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http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-doi-10_4064-fm175-2-3 | PL EN
Preferencje
Język
Widoczny [Schowaj] Abstrakt
Liczba wyników
Czasopismo
Fundamenta Mathematicae
2002 | 175 | 2 | 127-142
Tytuł artykułu
Potential isomorphism and semi-proper trees
Autorzy
Treść / Zawartość
Warianty tytułu
Języki publikacji
EN
Abstrakty
EN
We study a notion of potential isomorphism, where two structures are said to be potentially isomorphic if they are isomorphic in some generic extension that preserves stationary sets and does not add new sets of cardinality less than the cardinality of the models. We introduce the notion of weakly semi-proper trees, and note that there is a strong connection between the existence of potentially isomorphic models for a given complete theory and the existence of weakly semi-proper trees.
We show that the existence of weakly semi-proper trees is consistent relative to ZFC by proving the existence of weakly semi-proper trees under certain cardinal arithmetic assumptions. We also prove the consistency of the non-existence of weakly semi-proper trees assuming the consistency of some large cardinals.
Słowa kluczowe
Kategorie tematyczne
Czasopismo
Rocznik
Tom
Numer
Strony
127-142
Opis fizyczny
Daty
wydano
2002
Twórcy
autor
• Department of Mathematics, University of Helsinki, 00014 Helsinki, Finland
autor
• Department of Mathematics, University of Helsinki, 00014 Helsinki, Finland
autor
• Institute of Mathematics, The Hebrew University, 91904 Jerusalem, Israel
• Department of Mathematics, Rutgers University, New Brunswick, NJ 08903, U.S.A.
Bibliografia
Typ dokumentu
Bibliografia
Identyfikatory | {"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": 0.9001652598381042, "perplexity": 875.2879411886336}, "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-2021-04/segments/1610703531429.49/warc/CC-MAIN-20210122210653-20210123000653-00631.warc.gz"} |
http://www.ck12.org/physics/Pressure-and-Force/quiz/Pressure-and-Force-Quiz-PPB/r1/ | <meta http-equiv="refresh" content="1; url=/nojavascript/">
# Pressure and Force
Pressure is a force spread out over an area. A small force applied over a very small area can exert a large force.
%
Progress
Practice Pressure and Force
Progress
%
Pressure and Force Quiz - PPB
Teacher Contributed
Calculate force, pressure, and the area of basic geometric shapes. Explain the concept of pressure. Know that pressure is equal throughout a liquid. | {"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": 0.8729889392852783, "perplexity": 2793.3528191036125}, "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-2015-22/segments/1432207929205.63/warc/CC-MAIN-20150521113209-00189-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://forum.math.toronto.edu/index.php?PHPSESSID=45k12ld2qqchab36g4ojdlt2j7&topic=1408.0;wap2 | MAT334-2018F > Term Test 1
TT1 Problem 2 (night)
(1/1)
Victor Ivrii:
(a) $\displaystyle{\sum_{n=1}^\infty \frac{z^n}{2^n n^2}}$
(b) $\displaystyle{\sum_{n=1}^\infty \frac{z^{3n} (3n)!}{20^n (2n)! }}$
If the radius of convergence is $R$, $0<R< \infty$, determine for each $z\colon |z|=R$ if this series converges.
Heng Kan:
See the attached scanned picture.
Xiting Kuang:
Just a concern, it says in the problem that R should be positive.
Heng Kan:
I think the question means that if the radius of convergence is positive,you have to figure out whether the series is convergent at the radius of convergence. It doesn't mean the radius is always positive.
Victor Ivrii:
--- Quote from: Heng Kan on October 19, 2018, 09:45:26 AM ---I think the question means that if the radius of convergence is positive,you have to figure out whether the series is convergent at the radius of convergence. It doesn't mean the radius is always positive.
--- End quote ---
Indeed. | {"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": 0.9863683581352234, "perplexity": 1060.3099212437132}, "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/1634323584913.24/warc/CC-MAIN-20211016170013-20211016200013-00675.warc.gz"} |
https://stats.stackexchange.com/questions/389624/how-to-call-this-frequentist-interval-estimate-that-is-neither-a-prediction-inte | # How to call this frequentist interval estimate that is neither a prediction interval nor a confidence interval
This question is inspired by Confidence Interval on a random quantity?. That question introduces an interesting concept for a type of interval that is neither a prediction nor a confidence interval (possibly one could see it as a tolerance interval although I believe it is neither that).
### A frequentist interval estimate
In short: For pairs of (possibly multidimensional) variables $$x_i,y_i$$, which are both distributed according to a distribution parameterized by $$a$$, and where $$x_i|a \not\!\perp\!\!\!\perp y_i|a$$, we wish to perform interval estimation for the value of $$x_i$$ as function of $$y_i$$, where $$a$$ is unknown.
Given the following:
• Let $$X,Y$$ be random variables that are paired.
