wednesday, October 17, 2018 — 4:00 PM EDT
## David Sprott Distinguished Lecture Speaker: Dr. Emery Brown; Affiliation: Institute for Medical Engineering & Science
Uncovering the Mechanisms of General Anesthesia: Where Neuroscience Meets Statistics
General anesthesia is a drug-induced, reversible condition involving unconsciousness, amnesia (loss of memory), analgesia (loss of pain sensation), akinesia (immobility), and hemodynamic stability. I will describe a primary mechanism through which anesthetics create these altered states of arousal. Our studies have allowed us to give a detailed characterization of the neurophysiology of loss and recovery of consciousness, in the case of propofol, and we have demonstrated that the state of general anesthesia can be rapidly reversed by activating specific brain circuits. The success of our research has depended critically on tight coupling of experiments, statistical signal processing and mathematical modeling.
Thursday, October 11, 2018 — 4:00 PM EDT
## Department seminar by Michael Hahsler, Southern Methodist University
Probabilistic approaches to mine association rules
Mining association rules is an important and widely applied data mining technique for discovering patterns in large datasets. However, the used support-confidence framework has some often overlooked weaknesses. This talk introduces a simple stochastic model and shows how it can be used in association rule mining. We apply the model to simulate data for analyzing the behavior and shortcomings of confidence and other measures of interestingness (e.g., lift). Based on these findings, we develop a new model-driven approach to mine rules based on the notion of NB-frequent itemsets, and we define a measure of interestingness which controls for spurious rules and has a strong foundation in statistical testing theory.
Thursday, October 4, 2018 — 4:00 PM EDT
## Department seminar by Hongzhe Li, University of Pennsylvania
Methods for High Dimensional Compositional Data Analysis in Microbiome Studies
Human microbiome studies using high throughput DNA sequencing generate compositional data with the absolute abundances of microbes not recoverable from sequence data alone. In compositional data analysis, each sample consists of proportions of various organisms with a unit sum constraint. This simple feature can lead traditional statistical methods when naively applied to produce errant results and spurious associations. In addition, microbiome sequence data sets are typically high dimensional, with the number of taxa much greater than the number of samples. These important features require further development of methods for analysis of high dimensional compositional data. This talk presents several latest developments in this area, including methods for estimating the compositions based on sparse count data, two-sample test for compositional vectors and regression analysis with compositional covariates. Several microbiome studies at the University of Pennsylvania are used to illustrate these methods and several open questions will be discussed.
Saturday, September 29, 2018 — 8:00 AM to 6:00 PM EDT
## Waterloo Datathon - Fall 2018
We are extending an invitation to a select group of talented undergraduate, graduate and PhD students to participate in the upcoming University of Waterloo Datathon.
Thursday, September 27, 2018 — 4:00 PM EDT
## Department seminar by Yuanying Guan, Indiana University Northwest
Agent-based Asset Pricing, Learning, and Chaos
The Lucas asset pricing model is one of the most studied model in financial economics in the past decade. In our research, we relax the original assumptions in Lucas model of homogeneous agents and rational expectations. We populate an artificial economy with heterogeneous and boundedly rational agents. By defining a Correct Expectations Equilibrium, agents are able to compute their policy functions and the equilibrium pricing function without perfect information about the market. A natural adaptive learning scheme is given to agents to update their predictions. We examine the convergence of equilibrium with this learning scheme and show that the equilibrium is learnable (convergent) under certain parameter combinations. We also investigate the market dynamics when agents are out of equilibrium, including the cases where prices have excess volatility and the trading volume is high. Numerical simulations show that our system exhibits rich dynamics, including a whole cascade from period doubling bifurcations to chaos.
Thursday, September 20, 2018 — 4:00 PM EDT
## Department seminar by Michele Guindani, University of California
Bayesian Approaches to Dynamic Model Selection
In many applications, investigators monitor processes that vary in space and time, with the goal of identifying temporally persistent and spatially localized departures from a baseline or normal" behavior. In this talk, I will first discuss a principled Bayesian approach for estimating time varying functional connectivity networks from brain fMRI data. Dynamic functional connectivity, i.e., the study of how interactions among brain regions change dynamically over the course of an fMRI experiment, has recently received wide interest in the neuroimaging literature. Our method utilizes a hidden Markov model for classification of latent neurological states, achieving estimation of the connectivity networks in an integrated framework that borrows strength over the entire time course of the experiment. Furthermore, we assume that the graph structures, which define the connectivity states at each time point, are related within a super-graph, to encourage the selection of the same edges among related graphs. Then, I will propose a Bayesian nonparametric model selection approach with an application to the monitoring of pneumonia and influenza (P&I) mortality, to detect influenza outbreaks in the continental United States. More specifically, we introduce a zero-inflated conditionally identically distributed species sampling prior which allows borrowing information across time and to assign data to clusters associated to either a null or an alternate process. Spatial dependences are accounted for by means of a Markov random field prior, which allows to inform the selection based on inferences conducted at nearby locations. We show how the proposed modeling framework performs in an application to the P&I mortality data and in a simulation study, and compare with common threshold methods for detecting outbreaks over time, with more recent Markov switching based models, and with other Bayesian nonparametric priors that do not take into account spatio-temporal dependence.
Thursday, September 13, 2018 — 4:00 PM EDT
## Department seminar by Karen Kafadar, University of Virginia
The Critical Role of Statistics in Evaluating Forensic Evidence
Statisticians have been important contributors to many areas of science, including chemistry (chemometrics), biology (genomics), medicine (clinical trials), and agriculture (crop yield), leading to valuable advances in statistical research that have benefitted multiple fields (e.g., spectral analysis, penalized regression, sequential analysis, experimental design). Yet the involvement of statistics specifically in forensic science has not been nearly as extensive, especially in view of its importance (ensuring proper administration of justice) and the value it has demonstrated thus far (e.g., forensic DNA, assessment of bullet lead evidence, significance of findings in the U.S. anthrax investigation, reliability of eyewitness identification). Forensic methods in many areas remain unvalidated, as recent investigations have highlighted (notably, bite marks and hair analysis). In this talk, I will provide three examples (including one with data kindly provided by Canadian Forensic Service) where statistics plays a vital role in evaluating forensic evidence and motivate statistical research that can have both theoretical and practical value, and ultimately strengthen forensic evidence.
Thursday, August 9, 2018 — 4:00 PM EDT
## Department seminar by Lixing Zhu, Hong Kong Baptist University
An Adaptive-to-Model Test for Parametric Single-Index Errors-in-Variables Models
This seminar talks about some useful tests for fitting a parametric single-index regression model when covariates are measured with error and validation data is available. We propose two tests whose consistency rates do not depend on the dimension of the covariate vector when an adaptive-to-model strategy is applied. One of these tests has a bias term that becomes arbitrarily large with increasing sample size but its asymptotic variance is smaller, and the other is asymptotically unbiased with larger asymptotic variance. Compared with the existing local smoothing tests, the new tests behave like a classical local smoothing test with only one covariate, and still are omnibus against general alternatives.
This avoids the difficulty associated with the curse of dimensionality.
Further, a systematic study is conducted to give an insight on the effect of the values of the ratio between the sample size and the size of validation data on the asymptotic behavior of these tests. Simulations are conducted to examine the performance in several finite sample scenarios.
Wednesday, August 8, 2018 — 4:00 PM EDT
## Department seminar by Yang Li, Renmin University
Model Confidence Bounds for Variable Selection
In this article, we introduce the concept of model confidence bounds (MCBs) for variable selection in the context of nested models. Similarly to the endpoints in the familiar confidence interval for parameter estimation, the MCBs identify two nested models (upper and lower confidence bound models) containing the true model at a given level of confidence. Instead of trusting a single selected model obtained from a given model selection method, the MCBs proposes a group of nested models as candidates and the MCBs’ width and composition enable the practitioner to assess the overall model selection uncertainty. A new graphical tool — the model uncertainty curve (MUC) — is introduced to visualize the variability of model selection and to compare different model selection procedures. The MCBs methodology is implemented by a fast bootstrap algorithm that is shown to yield the correct asymptotic coverage under rather general conditions. Our Monte Carlo simulations and a real data example confirm the validity and illustrate the advantages of the proposed method.
Thursday, August 2, 2018 — 4:00 PM EDT
## Department seminar by Daniel Farewell, Cardiff University
No Such Thing as Missing Data
The phrase "missing data" has come to mean "information we really, really wish we had". But is it actually data, and is it actually missing? I will discuss the practical implications of taking a different philosophical perspective, and demonstrate the use of a simple model for informative observation in longitudinal studies that does not require any notion of missing data.
Thursday, July 19, 2018 — 4:00 PM EDT
## Department seminar by Geneviève Gauthier, HEC Montreal
Extracting Latent States from High Frequency Option Prices
We propose the realized option variance as a new observable variable to integrate high frequency option prices in the inference of option pricing models. Using simulation and empirical studies, this paper documents the incremental information offered by this realized measure. Our empirical results show that the information contained in the realized option variance improves the inference of model variables such as the instantaneous variance and variance jumps of the S&P 500 index. Parameter estimates indicate that the risk premium breakdown between jump and diffusive risks is affected by the omission of this information.
Tuesday, July 3, 2018 — 4:00 PM EDT
## Department seminar by Shui Feng, McMaster University
Dirichlet Process and Poisson-Dirichlet Distribution
Dirichlet process and Poisson-Dirichlet distribution are closely related random measures that arise in a wide range of subjects. The talk will focus on their constructions and asymptotic behaviour in different regimes including the law of large numbers, the fluctuation theorems, and large deviations.
Monday, June 25, 2018 — 4:00 PM EDT
## David Sprott Distinguished Lecture by Dr. Pauline Barrieu, London School of Economics and Political Science
Assessing financial model risk
Model risk has a huge impact on any financial or insurance risk measurement procedure and its quantification is therefore a crucial step. In this talk, we introduce three quantitative measures of model risk when choosing a particular reference model within a given class: the absolute measure of model risk, the relative measure of model risk and the local measure of model risk. Each of the measures has a specific purpose and so allows for flexibility. We illustrate the various notions by studying some relevant examples, so as to emphasize the practicability and tractability of our approach.
Thursday, May 24, 2018 — 4:00 PM EDT
## Department seminar by Steffen Klaere, University of Auckland
Does your phylogenetic tree fit your data?
Phylogenetic methods are used to infer ancestral relationships based on genetic and morphological data. What started as more sophisticated clustering has now become a more and more complex machinery of estimating ancestral processes and divergence times. One major branch of inference is maximum likelihood methods. Here, one selects the parameters from a given model class for which the data are more likely to occur than for any other set of parameters of the same model class. Most analysis of real data is executed using such methods.
However, one step of statistical inference that has little exposure to application is the goodness of fit test between inferred model and data. There seem to be various reasons for this behaviour, users are either content with using a bootstrap approach to obtain support for the inferred topology, are afraid that a goodness of fit test would find little or no support for their phylogeny thus demeaning their carefully assembled data, or they simply lack the statistical background to acknowledge this step.
Recently, methods to detect sections of the data which do not support the inferred model have been proposed, and strategies to explain these differences have been devised. In this talk I will present and discuss some of these methods, their shortcomings and possible ways of improving them.
Thursday, May 17, 2018 — 4:00 PM EDT
## Department seminar by Jinko Graham, Simon Fraser University
Combining phenotypes, genotypes and genealogies to find trait-influencing variants
A basic tenet of statistical genetics is that shared ancestry leads to trait similarities in individuals. Related individuals share segments of their genome, derived from a common ancestor. The coalescent is a popular mathematical model of the shared ancestry that represents the relationships amongst segments as a set of binary trees, or genealogies, along the genome. While these genealogies cannot be observed directly, the genetic-marker data enable us to sample from their posterior distribution. We may compare the clustering of trait values on each genealogical tree that is sampled to the clustering under the coalescent prior distribution. This comparison provides a latent p-value that reflects the degree of surprise about the trait clustering in the sampled tree. The distribution of these latent p-values is the fuzzy p-value as defined by Geyer and Thompson. The fuzzy p-value contrasts the posterior and prior distributions of trait clustering on the latent genealogies and is informative for mapping trait-influencing variants. In this talk, I will discuss these ideas with application to data from an immune-marker study, present results from preliminary analyses and highlight potential avenues for further research.
Saturday, May 12, 2018 — 8:00 AM to 6:00 PM EDT
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9. 2012 (44) | 2019-12-07 15:15:47 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40376776456832886, "perplexity": 1517.709605245676}, "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-2019-51/segments/1575540499439.6/warc/CC-MAIN-20191207132817-20191207160817-00285.warc.gz"} |
https://byjus.com/jee/area-under-the-curve-jee-advanced-previous-year-questions-with-solutions/ | Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# JEE Advanced Maths Area Under the Curves Previous Year Questions with Solutions
Area under the curves JEE Advanced previous year questions with solutions are given on this page. These solutions are designed by our team of experts. The area under a curve between two points is calculated by doing a definite integral between the two points. Students can expect questions from this topic for any entrance exam. So, students are advised to revise these solutions so that they can easily crack the JEE Advanced exam.
Question 1: Area of the region bounded by the curve y = ex and lines x = 0 and y = e is
(a) e-1
(b) 1
(c) e – ∫01ex dx
(d) ∫1e ln y dy
Solution:
Required area is A = ∫1e ln y dy = e – ∫01 ex dx
= e – (ex)01
= e – (e-1)
= 1
Hence option b, c, d are correct.
Question 2: A farmer F1 has a land in the shape of a triangle with vertices at P(0, 0), Q(1, 1) and R(2, 0). From this land, a neighbouring farmer F2 takes away the region which lies between the side PQ and a curve of the form y = xn (n > 1). If the area of the region taken away by the farmer F2 is exactly 30% of the area of ΔPQR, then the value of n is
Solution:
Area of ΔPQR = ½ bh = ½ × 2×1
= 1 unit
Given that area of the region taken away by the farmer F2 is exactly 30% of the area of ΔPQR.
So area taken away by F1 = 0.3
So ∫01 (x-xn)dx = 0.3
=> [x2/2 – xn+1/n+1)]01 = 0.3
=> [½ – 1/(n+1) = 0.3
=> 0.2 = 1/(n+1)
=> 2n+2 = 10
=> 2n = 8
=> n = 4
Question 3: If the line x = α divides the area of region R = {(x, y) ∈ R2, x3 ≤ y ≤ x, 0 ≤ x ≤ 1} into two equal parts, then
(a) ½ < α < 1
(b) α4 + 4α2 – 1 = 0
(c) 0 < α ≤ ½
(d) 2α4 – 4 α2 + 1 = 0
Solution:
Required area is the area of the shaded portion.
A = ∫01 (x-x3) dx
= [x2/2 – x4/4]01
= ½ – ¼
= 1/4
Given that line x = α divides the area in 2 equal parts,
Area (OPQ) = A/2 = 1/8
A/2 = ∫0α (x-x3) dx
= [x2/2 – x4/4]0α
=> (α2/2) – (α4/4) = ⅛
=> 2α4 – 4α2 + 1 = 0
=> (α2-1)2 = 1/2
=> α2 = (√2-1)/√2
=> ½ < α < 1
Hence option a and d are correct.
Question 4: The area (in sq.units) of the region {(x, y): 0 ≤ y ≤ x2+1, 0≤y≤x+1, ½ ≤ x≤ 2} is
(a) 23/16
(b) 79/24
(c) 79/16
(d) 23/6
Solution:
Required area = ∫1/21 (x2+1) dx + ∫12 (x+1) dx
= [x3/3 + x]11/2 + [x2/2 + x]12
= [4/3 – 13/24] + 5/2
= (32 – 13 + 60)/24
= 79/24
Hence option b is the answer.
Question 5: Let the functions f: R→ R and g: R→ R be defined by f(x) = ex-1 – e-|x-1| and g(x) = ½ (ex-1 + e1-x). Then the area of the region in the first quadrant bounded by the curves y = f(x), y = g(x) and x = 0 is
(a) (2 – √3) + ½ (e – e-1)
(b) (2 + √3) + ½ (e – e-1)
(c) (2 – √3) + ½ (e + e-1)
(d) (2 +√3) + ½ (e + e-1)
Solution:
Given that f(x) = ex-1 – e-|x-1|
So
$$\begin{array}{l}f(x)= \left\{\begin{matrix} 0 & x\leq 1\\ e^{x-1}-e^{1-x}& x\geq 1 \end{matrix}\right.\end{array}$$
g(x) = ½ (ex-1 + e1-x)
If f(x) = g(x)
=> ex-1 – e-(x-1) = (ex-1+e1-x)/2
=> e2(x-1) = 3
=> x = ½ ln 3 + 1
=> x = 1+ ln √3
Bounded area = ∫0(1/2)ln3 + 1g(x) dx – ∫1(1/2)ln3 + 1f(x) dx
= ½ [ex-1 – e1-x]o(1/2)ln3 + 1 – [ex-1+e1-x]1(1/2)ln3 + 1
= (2 – √3) + ½ (e – e-1)
Hence option a is the answer.
Question 6: If the area (in sq.units) bounded by the parabola y2 = 4λx and the line y = λx, λ >0 is 1/9, then λ is equal to
(a) 2√6
(b) 48
(c) 24
(d) 4√3
Solution:
Given parabola y2 = 4λx
And line y = λx
Putting y = λ in y2 = 4λx,
x = 0, 4/λ
Area = ∫04/λ (√(4λx) – λx) dx = 1/9
=> 2√λ (x3/2/3/2)04/λ – λ(x2/2)04/λ = 1/9
Solving we get λ = 24
Hence option c is the answer.
Question 7: The area (in sq.units) of the region bounded by the curves y = 2x and y = |x+1|, in the first quadrant is
(a) loge 2 + (3/2)
(b) 3/2
(c) 1/2
(d) (3/2) – 1/loge2
Solution:
Area = ∫01 ((x+1) – 2x)dx
= [x2/2 + x – 2x/ln 2]01
= (½ + 1 – 2/ln 2) – (-1/ln 2)
= (3/2) – 1/ln 2
Hence option d is the answer.
Question 8: The area (in sq.units) of the region A = {(x, y): y2/2 ≤ x ≤ y+4} is
(a) 53/3
(b) 30
(c) 16
(d) 18
Solution:
Given A = {(x, y): y2/2 ≤ x ≤ y+4}
Hence area = ∫-24 x dy
= ∫-24 (y+4- (y2/2)) dy
= [(y2/2) + 4y – y3/6]-24
= (8 + 16 – 64/6) -(2 – 8 + 8/6)
= 108/6
= 18
Hence option d is the answer.
Question 9: The area (in sq.units) of the region bounded by the parabola, y = x2 + 2 and the lines, y = x+1, x = 0, and x = 3 is
(a) 15/4
(b) 21/2
(c) 17/4
(d) 15/2
Solution:
Area of the bounded region = ∫03 [(x2+2) – (x+1)] dx
= [x3/3 – x2/2 + x]30
= 9 – 9/2 + 3
= 15/2
Hence option d is the answer.
Question 10: The area of the region {(x, y): xy ≤8, 1≤y≤x2} is
(a) 8 loge2 – 14/3
(b) 16 loge2 – 14/3
(c) 8 loge2 – 7/3
(d) 16 loge2 – 6
Solution:
Given xy ≤ 8, 1 ≤ y ≤ x2
Intersection points of xy = 8 and y = 1 is (8, 1): xy = 8 and y = x2 is (2, 4) and y = x2 and y = 1 is (1, 1).
Required area = ∫12 x2 dx + ∫28 (8/x) dx – ∫18 dx
= [x3/3]12 + [8 ln x]28 – [x]18
= (8/3) – (⅓) + 8 ln 8 – 8 ln 2 – (8-1)
= (7/3) + 24 ln 2 – 8 ln 2 – 7
= 16 ln 2 – 14/3
Hence option b is the answer.
Recommended Video | 2022-08-10 07:33:27 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5930874347686768, "perplexity": 4425.926931091205}, "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-2022-33/segments/1659882571150.88/warc/CC-MAIN-20220810070501-20220810100501-00521.warc.gz"} |
https://bitbucket.org/ddrake/sagetex/diff/example.tex?diff2=a47516edbb85&at=default | Diff from to
# File example.tex
\section{Plotting}
-Here's a plot of the elliptic curve $E$.
+Here's a very large plot of the elliptic curve $E$.
\sageplot{E.plot(-3,3)}
On second thought, use a size of $3/4$ the \verb|\textwidth| and don't
use axes:
-\sageplot{p, axes=False}
+\sageplot[width=.75\textwidth]{p, axes=False}
Remember, you're using Sage, and can therefore call upon any of the
software packages Sage is built out of.
\end{sageblock}
And here's the picture:
-\sageplot{matrixprogram}
+\sageplot[scale=.5]{matrixprogram}
Reset \texttt{x} in Sage so that it's not a generator for the polynomial
ring: \sage{var('x')} | 2015-08-05 10:09:41 | {"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": 0.5103453397750854, "perplexity": 2585.150065027647}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2015-32/segments/1438043062723.96/warc/CC-MAIN-20150728002422-00323-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/79194/misplaced-qed-symbol-after-displaymath-inside-item-of-inline-list | # misplaced qed symbol after displaymath inside item of inline list
My problem involes a qed symbol which appears too early. My observations are that it happens after a displaymath inside an item of an unboxed inline list. All of this seems to be necessary to produce the error.
The packages involved are ntheorem, thmtools and enumitem.
If I compile this code with lualatex using TeXLive 2011.
\documentclass{scrreprt}
\PassOptionsToPackage{thmmarks}{ntheorem}
\PassOptionsToPackage{inline}{enumitem}
\usepackage{amsmath,MnSymbol}
\usepackage{ntheorem,thmtools}
\usepackage{
enumitem,
}
\declaretheorem[style=plain,numbered=no,name=Proof]{proof}%
\begin{document}
\begin{proof}
\begin{enumerate*}[mode=unboxed]
$a$
Still proof.
\end{enumerate*}
\end{proof}
\end{document}
the output becomes:
The symbol is placed correctly if the enumerate* environment isn't used or if or \begin{displaymath and \end{displaymath} is used to mark displaymath mode instead of $ and $.
Clearly, the qed symbol should be at the end of the proof after “Still proof.” – What am I doing wrong? Couldn't find anything so far.
-
I also encountered similiar misplacements of the qed symbol in other contexts, but I couldn't reproduce them so far. If I find further such reproduceable misplacements, I'll post them, too. – k.stm Oct 26 '12 at 16:22
this problem doesn't occur with amsthm. i'm not a user of ntheorem, so you may have other reasons for using it. – barbara beeton Oct 26 '12 at 16:36
@barbarabeeton I use it to define certain theoremstyles. Maybe I can switch. But this, of course, still wouldn't really solve this particular problem. – k.stm Oct 26 '12 at 16:51
in the ams author faq, there's a question that deals with qed and other symbols in "nonstandard" places in and out of proofs, by reference to a linked example file. it's specific to amsthm, but maybe it would give you some ideas. sorry i can't help with ntheorem. – barbara beeton Oct 26 '12 at 17:04
ntheorem seems to add the \qed to the last display mode equation. So, a hackish solution would be to end the proof with \vspace{-\belowdisplayskip}. – Peter Grill Oct 26 '12 at 21:06
show 1 more comment
As per my earlier comment ntheorem seems to add the \qed to the very last display mode equation. So a hack is to to end a proof with
\vspace{-\belowdisplayskip}
## Code:
\documentclass{scrreprt}
\PassOptionsToPackage{thmmarks}{ntheorem}
\PassOptionsToPackage{inline}{enumitem}
\usepackage{amsmath,MnSymbol}
\usepackage{ntheorem,thmtools}
\usepackage{
enumitem,
}
\declaretheorem[style=plain,numbered=no,name=Proof]{proof}%
\begin{document}
\begin{proof}
\begin{enumerate*}[mode=unboxed]
$a$
Still proof.
\end{enumerate*}
\vspace{-\belowdisplayskip}
\end{proof}
\end{document}
-
I accepted the answer even though it is – as you said yourself – merely a hackish solution. I personally just switched to using the displaymath-environment. But thanks anyway! – k.stm Aug 25 '13 at 8:13 | 2014-07-24 18:01:48 | {"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": 2, "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": 0.8879597187042236, "perplexity": 2593.2090432498026}, "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-2014-23/segments/1405997890181.84/warc/CC-MAIN-20140722025810-00156-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://motls.blogspot.com/2019/06/sane-genuine-libertarian-system-wouldnt.html?m=1 | ## Sunday, June 30, 2019
### Sane, genuine libertarian system wouldn't enable peaceful mass migration
Last night, I accidentally watched Soph's gaming streaming channel – again, for an hour – and while I didn't find it too important that Kabbalah in Minecraft has frozen over, I did want to learn something new.
And it's just my conclusion that this 15-year-old girl is more likely to teach me new things about politics, society, and Generation Z than the likes of Ben Shapiro or even Jordan Peterson. First of all, she must share some rare genes with me – I would like to know what they are. It's not just about the identical views on political issues but also the authentic, natural individualism, lack of desire to integrate into herds, and the related pattern of smiles and non-smiles and many other things.
OK, fine, so she has often agreed to be called a libertarian. So have I. It's normal. We're supporters of the individual freedoms, abolition of the government, and stuff like that. It's somewhat accidental that this label gets appropriate at some point – classical liberalism has meant almost the same thing but the term "liberalism" has been hijacked by the far left, anti-liberal folks.
In the U.S., libertarians often look like a minority of fringe whackos which is strange because as I see it, much of America has been built on libertarianism.
Let's accept that we're libertarians in the axiomatic system of this blog post. We just want to know what it means. Libertarianism is often about a talk about fantasy lands with policies that seem impossible to be introduced in a foreseeable future of the real world. But we may still want to know: What would libertarianism say about important events that are actually taking place in the world around us? How would our perfect libertarian world deal with these things?
Libertarianism and immigration is obviously one of the first questions that we would consider in this context. OK, everyone who has watched at least a bunch of Soph's videos must understand that she's against mass immigration – and so am I – and you would expect the commenters on her DLive gaming channel to be leaning in the anti-immigration direction, too.
But your expectations would be wrong. Almost everybody – even in that room that is composed of people who might be considered Soph's fans – prefers to make pro-immigration, would-be libertarian statements such as
A stateless libertarian utopia is a global society without borders where everyone may move wherever he or she wants and employees would be hiring anybody, including lots of exotic immigrants etc. No government would be preventing them from doing so.
The same claims prevail on the Wikipedia page I have linked to, too. Clearly, "left-wing libertarians" are in control of that Wikipedia page. But holy cow, these assertions are so totally wrong.
In a world where people are really free (free of the government-mandated restrictions and bans), they may do lots of things but they will still face various hurdles that were created by other people – who were free to do things, too! And everyone would still be constrained by the laws of mathematics and physics, among others. "Libertarianism" claiming that "humans must be free to break the laws of Nature" is just a nuttiness. One of the manifestations of the freedom is that people start to define their ownership and defend it – we are talking both about mobile and immobile assets (and perhaps even intellectual property, patents, and copyrights).
Even if you don't include "the private ownership" among the defining axioms of an ideal libertarian society, the free people will surely invent this concept because it's good for them. It's good for a person to have a place to live – a home that can't be easily disabled others. People need or want to have many other kinds of assets, too. If they can't defend their ownership individually, they will hire someone else who does the work or they will spontaneously team up and create their own militias etc.
Why? Because private ownership is good for them and most (or at least many – many important) people know it.
Much of our world actually does include more or less free people, at least in some respects. And even without any government pressure, most people buy doors and locks for their apartments or build fences or walls around their houses – because they think it's a good idea to protect their property and their privacy from others. Would this behavior go away if there were no governments? It wouldn't. What would happen if there were no governments in the world and people were really free? You should need just a minute to realize that:
A libertarian world of truly free people is a world with billions of walls and fences.
They're absolutely unavoidable. I am not saying that fences and walls are good for everybody. I am not even saying that they're certainly good for the "whole" – e.g. for mankind. But what I am saying is that there is a huge fraction of humanity that finds walls and fences important. And in many cases, the importance seems vital or existentially needed and many people will do a lot to make sure that these walls and fences are built and preserved.
It follows that the stateless world composed of free humans isn't similar to "one country that covers the whole globe". Instead, the stateless world composed of free humans is similar to a world with a huge number of countries and alliances. To one extent or another, almost every house and almost every village is behaving as a country. They will defend their integrity and interests in various ways. There are good reasons for individuals to do so – and there are also good reasons for groups of individuals to team up and create shared defenses etc.
More generally, free people in the world of a libertarian utopia would create lots of structures that are analogous to structures in the world around us simply because they're needed. They're a good idea. And after all, some of these things in the world around us were actually created by free humans from the bottom up and their association with the "evil government" is a distortion of reality.
In a stateless libertarian world with the current location of the people, Africans and Muslims could still prefer to move to Europe or the wealthy Western world. But would they be able to do so? One part of the answer is clear. Most of the motivation would evaporate because there would be no governments that actually pump hundreds of billions of dollars into subsidies for the masses of immigrants. If that welfare were erased, most migrants that are moving towards Europe today would actually lose the motivation to do so. Don't forget that the unemployment rate is some 75% among the new exotic immigrants.
Some exotic migrants may still want to get to the Western world, get a serious job (or start to make a good kebab), and not rely on the government at all. And their potential employees may like the new – and probably cheaper – workers as well. And many of us, "far right" pundits, sometimes eat kebab – I surely do. ;-) But that doesn't mean that this migration would take place. Why? Even if you eliminate all the migrants who are moving mainly because of the subsidies that they will directly receive from the Western governments, the mass migration will still lead to extra expenses and disadvantages.
Even the subset of the exotic migrants who don't "expect any explicit subsidies" from the Western governments will have a higher crime rate, incompatible cultural habits, higher natality, higher frequency of diseases that are almost unknown in the West, greater need to get some education which would be more difficult, and more. Equally importantly, most of the current – mostly white etc. – inhabitants just don't want the demographics of their homes to dramatically change because they want their ownership to "sort of" continue after they die, too: most people want relatives or similar people to be in charge of the land and assets for decades or centuries in the future. It boils down to the selfish genes.
For all these reasons, such a world of free people will spontaneously create – from the bottom up – many anti-immigration policies that are often realized by governments in the real world. These policies aren't artificial acts that depend on an evil government. Instead, the governments do many of these things because there are good reasons for that – current inhabitants of some territory actually want them or demand them, at least in some territories (like Central Europe). So even if these portions of the government became non-existent, the people would reinvent and rebuild these structures again, from the bottom up!
So Africans or Arabs could be de facto prevented (and we won't discuss any "de iure" because in a stateless world, there aren't truly "totally, uniquely, and canonically official" courts, so there is also no "de iure") from stepping on some roads, streets, into the private schools, and lots of other things. Extra hurdles would be created by the people who consider the mass immigration to be a net damage – people who see too many disadvantages, costs, and risks induced by mass immigration. And the employers who would prefer new, cheaper workers could find out that the fight against these hurdles becomes too hard and they would give up their dreams.
Once again, if the people became free and the governments were abolished, there would be many hurdles preventing people from migration en masse. The left-wing libertarians seem to assume that all these hurdles slowing down migration are social constructs by the evil governments. But it just ain't so. What the governments are doing is just something that real people on the ground actually want to do – and they would be doing it, or something equivalent, even if no government existed!
It's possible that the fans of mass immigration in Sweden would manage to destroy all the walls and hurdles – even in a hypothetical Sweden without any government. But I assure you that in Czechia, even if the government didn't exist (and, like in most countries, about one-half of the nation would be happy if the prime minister suddenly evaporated), the people who prefer to build the fences – physical and administrative fences – would win. This victory in the domestic conversation (if not a civil war) has no easy relationship with the government. Governments really reflect the desires of the actual citizens – in most cases. Well, in some cases, governments deviate too far from the wishes and interests of the people – and it may mean that such governments are going to be removed pretty soon. But governments create not only visa requirements and other hurdles of migration. Governments are also inventing things like quotas for foreigners that encourage migration. So here in Czechia, we surely view the efforts to accelerate the Islamization of Europe to be the work by an evil government (primarily the European Commission and the government of Germany) that wouldn't exist if people (and we're thinking about Czech people who are still allowed to decide about their Czech basin) were really free and in control of things!
Finally, I want to make a more general point about the left-wing libertarians.
The assumption that the "truly free humans on Earth would enable unlimited mass migration" seems to be just one important example of a much more general misconception shared by the left-wing libertarians. They seem to mentally live in the vacuum and they assume that
most institutions and social structures that are linked with a government in the real present world just wouldn't exist if the people were really free.
But in almost all cases, this "vacuum" assumption about freedom is totally incorrect. I have mentioned that walls, fences, visas, and anti-immigration policies would be reinvented and rebuilt even if the governments were abolished – this part of government would be "reinvented" from the bottom up because it's a good idea according to very many people or many important people.
The anti-immigration policies are just example of the fallacy. Another example is the hatred towards fiat money and commercial banks that is also widespread among the left-wing libertarian vacuum-dwellers. I would argue that the would-be fanatical support for the cryptocurrencies that is justified by "libertarianism" is a manifestation of the same incorrect assumption that the free people surely want to live in the vacuum – an assumption that also makes most of them believe that the world of free humans would see huge mass immigration.
You know, there is nothing wrong with the fiat money and commercial banks. After all, we may argue that the monetary systems, banks, and other things were largely created by free people from the bottom up because those were a good idea. People barter traded. Then they discovered reserves in precious metals or marten skins or other commodities. Those weren't terribly useful so they invented coupons that gave one rights to get some amount of gold or a marten skin etc. These coupons, or banknotes, were more convenient for payments than anything that was used before. The banknotes remembered in trustworthy databases – electronic money – were even more convenient, but only in some scenarios. In recent centuries, kings and governments were responsible for the codification and propagation of the money – and for the enforcement of mechanisms that guarantee that the money keeps its value, plus minus an error margin, a value defined according some rules that make it credible.
But even if kings and governments didn't exist, people would eventually end up having low-inflation money, banknotes, commercial banks, loans, and many other things, too – simply because those are clearly useful. The modern world would break down rather quickly if people were prevented from using the fiat money or something really, really close to it. So in a world where the money wouldn't exist for a while but people's intelligence or experience would exist, the money would be reinvented by private companies. The trustworthy – probably strong – company guaranteeing the stability and freedom from risks that you will lose everything would probably be the most successful emitters of the private banknotes. These companies could need executioners, bodyguards, and maybe armies. The idea that "the best money is the money connected with the weakest player i.e. most liberated from the big players" would be falsified in the real world because this belief spread among the cryptocurrency cultists is dumb beyond imagination. Rational people prefer to hold the U.S. dollars, a currency of the strongest country in the world, over children banknotes in a village in Zimbabwe and they have very good, rational reasons for this preference.
Much of the Bitcoin religious cult is driven by this totally irrational hatred towards the "big or even central authorities", such as commercial banks and central banks, that guarantee some loans or the persistent value of banknotes, and many other things. But they work fine and have properties – like the stable value of the money – that make them much more usable than e.g. the Bitcoin and similar would-be replacements. The authority of the commercial and central banks is clearly a good thing, not a bad thing, that makes it safer for the people to use their products and services.
(Two weeks ago, Soph published her BTC and BCH – Bitcoin Cash – wallet addresses on Twitter. Replies included a sectarian war, well, mostly BTC fans' attacks on BCH. In a DLive broadcast, she complained that this reaction of the people was so stupid and she couldn't give a damn. I couldn't agree more with her. Needless to say, BCH is somewhat better for small payments because the BTC fees have returned to \$5 per transaction again and BCH fees are negligible. That's what could actually matter for her, not some general partisan lines of the two camps.)
In this text, I have described two important examples of structures – anti-immigration hurdles, policies, and walls; and monetary systems with commercial banks – that would be recreated by the free people in a government-less libertarian society simply because they're good ideas, and in some cases, almost necessary ideas for the survival in the modern world.
The left-wing libertarians aren't really planning a world where the humans are free. Instead, they are thinking about a global government that prohibits people from doing many things that people have good reasons to do – like installing doors and locks, building walls, defending their families and villages against invaders, joining a promising system with banknotes of a predictable value, and founding and using commercial banks that may lend and borrow.
My two examples – anti-immigration policies and monetary systems with banking – aren't the only ones. Clearly, a world with the free people would gradually reinvent lots of other things that we know from our world with the governments, too: armies, police, courts, and schools. Yes, maybe they would recreate some patent offices, NASA, centers for particle physics, trade unions, welfare systems, environmental inspections, and perhaps other "socialist" things, too. It's just silly to assume that the free people would prefer not to have most of these things. Well, I think that some of the newest creations by the governments (think about the EU commissioners responsible for the 26th gender etc.) are really counterproductive (and have expenses exceeding the benefits) but only nuts believe that the seemingly government-related structures that have existed in much of the world for 200+ years are "evil" and would be completely and forever abolished in a free world.
Left-wing libertarians seem to hate everything that is big – but free people often want to build big things and widespread networks. Programmers and entrepreneurs have good reasons why they want to offer their products or operating system in huge areas, perhaps globally, and the consumers often find out that the global accessibility of Microsoft products or McDonald's is good for them, too. No government is needed for that.
At the end, the left-wing libertarians don't really want the world of free people but rather a world where the people must obey some arbitrary, left-wing – and almost certainly harmful if not suicidal for life – ideological criteria such as the condition that people on Earth have the duty to float in the vacuum, welcome other people who float in the vacuum, and they must float without any cash. Given all these huge differences between left-wing libertarians and proper or conservative libertarians, I find the usefulness of the very term "libertarianism" questionable.
And that's the memo. | 2022-01-26 20:15:38 | {"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": 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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24696797132492065, "perplexity": 1799.1825006370593}, "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-2022-05/segments/1642320304961.89/warc/CC-MAIN-20220126192506-20220126222506-00541.warc.gz"} |
https://answerbun.com/mathematics/intersection-of-irreducible-hypersurface-with-tangent-hyperplane-in-a-non-singular-point-is-singular/ | # Intersection of irreducible hypersurface with tangent hyperplane in a non-singular point is singular
Mathematics Asked by lupidupi on December 23, 2020
Prove that the intersection of an irreducible hypersurface $$V(F) subseteq mathbb{A}^{n}$$ with the tangent hyperplane $$T(V)_p$$ in a non-singular point $$p in V$$ is singular at $$p$$. Note: Here we define the intersection by the ideal $$(F, L)$$ where $$L$$ is the linear equation for the tangent hyperplane.
Also show that this need not be true when one defines the intersection with reduced structure, that is, by the radical of the ideal $$(F, L)$$
I found a similar question and answer in this math.stack here. Since I don’t have enough reputation to comment, I cannot ask questions to the OP and the person who answered there.
The difference between my question and the one linked is that I am in affine space rather than in projective space. Is the solution the same for the affine case I am in? If yes, then there are certain things that I do not understand: What does it mean that the "problem is local at $$P$$" and how does this imply that the tangent hyperplane is given by $$x_1 = 0$$; or is it perhaps in my case $$X_1 = a_1$$? And how does one know that the polynomial defining the (affine) hypersurface is given by $$x_1 + f_2(x_2, ldots, x_n) + ldots + f_r(x_2, ldots, x_n)$$, where the $$f_i$$ are homogeneous of degree $$i$$; is it because any polynomial can be decomposed into a sum of homogenoues components? And lastly, how does then the intersection become the hypersurface given by the polynomial $$f_0$$, where the $$x_1$$ coordinate become 0; is this because the tangent hyperplane is given by coordinate $$x_1 = 0$$?
If no, then how do we solve this problem?
The definition of the tangent space to an affine hypersurface I am given is: The tangent space $$T_PV$$ to an affine hypersurface $$V = (f) subseteq mathbb{A}^{n}$$ at $$P = (a_1, ldots, a_n) in V$$ is the linear subvariety
$$V(frac{partial f}{partial x_1}(P)(X_1 – a_1) + ldots + frac{partial f}{partial x_n}(P)(X_n – a_n)).$$
And the point $$P$$ is non-singular if $$T_P V$$ is a hyperplane, i.e. if $$df(P) = (frac{partial f}{partial x_1}(P), ldots, frac{partial f}{partial x_n}(P)) neq 0$$.
What I know: $$p$$ is non-singular, i.e. $$frac{partial F}{partial X_j} neq 0$$ for some $$j$$.
What I have to show: That the polynomial $$f$$ defining the intersection is singular at $$p$$, i.e. $$frac{partial f}{partial X_i} = 0$$ for all $$i$$.
BTW: the question is taken from Reid’s Undergraduate algebraic geometry.
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https://prefetch.eu/know/concept/heaviside-step-function/ | Categories: Mathematics, Physics.
# Heaviside step function
The Heaviside step function $$\Theta(t)$$, is a discontinuous function used for enforcing causality or for representing a signal switched on at $$t = 0$$. It is defined as:
\begin{aligned} \boxed{ \Theta(t) = \begin{cases} 0 & \mathrm{if}\: t < 0 \\ 1 & \mathrm{if}\: t > 1 \end{cases} } \end{aligned}
The value of $$\Theta(t \!=\! 0)$$ varies between definitions; common choices are $$0$$, $$1$$ and $$1/2$$. In practice, this rarely matters, and some authors even change their definition on the fly for convenience. For physicists, $$\Theta(0) = 1$$ is generally best, such that:
\begin{aligned} \boxed{ \forall n \in \mathbb{R}: \Theta^n(t) = \Theta(t) } \end{aligned}
Unsurprisingly, the first-order derivative of $$\Theta(t)$$ is the Dirac delta function:
\begin{aligned} \boxed{ \Theta'(t) = \delta(t) } \end{aligned}
The Fourier transform of $$\Theta(t)$$ is noteworthy. In this case, it is easiest to use $$\Theta(0) = 1/2$$, such that the Heaviside step function can be expressed using the signum function $$\mathrm{sgn}(t)$$:
\begin{aligned} \Theta(t) = \frac{1}{2} + \frac{\mathrm{sgn}(t)}{2} \end{aligned}
We then take the Fourier transform, where $$A$$ and $$s$$ are constants from its definition:
\begin{aligned} \tilde{\Theta}(\omega) = \hat{\mathcal{F}}\{\Theta(t)\} = \frac{A}{2} \Big( \int_{-\infty}^\infty \exp(i s \omega t) \dd{t} + \int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t} \Big) \end{aligned}
The first term is proportional to the Dirac delta function. The second integral is problematic, so we take the Cauchy principal value $$\pv{}$$ and look up the integral:
\begin{aligned} \tilde{\Theta}(\omega) &= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}} = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}} \end{aligned}
The use of $$\pv{}$$ without an integral is an abuse of notation, and means that this result only makes sense when wrapped in an integral. Formally, $$\pv{\{1 / \omega\}}$$ is a Schwartz distribution. We thus have:
\begin{aligned} \boxed{ \tilde{\Theta}(\omega) = \frac{A}{|s|} \Big( \pi \delta(\omega) + i \: \mathrm{sgn}(s) \pv{\frac{1}{\omega}} \Big) } \end{aligned} | 2021-04-18 22:45:15 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9992426633834839, "perplexity": 1606.0218114142606}, "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-2021-17/segments/1618038862159.64/warc/CC-MAIN-20210418224306-20210419014306-00424.warc.gz"} |
https://www.physicsforums.com/threads/thermodynamics-atmospheric-pressure-question.776956/ | # Thermodynamics atmospheric pressure Question
Tags:
1. Oct 19, 2014
### benjibutton
1. The problem statement, all variables and given/known data
A liter of air, initially at room temperature and atmospheric pressure, is heated at constant pressure until it doubles in volume. Calculate the increase in its entropy during this process.
so Ti= 300K, Volume which is 2Vi=Vf; Pressure is constant
2. Relevant equations
ΔS @constant pressure is = ∫Cp/T dT (where Ti -> Tf
3. The attempt at a solution
so since I don't know what T is but I know how the volume changes is there a way I can relate it? is it just PV=nkT so use V as an analog to gauge the proportional change in T? so would that give me something like Cp*Ln[2]?
2. Oct 19, 2014
### Simon Bridge
But you do know what T is - it's the temperature. You mean you are not told what the final temperature is?
Then - yes - you would need to draw information in from another source - the equation you are using for entropy change is for an ideal gas, so it seems reasonable to use the ideal gas law to relate temperature and volume. Unless you have some notes about how air behaves?
http://www.thebigger.com/chemistry/...anges-of-an-ideal-gas-in-different-processes/
3. Oct 19, 2014
### benjibutton
Yeah, sorry; the final T is unknown. So do I write it as Ln[Tf/Ti]? where Tf= 2PVi/Nk ? how do I solve for N?
4. Oct 19, 2014
### benjibutton
OK, I think I figured it out. I got it to be Tf=600K so it becomes 5/2*R*Ln(600K/300K)
5. Oct 19, 2014
### rude man
What happened to n? Your answer is for 1 mole, not 1 liter.
6. Oct 19, 2014
### benjibutton
wouldn't n be the same for both sides of the ideal gas law, which would lead them to cancel out?
7. Oct 19, 2014
### benjibutton
I also never had n, so I assumed you hold is constant, which would still give me the same result, since all other variables are held constant. Unless there's something else I'm missing.