• The random variables $$X$$ and $$Y$$ follow a distribution function that is parameterized by $$a$$ $$f_{Y|a}(y|a) \equiv g_Y(y,a)$$ $$f_{X|a}(y|a) \equiv g_X(y,a)$$
• There is a known relationship between $$X$$ and $$Y$$ and $$a$$, that defines a conditional distribution for $$X$$ $$f_{X|y,a}(x|y,a) \equiv h(x,y,a)$$
• There is a sample of measured values $$y_i$$
We wish to compute:
for each $$x_i$$ a one-sided interval bound $$c(y_i,\alpha)$$ such that: $$\forall a : P(X or less strong $$\sup \lbrace P(X
That is, probability in a frequentist sense. If we would have a large sample with pairs $$x_i,y_i$$ (where we only measure $$y_i$$ and do not know $$a$$) then the frequency/fraction of 'failures' of the interval, $$x_i, should be around $$\alpha$$ independent from the true value of $$a$$ (or the smallest upper bound is $$\alpha$$).
### How do/should we call that sort of interval?
This is not a confidence interval, because the estimate is for $$X$$, which is not a (fixed) population parameter, but a random variable.
This is neither a prediction interval, because $$c(y_i,\alpha)$$ is only a region for the $$x_i$$ that is paired with $$y_i$$ and it is not a region for future values of $$X$$.
What is it?
### Example case problems
• (this one was mentioned by shabbychef in the comments and relates to the before mentioned question)
You observe returns from $$p$$ stocks in vector $$\vec{y}_i$$. Then from a sample of $$n$$ such observations, you form the Markowitz Portfolio, based on the sample mean and covariance. Then you wish to estimate the Sharpe Ratio of that sample Markowitz Portfolio.
• Say I have a batch of films for which I want to predict the strength $$X$$ of each film. Let the strength be a function of two parameters, say film thickness $$Y$$ and film density $$a$$.
Say I can not measure $$X$$ directly (would damage the film), and I do not know $$a$$ for every film, nor do I wish to measure it (say it is a costly measurement). I can, however, measure $$Y$$ for each film and I know that $$Y$$ is distributed according to some pdf that is parameterized by $$a$$.
So now the idea is to use measurements of film thickness $$Y$$, which carries information of $$a$$ to compute some confidence/prediction/tolerance/whatever interval for $$X$$ which I know depends on $$Y$$ and $$a$$. I want this interval to fail only $$\alpha$$ percent of the time.
• I think it's not useful to make the stipulation in the fourth (final) bullet, due to the dependence on the unknown parameter $a.$ You need to consider either the supremum or the infimum of the left hand side over the set of posited distributions of $Y,$ depending on your objective. – whuber Jan 28 at 22:37
• I agree. That is what I did in my answer here. Beyond that one may wonder whether there ain't better approaches for the problem in practice (but that is beyond the point of the question which is about the principle). – Martijn Weterings Jan 28 at 22:48
• Another example would be: you observe returns from $p$ stocks in vector $\vec{y_i}$. Then from a sample of $n$ such observations, you form the Markowitz Portfolio, based on the sample mean and covariance. Then you wish to estimate the Sharpe Ratio of that sample Markowitz Portfolio. – shabbychef Jan 29 at 5:42
We could describe the distribution of $$Y$$, conditional on $$X$$ and $$a$$, as a distribution parameterized by $$X$$ and $$a$$:
$$f_{Y|x,a}(y,x,a) = \frac{f_{X|y,a}(x,y,a)f_{Y,a}(y,a)}{f_{X,a}(x,a)}$$
In this view the random variable $$X$$ is a parameter in the (conditional) distribution of $$Y$$, and we could see the interval estimation of $$X$$ as a confidence interval for the parameter $$X$$.
Complications are that the estimate of $$X$$ is dependent on the value of the parameter $$a$$ which acts as a nuisance parameter, and in addition $$X$$ itselve is distributed according to distribution parameterized by $$a$$. So one may not tackle the interval estimation as a 'regular' confidence interval estimation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 65, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9739524126052856, "perplexity": 414.63114571103705}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999000.76/warc/CC-MAIN-20190619143832-20190619165832-00535.warc.gz"} |
https://wiseodd.github.io/techblog/2016/10/13/residual-net/ | $$\newcommand{\dint}{\mathrm{d}} \newcommand{\vphi}{\boldsymbol{\phi}} \newcommand{\vpi}{\boldsymbol{\pi}} \newcommand{\vpsi}{\boldsymbol{\psi}} \newcommand{\vomg}{\boldsymbol{\omega}} \newcommand{\vsigma}{\boldsymbol{\sigma}} \newcommand{\vzeta}{\boldsymbol{\zeta}} \renewcommand{\vx}{\mathbf{x}} \renewcommand{\vy}{\mathbf{y}} \renewcommand{\vz}{\mathbf{z}} \renewcommand{\vh}{\mathbf{h}} \renewcommand{\b}{\mathbf} \renewcommand{\vec}{\mathrm{vec}} \newcommand{\vecemph}{\mathrm{vec}} \newcommand{\mvn}{\mathcal{MN}} \newcommand{\G}{\mathcal{G}} \newcommand{\M}{\mathcal{M}} \newcommand{\N}{\mathcal{N}} \newcommand{\S}{\mathcal{S}} \newcommand{\diag}[1]{\mathrm{diag}(#1)} \newcommand{\diagemph}[1]{\mathrm{diag}(#1)} \newcommand{\tr}[1]{\text{tr}(#1)} \renewcommand{\C}{\mathbb{C}} \renewcommand{\R}{\mathbb{R}} \renewcommand{\E}{\mathbb{E}} \newcommand{\D}{\mathcal{D}} \newcommand{\inner}[1]{\langle #1 \rangle} \newcommand{\innerbig}[1]{\left \langle #1 \right \rangle} \newcommand{\abs}[1]{\lvert #1 \rvert} \newcommand{\norm}[1]{\lVert #1 \rVert} \newcommand{\two}{\mathrm{II}} \newcommand{\GL}{\mathrm{GL}} \newcommand{\Id}{\mathrm{Id}} \newcommand{\grad}[1]{\mathrm{grad} \, #1} \newcommand{\gradat}[2]{\mathrm{grad} \, #1 \, \vert_{#2}} \newcommand{\Hess}[1]{\mathrm{Hess} \, #1} \newcommand{\T}{\text{T}} \newcommand{\dim}[1]{\mathrm{dim} \, #1} \newcommand{\partder}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\rank}[1]{\mathrm{rank} \, #1}$$
# Residual Net
September 2015, at the ImageNet Large Scale Visual Recognition Challenge’s (ILSVRC) winners announcement, there was this one net by MSRA that dominated it all: Residual Net (ResNet) (He et al., 2015). The ensemble of ResNets crushing the classification task of other competitiors and almost halving the error rate of the 2014 winner.
Aside from winning the ILSVRC 2015 classification, ResNet also won the detection and localization challenge of the said competition. Additionally, it also won the MSCOCO detection and segmentation challenge. Quite a feat!
So, what makes ResNet so good? What’s the difference compared to the previous convnet models?
## ResNet: the intuition behind it
The authors of ResNet observed, no matter how deep a network is, it should not be any worse than the shallower network. That’s because if we argue that neural net could approximate any complicated function, then it could also learn identity function, i.e. input = output, effectively skipping the learning progress on some layers. But, in real world, this is not the case because of the vanishing gradient and curse of dimensionality problems.
Hence, it might be useful to explicitly force the network to learn an identity mapping, by learning the residual of input and output of some layers (or subnetworks). Suppose the input of the subnetwork is $x$, and the true output is $H(x)$. The residual is the difference between them: $F(x) = H(x) - x$. As we are interested in finding the true, underlying output of the subnetwork, we then rearrange that equation into $H(x) = F(x) + x$.
So that’s the difference between ResNet and traditional neural nets. Where traditional neural nets will learn $H(x)$ directly, ResNet instead models the layers to learn the residual of input and output of subnetworks. This will give the network an option to just skip subnetworks by making $F(x) = 0$, so that $H(x) = x$. In other words, the output of a particular subnetwork is just the output of the last subnetwork.
During backpropagation, learning residual gives us nice property. Because of the formulation, the network could choose to ignore the gradient of some subnetworks, and just forward the gradient from higher layers to lower layers without any modification. As an extreme example, this means that ResNet could just forward gradient from the last layer, e.g. layer 151, directly to the first layer. This gives ResNet additional nice to have option which might be useful, rather than just strictly doing computation in all layers.
## ResNet: implementation detail
He et al. experimented with 152 layers deep ResNet in their paper. But due to our (my) monitor budget, we will look at the 34 layers version instead. Furthermore, it’s easier to understand with not so many layers, isn’t it?
At the first layer, ResNet use 7x7 convolution with stride 2 to downsample the input by the order of 2, similar to pooling layer. Then it’s followed by three identity blocks before downsampling again by 2. The downsampling layer is also a convolution layer, but without the identity connection. It continues like that for several layer deep. The last layer is average pooling which creates 1000 feature maps (for ImageNet data), and average it for each feature map. The result would be 1000 dimensional vector which then fed into Softmax layer directly, so it’s fully convolutional.
In the paper, He et al. use bottleneck architecture for each the residual block. It means that the residual block consists of 3 layers in this order: 1x1 convolution - 3x3 convolution - 1x1 convolution. The first and the last convolution is the bottleneck. It mostly just for practical consideration, as the first 1x1 convolution is being used to reduce the dimensionality, and the last 1x1 convolution is to restore it. So, the same network is now become 50 layers.
Notice, in 50 layers and more ResNet, at each block, there are now two 1x1 convolution layers. | {"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": 0.8606759309768677, "perplexity": 1113.6587846833988}, "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-2020-34/segments/1596439738913.60/warc/CC-MAIN-20200812171125-20200812201125-00311.warc.gz"} |
https://learnbps.bismarckschools.org/mod/book/view.php?id=83229&chapterid=27606 | # [S] Statistics and Probability
Students’ prior knowledge includes:
• Students investigate patterns of association in bivariate data (grade 8)
• Students use random sampling to draw inferences about a population (grade 7)
• Students investigate chance processes and develop, use, and evaluate probability models (grade 7)
• Students develop an understanding of statistical variability (grade 6)
• Students summarize and describe distributions (grade 6)
##### Interpreting Categorical and Quantitative Data
• Summarize, represent, and interpret data on a single count or measurement variable.