8. Oct 20, 2014
### Simon Bridge
At constant pressure $$\frac{PV_f=nRT_f}{PV_i=nRT_i}\implies \frac{V_f}{V_i} = \frac{T_f}{T_i}$$ ... it's a law that got named after someone.
It means you don't actually have to calculate the temperatures here if you know the volumes.
Of course you can also look up the molar density of air "at room temperature and atmospheric pressure".
9. Oct 20, 2014
### rude man
Does it make sense to you that the change in entropy of 1 mole of gas is the same as the change in 1 liter? Your formula gave the change for 1 mole. So figure out how many moles in 1 liter of air at STP.
Your formula for ΔS = Cp ln(T2/T1) is correct. But Cp = ncp and you only hav cp. BTW air is essentially a diatomic gas for which cp ~ (7/2)R. | 2017-08-20 22:20:14 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6812677979469299, "perplexity": 969.9485104697346}, "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-2017-34/segments/1502886106990.33/warc/CC-MAIN-20170820204359-20170820224359-00661.warc.gz"} |
http://math.stackexchange.com/questions/11682/the-sheafification-of-a-constant-presheaf | # the sheafification of a constant presheaf
Let X be a topological space and A be an abelian group. Give A the discrete topology. For any open set U of X, Let $\cal A(U)$ be the group of all continuous aps of U into A. Thus with the usual restriction maps we obtain a sheaf $\cal A$.
So, why for every connected open set U, $\cal A(U)\cong A$ for all U? And, what is the sheafifcation of the presheaf $U \mapsto A$?
-
For your first question: $A$ has the discrete topology, so elements of $A$ are open and closed. Then, what properties does the preimage of a single element under a continuous map $U \to A$ have? Recall the definition of connectedness. What does this imply for continuous maps $U \to A$? | 2014-10-24 10:51:00 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9516624808311462, "perplexity": 132.0061560883409}, "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-2014-42/segments/1414119645845.57/warc/CC-MAIN-20141024030045-00056-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.qalaxia.com/questions/I-cant-explain-why-these-matrix-row-transformations-are-wrong | D
#### I cant explain why these matrix row transformations are wrong
35 viewed last edited 7 months ago
Anonymous
0
I was helping someone find the determinant of the matrix:
$\left[ \begin{array}{ccc} 1 & -a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{array} \right]$
Apparently, most students in the class applied the following row transformations:
R1 = R1 - R2 and
R2 = R2 - R3 and
R3 = R3 - R1
to get
$\left[ \begin{array}{ccc} 0 & (a-b) & (a^2-b^2) \\ 0 & (b-c) & (b^2-c^2) \\ 0 & (c-a) & (c^2-a^2) \\ \end{array} \right]$
The determinant of this matrix is 0. However, the determinant of the original matrix is (a-b)(b-c)(c-a).
What's wrong with the row transformations applied?
Vivekanand Vellanki
0
Row transformations have to be applied one at a time. In this case, if the three transformations were applied one after the other, they would look like this:
R1 = R1 - R2; R2 = R2; R3 = R3 after the first transformation; and
R1 = R1 - R2; R2 = R2 - R3; R3 = R3 after the second transformation; and
R1 = R1 - R2; R2 = R2 - R3; R3 = R3 - (R1 - R2) after the third transformation
I see that teachers apply multiple such transformations at the same time. While this is possible and saves time in class, it has to be done with caution.
I am working on figuring out simple rules that inform if multiple row transformations can be applied at the same time. | 2019-02-17 21:02:55 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.7804452180862427, "perplexity": 656.652929300546}, "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-2019-09/segments/1550247482478.14/warc/CC-MAIN-20190217192932-20190217214932-00322.warc.gz"} |
https://digitalcommons.wpi.edu/mathematicalsciences-pubs/76/ | ## Mathematical Sciences Faculty Publications
Article
9-1-2008
#### Publication Title
Mathematische Annalen
#### Abstract
We consider the mixed problem $$\left\{ \begin{array}{ll} \Delta u = 0 \quad & {\rm in }\, \Omega\\ \frac{\partial u }{\partial \nu} = f_N \quad & {\rm on }\, {\rm N} \\ u = f_D \quad & {\rm on}\,D \end{array} \right.$$ in a class of Lipschitz graph domains in two dimensions with Lipschitz constant at most 1. We suppose the Dirichlet data, f_D has one derivative in L^p(D) of the boundary and the Neumann data f_N, is in L^p(N). We find a p_0>1 so that for p in the interval (1,p_0), we may find a unique solution to the mixed problem and the gradient of the solution lies in L^p.
342
1
91
124
#### DOI
10.1007/s00208-008-0223-6
Preprint
#### Publisher Statement
This is a pre-print of an article published in Mathematische Annalen. The final authenticated version is available online at: https://doi.org/10.1007/s00208-008-0223-6. | 2019-03-19 21:55:15 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5580607056617737, "perplexity": 844.4481005722237}, "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-2019-13/segments/1552912202131.54/warc/CC-MAIN-20190319203912-20190319225912-00296.warc.gz"} |
https://www.physicsforums.com/threads/minimize-parameter-for-least-absolute-deviation-lad.409028/ | # Minimize parameter for Least Absolute Deviation LAD
1. Jun 9, 2010
### dabd
How to compute $$\beta = arg min_\beta \sum_{i=1}^N {|y_i - x_i^T \beta|$$
2. Jun 9, 2010
There is no closed-form solution for this (contrary to the situation with least squares). The software you use (R, SAS, etc) use a variety of methods. check the relevant documentation for those programs.
3. Jun 9, 2010
### dabd
I was just interested in the calculation not in applying it to real data.
The estimate is the median of the data x1,...,xn and I wanted to see how they derived that result.
4. Jun 9, 2010
$$\sum_{i=1}^n |x_i - a|$$
as a function of $a$, the solution is the sample median. This does not generalize to regression. | 2018-12-17 02:12:01 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6119752526283264, "perplexity": 1584.4355732732581}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2018-51/segments/1544376828056.99/warc/CC-MAIN-20181217020710-20181217042710-00384.warc.gz"} |
http://gatkforums.broadinstitute.org/gatk/discussion/3223/vcf-liftover-header-stringency | The current GATK version is 3.6-0
Examples: Monday, today, last week, Mar 26, 3/26/04
#### Howdy, Stranger!
It looks like you're new here. If you want to get involved, click one of these buttons!
Posts: 4Member
I am trying to liftover from NIST b37 to hg19. I have all the files I need and I can kick off the liftover just fine, but I keep running into problems because the NIST vcf has tags in the variant line INFO field that are not in the header.
##### ERROR MESSAGE: Key PLHSWG found in VariantContext field INFO at chr1:52238 but this key isn't defined in the VCFHeader. We require all VCFs to have complete VCF headers by default
I identified about 90 tags that are not properly documented in the header.
Is there a way to ignore all of these INFO header lapses?
Tagged:
You can probably work around this using --unsafe LENIENT_VCF_PROCESSING but keep in mind there's a good reason why this argument is called unsafe... And if you experience issues down the road with this VCF, we won't be able to provide any support (using an unsafe argument is a bit like voiding the warranty on an electronic device).
It would be much better to just fix your headers so they are valid.
Geraldine Van der Auwera, PhD | 2016-08-26 04:54:54 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2533346116542816, "perplexity": 2608.6424964470566}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2016-36/segments/1471982295264.30/warc/CC-MAIN-20160823195815-00070-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://energy-oil-gas.com/news/greener-diesel/ | # Greener diesel
UPS has confirmed agreements for up to 46 million gallons of renewable fuels over the next three years, constituting a 15-fold increase over prior contracts and making UPS one of the largest users of renewable diesel in the world.
The agreements with three leading suppliers of renewable fuels, secure access to an advanced renewable diesel fuel in order to meet the company’s objectives for alternative fuel utilisation. Neste, Renewable Energy Group (REG) and Solazyme will supply renewable diesel to UPS to help facilitate the company’s shift to move more than 12 per cent of its purchased ground fuel from conventional diesel and gasoline fuel to alternative fuels by the end of 2017. UPS has previously announced a goal of driving one billion miles with alternative fuel and advanced technology vehicles by the end of 2017.
“Advanced alternative fuels like renewable diesel are an important part of our strategy to reduce the carbon emissions impact of our fleet,” said Mark Wallace, UPS senior vice president, global engineering and sustainability. “Renewable diesel has a huge impact significantly reducing lifecycle greenhouse gas emissions by up to 90 per cent less versus conventional petroleum diesel. UPS believes these agreements are especially important because they will help stimulate demand for investment in refinery technologies and sustainable feedstocks needed to produce renewable fuels at a total cost that is comparable to more carbon-intensive petroleum fuels.”
Issue 124 September 2015 | 2022-07-01 13:08:56 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8711156845092773, "perplexity": 4079.997483973915}, "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-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00591.warc.gz"} |
https://forum.allaboutcircuits.com/threads/frequency-response-1st-order.117434/ | # Frequency response 1st order
Discussion in 'Homework Help' started by Montop, Nov 13, 2015.
1. ### Montop Thread Starter New Member
Oct 29, 2015
3
0
I am having a bit of trouble trying to figure this out. The equation I come up with keeps canceling Omega out.
Attempt:
Converting the circuit to the frequency domain the capacitor becomes $\frac{4}{j \omega}$.
I then used a current divider to find $\underline Y(j \omega) = \underline u(j \omega) * \frac{(2+(\frac{4}{j \omega}))}{6+(2+(\frac{4}{j \omega}))}$
Simplifying this I get the frequency response to be $\frac{1}{2}$. I am fairly certain that his is not correct.
2. ### RBR1317 Senior Member
Nov 13, 2010
472
90
I would tend to agree since Y=U*Z, where Z is the parallel combination of the R & RC branches. | 2019-08-18 21:59:24 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 3, "/images/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": 0.8699437379837036, "perplexity": 2184.4292731561163}, "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-2019-35/segments/1566027314130.7/warc/CC-MAIN-20190818205919-20190818231919-00266.warc.gz"} |
https://math.stackexchange.com/questions/1318372/is-kx-an-artinian-noetherian-kx-module | # Is $K[[x]]$ an Artinian/Noetherian $K[x]$-module?
Let $K$ be a field and consider $K[[x]]$ as a $K[x]$-module. Determine if it is Artinian/Noetherian.
I used the following propositions:
If $M$ is an $R$-module and $N\subseteq M$ a submodule, then $M$ is Artinian/Noetherian if and only if both $N$ and $M/N$ are Artinian/Noetherian.
A ring $R$ is Artinian if and only if $R$ is Noetherian and every prime ideal of $R$ is maximal.
First the ring $K[x]$. All its ideals are finitely generated, so it's Noetherian. But the ideal $(0)$ is prime, but not maximal. So $K[x]$ is not Artinian.
I had then planned to use that $K[x]/(x^n)\cong K[[x]]/(x^n)$. Using this relation we can see that either $K[[x]]/(x^n)$ or $(x^n)$ isn't Artinian, since $K[x]$ is not either. This means $K[[x]]$ is not Artinian.
Using the same relation we can see that both $K[[x]]/(x^n)$ and $(x^n)$ are Noetherian since $K[x]$ is. This would mean $K[[x]]$ is Noetherian too.
Could anyone verify this proof.
• In $K[x]/(x^n)$, $(x^n)$ means $x^nK[x]$, but in $K[[x]]/(x^n)$ it means $x^nK[[x]]$, and these are different (and non-isomorphic as $K[x]$-modules). But I think you're assuming they're the same. – Jeremy Rickard Jun 9 '15 at 12:03
• There is a problem: you are denoting with the same symbol $(x^n)$ two distinct objects. Note that $(x^n) \subset K[x] \subset K[[x]]$ is NOT na ideal of $K[[x]]$ since for instance $\sum_{k \ge n}x^k$ is not in $K[x]$. – Crostul Jun 9 '15 at 12:03
• @JeremyRickard That is indeed where I screwed up. – gebruiker Jun 9 '15 at 12:10
1. $K[[X]]$ is not a noetherian $K[X]$-module since it's not finitely generated.
2. $K[[X]]$ is not an artinian $K[X]$-module: $XK[X]\supset X^2k[X]\supset\cdots$ is a strictly descending chain of $K[X]$-submodules. (You can also argue as follows: every submodule of an artinian module is artinian, so if $K[[X]]$ is an artinian $K[X]$-module, then $K[X]$ is an artinian $K[X]$-module, that is, $K[X]$ is an artinian ring, a contradiction.) | 2019-07-21 09:04:48 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.8873421549797058, "perplexity": 182.9390917379123}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2019-30/segments/1563195526940.0/warc/CC-MAIN-20190721082354-20190721104354-00191.warc.gz"} |
http://openmx.ssri.psu.edu/thread/445?q=thread/445 | # Improving Mixture Model Specification
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Improving Mixture Model Specification
Tim and I were discussing my recent botched growth mixture code, and have a proposal for a future version of mixture modeling which should be a little bit easier on the user. It consists of two parts:
-A new matrix type, "Prop" or proportional. A (row-wise) proportional matrix consists of k rows which must sum to a constant (1, unless someone really wants to be able to change it). A 1 x k prop matrix would be used for model-wide class probabilities, while an n x k prop matrix would be used for individual class probabilities. Prop matrices also open us up to compositional data and associated models.
-A mixture optimizer, named something like "mxMixtureObjective." This objective function would take the names of submodels (or their objectives) as arguments, as well as a matrix indicating class probabilities. In addition to making mixture/LCA specification simpler, it also opens us up to optimization of the mixture routines above and beyond improvements to the FIML objective.
Assuming that R objects model1 and model2 are models representing two classes with omx names "M1" and "M2", this model would look something like this:
mxModel("Mixture Example",
mxData(myDataFrame, type="raw"),
model1, model2,
mxMatrix("Prop", 1, 2, free=TRUE, start=.5,
labels=c("classprob1", "classprob2"), name="class"),
mxMixtureObjective(c("M1", "M2"), "class")
)
This could be shortened more with a default class probability matrix, but that's probably too far. This proposal frees users from specifying their own constraints on class probabilities (i.e., to keep them between zero and one with a unit sum) and simplifies the objective function with possible speedups in the future. It may be too big for a 1.0 release, but that's part of why we have a wishlist.
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Joined: 07/31/2009 - 14:25
that looks like a really nice
that looks like a really nice approach Ryne... I find it very intuitive!
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Joined: 07/31/2009 - 15:14
Agreed, a nice approach. It
Agreed, a nice approach. It would be even better if "class" could refer to an algebra or a matrix.
Personally, I'd like to see this as feature #1 to implement post 1.0. It should not, I think, be too difficult. If it is, then perhaps adding an argument weight="myweightAlgebraorMatrix" to the mxFIMLObjective() function would be simpler. This would be applied to every row in the event that vector=TRUE is set.
A major piece of flexibility is added if it is possible to make the weight matrix/algebra a function of definition variables, which would require re-evaluation of the weight algebra for each data vector, in the event that it depends on definition variables.
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Joined: 07/31/2009 - 15:12
As the results of algebras
As the results of algebras are treated as matrices, I don't see a problem using them as arguments. The covariance and means matrix arguments in all of our other optimizers already accept algebras.
What is the difference mathematically between a weight and a class probability parameter? The class probability parameters are really just a set of freely estimated weights with some type of constraint on their sum that reduce their dimensionality.
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I don't see a mathematical
I don't see a mathematical difference. In principle, both could be functions of definition variables and/or parameters of the model.
Sample weights are sometimes applied even when the number of classes is 1, which is a very special case of a latent class or factor mixture model. For programing purposes, it would not be necessary to use a mxAlgebraObjective() if weights were available as an argument within the mxFIMLObjective() call. Conveniently, definition variables would likely already be part of the dataframe in the mxData() call, so their use in weight formulae would be facilitated if they were part of mxFIMLObjective() | 2017-07-27 04:32:29 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5805045366287231, "perplexity": 1395.8586821952647}, "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-2017-30/segments/1500549427429.9/warc/CC-MAIN-20170727042127-20170727062127-00583.warc.gz"} |
https://www.zbmath.org/?q=an%3A0918.33004 | # zbMATH — the first resource for mathematics
Gröbner bases and hypergeometric functions. (English) Zbl 0918.33004
Buchberger, Bruno (ed.) et al., Gröbner bases and applications. Based on a course for young researchers, January 1998, and the conference “33 years of Gröbner bases”, Linz, Austria, February 2–4, 1998. Cambridge: Cambridge University Press. Lond. Math. Soc. Lect. Note Ser. 251, 246-258 (1998).
This work illustrates the use of Gröbner bases and Buchberger’s algorithm in the algebraic study of linear partial differential equations. The $$A$$-hypergeometric system of Gel’fand, Kapranov and Zelevinsky (GKZ) for a function $$\varphi (x_1,x_2,x_3,x_4)$$ in four complex variables is considered as a reference example. It is shown that, for every open ball $$U$$ in $$\mathbb{C}^4$$, the dimension of the $$\mathbb{C}$$-vector space of holomorphic functions $$\varphi$$ on $$U$$ is at most five. The creation operators that allow to obtain new solutions from known solutions are also obtained. Finally, series solutions of GKZ systems are studied and an $$A$$-hypergeometric series is constructed as illustrative example.
For the entire collection see [Zbl 0883.00014].
Reviewer: G.Zet (Iaşi)
##### MSC:
33C20 Generalized hypergeometric series, $${}_pF_q$$ 68W30 Symbolic computation and algebraic computation 13P10 Gröbner bases; other bases for ideals and modules (e.g., Janet and border bases) | 2021-07-26 06:14:29 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7752783298492432, "perplexity": 750.084261449642}, "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-2021-31/segments/1627046152000.25/warc/CC-MAIN-20210726031942-20210726061942-00003.warc.gz"} |
http://www.ck12.org/book/CK-12-Geometry-Second-Edition/r1/section/5.1/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 5.1: Midsegments of a Triangle
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Identify the midsegments of a triangle.
• Use the Midsegment Theorem to solve problems involving side lengths, midsegments, and algebra.
## Review Queue
Find the midpoint between the given points.
1. (-4, 1) and (6, 7)
2. (5, -3) and (11, 5)
3. (0, -2) and (-4, 14)
4. Find the equation of the line between (-2, -3) and (-1, 1).
5. Find the equation of the line that is parallel to the line from #4 through (2, -7).
Know What? A fractal is a repeated design using the same shape (or shapes) of different sizes. Below, is an example of the first few steps of a fractal. What does the next figure look like? How many triangles are in each figure (green and white triangles)? Is there a pattern?
## Defining Midsegment
Midsegment: A line segment that connects two midpoints of adjacent sides of a triangle.
Example 1: Draw the midsegment \begin{align*}\overline{DF}\end{align*} between \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*}. Use appropriate tic marks.
Solution: Find the midpoints of \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*} using your ruler. Label these points \begin{align*}D\end{align*} and \begin{align*}F\end{align*}. Connect them to create the midsegment.
Don’t forget to put the tic marks, indicating that \begin{align*}D\end{align*} and \begin{align*}F\end{align*} are midpoints, \begin{align*}\overline{AD} \cong \overline{DB}\end{align*} and \begin{align*}\overline{BF} \cong \overline{FC}\end{align*}.
Example 2: Find the midpoint of \begin{align*}\overline{AC}\end{align*} from \begin{align*}\triangle ABC\end{align*}. Label it \begin{align*}E\end{align*} and find the other two midsegments of the triangle.
Solution:
For every triangle there are three midsegments.
Let’s transfer what we know about midpoints in the coordinate plane to midsegments in the coordinate plane. We will need to use the midpoint formula, \begin{align*}\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\end{align*}.
Example 3: The vertices of \begin{align*}\triangle LMN\end{align*} are \begin{align*}L(4, 5), M(-2, -7)\end{align*} and \begin{align*}N(-8, 3)\end{align*}. Find the midpoints of all three sides, label them \begin{align*}O, P\end{align*} and \begin{align*}Q\end{align*}. Then, graph the triangle, it’s midpoints and draw in the midsegments.
Solution: Use the midpoint formula 3 times to find all the midpoints.
\begin{align*}L\end{align*} and \begin{align*}M = \left (\frac{4+(-2)}{2}, \frac{5+(-7)}{2}\right)=(1,-1)\end{align*}, point \begin{align*}O\end{align*}
\begin{align*}L\end{align*} and \begin{align*}N = \left(\frac{4+(-8)}{2}, \frac{5+3}{2}\right)=(-2,4)\end{align*}, point \begin{align*}Q\end{align*}
\begin{align*}M\end{align*} and \begin{align*}N = \left(\frac{-2+(-8)}{2}, \frac{-7+3}{2}\right)=(-5,-2)\end{align*}, point \begin{align*}P\end{align*}
The graph would look like the graph to the right. We will use this graph to explore the properties of midsegments.
Example 4: Find the slopes of \begin{align*}\overline{NM}\end{align*} and \begin{align*}\overline{QO}\end{align*}.
Solution: The slope of \begin{align*}\overline{NM}\end{align*} is \begin{align*}\frac{-7-3}{-2-(-8)}=\frac{-10}{6}=-\frac{5}{3}\end{align*}.
The slope of \begin{align*}\overline{QO}\end{align*} is \begin{align*}\frac{-1-4}{1-(-2)}=-\frac{5}{3}\end{align*}.
From this we can conclude that \begin{align*}\overline{NM} \ || \ \overline{QO}\end{align*}. If we were to find the slopes of the other sides and midsegments, we would find \begin{align*}\overline{LM} \ || \ \overline{QP}\end{align*} and \begin{align*}\overline{NL} \ || \ \overline{PO}\end{align*}. This is a property of all midsegments.
Example 5: Find \begin{align*}NM\end{align*} and \begin{align*}QO\end{align*}.
Solution: Now, we need to find the lengths of \begin{align*}\overline{NM}\end{align*} and \begin{align*}\overline{QO}\end{align*}. Use the distance formula.
\begin{align*}NM &= \sqrt{(-7-3)^2+(-2-(-8))^2}=\sqrt{(-10)^2+6^2}=\sqrt{100+36}=\sqrt{136} \approx 11.66\\ QO &= \sqrt{(1-(-2))^2+(-1-4)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34} \approx 5.83\end{align*}
From this we can conclude that \begin{align*}QO\end{align*} is half of \begin{align*}NM\end{align*}. If we were to find the lengths of the other sides and midsegments, we would find that \begin{align*}OP\end{align*} is half of \begin{align*}NL\end{align*} and \begin{align*}QP\end{align*} is half of \begin{align*}LM\end{align*}. This is a property of all midsegments.
## The Midsegment Theorem
The conclusions drawn in Examples 4 and 5 can be generalized into the Midsegment Theorem.
Midsegment Theorem: The midsegment of a triangle is half the length of the side it is parallel to.
Example 6: Mark everything you have learned from the Midsegment Theorem on \begin{align*}\triangle ABC\end{align*} above.
Solution: Let’s draw two different triangles, one for the congruent sides, and one for the parallel lines.
Because the midsegments are half the length of the sides they are parallel to, they are congruent to half of each of those sides (as marked). Also, this means that all four of the triangles in \begin{align*}\triangle ABC\end{align*}, created by the midsegments are congruent by SSS.
As for the parallel midsegments and sides, several congruent angles are formed. In the picture to the right, the pink and teal angles are congruent because they are corresponding or alternate interior angles. Then, the purple angles are congruent by the \begin{align*}3^{rd}\end{align*} Angle Theorem.
To play with the properties of midsegments, go to http://www.mathopenref.com/trianglemidsegment.html.
Example 7: \begin{align*}M, N,\end{align*} and \begin{align*}O\end{align*} are the midpoints of the sides of the triangle.
Find
a) \begin{align*}MN\end{align*}
b) \begin{align*}XY\end{align*}
c) The perimeter of \begin{align*}\triangle XYZ\end{align*}
Solution: Use the Midsegment Theorem.
a) \begin{align*}MN = OZ = 5\end{align*}
b) \begin{align*}XY = 2(ON) = 2 \cdot 4 = 8\end{align*}
c) The perimeter is the sum of the three sides of \begin{align*}\triangle XYZ\end{align*}.
\begin{align*}XY + YZ + XZ = 2 \cdot 4 + 2 \cdot 3 + 2 \cdot 5 = 8 + 6 + 10 = 24\end{align*}
Example 8: Algebra Connection Find the value of \begin{align*}x\end{align*} and \begin{align*}AB\end{align*}.
Solution: First, \begin{align*}AB\end{align*} is half of 34, or 17. To find \begin{align*}x\end{align*}, set \begin{align*}3x-1\end{align*} equal to 17.
\begin{align*}3x - 1 &= 17\\ 3x &= 18\\ x &= 6\end{align*}
Let’s go back to the coordinate plane.
Example 9: If the midpoints of the sides of a triangle are \begin{align*}A(1, 5), B(4, -2)\end{align*}, and \begin{align*}C(-5, 1)\end{align*}, find the vertices of the triangle.
Solution: The easiest way to solve this problem is to graph the midpoints and then apply what we know from the Midpoint Theorem.
Now that the points are plotted, find the slopes between all three.
slope \begin{align*}AB = \frac{5+2}{1-4}=-\frac{7}{3}\end{align*}
slope \begin{align*}BC = \frac{-2-1}{4+5}=\frac{-3}{9}=-\frac{1}{3}\end{align*}
slope \begin{align*}AC = \frac{5-1}{1+5}=\frac{4}{6}=\frac{2}{3}\end{align*}
Using the slope between two of the points and the third, plot the slope triangle on either side of the third point and extend the line. Repeat this process for all three midpoints. For example, use the slope of \begin{align*}AB\end{align*} with point \begin{align*}C\end{align*}.
The green lines in the graph to the left represent the slope triangles of each midsegment. The three dotted lines represent the sides of the triangle. Where they intersect are the vertices of the triangle (the blue points), which are (-8, 8), (10, 2) and (-2, 6).
Know What? Revisited To the left is a picture of the \begin{align*}4^{th}\end{align*} figure in the fractal pattern. The number of triangles in each figure is 1, 4, 13, and 40. The pattern is that each term increase by the next power of 3.
## Review Questions
\begin{align*}R, S, T,\end{align*} and \begin{align*}U\end{align*} are midpoints of the sides of \begin{align*}\triangle XPO\end{align*} and \begin{align*}\triangle YPO\end{align*}.
1. If \begin{align*}OP = 12\end{align*}, find \begin{align*}RS\end{align*} and \begin{align*}TU\end{align*}.
2. If \begin{align*}RS = 8\end{align*}, find \begin{align*}TU\end{align*}.
3. If \begin{align*}RS = 2x\end{align*}, and \begin{align*}OP = 20\end{align*}, find \begin{align*}x\end{align*} and \begin{align*}TU\end{align*}.
4. If \begin{align*}OP = 4x\end{align*} and \begin{align*}RS = 6x - 8\end{align*}, find \begin{align*}x\end{align*}.
5. Is \begin{align*}\triangle XOP \cong \triangle YOP\end{align*}? Why or why not?
For questions 6-13, find the indicated variable(s). You may assume that all line segments within a triangle are midsegments.
1. The sides of \begin{align*}\triangle XYZ\end{align*} are 26, 38, and 42. \begin{align*}\triangle ABC\end{align*} is formed by joining the midpoints of \begin{align*}\triangle XYZ\end{align*}.
1. Find the perimeter of \begin{align*}\triangle ABC\end{align*}.
2. Find the perimeter of \begin{align*}\triangle XYZ\end{align*}.
3. What is the relationship between the perimeter of a triangle and the perimeter of the triangle formed by connecting its midpoints?
Coordinate Geometry Given the vertices of \begin{align*}\triangle ABC\end{align*} below, find the midpoints of each side.
1. \begin{align*}A(5, -2), B(9, 4)\end{align*} and \begin{align*}C(-3, 8)\end{align*}
2. \begin{align*}A(-10, 1), B(4, 11)\end{align*} and \begin{align*}C(0, -7)\end{align*}
3. \begin{align*}A(0, 5), B(4, -1)\end{align*} and \begin{align*}C(-2, -3)\end{align*}
4. \begin{align*}A(2, 4), B(8, -4)\end{align*} and \begin{align*}C(2, -4)\end{align*}
Multi-Step Problem The midpoints of the sides of \begin{align*}\triangle RST\end{align*} are \begin{align*}G(0, -2), H(9, 1)\end{align*}, and \begin{align*}I(6, -5)\end{align*}. Answer the following questions.
1. Find the slope of \begin{align*}GH, HI\end{align*}, and \begin{align*}GI\end{align*}.
2. Plot the three midpoints and connect them to form midsegment triangle, \begin{align*}\triangle GHI\end{align*}.
3. Using the slopes, find the coordinates of the vertices of \begin{align*}\triangle RST\end{align*}.
4. Find \begin{align*}GH\end{align*} using the distance formula. Then, find the length of the sides it is parallel to. What should happen?
More Coordinate Geometry Given the midpoints of the sides of a triangle, find the vertices of the triangle. Refer back to problems 19-21 for help.
1. (-2, 1), (0, -1) and (-2, -3)
2. (1, 4), (4, 1) and (2, 1)
Given the vertices of \begin{align*}\triangle ABC\end{align*}, find:
a) the midpoints of \begin{align*}M, N\end{align*} and \begin{align*}O\end{align*} where \begin{align*}M\end{align*} is the midpoint of \begin{align*}\overline{AB}\end{align*}, \begin{align*}N\end{align*} is the midpoint of \begin{align*}\overline{BC}\end{align*} and \begin{align*}C\end{align*} is the midpoint of \begin{align*}\overline{AC}\end{align*}.
b) Show that midsegments \begin{align*}\overline{MN}, \overline{NO}\end{align*} and \begin{align*}\overline{OM}\end{align*} are parallel to sides \begin{align*}\overline{AC}, \overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*} respectively.
c) Show that midsegments \begin{align*}\overline{MN}, \overline{NO}\end{align*} and \begin{align*}\overline{OM}\end{align*} are half the length of sides \begin{align*}\overline{AC}, \overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*} respectively.
1. \begin{align*}A(-3, 5), B(3, 1)\end{align*} and \begin{align*}C(-5, -5)\end{align*}
2. \begin{align*}A(-2, 2), B(4, 4)\end{align*} and \begin{align*}C(6, 0)\end{align*}
For questions 27-30, \begin{align*}\triangle CAT\end{align*} has vertices \begin{align*}C(x_1,y_1), A(x_2,y_2)\end{align*} and \begin{align*}T(x_3,y_3)\end{align*}.
1. Find the midpoints of sides \begin{align*}\overline{CA}\end{align*} and \begin{align*}\overline{CT}\end{align*}. Label them \begin{align*}L\end{align*} and \begin{align*}M\end{align*} respectively.
2. Find the slopes of \begin{align*}\overline{LM}\end{align*} and \begin{align*}\overline{AT}\end{align*}.
3. Find the lengths of \begin{align*}\overline{LM}\end{align*} and \begin{align*}\overline{AT}\end{align*}.
4. What have you just proven algebraically?
## Review Queue Answers
1. \begin{align*}\left ( \frac{-4+6}{2}, \frac{1+7}{2} \right ) = (1, 4)\end{align*}
2. \begin{align*}\left ( \frac{5+11}{2}, \frac{-3+5}{2} \right ) = (8, 1)\end{align*}
3. \begin{align*}\left (\frac{0 - 4}{2}, \frac{-2 + 14}{2} \right ) = (-2, 6)\end{align*}
4. \begin{align*}m=\frac{-3-1}{-2-(-1)} = \frac{-4}{-1}=4\!\\ y=mx+b\!\\ -3=4(-2)+b\!\\ b=5, \ y=4x+5\end{align*}
5. \begin{align*}-7=4(2)+b\!\\ {\;} \ b=-15, \ y=4x-15\end{align*}
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Subjects: | 2016-08-31 16:32:17 | {"extraction_info": {"found_math": true, "script_math_tex": 151, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9039669036865234, "perplexity": 3416.397568516389}, "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-2016-36/segments/1471982295966.49/warc/CC-MAIN-20160823195815-00291-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://www.astro.gla.ac.uk/users/eduard/cesra/?tag=type-iii-bursts&paged=5 | ### Critical Fluctuations in Beam-Plasma Systems and Solar Type III Radio Bursts by G. Thejappa and R. J. MacDowall
2017-12-05
The type III radio bursts are the most intense radio emissions from the sun. In Figure 1, we present a typical type III burst observed by the STEREO spacecraft. Ginzburg and Zheleznyakov (1958) were the first to suggest that Langmuir waves excited by the solar flare accelerated electrons are the source of these bursts. The in situ detection of electron beams and Langmuir waves in association with type III bursts […]
### Observations of solar radio burst fine structures with LOFAR by E. Kontar et al.*
2017-11-28
During solar flares, electrons are accelerated up to relativistic speeds. As they propagate upwards through the solar corona, they produce so-called type III radio bursts. These type III bursts often demonstrate fine structure, with their spectra consisting of multiple narrowband “striae”; most likely, these structures are caused by small-scale density inhomogeneities of the coronal plasma so that the subsequent fundamental plasma processes that produce the radio bursts is produced in […]
### Small electron acceleration episodes in the solar corona by T. James et al.
2017-11-07
Large solar flares are well known sites of prodigious particle acceleration. While these have deservedly attracted considerable attention, small episodes of electron acceleration and heating have been lately recognized as possible candidates for heating the quiet solar corona. We study the number, power and energy carried by nonthermal electrons produced by instances of small scale electron acceleration in the solar corona. Our primary focus is on small electron acceleration events […]
### How Electron Beams Produce Continuous Coherent Plasma Emission by H. Che, M. Goldstein, P. Diamond, and R. Sagdeev
2017-04-11
It is commonly accepted that energetic electron beams can produce drift frequency radio emission or Type III bursts since Ginzburg and Zhelezniakov first proposed the idea in 1958. However, the electron two-stream instability time (see reference 2) in the corona is fraction of a second, while the duration of coronal Type III bursts lasts several orders of magnitude longer. This problem is called the “Sturrock Dilemma” and remains a subject […]
### Large-scale simulations of Langmuir Wave Distributions Induced by Electron Beams by H. Reid and E. Kontar
2017-02-14
Langmuir waves that generate type III radio bursts are excited by high-energy electron beams streaming out from the corona through interplanetary space. Despite a smooth temporal distribution of electrons, the Langmuir waves are measured to occur in discrete clumps, commonly attributed to the turbulent nature of the solar wind electron density (e.g. Smith and Sime 1979, Melrose et al 1986). But how do fluctuations in the background plasma shape the […]
### Emission of radiation by plasmas with counter-streaming electron beams by L. F. Ziebell et al.*
2017-01-31
The phenomena of emission of radiation by the Sun, which are known as type II and type III solar radio bursts, have been known and investigated for more than sixty years. The bursts of radiation occur at a frequency corresponding to the plasma frequency at the source region, and harmonics are frequently observed. It is commonly accepted that the radiation is generated by a nonlinear mechanism, proposed by Ginzburg and […]
### Diagnosing the Source Region of a Solar Burst on 26 September 2011 by Using Microwave Type-III Pairs by Tan B. L. et al.*
2016-12-20
Accelerated electron beams are believed to be responsible for both hard X-ray (HXR) and strong coherent radio emission during solar flares. However, so far the location of the electron acceleration and its physical parameters are poorly known. The solar microwave Type-III pair burst is possibly the most sensitive signature of the primary energy release and electron accelerations in flares (Aschwanden & Benz, 1997, ApJ). A Type-III pair is composed of […]
### Acceleration of electrons in the solar wind by Langmuir waves produced by a decay cascade by Catherine Krafft and Alexander Volokitin
2016-12-13
It was recently reported that a significant part of the Langmuir waveforms observed by the STEREO satellite (Graham and Cairns, 2013) during type III solar radio bursts are likely consistent with the occurrence of electrostatic decay instabilities, when a Langmuir wave $\mathcal{L}$ resonantly interacts with another Langmuir wave $\mathcal{L}^{\prime}$ and an ion sound wave $\mathcal{S}^{\prime}$ through the decay channel $\mathcal{L} \rightarrow\mathcal{L}^{\prime}+\mathcal{S}^{\prime}$. Usually such wave-wave interactions occur in regions of the […]
1 3 4 5 6 | 2023-02-02 08:48:47 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.6039940714836121, "perplexity": 1830.9384719101818}, "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-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00756.warc.gz"} |
https://en.wikipedia.org/wiki/Karplus-Strong_string_synthesis | # Karplus–Strong string synthesis
(Redirected from Karplus-Strong string synthesis)
Karplus–Strong string synthesis is a method of physical modelling synthesis that loops a short waveform through a filtered delay line to simulate the sound of a hammered or plucked string or some types of percussion.
At first glance, this technique can be viewed as subtractive synthesis based on a feedback loop similar to that of a comb filter for z-transform analysis. However, it can also be viewed as the simplest class of wavetable-modification algorithms now known as digital waveguide synthesis, because the delay line acts to store one period of the signal.
Alexander Strong invented the algorithm, and Kevin Karplus did the first analysis of how it worked. Together they developed software and hardware implementations of the algorithm, including a custom VLSI chip. They named the algorithm "Digitar" synthesis, as a portmanteau for "digital guitar".
## How it works
1. A short excitation waveform (of length L samples) is generated. In the original algorithm, this was a burst of white noise, but it can also include any wideband signal, such as a rapid sine wave chirp or frequency sweep, or a single cycle of a sawtooth wave or square wave.
2. This excitation is output and simultaneously fed back into a delay line L samples long.
3. The output of the delay line is fed through a filter. The gain of the filter must be less than 1 at all frequencies, to maintain a stable positive feedback loop. The filter can be a first-order lowpass filter (as pictured). In the original algorithm, the filter consisted of averaging two adjacent samples, a particularly simple filter that can be implemented without a multiplier, requiring only shift and add operations. The filter characteristics are crucial in determining the harmonic structure of the decaying tone.
4. The filtered output is simultaneously mixed back into the output and fed back into the delay line.
## Tuning the string
The fundamental frequency (specifically, the lowest nonzero resonant frequency) of the resulting signal is the lowest frequency at which the unwrapped phase response of the delay and filter in cascade is ${\displaystyle -2\pi }$. The required phase delay D for a given fundamental frequency F0 is therefore calculated according to D = Fs/F0 where Fs is the sampling frequency.
The length of any digital delay line is a whole-number multiple of the sampling period. In order to obtain a fractional delay, interpolating filters are used with parameters selected to obtain an appropriate phase delay at the fundamental frequency. Either IIR or FIR filters may be used, but FIR have the advantage that transients are suppressed if the fractional delay is changed over time. The most elementary fractional delay is the linear interpolation between two samples (e.g., s(4.2) = 0.8s(4) + 0.2s(5)). If the phase delay varies with frequency, harmonics may be sharpened or flattened relative to the fundamental frequency. The original algorithm used equal weighting on two adjacent samples, as this can be achieved without multiplication hardware, allowing extremely cheap implementations.
Z-transform analysis can be used to get the pitches and decay times of the harmonics more precisely, as explained in the 1983 paper that introduced the algorithm.
A demonstration of the Karplus-Strong algorithm can be heard in the following Vorbis file. The algorithm used a loop gain of 0.98 with increasingly attenuating first order lowpass filters. The pitch of the note was A2, or 220 Hz.
Holding the period (= length of the delay line) constant produces vibrations similar to those of a string or bell. Increasing the period sharply after the transient input produces drum-like sounds.
## Refinements to the algorithm
Alex Strong and Kevin Karplus realized that the Karplus-Strong algorithm was physically analogous to a sampling of the transversal wave on a string instrument, with the filter in the feedback loop representing the total string losses over one period. Julius O. Smith III [1] and others generalized the algorithm to digital waveguide synthesis, which could also be used to model acoustic waves in tubes and on drum membranes. The first set of extensions and generalizations was presented in a paper in 1982 at the International Computer Music Conference in Venice, Italy, and published in more detail in 1983 in Computer Music Journal in an article entitled "Extensions of the Karplus Strong Plucked String Algorithm," by David A. Jaffe and Julius O. Smith.[1]
Alex Strong developed a superior wavetable-modification method for plucked-string synthesis, but only published it as a patent.
## Musical Applications
The first musical use of the algorithm was in the work May All Your Children Be Acrobats written in 1981 by David A. Jaffe, and scored for eight guitars, mezzo-soprano and computer-generated stereo tape, with a text based on Carl Sandburg's The People, Yes. Jaffe continued to explore the musical and technical possibilities of the algorithm in Silicon Valley Breakdown, for computer-generated plucked strings (1982), as well as in later works such as Telegram to the Higher elf-lord, 1984 for string quartet and tape, and Grass for female chorus and tape (1987).
The patent was licensed first to Mattel Electronics, which failed as a company before any product using the algorithm was developed, then to a startup company founded by some of the laid-off Mattel executives. They never got sufficient funding to finish development, and so never brought a product to market either. Eventually Yamaha licensed the patent, as part of the Sondius package of patents from Stanford. It is unknown whether any hardware using the algorithm was ever sold, though many software implementations (which did not pay any license fees to the inventors) have been released.
## References
Citations
Bibliography
• Moore, F. Richard (1990). Elements of Computer Music. Upper Saddle River: Prentice-Hall. ISBN 0-13-252552-6. | 2017-02-28 00:02:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4970508813858032, "perplexity": 1626.8933261805569}, "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-2017-09/segments/1487501173866.98/warc/CC-MAIN-20170219104613-00011-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://mathcentral.uregina.ca/QQ/database/QQ.09.12/h/chad1.html | SEARCH HOME
Math Central Quandaries & Queries
Question from Chad, a parent: I have 3 acres 100ft at the top and bottom what is the length of each side to make up the 3 acres
I am afraid that we need more information to answer your question. If your plot of land is a rectangle, that is all the angles are right angles then I can answer.
There are 43560 square feet in an acre so 3 acres would be $3 \times 43560 = 139590$ square feet. The area of a rectangle is the length times the width so if the length is L feet and the width is 100 feet then we have
$100 \times L = 139590$
and hence $L = 1395.9$ feet.
If the angles are not right angles the the plot might look like
or
or many other shapes. Without knowledge of the size of the angles the distance from the top to the bottom can't be determined.
Penny
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. | 2017-11-23 07:35:29 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.668502151966095, "perplexity": 312.76709833052195}, "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-2017-47/segments/1510934806760.43/warc/CC-MAIN-20171123070158-20171123090158-00091.warc.gz"} |
https://www.physicsforums.com/threads/little-issue-regarding-physical-states.932322/ | # I Little issue regarding physical states
1. Nov 21, 2017
### Tio Barnabe
Consider the QM postulate which states that physical states are represented by rays in a Hilbert space. Consider a ray $R$. An observer from other frame will have a correspoding $R'$ which can be either
- equal to $R$ or,
- not equal to $R$
Suppose the two frames are inertial frames. Consider the Relativity principle that "the laws of nature are the same in all inertial frames". This is translated to the statement that the rays are the same, i.e. $R' = R$, correct?
Then only the first scenario above would satisfy Relativity. What if it turns out that the second case is meet?
2. Nov 21, 2017
### Staff: Mentor
Well first consider physically what states are. They are used, with observable's, to predict probabilities. Do you expect probabilities to change with frame? This is used by Ballentine in his text to derive Schrodinger's equation etc, and yes of course it easily's implied states are frame independent.
Thanks
Bill
3. Nov 21, 2017
### Tio Barnabe
Then we should ask ourselves in what sense are they frame independent. To me they do change from one frame to the other, but they would go along the same ray in the Hilbert space. So they are not equal, but they are equivalent, because same ray means same physical state.
4. Nov 21, 2017
### Staff: Mentor
I explained the exact sense; remember they encode the probabilities of outcomes of observations and even imply it via Gleason's theorem. If they were frame dependent then the probabilities of observations would vary between frames. Different observers would register different probabilities of outcomes of observations. Imagine the outcome on a digital readout. All observers would observe the same readout and hence the same probabilities. That they would be different is physical non-sense. Strictly speaking you are invoking the POR, but aside from that it would be downright nonsensical and inconsistent even.
Thanks
Bill
5. Nov 21, 2017
### Staff: Mentor
You aren't just assuming that. You are assuming that the Hilbert space is a space of rays that correspond to "the state of some system at some time". In other words, something that gets transformed from frame to frame. (At least you are allowing for that possibility.)
But if we are talking about QM in the context of relativity, we are talking about quantum field theory, and in QFT, the Hilbert space is a space of rays that correspond to "quantum field operators at some spacetime event". In other words, there is nothing to transform because everything is at one event. If you are talking about observations made by different observers in different states of motion at that event, you are talking about different operators; that is how different "frames" are realized. So you don't "transform" anything when you go from frame to frame; you just switch which operators you are talking about.
6. Nov 22, 2017
### mikeyork
Not in the conventional construction of a Hilbert space. The reason is, as Wigner showed, that a frame transformation is represented by a unitary operator.
Sometimes people will argue that the transformed ray is in a distinct Hilbert space -- as if each Hilbert space had an associated frame/observer. But that is not consistent with a frame transformation being a unitary operator in a single space.
So, even though the physical state is observer-independent, the ray (state vector) representing it differs between frames/observers if we stick with a single Hilbert space picture.