• Summarize, represent, and interpret data on two categorical and quantitative variables.
• Interpret linear models.
##### Making Inferences and Justifying Conclusions
• Understand and evaluate random processes underlying statistical experiments.
• Make inferences and justify conclusions from sample surveys, experiments and observational studies.
##### Conditional Probability and the Rules of Probability
• Understand independence and conditional probability and use them to interpret data.
• Use the rules of probability to compute probabilities of compound events in a uniform probability model.
##### Using Probability to Make Decisions
• Calculate expected values and use them to solve problems.
• Use probability to evaluate outcomes of decisions.
## MAT-HS.S [S] Overview: Statistics and Probability
### MAT-HS.S-ID Domain: [S-ID] Interpreting Categorical and Quantitative Data
• MAT-HS.S-ID.01 Represent data with plots on the real number line
• MAT-HS.S-ID.02 Use statistics to compare center and spread of two or more data sets
• MAT-HS.S-ID.03 Interpret differences in shape/center/spread of data, accounting for outliers
• MAT-HS.S-ID.04 Use the mean and standard deviation of a data set for normal distribution
• MAT-HS.S-ID.05 Summarize categorical data for two categories in twoway frequency tables
• MAT-HS.S-ID.06 Represent data on two quantitative variables on a scatter plot
• S-ID.06.a Fit a function to the data to solve problems
• S-ID.06.b Informally assess the fit of a function by plotting and analyzing residuals
• S-ID.06.c Fit a linear function for a scatter plot that suggests a linear association
• MAT-HS.S-ID.07 interpret the Slope and the Intercept of a linear model in context of data
• MAT-HS.S-ID.08 Compute and interpret the correlation coefficient of a linear fit
• MAT-HS.S-ID.09 Distinguish between correlation and causation
### MAT-HS.S-IC Domain: [S-IC] Making Inferences and Justifying Conclusions
• MAT-HS.S-IC.01 Understand statistics as a process for making inferences based on random sample
• MAT-HS.S-IC.02 Decide if a model is consistent with results from a data-generating process
• MAT-HS.S-IC.03 Compare the purposes of surveys/experiments/observational studies
• MAT-HS.S-IC.04 Use data from a sample survey to estimate a population mean or proportion
• MAT-HS.S-IC.05 Use data from a randomized experiment to compare two treatments
• MAT-HS.S-IC.06 Evaluate reports based on data
### MAT-HS.S-CP Domain: [S-CP] Conditional Probability and the Rules of Probability
• MAT-HS.S-CP.01 Describe events as subsets of a sample space using characteristics
• MAT-HS.S-CP.02 Using probabilities, understand and show that two events are independent
• MAT-HS.S-CP.03 Understand the conditional probability of A given B as P(A and B)/P(B
• MAT-HS.S-CP.04 Construct and interpret two-way frequency tables of data
• MAT-HS.S-CP.05 Recognize and explain the concepts of conditional probability and independence
• MAT-HS.S-CP.06 Find the probability of A given B as the fraction of B??s outcomes belong to A
• MAT-HS.S-CP.07 Apply and interpret the Addition Rule of probability
• MAT-HS.S-CP.08 Apply the general Multiplication Rule in a uniform probability model
• MAT-HS.S-CP.09 Use permutations and combinations to compute probabilities
### MAT-HS.S-MD Domain: [S-MD] Using Probability to Make Decisions
• MAT-HS.S-MD.01 Define a random variable for a quantity of interest by assigning a numerica
• MAT-HS.S-MD.02 Calculate the expected value of a random variable; interpret it as the mean
• MAT-HS.S-MD.03 Develop a probability distribution for a random variable defined for a samp
• MAT-HS.S-MD.04 Develop a probability distribution for a random variable defined for a samp
• MAT-HS.S-MD.05 Weigh outcomes of a decision by assigning probabilities to payoff values
• S-MD.05.a Find the expected payoff for a game of chance
• S-MD.05.b Evaluate and compare strategies on the basis of expected values
• MAT-HS.S-MD.06 Use probabilities to make fair decisions
• MAT-HS.S-MD.07 Analyze decisions and strategies using probability concepts
### A Sample of HS Math Concept Statistics and Probability your child will be learning
#### Interpreting Categorical and Quantitative Data
• Summarize, represent, and interpret data on a single count or measurement variable.
• Summarize, represent, and interpret data on two categorical and quantitative variables.
• Interpret linear models.
#### Making Inferences and Justifying Conclusions
• Understand and evaluate random processes underlying statistical experiments.
• Make inferences and justify conclusions from sample surveys, experiments and observational studies.
#### Conditional Probability and the Rules of Probability
• Understand independence and conditional probability and use them to interpret data.