There is a way around this and that is to recognize that the change in an observable that corresponds to a frame transformation is represented by the inverse of the unitary operator representing the frame transformation. Consider a spatial translation in which the origin is shifted by $+a$. Then the change in x is given by $x\rightarrow x' = x-a$ because a positive shift in the origin results in a negative shift in the coordinate.
Now a coordinate base vector representing coordinate $x$ is usually written $|x\rangle$. But notice the frame of reference is not explicitly specified; it is just assumed as implicit. Now suppose we make a frame transformation such that $x\rightarrow x'$. Clearly $|x'\rangle$ is a different base vector. Now suppose we make the frame of reference $F$ explicit by specifying the base vector in $F$ as $|x,F\rangle$. Clearly then $|x,F\rangle \ne |x',F\rangle$ and $|x,F\rangle \ne |x,F'\rangle$. But now we are free to equate $|x,F\rangle = |x',F'\rangle$ so that we get an observer-independent state vector. In effect, the unitary operator representing $x\rightarrow x'$ is cancelled by the unitary operator representing $F\rightarrow F'$
7. Nov 22, 2017
### A. Neumaier
The situation is complicated because ordinary quantum mechanics is nonrelativistic, and relativistic elements can only be introduced in a heuristic way. See papers by Peres and Terno (e.g., https://arxiv.org/abs/quant-ph/0212023). In particular, the Hilbert spaces of observers related by a Lorentz boost are not directly comparable, hence one cannot talk about the same ray!
A fully correct treatment should involve quantum field theory. But observers break Lorentz invariance, due to their own preferred frame, and in quantum field theory, the observer cannot be represented, only the change of frame, which is a Lorentz transformation applied to the field according to its transformation properties (which depend on its spin).
8. Nov 22, 2017
### mikeyork
Wigner treats a Lorentz boost as a translation of a momentum frame. It can therefore be treated in the same way as a spatial translation in my post #6 simply by employing a momentum basis instead of a coordinate basis and substituting $p$ for $x$.
9. Nov 22, 2017
### A. Neumaier
A boost is _not_ a translation in the momentum frame, but a more complicated transformation!
10. Nov 23, 2017
### mikeyork
Technically, a Lorentz boost is a translation in velocity space not momentum space. But in the case of a momentum basis it is a distinction without a difference because for a mass that is a single-valued function of velocity, the application of such a boost to a state of definite momentum is exactly equivalent to a translation in momentum space. (And yes, of course, in the case of spin it will also involve a Wigner rotation. But the result in terms of state vectors in a fixed basis is still a unitary operator that is the inverse of the operator performed on the frame.)
Last edited: Nov 23, 2017
11. Nov 23, 2017
### Staff: Mentor
The other thing that needs to be mentioned is the Heisenberg and Schrodinger pictures.
I will let the OP look them up and see the obvious relation to his question and how it 'mucks up' even unambiguously making sense of it.
And to make matters worse there is even the Dirac picture which is sort of a combination of the two - its maddening these foundational things.
Guess which picture my answer referred to?
Thanks
Bill
Last edited: Nov 23, 2017
12. Nov 23, 2017
### Tio Barnabe
Thank you all.
I need to carefully read the answers and I will do so as soon as possible.
13. Nov 26, 2017
### vanhees71
Well, only because you have an observer you don't break Lorentz (or Poincare) invariance. You can't also break Lorentz invariance simply because there is an equilibrated medium. Unfortunately it's sometimes claimed that this is the case in some textbooks on thermal field theory, but you can very easily formulate everything in a manifestly covariant way by e.g., writing the standard grand-canonical statistical operator in a covariant way,
$$\hat{\rho}=\frac{1}{Z} \exp(-\beta u \cdot \hat{P}-\beta \sum_i \mu_i \hat{Q}_i), \quad Z=\mathrm{Tr} \exp(\dots).$$
Here $u=(u^{\mu})$ is the four-velocity of the rest frame of the heat bath, $\beta$ the inverse temperature, and $\mu_i$ the chemical potentials with respect to a set of conserved charges $\hat{Q}_i$.
So far there's no hint at violation of Poincare symmetry (as long as gravity is neglected).
14. Nov 26, 2017
### vanhees71
Lorentz boosts are NOT translations in velocity space. They do not even commute with each other, nor do they build a group! Only together with the rotations they build the proper orthochronous Lorentz group. Restricting to one dimension it's a translation in rapidity space.
15. Nov 26, 2017
### A. Neumaier
Of course the whole theory is covariant, but how do you define an observer in QFT? It is not just a frame, because a frame cannot observe anything. Thus the observer must be a part of the interacting system, just like in QM when one models the observation process as for the nonrelativistic case in the work by Allahverdyan et al. reviewed by me on PO and discussed here on PF.
16. Nov 26, 2017
### mikeyork
And the physical significance of your quibble is?
Regardless, my point stands: the frame transformation is a unitary transformation which can be viewed either as a change in observables such as momentum (as per Wigner) or the inverse transformation of the frame itself and, if you include both the observable and the frame in specifying the basis, then you have state vectors that are unchanged -- which I believe is what the OP was asking about.
And, BTW, this true for any frame transformation; the unitary operator and its inverse will always cancel each other. Only the state description (observable, frame) changes.
17. Nov 27, 2017
### vanhees71
Well, I simply wanted to correct a misconception. If you are not interested, just ignore it. Of course, what you wrote, is correct. Invariant objects are, big surprise, invariant ;-).
18. Nov 27, 2017
### vanhees71
Of course, "an observer" is some device interacting with the system to provide finally information about the measured system to us. It still doesn't break the fundamental symmetries of nature, because it's part of nature (assumed the symmetries are really realized in nature, of course).
19. Nov 27, 2017
### mikeyork
How often do you see it explicitly stated in QM textbooks that if you explicitly include the frame in the basis specification (so that $|x,F\rangle and |x,F'\rangle$ are each base vectors in differing bases in the same Hilbert space) then every state can be ascribed a unique frame-independent state vector? I have never seen a textbook that says this. In fact, since a frame rotation about the spin projection axis is a simple phase change, this observation enables us to remove arbitrary phase factors from state vectors and reduce them to a single global phase factor for the entire Hilbert space. And this alone enables us to deduce the spin-statistics theorem whereas just about every textbook treats it as some arbitrary rule imposed by nature. So its hardly as trivial as you suggest.
20. Nov 27, 2017
### Staff: Mentor
Errr - ever heard of the Heisenberg picture where the state remains the same and the observable varies? Its in just about every single textbook.
In standard QM the Galilean transformations are assumed (see chapter 3 - Ballentine) so this implies very simply, the state vector stays the same between frames - in that picture.
As I said in a previous post this is not a simple yes/no answer because of the Schrodinger, Dirac and Heisenberg pictures. In my original answer I implicitly assumed the Heisenberg picture - I really should have made that explicit - hence my second post. I didn't precisely spell it out because I wanted the OP to think about it a bit - maybe that's my bad - I don't know - but what you figure out for yourself you understand better and in my answers I try to get people thinking a bit.
This however is very very basic QM.
Thanks
Bill
Last edited: Nov 27, 2017 | 2018-10-19 07:07:35 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8252285718917847, "perplexity": 510.5880080675308}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2018-43/segments/1539583512332.36/warc/CC-MAIN-20181019062113-20181019083613-00297.warc.gz"} |
http://mathhelpforum.com/calculus/179000-taylor-development.html | # Math Help - Taylor development
1. ## Taylor development
Hi there. I have this exercise which I'm trying to solve now. It says:
Using that $\displaystyle\sum_{n=0}^{\infty}x^n=(1-x)^{-1}$ find one Taylor development for the function $f(x)=\ln(1-x)$
$f^1(x)=\displaystyle\frac{-1}{(1-x)},f^2(x)=\displaystyle\frac{-1}{(1-x)^2},f^3(x)=\displaystyle\frac{-2}{(1-x)^3},f^4(x)=\displaystyle\frac{-6}{(1-x)^4},f^5(x)=\displaystyle\frac{-24}{(1-x)^5}$
And then:
$\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac {f^b(x_0)(x-x_0)^n}{n!}=-\displaystyle\frac{(x-x_0)}{(1-x_0)}-\displaystyle\frac{(x-x_0)^2}{2(1-x_0)^2}-\displaystyle\frac{2(x-x_0)^3}{6(1-x_0)^3}-\displaystyle\frac{6(x-x_0)^4}{24(1-x_0)^4}-\displaystyle\frac{24(x-x_0)^5}{120(1-x_0)^5}+\ldots+-\displaystyle\frac{(x-x_0)^n}{n(1-x_0)^n}$
I have two problems with this. In the first place, the general expression that I've found (which is probably wrong) doesn't work for n=0, it does for the others values of n. I thought of starting the summation at 1, but I'm not sure if this is valid. In the second place I don't know how to use the relation the problem gives at the beginning. I can see that I have (1-x_0) for every term, but I couldn't make it fit inside the summation.
So this is what I got: $\displaystyle\sum_{n=1}^{\infty}-\displaystyle\frac{(x-x_0)^n}{n(1-x_0)^n}$
Bye there, thanks for your help and suggestions.
2. You didn't use the hint
Notice that
$f(x)=\ln(1-x) \implies f'(x)=\frac{-1}{1-x}$
But now by the hint we know this is a geometric series and gives
$f'(x)=\frac{-1}{1-x}=-\sum_{n=0}^{\infty}x^n$
But we want the power series for f not its derivative so lets integrate to get
$\int f'(x)dx =-\int \sum_{n=0}^{\infty}x^ndx=-\sum_{n=0}^{\infty}\int x^ndx$
This gives
$f(x)=C-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$
and since $f(0)=0 \implies C=0$
$f(x)=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$
3. Very nice. Thank you. I would never realized of it :P
BTW, the problem is implicitly telling that the function is taken on an interval where its uniformly convergent, right?
4. Oops, I thought wrong; sorry. :)
5. I think not but now I don't need 'em anyway :P
6. Originally Posted by Ulysses
I think not but now I don't need 'em anyway :P
Yeah, I missed that you accounted for the sign alternations in the denominators! :P
7. Originally Posted by Ulysses
Very nice. Thank you. I would never realized of it :P
BTW, the problem is implicitly telling that the function is taken on an interval where its uniformly convergent, right?
Yes the geometric series is uniformly convergent on |x|<1
8. Thank to all of you. Now I'm trying to solve the same exercise, but with the function $f(x)=(1+4x^2)^{-1}$ I've tried to make some algebraic work to get the expression I'm looking for, $(1-x)^{-1}$ but I didn't get it.
I've made the first derivative, and a few more, but I think its just necessary the first, as before. I get:
$f'(x)=\diplaystyle\frac{-8x}{(1+4x^2)^2}$
I thought of making $f(x)=\diplaystyle\frac{1}{(1+4x^2)}= \diplaystyle\frac{1-4x^2}{(1+4x^2)(1-4x^2)}=\diplaystyle\frac{1-4x^2}{(1-16x^2)}$
So, when I take the derivative in this last form I get:
$f'(x)=\diplaystyle\frac{-256x^3}{(1-16x^4)^2}$
I'm not sure how to proceed.
9. Originally Posted by Ulysses
Thank to all of you. Now I'm trying to solve the same exercise, but with the function $f(x)=(1+4x^2)^{-1}$ I've tried to make some algebraic work to get the expression I'm looking for, $(1-x)^{-1}$ but I didn't get it.
I've made the first derivative, and a few more, but I think its just necessary the first, as before. I get:
$f'(x)=\diplaystyle\frac{-8x}{(1+4x^2)^2}$
I thought of making $f(x)=\diplaystyle\frac{1}{(1+4x^2)}= \diplaystyle\frac{1-4x^2}{(1+4x^2)(1-4x^2)}=\diplaystyle\frac{1-4x^2}{(1-16x^2)}$
So, when I take the derivative in this last form I get:
$f'(x)=\diplaystyle\frac{-256x^3}{(1-16x^4)^2}$
I'm not sure how to proceed.
It is just function composition.
Let $y=-4x^2$
Then we know that
$f(x)=\frac{1}{1+4x^2}=\frac{1}{1-y}=f(y)$
But this is just a geometric series so we get
$f(y)=\frac{1}{1-y}=\sum_{n=0}^{\infty}y^n$
Now just plug back in $y=-4x^2$
$f(x)=\sum_{n=0}^{\infty}(-4x^2)^n=\sum_{n=0}^{\infty}(-1)^n4^nx^n$ | 2015-07-04 08:06:22 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 26, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.925912082195282, "perplexity": 327.94785565505055}, "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-2015-27/segments/1435375096579.52/warc/CC-MAIN-20150627031816-00248-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://ask.sagemath.org/questions/33317/revisions/ | I am in the situation where I have groups $A,G$ with $G$ normal in $A$ such that $A/G$ is cyclic. I would like to find the coset of $G$ corresponding to a generator of $A/G$. | 2021-09-23 05:52:31 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.8902477025985718, "perplexity": 32.18626578644284}, "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-2021-39/segments/1631780057417.10/warc/CC-MAIN-20210923044248-20210923074248-00511.warc.gz"} |
https://www.tutorialspoint.com/flip-and-invert-matrix-in-python | # Flip and Invert Matrix in Python
PythonServer Side ProgrammingProgramming
Suppose we have a binary matrix mat. We have to select each row in matrix, then reverse the row. After that, flip each bit (0 to 1 and 1 to 0).
So, if the input is like
1 1 0 0 1 0 0 0 1
then the output will be
1 0 0 1 0 1 0 1 1
To solve this, we will follow these steps −
• track:= 0
• for each row in mat, do
• reverse the row
• tracker := 0
• for each val in row, do
• if val is 1, then
• mat[track, tracker] := 0
• otherwise,
• mat[track, tracker] := 1
• tracker := tracker + 1
• track := track + 1
• return mat
Let us see the following implementation to get better understanding −
## Example
Live Demo
class Solution:
def solve(self, mat):
track=0
for row in mat:
row.reverse()
tracker = 0
for val in row:
if val == 1:
mat[track][tracker] = 0
else:
mat[track][tracker] = 1
tracker += 1
track += 1
return mat
ob = Solution()
mat = [[1,1,0],[0,1,0],[0,0,1]]
print(ob.solve(mat))
## Input
[[1,1,0],[0,1,0],[0,0,1]]
## Output
[[1, 0, 0], [1, 0, 1], [0, 1, 1]]
Published on 23-Sep-2020 06:37:41 | 2022-05-23 19:41:30 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23332332074642181, "perplexity": 6430.261038854348}, "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-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00275.warc.gz"} |
https://brilliant.org/discussions/thread/chemistry-in-brilliant/ | ×
# Chemistry in Brilliant
Sir can you start chemistry in Brilliant
Note by Shriram S B
4 years, 12 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
Unfortunately, starting to offer chemistry problem sets is not as simple as pushing a button for us. We acknowledge that there is a strong interest in chemistry among people who solve Brilliant's problem sets. One of our long range goals is to offer chemistry content. In the short and medium term, our efforts will be spent on improving the math experience and expanding the physics one.
Staff - 4 years, 12 months ago
I think there are quite a few people who would enjoy computer science as well. Not programming per se but more theoretical computer science which is almost like a different flavor of math but reasoning about data structures and runtimes. I'm excited for any new subjects.
- 4 years, 12 months ago
they can give problems in physical chemistry
- 4 years, 12 months ago
Example of Covalent Bond India + Pakistan ---> Sub Continent
- 4 years, 12 months ago
Yes, chemistry in brilliant will be very beneficial.
- 4 years, 11 months ago
I think that have a lot of subject better than chemistry,so chemistry should be later. This is just a thought :) All best Arbër
- 4 years, 12 months ago
Chemistry is just as good as physics or math, but school chemistry has to offer substantially less computational exercises. Plus, sometimes you can't pull a reaction out of nowhere, just by using common sense, like in math, knowing only a few things about reaction mechanisms etc. University chemistry is another story.
- 4 years, 12 months ago
- 4 years, 12 months ago
yes i also want chemistry
- 4 years, 12 months ago
For those of you who are interested in building out Chemistry on Brilliant, and would be willing to spend time writing up great wiki pages over the next month, please send me an email (Calvin@Brilliant.org).
Staff - 2 years, 6 months ago
i think the answer can be just symbols and numbers
- 4 years, 11 months ago
It would be interesting if chemistry was started...but the answers for questions will have to be in sentences
- 4 years, 12 months ago | 2018-01-23 11:56:29 | {"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": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9550390243530273, "perplexity": 4117.4014848634715}, "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-2018-05/segments/1516084891926.62/warc/CC-MAIN-20180123111826-20180123131826-00040.warc.gz"} |
https://tex.stackexchange.com/questions/419229/titlesec-seems-not-to-work-numbering | # titlesec seems not to work (numbering)
I want to write a document of the book class. I removed the word "Chapter" from every chapter title. To add a roman chapter number I used titlesec. But since I installed a new Linux distro (Mint 17.3 -> Mint 18.3) it seems to not work anymore. I use TeXStudio with the standard configuration (compiler: pdflatex).
In the minimal working example you can see that the chapter numbering still works for the header and the ToC, but not for the chapter title itself.
\documentclass[11pt,a4paper]{book}
\usepackage {dsfont}
\usepackage {emptypage}
\usepackage {fancyhdr}
\usepackage {geometry}
\usepackage[latin1] {inputenc}
\usepackage {lipsum}
\usepackage {lmodern}
\usepackage {subfig}
\usepackage {titlesec}
\usepackage[nottoc,chapter,numbib] {tocbibind}
\usepackage[subfigure] {tocloft}
\usepackage {ucs}
% % % Title configuration % % %
\renewcommand{\thechapter}{\Roman{chapter}}
\setcounter{secnumdepth}{1}
\titleformat{\chapter}[hang]
{\Huge\bfseries}{\thechapter.\ }{0pt}{}
\titleformat{\section}[hang]
{\Large\bfseries}{\thesection\ }{0pt}{}
\titlespacing*{\chapter}{0pt}{-15pt}{10pt}
\titlespacing*{\section}{0pt}{10pt}{5pt}
% % % more changes % % %
\parindent0pt
\geometry{top=20mm, outer=10mm, inner=20mm, bottom=20mm}
% % % Header % % %
\fancypagestyle{plain}{
\fancyhf{}
}
\pagestyle{plain}
\renewcommand{\chaptermark}[1]{ \markboth{\thechapter.\ #1}{} }
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
%Contents
\setcounter{tocdepth}{1}
\setlength{\cftchapnumwidth}{3em}
\setlength{\cftsecnumwidth}{3em}
\tableofcontents
\newpage
\chapter{Chapter name}
\section{Section name}
\lipsum[1-40]
\end{document}
• Welcome! Try replacing \newpage with \mainmatter. – cfr Mar 9 '18 at 0:28
• No problem here. Chapter title is I. Chapter name. – Sigur Mar 9 '18 at 0:28
• if your document is in latin1 don't load ucs package (actually in general I'd avoid using ucs and just using the standard [utf8] option to inputenc even if your documemt is utf-8 encoded) but specifying latin1 then loading ucs looks wrong – David Carlisle Mar 9 '18 at 0:39
• OK so for you it looks correct? Then it is probably a compiler issue. But how do I solve it? – Moritz Mar 9 '18 at 9:37
It seems like this is a well known bug. Solutions can be found here:
sudo add-apt-repository ppa:jonathonf/texlive-2017 | 2019-08-17 18:15:26 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.850961446762085, "perplexity": 8277.560120559881}, "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-2019-35/segments/1566027313436.2/warc/CC-MAIN-20190817164742-20190817190742-00165.warc.gz"} |
https://math.stackexchange.com/questions/3385593/find-the-volume-of-given-region-using-cylinderical-coordinates | # Find the Volume of Given Region using Cylinderical Coordinates
I need to find the volume of solid enclosed by the cone $$z =\sqrt{x^2+ y^2}$$ between the planes $$z =1$$ and $$z =2$$
Now using Spherical Coordinates I can set up the integral as:
$$\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{sec\phi}^{2sec\phi} {\rho}^2sin\phi\text{ } d{\rho}\text{ } d{\phi}\text{ } d\theta$$
Just for my practice I also want to find this via cylindrical Coordinates. But, I don't understand how should I express the region in terms of $$dz$$ $$dr$$
Can anyone please explain to me step by step how should I express this in cylindrical coordinates ?
Thank You.
• If you do the order $dz\,dr$, you will need to break it up into two separate integrals ($0\le r\le 1$ and $1\le r\le 2$). But if you do the order $dr\,dz$ then $z$ goes from $1$ to $2$ and $r$ goes from $0$ to $z$ ... – Ted Shifrin Oct 8 '19 at 16:25
• @TedShifrin: so, if I do it by the order $dzdr$ the integral setup will be : $\displaystyle\int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{1} rdrdzd\theta + \int_{0}^{2\pi}\int_{1}^{2}\int_{0}^{2} rdrdzd\theta$ Is this correct ? – sat091 Oct 8 '19 at 17:00
• That isn't the order $dz\,dr$, is it?!! You want to edit and proofread to make sure? – Ted Shifrin Oct 8 '19 at 17:01
• @TedShifrin:: Ah sorry , I actually meant to do it by the order $dr dz$ and not $dzdr$ – sat091 Oct 8 '19 at 17:15
• No, as I already said, we are told that $z$ goes from $1$ to $2$, so you're just going to have a single integral. – Ted Shifrin Oct 8 '19 at 17:18
If you insist in cylindrical coordinates you need two integrals $$\int _0^{2\pi} \int _0^1\int_1^2dzrdrd\theta +\int _0^{2\pi} \int _1^2\int_r^2dzrdrd\theta=$$
$$\pi + 4\pi/3 = 7\pi/3$$
where the first integral evaluates the middle cylinder and the second one evaluates the rest of the volume.
However the easier way is to use the solid of revolution formula.
The volume is found by the integral $$\int _1^2 \pi r^2 dz$$ where $$r^2 = z^2$$
Therefore the answer is $$\int _1^2 \pi r^2 dz =\int _1^2 \pi z^2 dz =\frac {7\pi }{3}$$
• Although the OP accepted the answer, from my perspective it did not begin to address the question as asked. – Ted Shifrin Oct 8 '19 at 17:04
• @TedShifrin Thanks for the informative comment. Please check my editted version. – Mohammad Riazi-Kermani Oct 8 '19 at 19:15
• Yes, thanks. As I commented to the OP, it is actually easier to do the integral in the (less standard) order $dr\,dz$; then only one integral is required. – Ted Shifrin Oct 8 '19 at 19:22 | 2020-02-23 10:55:01 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8280777931213379, "perplexity": 219.58458863901654}, "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-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00254.warc.gz"} |
http://inyk.ajte.pw/generate-exponential-random-variable-in-r.html | To generate an exponential random number, we use the formula-rate * log(U) where U is a U(0,1) random number. The poisson_sp. If we can generate H and qeasily, we can use. ference), exponential random variables can be used to simulate beta random variables via the transformation method. The exponential distribution can be parameterized by its mean αwith the probability density function f(x)= 1 α e−x/α x >0, for α>0. If the distribution you need to use in your worksheet models is not supported, you can create the inverse CDF formula yourself, based on the analytic expression for the cumulative distribution function. 2 Random Variables 21 2. P(X≥14), n=16, p=0. Continuous Random Variables. Eventbrite - Simplykart Inc presents Data Science Certification Training in Oshawa, ON - Tuesday, November 26, 2019 | Friday, October 29, 2021 at Business Hotel / Regus Business Centre, Oshawa, ON, ON. 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Exponential random variables are commonly encountered in the study of queueing systems. Bernoulli trials An experiment, or trial, whose outcome can be classified as either a success or failure is A random variable X, taking on the values 0, 1, 2. Since most computer languages come with a method of generating uniform random numbers, we can use these to generate exponential random quantities. Review Status. Let Z be standard normal. One very flexible but memory-intensive approach is to use look-up tables to convert them. Show thatX is memoryless. Thus, r is a sample value of the random variable R with pdf Inversion method. seed Random numbers There is a philosophical problem here. random variables, such as in Section 2. Learn more about exponential, random, variable, multiple. Campbell’s Theorem c. iare Gaussian random variables Density function analogous to 1-D case, but note covariances! p(x) = 1 (2ˇ)12j j exp (x )T 1(x ) 2 Probability density for a 2-D Gaussian random vector. Over 80 continuous random variables (RVs) and 10 discrete random variables have been implemented using these classes. I save in order to generate. This lesson explains how to make a linear transformation and how to compute the mean and variance of the result. The simple exponential smoothing model can be generalized to obtain a linear exponential smoothing (LES) model that computes local estimates of both level and trend. Unlike Negative Binomial distribution, there is no function for generating Quasi-Poisson distributed random variable in R. Most of them start with r. In this post, I would like to discuss how to generate Gamma distributed random variables. Manipulation of Discrete Random Variables with discreteRV by Eric Hare, Andreas Buja and Heike Hofmann Abstract A prominent issue in statistics education is the sometimes large disparity between the theoretical and the computational coursework. ρ>0 and along the line y = −x if ρ<0. The default value is max=1. An example of such random variable would be the distance of darts from the target center in a dart-throwing game where the deviations in the two dimensions of the target plane are independent and normally distributed. The long-run rate at which events occur is the reciprocal of the expectation of , that is, /. This exponential regression model is relatively easy to understand and serves as our starting point. 5, and a lambda of 0. A Convenient Way of Generating Gamma Random Variables Using Generalized Exponential Distribution Debasis Kundu1 & Rameshwar D. glorot_normal_initializer(seed=ms. Manipulation of Discrete Random Variables with discreteRV by Eric Hare, Andreas Buja and Heike Hofmann Abstract A prominent issue in statistics education is the sometimes large disparity between the theoretical and the computational coursework. Consider an exponentially-distributed random variable, characterized by a CDF F )(x = 1 −e−x/θ Exponential distributions often arise in credit models. The distribution function can be de ned in terms of the probability density function fas follows: F(x) = P(Z x) = Z. The exponential distribution is a continuous analogue of the geometric distribution. We first consider the most fundamental of the techniques for generating sample values of random variables. Generate a random variable X with distribution function FI. • f X (x) is the (Probability) Density Function of X. Using characteristic functions, show that as n!1and p!0 such that np! , the binomial distribution with parameters nand ptends to the Poisson distribution. We say that this random variable x has the one-parameter exponential distribution of probability density function f(x) described by (3. Eventbrite - Zillion Venture presents Data Science Online Training in Saint Anthony, NL - Tuesday, October 22, 2019 | Friday, October 1, 2021 at Regus Business Hotel, Saint Anthony, NL, NL. Correlation and Simple Linear Regression Pearson s Product Moment Correlation (sample correlation r estimates population correlation r) Measures the strength of – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. Both x and h can take values in either the continuous or the discrete domain. The pdf contains a parameter γ > 0 and it is defined as P(x) = γexp[−γx] Confirm that R∞ 0 P(x)dx = 1. The mean and variance of Xare E(X) = νand var(X. On the (Z 1, Z 2) coordinate system, we have Z 1 = H cos(q) Z 2 = H sin(q) H 2= Z 1 + Z 2 2 has c2 distribution with 2 degrees of freedom which is equivalent to an exponential distribution with mean 2 (λ=1/2). Or copy & paste this link into an email or IM:. list make foreign make foreign 1. What can we say about the distribution of random variables that form a Markov random field with re-spect to a graph G and that are at the same time member of an exponential family? It turns out that the Gibbs form translates into a simple decomposition of the sufficient statistics and in turn the kernel func-tion. The univariate exponential distribution is also (sort of) closed under convolution. , when we nd out how Zbehaves. Those include the Cauchy, Weibull, normal, log-normal, logistic, exponential, uniform, gamma distributions, the central and noncentral beta, chi-squared, Fisher's F-distribution, Student's t-distribution, as well as the discrete binomial and negative binomial distributions. Set R = F(X) on the range of. Generate Normal Random Variable --- Polar method The theory is complicated, we only list the algorithm here: Objective: Generate a pair of independent standard normal r. 10 Negative Binomial Distribution. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The latest W3C recommendation is XHTML 1. discrete random variables take on either a nite our countable number of possible values continuous random variables take on a continuum of possible values cumulative distribution function (cdf) for random variable Xis de ned for any real number b, 1 0. A fair die is tossed. The univariate exponential distribution is also (sort of) closed under convolution. Example: Assume that X has an exponential distribution with = 2. The mean and the variance of exponential distribution are = and ˙2 = 2. In addition, the upper bound in ( 5 ) is just one example. The exponential distribution is a continuous analogue of the geometric distribution. Besides, we seek to know if. 1 Random number generators in R-- the r'' functions. The exponential distribution with rate λ has density. The purpose is to get an idea about result of a particular situation where we are given probabilities of different outcomes. If a random variable X follows the normal distribution, then we write: In particular, the normal distribution with μ = 0 and σ = 1 is called the standard normal distribution, and is denoted as N (0, 1). The time between arrivals of customers at a bank, for example, is commonly modeled as an exponential random variable, as is the duration of voice conversations in a telephone network. Campbell’s Theorem c. It is essentially a chi distribution with two degrees of freedom. Notation: X~Exp (m). [forget about the mean. New users of R will find the book’s simple approach easy to under-. Discrete Random Variables series gives overview of the most important discrete probability distributions together with methods of generating them in R. be independent exponential random variables with mean 1, Simulation Lecture 8. Weibull Distribution In practical situations, = min(X) >0 and X has a Weibull distribution. The ratio of a standard normal random variable Z to the square root of an independent chi-square(n) random variable divided by its degrees of freedom has the t distribution with n degrees of freedom. \+,œTÐ+Ÿ\Ÿ,Ñœ0ÐBÑ. 0 application. Why is a Lockheed MC-130J Commando II creating such a. Terms are specified by an R formula object, giving the network and network statistics, of the form. It can be applied, at least in principle, in all cases where an explicit expression exists for the cumulative distribution function of the random variable. Compute the standard normal random variable as Z 2. Or copy & paste this link into an email or IM:. Monte Carlo Simulation The Monte Carlo method uses a pseudorandom number generator to generate numbers uniformly between zero and one. Find X such that F(X) = U and return this value X. ¾ Develop an algorithm to generate random variates from the standard normal r. 15 A web site experiences traffic during normal working hours at a rate of 12 visits per hour. Transformations of Variables When a residual plot reveals a data set to be nonlinear, it is often possible to "transform" the raw data to make it more linear. exponential distribution must have a positive parameter i). Those include the Cauchy, Weibull, normal, log-normal, logistic, exponential, uniform, gamma distributions, the central and noncentral beta, chi-squared, Fisher's F-distribution, Student's t-distribution, as well as the discrete binomial and negative binomial distributions. As a language for statistical analysis, R has a comprehensive library of functions for generating random numbers from various statistical distributions. Generate Normal Random Variable --- Polar method The theory is complicated, we only list the algorithm here: Objective: Generate a pair of independent standard normal r. We would simply continue the same process — that is, generating y, a random U(0,1) number, inserting y into the above equation, and solving for x — 997 more times if we wanted to generate 1000 exponential(5) random numbers. Most random number generators simulate independent copies of this random variable. ference), exponential random variables can be used to simulate beta random variables via the transformation method. This is a general fact about continuous random variables that helps to distinguish them from discrete random variables. For the exponential distribution, on the range of. So I set the random seed in tensorflow through: tf. For the purposes of this document, nodes may not be connected to themselves. Set R = F(X) on the range of. Implementation in R R is the interactive language for statistical computing we are most interested in using in this course. 1 Exponential distribution, Extreme Value and Weibull Distribution 1. # per min) then its probability density function is given by EXPONENTIAL DISTRIBUTION THE PROBABILITY DENSITY. In particular cases, there can be clever ways to simulate random variables. Learning Outcomes 4Generating Continuous Random Variables Generate random variables using theInverse-Transform and Acceptance-RejectionMethod Develop algorithms for simulatingExponential, Normal, Poisson andNonhomogeneous Poisson distributions Perform simulations using R 5. Eventbrite - Zillion Venture presents Data Science Online Training in Etobicoke, ON - Tuesday, November 26, 2019 | Friday, November 29, 2019 at Regus Business Hotel, Etobicoke, ON, ON. Exponential Distribution Excel. If is an exponential random variable,. Exponential random variables are often used to model waiting times between events. Exponential random variables are commonly encountered in the study of queueing systems. CHAPTER 1 Introducing Exponential Family Random Graph Models. Generate X -ln. , multiplying by a constant, or a parameter that corresponds to shifting the random variable, i. Draw any number of variables from a joint normal distribution. Exponential correlated random variables are generated by stochastic differential equations (SDEs), which are described by Markov diffusion processes. 2 Lecture Notes – Part A Simulation – Oxford TT 2011 of view, the efficiency of such generation of random variables can be analysed. In a random uniform sample of size 10 5 , the chance of drawing at least one duplicate is greater than 50%. One is , which is the time that a ‘significant’ rainfall begins. Gamma random variate has a number of applications. The variables also have a > known correlation, so I can represent their correlations in a matrix > like so: > > a <- array(c(0. exponential taken from open source projects. That is, some function which specifies the probability that a random number is in some range. 1 A rst example: Gaussian with linear su cient statistic Consider the standard normal distribution P 0(A) = Z A e 2z =2 p 2ˇ dz and. A simple random number generator uses Lehmer’s recursion, i. The rate parameter is an alternative, widely used parameterization of the exponential distribution. Generating Sequence of Random Numbers. Often, the rst thing we do with them is construct the non-uniform random variables that our problem requires. As we know, random numbers are described by a distribution. I Formula PfX >ag= e a is very important in practice. So for instance, when I taught an undergraduate modeling course, I had one student who went to the Mathematics Help Room and had a stopwatch and kept track of the t. How to generate the exponential random numbers from uniform random number generator? If you search "generate random. Then X = R cos(T) and Y = R sin(T). Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter = 1 12. Eventbrite - Zillion Venture presents Data Science Online Training in Tuktoyaktuk, NT - Tuesday, November 26, 2019 | Friday, November 29, 2019 at Regus Business Hotel, Tuktoyaktuk, NT, NT. 5$, I can generate 500 samples and then I want to plot the poisson process path on a time interval of [0,10] for example, how can I do this in R?. When = 1 we call Xthe standard exponential random variable. It turns out that a Pareto random variable is simply b*exp(X), where X is an exponential random variable with rate=a (i. If we can generate H and qeasily, we can use. For the exponential distribution, on the range of. f(x) = λ {e}^{- λ x} for x ≥ 0. ¾ Develop an algorithm to generate random variates from the standard normal r. The tutorial describes a method to generate uniformly distributed random variates that exhibit a prespecified linear correlation. Something neat happens when we study the distribution of Z, i. For example consider the exponential random variable which has density. A Rayleigh distribution is often observed when the overall magnitude of a vector is related to its directional components. If U f(Y) cg(Y) set X= Y. A random variable Xis said to be continuous if its distri-bution function can be written as P[X x] = Z x 1 f X(u)du; for some integrable f X: R ![0;1), which is called density function of X. Now, x is a random number with an exponential distribution.$\endgroup$– Sasho Nikolov Jun 29 '13 at 2:27$\begingroup$I have tried to repeat the proof of chernoff. NormalGamma provides the density of the sum of a gaussian and a gamma random variables. Gaussian Random Number Generator. Find the probability density function of the random variable Z = min(X;Y) and the probabilityP(X > Y). It's more about feeding the right set of features into the training models. stats and a fairly complete listing. If you were given machine to generate samples of U, how would you go about estimating E[V]. Discrete Random Variables series gives overview of the most important discrete probability distributions together with methods of generating them in R. The Laplace distribution is the distribution of the difference of two independent random variables with identical exponential distributions (Leemis, n. Loading Unsubscribe from Katie Ann Jager? Creating and Graphing Mathematical Functions in R - Duration: 8:01. 4 Creating new variables in R Many research studies involve some data management before the data are ready for statistical analysis. Presented model will be able us to do probability and statistical inferences in efficient and different way by modern computers. 2 2 1) = ˇ 4 : For a sequence of such i. Over 80 continuous random variables (RVs) and 10 discrete random variables have been implemented using these classes. If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut. Here’s the density function for an exponential distribution random variable: Suppose you have a sample from a variable and want to see if it can be modelled with an Exponential distribution Variable. Therefore, for every number genera ted with a uniform random number gen-erator, compute as where. Dear R users I'd like to generate two sets of random numbers with a fixed correlation coefficient, say. We won't be using the "r" functions (such as rnorm) much. For example, define X to be 1 if a bead is blue and red otherwise:. Then the null distribution of is the exponential with rate 1, and, based on the observed, the p-value is. 2 Random Variables 21 2. The Method of Norms requires the ability to generate random variables distributed according to the norm of the desired random variable and the generation of random vectors from a different SIRV distribution. Use your program to generate NOBS = 10,000 values of an exponential random variable with λ = 0. If Y 1 and Y 2 are independent exponential random variables, both with mean β , find the density function for their sum. We would simply continue the same process — that is, generating y, a random U(0,1) number, inserting y into the above equation, and solving for x — 997 more times if we wanted to generate 1000 exponential(5) random numbers. 5$, I can generate 500 samples and then I want to plot the poisson process path on a time interval of [0,10] for example, how can I do this in R?. Wcan take on values from 1 to 1. That is, the joint probability distribution for. Log in Sign up. Otherwise, return to (a). In this post I will demonstrate in R how to draw correlated random variables from any distribution. This means that the probability that X exceeds x + k , given that it has exceeded x , is the same as the probability that X would exceed k if we had no knowledge about it. glorot_normal_initializer(seed=ms. Expected Value of Transformed Random Variable Given random variable X, with density fX(x), and a function g(x), we form the random. Let X 1;X 2; ;X nbe independent random variables with X i. The tutorial describes a method to generate uniformly distributed random variates that exhibit a prespecified linear correlation. Create a new variable based on existing data in Stata. Think about this for a moment; the rest of the continuous random variables that we have worked with are unbounded on one end of their supports (i. How to's on indicators, trading methods, and alike Wed, 30 Oct 2019 14:29:48 +0000 Wed, 30 Oct 2019 14:29:48 +0000 ThinkScripter Community Forum - Give Help, Get Help. The exponential distribution is a continuous analogue of the geometric distribution. The variables also have a known correlation, so I can represent their correlations in a matrix like so: a <- array(c(0. Suppose u is generated according to a uniformly distributed in (0,1). The distribution of the squared values is given by the random variable Q = Z 2. To nd percentiles, we must nd the cumulative density function. Hence, the collections and are called random processes. To create an N by M matrix of iid normal random variables type this:. The exponential distribution can be simulated in R with rexp(n, lambda) where lambda is the rate parameter. c ===== //= Program to generate exponentially distributed random variables = //===== //= Notes: 1) Writes to a user specified output file. We use the notation E (X) and E(X2) to denote these expected values. runif will not generate either of the extreme values unless max = min or max-min is small compared to min, and in particular not for the default arguments. Eventbrite - Simplykart Inc presents Data Science Certification Training in Oshawa, ON - Tuesday, November 26, 2019 | Friday, October 29, 2021 at Business Hotel / Regus Business Centre, Oshawa, ON, ON. ference), exponential random variables can be used to simulate beta random variables via the transformation method. In this lab, we'll learn how to simulate data with R using random number generators of different kinds of mixture variables we control. Exponential (Random Variable) Good for modeling variables that are the multiplicative product of other random variables. The Uniform is interesting because it is a continuous random variable that is also bounded on a set interval. The mean is a measure of the “center” or “location” of a distribution. [code] import random lam = (mean of exponential distribution) a = (left end point of interval) b = (right end point of interval) X = random. Joint probability distributions are defined in the form below: where by the above represents the probability that events x and y occur at the same time. Our results about exponential random graph models are actually special cases of more general results about exponential families of dependent random variables, and are just as easy to state and prove in the general context as for graphs. 1 Random variables. 1 The Uniform Random Variable 32 2. 6, label = 'norm pdf') Alternatively, the distribution object can be called (as a function) to fix the shape, location and scale parameters. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Most widely used mathematical computing environments include generators for a wide selection of non-uniform distribu-tions. So I set the random seed in tensorflow through: tf. The Gumbel extreme value distribution is related to the exponential distribution as. (ii) Let X be the volume of coke in a can marketed as 12oz. (I am working in R. Example (Problem 74): Let X = the time (in 10 1 weeks) from shipment of a defective product until the customer returns the. X is the Random Variable "The sum of the scores on the two dice". I am trying to generate exponential random variables that meet a certain condition in R. In simulation we often have to generate correlated random variables by giving a reference intercorrelation matrix, R or Q. Apply the univariate normal CDF of variables to derive probabilities for each variable. generate two sets of random numbers that are correlated. Random variables are numeric outcomes resulting from random processes. NormalLaplace provides d, p, q, r functions for the sum of a normal and a Laplace random variables, while LaplacesDemon provides d, r function of the sum of a normal and a Laplace random variables. The moment generating function of a random variable is defined in terms of an expected value. This form allows you to generate random integers. Generate two uniform random variables, U;V. 1 A rst example: Gaussian with linear su cient statistic Consider the standard normal distribution P 0(A) = Z A e 2z =2 p 2ˇ dz and. A Convenient Way of Generating Normal Random Variables Using Generalized Exponential Distribution Debasis Kundu1, Rameshwar D. Random generation for the Gamma distribution with parameters shape and rate. They are used to model physical characteristics such as time, length, position, etc. For example, to generate a sample of 6 independent random variables from the normal distribution with mean 3 and standard deviation 2, we would type:. So, essentially, the moments of the distribution are these expectations of the random variable to integer powers, and often they help to give valuable information about the random variable itself (we've already seen how moments relate to the mean and variance). Let Z be standard normal. Comparisons The CALL RANEXP routine, an alternative to the RANEXP function, gives greater control of the seed and random number streams. Therefore, we can generate Xby summing the independent and identically distributed. \label{sec:intro} The methods described in this article mostly rely on the possibility of producing (with a computer) a supposedly endless flow of random variables (usually iid) for well-known distributions. We further assume that this noise, n, is random and sampled from a zero-mean Gaussian distribution with. In this chapter, we explain how to generate discrete distributions. , uni-formly distributed between 0 and 1, and R = U2/U1. The (central) chi-squared random variable with k degroes of freedom has the fol- lowing probability de nsity function f(rk) r for x> 0. In simple terms, the idea is this: if U1 and U2 are i. Introduction to Statistical Methodology Random Variables and Distribution Functions Even though the cumulative distribution function is defined for every random variable, we will often use other characterizations, namely, the mass function for discrete random variable and the density function for continuous random variables. For example consider the exponential random variable which has density. sample size. In other words, given observations, these exponential family distributions define the likelihood functions of the latent variables, i. If , accept X = R, goto Step 3 Step 2b. Learn more about exponential, random, variable, multiple. Any suggestion will be. The mean and the variance of exponential distribution are = and ˙2 = 2. - Generate an array R of n (= 100) random numbers from sequence Xk - Generate an additional number X to start the process • Each time generator is called - Use X to find an index into the array R j ←X * n - X ←R[ j ] - R[ j ] ←a new random number - Return X as random number for call Shuffling with two generators. 1 day ago · where the noise, n, arises due to the imperfect representation of the decision variable. which is the density for an exponential random variable with parameter = 1/(2 2a), as can be seen from inspection of (2-27). Functions that generate random deviates start with the letter r. To generate a Negative Binomial random variable we make use of the fact that a Negative Binomial random variable is sum of r independent Geometric random variables, where r is the of trials required to observe the r th success and p is the probability of a success. Suppose u is generated according to a uniformly distributed in (0,1). Generating Random Variates • Overview ¾We will discuss algorithms for generating observations ("variates") from non-uniform distributions (e. Find the median of Xif Xis exponential with rate. 3 Continuous Random Variables 31 2. The function rgpd generates Generalized Pareto Random Variables. covariance matrices, Matlab, R, random numbers, random variables. The mean and variance of Xare E(X) = νand var(X. The most important of these properties is that the exponential distribution is memoryless. Ashour, Samir K; Eltehiwy, Mahmoud A. 44 CHAPTER 3. Thanks a lot. Generating spatially correlated random fields with R In several occasions I needed to generate synthetic data with a desired level of spatial autocorrelation. I am trying to generate exponential random variables that meet a certain condition in R. I am trying to simulate a poisson process sample path in R by starting off with exponentially distributed random variables. Characteristic functions and central limit I Let X be a random variable. The functions for working with the basic statistical distributions implemented in the R language are considered. The binornd function uses a modified direct method, based on the definition of a binomial random variable as the sum of Bernoulli random variables. (If you create one, please contribute it). 3 Continuous Random Variables 31 2. 2015-05-01. It is essentially a chi distribution with two degrees of freedom. Description. This process of feeding the right set of features into the model mainly take place after the data collection process. 1 (Two independent normals). If U < exp(1 2 (Y 1)2) output Y. If we can generate H and qeasily, we can use. 6 Transformations of Random Variables 107 A new random variable is to be formed according to the square law transfor-mation Y = X 2. One of the most important application is to generate Dirichlet distributed random vectors, which plays a key role in topic modeling and other Bayesian algorithms. For instance, if you want to simulate from a standard normal distribution, you can simulate from a standard uniform and transform it using the quantile function of the normal distribution. s, since their α-moments equal λ α Γ (α ∕ p + 1). Scaling and shifting For many families of random variables there is a parameter that just corresponds to scaling the random variable, i. Speeding up repeated generation of Exponential random variables in loop I am implementing an algorithm, and as part of that, I need to generate exponential random variables. Of course, we wouldn't really do it by hand, but rather let statistical software do it for us. If you are measuring, the distribution of the result will almost always be continuous. Find X such that F(X) = U and return this value X. m file uses the rand function call in matlab to generate uniformly distributed ran-dom numbers. Generating Sequence of Random Numbers. Then conditioned on the event A = {U1 ≤ f(U2/U1)}, the random variable R has the density function f. 2 Lecture Notes – Part A Simulation – Oxford TT 2011 of view, the efficiency of such generation of random variables can be analysed. Generate 625 samples of size 961 random numbers from U(1, 9). Generate U˘Uniform(0;1). In R, there are many functions to generate random deviates. You can : 1) Create 1001 U(0,1) random variables. Generate a random number R Step 2a. Then the null distribution of is the exponential with rate 1, and, based on the observed, the p-value is. gi are not correlated with the number rm, since i > m. | 2020-01-19 13:51:55 | {"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": 0.6184329390525818, "perplexity": 637.2299498832142}, "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-2020-05/segments/1579250594603.8/warc/CC-MAIN-20200119122744-20200119150744-00178.warc.gz"} |
https://math.stackexchange.com/questions/2783393/why-wolframalpha-does-lu-decomposition-with-pivoting-even-when-it-isnt-needed | # Why WolframAlpha does LU decomposition with pivoting even when it isn't needed?
I want to do LU decomposition with $A$:
$A= \left[ {\begin{array}{cc} 1 & 2 & 3 & 4 \\ -9 & 8 & -15 & 4 \\ 2 & 13 & -21 & 7 \\ 4 & -5 & 5 & 3 \end{array} } \right]$
My answer is this, which the result is approximation because of limitations of floating point data. $A$ can be done LU decomposition without pivoting.
But WolframAlpha gives me different solution. They do LU decomposition with pivoting. I don't know why WolframAlpha calculate LU decomposition with pivoting even pivoting isn't needed.
• Try this calculator: gregthatcher.com/Mathematics/LU_Factorization.aspx – Moo May 16 '18 at 17:26
• $$\left( \begin{array}{ccc} 1 & 0 & 0 \\ -9 & 1 & 0 \\ 2 & \frac{9}{26} & 1 \\ \end{array} \right).\left( \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 0 & 26 & 12 & 40 \\ 0 & 0 & -\frac{405}{13} & -\frac{193}{13} \\ 4 & -5 & 5 & 3 \\ \end{array} \right)=\left( \begin{array}{cccc} 1 & 2 & 3 & 4 \\ -9 & 8 & -15 & 4 \\ 2 & 13 & -21 & 7 \\ 4 & -5 & 5 & 3 \\ \end{array} \right)$$ – Moo May 16 '18 at 17:31
• Oh this calculates LU rather than PLU. And.. no reason why WolframAlpha using pivot without any ground? – oddman621 May 17 '18 at 1:24
• It is just the way they coded up their routine - even the professional Mathematica does it that way. I coded my own variant and produced the results above - as well as the site I provided. This is not bad to calculate by hand for such a small matrix. – Moo May 17 '18 at 1:35
• Seems unclear. You mean the result(LU or PLU) depends on what algorithms are used? – oddman621 May 17 '18 at 2:47 | 2019-05-19 20:53:17 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4155636727809906, "perplexity": 1527.4842949594422}, "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-2019-22/segments/1558232255165.2/warc/CC-MAIN-20190519201521-20190519223521-00485.warc.gz"} |
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# Tag Info
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Runic Enchantments, Score: 1091 "Hell0 W0rld!"@ Try it online! I honestly can't find a way to do this cheaper. The costs of string manipulation are just not worth the savings that can be gained. Eg. subtrating 30 from each character the output string saves 360 but costs 289 to undo plus another 165 in control logic (454 total). Encoding the operational ...
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7, 41 characters, code point sum 81 32002453003001200522231203103002440537403 Try it online! Unlike with many of my 7 programs, there's nothing particularly clever going on here; this basically just prints a string in the simplest possible way (which in 7 involves encoding the string into a piece of source code, evaluating it, analysing the resulting ...
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Japt, Score: 863 ")FMM8SME"c+# Test it
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str, Score: 1081 (ELL7RLD32+o; There are a lot of unprintable characters that don't show up above, see the TIO link for the full code. Adds 32 to each character to get a lower score (the naive solution has a score of 1321) Try it online!
6
05AB1E, score 513 Posting this separately from my other answer, as this one is arguably less cheating. I've been waiting for somebody else to do this for a while, but nobody did. I tried multiple languages. Pyth has extremely short code for this (Cl"...", if I remember correctly), but it turned out it can't input null bytes, and most other golfing languages ...
1
Bash, score: 1406 echo Hell0 W0rld! Try it online!
0
ink, Score: 959 Hell0 W0rld! Try it online!
1
Turing Machine Code, Score: 7105 0 _ H r 1 1 _ e r 2 2 _ l r 3 3 _ l r 4 4 _ 0 r 5 5 _ _ r 6 6 _ W r 7 7 _ 0 r 8 8 _ r r 9 9 _ l r a a _ d r ! ! _ ! r halt Try it online! Turing Machine But Way Worse, Score: 69695 0 0 0 1 1 0 0 0 1 1 1 2 0 0 0 2 0 1 3 0 0 0 3 0 1 4 0 0 0 4 1 1 5 0 0 0 5 0 1 6 0 0 0 6 0 1 7 0 0 0 7 0 1 8 1 0 0 8 0 1 9 0 0 0 9 1 1 10 0 0 0 ...
2
brainfuck, score: 5822 ++++++++++[>+++++++>++++++++++>+++++++++++>+++++>+++>+++++++++<<<<<<-]>++.>+.>--..>--.>++.>---.<<.<++++++.------.<-.>>>+. Try it online!
5
Keg, sum = 966 868 853 �GOU�� +��OOH� +(�+, Try it online! -15 bytes from cutting out the 0 (with a bit of help from the awful old Python script). Based on a comment by Night2, as well as Jo King's old 32-byte quine. Old explanation: �DLR�� +0�LLE� + Push every codepoint of "Hell0 W0rld!", minus 32. � + (This includes pushing ...
1
MSM, Score: 1465 !dlr0W 0lleH........... It is great that ! is not an MSM instruction. Half of the program pushes onto stack, half of it concatenates them.
6
COW, Score: 117799 MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO Moo MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO ...
0
Scratch 3, score: 2908 2012 In ScratchBlocks Syntax. define say[Hell0 W0rld! Defines an anonymous function. When called, it returns Hell0 W0rld!. http://scratchblocks.github.io/#?style=scratch3&script=define%0Asay%5BHell0%20W0rld!%5D
1
Labyrinth, Score:2151 72.101.108.108.48.32.87.48.114.108.100.33.@ Explanation 72 Push ord code of 'H' . Output 101.108.108.48.32.87.48.114.108.100.33. And so on @ And exit. Try it online!
1
evil, Score: 12853 Hard-code the characters zaeeeaeeewzaeeaeeaeawzuueeueueawzuueeueueawzaeeeaeewzaeeeeewzaeeeeeuewzaeeeaeewzuueeueeaaawzuueeueueawzaeeaeeaewzaeeeeeaw Try it online!
5
MineFriff, Score: 3725 Cff3,a*a,9*c,a*b4,6*8,8*a7,ff2,8*6,9*c,:a*a1,9*8,` >l?#;o~ Try it online! This ain't winning any competitions any time soon I tell you (I dread seeing what Scratch's score would be...). I'll post an image of what this looks like in-game soon, but first I have to actually re-install the python-in-minecraft mod and actually get it ...
1
J, Score: 1452 echo'Hell0 W0rld!' Try it online!
3
bit**, Score: 2519 #72/#101/#108/#108/#48/#32/#87/#48/#114/#108/#100/#33/ Try it online!
10
Lenguage, score 0 Well, this is quite boring and obvious, but it seems to achieve the perfect score (assuming I understand the challenge correctly). To fix this loophole, you should probably add the program's length to the sum of the code points; I'm pretty sure that won't break the relative order of the current answers. The program is ...
2
Forth (gforth), Score: 1105 ." Hell0 W0rld!" Try it online! Prints the given string to stdout.
1
shortC, Score:1166 ShortC version of this. AJ"Hell0 W0rld!" Try it online!
1
Ruby, Score:1139 p"Hell0 W0rld!" Try it online!
1
Python 2, Score:1584 Prints the string... print"Hell0 W0rld!" Try it online!
1
Zsh, Score 1207 <<<"Hell0 W0rld!" Try it online, with a self-scorer! Using <<<Hell0\ W0rld! is shorter, but more expensive: \ = 92, but 2⨯" = 68.
4
HTML, Score: 959 Hell0 W0rld! If newlines (\r and \n) would have been count as their ASCII code (10 and 13) the score would be 937 or 940 because in HTML, new lines in "inline" elements are rendered as space.
3
05AB1E, Score:993 Just the string "Hell0 W0rld! Try it online! Polyglot with Actually and Stax and Foo and Pyth
3
Jelly, Score:1213 Just the string. Nothing interesting. “Hell0 W0rld! Score: 1744 Simply a compression. “ØƲṡhḍḊF®Ḍ?» Try it online! Code page summator
1
C (gcc), Score: 2021 Thanks to GammaFunction A(){puts("Hell0 W0rld!");} Also, just for fun, here is a C program that has a score of 1 using overflow. It's a base64 encoded rar file https://pastebin.com/SJMcRnLj Try it online!
2
MathGolf, Score: 1027 "Hell0 W0rld!" 2218 is not short at all. 'H╖○ƒ0 'W0╖yü'! Try it online! Polyglot with CJam and GolfScript
3
PHP, Score: 959 "Classic" answer. PHP outputs by default. Hell0 W0rld! Try it online! Polyglot with Carrot and Canvas and Charcoal
3
Keg, Score: 1051 Hell0 W0rld\! Try it online!
2
Wolfram Language (Mathematica), score 31 + 2 = 33 & ) # @ / ] & 2 # > - " " . / # [ d l o F ( @ e ...
0
F is for Forth (gforth), 26 + 5 = 31 '{ 'a [do] [i] emit [loop] Try it online! Two improvements over the submission by leancz 6 years ago: An ASCII char prefixed with a quote gives the ASCII value of that char. So '{ saves a byte over 123. 'a is just for consistency. The looping words including do and loop can't be used in interpreted mode, but gforth ...
2
Charcoal, score 22 18 κ→ Ｐ ! Ｋ Ｋ ‹ ¿ ι Ｆ ← «Ｊ⁰¬¹ Ｓ Ｗ Note: This is designed for a pleasant input format i.e. at least one character on each line, then a completely empty line to terminate the input. Replace Ｗ with Ｆ and Ｓ with Ａ for a JSON-like input ...
4
Jelly, score 9 + 0 = 9 Z ⁶ € ɗ o Ḣ ḟ Ṛ Ỵ From the top: ỴZḟḢoɗ€⁶ Try it online! (or see it self-operate) From the side: ỴṚḟḢoɗ€⁶Z Try it online! (or see it self-operate) How? Ỵ Z|Ṛ ḟḢoɗ€⁶ [Z] - Link list of characters: Ỵ - split at newline characters Z - | top: transpose to get columns ...
3
Pyth, 15 score z .T. ; d + ! \ D < h m s _ The two programs are: smh<D\!d.T.z Try it online! _smh<D\!+d;.z Try it online! I attempted to make the code as similar as possible between the two programs, hence a few inefficiencies in each. Both programs filter out unwanted spaces by ...
9
An extension to @Grimy's code gets N=8 This just underlines that @Grimy deserves the bounty: I could prune the search tree by extending the code to check, after each finished polyomino, that the remaining free space is not partitioned into components of size not divisible by N. On a machine where the original code took 2m11s for N=7, this takes 1m4s, and ...
14
C, 7 terms The seventh term is 19846102. (The first six are 1, 1, 2, 22, 515, 56734, as stated in the question). #include <stdio.h> #include <string.h> #include <stdint.h> #define N 7 #define ctz __builtin_ctzl typedef uint64_t u64; static u64 board[N*N] = { 0 }; static u64 adjacency_matrix[N*N] = { 0 }; static u64 count = 0; static ...
1
X is for x86, 9 bytes + 3 -> score 12 (or "x86 machine code" for 9 + 16 = 25) b0 61 aa 40 3c 7a 76 fa c3 This is a function, callable with void alpha(char edi[26]), like the x86-64 System V calling convention but in 32-bit mode. Or 16-bit mode (where 40 decodes as inc ax, otherwise the same). The question says "program" that "prints", and this is ...
1
F is for Fish, 16 bytes; score 20 "a"r:o1+:"{"=?;r
2
53,268 29,596 29,306 total precision Private communication with @A.Rex led to this solution, in which we construct a neural net that memorizes the answers. The core idea is that every function $f\colon S\to\mathbf{R}$ over a finite set $S$ enjoys the decomposition $$f(x) = \sum_{s\in S}f(s)\cdot \left\{\begin{array}{cl}1&\text{if }x=s\\0&\... 0 Go, 84426 func f(){print("",'')} The character in the single quotes is the Unicode character with code point U+149CA, encoded as 4 bytes in UTF-8. The Unicode standard does not currently assign it a value, but it is nevertheless valid to place it in Go source code, earning me only 18 cents below the maximum for 27 bytes. If you prefer, the character U+... 1 Zsh, 127569, 15 bytes ];<<<$?{-%?} Try it online! Squeezing out a bit more than the Bash answer by abusing the default flags. By default, the - parameter is set to 569X. { %?} removes the last character. We can actually get a lot further by manually controlling the flags: zsh -178, 156789, 11 bytes <<<${-%?} Try it online! ... 0 Cascade, 177827, 10 bytes ^𫚣 #" Try it online! I'm lucky here in that 𫚣 counts as a letter variable, which allows me to fetch its ordinal value, with the cost that it is a multi-byte character. 5 14,674,000,667 5,436,050 5,403,448 10,385 5,994 4,447 3,806 total precision For a baseline, I investigated the following approach: Select $M,\delta,\epsilon>0$ such that if we sample the polynomial $p(x)=x^3+ax^2+bx+c$ at$$ S:=\{-M,-M+\delta,-M+2\delta,\ldots,M\}, then the largest sample point $s^\star\in S$ satisfying \$p(s^\star)<\... 1 182 white squares Inspired by Robin Ryder's answer, I tried to squeeze in a couple more white squares. I believe this solution is unique, and I will soon post verification code accordingly. Computer-readable grid: HETERONORMATIVE OVEROPINIONATED POSSESSEDNESSES B############## INCOMMUNICATIVE NEUROANATOMICAL DETERMINATENESS ############### ... 12 Octave, 96 88 87 84 76 54 50 weights & biases This 6-layer neural net is essentially a 3-step sorting network built from a very simple min/max network as a component. It is basically the example network from wikipedia as shown below, with a small modification: The first two comparisons are done in parallel. To bypass negative numbers though the ReLU, ... 9 180 white squares My strategy was simply to find a smaller rectangle with no black squares, such that it can be filled in uniquely. All 2×k rectangles have multiple solutions. For 3×k rectangles, there are multiple solutions for k between 3 and 14, but there is a exactly one solution for k=15. I then fit 4 such rectangles in the grid. This means that ... 0 [Python 3] (644449*2+0) 1288898 points Perfect accuracy in only 644449 bytes import zlib,base64 as s t=enumerate(zlib.decompress(s.b64decode(b'###')).decode());a=lambda c:next(t)[1] The full code cannot fit in an answer, so I have put it here and replaced the large binary string literal with b'###' in the answer text. This is generated with the ... 1 I is for Io, 31 characters -> Score: 33 for($,65,90,\$asCharacter print)
Top 50 recent answers are included | 2019-10-20 19:46:17 | {"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": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38711827993392944, "perplexity": 5161.894858730401}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2019-43/segments/1570986718918.77/warc/CC-MAIN-20191020183709-20191020211209-00170.warc.gz"} |
https://www.physicsforums.com/threads/name-of-function.13569/ | Name of Function
1. Jan 31, 2004
Lonewolf
The function defined over the real numbers as
Code (Text):
f(x) = {0, x rational
{1, x irrational
appears quite a lot in mathematics textbooks. Does anybody know if this function has a commonly accepted name?
2. Jan 31, 2004
Staff Emeritus
Well, it's the characteristc function of the set of irrationals, $$\chi (R - Q)$$.
3. Jan 31, 2004
Hurkyl
Staff Emeritus
I've heard it called the "salt and pepper" function.
4. Jan 31, 2004
Lonewolf
5. Jan 31, 2004
phoenixthoth
i've got a simple name and a mnemoic to remember it's name.
name: the 0 if rational and 1 if irrational function.
simple mnemoic: tziraoiif
6. Feb 2, 2004
vepore2
that doesn't sound simple
7. Feb 3, 2004
Lonewolf
What do you mean? I think it rolls right off the tounge.
8. Feb 3, 2004
himanshu121
Probably Dirchlet Function(Check Spelling)
9. Feb 3, 2004
Kalimaa23
Sure about that? The name Dirichlet has popped quite a number of time in my textbooks. I've never heard of Dirchlet. | 2017-03-30 17:14:08 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5273274779319763, "perplexity": 6868.993986337052}, "config": {"markdown_headings": false, "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-2017-13/segments/1490218195419.89/warc/CC-MAIN-20170322212955-00128-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://plainmath.net/10572/solve-following-system-equations-using-elimination-equal-equal-solve | # 1. Solve the following system of equations using elimination. 5x+y=24 -5x-4y=-56 Solve for y: Solve for x
Question
Equations and inequalities
1.
Solve the following system of equations using elimination.
5x+y=24
-5x-4y=-56
Solve for y:
Solve for x
2021-01-05
5x+y=24---1
-5x-4y=-56---2
X=?
Y=?
Adding equation 1 & 2
(5x+y)+(-5x-4y)=24+(-56)
-3y=-32
$$\displaystyle{Y}=\frac{{32}}{{3}}$$
Substitute the value of y in 1st eq
5x+y=24
$$\displaystyle{5}{x}+{\left(\frac{{32}}{{3}}\right)}={24}$$
$$\displaystyle{5}{x}={24}-\frac{{32}}{{3}}=\frac{{40}}{{3}}$$
$$\displaystyle{X}=\frac{{8}}{{3}}$$
### Relevant Questions
Solve the system of equations using the addition/elimination method.
4x+3y=15
2x-5y=1
Find a general solution to $$\displaystyle{y}{''}+{4}{y}'+{3.75}{y}={109}{\cos{{5}}}{x}$$
To solve this, the first thing I did was find the general solutionto the homogeneous equivalent, and got
$$\displaystyle{c}_{{1}}{e}^{{-{5}\frac{{x}}{{2}}}}+{c}_{{2}}{e}^{{{3}\frac{{x}}{{2}}}}$$
Then i used the form $$\displaystyle{K}{\cos{{\left({w}{x}\right)}}}+{M}{\sin{{\left({w}{x}\right)}}}$$ and got $$\displaystyle-{2.72}{\cos{{\left({5}{x}\right)}}}+{2.56}{\sin{{\left({5}{x}\right)}}}$$ as a solution of the nonhomogeneous ODE
Solve the systems of equations using matrices.
$$4x+5y=8$$
$$3x-4y=3$$
Leave answer in fraction form.
$$4x+y+z=3$$
$$-x+y=-11+2z$$
$$2y+2z=-1-x$$
Solve the given system of equations by matrix equation.
5x-4y=4
3x-2y=3
Solve the system of equations x+y=-1 and 5x-7y=79 by combining the equations.
Solve the following system of linear equations in two variables by Substitution method.
$$5x-2y=-7$$
$$x=-2y+1$$
$$\displaystyle{\left|{5}{x}-{1}\right|}={\left|{3}-{4}{x}\right|}$$
Solve the equations and inequalities. Find the solution sets to the inequalities in interval notation. $$\displaystyle\sqrt{{{t}+{3}}}+{4}={t}+{1}$$ | 2021-06-24 09:15:27 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8701251745223999, "perplexity": 827.4983252990686}, "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-2021-25/segments/1623488552937.93/warc/CC-MAIN-20210624075940-20210624105940-00350.warc.gz"} |
https://lgw2.github.io/teaching/csci132-fall-2022/classwork/classwork2/ | # Classwork 2
## Logistics
• Due: Tuesday, May 11th no later than 1:35pm.
• Submission instructions: push a commit with the tag classwork2 to your git repository.
• Deadline reminder: after the deadline passes, you cannot earn any points for this assignment. If the deadline is approaching, submit what you have in order to earn partial credit.
## Learning outcomes
• Write your first C program.
• Familiarize yourself with C’s I/O commands.
• Write a program that interacts with a human user.
• Write expressions using arithmetic operators.
## Assignment
This comes from problem 1, chapter 2 on page 102.
Write a program that calculates mileage reimbursement for a salesperson at a rate of $.35 per mile. Your program should interact with the user in this manner: MILEAGE REIMBURSEMENT CALCULATOR Enter beginning odometer reading=> 13505.2 Enter ending odometer reading=> 13810.6 You traveled 305.4 miles. At$0.35 per mile, your reimbursement is \$106.89.
### Requirements
Write your program in the classwork directory, inside another directory called classwork2, in a file called calculator.c.
• Compile your program into an executable called calculator and run it using ./calculator.
### Hints
• Develop your program in steps, checking that it compiles and runs after every change.
• Format your output to print only one or two decimal places using %.mf instead of %f, where m is the number of decimal places.
• 1 point - there is a file called calculator.c in your classwork/classwork2 directory.
Note: you will not earn any points if your work is not committed and pushed to Github with the tag classwork2. | 2022-08-13 19:29:58 | {"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": 0.20598353445529938, "perplexity": 4812.35724542933}, "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-2022-33/segments/1659882571982.99/warc/CC-MAIN-20220813172349-20220813202349-00466.warc.gz"} |
https://electronics.stackexchange.com/questions/573785/is-it-a-bad-idea-to-use-2-vias-to-change-layers-in-power-tracks | # Is it a bad idea to use 2 vias to change layers in power tracks?
I have some power traces (net VIN) in my PCB design that might (in an extreme scenario) carry up to 1A of current. Although nominally, I wouldn't expect to see more than 250 mA. These are 50 mil traces, just to be safe, and since I have to switch layers, I have put two 28 mil hole vias. Is it a bad idea to use double vias like this?
• Just a note that a 50mil track, external, 2oz copper has a resistance of 5.05mΩ/in. For 10°C temp rise, it is good for 4.7A. At 4.7A, it would drop 23.7mV/in (likely much thinner tracks can be used, unless very long.) Jul 7 at 20:15
• Thank you for the heads up. I probably oversized it too much. Since I don't expect very large currents (<1A), I wouldn't expect the voltage drop to reach 23 mV/in. But still, could've made it thinner. Thx. Jul 7 at 23:25
I use multiple vias to connect power traces to other layers all the time on many projects: -
Is it a bad idea to use double vias like this?
No, it's a valid way of doing it. Many of the connections in the picture above use 24 vias. Mind you, the tracks are carrying over one hundred amps and are multi-layer connected.
• That's a great example, Andy! Jul 2 at 17:05
Not a bad idea. In fact it's recommended when you have high currents and you're changing layers as it reduces track impedance.
There are other considerations like whether to use a couple of large vias as you have, or multiple smaller vias. Ask your fabricator whether this is worth worrying about. Personally, I tend to use the same size via and place them at about 1mm pitch for stitching large current tracks or polygons.
If you have solid planes (ground or otherwise), you may want to try to space the vias in such a way that the planes have connectivity between the vias on the other layers. In other words, avoid creating a "slot" in the plane. | 2021-12-06 12:44:16 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5389295220375061, "perplexity": 1967.5581466222861}, "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-2021-49/segments/1637964363292.82/warc/CC-MAIN-20211206103243-20211206133243-00350.warc.gz"} |
https://activecalculus.org/multi/preface-1.html | ## PrefaceFeatures of the Text
Similar to the presentation of the single-variable Active Calculus, instructors and students alike will find several consistent features in the presentation, including:
Motivating Questions.
At the start of each section, we list motivating questions that provide motivation for why the following material is of interest to us. One goal of each section is to answer each of the motivating questions.
Preview Activities.
Each section of the text begins with a short introduction, followed by a preview activity. This brief reading and the preview activity are designed to foreshadow the upcoming ideas in the remainder of the section; both the reading and preview activity are intended to be accessible to students in advance of class, and indeed to be completed by students before a day on which a particular section is to be considered.
Activities.
Every section in the text contains several activities. These are designed to engage students in an inquiry-based style that encourages them to construct solutions to key examples on their own, working either individually or in small groups.
Exercises.
There are dozens of calculus texts with (collectively) tens of thousands of exercises. Rather than repeat a large list of standard and routine exercises in this text, we recommend the use of WeBWorK with its access to the National Problem Library and its many multivariable calculus problems. In this text, each section begins with several anonymous WeBWorK exercises, and follows with several challenging problems. The WeBWorK exercises are best completed in the .html version of the text. Almost every non-WeBWorK problem has multiple parts, requires the student to connect several key ideas, and expects that the student will do at least a modest amount of writing to answer the questions and explain their findings. For instructors interested in a more conventional source of exercises, consider the freely available APEX Calculus text by Greg Hartmann et al.
Graphics.
As much as possible, we strive to demonstrate key fundamental ideas visually, and to encourage students to do the same. Throughout the text, we use full-color graphics to exemplify and magnify key ideas, and to use this graphical perspective alongside both numerical and algebraic representations of calculus. When the text itself refers to color in images, one needs to view the .html or .pdf electronically. The figures and the software to generate them have been created by David Austin.
Summary of Key Ideas.
Each section concludes with a summary of the key ideas encountered in the preceding section; this summary normally reflects responses to the motivating questions that began the section. | 2020-01-25 06:35:22 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30216479301452637, "perplexity": 1017.62812046245}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2020-05/segments/1579251669967.70/warc/CC-MAIN-20200125041318-20200125070318-00276.warc.gz"} |
http://physics.stackexchange.com/tags/spectroscopy/hot | # Tag Info
30
If I understand you right, you're referring to the phenomenon seen in this picture (from the first Google hit), that near the horison the color of the sky is more light-blue (not exactly white): Rayleigh scattering The scattering in the atmosphere is for a large part Rayleigh scattering off of nitrogen and oxygen molecules, which are much smaller than ...
24
To be a little pedantic, nobody has yet done precision spectroscopy of antihydrogen, though the recent success in trapping it at CERN (all over the news this week, paper here) is an early step toward that. It's possible that there are small differences in the spectrum of antihydrogen and hydrogen, though these differences can't be all that large, or they ...
20
Suppose you are analysing the weights of people in the UK to see what the distribution of weights looks like. Suppose also you can measure the weight to arbitrary precision, so that no two people's weights will be exactly the same. When you're finished you plot your data on a histogram, but the trouble is that because everyone has a different weight you get ...
18
The full quantum analysis of the hydrogen atom is a quantum two body problem, however, one of those bodies is extremely massive compared to the other, so that this problem, as a first approximation, is analysed by solving either the first quantised (i.e. for one quantum particle in a classical environment) Schrödinger or Dirac equations for inverse square ...
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When astronomers started to get spectra of stars and began classifying them, the initial classification was based on the strength of the Balmer absorption lines in the spectra. The Balmer lines are created by electons in hydrogen atoms that are currently in the second energy level (N=2) absorbing energy and jumping up to higher levels. The stars with the ...
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There's a few ways the temperature can be measured remotely. The easiest way is to measure the amount, and for bonus, spectrum, of the radiated heat. All objects greater than Absolute Zero radiate a certain amount of energy. The wavelength spectrum can be determined by Planck's Law, and the amount of energy by the Stefan-Boltzmann law, both of which are ...
16
The resolving power of a prism is given by the formula $$\frac{\lambda}{\Delta \lambda} = b\ \frac{dn}{d\lambda},$$ where $b$ is the base length of the prism, $\lambda$ is the wavelength and $n(\lambda)$ is the refractive index. You don't say, but let's assume you are using a crown glass prism. According to this useful document, crown class has $dn/d\... 14 Given that most green pointers are frequency doubled from a 281.8 THz infrared laser ($c$/1064 nm), it's possible that you have a two frequencies$f_1$and$f_2$in the original infrared laser (i.e., it is multimode). After passing through the "frequency doubling" nonlinear crystal you see three frequencies:$2 f_1$,$2f_2$, and$f_1 + f_2$. It looks like ... 13 Almost all exoplanets observed are near F, G, and K stars. In part, this is because astronomers are looking for earth-like planets, so they look at stars similar to our Sun, but there are also some physical reasons. Sahu et al (2006) have provided some evidence that red dwarfs (class M) are more likely to have planets than other spectral types, though it is ... 12 Good quantum number are associated with operators that commute with the Hamiltonian. They correspond to conseved quantities. Overall angular momentum is conserved, but the portions of it due to orbital motion and due to spins are not themselves conserved. n is 'bad' in that there's no conserved physical quantity related to radial motion. A decently ... 12 What you're describing does happen to some extent. It's called "Doppler broadening": absorption and emission lines in hot materials are wider (in wavelength) than absorption and emission lines in cool materials, because atoms in the hot material occupy a broader range of velocities. An atom at rest can't absorb any old photon and convert the extra energy ... 11 I think that it is important to recognize the practical difference between Raman scattering and fluorescence. If the energy of the photon is resonant with some molecular transition (meaning that it is equal to the energy difference between ground energy state and one of the excited states of the molecule), that the molecule can absorb this photon undergoing ... 11 It's because you're not looking far enough. From personal experience, it takes at least 10 km of atmosphere to build up a really obvious blue (see, for example, this picture), and if you're not in hilly country, the horizon is only 5km away. In contrast, most of the sky has distances to space on the order of hundreds of kilometers. 11 The broadening of emission lines is not due to something that is happening to each individual star, but rather something that affects the whole population. As stars in a galaxy get older, their orbits change relative to the orbits of other stars of the same age. Most relevant for the discussion here, the "velocity dispersion" of a stellar population ... 10 Just compare the resolution of the two: Prism depending on n, there is no good material n>1.7 (besides diamond) depending on base length if you use a equilateral triangle have to use more than one to overcome this prism absorb light, you have got scattering (stray light) too Now a grating: optimize it for your wavelength choose lines per milimeter ... 10 Given what we know about planetary formation (Link 1, Link 2, Link 3 and Link 4), and the theories around it, it would probably be a safe bet to say that ALL stars end up having some left over material that might become planets. I think the bigger question is how many of those planetary orbits stay stable enough throughout the life of the star? All these ... 9 In theory, perhaps. It is possible, using multilayer dielectric coatings, to produce a surface which is reflective in very narrow bands (in this case, the Sun's dark lines)and transmissive (or absorptive) elsewhere. In practice, the spectral "blurring" caused by atmospheric transmission/absorption/re-emission effects would make this effect pretty much ... 9 The Aethrioscope (see Wiki page with this name) was invented in 1818 by Sir John Leslie and the basic idea for a pyrometer (see Wiki pahe with this name) was conceived in the late 1700s by Josiah Wedgewood. These were calibrated by comparing observed colour with that of hot metals / clays (as appropriate) of known temperature. The idea was to heat a small ... 9 Lasers by definition only emit a single wavelength of light. You use one if you want that wavelength or if you want your photons to be in phase. You don't care about the photon phases, and you want to sample all wavelengths, so a laser is very much the wrong tool. If you just want collimation of the light, mirrors, lenses, or even just well-separated ... 8 Blackbody radiation for a white hot object emits the spectrum from infrared to ultraviolet. See: http://en.wikipedia.org/wiki/Black_body Graphite is a decent approximation of a blackbody radiator. So graphite heated to white hot will emit the full spectrum of visible light. Note however, that the spectrum will not be flat. There will be more energy on ... 8 You've asked a lot of questions there, and I'll try to answer them one by one. First, though, I want to ask what post you're reading about metallicity in the core vs. out here in the 'burbs because I don't think it is correct. Obviously, for example, we exist and we're ~26,000 light-years (half-way out) from the galactic center and we have a fair amount of ... 8 The major tool in investigating the composition of an astronomical body is spectroscopy. This technique makes use of the fact that a body composing certain elements/compounds will shine more brightly(or less) at particular wavelengths. The pattern of 'lines' taken by a spectrograph can then be used to infer the original composition. This technique is used ... 8 A galaxy spectrum is a quite complex and complicated topic, and many entire careers are fully devoted to understanding them, so this can only be a simplified answer. It is still quite lengthy, though, so if you're impatient, I've summarised it at the bottom. A blend of starlight of different spectral types makes up the continuum. The light is emitted by ... 7 the physical reason why this is happening is that absorption of a medium is frequency dependent. Mathematical description The most prominent example of a natural description might be the Lambert-Beer law that states that the change of a quantity$q$,$dq/dx$is related to its value at$x$multiplied by a scalar factor$\lambda\$, which might depend on some ...
7
While most measurements in astronomy are better in space, precision spectroscopy can actually do quite well on the ground. One of the best spectrographs (some would say the best) is HARPS, the High-Accuracy Radial Velocity Planetary Searcher used for finding extrasolar planets. As described in its instrument paper (pdf; note that the sole purpose of this ...
7
The atmosphere obscures data in three main ways; it absorbs light, it emits light in the infrared, and finally it diffracts light leading to distorted images. Observers have ways to deal with all three things, but I'll focus on the first two since they are more directly related to your question: 1) Atmospheric absorption. This plot gives a rough idea of ...
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According to Bohr model, the absorption and emission lines should be infinitely narrow, because there is only one discrete value for the energy. There are few mechanism on broadening the line width - natural line width, Lorentz pressure broadening, Doppler broadening, Stark and Zeeman broadening etc. Only the first one isn't described in Bohr theory - it's ...
6
This question was asked a couple of years ago and things have changed since then. We now know that small planets are found around stars across a broad range of metallicities and that it is only the existence of giant planets that are affected by low metallicity. Nature article here. It was previously thought that small planets were more common around small ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2016-06-30 14:00:18 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6937488317489624, "perplexity": 502.669441447326}, "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-2016-26/segments/1466783398869.97/warc/CC-MAIN-20160624154958-00033-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://cris.ulima.edu.pe/es/publications/the-effect-of-normality-and-outliers-on-bivariate-correlation-coe | # The effect of normality and outliers on bivariate correlation coefficients in psychology: A Monte Carlo simulation
José Ventura-León, Brian Norman Peña-Calero, Andrés Burga-León
Resultado de la investigación: Contribución a una revistaArtículo (Contribución a Revista)revisión exhaustiva
## Resumen
This study aims to examine the effects of the underlying population distribution (normal, non-normal) and OLs on the magnitude of Pearson, Spearman and Pearson Winzorized correlation coefficients through Monte Carlo simulation. The study is conducted using Monte Carlo simulation methodology, with sample sizes of 50, 100, 250, 250, 500 and 1000 observations. Each, underlying population correlations of 0.12, 0.20, 0.31 and 0.50 under conditions of bivariate Normality, bivariate Normality with Outliers (discordant, contaminants) and Non-normal with different values of skewness and kurtosis. The results show that outliers have a greater effect compared to the data distributions; specifically, a substantial effect occurs in Pearson and a smaller one in Spearman and Pearson Winzorized. Additionally, the outliers are shown to have an impact on the assessment of bivariate normality using Mardia’s test and problems with decisions based on skewness and kurtosis for univariate normality. Implications of the results obtained are discussed.
Idioma original Inglés Journal of General Psychology 5 jul. 2022 https://doi.org/10.1080/00221309.2022.2094310 Publicada - 6 jul. 2022
## Huella
Profundice en los temas de investigación de 'The effect of normality and outliers on bivariate correlation coefficients in psychology: A Monte Carlo simulation'. En conjunto forman una huella única. | 2023-03-23 11:06:42 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8088491559028625, "perplexity": 3506.3841577460853}, "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-2023-14/segments/1679296945144.17/warc/CC-MAIN-20230323100829-20230323130829-00768.warc.gz"} |
https://zbmath.org/?q=an:0871.32008 | # zbMATH — the first resource for mathematics
Open sets with Stein hypersurface sections in Stein spaces. (English) Zbl 0871.32008
Let $$D\subset \mathbb C^n, n\geq3,$$ be an open set such that for any linear hyperplane $$H\subset \mathbb C^n$$ the intersection $$H\cap D$$ is Stein. It is natural to raise the following problem of hypersurface sections. Let $$X$$ be a Stein space of dimension $$n\geq3$$ and $$D\subset X$$ an open subset such that $$H\cap D$$ is Stein for every hypersurface $$H\subset X.$$ Does it follow that $$D$$ is Stein? The authors produce a counter-example to this problem. There is a normal Stein space $$X$$ of pure dimension 3 with only one singular point, and a closed connected analytic subset $$A\subset X$$ of pure dimension 2, such that $$D:=X\backslash A$$ is not Stein, and for every hypersurface $$H\subset X$$ (i.e. closed analytic subset of $$X$$ of pure codimension 1) the intersection $$H\cap D$$ is Stein.
##### MSC:
3.2e+11 Stein spaces
Full Text: | 2021-10-16 22:07:36 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8740935921669006, "perplexity": 234.09378703352772}, "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-2021-43/segments/1634323585025.23/warc/CC-MAIN-20211016200444-20211016230444-00536.warc.gz"} |
https://web2.0calc.com/questions/a-simple-inequality-equation-that-i-cannot-solve | +0
# A simple Inequality equation that I cannot solve
0
215
2
(x+1)/(2-x) < x/(3+x)
Guest Mar 16, 2015
#2
+91972
+10
$$\\\frac{(x+1)}{(2-x) }< \frac{x}{(3+x)}\\\\ I am going to multiply both sides by positive numbers and see if that will help\\\\ (2-x)^2(3+x)^2*\frac{(x+1)}{(2-x) }<(2-x)^2(3+x)^2 \frac{x}{(3+x)}\\\\ (2-x)(3+x)^2(x+1)<(2-x)^2(3+x) x\\\\ (2-x)(3+x)^2(x+1)-(2-x)^2(3+x) x<0\\\\ (2-x)(3+x)\;[(3+x)(x+1)-(2-x) x]<0\\\\ -(x-2)(x+3)\;[(x+3)(x+1)-x(-x+2)]<0\\\\ -(x-2)(x+3)\;[x^2+x+3x+3+x^2-2x]<0\\\\ -(x-2)(x+3)\;[2x^2+2x+3]<0\\\\$$
consider
$$\\2x^2+2x+3=0\\\\ \triangle=4-24<0\\\\ since the discriminant is 0 there are no roots for this.\\\\ the axis of symmetry for y=2x^2+2x+3\;\;is\;\; \frac{-b}{2a}=\frac{-2}{4}=\frac{-1}{2}$$
the roots are x=2 AND x=-3
the polynomial if set to y and graphed will finish in the bottom right corner because of the - out to front.
I can see that the polynomial will be less than 0 when x>2 and when x<-3
I am sorry i probably have not explained this very well, it was a difficult question for this type.
I did not use the graph but I will draw it now to show you.
https://www.desmos.com/calculator/5xfsalfbbg
CPhill's way was probably better for this one. :)
Melody Mar 16, 2015
Sort:
#1
+84385
+10
(x+1)/(2-x) < x/(3+x)
This might be solved easiest by some analysis
Let's divide the inequality into three iintervals
(-∞, -3), (-3, 2) and (2, ∞)
Notice that, on the first interval, the function on the left is always less than the function on the right
We can't say what happens at x = -3 because the function on the right isn't defined there
On the second interval, the function on the left is always greater than the function on the right
And at x = 2 the function on the left is undefined
On the third interval, the function on the left is again always less than the function on the right
So, the solution to this problem is that the intervals (-∞, -3) and (2, ∞) make the inequality true and the interval (-3, 2) makes it false
Here's a graph that confirms our suspicions.... https://www.desmos.com/calculator/ymvoluxlsa
CPhill Mar 16, 2015
#2
+91972
+10
$$\\\frac{(x+1)}{(2-x) }< \frac{x}{(3+x)}\\\\ I am going to multiply both sides by positive numbers and see if that will help\\\\ (2-x)^2(3+x)^2*\frac{(x+1)}{(2-x) }<(2-x)^2(3+x)^2 \frac{x}{(3+x)}\\\\ (2-x)(3+x)^2(x+1)<(2-x)^2(3+x) x\\\\ (2-x)(3+x)^2(x+1)-(2-x)^2(3+x) x<0\\\\ (2-x)(3+x)\;[(3+x)(x+1)-(2-x) x]<0\\\\ -(x-2)(x+3)\;[(x+3)(x+1)-x(-x+2)]<0\\\\ -(x-2)(x+3)\;[x^2+x+3x+3+x^2-2x]<0\\\\ -(x-2)(x+3)\;[2x^2+2x+3]<0\\\\$$
consider
$$\\2x^2+2x+3=0\\\\ \triangle=4-24<0\\\\ since the discriminant is 0 there are no roots for this.\\\\ the axis of symmetry for y=2x^2+2x+3\;\;is\;\; \frac{-b}{2a}=\frac{-2}{4}=\frac{-1}{2}$$
the roots are x=2 AND x=-3
the polynomial if set to y and graphed will finish in the bottom right corner because of the - out to front.
I can see that the polynomial will be less than 0 when x>2 and when x<-3
I am sorry i probably have not explained this very well, it was a difficult question for this type.
I did not use the graph but I will draw it now to show you.
https://www.desmos.com/calculator/5xfsalfbbg | 2018-03-24 06:28:35 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8231288194656372, "perplexity": 739.2091365974411}, "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-2018-13/segments/1521257649931.17/warc/CC-MAIN-20180324054204-20180324074204-00150.warc.gz"} |
http://clay6.com/qa/4852/prove-that- | Home >> CBSE XII >> Math
Prove that : $2tan^{-1} \bigg( \frac{1}{2} \bigg) + tan^{-1} \bigg( \frac{1}{7} \bigg) = tan^{-1} \bigg( \frac{31}{17} \bigg)$
Can you answer this question?
Toolbox:
• $2tan^{-1}x=tan^{-1}\frac{2x}{1-x^2}$
• $tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy}, xy <1$
L.H.S $tan^{-1}\frac{4}{3}=tan^{-1}\frac{1}{7}$
$= tan^{-1}\frac{\frac{4}{3}=\frac{1}{7}}{1-\frac{4}{21}}=tan^{-1}\frac{31}{17}$ = R.H.S
answered Feb 28, 2013 | 2017-01-18 18:18:51 | {"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": 3, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5241211652755737, "perplexity": 5142.294390922059}, "config": {"markdown_headings": false, "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-2017-04/segments/1484560280310.48/warc/CC-MAIN-20170116095120-00208-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://scicomp.stackexchange.com/tags/hyperbolic-pde/hot | # Tag Info
30
In the solution of nonlinear hyperbolic PDEs, discontinuities ("shocks") appear even when the initial condition is smooth. In the presence of discontinuities, the notion of solution can only be defined in the weak sense. The numerical velocity of a shock depends on the correct Rankine-Hugoniot conditions being imposed, which in turn depends on numerically ...