• Use the rules of probability to compute probabilities of compound events in a uniform probability model.
#### Using Probability to Make Decisions
• Calculate expected values and use them to solve problems.
• Use probability to evaluate outcomes of decisions. | {"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": 0.8497014045715332, "perplexity": 4075.9343727605437}, "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-2023-06/segments/1674764500983.76/warc/CC-MAIN-20230208222635-20230209012635-00416.warc.gz"} |
http://math.stackexchange.com/users/17338/sean-gomes?tab=activity | Sean Gomes
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Mar6 comment Piecing together full density subsequences Perfect, thank you! My attempts were along similar lines, but I missed the trick of working with complements (and the consequently logical choice for our $N_k$). Mar6 accepted Piecing together full density subsequences Mar6 asked Piecing together full density subsequences Mar3 awarded Nice Answer Jan8 awarded Tumbleweed Jan1 asked The heat kernel as the fundamental solution of the heat equation Oct9 awarded Yearling Sep24 awarded Autobiographer Jul2 awarded Curious Apr15 comment Regularity of Dirichlet Eigenvalues on Lipschitz Domain At least interior $C^2$ and continuous up to boundary. My problem only has a piecewise smooth boundary though (but the corners are not too bad, so the domain is still Lipschitz). Apr15 comment Regularity of Dirichlet Eigenvalues on Lipschitz Domain Thanks for the reference, this book should be quite useful in general. It seems the Dirichlet regularity result in this section assumes at least a $\mathcal{C}^2$ boundary though. Apr15 asked Regularity of Dirichlet Eigenvalues on Lipschitz Domain Feb18 comment show that $f$ is not integrable on $[0,1]$ It is equal to cos a.e., not sin. And it is discontinuous at every point in the interval, not just the rationals. ie it is not Riemann integrable. Feb18 comment What is wrong with this equations? (5-5) and (x-y) are both zero. Feb10 answered Showing that the square root is monotone Jan7 comment How to prove the inequality: $\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$ Thanks. Out of interest, where did the motivation for looking at: $(4a+1)(a-1/3)^2$ come from? Jan7 accepted How to prove the inequality: $\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$ Jan7 revised How to prove the inequality: $\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$ edited body; edited title Jan7 asked How to prove the inequality: $\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$ Nov28 revised restriction of functions of several variables added 1 characters in body | {"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": 0.9310117959976196, "perplexity": 1646.8185121531715}, "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-2015-22/segments/1432207927843.59/warc/CC-MAIN-20150521113207-00129-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://archive-ouverte.unige.ch/unige:11957 | Title
# Oblique poles of $\int_X| {f}| ^{2\lambda}| {g}|^{2\mu} \square$
Authors
Barlet, Daniel
Year 2009
Abstract Existence of oblique polar lines for the meromorphic extension of the current valued function $\int |f|^{2\lambda}|g|^{2\mu}\square$ is given under the following hypotheses: $f$ and $g$ are holomorphic function germs in $\CC^{n+1}$ such that $g$ is non-singular, the germ $S:=\ens{\d f\wedge \d g =0}$ is one dimensional, and $g|_S$ is proper and finite. The main tools we use are interaction of strata for $f$ (see \cite{B:91}), monodromy of the local system $H^{n-1}(u)$ on $S$ for a given eigenvalue $\exp(-2i\pi u)$ of the monodromy of $f$, and the monodromy of the cover $g|_S$. Two non-trivial examples are completely worked out.
Identifiers
Note Texte publié dans les actes de la conférence "Complex analysis : several complex variables and connections with PDE theory and geometry", Fribourg (Suisse), 2008. Birkhäuser, 2010, p. 1-23
Full text
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BARLET, Daniel, MAIRE, Henri Michel. Oblique poles of $\int_X| {f}| ^{2\lambda}| {g}|^{2\mu} \square$. 2009. https://archive-ouverte.unige.ch/unige:11957
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https://www.physicsforums.com/threads/different-metrics-in-different-dimensions.976101/ | Different metrics in different dimensions
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• #1
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Summary:
Given a space in R_n = R_1 X R_2 X R_3 X R_4 ... can the metric for the R_1 x R_2 subspace be different from the metric for the R_3 X R_4 subspace?
I'm trying to get a handle on how general a space in R_n can be. Part of my motivation is the curled up dimensions physicists talk about. How does one dimension work differently than another dimension? Can one part of the dimensional structure follow one metric and another part follow a different metric?
I rather think it should be possible. That raises questions about the combinations of subspaces. Can R_1 X R_2 be different (say, taxicab geometry) from R_1 X R_3 (say, Euclidean) as long as R_1 X R_3 is consistent (um, somewhere in between maybe)?
Sorry if this is worded poorly, and if it's in an inappropriate folder. And how does one access the proper notation symbols?
Thanks.
• #2
fresh_42
Mentor
13,457
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Summary: Given a space in R_n = R_1 X R_2 X R_3 X R_4 ... can the metric for the R_1 x R_2 subspace be different from the metric for the R_3 X R_4 subspace?