17
The central question is which physical processes (waves or source terms) have time scales that you are interested in resolving and which you would prefer to step over. If you are not interested in the fastest time scale in the system, then the equations are called "stiff". Hyperbolic conservation laws are typically written as first-order systems $$u_t + \... 14 The most common method is to reset negative values to some small, positive number. Of course, this is not a mathematically sound solution. A better general approach that may work and is easy, is to reduce the size of your time step. Negative values often arise in the solution of hyperbolic PDEs, because the appearance of shocks can lead to oscillations, ... 11 The principal numerical difficulty in solving a nonlinear first-order system of hyperbolic PDEs like the Euler equations (for compressible, inviscid flow) is that discontinuities (shock waves) appear in the solution after finite time, even if the initial data are smooth. In order to deal with this, most modern codes use both slope (or flux) limiters, which ... 11 If it's shock-capturing that you're interested in, I would suggest you use the finite volume method instead of the finite element method. When applied naively, FEM is actually notoriously bad at resolving shocks -- usually there are spurious oscillations or unwanted diffusion. Provided your original PDE is a conservation law, the FVM method will preserve ... 10 Here is a 97-line example of solving a simple multivariate PDE using finite difference methods, contributed by Prof. David Ketcheson, from the py4sci repository I maintain. For more complicated problems where you need to handle shocks or conservation in a finite-volume discretization, I recommend looking at pyclaw, a software package that I help develop. ""... 8 You could have a look at Fenics, which is a python/C framwork which allows quite general equations to be solved using a special markup language. It mostly uses finite elements though, but worth a look. The tutorial should give you an impression of how easy it can be to solve problems. 7 The short answer is: it requires specific work for different equations, but there are some general techniques that suggest how to do it. Essentially, given a first order evolution PDE$$u_t = Au + Bu$$where A,B are some (possibly differential) operators, the steady states are those for which$$Au + Bu=0.$$It is common to use a splitting approach in ... 6 Assuming we are solving hyperbolic equations without any source terms and assuming we provide physical initial conditions, making sure the numerical scheme we use is Total Variation Diminishing is a good way of ensuring the "physicality" of the computed solution. Since a TVD scheme preserves monotonicity, no new minima or maxima will be created and the ... 6 I think one answer to your question is that certain communities simply always used conservative schemes and so it has become part of "the way it's done". One may argue whether that's the best way to do it, but that's about as fruitful as asking the British to drive on the right because it would simply be more convenient to have only on standard side. That ... 6 There is not much point using an implicit method for pure wave propagation because you have to resolve phase to have an accurate method. If you have a hyperbolic system in which some waves are very stiff (not interesting except for their influence on evolution of a slow manifold), you might want an implicit method. It is fairly problem-dependent whether you ... 6 There is definitely literature on schemes like this. Two keywords are Modified method of characteristics Semi-Lagrangian schemes After 20 minutes of googling: some possibly important papers are http://dx.doi.org/10.1137/0719063 and http://dx.doi.org/10.1137/0728024 (search forward from there). Those probably aren't the best references out there, but ... 6 A nonconvex flux function means that the characteristic velocity associated with a given characteristic field may not be a monotone function of the conserved variables. This can lead to non-classical Riemann solutions; for instance the total number of (shock, rarefaction, and contact) waves arising in the solution may be different from the number of ... 6 This is an important and challenging issue. Yes, using quadratic interpolation means that your solution values may lie outside the interval in which the initial data lie. This is not what we usually mean when we refer to numerical instability, but it is a potentially undesirable feature. Yes, forcing the interpolated values to lie in an interval destroys ... 6 Let me give a part of your answer, I would need some more indications from your side to answer you fully. So please read, and write some comments so that I can complete my answer. About notations in numerical methods There are a few mistakes in the way you write your equations. These are only details, but for someone who is used to it, it can a bit ... 6 Such problems (sometimes called lateral Cauchy problems) are in general not well-posed (meaning they either lack a solution, or there are infinitely many of them, or the solution is unstable under perturbations of the boundary conditions). For parabolic (or dissipative) equations, it makes sense to study the stationary limit (simply omit the term u_t in ... 6 The stationary equation you show transports information from the right to the left via the advection term; it also diffuses slightly. If you switch off the diffusion term altogether, then you only have transport from the right to the left, and you need to also drop the boundary condition at the left: because information is from the right to the left, nothing ... 5 Godunov's method has an exact Riemann solver so no entropy fix is needed. A Roe solver (of which there are a few variants) uses a local linearization which has no diffusion to "fill in" the rarefaction fan, so it needs an entropy fix. Other approximate Riemann solvers, including Lax-Friedrichs, Rusanov, and the HLL family are inherently diffusive and do not ... 5 Starting from where David Ketcheson left me in his answer, a little bit more search revealed some historical notes. The scheme I outlined above was considered already back in 1900 by J. Massau, in Mémoire sur l'intégration graphique des équations aux dérivées partielles. The work is republished in 1952 by G. Delporte, Mons. The first (albeit brief) ... 5 As with many high order methods, the accuracy of the scheme is often less sensitive to the Riemann solver. None of the DG papers for hyperbolic problems will actually be using averages, however. The most common choice is a Rusanov (aka. Local Lax-Friedrichs) flux, which is very simple if you have an upper bound for the fastest wave speed. 5 I suggest looking at the literature on DG methods for incompressible flow, which has the mixed hyperbolic-elliptic character you mention. There are a lot of approaches. This paper, for instance, even uses an exact Riemann solver. This one suggests using a discontinuous space for the hyperbolic part and a continuous one for the elliptic part. 5 If the amplification factor is \le 1, then your scheme is stable; in your case it is exactly one, so, for this PDE, you won't have unbounded growth of errors. The Lax theorem then guarantees convergence of the method. In this narrow sense, the Crank-Nicholson method will work for you. (Have a look at Leveque's book if you want a good reference.) But, your ... 5 I have implemented the solution derived below using PyClaw in an IPython notebook. If you download that, you can adjust the initial values and see the computed solutions. General setup In the solution of a Riemann problem for the 1D Euler equations, there are generally 3 waves. Two of them are genuinely nonlinear; the other is a contact discontinuity, ... 5 Eigenvalues with zero eigenvalue correspond to purely oscillatory modes. You can see it by diagonalising the system. Your matrix A can be written as $$A = P \Sigma P^{-1}$$ where \Sigma is a diagonal matrix with entries \lambda_n corresponding to the eigenvalues of Q. You can now transform your ODE to dQ/... 5 First-order hyperbolic equations model conservation laws; as the alternative name "transport equations" suggests, they transport information along so-called "characteristic curves" with a finite speed of propagation. For the simple model equation u_t + u_x = 0, you can show (e.g., by separation of variables) that the solution should be of the form$$u(t,x)...
5
Your equation can be written in the following fashion (any spatial derivative approximation is valid), once space is discretised: $$\frac{1}{c}\frac{du_i}{dt}=-\left(\frac{\partial u}{\partial x}\right)_i(t) + v_i(t) \tag{*}$$ Keep in mind that $v_i(t) = v(x_i,t)$. Now the system of equations depends only on time $t$ you can apply Crack Nicholson method to ...
4
For the numerical solution of hyperbolic PDEs the use of Riemann solvers are essential components of conservative shock capturing methods for accurate simulation of wave problems which may have shocks (discontinuous jumps in conserved variables). To be able to obtain accurate solutions to such problems, we need to use proper upwinding techniques -- the ...
4
The above answers apply to time-dependent problems, but you could also demand positivity in a simple elliptic equation. In this case, you could formulate it as a variational inequality, giving bounds for the variables. In PETSc, there are two VI solvers. One uses a reduced-space method, where variables in active constraints are removed from the system to be ...
4
It is appears to be a standard finite difference discretization via method of lines of the wave equation $u_{tt} + c^2 \nabla^2u = 0, \quad (x,y)\in[0,L_x]\times[0,L_y], \quad t\geq0$ Standard finite difference approximations can be used to formulate a discrete problem by introducing the formally second-order accurate approximations \$u_{tt}\approx \frac{u^...
4
PCHIP is not a conservative reconstruction, making it inappropriate for conservation laws. Furthermore, hyperbolic problems have discontinuous solutions so there is generally no benefit to a continuous reconstruction. Conservative monotone spline reconstructions are being investigated by the UK Met Office for use in tracer advection for atmosphere modeling, ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2019-12-15 13:46:47 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.7139387726783752, "perplexity": 631.6268917501136}, "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-2019-51/segments/1575541308149.76/warc/CC-MAIN-20191215122056-20191215150056-00430.warc.gz"} |
http://mathhelpforum.com/calculus/94776-physical-meaning.html | # Math Help - physical meaning
1. ## physical meaning
hi
i have been learning calculus for abt a year now....
usually most topics learnt in maths are used in physics. and so is calculus.
can someone tell me wat is the physical meaning of integration?
differentiation is basically rates.
wat abt integration?>
specifically integration and differentiation of trigonometric functions?
2. ## integration means sum
Hi Darachit,
Generally Integeration means sum(adding) or total...
so for example,
say to calculate the mass flow rate across a cross section, where density, velocity is not uniform in that cross section.. then you can calculate the mass flow rate in a small region dA as = velocity * density * dA
so, So the total mass flow rate across a given area can be computed by adding the mass flow across many such small elemental areas(dA). so to represent this in mathametics we can write integration ( velocity * denstiy * dA) if we know the variation of velocity and density in that area then you can compute the mass flow rate as integration ( velocity * denstiy * dA) where limits of the integration will be your area of interst across which you want to caluclate the total mass flow rate
3. thanx for the info...
any idea abt the second question? physical significance of trig functions?
4. sorry !!
not sure....
5. Originally Posted by vinay
sorry !!
not sure....
no probs man...
6. analysis of wave motion and my name " pendulum "
also theory of diffraction , Fresnel integral $\int_0^{\infty} \sin{x^2} ~dx$
7. Originally Posted by darachit
thanx for the info...
any idea abt the second question? physical significance of trig functions?
Anything related to circular motion and/or periodic phenomena.
CB | 2015-07-02 22:50:41 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9479057192802429, "perplexity": 2511.244294021493}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2015-27/segments/1435375095671.53/warc/CC-MAIN-20150627031815-00230-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://questioncove.com/updates/51e41a02e4b0d9a104dc6a66 | OpenStudy (anonymous):
How do i solve one of these problems? {Which of the following lines is perpendicular to y=4x+2?}
5 years ago
OpenStudy (anonymous):
I think i have the answer, although ive never solved one of these before y=-1/4x + 1.
5 years ago
OpenStudy (anonymous):
isnt there a picture?
5 years ago
OpenStudy (anonymous):
Sure, i'll post it..
5 years ago
OpenStudy (anonymous):
5 years ago
OpenStudy (anonymous):
That's what i got.
5 years ago
OpenStudy (anonymous):
yeah there is only one answer that fits because when it is perpendicular the slope has to be the complete opposite
5 years ago
OpenStudy (anonymous): | 2018-11-21 01:51:25 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8210293054580688, "perplexity": 8595.438615709172}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2018-47/segments/1542039746926.93/warc/CC-MAIN-20181121011923-20181121033923-00525.warc.gz"} |
https://www.semanticscholar.org/paper/Optimal-network-online-change-point-localisation-Yu-Padilla/d5f1744312b77c385db7c2248c382e51cabe3946 | • Corpus ID: 231603213
# Optimal network online change point localisation
@article{Yu2021OptimalNO,
title={Optimal network online change point localisation},
author={Yi Yu and Oscar Hernan Madrid Padilla and Daren Wang and Alessandro Rinaldo},
journal={ArXiv},
year={2021},
volume={abs/2101.05477}
}
• Published 14 January 2021
• Computer Science, Mathematics
• ArXiv
We study the problem of online network change point detection. In this setting, a collection of independent Bernoulli networks is collected sequentially, and the underlying distributions change when a change point occurs. The goal is to detect the change point as quickly as possible, if it exists, subject to a constraint on the number or probability of false alarms. In this paper, on the detection delay, we establish a minimax lower bound and two upper bounds based on NP-hard algorithms and…
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## References
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Optimal change point detection and localization in sparse dynamic networks
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The Annals of Statistics
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This work proposes a computationally simple novel algorithm for network change point localization, called Network Binary Segmentation, which relies on weighted averages of the adjacency matrices, and devise a more sophisticated algorithm based on singular value thresholding, called Local Refinement, that delivers more accurate estimates of the change point locations.
A Note on Online Change Point Detection
• Mathematics
• 2020
Online change point detection is originated in sequential analysis, which has been thoroughly studied for more than half century. A variety of methods and optimality results have been established
Optimal Change-Point Detection and Localization
• Computer Science, Mathematics
• 2020
This work establishes the energy detection threshold and shows similarly that the optimal localization error of a specific change-point becomes purely parametric, and tightly characterize optimal rates for both problems.
Sequential change-point detection based on nearest neighbors
• Hao Chen
• Computer Science
The Annals of Statistics
• 2019
We propose a new framework for the detection of change-points in online, sequential data analysis. The approach utilizes nearest neighbor information and can be applied to sequences of multivariate
Online detection of local abrupt changes in high-dimensional Gaussian graphical models
• Mathematics
ArXiv
• 2020
A novel test is developed that is based on the $\ell_\infty$ norm of the normalized covariance matrix of an appropriately selected portion of incoming data and illustrates the good performance of the proposed detection procedure both in terms of computational and statistical efficiency across numerous experimental settings.
Sequential change-point detection in high-dimensional Gaussian graphical models
• Computer Science
J. Mach. Learn. Res.
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This work introduces a novel scalable online algorithm for detecting an unknown number of abrupt changes in the inverse covariance matrix of sparse Gaussian graphical models with small delay and can be extended to a large class of continuous and discrete graphical models.
Sequential Graph Scanning Statistic for Change-point Detection
• Computer Science, Mathematics
2018 52nd Asilomar Conference on Signals, Systems, and Computers
• 2018
This work presents two graph scanning statistics that can detect local changes in the distribution of edges in a subset of the graph and demonstrates the efficiency of the detection statistics for ambient noise imaging, using a real dataset that records real-time seismic signals around the Old Faithful Geyser in the Yellowstone National Park.
Information Bounds and Quick Detection of Parameter Changes in Stochastic Systems
• T. Lai
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By using information-theoretic bounds and sequential hypothesis testing theory, this paper provides a new approach to optimal detection of abrupt changes in stochastic systems which leads to detection rules which have manageable computational complexity for on-line implementation and yet are nearly optimal under the different performance criteria considered.
Algorithms and Models for the Web Graph
• Mathematics, Computer Science
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• 2016
It is argued that the network is robust if τ < 2+ 1 δ , but fails to be robust ifτ > 2 + 1 ε−1 .
Rate-optimal graphon estimation
• Computer Science, Mathematics
• 2015
This paper establishes optimal rate of convergence for graphon estimation in a H\"{o}lder class with smoothness $\alpha$, which is, to the surprise, identical to the classical nonparametric rate. | 2022-08-11 03:01:22 | {"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": 0.5953166484832764, "perplexity": 1721.5334842987165}, "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-2022-33/segments/1659882571232.43/warc/CC-MAIN-20220811012302-20220811042302-00497.warc.gz"} |
http://math.stackexchange.com/questions/280800/what-is-the-formula-to-predict-a-high-probability-of-range-09-given-by-past-re?answertab=votes | # What is the formula to predict a high probability of range 0~9, given by past results?
I'm very new and i'm below average for my math.
But there's this thing about a lucky draw game that bugs me to think of a probability.
Say the host of a party has already drawn several luck draw numbers for each 2 hours, such as below:
$$\begin{array}{c|c|c} \text{Time} & \text{1st Draw} & \text{2nd Draw} & \text{3rd Draw}\\ \hline \\08 AM & 1516 & 4865 & 9876 \\10 AM & 0513 & 7805 & 9843 \\12 PM & 1124 & 0350 & 8790 \\02 PM & 9802 & 7967 & 3210 \\04 PM & 8794 & 6350 & 7842 \\06 PM & ???? & ???? & ???? \end{array}$$
Now, at $06:00 PM$ I would like to find out the [????].
I wish to know from the range of [0~9], what will be the 4 digits highly probable number to appear.
It does not need to be sorted in any order.
e.g, if you think the result is $0531$, then it doesn't matter if its $0351$ or $5130$ or $3150$ or etc...
It does not need to be in any specific draws, meaning dont care if its [1st draw] or [3rd draw]...
All I need to know is the 4 high probability digits. | 2014-12-18 20:57:57 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23166599869728088, "perplexity": 701.1274101663627}, "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-2014-52/segments/1418802767873.65/warc/CC-MAIN-20141217075247-00097-ip-10-231-17-201.ec2.internal.warc.gz"} |
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# Is x > y ? (1) x^(1/2)>y (2) x^3>y
Author Message
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Senior Manager
Joined: 22 Dec 2011
Posts: 298
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Re: Is x > y ? [#permalink] 09 Oct 2012, 04:27
Bunuel wrote:
satishreddy wrote:
hey bunnel.....can we also rewrite sq root X>M as X>M2 , by squaring both sides to make things simple,,,,,,,,,
We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
2<4 --> we can square both sides and write: 2^2<4^2;
0\leq{x}<{y} --> we can square both sides and write: x^2<y^2;
But if either of side is negative then raising to even power doesn't always work.
For example: 1>-2 if we square we'll get 1>4 which is not right. So if given that x>y then we can not square both sides and write x^2>y^2 if we are not certain that both x and y are non-negative.
We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
-2<-1 --> we can raise both sides to third power and write: -2^3=-8<-1=-1^3 or -5<1 --> -5^2=-125<1=1^3;
x<y --> we can raise both sides to third power and write: x^3<y^3.
Hope it helps.
Hi few observations please correct me if Im wrong ->
\sqrt{x} > y -> cannot square this but I can always cube both sides
y > \sqrt{x} - Can i square this? Rational behind this is right hand side of the inequality is +ve so left must be to, hence this can be written as y^2>x ??
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WE: Science (Education)
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Re: Is x > y ? [#permalink] 09 Oct 2012, 05:37
1
KUDOS
Jp27 wrote:
Bunuel wrote:
satishreddy wrote:
hey bunnel.....can we also rewrite sq root X>M as X>M2 , by squaring both sides to make things simple,,,,,,,,,
We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
2<4 --> we can square both sides and write: 2^2<4^2;
0\leq{x}<{y} --> we can square both sides and write: x^2<y^2;
But if either of side is negative then raising to even power doesn't always work.
For example: 1>-2 if we square we'll get 1>4 which is not right. So if given that x>y then we can not square both sides and write x^2>y^2 if we are not certain that both x and y are non-negative.
We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
-2<-1 --> we can raise both sides to third power and write: -2^3=-8<-1=-1^3 or -5<1 --> -5^2=-125<1=1^3;
x<y --> we can raise both sides to third power and write: x^3<y^3.
Hope it helps.
Hi few observations please correct me if Im wrong ->
\sqrt{x} > y -> cannot square this but I can always cube both sides
y > \sqrt{x} - Can i square this? Rational behind this is right hand side of the inequality is +ve so left must be to, hence this can be written as y^2>x ??
\sqrt{x} > y -> cannot square this but I can always cube both sides - YES (although it won't help much, the square root will stay)
y > \sqrt{x} - Can i square this? Rational behind this is right hand side of the inequality is +ve so left must be to, hence this can be written as y^2>x ?? YES, but the right hand side always non-negative (it can be 0)
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] 05 Nov 2012, 09:00
Is x > y ?
a). x2 > y
Gives 2 conditions x > y & x < y
eg:- Consider x=2 and y=1, then x2 > y and x > y
Consider x= -4 and y = 1, then x2 > y and x < y
Insufficient.
b) sqrt x < y
ie., x < y2, again as above gives 2 conditions x > y & x < y
Insufficient.
C) Form a we have x > y & x < y, and form b we have x > y & x < y, literally both are same
Insufficient.
Ans . E
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] 05 Nov 2012, 20:58
Expert's post
Maurice wrote:
Is x > y ?
a). x2 > y
Gives 2 conditions x > y & x < y
eg:- Consider x=2 and y=1, then x2 > y and x > y
Consider x= -4 and y = 1, then x2 > y and x < y
Insufficient.
b) sqrt x < y
ie., x < y2, again as above gives 2 conditions x > y & x < y
Insufficient.
C) Form a we have x > y & x < y, and form b we have x > y & x < y, literally both are same
Insufficient.
Ans . E
You might want to re-consider what happens when you take both statements together. Check out the solutions on page 1 or check this link:
http://www.veritasprep.com/blog/2011/08 ... -question/
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 17321 Followers: 2875 Kudos [?]: 18401 [0], given: 2350 Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] 06 Nov 2012, 04:49 Expert's post Maurice wrote: Is x > y ? a). x2 > y Gives 2 conditions x > y & x < y eg:- Consider x=2 and y=1, then x2 > y and x > y Consider x= -4 and y = 1, then x2 > y and x < y Insufficient. b) sqrt x < y ie., x < y2, again as above gives 2 conditions x > y & x < y Insufficient. C) Form a we have x > y & x < y, and form b we have x > y & x < y, literally both are same Insufficient. Ans . E Answer to the question is C, not E. Check the solutions on page 1 and 2. For example: is-x-y-1-x-1-2-y-2-x-3-y-100636.html#p777179 Hope it helps. _________________ Senior Manager Joined: 13 Aug 2012 Posts: 465 Concentration: Marketing, Finance GMAT 1: Q V0 GPA: 3.23 Followers: 14 Kudos [?]: 152 [0], given: 11 Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] 16 Jan 2013, 04:36 It totally helped me to setup a chart to test the values out: y < x (a) increasing integer (b) increasing fraction y > x (c) decreasing fraction (d) decreasing integer Test this out on statement (1) and statement (2)... _________________ Impossible is nothing to God. Senior Manager Joined: 13 Aug 2012 Posts: 465 Concentration: Marketing, Finance GMAT 1: Q V0 GPA: 3.23 Followers: 14 Kudos [?]: 152 [0], given: 11 Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] 22 Jan 2013, 00:35 shrouded1 wrote: I thought this was a really tough question ! Is x > y ? (1) \sqrt{x} > y (2) x^3 > y I notice usually these types test the fractions... if x and y are swapping signs... 1. \sqrt{x} > y==>x > y^2 Let x>y: 2 > 1 (It works!) Let x<y: 1/3 > (1/4)^2 (It works!) Let x<y: -1/3 ? 1/16 (It doesn't work) INSUFFICIENT! 2. x^3 > y Let x>y: (2)^3 > 1 (It works!) Let x<y: 1/27 ? 1/4 (It doesn't work!) Let x<y: -1/27 > -1/4 (It works) INSUFFICIENT! Combine: eq in Statement 1 and 2 works for x>y Answer: C _________________ Impossible is nothing to God. Verbal Forum Moderator Status: Preparing for the another shot...! Joined: 03 Feb 2011 Posts: 1427 Location: India Concentration: Finance, Marketing GPA: 3.75 Followers: 108 Kudos [?]: 491 [0], given: 62 Re: DS Inequality [#permalink] 05 Jun 2013, 07:48 Expert's post Manhnip wrote: Is x > y? (1) √ x > y (2) x^3 > y Please provide an easy method to solve this problem in less than 2 mins The number properties change when the numbers move beyond the range of 0 and 1. Draw a number line and try to check the values of \sqrt{x} and x^3. Both the statements explain only about one range, but as soon as one combines both the statements, the range becomes quite clear and hence C becomes the answer. _________________ VP Status: I'm back and not stopping until I hit 760+ Joined: 06 Sep 2013 Posts: 1345 Location: United States Concentration: Finance, General Management Schools: Wharton '17 GPA: 3.5 WE: Corporate Finance (Investment Banking) Followers: 7 Kudos [?]: 87 [0], given: 177 Re: Is x > y ? [#permalink] 27 Dec 2013, 08:42 gurpreetsingh wrote: Statement 1: \sqrt{x} > y ....take x = 4, y = 1 (yes); take x = 1/4 and y = 1/3 (no) hence not sufficient. Statement 2: x^3 > y take x = 2 and y = 3 (no) , take x = 2, y = 1 (yes) hence not sufficient. take statement 1 and 2 together. Now the answer could be either C or E. Either y is -ve or postive. If y is -ve then x >y always holds true. If y is +ve then x>y^2 and x^3>y => x^6 > y^2 divide both the inequalities: x^5 >1 => x >1. Since square root of x (> 1) is greater than y => x>y. Hence C. The graphical approach is good, but its essential to understand that how the inequalities behaves in different domains. Yeah kinda tough indeed Hey gupreet or anyone how did you get to this? If y is +ve then x>y^2 and x^3>y => x^6 > y^2 Like from both statements to the x^6 > y^2? Thanks Cheers! J Kudos rain! Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4178 Location: Pune, India Followers: 895 Kudos [?]: 3795 [0], given: 148 Re: Is x > y ? [#permalink] 01 Jan 2014, 22:24 Expert's post jlgdr wrote: gurpreetsingh wrote: Statement 1: \sqrt{x} > y ....take x = 4, y = 1 (yes); take x = 1/4 and y = 1/3 (no) hence not sufficient. Statement 2: x^3 > y take x = 2 and y = 3 (no) , take x = 2, y = 1 (yes) hence not sufficient. take statement 1 and 2 together. Now the answer could be either C or E. Either y is -ve or postive. If y is -ve then x >y always holds true. If y is +ve then x>y^2 and x^3>y => x^6 > y^2 divide both the inequalities: x^5 >1 => x >1. Since square root of x (> 1) is greater than y => x>y. Hence C. The graphical approach is good, but its essential to understand that how the inequalities behaves in different domains. Yeah kinda tough indeed Hey gupreet or anyone how did you get to this? If y is +ve then x>y^2 and x^3>y => x^6 > y^2 Like from both statements to the x^6 > y^2? Thanks Cheers! J Kudos rain! If y is positive and x^3>y, it means x^3 is also positive (since it is greater than y which is positive). If both sides of an inequality are positive, you can square the inequality. x^3>y (x^3)^2>y^2 x^6 > y^2 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Is x > y? (1): sqrt(x) > y, (2): x^3 > y [#permalink] 19 Mar 2014, 14:25
Is x > y?
(1) sqrt(x) > y
(2) x^3 > y
Hi, I want to know if we can solving this without picking numbers, please.
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Re: Is x > y? (1): sqrt(x) > y, (2): x^3 > y [#permalink] 20 Mar 2014, 00:37
Expert's post
goodyear2013 wrote:
Is x > y?
(1) sqrt(x) > y
(2) x^3 > y
Hi, I want to know if we can solving this without picking numbers, please.
Merging similar topics. Please refer to the solutions on pages 1 and 2.
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] 18 Apr 2014, 03:04
shrouded1 wrote:
I thought this was a really tough question !
Is x > y ?
(1) \sqrt{x} > y
(2) x^3 > y
Statement I is insufficient:
x = 4 and y = 3 then x is greater than y
x = 1/4 and y = 1/3 then x is not greater than y
Statement II is insufficient:
x = 4 and y = 2 then x is greater than y
x = -1/4 and y = -1/2 then x is not greater than y
Combining both we can clearly see that:
x^1/2 > y
x^3 > y
If we plug in positive fractions or positive numbers each time x is greater than y. Hence answer is C.
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] 18 Apr 2014, 03:04
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Display posts from previous: Sort by | 2014-04-19 22:37:25 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.7346596121788025, "perplexity": 3196.2560354882794}, "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-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00554-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://www.intmath.com/blog/mathematics/sine-wave-in-the-sky-over-singapore-994 | # Sine wave in the sky over Singapore
By Murray Bourne, 23 Feb 2008
I had a friend who was heavily into gliding. He would become very excited when he observed "wave" in the sky.
Gliders (aka sailplanes) don't have engines, of course, so you spend all your time looking for lift.
[Image source: Boeing]
The lift comes from thermals (rising hot air, usually over dark-colored ground) or waves that are the result of wind blowing over mountains. When airflow is lifted by a mountain range, it is called 'orographic uplift'.
[Image source, from Natural History of the White-Inyo Range, Eastern California.]
So in gliding, "wave" is a standing sine curve on the leeward side of a mountain.
I saw the following wave pattern over Singapore recently. The clouds are much higher than where gliders can reach, but there is evidence of alternating rise and fall of the moist airmass. In this case, the wave could be the result of air passing over mountains in Indonesia (there are no mountains in Singapore).
[Photo: LimPH].
### 4 Comments on “Sine wave in the sky over Singapore”
1. Johan de Nijs says:
My nephew Harald Boetz, who lives in The Hague Holland, like to sail plane in Germany around the high mountains. He knows about the rising winds and how to use them. Here where I live at the North Carolina coast in USA I always observe with great interest how the eagles and other big birds can just sail de sky, without flying, for hours. Nature has that all figured out.
2. Murray says:
Yes, Johan, and the carbon footprint of that sail plane (and of the eagles) is very low.
Not so with your average jet...
3. Johan de Nijs says:
I suddenly remember. We lived on the sloop of the Tangubanprau north of Bandung. age 13 (1937). We had an original English tennis court. I loved to fly kites. Also to make all kinds of little airplains out of paper. Often the airplanes would take to the air and rose and rose and flew by themself sometimes as long as ten to 15 minutes. The wind striking over Bandung (Preanger valley) Would rise up the mountain slope and caused that feat. The wind touching you, however, would not give that indication. Flying kites in Indonesia means fighting kites. One would fly ones kite far away over a village and then the kite gets attacked by several enemy kites. Very exiting. Complicated manoeuvring and trying to cut the enemies glass coated line with ones own. Good memories.
4. jennifer62 says:
I had a friend who was heavily into gliding. He would become very excited when he observed
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NOTE: You can't mix both types of math entry in your comment. | 2016-09-29 22:04:01 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5746580958366394, "perplexity": 5779.085161056081}, "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-2016-40/segments/1474738661953.95/warc/CC-MAIN-20160924173741-00205-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://socratic.org/questions/a-student-drops-a-0-011-kg-super-ball-from-a-height-of-1-m-if-the-ball-returns-t#558483 | # A student drops a 0.011 kg super ball from a height of 1 m. If the ball returns to a height of .75 m, how much mechanical energy was lost to dissipative forces?
mg$\Delta$h | 2022-08-16 01:15:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3873167634010315, "perplexity": 604.5662185582989}, "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-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00674.warc.gz"} |
https://physics.aps.org/synopsis-for/10.1103/PhysRevX.6.041054 | # Synopsis: Chemical Echo
A set of over 1000 tiny, parallel chemical reactions demonstrates the first example of an echo phenomenon in a chemical system.
Echoes are not limited to sound reflecting off cave walls. A similar phenomenon—a delayed response following an immediate response to some stimulus—can occur after coupled oscillators are stimulated by a sequence of two input pulses. Researchers have now observed such an echo phenomenon in a system of coupled chemical oscillators.
Edward Ott of the University of Maryland in College Park, Kenneth Showalter of West Virginia University in Morgantown, and their colleagues studied a standard oscillating chemical system known as the Belousov-Zhabotinsky reaction. This light sensitive reaction involves transitions between an opaque state and one that transmits light. The team fixed more than 1000 tiny beads containing the relevant chemicals in a setup that allowed the beads—each with its own oscillation frequency—to be individually illuminated and the light transmitted through each bead to be separately detected.
A strong light pulse synchronized the chemical reactions in beads that were past a certain point in their oscillation cycle. The degree of synchronization of the reaction, quantified by an order parameter, decayed rapidly and then rose when a second pulse was applied a time $𝜏$ later (typically several hundred seconds). After an additional time $𝜏$, a smaller increase in the order parameter (the echo) appeared without any pulse stimulation.
The team’s simulations show that the oscillators re-synch to create the echo. The phase relationships among oscillators, which are maintained even when the oscillators are completely out of synch, are critical to seeing an echo. The researchers suggest that echoes could be used to probe the underlying oscillatory cells in colonies of yeast or bacteria.
This research is published in Physical Review X.
–David Ehrenstein
David Ehrenstein is the Focus Editor for Physics.
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Vulnerabilities in a city’s public transport system are identified through a network analysis that accounts for the number of passengers and vehicles at any given time. Read More »
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### Synopsis: Straying from the Norm in Pedestrian Movements
Experiments tracking people as they walk down a corridor reveal universal behaviors that, if incorporated into models, could ensure safe flow in large crowds. Read More » | 2017-11-20 20:58:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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": 0.36152568459510803, "perplexity": 3008.69790620313}, "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-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00113.warc.gz"} |
http://mathhelpforum.com/differential-geometry/112182-sequence-intervals.html | Math Help - Sequence of intervals
1. Sequence of intervals
Hi all,
I'm having a spot of bother on the following question; I simply do not know where to begin. Any help would be massively appreciated.
Cheers,
Kef
2. assume there are two such points such that $p,q\in J_n$ for all n then show that these two points will be the same. it is not difficult since $x_n\rightarrow 0 \text{ as } n\rightarrow\infty$
you can also show that there has to be at least one point and by the previously mentioned part there can then be only one
3. I see, so I should go for the proof by contradiction method? Could you start me off on how to show that the points are the same? I'm having a lot of difficulty with this one.
4. Originally Posted by kef
I see, so I should go for the proof by contradiction method? Could you start me off on how to show that the points are the same? I'm having a lot of difficulty with this one.
By all means you can use contradiction.
But, first of all how do you know the intersection is not empty?
Do you know that monotone bounded sequences converge?
Then how could there be two points in the intersection?
5. Sorry, I don't quite follow...
6. Originally Posted by kef
Sorry, I don't quite follow...
Can you prove that $(b_n)$ is a decreasing sequence?
Can you prove that $(a_n)$ is a increasing sequence?
Can you prove that $\left( {\forall n} \right)\left( {\forall m} \right)\left[ {a_n < b_m } \right]$?
Does that mean that both sequences converge? WHY?
7. Thanks a lot for the help guys, it's done now. Took me a while to understand what the question was actually asking.
Cheers. | 2015-05-06 18:42:00 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.7743853330612183, "perplexity": 214.75204744267043}, "config": {"markdown_headings": false, "markdown_code": false, "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-2015-18/segments/1430459005968.79/warc/CC-MAIN-20150501054325-00035-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/following-combustion-reaction-octane-c8h18-l-125-o2-g-8-co2-g-9-h2o-g-using-following-enth-q886651 | The following is the combustion reaction of octane.
C8H18 (l) + 12.5 O2 (g) -> 8 CO2 (g) + 9 H2O (g)
a) Using the following enthalpies of formation, find the enthalpy change for this
combustion reaction.
?Hf(C8H18 (l)) = -249.95 kJ/mol
?Hf(CO2 (g)) = -393.52 kJ/mol
?Hf(H2O (l)) = -285,82 kJ/mol
?Hf(H2O (g)) = -241.82 kJ/mol
b) Assuming that the reaction is carried out at constant pressure, find q for the
combustion of 10 g of octane. (MMc = 12.01 g/mol, MMH = 1.00 g/mol).
c) How much work will have to be done on the system in b) so that ?E = 0?
d) If the reaction in b) is carried out in a calorimeter (Ccal = 6.43 J/°C), what would be the temperature change of the calorimeter? | 2014-09-03 07:13:54 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8002703189849854, "perplexity": 3810.8523130891263}, "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-2014-35/segments/1409535925433.20/warc/CC-MAIN-20140901014525-00133-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://stats.stackexchange.com/questions/461375/fourier-analysis-to-retrieve-components-of-individual-spectra | # Fourier analysis to retrieve components of individual spectra
I have a basic, simple question, I am a physics student, and searching internet gives me a lot of signal processing theory but couldn't find this basic answer, which I plan to implement in my speech detection algorithm (if the answer to this question is yes) so :
Imagine I have two sine waves :
G1 = A1*sin(t)
G2 = A2*sin(t+dt)
We dont know about this two waves however. What we know is a signal superposition of this two waves measured in terms of amplitude over time.
So we can have a measurement data like :
time Amplitude
t1. a1
t2. a2
. .
. .
tn an
Now, We know using Fourier transformation, we con obtain the frequency of this individual components, but I have another question, is it possible to retrieve the amplitudes and the phase shift A1, A2, dt of the individual input components ? So we can reconstruct the individual input signals fully ?
As I understand, you observe $$G_t = A_1\sin(t) + A_2\sin(t+dt)$$. Assume first that you know $$dt$$, the time delay of the second signal. If you have observations over a stretch of time $$T$$ you could multiply $$G$$ by $$\sin(t)$$ and $$\sin(t+dt)$$ and add over time to obtain: $$\sum_tG_t\sin(t) = A_1\sum_t\sin(t)^2 + A_2\sum_t\sin(t)\sin(t+dt)$$ $$\sum_tG_t\sin(t+dt) = A_1\sum_t\sin(t+dt)\sin(t) + A_2\sum_t\sin(t+dt)^2$$ This yields a system of two equations which you can solve for $$A_1$$ and $$A_2$$. It is clear that if $$dt$$ is small, the system above is very nearly singular, and your chances of singling out both signals are slim.
If you do not know the time delay, you could repeat the procedure above for different tentative values of $$dt$$, solve for $$A_1$$, $$A_2$$ in each case, and retain the solution which gives the best fit to $$G_t$$, i.e., the value of $$dt$$ for which $$\sum_t(G_t - A_1\sin(t) - A_2\sin(t+dt))^2$$ is minimum. | 2021-04-17 06:27:24 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6699067950248718, "perplexity": 248.2004638223173}, "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-2021-17/segments/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00459.warc.gz"} |
https://www.geteasysolution.com/-8c+10=-8c-4 | # -8c+10=-8c-4
## Simple and best practice solution for -8c+10=-8c-4 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.
If it's not what You are looking for type in the equation solver your own equation and let us solve it.
## Solution for -8c+10=-8c-4 equation:
Simplifying
-8c + 10 = -8c + -4
Reorder the terms:
10 + -8c = -8c + -4
Reorder the terms:
10 + -8c = -4 + -8c
Add '8c' to each side of the equation.
10 + -8c + 8c = -4 + -8c + 8c
Combine like terms: -8c + 8c = 0
10 + 0 = -4 + -8c + 8c
10 = -4 + -8c + 8c
Combine like terms: -8c + 8c = 0
10 = -4 + 0
10 = -4
Solving
10 = -4
Couldn't find a variable to solve for.
This equation is invalid, the left and right sides are not equal, therefore there is no solution.` | 2018-11-14 22:47:04 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24920758605003357, "perplexity": 1225.0887274656952}, "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-2018-47/segments/1542039742316.5/warc/CC-MAIN-20181114211915-20181114233915-00214.warc.gz"} |
https://agentfoundations.org/item?id=1360 | by Stuart Armstrong 568 days ago | Scott Garrabrant likes this | link | parent Can you argue that $$X$$ must have a semi-metric compatible with the topology by using $$d(x,y)=sup_{z\in X}|h(x,z)-h(y,z)|$$? I’m wondering if you can generalise this to some sort of argument that goes like this. Using X, project down via $$\pi$$ from $$X^0=X$$ to $$X^1=X^0/d$$. Let $$\phi$$ be our initial surjection; it’s now a bijection between $$X^1$$ and maps from $$X^0$$ to $$[0,1]$$. If the projection is continuous, then every map from $$X^1$$ to $$[0,1]$$ lifts to a map from $$X^0$$ to $$[0,1]$$. Restricting to the subset of maps that are lifts like this, and applying $$\phi^{-1}$$, gives a subset $$X^2 \subset X^1$$. We now have a new equivalence relationship, maps from $$X^1$$ that are equal to each other on $$X^2$$. Project down from $$X^2$$ by this relationship, to generate $$X^3$$. Continue this transfinitely often (?) to generate a space $$X'$$ where $$\phi$$ is a homeomorphism, and find a contradiction? This feels dubious, but maybe worth mentioning…
by Alex Mennen 567 days ago | Scott Garrabrant and Stuart Armstrong like this | link I haven’t checked that argument carefully, but that sounds like it should give you $$X'$$ with a continuous bijection $$\phi:X'\rightarrow[0,1]^{X'}$$, which might not necessarily be a homeomorphism. reply
by Stuart Armstrong 567 days ago | link Yes, you’re right. reply
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https://maker.pro/forums/threads/improving-phase-noise.18655/ | Login Join Maker Pro
# improving phase noise
R
#### Rene Tschaggelar
Jan 1, 1970
0
I found it rather hard to reach 300MHz or so from
20MHz with low phase noise. A singlechip PLL appears
not to be able to do better than 120dBc/Hz through
to 100kHz
I haven't checked on a discrete solution yet.
That means, I'll have to find an odd divider first.
Is there a way to improve the phase noise performance
of a PLL ? Perhaps with a resonator ?
Rene
F
#### Fred Bloggs
Jan 1, 1970
0
Rene said:
I found it rather hard to reach 300MHz or so from
20MHz with low phase noise. A singlechip PLL appears
not to be able to do better than 120dBc/Hz through
to 100kHz
I haven't checked on a discrete solution yet.
That means, I'll have to find an odd divider first.
Is there a way to improve the phase noise performance
of a PLL ? Perhaps with a resonator ?
Rene
No- the next step is up, and this will be only 20-30dB improvement, is
to move to what is called "phase-locked frequency source"- see MITEQ for
some examples. The 20MHz is not a common reference and will be divided
by 2-4 to work with most standard product lines.
M
#### Mike
Jan 1, 1970
0
I found it rather hard to reach 300MHz or so from
20MHz with low phase noise. A singlechip PLL appears
not to be able to do better than 120dBc/Hz through
to 100kHz
I haven't checked on a discrete solution yet.
That means, I'll have to find an odd divider first.
Is there a way to improve the phase noise performance
of a PLL ? Perhaps with a resonator ?
Rene
Quite possibly - it depends on what your starting point is.
Resonator-based VCO's typically have better noise performance than
non-resonator VCO's. Increasing your damping factor may help - if your
damping factor is too low, peaking in the frequency response will increase
the phase noise. Depending on the dominant noise source, increasing or
decreasing the PLL bandwidth may reduce noise further. If you're using an
active loop filter, you will probably get lower noise by converting to a
passive filter.
A tool like Mathcad can be invaluable for problems like this. The PLL loop
can be modeled relatively easily, and the effects of changes to the loop
can be viewed almost instantly.
-- Mike --
R
#### Rene Tschaggelar
Jan 1, 1970
0
Mike said:
Quite possibly - it depends on what your starting point is.
Resonator-based VCO's typically have better noise performance than
non-resonator VCO's. Increasing your damping factor may help - if your
damping factor is too low, peaking in the frequency response will increase
the phase noise. Depending on the dominant noise source, increasing or
decreasing the PLL bandwidth may reduce noise further. If you're using an
active loop filter, you will probably get lower noise by converting to a
passive filter.
A tool like Mathcad can be invaluable for problems like this. The PLL loop
can be modeled relatively easily, and the effects of changes to the loop
can be viewed almost instantly.
I had a look at SimPLL from AnalogDevices. It just supports their
PLLs though. With one of theirs I wasn't able to get better than
-120 dBc/Hz.
On my own, I'd have a tough time in simulating my own PLL.
How does a divider come in ?
I tend to doubt rms jitter numbers of 3ps, they are likely not
able to measure it then.
Rene
M
#### Mike
Jan 1, 1970
0
I had a look at SimPLL from AnalogDevices. It just supports their
PLLs though. With one of theirs I wasn't able to get better than
-120 dBc/Hz.
On my own, I'd have a tough time in simulating my own PLL.
How does a divider come in ?
I tend to doubt rms jitter numbers of 3ps, they are likely not
able to measure it then.
Rene
Measuring 3ps RMS isn't difficult - the OC-48 parts I've designed (at
2.488GHz) come in at around 1.5ps RMS. We measure them with a Tek
Communications Signal Analyzer. Even so, my noise is over 120dBc/Hz until
around 15MHz (it goes below 110dBc at roughly 10kHz).
The divider introduces a gain factor in the loop, and effectively
multiplies the noise from your reference (imagine that the loop keeps the
noise at the summing node constant - if that's the case, then the noise at
the VCO side of the divider must be scaled by the divider ratio).
To model the divider, consider the effect it has on the noise in the time
domain. Assume the noise is white, and in the time domain it causes t
seconds of jitter (RMS). The equivalent phase jitter is 2*pi*t/T, where T
is the signal period. At the divider output, the time domain jitter is
unchanged - it is still t seconds RMS, assuming the divider doesn't
contribute significant amounts of noise. However, since the period of the
divider output is M*T, where M is the divider ratio, the equivalent phase
noise at the divider output is t/M. In the ideal PLL loop equations, a
divide by M becomes a 1/M scale factor.
-- Mike --
M
#### Mike
Jan 1, 1970
0
However, since the period of the > divider output is M*T, where M is
the divider ratio, the equivalent phase noise at the divider output
is t/M.
Not quite. What I meant to say was that it's either p/M, where p is the
phase noise, or 2*pi*t/(M*T). The effect of the divider is to reduce the
phase noise by a factor of M.
-- Mike --
M
#### maxfoo
Jan 1, 1970
0
I found it rather hard to reach 300MHz or so from
20MHz with low phase noise. A singlechip PLL appears
not to be able to do better than 120dBc/Hz through
to 100kHz
I haven't checked on a discrete solution yet.
That means, I'll have to find an odd divider first.
Is there a way to improve the phase noise performance
of a PLL ? Perhaps with a resonator ?
Rene
Try a better phase detector with phase noise of -153dbc/hz.
http://www.hittite.com/product_info/product_specs/vcosprescaler/dividersdetectors/hmc439qs16g.pdf
R
#### Rene Tschaggelar
Jan 1, 1970
0
No- the next step is up, and this will be only 20-30dB improvement, is
to move to what is called "phase-locked frequency source"- see MITEQ for
some examples. The 20MHz is not a common reference and will be divided
by 2-4 to work with most standard product lines.
Regarding the 20MHz, it is rather hard to get anything at all.
At any frequency almost.
While the manufacturers *can* do almost everything, there
is hardly anything on the shelves. And when you only need a few,
most are not interested. Well, I found a part that looked good.
I consider this 20MHz, 8ppb from 0 to 70degC, with -120dBc/Hz @10Hz
and -145dBc/Hz @10kHz for approx 350$a seldom chance for a reasonable price. Now I should go from this one to 300MHz fixed. Divide by 15 is a bit odd, yes. And that should cost me 12dB then. The PLL should flatten the lower frequency part down though. Rene R #### Rene Tschaggelar Jan 1, 1970 0 Mike said: Not quite. What I meant to say was that it's either p/M, where p is the phase noise, or 2*pi*t/(M*T). The effect of the divider is to reduce the phase noise by a factor of M. You mean by dividing the phase noise is reduced. This means, since the control loop of the PLL works at the lower frequency, that the noise there is M times smaller than at the M times higher frequency. That's what the PLL tools tell me too: 10 times the frequency makes 10db more noise. Thats what made me wonder too : a Miteq low-noise 15GHz source has phase noise at -70dBc. A wideband VCO in a PLL appears not to be it. I may have to consider a two oscillator approach : Use a VCXO at 300MHz and lock it to the OCXO at 20MHz. Rene M #### maxfoo Jan 1, 1970 0 Not quite. What I meant to say was that it's either p/M, where p is the phase noise, or 2*pi*t/(M*T). The effect of the divider is to reduce the phase noise by a factor of M. -- Mike -- The phase noise will be reduced by 10*log(N) if the divider is on the Ref of the detector and increase if its on the N inside the loop. M #### maxfoo Jan 1, 1970 0 The phase noise will be reduced by 10*log(N) if the divider is on the Ref of the detector and increase if its on the N inside the loop. oops... should be 20*log(N) in dB, where N is the Division Ratio of the divider. D #### ddwyer Jan 1, 1970 0 Rene said: Regarding the 20MHz, it is rather hard to get anything at all. At any frequency almost. While the manufacturers *can* do almost everything, there is hardly anything on the shelves. And when you only need a few, most are not interested. Well, I found a part that looked good. I consider this 20MHz, 8ppb from 0 to 70degC, with -120dBc/Hz @10Hz and -145dBc/Hz @10kHz for approx 350$ a seldom chance for a reasonable
price.