I'm trying to get a handle on how general a space in R_n can be. Part of my motivation is the curled up dimensions physicists talk about. How does one dimension work differently than another dimension? Can one part of the dimensional structure follow one metric and another part follow a different metric?
Sure. You can build direct products of different topological spaces.
I rather think it should be possible. That raises questions about the combinations of subspaces. Can R_1 X R_2 be different (say, taxicab geometry) from R_1 X R_3 (say, Euclidean) as long as R_1 X R_3 is consistent (um, somewhere in between maybe)?
Sorry if this is worded poorly, and if it's in an inappropriate folder. And how does one access the proper notation symbols?
Thanks.
Phase spaces are considered In stochastic and physics which cover all possible states, i.e. their description. This leads to different dimensions in the components and thus different units and scales.
The actual question is not whether it can be defined rather what should it be good for, i.e. what do you want to do?
• #3
jbriggs444
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Summary: Given a space in R_n = R_1 X R_2 X R_3 X R_4 ... can the metric for the R_1 x R_2 subspace be different from the metric for the R_3 X R_4 subspace?
One example would be the four dimensional space-time we live in. You can pick out a two dimensional space-like slice using x and y coordinates and you can pick out an orthogonal two dimensional Minkowski slice using z and t coordinates.
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https://dsp.stackexchange.com/questions/83207/what-does-the-two-following-formulas-mean | # What does the two following formulas mean?
I am a bit confused between the following two formulas:
$$f\le \frac{f_s}{2} \tag{1}$$
and:
$$f = \frac{f_s}{N} \tag{2}$$
Now I am given an ACF Plot of a sine/cosine wave and I am asked to find the frequency of the signal if the signal is sampled at 1,000 samples/second. I am asked to use the information given in the ACF plot and the sampling rate of 1,000 samples/second.
In the book "Probability and Stochastic Processes" I find the formula (1) i.e., Nyquist Theorem and at another site, here, I find the formula (2) in use.
I have the above two formulas but I am confused in which formula to use. I tried asking this question before but could not explain it clearly. I hope I have explained it properly this time.
As you say,
$$f\le \frac{f_s}{2} \tag{1}$$
is the Nyquist criterion or frequency. It means that, to avoid aliasing (frequency domain folding), a signal must have frequencies below $$\frac{f_s}{2}$$. This probably won't help you determine the frequency of the ACF.
This equation $$f = \frac{f_s}{N} \tag{2}$$ is saying something different. The value $$\frac{1}{f_s}$$ is the time between samples, or the sampling period. When a signal has $$N$$ samples in one period, that means the period of the signal is $$\frac{N}{f_s}$$.
That means that the frequency of the signal is given by formula (2).
With reference to your (now added) ACF: this is not just a sinusoid. It's a damped sinusoid (it is a sinusoid with, probably, a linear decay).
That means it doesn't make sense to look at the peaks because the peak position will change because of the decay.
The peak that won't change is at Lag $$0$$. Start there.
Then look at the zeros. The zero positions won't change because of the decay envelope. I see three zeros: 10, 30, and 50.
All of these give a period of $$10/0.25 = 30/0.75 = 50/1.25 = 40\ \text{samples}.$$
• So I count the number of samples from the graph and along with the value $1,000$ put the values in second formula in order to find the frequency of the signal. right? May 26 at 13:15
• @MuhammadAhmad Yes, pretty much! Often the period won't be an integer number of samples long. Then you need to do something much more complicated to get the frequency.
– Peter K.
May 26 at 13:19
• check my edit and kindly clarify it May 26 at 13:58 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8053204417228699, "perplexity": 342.49314979810373}, "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/1656103877410.46/warc/CC-MAIN-20220630183616-20220630213616-00352.warc.gz"} |
http://math.stackexchange.com/questions/10916/can-somebody-explain-the-plate-trick-to-me?answertab=votes | # Can somebody explain the plate trick to me?
I learned of the plate trick via Wikipedia, which states that this is a demonstration of the fact that SU(2) double-covers SO(3). It also offers a link to an animation of the "belt trick" which is apparently equivalent to the plate trick. Since I've thought most about the belt version, I'll phrase my question in terms of the belt trick.
I am not clear on how the plate/belt trick relates to the double covering. Specifically, I am looking for a sort of translation of each step of the belt trick into the Lie group setting. For example, am I correct in interpreting the initial twisting of the belt as corresponding to the action of a point in SU(2)? Which point? Do I have the group right?
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I thought somebody talked about the plate trick in an answer or a comment on this question, but I don't see it now. Anyway, the answers there are worth looking over. – yasmar Nov 19 '10 at 3:44
@yasmar: Perhaps it was this question you were thinking of. – Rahul Nov 19 '10 at 17:09
@Rahul Thanks. That is the one. – yasmar Nov 19 '10 at 18:13
There is a decent video at youtube.com/watch?v=Rzt_byhgujg – Jeff Mar 18 at 18:36
The diagram in the lower right shows the paths in the space $\operatorname{SO}(3)$ which is the $3$-ball with its antipodal points on its boundary identified. That is why the doubled path appears to be broken; it is passing through antipodal points. | {"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": 0.8743342161178589, "perplexity": 314.7934119298375}, "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-2014-41/segments/1410657129409.8/warc/CC-MAIN-20140914011209-00344-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://www.physicsforums.com/threads/taylor-polynomial.90602/ | # Taylor Polynomial
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Find the thrid taylor polynomial P3(x) for the function $f(x) = \sqrt{x+1}$ about a=0. Approximate f(0.5) using P3(x) and find actual error
thus Maclaurin series
$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2} x^2 + \frac{f^{3}(0)}{6} x^3$$
$$f(x) = x + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{3}{48} x^3$$
am i right so far?
To approximate f(0.5) i simply put x=0.5 in the above equation?
How do i fin the actual error, though?
Thank you
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• #2
Tom Mattson
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stunner5000pt said:
am i right so far?
All but the first term is right. $f(0)\neq x$
To approximate f(0.5) i simply put x=0.5 in the above equation?
After you fix it, yes.
How do i fin the actual error, though?
Plug x=0.5 into f(x) on a calculator, and subtract your result from it. You won't exactly get the "actual" error because your calculator approximates, too. But it will be a very good estimate.
That depends on what is asked for. The remainder doesn't give you the actual error, but rather the maximum of the actual error. So unless you were asked to put bounds on the error, I would think that you would not have to use the remainder.
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1,444
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$$f(x) = 1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{3}{48} x^3$$
i see the problem, its fixed now
im being cautious so im goingto put hte upper limits
$$R_{4} = \frac{15}{384} (c+1)^{\frac{-7}{2}} x^4$$
so the error must be lesser than or equal to this R4 value. THat c value lies between 0.5 and x?
Is this right?
• #4
1,444
2
is this how one would solve for the maximum possible error as stated in the above post? Please do advise
Thank you for your help and input
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http://www.physicsforums.com/showthread.php?p=3529392 | # Expectation of x^2
P: 2 How to calculate E(x^2) given that x are i.i.d random variables distributed as a standard normal i.e. N(0,1) ? Thank you.
P: 4,572
Quote by James1990 How to calculate E(x^2) given that x are i.i.d random variables distributed as a standard normal i.e. N(0,1) ? Thank you.
Hey James1990 and welcome to the forums.
Do you know the relationship for Variance to second and first order moments?
[HINT: Var(X) = E[X^2] - {E[X]}^2].
What do you know about the mean and variance of your distribution?
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https://www.physicsforums.com/threads/fluids-problem.151237/ | # Fluids problem
1. Jan 13, 2007
### kitty9035
1. The problem statement, all variables and given/known data
Water is flowing at 1 m/s in a circular pipe. If the diameter of the pipe decreases to four-ninth its former value, what is the velocity fo the water downstream?
2. Relevant equations
AV=AV
3. The attempt at a solution[/b
(Area1)(Velocity1)=(Area2)(Velocity2)
then...
(A1)(1 m/s)=(4/9)A1(V2) ?
or
A1=(4/9)(V2)
Is that right what would u do next??
2. Jan 13, 2007
### durt
Rewrite the area in terms of its diameter, and then try again.
3. Jan 13, 2007
### AngeloG
Indeed, remember the area for a circle? *circular* being the key word =).
4. Jan 13, 2007
### snowJT
couldn't you just with what your given calculate the volumetric flow rate then, then apply Q = AV to get the velocity in the second pipe.. thats how I would do it
5. Jan 13, 2007
### AngeloG
A1V1 = A2V2, since the density is uniform/incompressible.
You could do as you say, but then you'd have the equation:
A1 * (delta L1) / delta t = A2 * (delta L2) / delta t
A1V1 = A2V2 is a better one to use; well more "simple" one.
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http://tex.stackexchange.com/questions/38427/custom-alignment-of-columns-in-align-environment | # Custom alignment of columns in align environment
I need to typeset two columns of equations, each row typeset with an equation number to the right as with an ordinary equation. The first column must be centered, and the second column must be left-aligned. There will be many rows, so it must automatically break at a page break as appropriate.
There are plenty of related questions, but I can't find any existing solutions that will let me do exactly what I want. (It's a real pain the align environment can't take optional column alignment specifiers.)
-
How are the equations going to be numbered? For instance, will all equations in the left-hand column be numbered independently from those in the right-hand column, or will they share the same counter? – Mico Dec 15 '11 at 21:04
The idea is that each line would have one number, typeset to the right, like an ordinary align environment. The only difference to align is that instead of rl column alignment, I need cl column alignment. – Jamie Vicary Dec 15 '11 at 21:15
Would it be acceptable if the first column were left-aligned or aligned at an operator? – rdhs Dec 15 '11 at 22:06
Hi, thanks for your comment. Unfortunately the first column does need to be centre-aligned. And entries in the first column are actually going to be graphical diagrams, so there are no characters to align at. – Jamie Vicary Dec 15 '11 at 22:14
Making align break across pages is simple: add \allowdisplaybreaks[1] to your preamble.