Now I should go from this one to 300MHz fixed. Divide by 15 is a bit
odd, yes. And that should cost me 12dB then.
The PLL should flatten the lower frequency part down though.
Rene
I bought a bunch of the best crystal SC OCXOs listed as $500+ at 16.364MHZ note the SC cut can be driven harder so the design gives - 170dbc floor. CAN BE USED TO DERIVE 1Khz reference; for sale at a very modest price. Reply to [email protected] and avoid my spam congestion. F #### Fred Bloggs Jan 1, 1970 0 Rene said: Regarding the 20MHz, it is rather hard to get anything at all. At any frequency almost. While the manufacturers *can* do almost everything, there is hardly anything on the shelves. And when you only need a few, most are not interested. Well, I found a part that looked good. I consider this 20MHz, 8ppb from 0 to 70degC, with -120dBc/Hz @10Hz and -145dBc/Hz @10kHz for approx 350$ a seldom chance for a reasonable
price.
Now I should go from this one to 300MHz fixed. Divide by 15 is a bit
odd, yes. And that should cost me 12dB then.
The PLL should flatten the lower frequency part down though.
Rene
Take a look at the "jitter cleaner" programmable PLL's/buffers such as
the CDC7005 from Texas Instruments in SiGe with fundamental time jitter
of 0.5fs- http://focus.ti.com/lit/ds/symlink/cdc7005.pdf .
R
#### Rene Tschaggelar
Jan 1, 1970
0
Fred said:
Take a look at the "jitter cleaner" programmable PLL's/buffers such as
the CDC7005 from Texas Instruments in SiGe with fundamental time jitter
of 0.5fs- http://focus.ti.com/lit/ds/symlink/cdc7005.pdf .
Thanks for the hint. I had a look at it.
Another paper, the SCAA067 at :
http://focus.ti.com/docs/apps/catalog/resources/appnoteabstract.jhtml?abstractName=scaa067
shows the noise. The input clocks can be considered lousy though.
The shown graphs with -60dBc @10Hz and -100dBc @100Hz both @245MHz are
way above what I expect. But perhaps with a better reference ...
I might have a closer look at it.
Rene
I
#### Ian Buckner
Jan 1, 1970
0
Rene Tschaggelar said:
You mean by dividing the phase noise is reduced. This means,
since the control loop of the PLL works at the lower frequency,
that the noise there is M times smaller than at the M times higher
frequency.
That's what the PLL tools tell me too: 10 times the frequency
makes 10db more noise. Thats what made me wonder too : a Miteq
low-noise 15GHz source has phase noise at -70dBc.
A wideband VCO in a PLL appears not to be it.
I may have to consider a two oscillator approach :
Use a VCXO at 300MHz and lock it to the OCXO at 20MHz.
Rene
Hi Rene:
Does it need to be exactly 300MHz? Could you use, say, one of the
Sonet
frequencies divided to 311MHz?
You may find a SAW filter at a convenient frequency that you could use
in a VCO, for example 150MHz and a doubler.
Regards
Ian
R
#### Rene Tschaggelar
Jan 1, 1970
0
Ian said:
Hi Rene:
Does it need to be exactly 300MHz? Could you use, say, one of the
Sonet
frequencies divided to 311MHz?
You may find a SAW filter at a convenient frequency that you could use
in a VCO, for example 150MHz and a doubler.
Thanks Ian,
I should have 300MHz or somewhat below. They are used for the AD9854
DDS synth, to achieve very low phase noise over a wide frequency range.
Upon studing the appnote to the mentioned CDC7005 chip, I found
that a PLL (also the CDC7005) takes the stability of the reference
and the phase noise of the VCO. This means I'd now just need a low noise
VCO, or rather VCXO, this time at an integer multiple of the 20MHz,
meaning 260, 280 or 300MHz
Rene
I
#### Ian Buckner
Jan 1, 1970
0
Rene Tschaggelar said:
Ian Buckner wrote:
Thanks Ian,
I should have 300MHz or somewhat below. They are used for the AD9854
DDS synth, to achieve very low phase noise over a wide frequency range.
Upon studing the appnote to the mentioned CDC7005 chip, I found
that a PLL (also the CDC7005) takes the stability of the reference
and the phase noise of the VCO. This means I'd now just need a low noise
VCO, or rather VCXO, this time at an integer multiple of the 20MHz,
meaning 260, 280 or 300MHz
Rene
There are standard narrow band SAW filters around 280MHz that look
like
they could be used for this.
The phase noise out of a DDS is proportional to the ratio of the
output
frequency to the clock frequency.
For example, if your master clock frequency source had a phase noise
of
-120dBc at 10kHz offset at 280MHz (I would expect a SAW based VCO to
be better than
this), the phase noise at 10kHz offset on a 28MHz signal from the DDS
would be
-140dBc, unless other limitations took effect. Spur levels, for
example, will be much
higher than this.
Regards
Ian
M
#### maxfoo
Jan 1, 1970
0
the better way to do it is use a 3ghz dro then divide by 10 to get a
20dB phase noise improvement.
the dro has a pn of -140dbc/hz @10khz with the divide by 10
thats -160dbc/hz.....
the dds pn is 135-140dbc/hz...iirc
J
#### John Miles
Jan 1, 1970
0
the better way to do it is use a 3ghz dro then divide by 10 to get a
20dB phase noise improvement.
the dro has a pn of -140dbc/hz @10khz with the divide by 10
thats -160dbc/hz.....
the dds pn is 135-140dbc/hz...iirc
Who sells DROs this clean? Got any links?
-- jm
R
#### Rene Tschaggelar
Jan 1, 1970
0
Ian said:
There are standard narrow band SAW filters around 280MHz that look
like
they could be used for this.
The phase noise out of a DDS is proportional to the ratio of the
output
frequency to the clock frequency.
For example, if your master clock frequency source had a phase noise
of
-120dBc at 10kHz offset at 280MHz (I would expect a SAW based VCO to
be better than
this), the phase noise at 10kHz offset on a 28MHz signal from the DDS
would be
-140dBc, unless other limitations took effect. Spur levels, for
example, will be much
higher than this.
I've found SAW with bandwidths as small as 70kHz or so.
That'd be the SAW between the cheap VCO and the DDS. That would only
lower the wideband noise, but probably not the sidebands of the VCO.
Rene
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1K | 2022-10-05 12:55:20 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.6433976888656616, "perplexity": 5616.902563811893}, "config": {"markdown_headings": true, "markdown_code": false, "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-2022-40/segments/1664030337625.5/warc/CC-MAIN-20221005105356-20221005135356-00389.warc.gz"} |
http://stats.stackexchange.com/questions/28354/numeric-methods-to-estimate-curve-parameter-from-2-input-data-with-gaussian-nois | # Numeric methods to estimate curve parameter from 2 input data with Gaussian noise
I have:
• a function of the form
$S=M f(t, \theta)$
where $\theta$ is a variable, $t$ and $M$ are parameters.
• two observations
$(\theta_1, S_1)$ and $(\theta_2, S_2)$
in which $S_1$ and $S_2$ are observed values for $t$, and $\theta_1$ and $\theta_2$ are the observed values of $\theta$.
Furthermore, we know that my $S_1$, $S_2$ contains noise, which are normal distribution $N(0, d_1)$ $N(0,d_2)$ with unknown variance $d_1$, $d_2$.
Now I want to estimate $t$.
Mathematically, if $S_1$ and $S_2$ are observed precisely, we have
$$S_1= M f( t, \theta_1)$$ $$S_2= M f( t, \theta_2)$$
Then, by division we have $\frac{S_1}{S_2}=g(t,\theta_1,\theta_2)$ following which the parameter $t$ can be estimated by a Mathematica tool like Findroot w.r.t. the parameter $t$.
However, the problem here is the noise of normal distribution. The goal is to find a $t$ as precise as possible, in the sense of some probability-theoretic characteristics. The preferred way is to use a Matlab or Mathematica function, although I don't think they have a direct solution.
I am totally new to numeric analysis, and totally new to Matlab. My naive idea is that:
1. Should the first step be removing the parameter $M$?
2. If so, there are infinitely many ways to do that, among which we can list
• $\frac{S_1}{S_2} = \frac{f(t, \theta_1)}{f(t,\theta_2)}$
• $\frac{S_1+S_2}{S_1-S_2} = \frac{f(t, \theta_1)+f(t,\theta_2)}{f(t, \theta_1)-f(t,\theta_2)}$
• $\frac{S_1^2 }{S_2^2} = \frac{f(t, \theta_1) ^2 }{f(t,\theta_2) ^2}$
It is clear that all these variations are equivalent mathematically if $S_1$, $S_2$ are observed precisely. However, they should yield different $t$ (because of the Gaussian noise in $S_1$ and $S_2$)
My questions are
1. Do you think my first step should be to remove $M$ using the foregoing approach?
2. Is there some kind of combination of $S_1$, $S_2$ that gives better results than others?
3. What would be the best way to estimate $t$ of this problem? (Maybe in Matlab, or Mathematica)
Thank you. I hope I was clear.
-
I don't want to edit your question but it is a little confusing. since you are talking to statisticians. You should say that S1 and S2 are observed values for t and theta1 and theta2 are the observed values of theta. You misuse the term precise. To a statistician precise means small variability where I am pretty sure you mean to say that S1 and S2 are observed exactly (without any noise). You also should not say that after division t and be calculated rather you should say it can be estimated. – Michael Chernick May 12 '12 at 17:19
Also English can be a confusing language with plurals, the plural of dog is dogs but the plural of noise is noise not noises. – Michael Chernick May 12 '12 at 17:20
Nice remarks. Thank you. – zell May 12 '12 at 18:02
A mistake in my comment that I am too late to edit was saying that S1 and S2 were sample values of t. – Michael Chernick May 12 '12 at 18:10 | 2013-05-21 07:20:38 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6391332745552063, "perplexity": 361.1359257185317}, "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-2013-20/segments/1368699776315/warc/CC-MAIN-20130516102256-00002-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/79353/what-is-the-simplest-way-to-represent-a-d-5-singularity/79439 | ## What is the simplest way to represent a $D_5$ singularity?
Consider this curve $f(x,y)=0$ given by
$$f(x,y) := y^3 + y^2 x + x^4 =0.$$ Is it obvious that after a change of coordinates near the origin, this curve is equivalent to
$$\hat{y}^2 \hat{x} + \hat{x}^4 = 0$$
I think, these are both $D_5$ singularities. It seems like the change of coordinates that would achieve this is of the form
$$x = \hat{x} - y + c_2 y^2 + c_3 y^3 + \ldots$$
where $x$ is an infinite power series. We can kill off the coefficients of $y^n$, for all $n$. This would give us a factor of $\hat{x}$, i.e we get something of the form $$f = \hat{x}\cdot g$$ And then we can make another change of coordinate, so that $g$ becomes $$g = \hat{y}^2 + \hat{x}^3.$$ Is there a simpler way to prove this? And aside from proving the power series converges, is there anything missing in the proof? Everything is over the complex numbers.
-
Ritwik -- Maybe you should talk to Patricio. He lectured about related things in the student seminar a couple of weeks ago. – Jason Starr Oct 28 2011 at 15:14
For classifying plane curves singularities, the "coordinate approach" is not always the better one*. For your question, the general case is in table 1, page 3, C.T.C Wall article" "sextic curves and quartic surfaces with higher multiplicity". For general methods, I recommend, C.T.C Wall article: "Notes in the classification of singularities"
In general, the singularities $J_{r,i}$ or $E_{r,i}$ (the notation is not uniform) are given by $y^3+y^2x^r+x^{3r+i}$ with $r \geq 1$,and $i \geq 0$. In your case $r=i=1$ and the singularity is actually $D_5$. You can recognized it because it has two branches: one smooth, and another one with an $A_2$ singularity. Those branches separates after one blowing up. (see Table A, from the latest reference), and they are the factors that you see in your calculation. This "branch behavior" is the* definition of the $D_5$ singularity, and the normal form is deduced from it. A detailed discussion is in Barth's book in compact complex surfaces, page 79.
I hope it helps!
Psd: I don't see anything missing in your argument, but it is "simpler" by using $D_5$'s resolution.
*to my knowledge/opinion
- | 2013-06-18 05:32:56 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9075661897659302, "perplexity": 171.35139243967663}, "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-2013-20/segments/1368706933615/warc/CC-MAIN-20130516122213-00019-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://subjectcoach.com/tutorials/math/topic/prekinder-to-grade-2-mathematics/chapter/minus-numbers | # Minus Numbers
## Alice Visits the Snow
Alice the cow has decided that she'd like to learn to ski. So, she heads off to Perisher Valley with her friend, Annabelle. When she gets there, she realises that she's never been so cold in her life. There's all this white stuff around and it's freezing cold to touch. What's going on?
Well, Alice, that white stuff is called snow. We need snow to ski on. It's made out of frozen water, and for water to freeze, it has to be cold.
Water freezes at $0^\circ$ Celsius. For snow to stick around, the temperature has to be less than $0^\circ$ Celsius. Temperatures less than $0^\circ$ Celsius are very cold indeed.
Have a look at the thermometer. Do you notice how the alcohol (the red stuff) is actually below the zero mark? It looks like it's halfway between the zero and the mark that says $-15$, so the temperature must be about $-8^\circ$ Celsius. That's pretty cold!
Did you notice the "minus" in front of the $8$ in that temperature? It's there because we're measuring a temperature below zero. If you look at the thermometer in the picture, you can actually see minus signs in front of the $15$ and the $30$. $-8$, $-15$ and $-30$ are all examples of minus numbers. Let's find out more about them.
## The Numbers Less than Zero
The number line above looks a little bit different from the number lines that we've seen until now. It goes on to the left of zero, and it is labelled with numbers like $-1, -2, -3$ and so on. The left part of the number line looks like a reflection of the right part, except that the numbers all have minus signs in front of them. The numbers on the left hand part of the number line are the minus numbers. Another name for them is the negative numbers.
The minus numbers are the numbers that are less than zero. They might be whole numbers, fractions, or decimals. Some examples are $-8$, $-15$, and $-20$. Other examples are $-12 \dfrac{1}{2}$ and $- 2.7$.
## Uses for Minus Numbers
We've just seen that they're useful for measuring very cold temperatures like the ones that Alice is experiencing at Perisher Valley. They tell us that the temperature is below the temperature at which water freezes, so it must be very cold indeed. Negative numbers have other uses as well.
Have a think about when you do subtraction with trading. You might say something like "$3 - 5$ can't be done, so we trade $10$ with the tens column and find $13 -5 = 8$". Sometimes there isn't a tens column to trade with, but we still need to give an answer. The answer will be a minus number.
For example, suppose you need to give your friend $7$ lollies, but you only have $5$. The number of lollies you have left will be $5 - 7 = -2$, a minus number. This minus number means that you owe your friend $2$ lollies. Next time you have lollies, don't forget to give your friend the $2$ you owe them! Banks use minus numbers to tell people that they owe them money.
### Conclusion
Don't worry too much about the minus numbers at this stage. For now, it's enough to know that they exist, and that they are useful. Alice needs to know that the minus numbers exist, and that they're less than zero so that she can tell how cold it is at Perisher Valley.
Eventually, you'll learn to compare, add, subtract, multiply and divide minus numbers. Arithmetic with minus numbers is much the same as normal arithmetic, but there are some new rules to follow. It's a good idea to get more used to working with the numbers that aren't minus numbers first. We call the numbers that we've been working with until now the positive numbers.
### Description
This mini book covers the core of Math for Foundation, Grade 1 and Grade 2 mathematics including
1. Numbers
3. Subtraction
4. Division
5. Algebra
6. Geometry
7. Data
8. Estimation
9. Probability/Chance
10. Measurement
11. Time
12. Money
13. and much more
This material is provided free of cost for Parent looking for some tricks for their Prekinder, Kinder, Prep, Year 1 and Year 2 children
### Audience
You must be logged in as Student to ask a Question. | 2021-11-29 12:23:15 | {"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": 0.493263304233551, "perplexity": 688.0796818931112}, "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-2021-49/segments/1637964358705.61/warc/CC-MAIN-20211129104236-20211129134236-00039.warc.gz"} |
https://frontistr-commons.gitlab.io/FrontISTR_manual/en/tutorial/tutorial_17.html | # Frequency Response Analysis
## Frequency Response Analysis
This analysis uses the data of tutorial/17_freq_beam. The first step of the analysis is to change the overall control data for eigenvalue analysis, hecmw_ctrl_eigen.dat, to hecmw_ctrl.dat and perform eigenvalue analysis. Further, change the overall control data for frequency response analysis, hecmw_ctrl_freq.dat, to hecmw_ctrl.dat, and the eigenvalue analysis result log file, 0.log, to eigen_0.log (which is specified within the analysis control data for frequency response analysis.) Finally, frequency response analysis is performed.
### Analysis target
The target of this analysis is a cantilever whose shape and mesh data are shown in Figs. 4.17.1 and 4.17.2, respectively. The mesh is a tetrahedral primary element with 126 element and 55 nodes.
Fig. 4.17.1 : Shape of the cantilever
Fig. 4.17.2 : Mesh data of the cantilever
### Analysis content
This is a frequency response analysis in which the edge of the cantilever was fully restrained, and concentrated load was added to two nodes on the opposite edge. After performing eigenvalue analysis up to the tenth order with the same boundary conditions, the frequency response analysis was conducted with the eigenvalues and eigenvectors up to the fifth order. The analysis control data for frequency response analysis is shown below.
# Control File for FISTR
!VERSION
3
!WRITE,RESULT
!WRITE,VISUAL
!SOLUTION, TYPE=DYNAMIC
!DYNAMIC
11 , 2
14000, 16000, 20, 15000.0
0.0, 6.6e-5
1, 1, 0.0, 7.2E-7
10, 2, 1
1, 1, 1, 1, 1, 1
eigen_0.log
1, 5
!BOUNDARY
_PickedSet4, 1, 3, 0.0
_PickedSet5, 2, 1.
_PickedSet6, 2, 1.
!SOLVER,METHOD=CG,PRECOND=1,ITERLOG=NO,TIMELOG=YES
10000, 2
1.0e-8, 1.0, 0.0
### Analysis results
The relationship between frequency and displacement amplitude of the monitoring nodes, specified with analysis control data (nodal number 1) and created with Microsoft Excel, is shown in Fig. 4.17.3. Furthermore, a part of the log files of the analysis results is shown below as numerical data of the analysis.
Fig.4.17.3 Relationship between frequency and displacement amplitude of the monitoring nodes
Rayleigh alpha: 0.0000000000000000
Rayleigh beta: 7.1999999999999999E-007
start mode= 1
end mode= 5
start frequency: 14000.000000000000
end frequency: 16000.000000000000
number of the sampling points 20
monitor nodeid= 1
14100.000000000000 [Hz] : 8.3957630463152688E-002
14100.000000000000 [Hz] : 1 .res
14200.000000000000 [Hz] : 9.1237051959262350E-002
14200.000000000000 [Hz] : 2 .res
14300.000000000000 [Hz] : 9.9610213760033539E-002
14300.000000000000 [Hz] : 3 .res
14400.000000000000 [Hz] : 0.10918495323840580
14400.000000000000 [Hz] : 4 .res
14500.000000000000 [Hz] : 0.11996212788265602
14500.000000000000 [Hz] : 5 .res
14600.000000000000 [Hz] : 0.13169277524043285
14600.000000000000 [Hz] : 6 .res
14700.000000000000 [Hz] : 0.14365135321213662
14700.000000000000 [Hz] : 7 .res
14800.000000000000 [Hz] : 0.15439888482329628
14800.000000000000 [Hz] : 8 .res
14900.000000000000 [Hz] : 0.16182392983620905
14900.000000000000 [Hz] : 9 .res | 2020-09-27 20:04:37 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4338754117488861, "perplexity": 8203.349924352862}, "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-2020-40/segments/1600401578485.67/warc/CC-MAIN-20200927183616-20200927213616-00528.warc.gz"} |
https://www.tutorialspoint.com/choose-the-correct-answer-from-the-given-four-options-in-the-following-questions-the-zeroes-of-the-quadratic-polynomial-x-2-plusk-xplusk-k-0-a-c | # Choose the correct answer from the given four options in the following questions:The zeroes of the quadratic polynomial $x^{2}+k x+k, k 0$,(A) cannot both be positive(B) cannot both be negative(C) are always unequal(D) are always equal
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Given:
Quadratic polynomial $x^{2}+k x+k, k ≠ 0$.
To do:
We have to find the nature of the zeroes.
Solution:
Let $p(x)=x^{2}+k x+k$
Here,
Product of zeroes $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$
$=\frac{k}{1}$
$=k$
The sign is positive it means both the zeroes should have the same sign(both positive or both negative).
Sum of zeroes $=-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$
$=-\frac{k}{1}$
$=-k$
The sign is negative and both have the same sign.
This implies, the zeroes are both negative.
Updated on 10-Oct-2022 13:27:08 | 2022-12-08 03:28:09 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2045048475265503, "perplexity": 7132.140690006299}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2022-49/segments/1669446711232.54/warc/CC-MAIN-20221208014204-20221208044204-00696.warc.gz"} |
https://www.physicsforums.com/threads/ode-help-not-sure-what-method-to-use.303027/ | # ODE help, not sure what method to use
1. Mar 27, 2009
I have the ODE:
$$y^{(4)}+2y''+y = 3 + cos(2x)$$
I believe I can use undetermined coefficients for the particular, but I'm not sure and it isn't working well for me so far, and the homogeneous looks nasty and I'm not sure what to attempt with.
Thanks!
2. Mar 27, 2009
### arildno
Now, let's take the homogenous trial solution, $$y_{h}(x)=Ce^{kx}$$
Thus, the characteristic equation can be written as:
$$k^{4}+2k^{2}+1=0\to(k^{2}+1)^{2}=0\to{k}^{2}+1=0$$
This ought to be readily solvable for two of the roots.
Don't give up even before you had tried! | 2017-08-19 08:49:51 | {"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8654604554176331, "perplexity": 994.3259967450986}, "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-2017-34/segments/1502886105326.6/warc/CC-MAIN-20170819070335-20170819090335-00537.warc.gz"} |
https://www.numerade.com/questions/make-a-careful-sketch-of-the-graph-of-f-and-below-it-sketch-the-graph-of-f-in-the-same-manner-as-i-2/ | 💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!
Make a careful sketch of the graph of $f$ and below it sketch the graph of $f'$ in the same manner as in Exercises 4-11. Can you guess a formula for $f'(x)$ from its graph?$f(x) = e^x$
The slope at 0 appcars to be 1 and the slope at 1 appears 0. Since to be 2.7 . As $x$ decreases, the slope gets closer to the graphs are so similar, we might gacss that $f^{\prime}(x)=e^{x}$.
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this problem for seventeen of the Stuart Calculus eighth edition, Section two point eight make a careful sketch of the graph of F and blew it sketch the graph of every time in the same manner as an exercise is for true love. Can you guess a formula for halftime from its crap? So first, let's plot after of X equals e to the X and the top craft were were sketching Ah, good. The graph of F and looks roughly like this One of the key points is this point zero one, not a word that is the white intercept of the function. Because at X equals zero, it is there is one. Ah, and we see that the limit is expertise needed Infinity a zero and that limit is exported infinities infinity. So this increases. Ah, as a X goes towards infinity. Now I may be a bit difficult, but we're going to attempt to drop the graph of time by using the slopes of the tangent lines that our attention to this function f the easy are one of the easier ones to look at is as X approaches negative infinity. We see that dysfunction that pulls out. Search them. The slopes of the tension line decrease in Stevenage and in value until it becomes almost flat. A sex person. Negative infinity. So we're going to get this Ah, decrease of the slope towards zero. Because the slopes of the tension lines approach zero. As we approach any other affinity arm, we can estimate the slope of the tangent line at X equals zero here and where we end up. If we were to fear that's not exactly the slope of the tension line at X equals zero ends up being exactly one. So there's an interesting result. And what is more interesting is that if we continue to find the slopes of each of the tendon lines as X increases, we will begin. We will continue to plot what is exactly the same graph as F arms. So in reality, what we have just plotted is the functioning of the ex. And this is consistent with our definition for this function, any of the ex because we know that the definition of either the ex is indeed itself even the X and so they're dirt of craft is exactly the same esteem original graph, and it at every point is the slope of the tangent line to dysfunction at every point
Topics
Limits
Derivatives
Anna Marie V.
Campbell University
Kayleah T.
Harvey Mudd College
Caleb E.
Baylor University
Kristen K.
University of Michigan - Ann Arbor
Lectures
Join Bootcamp | 2021-10-25 06:26:35 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.7252746820449829, "perplexity": 769.4572986170655}, "config": {"markdown_headings": false, "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-2021-43/segments/1634323587655.10/warc/CC-MAIN-20211025061300-20211025091300-00267.warc.gz"} |
https://plainmath.net/83798/explain-how-absorption-and-emission-line | # Explain how absorption and emission lines are created in the atom.
Explain how absorption and emission lines are created in the atom.
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An emission line is produced by an atom in an excited energy state. When an electron jumps from a higher energy level to a lower energy level, energy will be released. And that is when absorption lines are produced. In order to go to a lower energy orbit, the electron must lose energy of a certain specific amount. The atom releases the energy in the form of a photon with that particular energy.
If an atom with an electron at the ${E}_{2}$ orbit and wants to get to the lower ${E}_{1}$ energy orbit, it gives off a photon with energy,
$E=hv\phantom{\rule{0ex}{0ex}}={E}_{2}-{E}_{1}$
The electron may reach the ground state in one jump or it may temporarily stop at one or more energy levels on the way, but it cannot stop somewhere between the energy levels.
An absorption line is produced when a photon of just the right energy is absorbed by an atom. When an electron absorbs enough amount of energy, then it will jump to a higher energy orbit. The size of the outward jumps made by the electrons is the same as that of the inward jumps. Therefore, the pattem of absorption lines is the same as the pattem of emission lines.
If an atom with electron in the E, orbit sees a photon with energy ${E}_{photon}={E}_{2}-{E}_{1}$ , the photon is absorbed and electron moves to ${E}_{2}$. | 2022-08-09 12:56:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "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": 0.6062472462654114, "perplexity": 270.1055711540024}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2022-33/segments/1659882570977.50/warc/CC-MAIN-20220809124724-20220809154724-00747.warc.gz"} |
https://py-pde.readthedocs.io/en/latest/packages/pde.pdes.kuramoto_sivashinsky.html | # 4.3.6. pde.pdes.kuramoto_sivashinsky module
The Kardar–Parisi–Zhang (KPZ) equation describing the evolution of an interface
class KuramotoSivashinskyPDE(nu=1, *, noise=0, bc='auto_periodic_neumann', bc_lap=None)[source]
Bases: PDEBase
The Kuramoto-Sivashinsky equation
The mathematical definition is
$\partial_t u = -\nu \nabla^4 u - \nabla^2 u - \frac{1}{2} \left(\nabla h\right)^2 + \eta(\boldsymbol r, t)$
where $$u$$ is the height of the interface in Monge parameterization. The dynamics are governed by the parameters $$\nu$$ , while $$\eta$$ is Gaussian white noise, whose strength is controlled by the noise argument.
Parameters:
diagnostics: Dict[str, Any]
Diagnostic information (available after the PDE has been solved)
Type:
dict
evolution_rate(state, t=0)[source]
evaluate the right hand side of the PDE
Parameters:
• state (ScalarField) – The scalar field describing the concentration distribution
• t (float) – The current time point
Returns:
Scalar field describing the evolution rate of the PDE
Return type:
ScalarField
explicit_time_dependence: bool | None = False
Flag indicating whether the right hand side of the PDE has an explicit time dependence.
Type:
bool
property expression: str
the expression of the right hand side of this PDE
Type:
str | 2023-03-20 15:17:04 | {"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": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4107428193092346, "perplexity": 2298.901247013306}, "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-2023-14/segments/1679296943484.34/warc/CC-MAIN-20230320144934-20230320174934-00493.warc.gz"} |
https://docs.mdanalysis.org/2.0.0-dev0/documentation_pages/analysis/polymer.html | # 4.6.1. Polymer analysis — MDAnalysis.analysis.polymer¶
Author: Richard J. Gowers 2015, 2018 GNU Public License v3
This module contains various commonly used tools in analysing polymers.
class MDAnalysis.analysis.polymer.PersistenceLength(atomgroups, **kwargs)[source]
Calculate the persistence length for polymer chains
The persistence length is the length at which two points on the polymer chain become decorrelated. This is determined by first measuring the autocorrelation ($$C(n)$$) of two bond vectors ($$\mathbf{a}_i, \mathbf{a}_{i + n}$$) separated by $$n$$ bonds
$C(n) = \langle \cos\theta_{i, i+n} \rangle = \langle \mathbf{a_i} \cdot \mathbf{a_{i+n}} \rangle$
An exponential decay is then fitted to this, which yields the persistence length
$C(n) \approx \exp\left( - \frac{n \bar{l_B}}{l_P} \right)$
where $$\bar{l_B}$$ is the average bond length, and $$l_P$$ is the persistence length which is fitted
Parameters: atomgroups (iterable) – List of AtomGroups. Each should represent a single polymer chain, ordered in the correct order. verbose (bool, optional) – Show detailed progress of the calculation if set to True.
results.bond_autocorrelation
the measured bond autocorrelation
Type: numpy.ndarray
results.lb
the average bond length
New in version 2.0.0.
Type: float
lb
Alias to the results.lb.
Deprecated since version 2.0.0: Will be removed in MDAnalysis 3.0.0. Please use results.lb instead.
Type: float
results.x
length of the decorrelation predicted by lp
New in version 2.0.0.
Type: numpy.ndarray
results.lp
calculated persistence length
New in version 2.0.0.
Type: float
lp
Alias to the results.lp.
Deprecated since version 2.0.0: Will be removed in MDAnalysis 3.0.0. Please use results.lp instead.
Type: float
results.fit
the modelled backbone decorrelation predicted by lp
New in version 2.0.0.
Type: numpy.ndarray
fit
Alias to the results.fit.
Deprecated since version 2.0.0: Will be removed in MDAnalysis 3.0.0. Please use results.fit instead.
Type: float
sort_backbone()
for producing the sorted AtomGroup required for input.
Example
from MDAnalysis.tests.datafiles import TRZ_psf, TRZ
import MDAnalysis as mda
from MDAnalysis.analysis import polymer
u = mda.Universe(TRZ_psf, TRZ)
# this system is a pure polymer melt of polyamide,
# so we can select the chains by using the .fragments attribute
chains = u.atoms.fragments
# select only the backbone atoms for each chain
backbones = [chain.select_atoms('not name O* H*') for chain in chains]
# sort the chains, removing any non-backbone atoms
sorted_backbones = [polymer.sort_backbone(bb) for bb in backbones]
persistence_length = polymer.PersistenceLength(sorted_backbones)
# Run the analysis, this will average over all polymer chains
# and all timesteps in trajectory
persistence_length = persistence_length.run()
print(f'The persistence length is: {persistence_length.results.lp}')
# always check the visualisation of this:
persistence_length.plot()
New in version 0.13.0.
Changed in version 0.20.0: The run method now automatically performs the exponential fit
Changed in version 1.0.0: Deprecated PersistenceLength.perform_fit() has now been removed.
Changed in version 2.0.0: Former results are now stored as results.bond_autocorrelation. lb, lp, fit are now stored in a MDAnalysis.analysis.base.Results instance.
plot(ax=None)[source]
Visualise the results and fit
Parameters: ax (matplotlib.Axes, optional) – if provided, the graph is plotted on this axis ax the axis that the graph was plotted on
MDAnalysis.analysis.polymer.fit_exponential_decay(x, y)[source]
Fit a function to an exponential decay
$y = \exp\left(- \frac{x}{a}\right)$
Parameters: y (x,) – The two arrays of data a – The coefficient a for this decay float
Notes
This function assumes that data starts at 1.0 and decays to 0.0
MDAnalysis.analysis.polymer.sort_backbone(backbone)[source]
Rearrange a linear AtomGroup into backbone order
Requires that the backbone has bond information, and that only backbone atoms are provided (ie no side chains or hydrogens).
Parameters: backbone (AtomGroup) – the backbone atoms, not necessarily in order sorted_backbone – backbone in order, so sorted_backbone[i] is bonded to sorted_backbone[i - 1] and sorted_backbone[i + 1] AtomGroup
New in version 0.20.0. | 2022-08-08 03:29:39 | {"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": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3950287401676178, "perplexity": 12864.907752906165}, "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-2022-33/segments/1659882570765.6/warc/CC-MAIN-20220808031623-20220808061623-00404.warc.gz"} |
https://www.hendrik-erz.de/post/goodbye-october | Goodbye, October | Hendrik Erz
# Goodbye, October
For the better part of the last decade, I built all my websites using October CMS. However, due to a change in their policy, that won't work anymore. So I need to migrate all my pages to a new system. After some fiddling around, I settled with Jekyll. In this post I just want to quickly summarise the why, the how, and the next steps.
If you wanted to “be” on the internet around 1995, it was both easy and difficult at the same time. Easy because writing a website was possible using Open Source tools even back then, and since HTML and CSS were much less powerful, it didn’t took ages until you got something nice (compared to those standards). However, it was also difficult because the knowledge of how to take that HTML and put it somewhere where other people could find it was not widely known (but maybe that’s just my impression since I was, like, 5 or 6 years old back then).
## CMS and October
When HTML became more complicated and more people wanted to put their faces out there – that is, when the trend began that people wanted to have their own personal website – they didn’t want to spend too much time writing code and just wanted to blog. So people who knew how to write code came up with solutions for those who didn’t: Content Management Systems, CMS. The first really popular was possibly MySpace. But soon thereafter, stuff like Tumblr and Wordpress gained traction and that’s more or less where we still are today.
Content Management Systems are legion on the internet; counting just the ones I remember from the top of my head, I get to 10. However, depending on what you need a lot of these systems fall out of the equation quickly. If you don’t have a big e-commerce page or some large corporate website, you probably wont need Drupal or Typo3. And if you’re not from Germany you probably never heard of smaller niche-systems such as Contao. Furthermore, if you would like to keep your site lightweight and/or modify the theme, Wordpress is not an option since theming this CMS is quite the task.
After years of fiddling around; even writing my own solutions, I finally settled with one called October CMS. October CMS ticks all the boxes: It’s a system where I can quickly change this or that HTML code so that my website does something it didn’t do previously. Or change some variable in the styling, so that the font looks a tad nicer. But I also don’t want to write my blogposts in literal HTML. And then the images! October CMS was the nice middle ground between customisability and ease of use. In fact, I used October CMS so much that I even wrote two plugins, one of which got relatively popular.
But all that changed a few days ago. At first, I stumbled upon a message in one of the updates of the system that said it was “the last patch for October 1.0”. At first I thought “Oh, cool, a big update?! Let’s have a look.” However, I was already suspicious that there was an update coming and I didn’t hear anything about that. And then, I stumbled upon this blogpost. “October CMS moves to become a paid platform.” Are you f***ing kidding me?!
## Why Open Source Matters
If you know me just a little bit, you know that I’m a fervent defender of Open Source software. I cannot really trust closed source, and let’s be really honest: Most paid software is not worth the price. Or do you think that last mobile game really was innovative enough to justify the three Euros you paid to get it? Right? I thought so. The thing is, companies have become quite versatile at pulling out money from employee’s wallets. Many have already switched to subscription services, that is: You pay more for Adobe per month than for the internet access required to run the software in the first place, so it’s out of the question for anyone who just wants to learn it. It’s similar with other software.
And right now, at this very moment, there is a large gap between completely free of charge Open Source software whose code is publicly visible, and some atrociously expensive proprietary software that literally drains the money out of your bank account. But I digress. If you want to know more about my stance on Software and why I think closed source software is bad for your health, have a seat and watch a video I made about this.
Long story short: I cannot host my websites and projects on closed source services, because that means that I will vanish from the internet once I cannot pay for the service anymore. Any paid solution requires I never run out of a job and/or money, and I won’t risk that1 — in today’s world and in my position I can’t afford to not be on the internet. So I had to switch.
But the internet 2021 is a different one than it was in 2010: CMS systems are on the retreat. Wordpress won’t get any better no matter how hard their devs might try, Drupal and Typo3 are still too big for most projects, and now that one of the few good alternatives hides their source code from me, I feel a little bit like standing in the rain. In fact, the internet has become so corporate that it’s almost impossible to find anything “click and write” that doesn’t try to rob you.
So I had to change my mindset. And, funny enough, the internet seems to move back into a direction it came from in 1995. Today, the best shot you got at highly customisable, highly available, and scalable homepages is to not use a CMS system, but to divide these tasks up again; between what is called a “headless CMS” that hosts your data, and a static site that hosts your frontend. I won’t go into the distinction between frontend and backend here, but suffice to say that this blog now runs on what is called a static site generator, more specifically, on Jekyll.
A static site generator functions as follows: You have some source code for a website, code that includes your HTML, some CSS to style the page, and content, mostly using Markdown. The static site generator is a small program that takes all of this, processes it and outputs a folder on your computer that contains static HTML pages. These you can then upload somewhere, just like in 1995 and you have a website.
## Conclusion
I don’t have too much time today to dive into what this all means, and why – viewed from a certain angle – this is even better than CMS systems, but it does have many benefits. Let me close today’s post with a few comments on what has changed, and where you might need to adapt something on your site:
• Most information should still be exactly where it was previously. However, the blogposts now require an .html appended to the URL. I’m trying to fix that over the weekend, but for now that’s how I got it to work
• I took the liberty of exchanging the main font. While I do like Crimson (the previous font), I always felt it was a little bit “fat.” While migrating my page to Jekyll, I stumbled upon the page from Yehuda Katz, and I really like the font he uses, Cardo
• A few other CSS improvements here and there have also happened, since it’s much easier editing CSS files using VS Code than in some online editor.
• The feed is now reachable at a different location. It’s now https://www.hendrik-erz.de/feed.xml, not https://www.hendrik-erz.de/blog/feed.rss (the latter link now redirects, but I found that RSS readers don’t recognise HTML redirect requests, so you might want to exchange that one).
So much for this week; I hope I can finally continue my How I Work series next week. See you then!
1. To be fair: Of course it costs money to put your face out on the internet. However, there is a large difference in paying $10 _per website_ you want to make, and the$15 per month I pay for all my websites in total since that is how much my webserver costs. | 2021-06-23 18:15:42 | {"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": 0.25816237926483154, "perplexity": 1166.4303750597355}, "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-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00335.warc.gz"} |
https://www.birs.ca/events/2014/5-day-workshops/14w5001/videos/watch/201406031101-Waldspurger.html | ## Video From 14w5001: The Future of Trace Formulas
Tuesday, June 3, 2014 11:01 - 12:05
Stabilization of the twisted trace formula | 2020-09-25 14:31:16 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8980363011360168, "perplexity": 4665.809488306168}, "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-2020-40/segments/1600400226381.66/warc/CC-MAIN-20200925115553-20200925145553-00353.warc.gz"} |
http://math.stackexchange.com/questions/219822/count-the-number-of-integer-solution-to-sum-i-1-4a-i-times-b-i-geq-8 | # Count the number of integer solution to $\sum_{i=1}^ {4}{a_i\times b_i} \geq 8$? [closed]
Count the number of integer solution to $\sum_{i=1}^ {4}{a_i\times b_i} \geq 8$ such that
condition 1: $1 \leq a_i \leq 7$
condition 2: $1 \leq b_i \leq 4$
condition 3: $\sum_{i=1}^{4} {a_i} = 10$
condition 4: $\sum_{i=1}^{4} {b_i} = 7$
Is there a general solution to find the number of integer solutions for an inequality like this given conditions? The way I'm finding the number of solution is by generating all possible solutions and check to see if they satisfy all the conditions or not. This solution is very time-consuming. A computer program which can find the number of solutions is very much appreciated.
-
## closed as not a real question by Andres Caicedo, Norbert, Noah Snyder, Thomas, ArkamisOct 24 '12 at 13:50
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.
After taking the time to answer your question, I discovered your other questions: here, here and here. Hmm... – Douglas S. Stones Oct 24 '12 at 2:59
@DouglasS.Stones so how do we call this, quadruplicate? – Jean-Sébastien Oct 24 '12 at 3:09
possible duplicate of Count the number of integer solutions for $a \times b \geq k$? – Noah Snyder Oct 24 '12 at 11:58
There's several ways to answer this problem computationally, and which is best depends on (a) personal preference, and (b) how, if at all, you're planning on re-using the code.
A brute force method is to have 8 nested for loops and, inside these for loops, have an if statement for the three conditions ($\sum_i a_i b_i \geq 8$, $\sum_i a_i=10$, and $\sum_i b_i=7$). On modern computers, this would take an almost infinitesimal amount of time to run (it'd probably take you longer to hit the "enter" key than the program takes to run). However, this approach would not have "scalability", in that, if you next asked for $9$ variables we would need to rewrite the code.
I wrote a solution in GAP, which has a bit better scalability (although, the code will still need to be edited, it can be edited rather easily). It's still quite brute-force and is not optimised.
Clearly, given a solution, we can permute the $a_i$'s and $b_i$'s to obtain another (not necessarily distinct) solution. So, this code iterates through the $a_i$'s and $b_i$'s in non-increasing order, and finds the number of permutations of the lists $(a_i)$ and $(b_i)$.
A:=RestrictedPartitions(10,[1..7],4);
B:=RestrictedPartitions(7,[1..4],4);
nr_orbits:=0;
count:=0;
for a in A do
a_orbit_size:=OrbitLength(SymmetricGroup(4),a,Permuted);
for b in B do
b_orbit_size:=OrbitLength(SymmetricGroup(4),b,Permuted);
if(a*b>=8) then
nr_orbits:=nr_orbits+1;
orbit_size:=a_orbit_size*b_orbit_size;
count:=count+orbit_size;
Print(nr_orbits," ",a," ",b," orbit size: ",orbit_size,"\n");
fi;
od;
od;
Print("Counted ",count," solutions to the inequality.\n");
which returns:
1 [ 3, 3, 2, 2 ] [ 2, 2, 2, 1 ] orbit size: 24
2 [ 3, 3, 2, 2 ] [ 3, 2, 1, 1 ] orbit size: 72
3 [ 3, 3, 2, 2 ] [ 4, 1, 1, 1 ] orbit size: 24
4 [ 3, 3, 3, 1 ] [ 2, 2, 2, 1 ] orbit size: 16
5 [ 3, 3, 3, 1 ] [ 3, 2, 1, 1 ] orbit size: 48
6 [ 3, 3, 3, 1 ] [ 4, 1, 1, 1 ] orbit size: 16
7 [ 4, 2, 2, 2 ] [ 2, 2, 2, 1 ] orbit size: 16
8 [ 4, 2, 2, 2 ] [ 3, 2, 1, 1 ] orbit size: 48
9 [ 4, 2, 2, 2 ] [ 4, 1, 1, 1 ] orbit size: 16
10 [ 4, 3, 2, 1 ] [ 2, 2, 2, 1 ] orbit size: 96
11 [ 4, 3, 2, 1 ] [ 3, 2, 1, 1 ] orbit size: 288
12 [ 4, 3, 2, 1 ] [ 4, 1, 1, 1 ] orbit size: 96
13 [ 4, 4, 1, 1 ] [ 2, 2, 2, 1 ] orbit size: 24
14 [ 4, 4, 1, 1 ] [ 3, 2, 1, 1 ] orbit size: 72
15 [ 4, 4, 1, 1 ] [ 4, 1, 1, 1 ] orbit size: 24
16 [ 5, 2, 2, 1 ] [ 2, 2, 2, 1 ] orbit size: 48
17 [ 5, 2, 2, 1 ] [ 3, 2, 1, 1 ] orbit size: 144
18 [ 5, 2, 2, 1 ] [ 4, 1, 1, 1 ] orbit size: 48
19 [ 5, 3, 1, 1 ] [ 2, 2, 2, 1 ] orbit size: 48
20 [ 5, 3, 1, 1 ] [ 3, 2, 1, 1 ] orbit size: 144
21 [ 5, 3, 1, 1 ] [ 4, 1, 1, 1 ] orbit size: 48
22 [ 6, 2, 1, 1 ] [ 2, 2, 2, 1 ] orbit size: 48
23 [ 6, 2, 1, 1 ] [ 3, 2, 1, 1 ] orbit size: 144
24 [ 6, 2, 1, 1 ] [ 4, 1, 1, 1 ] orbit size: 48
25 [ 7, 1, 1, 1 ] [ 2, 2, 2, 1 ] orbit size: 16
26 [ 7, 1, 1, 1 ] [ 3, 2, 1, 1 ] orbit size: 48
27 [ 7, 1, 1, 1 ] [ 4, 1, 1, 1 ] orbit size: 16
Counted 1680 solutions to the inequality.
-
You only need 6 nested loops since once three of the a-s or b-s are chosen, the fourth is determined. – marty cohen Oct 24 '12 at 3:45 | 2014-07-22 07:50:33 | {"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": 0.6087136268615723, "perplexity": 221.99668996978815}, "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-2014-23/segments/1405997857710.17/warc/CC-MAIN-20140722025737-00228-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://www.math.princeton.edu/events/instability-anti-de-sitter-spacetime-einstein-scalar-field-systemin-person-talk-2022-05 | # The instability of Anti-de Sitter spacetime for the Einstein-scalar field system(in-person talk)
-
Georgios Moschidis, Princeton University
The AdS instability conjecture provides an example of weak turbulence appearing in the dynamics of the Einstein equations in the presence of a negative cosmological constant. The conjecture claims the existence of arbitrarily small perturbations to the initial data of Anti-de Sitter spacetime which, under evolution by the vacuum Einstein equations with reflecting boundary conditions at conformal infinity, lead to the formation of black holes after sufficiently long time.
In this talk, I will present a rigorous proof of the AdS instability conjecture in the setting of the spherically symmetric Einstein-scalar field system. The construction of the unstable initial data will require carefully designing a family of initial configurations of localized matter beams and estimating the exchange of energy taking place between interacting beams over long periods of time, as well as estimating the decoherence rate of those beams. | 2022-05-28 20:05:09 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8051725625991821, "perplexity": 358.89000239920495}, "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-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00184.warc.gz"} |
https://socratic.org/questions/8-is-increased-to-22-what-is-the-percent-change | # 8 is increased to 22, what is the percent change?
Oct 28, 2016
The precent change was 175%
#### Explanation:
The formula for percent change is $p = \frac{N - O}{O} \cdot 100$ where $p$ is the percentage change, $N$ is the new value and $O$ is the old value.
Substituting what we know gives:
$p = \frac{22 - 8}{8} \cdot 100$
$p = \frac{14}{8} \cdot 100$
$p = 1.75 \cdot 100$
$p = 175$ | 2019-10-14 01:24:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6176044344902039, "perplexity": 997.075615244965}, "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-2019-43/segments/1570986648481.7/warc/CC-MAIN-20191014003258-20191014030258-00463.warc.gz"} |
https://plainmath.net/97637/in-a-simple-linear-regression-analysis | # In a Simple Linear Regression analysis, independent variable is weekly income and dependent variable is weekly consumption expenditure. Here 95% confidence interval of regression coefficient, β_1 is (.4268,.5914).
Kayley Dickson 2022-11-18 Answered
In a Simple Linear Regression analysis, independent variable is weekly income and dependent variable is weekly consumption expenditure. Here $95$% confidence interval of regression coefficient, ${\beta }_{1}$ is $\left(.4268,.5914\right)$.
You can still ask an expert for help
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## Answers (1)
luthersavage6lm
Answered 2022-11-19 Author has 22 answers
First in statistical speak:
Our model is ${Y}_{i}={\beta }_{0}+{\beta }_{1}{X}_{i}+{\epsilon }_{i}$. This means that ${\beta }_{1}$ is the amount that we expect $Y$ to increase by when $X$ increases by $1$ [or decrease if ${\beta }_{1}<0$].
Now in terms of the problem:
In this problem $X$ is weekly income and $Y$ is weekly consumption expenditure so $\stackrel{^}{{\beta }_{1}}$ is our estimate of the amount that weekly consumption expenditure increases for every $\mathrm{}1$ increase in weekly income. We are $95$% confidence that it is between $0.43$ and $0.59$ [where by "$95$%" confidence we mean that if we were to collect new data generated from the same distribution then in $19$ out of every $20$ experiments we'd get $\stackrel{^}{{\beta }_{1}}$ in this interval].
###### Did you like this example?
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• Pay only if we can solve it | 2022-12-05 14:30:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 35, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.41967007517814636, "perplexity": 2026.2756180367217}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "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-2022-49/segments/1669446711017.45/warc/CC-MAIN-20221205132617-20221205162617-00488.warc.gz"} |
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http://www.maa.org/publications/periodicals/convergence/convergence-articles?term_node_tid_depth=All&term_node_tid_depth_2=All&term_node_tid_depth_1=All&page=61 | # Convergence articles
Displaying 611 - 620 of 651
Page 10: Courant through D'Ambrosio
Page 11: Dantzig to Dieudonne
A wonderful book about the square root of 2, beginning with the search for the side of a square double a given square.
This is a page from an early printed edition of the Arithmetica of Jordanus de Nemore (early 13th century).
A collection of articles about historical and contemporary women in mathematics.
A man bought a number of sheep for $225; 10 of them having died, he sold 4/5 of the remainder for the same cost and received$150 for them. How many did he buy?
When knowing the sum of their ages along with another equation, determine how old a father and son are.
Two travelers, starting at the same time from the same point, travel in opposite directions round a circular railway.
I owe a man the following notes: one of $800 due May 16; one of$660 due on July 1; one of $940 due Sept. 29. He wishes to exchange them for two notes of$1200 each and wants one to fall due June 1. When should the other be due?
Find the area of the elliptical segment cut off parallel to the shorter axis; | 2014-07-30 12:28:11 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.23477447032928467, "perplexity": 1984.1670403455541}, "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-2014-23/segments/1406510270399.7/warc/CC-MAIN-20140728011750-00352-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://smerity.com/articles/2008/py2java.html | # Python to Java - A Survivalist's Guide
## Aim
This guide was written originally for Info1903 students at the University of Sydney who after two terms of Python needed to quickly begin coding in Java. This is in no way a replacement for a full Java textbook however and is merely here to allow an easy transition from Python to simple Java programs. In here we'll walk you through a few simple tasks - first a way to print primes.
## Basic Syntax and Semantics
The name of the file (Name.java) and class (public class Name) must be equal or else the code will not compile. It's convention to capitalise classes in Java.
Java's syntax is quite different to Python. First, all lines must end with a semicolon (;) and instead of specifying parts of the code with indentation like tabbing Java instead uses curly brackets ({}). Using indentation to mark the beginning and end of code blocks is still considered good practice however and increase readability.
//Python
for ...:
if ...:
do stuff
//Java
for ... {
if ... {
do stuff;
}
}
In Java everything is a class. When you run Java source code the class will automatically be searched for public static void main to execute. It also can accept command line input, in this case an array of strings called args - we'll return to exactly what that means later.
//from Primes.java
public class Primes {
public static void main (String[] args) {
...
}
}
The ellipsis (...) is where our main code will go. For our hello world we'll be using Primes.java as a base. Java implements things slightly differently in regards to printing.
System.out.print(x) // Prints x by itself
System.out.println(x) // Prints x + "\n", equivalent of Python's print function
System.out.printf(x, var) // Used in string formatting like C's printf, which we'll reference later
To print our hello world, just inject the println statement into our above code.
/from Primes.java
public class Primes {
public static void main (String[] args) {
System.out.println("Hello world");
}
}
Congratulations - you've written Hello World in Java. Now the next challenge - compiling it. As opposed to Python, which is an interpreted language (ie, translated as it's run), Java is instead a compiled language, and does all the 'translation' in a separate stage. To compile and run Java in a Unix environment such as Usyd's Congo server -
smerity@Loki:~/Coding/Courses/Java Prep/work$javac Primes.java smerity@Loki:~/Coding/Courses/Java Prep/work$ java Primes
Hello world
smerity@Loki:~/Coding/Courses/Java Prep/work\$
You've now compiled and executed Hello World. As you progress in Java you'll find it's much easier to use an IDE (integrated development editor) instead of doing everything from the command line. An IDE performs a number of helpful things like syntax/error highlighting, simple navigation between files and auto-completion.
## Simple Iteration
The For loop in Java is quite different to that of Python's. If you've done C, C++ or Javascript however you'll be familiar with it.
//Python
for i in xrange(10):
print i
//Java
for (int i=0; i<10; i++) {
System.out.println(i);
}
In the case of Java a for loop is split into three separate parts. Using layman's terms, there's the initialiser (run when the loop is first begun which in this case sets i to 0), the while (while i < 10 is true, keep looping) and the iterator (add one to i (i++, the same as i += 1 or i = i + 1) each time we loop).
## Variables
Next important tidbit - Java is a statically typed language. As opposed to Python where x can equal anything (such as a number, a string, a list...), once you say what a variable is in Java it must remain that type. Variables are declared by typing their (type) (name) = (value).
//Java
String name = "Stephen";
int age = 18;
Boolean male = true; //Booleans only allow true or false - just like Python's True and False
//This would raise a compilation error as you can't put a number into a
String name = 15; //EXPLODE
## Naive Prime Numbers
Excellent - if you've kept up with me then we're ready to find primes using Java. To do this, we simply combine the simple loops and variables introduced above.
//from Primes.java
public class Primes {
public static void main (String[] args) {
for (int num=2; num<40; num++) {
Boolean prime = true; //We'll assume the number's a prime to start with
for (int j=2; j <= Math.sqrt(num); j++) {
if (num%j == 0) {
prime = false;
break; //We found it's not a prime, so we can bail out of the loop early and save us work
}
}
if (prime) System.out.println(num);
/*Notice the above doesn't use curly brackets? It's the same as
if (prime) {
System.out.println(num);
}
If you don't use curly brackets Java assumes it ends at the next semicolon (;)*/
}
}
}
Shake, bake, compile and you have yourself all the primes up to 40. Let's look at the same code in Python, shall we?
//Python primes
import math
for num in xrange(2, 40):
prime = True
for j in xrange(2, int(math.sqrt(num)+1)):
if num%j == 0:
prime = False
break
if prime:
print num
## Primitives and Autoboxing
If you didn't know, all the variables in Python are classes - string for example is a class containing your string and also a bunch of functions to use with your string (such as startswith, upper etc). This isn't true in Java however, as there are things called primitives that store just what you put in. Why is this a problem? Well, in things like Collections (such as ArrayLists) they only accept classes and it won't accept primitives. How are we to fix this? For every primitive there is something called a wrapper class which is a class version of the primitive. To differentiate between the two the primitives are always lower case whilst their wrappers are always uppercase (as classes are by convention uppercase in Java). Autoboxing was introduced to Java to make the transition between primitives and their wrappers a simple process - you'll see an example below.
int => Integer
char => Character
none => String
boolean => Boolean
//Autoboxing
Integer x = 42;
//Without autoboxing the above becomes ...
Integer y = new Integer(42);
//You can also do all your normal operations with ints instead of Integers
x -= 4;
y -= new Integer(4);
Most of the time you don't have to worry about primitives and wrapper classes but when using certain datastructures it becomes very important to have an understanding of them.
## Data Structures
In this section we'll introduce Java's data structures and where they are equivalent to Python's.
### List
Python's list is essentially an array - Java has two similar data structures. First is an array, which is a list with only a set number of 'slots', and second is an ArrayList, which is the closest to Python's list. Notice that for both you need to declare what type of variables it's going to be holding.
/* Array */
String[] numbers = {"one", "two", "three"}; //Creates an array of strings that holds one to three
String[] names = new String[10]; //Creates an array named 'names' with enough space for ten strings
names[0] = "Smerity";
names[1] = "Josh";
int i = 0;
//Any unused slots are by default null, which is like Python's None
while(names[i] != null){
System.out.println(names[i]);
i++;
}
//Can also use the length of an array
for(i=0; i<numbers.length; i++)
System.out.println(numbers[i]);
/* ArrayList */
ArrayList<Integer> primes = new ArrayList<Integer>();
for(i=0; i<primes.size(); i++)
System.out.println(primes.get(i));
//Similar to Python's "for num in numbers:" syntax
for(Integer prime : primes)
System.out.println(prime);
### Dictionary
Dictionaries are one of the most useful of Python's default data structures. Java of course has similar structures.
/*HashMap<key_type, value_type>*/
//Once again you actually have to say what the types are due to static typing
HashMap<Character, Integer> letters = new HashMap<Character, Integer>();
//.toCharArray turns "tree" into a Character array of {'t', 'r', 'e', 'e'}
for(Character c : "tree".toCharArray()){
// The bang (!) means NOT - so if letters does NOT contain the key c ...
if (!letters.containsKey(c)) letters.put(c, 0);
letters.put(c, letters.get(c)+1);
}
for(Character c : letters.keySet())
System.out.println("Letter "+c+" used "+letters.get(c)+" times.");
/*Output -
Letter e used 2 times.
Letter t used 1 times.
Letter r used 1 times.
*/ | 2018-08-15 19:22:55 | {"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": 0.2045048326253891, "perplexity": 4126.749353097674}, "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-2018-34/segments/1534221210249.41/warc/CC-MAIN-20180815181003-20180815201003-00539.warc.gz"} |
https://www.qb365.in/materials/stateboard/11th-standard-maths-english-medium-free-online-test-book-back-1-mark-questions-with-answer-key-6216.html | #### 11th Standard Maths English Medium Free Online Test Book Back 1 Mark Questions with Answer Key
11th Standard
Reg.No. :
•
•
•
•
•
•
Maths
Time : 00:10:00 Hrs
Total Marks : 10
10 x 1 = 10
1. The function f:[0,2π]➝[-1,1] defined by f(x)=sin x is
(a)
one-to-one
(b)
on to
(c)
bijection
(d)
cannot be defined
2. Given that x, y and b are real numbers x<y, b>0, then
(a)
xb < yb
(b)
xb > yb
(c)
xb ≤ yb
(d)
$\frac { x }{ b } \ge \frac { y }{ b }$
3. If cos280+sin280=k3, then cos 170 is equal to
(a)
$\frac { { k }^{ 3 } }{ \sqrt { 2 } }$
(b)
-$\frac { { k }^{ 3 } }{ \sqrt { 2 } }$
(c)
±$\frac { { k }^{ 3 } }{ \sqrt { 2 } }$
(d)
-$\frac { { k }^{ 3 } }{ \sqrt { 3 } }$
4. In an examination there are three multiple choice questions and each question has 5 choices. Number of ways in which a student can fail to get all answer correct is
(a)
125
(b)
124
(c)
64
(d)
63
5. The sequence$\frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } +\sqrt { 2 } } \frac { 1 }{ \sqrt { 3 } +2\sqrt { 2 } }$...form an
(a)
AP
(b)
GP
(c)
HP
(d)
AGP
6. Which of the following equation is the locus of (at2; 2at)
(a)
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
(b)
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
(c)
x2+y2=a2
(d)
y2=4ax
7. What must be the matrix X, if 2x+$\begin{bmatrix} 1& 2 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix}?$
(a)
$\begin{bmatrix} 1& 3 \\ 2 &-1 \end{bmatrix}$
(b)
$\begin{bmatrix} 1& -3 \\ 2 &-1 \end{bmatrix}$
(c)
$\begin{bmatrix} 2& 6 \\ 4 &-2 \end{bmatrix}$
(d)
$\begin{bmatrix} 2& -6 \\ 4 &-2 \end{bmatrix}$
8. If y = f(x2+2) and f '(3) = 5,then ${dy\over dx}$ at x = 1 is
(a)
5
(b)
25
(c)
15
(d)
10
9. $\int {x+2\over \sqrt{x^2+1}}dx$is
(a)
$\sqrt{x^2-1}-2 log|x+\sqrt{x^2-1}|+c$
(b)
$sin^{-1}x-2 log|x+\sqrt{x^2-1}|+c$
(c)
$2 log|x+\sqrt{x^2-1}|-sin^{-1}x+c$
(d)
$\sqrt{x^2-1}+2log|x+\sqrt{x^2-1}|+c$
10. If m is a number such that m $\le$ 5, then the probability that quadratic equation 2x2+2mx+m+1=0 has real roots is
(a)
${1\over 5}$
(b)
${2\over 5}$
(c)
${3\over 5}$
(d)
${4\over 5}$ | 2021-05-08 19:22:11 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.7293533682823181, "perplexity": 3246.8984268484865}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00321.warc.gz"} |
https://math.stackexchange.com/questions/1433216/simple-limit-of-a-sequence | # Simple limit of a sequence
Need to solve this very simple limit $$\lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right)$$
I know how to solve these limits: by using $a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - very tedious, boring and tiring. I hope there is some artful and elegant solution. Thank you!
• What about the binomial expansion? Sep 13 '15 at 7:51
• I would say the standard way (not by using L'Hospital's rule) to solve this limit is not very tedious, boring and tiring. In the expression for 'a-b' divide the numerator and denominator of 'x' and the result is obvious. Sep 13 '15 at 8:05
• @georg but not so elegant as below :)) Sep 13 '15 at 9:40
• Apparently it will ;-) Sep 13 '15 at 13:51
$$\lim _{ x\to \infty \: } \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 } \right) =$$ $\lim _{ x\to \infty \: } \frac { \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 } \right) \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) } =$$\ =\lim _{ x\to \infty : } \frac { 3x^{ 2 }+4x+1-3x^{ 2 }-9x-2 }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) } =\lim _{ x\rightarrow \infty }{ \frac { -5x-1 }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) } }$$ now ,the power of denominator of polynomial is higher than the numerator so the limit is equal to$0$• This is the best way to calculate the limit. – A.Γ. Sep 13 '15 at 9:13 You have $$f(x)=\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}=\sqrt[3]{3x^2}\left(\sqrt[3]{1+\frac{4}{3x}+\frac{1}{3x^2}}-\sqrt[3]{1+\frac{3}{x}+\frac{2}{3x^2}}\right)$$ Using Taylor expansion at order one of the cubic roots$\sqrt[3]{1+y}=1+\frac{y}{3}+o(y)$at the neighborhood of$0, you get: $$f(x)=\sqrt[3]{3x^2}\left(\frac{4}{9x}-\frac{1}{x}+o\left(\frac{1}{x}\right)\right)$$ hence $$\lim\limits_{x \to \infty} f(x)=0$$ • Thanks! Can you explain me plese your second expression, where you using Taylor expansion? Sep 13 '15 at 9:01 • Yes I was using Taylor expansion and I edited the post to provide more details. Sep 13 '15 at 9:36 • Oh, now I am in clear! Thanks!! Sep 13 '15 at 9:42 You can use the binomial theorem to expand this. $$\lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right)$$ Here 1 and 2 are the smallest terms and they can be ignored. \begin{align} &\lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) \\=& \lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x}-\sqrt[3]{3x^2+9x}\right) \\=& \lim _{x\to \infty \:}\left(\sqrt[3]{3x^2\left(1+\frac{4}{3x}\right)}-\sqrt[3]{3x^2\left(1+\frac3x\right)}\right) \\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left(\sqrt[3]{1+\frac{4}{3x}}-\sqrt[3]{1+\frac3x}\right) \\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left({1+\frac{4}{9x}}-{1-\frac1x}\right) \\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left({\frac{4-9}{9x}}\right) \\=& \lim _{x\to \infty \:}\sqrt[3]{3}x^{2/3}\left({\frac{-5}{9x}}\right) \\=& \lim _{x\to \infty \:}\sqrt[3]{3}x^{-1/3}\left({\frac{-5}{9}}\right) \\=& \,0 \end{align} • Not clear why you can ignore constants and cannot ignore linear terms which are also of smaller order. – A.Γ. Sep 13 '15 at 9:05 • @A.G. Hm. Can we ignore constants? Sep 13 '15 at 9:39 • @PersonaNonGrata Well, in this particular case yes, but the explanation why it is possible gonna cost you as much efforts as to solve the problem without ignoring anything. – A.Γ. Sep 13 '15 at 9:44 • @A.G. Oh. So better don't ignore constants? :)) Sep 13 '15 at 9:46 • @PersonaNonGrata Better is to do first the Taylor expansion of the original function and then ignore properly lower order terms. – A.Γ. Sep 13 '15 at 9:52 Actually, this question can be solved without calculation if you're familiar with power. If the first term(in the square root), the biggest term is3x^2$and so does the second term. And any term has power below 2 are negligible. So you can just forget them and get your answers, which is$0$. • "Any term of lower order is negligible" is not correct. Example:$\sqrt{x^2+x+1}-\sqrt{x^2-x+1}$. Neglecting lower order terms gives the wrong limit$0$at infinity. (Correct limit is$1\$ in my example).
Can you help with O-symbols? It's all right here? $$f(x) = \sqrt[3]{3x^2}\left(1 + \frac{4}{9x} + O\left(\frac{1}{x^2}\right) - 1 - \frac{1}{x} -O \left(\frac{1}{x^2}\right)\right)= \sqrt[3]{3x^2} \left(\frac{-5}{9x} + \frac{1}{18x^2} \right).$$
Hence $$\lim _{x\to \infty }\sqrt[3]{3x^2} \left(\frac{-5}{9x} + \frac{1}{18x^2} \right)= \lim_{x \to \infty} \sqrt[3]{3}{x^{-1/3}}^{\to 0} \left( - \frac{5}{9}+\frac{1}{18x}^{\to 0} \right) = 0.$$ | 2021-12-02 09:22:40 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9983549118041992, "perplexity": 1615.2617252482644}, "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-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00506.warc.gz"} |
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# Analysis and Design of a Compact Leaky-Wave Antenna for Wide-Band Broadside Radiation
## Abstract
A low-cost compact planar leaky-wave antenna (LWA) is proposed offering directive broadside radiation over a significantly wide bandwidth. The design is based on an annular metallic strip grating (MSG) configuration, placed on top of a dual-layer grounded dielectric substrate. This defines a new two-layer parallel-plate open waveguide, whose operational principles are accurately investigated. To assist in our antenna design, a method-of-moments dispersion analysis has been developed to characterize the relevant TM and TE modes of the perturbed guiding structure. By proper selection of the MSG for a fabricated prototype and its supporting dielectric layers as well as the practical TM antenna feed embedded in the bottom ground plane, far-field pencil-beam patterns are observed at broadside and over a wide frequency range, i.e., from 21.9 GHz to 23.9 GHz, defining a radiating percentage bandwidth of more than 8.5%. This can be explained by a dominantly excited TM mode, with low dispersion, employed to generate a two-sided far-field beam pattern which combines to produce a single beam at broadside over frequency. Some applications of this planar antenna include radar and satellite communications at microwave and millimeter-wave frequencies as well as future 5G communication devices and wireless power transmission systems.
## Introduction
Planar printed antennas have been receiving remarkable interest in the last few decades thanks to their ease of realization, cost effectiveness, and integrability with active circuitry1. As is well-known, the most common type, the resonant microstrip patch antenna, has consolidated design procedures but typically provides broad, far-field patterns with low to moderate gain and narrow operational bandwidths. Phased arrays using such patch antennas have to be designed in order to obtain more directive as well as scannable patterns, although at the expense of a considerable increase in design complexity and cost. This is because bulky and expensive feeding networks and phase shifters are typically required.
Printed leaky-wave antennas (LWAs) offer an attractive alternative to phased arrays for the synthesis of directive beams with a variety of pattern shapes and steering capabilities2,3. In particular, pencil beams scannable in the elevation and azimuth planes can be obtained with linear arrays of one-dimensional (1-D) LWAs4, whereas either conical scanned beams or broadside pencil beams are possible with two-dimensional (2-D) LWAs3. In both cases, the guided-wave (GW) and the nonresonant nature of the radiation mechanism can provide a wide operational bandwidth. However, the main-beam angle typically scans with frequency, a feature which may or may not be desired and depends on the application.
As concerns 2-D LWAs, an interesting class of annular structures is the so-called ‘bull-eye’ antenna, first introduced in5 and carefully examined in6,7 considering operation in the microwave range, and in8,9,10 at millimeter waves. A prototype working in the terahertz range has also been proposed in11. In general, these single-layer structures are constituted by an arrangement of concentric microstrip rings driven by a suitably designed surface-wave launcher (SWL) positioned at the centre of the antenna12.
The cylindrical TM0 surface-wave (SW) field excited by the SWL travels radially, and, due to the perturbation of the radially periodic metallic grating, transforms into a cylindrical leaky wave (LW). The annular grating is usually printed on a single-layer grounded dielectric slab (GDS) and the antenna synthesis is essentially based on a dispersion analysis of the cylindrical LWs supported by the structure. Due to the lack of translational invariance, which prevents a direct modal characterization of the entire structure, such a dispersion analysis is performed on a linearized version of the 2-D radial annular structure, i.e., on the corresponding 1-D (periodic) linear metal strip grating (MSG). The MSG strips are normal to the propagation direction of the relevant leaky mode, whose radiation features can be described in terms of a fast spatial harmonic. A detailed discussion about the effectiveness of this approach can be found in7,13,14, based on both far-field and near-field arguments.
Following these developments, a half-annular version of a bull-eye antenna was recently reported by the authors in15, with the original structure briefly examined in16. Both designs were based on an annular microstrip grating placed on top of a dual- or two-layer (2L) grounded dielectric substrate (a 2L-GDS, as illustrated in Fig. 1(a,b)). A contrasting high-low profile for the dielectric constants was employed using commercial substrates. Due to the top metallic covering, the unperturbed (closed) guiding structure can be also described as a two-layer parallel-plate waveguide (2L-PPW). By this GDS and dielectric-superstrate configuration with top microstrip rings or slots for perturbation and radiation of the dominantly excited TM mode, and practically fed by a directive SWL integrated in the ground plane (as in12,17), a one-sided conical-sector and pencil-beam pattern was realized in15 with continuous frequency-scanning through broadside. Moreover, the antenna reported in15 can be described as a quasi-1-D LWA but with cylindrical-wave propagation within the low-profile guiding structure, due to the truncated and half-annular aperture of the antenna.
It is worth noting that the far-field radiation pattern and modal behavior for the 2-D planar periodic LWA in15 is similar to that of a one-sided 1-D periodic LWA with a mitigated open stopband. In fact, a single far-field pencil-beam pattern was achieved which continuously scanned, in the E(xz) plane (see Fig. 1(b)), from the backward to the forward quadrant with an increase in frequency. This one-sided LWA allowed for continuous radiation through broadside with a minor reduction in the realized gain at broadside. This is mainly due to a double-symmetric bump of the normalized LW attenuation constant (α/k0) centered around the broadside radiating frequency15. Other works, focusing on 1-D periodic and quasi-uniform LWAs, have also studied such a desired scanning behavior (see e.g.3,18,19,20,21,22,23); whereas 2-D scanning LWAs based on metasurfaces have been proposed in24,25,26. Different LWA designs able to achieve broadside or frequency-scanning radiation with directive beam patterns in the far-field have been proposed in the last decade from microwave to optical frequencies (see e.g.27,28,29,30,31,32,33,34,35).
In this paper the 2L-PPW guiding structure from15,16 is employed to achieve persistent (i.e., wideband) and highly directional radiation at broadside whilst employing a bi-directional and integrated TM SWL feed system7. To this aim, the unperturbed and perturbed nature of the relevant bound and leaky modes of the 2-D ‘bull-eye’ responsible for radiation are fully reported and accurately analysed. Moreover, our proposed antenna design takes advantage of the preliminary discussions and supporting theory presented in15,16, and which are further developed here to describe the complete modal analysis and design of the proposed planar LWA offering wideband radiation. In this frame, a method-of-moments (MoM) formulation is also suitably adapted to describe the relevant radiating and guided TM and TE modes that can be supported by the structure. Relevant results for the background closed waveguide (i.e., the 2L-PPW and the 2L-GDS) are also discussed. All this makes the present work new and original with respect to15, and, with a unique design motivation. For example, in15 a different 2L-LWA was used to understand and optimize the dispersion of the relevant LW mode while also reporting the radiation performances. In particular, a directive pencil beam was observed in the far-field in15 which scanned through broadside as a function of frequency and where LWA feeding was realized by a uni-directional SWL.
The design methodology is based on the MoM in the spectral domain applied to an electric-field integral-equation (EFIE) formulation within the unit cell. In particular, we have employed a spectral-domain formulation of the MoM, in which the resulting matrix elements are expressed by integrals involving the planar components of the spectral dyadic Green’s function of the 2L-GDS. To this aim a suitable transverse network formalism is employed to describe the multi-layer structure (see, e.g.36, for all the relevant details on the approach). To our understanding, this has never been done before for this specific and low-cost dual-layer LWA configuration and this approach can also be applied to other types of multi-layer metal-strip-grating LWAs (consisting of two or more dielectric layers) as well as Fabry-Perot antennas.
Numerical full-wave results using a commercial simulator and measurements of a prototype are also newly reported in this paper to assess the performance of the proposed LWA. Mainly to ensure that the employed TM mode for leaky-wave (LW) radiation has a zero cutoff frequency, is moderately dispersive, and operates within a unimodal regime over a significantly wide operating bandwidth. In addition, experiments confirm for the first time that enhanced broadside radiation characteristics are possible for such a simple and low-cost LWA. In particular, our fabricated prototype (see Fig. 2(a)) is capable of radiating a fixed-angle pencil beam at broadside over a significantly wide bandwidth of 8.7%, defining a new two-sided planar LWA. It should be made clear that, due to the combination of frequency-dependent conical-sector beam patterns, the physical operation of the antenna is still based on a two-sided frequency-scanned beam, with continued pencil-beam radiation at broadside.
For the first time numerical and experimental validations are reported on the role of the beam-splitting condition for such a compact (i.e., truncated) 2-D LWA. Thus we now bridge the connection with LW theory and the effects of a practically sized aperture. This further explains the achieved broadside radiating beam with a percentage bandwidth of more than 8.5%. To the best of the authors’ knowledge no similar 2L-LWA, with a rigorous analysis and the relevant supporting theory has been reported for this class of 2-D travelling-wave planar antenna structure which can offer simple and integrated feeding and efficient TM wave excitation for radiation.
## Methods
Scanning through broadside is typically problematic in more standard one-sided, 1-D periodic LWAs, due to the LW open stopband region3,18,23. However in15, broadside radiation was made possible by employing a unidirectional SWL positioned at the substrate periphery, which can be modeled as a horizontal magnetic dipole (HMD) antenna source in the ground plane. This HMD allows for broadside radiation provided that leakage from the antenna is optimized by removing or, at least, reducing the LW open stopband. To this aim, the top MSG aperture of the antenna in15, as well as its directive TM SWL and half-annular two-layer dielectric configuration, with the additional degree of freedom provided by the proper sizing of the dielectric superstrate layer, were suitably employed for antenna synthesis. On this basis a compact LWA offering a one-sided beam pattern scanning with frequency through broadside was obtained in15.
In more conventional periodic (or uniform) 2-D LWAs, the dominant cylindrical LW on the radial aperture generates a conical-sector beam pattern in the far-field where the main beam angle, $${\theta }_{p}\approx {\sin }^{-1}(\beta /{k}_{0})$$ (being β the LW phase constant), scans with an increase in frequency towards broadside, as illustrated in Fig. 1(b). Directive radiation at broadside (θp = 0°) can be realized in the far-field within a frequency range (fc1, fc2) centered at fc. For such a periodic 2-D structure working on the n = −1 spatial harmonic, by increasing the frequency, the beam angle of the two-sided beam pattern first starts to reduce for f < fc1 until it coalesces into a single broadside pencil beam at the beam splitting frequency fsp (given by α ≈ β−1)34, around fc. Typically radiation at broadside is obtained over a very narrow frequency range and can strongly deteriorate, mainly, due to the presence of a LW open stopband. This can introduce a considerable reduction of the realized gain for the LWA7. By further increasing the frequency above the open-stopband region for f > fc2, the beam splits again into a conical pattern (see Fig. 1(b)), gradually pointing far from broadside.
A similar narrow frequency range for broadside radiation is observed in uniform (or quasi-uniform) 2-D LWAs, where the beam angle of the two-sided beam pattern coalesces by decreasing the frequency until f = fsp and then the gain quickly deteriorates by further decreasing the frequency below the cutoff of the leaky mode34. In contrast, the proposed 2-D LWA design under study overcomes these conventional limitations, being able to radiate a single pencil-beam consistently pointing at broadside (θp = 0°) over a wide radiating bandwidth, while also demonstrating more conventional frequency beam scanning off broadside. As discussed in the following sections, the main reason for this physical response is related to the mitigation of the open-stopband effects of the dominant TM leaky mode in the periodic 2L-PPW which has low dispersion, and to the existence of a less stringent beam-splitting condition for LWAs of finite length, as theoretically discussed in35.
### Theoretical Formulation
The reference planar periodic structure is a linear equi-spaced array of slots etched on the top surface of a 2L-GDS, i.e., a locally linearized version of the annular structure as illustrated in Fig. 2(c). The spatial period of the linearized structure is p, the width of each slot is w (or the width of each strip is s = p − w), the thicknesses and relative permittivities of the bottom substrate-dielectric and top superstrate-dielectric layers are h1, εr1 and h2, εr2, respectively.
Thanks to the 2-D nature of the problem, the spectrum of the propagating Bloch waves across the slots can be divided into both TM and TE modes. Each mode is characterized by a Floquet representation in terms of an infinite number of space harmonics with (generally complex) wavenumbers kxn = βn − = β0 + 2πn/p − (see the reference system in Fig. 2(c)), where, typically, the n = −1 spatial harmonic mainly contributes to radiation. In particular, the LW mode responsible for radiation from the proposed LWA is the dominant TM mode of the perturbed 2L-PPW, in a frequency range where the n = −1 space harmonic is fast, i.e., −k0 < β−1 < +k0. With ‘low’ attenuation rates (i.e., α/k0 < 0.1), directive beam patterns can be observed at beam angles defined by $${\theta }_{p}\approx {\sin }^{-1}(\sqrt{{\hat{\beta }}_{-1}^{2}-{\hat{\alpha }}^{2}})$$, with |$${\hat{\beta }}_{-1}$$| ≥ $$\hat{\alpha }$$, where the hat $$\hat{\cdot }$$ indicates normalization with respect to k0.
### Design Guidelines
The 2L-GDS that constitutes the antenna substrate has to be properly designed to support the dominant TM mode for radiation. To this aim the permittivity of the substrates, their heights and the dimensions of the MSG should be properly sized. As concerns the substrate permittivity and thickness, their choice is mainly constrained by the SWL used to feed the proposed antenna. Such an antenna feeder, is fully planar and integrated into the bottom substrate and requires high values for the relative permittivity ($${\varepsilon }_{{{\rm{r}}}_{1}}\approx 10$$) and appropriate thickness ($${h}_{1}\sqrt{{\varepsilon }_{{\rm{r}}}}/{\lambda }_{0}\approx 1/4$$) for proper operation7. Moreover, the combined thickness and relative permittivities of the two dielectric layers has to be properly selected to generate an evanescent TM field in the top dielectric layer (or the dielectric superstrate region, defined by $${\varepsilon }_{{{\rm{r}}}_{2}}\le 3$$ and $${h}_{2}\approx {\lambda }_{0}/2\sqrt{{\varepsilon }_{{{\rm{r}}}_{2}}}$$). This ensures radial propagation within the bottom guide at the h1 and h2 interface of the 2L-PPW, similar to a TM0 SW mode that radially propagates at the air-dielectric interface of a GDS with an evanescent field component in the air region.
Once the two-layer structure is set, the dispersive behavior of the dominant TM mode can be determined while also analyzing its perturbed propagation due to an added MSG. This MSG can transform the TM mode into a fast LW that is responsible for directive radiation in the far-field. The slot width and the periodicity of the metal strips, which define the MSG, can be suitably tuned to obtain broadside radiation around a specific frequency and to provide sustained leakage, for example, such that α/k0 > 0.01 and where a double-symmetric bump around the open-stopband frequency, fc, is observed15. Furthermore, once the LW phase and attenuation constants are determined, one can further examine the Brillouin dispersion diagram of the guiding structure as well as its background waveguides, i.e., the relevant 2L-PPW and the 2L-GDS, whose dispersive features are investigated in the following sections. Following these developments the beam pointing angle in the far-field can be further characterized as well as the radiation performances of the developed LWA.
### Full-Wave Analysis of the Structure
To fully characterize the modal properties of the proposed structure, an efficient MoM code already developed by some of the authors37,38 has been modified to account for the presence of the two-substrate layers. This is achieved by exploiting the flexibility of the transverse network formalism. The approach is described as follows.
The periodicity allows for studying one single spatial period (unit cell). The modal surface density current Js on the top single strip section within such unit cell can be represented as a linear combination of transverse and longitudinal components, Jy(x) and Jx(x), respectively. Hence we can write
$$\begin{array}{rcl}{{\bf{J}}}_{s}(\rho ) & \cong & {{\bf{J}}}_{s}(x)={J}_{x}(x)\hat{{\bf{x}}}+{J}_{y}(x)\hat{{\bf{y}}}\\ & = & \sum _{q\mathrm{=0}}^{{N}_{x}-1}\,{A}_{q}{J}_{xq}(x)\hat{{\bf{x}}}+\sum _{r=0}^{{N}_{y}-1}\,{B}_{r}{J}_{yr}(x)\hat{{\bf{y}}}\end{array}$$
(1)
where $$\hat{{\bf{x}}}$$ and $$\hat{{\bf{y}}}$$ are the unit vectors of the Cartesian axes x and y, Nx and Ny are the number of basis functions used to represent the x and y components in the MoM formulation, and the complex coefficients Aq and Br are the unknowns of the problem. The entire-domain basis functions adopted here, in particular, are38
$$\begin{array}{c}{J}_{x{\rm{q}}}(x)=j{U}_{{\rm{q}}}(\frac{2x}{s})\sqrt{1-{(\frac{2x}{s})}^{2}}\\ {J}_{y{\rm{r}}}(x)={T}_{{\rm{r}}}(\frac{2x}{s})\frac{1}{\sqrt{1-{(\frac{2x}{s})}^{2}}}\end{array}$$
(2)
where the functions T and U are Chebyshev polynomials of the first and second kind, respectively, and the square-root functions have been included in order to take into account the behavior of the current components near the edges at x = ±s/2 of the metal strip.
An integral equation can be obtained by enforcing that the tangential electric field vanishes on the strip within the unit cell. This can be completed by representing the electric field integral equation (EFIE) for the modal currents in the space domain, transforming the result into the spectral domain by using the Fourier transform, and then accommodating for an infinite number of n spatial harmonics. The integral equation and the corresponding electric-field expansion in the space domain, E(x, z), are as follows:
$$\hat{{\bf{z}}}\times \sum _{n=-\infty }^{+\infty }\,{\underline{\tilde{{\bf{G}}}}}^{ee}({k}_{xn})\cdot {\tilde{{\bf{J}}}}_{s}({k}_{xn})\,{e}^{-j{k}_{xn}x}=0\,\,{\rm{for}}\,\,|x| < w/2,$$
(3)
$${\bf{E}}(x,z)=\frac{1}{2\pi p}\sum _{n=-\infty }^{+\infty }\,{\underline{\tilde{{\bf{G}}}}}^{ee}(z,{k}_{xn})\cdot {\tilde{{\bf{J}}}}_{s}({k}_{xn}){e}^{-j{k}_{xn}x}$$
(4)
where $${\underline{\tilde{{\bf{G}}}}}^{ee}$$ is the spectral dyadic Green’s function of the 2L-GDS for the electric field produced by an electric current source36,39,40 as illustrated in Fig. 2(c), and the tilde represents a Fourier transform with respect to x. The elements of the spectral Green’s function can customarily be determined in terms of the equivalent voltages and currents using the relevant transverse equivalent network. This is shown explicitly in Fig. 2(c) for our two-layer guiding structure under analysis.
By discretizing the integral equation within the unit cell (|x| < p/2), for both the transverse and longitudinal currents defined in Eq. (1), we get
$${\int }_{-s\mathrm{/2}}^{s\mathrm{/2}}[{J}_{xl}(x)\hat{{\bf{x}}}]\cdot \sum _{q=0}^{{N}_{x}-1}\,{A}_{q}\,\sum _{n=-\infty }^{+\infty }\,{\underline{\tilde{{\bf{G}}}}}^{ee}\,({k}_{xn})\cdot [{\tilde{J}}_{xq}({k}_{xn})\hat{{\bf{x}}}]\,{e}^{-j{k}_{xn}x}dx=0$$
(5)
for TM waves with l = 0, …, Nx − 1. Likewise for TE waves:
$${\int }_{-s\mathrm{/2}}^{s\mathrm{/2}}\,[{J}_{ym}(x)\hat{{\bf{y}}}]\cdot \sum _{r=0}^{{N}_{y}-1}\,{B}_{r}\,\sum _{n=-\infty }^{+\infty }\,{\underline{\tilde{{\bf{G}}}}}^{ee}\,({k}_{xn})\cdot [{\tilde{J}}_{yr}({k}_{xn})\hat{{\bf{y}}}]\,{e}^{-j{k}_{xn}x}dx=0$$
(6)
with m = 0, …, Ny − 1. Now Eqs (5) and (6) can be cast as a matrix linear system
$$[{Z}^{{\rm{TM}}}({k}_{x0})][A]=0\,{\rm{and}}\,[{Z}^{{\rm{TE}}}({k}_{x0})][B]=0,$$
(7)
by using the defined spectral currents $${\tilde{J}}_{x}$$ and $${\tilde{J}}_{y}$$ for both the TM and TE modes, respectively. The unknown complex wavenumber $${k}_{{x}_{0}}$$ can be determined by calculation of the zero of the determinant for these matrices representing the eigenvalues of the linear system. The column matrices [A] and [B] contain the unknown coefficients for Aq and Br, respectively, whereas the MoM-matrix elements are defined as follows
$${Z}_{lq}^{{\rm{TM}}}({k}_{x0})=\sum _{n=-\infty }^{+\infty }\,{\tilde{J}}_{xl}(\,-\,{k}_{xn}){\tilde{G}}^{ee,xx}{\tilde{J}}_{xq}({k}_{xn})$$
(8)
$${Z}_{mr}^{{\rm{TE}}}({k}_{x0})=\sum _{n=-\infty }^{+\infty }\,{\tilde{J}}_{ym}(\,-\,{k}_{xn}){\tilde{G}}^{ee,yy}{\tilde{J}}_{yr}({k}_{xn})$$
(9)
for l, q[m, r] = 0, …, Nx − 1[Ny − 1]. The numerical evaluation of these slowly-converging spectral series can be effectively accelerated through the extraction and subsequent closed-form evaluation of their asymptotic values (for further details see38). Moreover, the unknown complex wavenumber for the fundamental mode kx0 = β0 − can finally be determined by locating the zeros of the determinant of the matrices ZTM/TE in the complex plane, by suitably selecting the proper ({kzn} < 0) or improper ({kzn} > 0) nature of the relevant space harmonics. The vertical wavenumber in the air region kzn is related to kxn by the conventional separation condition.
### Transmission-Line Representation
As is known, the vertical propagation in the two-layer structure can be analyzed by reducing Maxwell’s equations to representative transmission-line equations39. Specifically, the electric and magnetic fields produced by the source can be expressed by means of voltages and currents on the transmission lines suitably excited by a unit amplitude, as shown in the dual-layer transverse equivalent network formulation (see Fig. 2(c)). By exploiting the spectral decomposition of the field for TM waves one can write39:
$$\begin{array}{ll}\frac{d}{dz}{V}^{TM} & =\,-\,j{k}_{z}{Z}^{TM}{I}^{TM}+{v}^{TM}\\ \frac{d}{dz}{I}^{TM} & =\,-\,j{k}_{z}{Y}^{TM}{V}^{TM}+{i}^{TM}\end{array}$$
(10)
where $${V}^{TM}={\tilde{E}}_{x}$$, $${I}^{TM}={\tilde{H}}_{y}$$, ZTM = 1/YTM = kz/ωε, $${v}^{TM}=\,-\,{\tilde{M}}_{ye}$$, $${i}^{TM}=\,-\,{\tilde{J}}_{xe}$$, and $${\tilde{M}}_{ye}={\tilde{M}}_{y}-{k}_{t}/(\omega \varepsilon ){\tilde{J}}_{z}$$, where kt is the transverse wavenumber, i.e., normal to the z direction. The expression for the TE waves are omitted here for brevity. The relevant quantities can be described independently for each of the two dielectric layers and the air region for z > 0 as follows36,40
$$[\begin{array}{c}{\tilde{E}}_{x}\\ {\tilde{E}}_{y}\\ {\tilde{E}}_{z}\end{array}]=[\begin{array}{ccc}-\frac{{k}_{x}^{2}{\hat{V}}_{i}^{TM}+{k}_{y}^{2}{\hat{V}}_{i}^{TE}}{{k}_{t}^{2}} & \frac{{k}_{x}{k}_{y}({\hat{V}}_{i}^{TE}-{\hat{V}}_{i}^{TM})}{{k}_{t}^{2}} & \frac{{k}_{x}{\hat{V}}_{v}^{TM}}{\omega \varepsilon ({z}_{0})}\\ \frac{{k}_{x}{k}_{y}({\hat{V}}_{i}^{TE}-{\hat{V}}_{i}^{TM})}{{k}_{t}^{2}} & -\frac{{k}_{x}^{2}{\hat{V}}_{i}^{TE}+{k}_{y}^{2}{\hat{V}}_{i}^{TM}}{{k}_{t}^{2}} & \frac{{k}_{y}{\hat{V}}_{v}^{TM}}{\omega \varepsilon ({z}_{0})}\\ \frac{{k}_{x}{\hat{I}}_{i}^{TM}}{\omega \varepsilon (z)} & \frac{{k}_{y}{\hat{I}}_{i}^{TM}}{\omega \varepsilon (z)} & -\frac{1}{\omega \varepsilon (z)}[\frac{{k}_{t}^{2}{\hat{I}}_{v}^{TM}}{\omega \varepsilon ({z}_{0})}-j\delta (z-{z}_{0})]\end{array}]\,[\begin{array}{c}{\tilde{J}}_{x}\\ {\tilde{J}}_{y}\\ {\tilde{J}}_{z}\end{array}]\mathrm{.}$$
(11)
where z, z0 are the vertical abscissas of the field and source points, respectively.
On this basis, the evaluation of the spectral component of the relevant field quantity within each layer can be reduced to the calculation of voltages and currents produced on the equivalent transmission-line network. If only electric current densities are present (i.e., the currents on the metalizations), the electric field radiated by the structure is given by the matrix in Eq. (11). By means of the network formalism, the spectral dyadic Green’s functions $${\underline{\tilde{{\bf{G}}}}}_{ee}$$, $${\underline{\tilde{{\bf{G}}}}}_{eh}$$, $${\underline{\tilde{{\bf{G}}}}}_{he}$$, $${\underline{\tilde{{\bf{G}}}}}_{hh}$$ can be also determined36. We also note that an alternative approach would be to discretize the equivalent magnetic currents associated with the electric fields in the slots. In this case, an integral equation would be obtained by enforcing the continuity of the magnetic field across the slots and the resulting MoM matrix elements would involve the spectral Green’s dyadic $${\underline{\tilde{{\bf{G}}}}}_{hh}$$. In any case, the only assumption for the two-layer structure relies on the homogeneous and isotropic nature of the considered dielectric materials.
For the 2L-GDS under analysis, excited by an electric line current of unit amplitude directed along $$\hat{{\bf{y}}}$$ (see the reference system in Fig. 1(b)) and placed on the metallic strip at z = z0 = 0, the impressed electric density current can be written as J(r) = $$\hat{{\bf{y}}}$$δ(x)δ(z). The nature of the source along the vertical z-direction allows one to associate a 1 A current generator as modeled in Fig. 2(d), where by solving the model through circuit theory one get $$\hat{V}$$(0) = 1/(Y+(kz0) + Y(kz1, kz2)). Y+(kz0) and Y(kz1,kz2) are the input admittances at the horizontal section z = 0 looking up and looking down, respectively, to be calculated for both the TE and TM modes. Here Y is a function of kz1 and kz2, which are related to the parameters of the two substrate layers defined by h1, h2, $${\varepsilon }_{{{\rm{r}}}_{1}}$$, and $${\varepsilon }_{{{\rm{r}}}_{2}}$$. A closed-form expression can be easily determined for $${Z}_{-}^{TM/TE}({k}_{z1},{k}_{z2})$$ following standard transmission line theory. For the case at hand, i.e., for TM waves and recalling that $${\hat{V}}_{i}^{TM}=V\mathrm{(0)}$$ in Eq. (11), the expression of the relevant Green’s function, $${\underline{\tilde{{\bf{G}}}}}_{ee}$$, for the 2L-GDS can be obtained. A similar procedure can be applied to determine $${\underline{\tilde{{\bf{G}}}}}_{eh}$$, $${\underline{\tilde{{\bf{G}}}}}_{he}$$, $${\underline{\tilde{{\bf{G}}}}}_{hh}$$, whose exact formulation is not required for the LWA under design in this paper.
## Discussion
To better understand the complex dispersion properties of the proposed 2L-PPW LWA, guided-wave (GW) propagation in different planar unperturbed (i.e., nonperiodic) structures were first studied as shown in Fig. 3, namely: (i) a single-layer grounded dielectric substrate (GDS or a 1L-GDS), (ii) a GDS with a dielectric superstrate having an air region above (a 2L-GDS), (iii) a parallel-plate waveguide (PPW) filled by two dielectrics (a 2L-PPW), and (iv) a PPW completely filled by a dielectric medium.
Possible modes are shown in Fig. 3(a) as well as the cross-sectional views of the unperturbed structures (see Fig. 3(b–e)). In the analysis, all values for the bottom layer were held constant ($${\varepsilon }_{{{\rm{r}}}_{1}}=10.2$$ and thickness h1 = 1.27 mm) while an air-dielectric interface, or a dielectric-dielectric interface and metal, was positioned on top when relevant (with thickness h2 = 1.524 mm and $${\varepsilon }_{{{\rm{r}}}_{2}}=3$$). It should also be mentioned that both the 2L-GDS and the 2L-PPW can excite an evanescent field in the top superstrate layer, allowing for design control of the vertical attenuation constant of the TM guided wave, which can be described as a TM SW-like mode. This mode has been exploited for LW excitation and antenna radiation here and in15,16.
The TM SW-like mode can also be more formally defined as the quasi-TEM mode of the 2L-PPW (see Fig. 3): the simulated magnitude and phase of the electric field for this mode is shown in Fig. 4(a) (for comparison see also Fig. 6(b) in16, where the same distribution of the TM0 SW of a single layer GDS is reported). This (unperturbed) TM SW-like mode of the 2L-PPW is the fundamental mode of the supporting two-layer structure, has a zero cutoff frequency, and is moderately dispersive: its normalized phase-constant varies from about 2.1 to 2.9 over a 40 GHz bandwidth (see black curve in Fig. 3(a)). Conversely, all other comparative modes, i.e., the TM0 GDS and the TM1 PPW, vary from 1 and 0, respectively, to about 2.9 over the same frequency region.
We stress that the physical modal behavior of the 2L-PPW is considerably advantageous when designing the proposed 2L-LWA. In particular, by suitable sizing of the perturbing annular slots, with an increase in frequency, a slowly scanning beam can be realized. The simulated electric-field transverse distribution for this structure is shown in Fig. 4(a) (top and bottom panels indicating amplitude and phase, respectively) and compared with that of the LWA under analysis, whereas the dispersive behavior of the relevant n = −1 spatial harmonic is presented in the next section.
It is important to note that the TM SW-like modes, i.e., both the TM0 mode of the 2L-GDS and the quasi-TEM mode of the 2L-PPW are the dominant modes for these kind of structures (as shown in Fig. 3(a)). This is important when considering the operational frequency bandwidth for the practically designed nondirective TM0 SWL that was optimized to have more than a 13% impedance bandwidth (|S11| < −10 dB) centered at 23 GHz, when considering a single-layer GDS implementation7. This suggests that efficient coupling into both the 2L-GDS and the 2L-PPW is also possible for the considered bi-directional TM SWL (see Fig. 2(a) inset), mainly because the phase constants are of similar value and since the majority of the fields are contained at the dielectric-dielectric interface for the 2L-PPW. Overall, the modal behavior for this quasi-TEM mode of the 2L-PPW (see Fig. 3(a)), suggests that one can introduce a small unit -cell perturbation (i.e., w < p/2) within the top metallic sheet for TM LW excitation. Futhermore, this perturbation should also be large enough to generate appreciable values of the leakage rate for antenna radiation. A parametric analysis on the period p for the considered 2L-PPW will be presented in the next subsections.
The optimal design frequency for the considered two-layer antenna, dictated by the employed SWL, is fixed to 23 GHz. As discussed next, this frequency lies within a stopband region for the perturbed version of the TE1 mode of the employed 2L-PPW, which was a requirement for efficient TM0 SW excitation and to avoid spurious radiation. Therefore, this further suggests that similar dominant-mode coupling efficiencies are expected for the 2L-LWA, since the normalized phase constant behavior for the TE1 mode of the single-layer GDS is also very similar to the TE1 mode of the 2L-GDS at 23 GHz.
By selecting a proper design frequency, and employing commercially available dielectric substrates (see Fig. 2(a)) along with a practical SW feed system with a 50 -Ω connecting transmission line (i.e., the nondirective SWL), one ensures that the bottom dielectric layer can strongly support the selected and dominant TM mode of the guiding structure (i.e., the 2L-GDS or the 2L-PPW) for efficient LW excitation and radiation.
## Results
### LW Analysis of the Periodically-Loaded Guiding Structure
As is well-known, based on LW theory, the main properties of the antenna radiation pattern can be predicted through a careful inspection of the leaky-mode dispersion behavior of the periodically-perturbed guiding structure. As shown in Figs 4(b) and 5(b), the normalized phase constant of the considered n = −1 spatial harmonic increases linearly with frequency passing through zero, i.e., β−1/k0 = 0, at a frequency value fsb around 24.7 GHz and defining a proper LW with a far-field pointing angle that will scan from backward endfire to broadside and an improper LW from broadside to forward endfire.
The LW dispersion analysis starts from the single-layer ‘bull-eye’ LWA design discussed in7 where a substrate having $${\varepsilon }_{{{\rm{r}}}_{1}}=10.2$$ and thickness h1 = 1.27 mm were chosen, and then different superstrates having variable thickness were added on the top of the single-layer GDS. This starting point ensures that the impedance matching features of the aforementioned TM SWL7 are preserved for the 2L-LWA under study. Following this added superstrate variation, a parametric analysis is provided in Fig. 4(b) for a selection of two-layer structures capable of providing the required behavior for the LW attenuation constant around the stopband frequency fsb. As observed in Fig. 4(b), a fairly symmetric bump for α/k0 around fsb is possible for the TM LW mode using a top substrate thickness of 1.48 mm and dielectric constant $${\varepsilon }_{{{\rm{r}}}_{2}}=3$$. Fortunately, a dielectric thickness of 1.52 mm is commercially available and the TM LW mode for this structure can provide similar modal behavior.
Around the broadside frequency fsb an open-stopband behavior is observed, where the attenuation constant α has a null point preceded or followed by a significant maximum. This behavior is typically responsible for a deterioration of the radiation performance for 1-D LWAs and the onset of undesired reactive effects. However, in most of the cases shown in Fig. 4(b) for our examined 2L-configuration, it can be observed that the maximum value of the normalized leakage constant is considerably lower than that obtained in the single-layer MSG (as commented in15, Fig. 3), and still allows for efficient radiation for the 2-D LWA. As shown in Fig. 4(b), for all other dielectric superstrate thicknesses, a symmetric bump around fsb was not observed.
In Fig. 5(a) results obtained with the modal Bloch approach based on full-wave CST simulations of a finite number of unit cells are also provided. Good agreement is observed with the MoM dispersion analysis and CST (see, e.g.15, and references therein). Typically, the number of unit cells simulated depends on the complexity of the structure; for the case at hand, good results have been obtained with 15 cells. The agreement between the MoM and the hybrid Bloch-wave approaches is very good both for the proper and improper branches. Similar results are also shown for α/k0 when the substrate losses are included.
We note that the presence of a symmetric bump around fc, when also considering dielectric losses, permits to eliminate the null point of the attenuation constant. This allows for the mitigation of the open-stopband behavior, as also commented in15, and in addition determines a wide frequency band where |β−1| < α or |β−1| ≈ α. In particular, the possibility of almost equalizing the value of the attenuation constant (having values ranging from 0.025k0 to 0.05k0) around the phase constant null (i.e., around fc) for the n = −1 spatial harmonic, when also the open stopband is mitigated or possibly suppressed, can be suitable for obtaining continued broadside radiation in a wide frequency band for the two-sided 2-D LWA as proposed here. Dispersion curves for β−1/k0 and α/k0 for a superstrate having permittivity $${\varepsilon }_{{{\rm{r}}}_{2}}=3$$ and different thickness h2 for the superstrate, as well as different periodicities for the MSG, are also shown in the parametric analysis of Fig. 5(b). Again, as in Fig. 4(b), it can be observed that the required double-symmetric bump for α/k0 is not obtained for any of these alternative configurations.
The corresponding Brillouin diagrams for perturbed TM and TE modes are presented in Fig. 6(a,b). The periodicity for this MSG and the substrate values are representative of the fabricated 2L-LWA. In particular, the perturbed fundamental TM spatial harmonic (n = −1) and the phase constants of the two related spatial harmonics (unperturbed), supported by the insightful cases (i.e. the 2L-GDS and the 2L-PPW) for the 2L-LWA, are shown in Fig. 6(a), as was done in Fig. 3 for the dispersion curves of the unperturbed cases. The almost perfect linear scanning behavior inside the fast-wave region (FWR, depicted with a light green background) is clearly observable and confirms the effectiveness of the proposed design. In addition, the Brillouin diagram for the related TE cases is shown in Fig. 6(b). A confined range relevant to the TM broadside radiating frequency range for our proposed LWA (i.e., around 23 GHz) is also shown in Fig. 6(b). It can be observed that the perturbed TE1 mode (TE1 2 L LW-GW) is in a stopband regime and outside the FWR when the dominant TM LW radiates5. This defines a reactive TE mode which does not contribute to antenna radiation.
Impressive results were also recently reported for an E-band corporate-fed slot array with a 17.2% broadside radiating bandwidth in47. That work was based on an involved and vertically stacked (multi-layer) corporate-feed slot array system which could be considered significantly involved to design, simulate, and optimize, as well as to numerically model. Moreover, the antenna fabrication and assembly process for this W-band slot antenna array and cavity-based structure might introduce some significant tolerance variations and thus cause some discrepancies between the simulated and measured performance. Regardless, the results in47 are impressive and suggest that with more layering and careful design of our proposed 2L-LWA, improved bandwidth may be possible.
Following these above discussions, we do feel that our proposed dual-layer bull-eye LWA represents a very good alternative with respect to the structures proposed in43,44. In fact, our design provides a pencil beam consistently pointing at broadside, in contrast with the fan beam or the omnidirectional beam provided by43,44, respectively. We would also like to stress that, since our 2-D LWA is based on a GDS with a fully integrated SWL feeding system, it can be considered more convenient when compared to43,44 for applications requiring integrated RF circuity and EM shielding effectiveness from the radiating aperture. This is because our SWL feed system is incorporated into the ground plane and on the backside of the antenna at its center and removed from any radiating element. This feed placement allows for simple RF circuit and IC ground plane integration for amplifiers, mixers, chip filters, etc. for communication applications, radar, and wireless power transmission systems.
### Antenna Simulations and Measurements
Figure 7(a) reports a comparison of the simulated input impedance matching of a 1L-GDS, a 2L-PPW, a 1L-LWA (with two different configurations of the MSG, as previously examined by the authors in7 and described in the figure inset), and the 2L-LWA of this work. All the structures are fed by the same non-directive SWL, which provided very good matching over a wide impedance bandwidth, and regardless of the top structure. To better appreciate the improved broadside radiation, Fig. 7(b) reports a comparison between the directivity and the realized gain of the 1L-LWA and the 2L-LWA versus the normalized frequency at broadside. As expected the 2L-LWA design of this work provides improved performance, in particular, an enhanced radiation bandwidth at broadside (i.e. θ = ϕ = 0°) when compared to the 1L-LWA. Finally, 7(c) reports the radiation efficiency, at broadside (again for θ = ϕ = 0°), versus the normalized frequency, for both the 1L-LWA (for two different configurations of the MSG) and the 2L-LWA. The latter shows more persistent (i.e., wide-band) broadside radiation, which can also be physically explained by comparing the LW attenuation constants (see the Fig. 7(c) inset in the bottom right corner) and observing the double-symmetric ‘bump’ provided by the 2L-LWA implementation. This configuration is able to provide sustained TM leakage and radiation over a wider frequency range when compared to the single-layer topologies.
The 2L-LWA prototype presented in Fig. 2(a) and simulated in Fig. 7 was also measured in a calibrated anechoic chamber. The measured and simulated maximum realized gain and the beam pointing angle versus frequency are shown and discussed. As clearly visible in Fig. 8(a,b), a frequency shift can be observed between the simulated and measured curves when considering a dielectric constant of the bottom GDS equal to $${\varepsilon }_{{{\rm{r}}}_{1}}=10.2$$, whereas a very good agreement is obtained when $${\varepsilon }_{{{\rm{r}}}_{{\rm{2}}}}=11.5$$ is used in the full-wave simulation. This is due to the tolerance and anisotropy for the relative dielectric constant for the commercial substrate48,49,50 and is consistent with the results for the single-layer bull-eye LWA previously reported by some of the authors (for example, see Fig. 17 from7), since the exact same substrate was employed again for our new 2L-LWA (i.e., by removing the radial microstrip top rings by wet chemical etching). Specifically, the same bottom dielectric slab and ground plane, and thus the same TM SWL from7, were explicitly employed for the bottom layer of the antenna under study in this paper. Then, the top dielectric-superstrate and MSG were affixed to this original GDS.
Regardless of these features, measurements and full-wave simulations generally show a consistent gain and pointing angle profile versus frequency in Fig. 8(a,b). Also, in the open stopband frequency range, a minor reduction in the realized gain is observed at broadside in both the measurements as well as the simulations (see Fig. 8(a)) demonstrating consistent results. However, it should be noted that the experimental results in Fig. 8(a,b) show a minor discrepancy with the full-wave simulations for frequencies around 23 GHz. This could be related to some practical variations in the relative dielectric constant for the top dielectric layer as well as some minor fabrication and assembly tolerance errors for the measured prototype. Thus some minor discrepancies between the measurements and the full-wave simulations, due to these practicalities, can generally be expected when operating at microwave and millimeter-wave frequencies.
Measured beam patterns normalized to the observed maximum at 22.8 GHz as well as 2-D contour gain patterns in the azimuth and elevation planes are reported in Figs 9 and 10, respectively. It is possible to appreciate the single pencil-beam pattern observed at broadside from about 22 GHz to about 23.7 GHz (see Fig. 10), confirming the noted bandwidth of about 8.7% as described in Fig. 8(a). For this broadside frequency range, and over the operating bandwidth of the antenna, sidelobe levels are generally less than 10 dB below the main beam maximum (but in a worst case about 7 dB) which may be acceptable for certain communication applications. Additional measurements and simulations for the fabricated 2L-LWA can be found in16 where measured 1-D and 2-D realized gain plots were provided for other frequencies along with additional comparisons to full-wave simulations.
We note in Fig. 8(a,b) that the obtained radiation bandwidth at broadside extends over almost 2 GHz. This result exceeds what is expected on the basis of the modal dispersion analysis only (see Fig. 5(a), where β ≤ α from 23.2 GHz to around 24 GHz with $${\varepsilon }_{{{\rm{r}}}_{1}}=10.2$$ and from 22.3 GHz to around 22.8 GHz with $${\varepsilon }_{{{\rm{r}}}_{1}}=11.5$$, whose relevant dispersion curve is essentially a down-shifted version of the same curve and it is not shown here for brevity). Interestingly, in35 it has been observed that for LWAs of finite length the beam-splitting condition is not strictly given by the condition β ≈ α, valid in the case of a LWA with infinite length, but by β ≈ nsα, with ns ≥ 1 and ns reaching values of 6 or more for pratical LWAs.
In Fig. 11(a), the theoretical curve showing the behavior of β/α versus F, the ratio of the power remaining at the ends of the LWA and the input power (indicated here with F as in35) is reported in red: the intersection points (blue on the proper branch and green on the improper one, 21.8 GHz and 23.5 GHz, respectively) of this curve with that relevant to the proposed design allow us to predict the frequency range for which broadside radiation is generated by the finite-length (i.e., truncated) LWA. The theoretical range (i.e., 22.8 GHz to 24.7 GHz) obtained with the ‘ideal’ value for the relative dielectric constant ($${\varepsilon }_{{{\rm{r}}}_{1}}=10.2$$) revealed by the dotted gray curve in Fig. 11(a) is in very good agreement with the simulated result shown in Fig. 8(b) (see the corresponding dotted gray curve). However, the theoretical frequency range (i.e., 21.8 GHz to 23.5 GHz) obtained with the ‘actual’ ($${\varepsilon }_{{{\rm{r}}}_{1}}=11.5$$) permittivity revealed by the solid black curve in Fig. 11(a) are in good agreement with both the simulated and measured results in Fig. 8(b) for the absolute value of the beam pointing angle (dashed dark gray and solid red curves).
By comparing the results presented in Figs 10 and 11(a), it is possible to observe a small frequency shift between the experimental and theoretical limit range of frequencies for broadside radiation. This is mostly likely due to the dielectric constant in the vertical direction of the practical substrates, which can be different than in the horizontal direction48,49,50. This anisotropy, which can be significant for thick substrates, is a result of manufacturing and shows a frequency dependence (see48, p. 758, Appendix A for an quite exhaustive discussion on these aspects). As discussed in48 by increasing the value for the dielectric constant (a similar procedure is reported in7) to achieve better agreement between the simulations and the measurements (see Fig. 8(a,b)). Regardless of these studies, the relevant broadband behavior is very well predicted by the extended beam-splitting condition for the considered and truncated LWA. We further stress that the condition β ≈ α is still valid for LWA design since it predicts the peak of the maximum realized gain35, as is confirmed by the maximum value of the gray curve in Fig. 8(a), obtained at around 23.3 GHz. This is in agreement with the condition β ≈ α observed in Fig. 5(a).
It is interesting to note that a ‘staircase-like’ function for the beam pointing angle is observed in both the measurements and simulations for Fig. 8(b), similar to7. However, the nonlinear scanning behavior is observed for off-broadside frequencies only. For example, from about 20.2 GHz to 20.5 GHz the beam pointing angle in Fig. 8(b) is fixed at ±40° (considering $${\varepsilon }_{{{\rm{r}}}_{1}}=10.2$$); moreover, the normalized LW attenuation constant is small, i.e., α/k0 < 0.01, which implies that the LW field may not be the dominant field on the antenna aperture. The mentioned ‘staircase-like’ function in the beam pointing angle can in fact be related to the presence of azimuthal current distributions generated by the slot ring modes, as depicted in Fig. 11(b) where such surface currents are plotted at 20.5 GHz on the top metallic aperture of 2L-LWA.
A very similar response was observed in7 for the constituent microstrip rings. We stress that these resonances for the 2L-LWA under study are related to the presence of the coplanar waveguide feeding line connected to the nondirective SWL. This is because a TEM mode is generated on the feedline (from the substrate periphery) with power guided to the planar TM source positioned at the origin. More specifically, the Ez field lines of the transmission line can be aligned with that of the relevant field component for the radial slot ring modes. However, their contributions to the radiated far-field are negligible (see also7). This can be observed in the measurements and simulations at 20.5 GHz as the realized gain is below 2 dBi for all cases in Fig. 8(a), and, less than −5 dBi for the simulations when $${\varepsilon }_{{{\rm{r}}}_{1}}=10.2$$.
## Conclusion
A dual-layer radial metal slot-grating planar antenna providing two-sided conical-sector and pencil beam patterns with a wide bandwidth for broadside radiation has been proposed. By means of a full-wave dispersion analysis for the reference structure, the complex modal behavior has been described. Through this modeling, optimized parameters for the 2L-PPW and MSG have been selected in order to mitigate the open stopband effects of the leaky mode responsible for radiation. The capabilities of the finite-length LWA, in providing persistent and continued broadside radiation over a wide frequency range, have been experimentally assessed and related to the more relaxed beam splitting conditions which characterize truncated LWA structures of practical size. Measured maximum gain values greater than 15 dBi are observed at broadside. The final design results in a compact, low-cost, and low-profile 2L-LWA prototype demonstrating consistent broadside radiation over more than an 8.5% wide-bandwidth.
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## Author information
Authors
### Contributions
S.P., D.C., P.Ba., and P.Bu. conceived the design, D.C. and S.K. wrote the manuscripts and performed full-wave simulations, S.K. and A.F., conducted the experiments, D.C., S.P., P.Ba., and P.Bu. conducted the dispersive analyses, A.G. and Y.A. contributed to the discussions on theoretical feasibility and design improvements. All authors reviewed the manuscript.
### Corresponding author
Correspondence to Davide Comite.
## Ethics declarations
### Competing Interests
The authors declare no competing interests.
Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
Reprints and Permissions
Comite, D., Podilchak, S.K., Baccarelli, P. et al. Analysis and Design of a Compact Leaky-Wave Antenna for Wide-Band Broadside Radiation. Sci Rep 8, 17741 (2018). https://doi.org/10.1038/s41598-018-35480-7
• Accepted:
• Published:
### Keywords
• Leaky-wave Antenna (LWA)
• Pencil Beam Pattern
• Structural Guidance
• Wireless Power Transmission
• ### Mechanically scanned leaky-wave antenna based on a topological one-way waveguide
• Qian Shen
• Yun You
• Sanshui Xiao
Frontiers of Physics (2020) | 2021-12-02 17:26:14 | {"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": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.7795735597610474, "perplexity": 2537.1184747058587}, "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-2021-49/segments/1637964362230.18/warc/CC-MAIN-20211202145130-20211202175130-00149.warc.gz"} |
https://www.physicsforums.com/threads/scalar-and-tensor-product.830393/ | # Scalar and tensor product
1. Sep 1, 2015
### jk22
Just a question : do we have in Dirac notation $$\langle u|A|u\rangle\langle u|B|u\rangle=\langle u|\langle u|A\otimes B|u\rangle |u\rangle$$ ?
2. Sep 1, 2015
### Geofleur
That doesn't look right to me. If we translate the Dirac notation into tensor notation then $\langle u |$ corresponds to $u^1$, a linear functional and $|u\rangle$ corresponds to $\textbf{u}$, a vector. Then $\langle u | A | u \rangle$ is the same as $u^1(A\textbf{u})$, the functional $u^1$ acting on the vector $A\textbf{u}$. Similarly, $\langle u | B | u \rangle$ corresponds to $u^1(B\textbf{u})$. Then I guess we would write $u^1(A\textbf{u})u^1(B\textbf{u})$ in Dirac notation as $\langle u | \otimes \langle u | (A| u \rangle, B| u \rangle)$. The first $\langle u |$ would act on the $A | u \rangle$ and the second $\langle u |$ would act on the $B | u \rangle$.
3. Sep 1, 2015
### DEvens
You have not defined what you mean by $\otimes$ in this case. As well, any kind of product of two operators is not necessarily going to be the same as the product of their results in brackets of this kind. How do the two operators act through each other? If they do.
4. Sep 2, 2015
### jk22
I was thinkg of A and B be square matrices and $$\otimes$$ the kronecker product.
5. Sep 2, 2015
### kith
So $A$ and $B$ act on the same space? Without context, the right-hand side looks like an unnecessary inflation of the state space but technically correct to me.
6. Sep 2, 2015
### kith
Your use of Dirac notation seems quite non-standard to me. I haven't seen it in QM texts. Instead of your $(\langle a| \otimes \langle b|) (|c\rangle, |d\rangle)$ I would write $(\langle a| \otimes \langle b|) (|c\rangle \otimes |d\rangle)$ which has the usual symmetry between bra and ket vectors.
7. Sep 2, 2015
### Geofleur
Ah, I see. In tensor analysis, particularly in the modern sort where tensors are viewed as multilinear maps, that kind of notation is common. But the bra-ket notation is not used in that context, at least not in the books I have been reading. I'm just happy that what I wrote down actually does make sense! | 2018-05-28 01:59:39 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9474446177482605, "perplexity": 359.054581045174}, "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-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00331.warc.gz"} |
https://socratic.org/questions/how-do-you-solve-the-inequality-9-2x-6-21 | # How do you solve the inequality 9<2x+6<21?
$3 < 2 x < 15$
$\frac{3}{2} < x < \frac{15}{2}$
$\therefore x \in \left(\frac{3}{2} , \frac{15}{2}\right)$. | 2019-10-20 08:40:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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": 0.7524182796478271, "perplexity": 1971.9477823602497}, "config": {"markdown_headings": true, "markdown_code": false, "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-2019-43/segments/1570986705411.60/warc/CC-MAIN-20191020081806-20191020105306-00293.warc.gz"} |
https://groupprops.subwiki.org/wiki/Collection_of_groups_satisfying_a_strong_normal_replacement_condition | # Collection of groups satisfying a strong normal replacement condition
## Definition
Suppose $\mathcal{S}$ is a finite collection of finite $p$-groups, i.e., groups of prime power order where the prime is $p$. We say that $\mathcal{S}$ satisfies a strong normal replacement condition if it satisfies the following equivalent conditions:
1. For any finite $p$-group $P$ that contains a subgroup $H$ isomorphic to an element of $\mathcal{S}$, $P$ contains a normal subgroup $K$, also isomorphic to an element of $\mathcal{S}$, such that $K$ is contained in the normal closure of $H$ in $P$.
2. For any finite $p$-group $P$ that contains a 2-subnormal subgroup $H$ isomorphic to an element of $\mathcal{S}$, $P$ contains a normal subgroup $K$, also isomorphic to an element of $\mathcal{S}$, such that $K$ is contained in the normal closure of $H$ in $P$.# For any finite $p$-group $Q$ and normal subgroup $P$ of $Q$, if there exists a subgroup of $P$ isomorphic to an element of $\mathcal{S}$, there exists a subgroup of $P$ that is normal in $Q$ and is isomorphic to an element of $\mathcal{S}$. | 2020-09-30 08:35:38 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 33, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9019353985786438, "perplexity": 54.426626791398014}, "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-2020-40/segments/1600402123173.74/warc/CC-MAIN-20200930075754-20200930105754-00059.warc.gz"} |
http://math.stackexchange.com/questions/286878/derivative-of-a-function-defined-by-conditionals | # Derivative of a function defined by conditionals
The question is:
Which values $a$ and $b$ can assume to be possible to derive the function $f$ at $x = 1$
$f(x) = \left\{ \begin{array}{rl} x^2 &\mbox{$x < 1$} \\ ax + b &\mbox{$x \geq 1$} \end{array} \right.$
Here is my progress
The one way i can wonder to solve this is using the definition of derivatives. So i started by verifying the following limit
$$\lim_{x \to 1^{+}} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^{+}} \frac{ax + b - (a + b)}{x - 1}$$ $$= \lim_{x \to 1^{+}} \frac{ax - a}{x - 1} = \lim_{x \to 1^{+}} \frac{a(x - 1)}{x - 1} = \lim_{x \to 1^{+}} a = a$$
And then i verify the limit by the left side
$$\lim_{x \to 1^{-}} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^{-}} \frac{x^2 - (a + b)}{x - 1}$$
From this point i can't go ahead.
-
You're almost there! If $a+b=1$, then your left limit is... If $a+b\neq 1$, then... – 1015 Jan 25 '13 at 19:24
Very roughly: you want $f$ continuous and differentiable at $x=1$. Continuity means setting the two expressions equal to each other (why?):
$$1^2 = a(1) + b$$
differentiability means matching the value of the derivative of each expression at $x=1$; that is
$$2 (1) = a$$
-
I guess i found the answer
If the function is differentiable, then it's continuous. So, let's verify if it's continuos
To be continuos it must satisfy the following conditions
- the limit must exists when $x$ tends to $a$
- $f(a)$ must be equal to the limit when $x$ tends to $a$
So, lets begin
$$f(1) = a + b$$ $$\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} x^2 = 1$$ $$\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} ax + b = a + b$$
So, we need the equality $\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{+}} f(x) = f(1)$ to be true. We get $a + b = 1$
Verifying if the function is differentiable
$$\lim_{x \to 1^{-}} \frac{f(x) - f(1)}{x - 1}$$ We have $f(1) = a + b = 1$ , so
$$\lim_{x \to 1^{-}} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1^{-}} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1^{-}} x + 1 = 2$$
Now the limit by the right side
$$\lim_{x \to 1^{+}} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^{+}} \frac{ax + b - (a + b)}{x - 1} = \lim_{x \to 1^{+}} \frac{ax - a}{x - 1} = \lim_{x \to 1^{+}} \frac{a(x - 1)}{x - 1} = a$$
As we need this limits to be equal, so $a = 2$
We had the condition $a + b = 1$, and $a = 2$, so we can substitute $2$ in $a$, then $2 + b = 1 \leftrightarrow b = 1 - 2 \leftrightarrow b = -1$
Finally we have that if $a = 2$ and $b = -1$ the function $f$ is differentiable at $x = 1$
- | 2015-04-19 05:25:00 | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9837821125984192, "perplexity": 313.57942146064283}, "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-2015-18/segments/1429246637445.19/warc/CC-MAIN-20150417045717-00249-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.biostars.org/p/9471954/ | How to find the names of elements while parsing PDB files via biopython?
1
0
Entering edit mode
4 months ago
Tzunami ▴ 10
Hi all! Biopython has methods to get the atoms' names but the returned values are of the form CA, CB, HD etc. Is there a way to get just the elements' name (for example instead of CA, CB I need to get C, C indicating two carbon atoms). Thank you!
Biopython PDB • 188 views
0
Entering edit mode
12 weeks ago
Tzunami ▴ 10
Found the solution (atom entries have an attribute called element which returns the element of the corresponding atom) -
for model in structure:
for chain in model:
for residue in chain:
for atom in residue:
print(atom.element) # atom.element can be used to access the element | 2021-10-21 00:24:49 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19444477558135986, "perplexity": 7019.086774550085}, "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-2021-43/segments/1634323585353.52/warc/CC-MAIN-20211020214358-20211021004358-00340.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/174064-show-columns-linearly-independent-if-homogeneous-system-has-only-trivial-solution-print.html | # show columns are linearly independent if homogeneous system has only trivial solution
You only need to express the system $Ax=0$ in the form $x_1C_1+\ldots+x_nC_n=0$ and apply the definitions of linearly independence and solution of a system. | 2015-08-05 13:42:53 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.7152516841888428, "perplexity": 199.07831037468063}, "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-2015-32/segments/1438044271733.81/warc/CC-MAIN-20150728004431-00125-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.107.240503 | # Synopsis: The Bound and the Free
Precisely prepared photon states can probe quantum statistical phenomena and generate intriguing forms of quantum entanglement.
One of the strangest of the strange manifestations of quantum mechanics is entanglement, a condition in which the states of distant objects can be intimately correlated. In practical terms, entanglement is viewed as a means to rapid solution of some hard computational problems by quantum computing. During the 1990s, theorists proposed that entanglement actually comes in two flavors: “bound” entanglement, such as the entangled singlet state of two spin-$1/2$ particles that cannot be reduced to any simpler form, and “free” entanglement, in which a complex entangled state can be distilled down into a more basic set of states. In recent years, claims of experimental confirmation of bound entanglement have been made, but these are controversial. Writing in Physical Review Letters, James DiGuglielmo at Leibniz University, Germany, and colleagues report their experiments on unconditional preparation of bound states of light.
Previous experiments have typically examined correlations with “postselection” methods to filter desired events from an initial distribution, however, DiGuglielmo et al. have designed a system to deterministically and precisely prepare their entangled states. The authors create four continuous-variable entangled laser fields with optical parametric amplifiers and verify that they have created bound entangled states by means of high-efficiency detectors to measure the correlations. The system offers not only technological utility in preparing exact states for future experiments, but the research team also provides a tool for studying irreversibility at the quantum level to better characterize the connections between quantum information and thermodynamics. – David Voss
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## Subject Areas
Quantum Information
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A proposed method of generating phonon states for quantum applications uses a single electron trapped in a suspended carbon nanotube. Read More » | 2016-12-03 02:32:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.4649180769920349, "perplexity": 2024.685547648813}, "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-2016-50/segments/1480698540804.14/warc/CC-MAIN-20161202170900-00147-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://www.speedsolving.com/threads/what-cube-should-i-get-the-puzzle-choice-thread.20726/page-644 | #### ObscureCuber
##### Member
I don't think it's a conspiracy so much as top flagship cubes mainly differentiate themselves by feel, and I can believe that if you are a top cuber who averages sub 7 or something than the feel does matter enough for practicing and competitions to justify the cost. I'm slow but I find myself practicing a lot more with cubes I like the feel of.
i said slower cubers,
not fast ones
also that was a simi-joke
and if you're slower and use a gan(and have used other flagships before)
than 90% of the time you wont like the feel,
but people in videos even though they aren't sub 10 or sub 15 or whatever like gan cubes a lot more than you would think.
#### EliteCuber
##### Member
how come a lot of beginner cubers (1min to 1min30sec) have expensive cubes lol
#### Skewb_Cube
##### Member
how come a lot of beginner cubers (1min to 1min30sec) have expensive cubes lol
They probably have the mindset that if something is expensive, then it's good.
#### JP cubing
##### Member
full disclosure: I don't own a GAN cube. I tried my friend's Angstrom GAN X and it was very light and papery. I couldn't get a full opinion on it because he had his magnets set really strong but I don't think they are significantly better than other top cubes. I liked his GTS3M better
I couldn't help notice, but very nice pic of trump holding a curvy copter.
#### ZB2op
##### Member
we don't know the price yet tho
ppl think its gonna be almost $100 so if u want a rlly good gan cube just buy the xs. if i were u i'd stick with ur current cube #### ObscureCuber ##### Member ppl think its gonna be almost$100 so if u want a really good gan cube just buy the xs. if i were u i'd stick with your current cube
People are literally gonna spend 100 dollars on a single ROOBIKS COOBE
i mean that's ridiculous, the xs was 60(or 70 i dont remember) when i released i thought that was expensive but still reasonable.
100? even 80 seems crazy to me, but a wii instead they sell around 90 today(much better value)
#### zslane
People have spent more than $100 on flagship cubes with "pro shop" setups, so I guess spending that kind of money on a single 3x3 puzzle is not that crazy, at least not to everyone. I think I would have difficulty spending that much on a single cube unless it had some kind of high collectability factor or something, but to each their own. #### Nmile7300 ##### Member People have spent more than$100 on flagship cubes with "pro shop" setups, so I guess spending that kind of money on a single 3x3 puzzle is not that crazy, at least not to everyone. I think I would have difficulty spending that much on a single cube unless it had some kind of high collectability factor or something, but to each their own.
I can't find a single premium cube on The Cubicle for more than 70 dollars.
#### ObscureCuber
##### Member
I can't find a single premium cube on The Cubicle for more than 70 dollars.
also there ripoffs because just getting the same lubes and setting it up your self is like 30$cheaper the xs supernove is like 110$ i think
#### zslane
FWIW, a Cosmic GAN 356 X Magnetic 3x3 (Numerical IPG) cube with PVC coating comes to $113.90. #### ObscureCuber ##### Member FWIW, a Cosmic GAN 356 X Magnetic 3x3 (Numerical IPG) cube with PVC coating comes to$113.90.
the gan x? not even an Xs? jeez...
#### MarkA64
I probably won't buy the 11 just for corner magnets and a new feel; the X and XS are fine and I main the Qiyi MS right now which you can get set-up for like $25. #### ObscureCuber ##### Member I probably won't buy the 11 just for corner magnets and a new feel; the X and XS are fine and I main the Qiyi MS right now which you can get set-up for like$25.
qiyi MS and rs3m are amazing,
i main the MS right now
and am gonna switch to the rs3m since i need a less harsh/strong magnet cube for roux
#### I'm A Cuber
##### Member
Do people actually think they are going to sell it for $100, or are they just joking? I don’t think that kind of price hike would happen in such a short amount of time. I think that$70 would be the most they would increase the price to. Anything more, and their sales would absolutely plummet.
#### EliteCuber
##### Member
Do people actually think they are going to sell it for $100, or are they just joking? I don’t think that kind of price hike would happen in such a short amount of time. I think that$70 would be the most they would increase the price to. Anything more, and their sales would absolutely plummet.
no they are not joking
#### SpeedyCube
##### Member
Official price is out: $65-$70 for the new Gan 11 M Pro, depending on what finish you choose.
#### Sub1Hour
##### Member
Official price is out: $65-$70 for the new Gan 11 M Pro, depending on what finish you choose.
Wait a minute, does this mean that the cubing companies finally...
actually listened to us about glossy plastic?
They probably have the mindset that if something is expensive, then it's good.
Very true. At comps I’ve been to, the bottom 25% of competitors used around 85-90% of all the Gan cubes used in competition. It’s basically like seeing a supreme logo on a cube and assuming that, since it’s “hype”, that it’s the best. I’ve only used 1 Gan cube in competition, the 354 for oh, and aside from that one round I’ve never used one since. The 354 v1 IMO was the last good Gan cube but I’ll stop myself there before I go off on a tangent. Back to the slow cubers and Gan cubes, most of the slower cubers don’t try other cubes to see what fits them best, they see what people like faz use and assume it’s automatically the best for everyone
New Cubers, don’t just buy the most expensive or most popular cube and assume it’s the best, go to a competition once they return and try some cubes there to see what you like.
#### SpeedyCube
##### Member
At comps I’ve been to, the bottom 25% of competitors used around 85-90% of all the Gan cubes used in competition. It’s basically like seeing a supreme logo on a cube and assuming that, since it’s “hype”, that it’s the best. I’ve only used 1 Gan cube in competition, the 354 for oh, and aside from that one round I’ve never used one since. The 354 v1 IMO was the last good Gan cube but I’ll stop myself there before I go off on a tangent. Back to the slow cubers and Gan cubes, most of the slower cubers don’t try other cubes to see what fits them best, they see what people like faz use and assume it’s automatically the best for everyone
New Cubers, don’t just buy the most expensive or most popular cube and assume it’s the best, go to a competition once they return and try some cubes there to see what you like.
I completely agree, but with one comment to add. There’s a reason why name-brand stuff is name-brand, in any industry. It generally means it’s a really good product! In this case I don’t think anyone would debate the fact that a $65 dollar Gan cube is a good cube. The debate comes from, is it the best cube for your particular situation? Your point of not automatically jumping on the name-brand bandwagon is a good one. But... for someone who wants a good cube and is not able to go to a comp or club, buying the name-brand cube (Gan) is a reasonable bet that you’ll get a very nice cube. #### Nmile7300 ##### Member I completely agree, but with one comment to add. There’s a reason why name-brand stuff is name-brand, in any industry. It generally means it’s a really good product! In this case I don’t think anyone would debate the fact that a$65 dollar Gan cube is a good cube. The debate comes from, is it the best cube for your particular situation? Your point of not automatically jumping on the name-brand bandwagon is a good one. But... for someone who wants a good cube and is not able to go to a comp or club, buying the name-brand cube (Gan) is a reasonable bet that you’ll get a very nice cube.
I'm not sure you get the idea that Gan is "name brand" and other companies are not. Technically, Rubik's is name brand. | 2020-10-27 06:16:19 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.3294002115726471, "perplexity": 3210.7923230045812}, "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-2020-45/segments/1603107893402.83/warc/CC-MAIN-20201027052750-20201027082750-00629.warc.gz"} |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-and-chemical-reactivity-9th-edition/chapter-17-principles-of-chemical-reactivity-other-aspects-of-aqueous-equilibria-study-questions-page-677c/58 | ## Chemistry and Chemical Reactivity (9th Edition)
Pure water: $Ksp=[Ag^+][Br^-]=s^2$ $s=\sqrt{5.4\dot{}10^{-13}}=7.35\dot{}10^{-7}\ M$ With NaBr: $[NaBr]=(0.15\ g\div 102.89\ g/mol)/(22.5/1000\ L)=0.065\ M$ $Ksp=s\dot{}(0.065+s)$ $s=8.334\dot{}10^{-12}\ M$ | 2020-07-09 18:08:46 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.8933841586112976, "perplexity": 10478.798342932645}, "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-2020-29/segments/1593655900614.47/warc/CC-MAIN-20200709162634-20200709192634-00100.warc.gz"} |
https://hal.univ-reunion.fr/hal-01477258 | # Length of an intersection
Abstract : A poset $\bfp$ is well-partially ordered (WPO) if all its linear extensions are well orders~; the supremum of ordered types of these linear extensions is the {\em length}, $\ell(\bfp)$ of $\bfp$. We prove that if the vertex set $X$ of $\bfp$ is infinite, of cardinality $\kappa$, and the ordering $\leq$ is the intersection of finitely many partial orderings $\leq_i$ on $X$, $1\leq i\leq n$, then, letting $\ell(X,\leq_i)=\kappa\multordby q_i+r_i$, with $r_i<\kappa$, denote the euclidian division by $\kappa$ (seen as an initial ordinal) of the length of the corresponding poset~:$\ell(\bfp)< \kappa\multordby\bigotimes_{1\leq i\leq n}q_i+ \Big|\sum_{1\leq i\leq n} r_i\Big|^+$ where $|\sum r_i|^+$ denotes the least initial ordinal greater than the ordinal $\sum r_i$. This inequality is optimal (for $n\geq 2$).
Document type :
Preprints, Working Papers, ...
Domain :
http://hal.univ-reunion.fr/hal-01477258
Contributor : Réunion Univ <>
Submitted on : Monday, February 27, 2017 - 11:31:59 AM
Last modification on : Thursday, March 28, 2019 - 11:24:10 AM
### Identifiers
• HAL Id : hal-01477258, version 1
• ARXIV : 1510.00596
### Citation
Christian Delhommé, Maurice Pouzet. Length of an intersection. 2015. ⟨hal-01477258⟩
Record views | 2019-08-26 06:16:15 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9508434534072876, "perplexity": 1905.525258337649}, "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-2019-35/segments/1566027330968.54/warc/CC-MAIN-20190826042816-20190826064816-00225.warc.gz"} |