And since your first column is text, it's also trivial to center it within align: just wrap it in a \makebox larger than all the images (which centers its contents) and use align as normal.
\documentclass{article}
\usepackage{amsmath}
\allowdisplaybreaks[1]
\newcommand{\centerdia}[1]{\makebox[2in]{\includegraphics{#1}}}
\begin{document}
\begin{align}
\centerdia{dia1} & c=20x^2+5x-10+\frac{x^3-4x^2+500x-f(240)}{50} \\
\centerdia{dia2} & \mu=10*\epsilon\\
\centerdia{dia3} & \mu=10*\epsilon
\end{align}
\end{document}
-
Use a modified version of align; change 2\tabcolsep into the separation you prefer.
\documentclass{article}
\usepackage{amsmath,environ}
\makeatletter
\NewEnviron{specialalign}
{\def\align@preamble{%
&\hfil
\strut@
\setboxz@h{\@lign$\m@th\displaystyle{####}$}%
\ifmeasuring@\savefieldlength@\fi
\set@field
\hfil
\tabskip2\tabcolsep
&\setboxz@h{\@lign$\m@th\displaystyle{{}####}$}%
\ifmeasuring@\savefieldlength@\fi
\set@field
\hfil
\tabskip\alignsep@
}%
\begin{align}\BODY\end{align}}
\makeatother
\begin{document}
\begin{specialalign}
x & a=b \\
yyy & c=d+e+f \\
zzzzzz & 1\ne0
\end{specialalign}
\end{document}
-
Just add \allowdisplaybreaks[1] if needed to break across pages. Nice solution! – Werner Dec 15 '11 at 23:24
It's better to avoid \allowdisplaybreaks and add \displaybreak where really needed (before the \\ after which the break should take place). \allowdisplaybreaks is useful during document preparation. – egreg Dec 15 '11 at 23:27 | {"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": 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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9602036476135254, "perplexity": 2435.835083769133}, "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-2015-11/segments/1424936462982.10/warc/CC-MAIN-20150226074102-00154-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/calculate-heat-of-vaporization-and-normal-boing-point-in-degree-c.673231/ | # Calculate Heat of Vaporization and normal boing point (in degree C).?
1. Feb 20, 2013
### laughingnahga
Using line equation y = -4058.7x + 16.10 with the experiment pressure measured in kPa instead of atm.
I already solved for heat of vaporization thus:
(-4058.7K)(-8.314 J/mol $\ast$K)=heat of vaporization = 33444 J/mol = 33.44 kJ/mol.
Also, the vapor pressure for normal boiling point:
(1atm)(101325 Pa/1atm)(1 kPa/1000 Pa)=101.32 kPa
I attempted to solve for T by rearranging the linear form of the Clausius-Clapeyron equation, ln Pvap = (-$\Delta$H/R)(1/T) + ln $\beta$ into something like this (ln101.32 kPa)(-33444 J/mol $\div$ 8.413 J/mol * K)$\div$1 - ln16.10 = T but I got something insane like 21857.7 K which even with subtracting 273 to get C is no where near close to the answer choices provided.
Any help with the last part would be greatly appreciated.
2. Feb 20, 2013
### Staff: Mentor
Re: Calculate Heat of Vaporization and normal boing point (in degree C
Telling us what is x and what is y should slightly increase chances that someone will try to understand the problem and what you did.
3. Feb 20, 2013
### laughingnahga
Re: Calculate Heat of Vaporization and normal boing point (in degree C
I had to turn the assignment in this afternoon so it really doesn't matter at this point.
Thanks anyway...
Similar Discussions: Calculate Heat of Vaporization and normal boing point (in degree C).? | {"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": 0.8894526362419128, "perplexity": 4234.83407371355}, "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-2017-43/segments/1508187825057.91/warc/CC-MAIN-20171022022540-20171022042540-00269.warc.gz"} |
http://mathhelpforum.com/calculus/201838-natural-log-equations-question.html | # Math Help - Natural log equations question
1. ## Natural log equations question
The equation 3lnx=lb3x is solved by the sqaure root of 3.
Thanks
2. ## Re: Natural log equations question
3ln(x) = ln(3x)
ln(x3) = ln(3x)
x3 = 3x
x3 - 3x = 0
x (x2 - 3) = 0
x = 0 or else x2 - 3 = 0
Edited...
4. ## Re: Natural log equations question
Originally Posted by tom@ballooncalculus
3ln(x) = ln(3x)
ln(x3) = ln(3x)
x3 = 3x
x3 - 3x = 0
x (x2 - 3) = 0
x = 0 or else x2 - 3 = 0
Note that \displaystyle \begin{align*} x = 0 \end{align*} and \displaystyle \begin{align*} x = -\sqrt{3} \end{align*} are NOT acceptable solutions, because \displaystyle \begin{align*} \ln{x} \end{align*} is only defined for \displaystyle \begin{align*} x > 0 \end{align*}. | {"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": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.967976987361908, "perplexity": 3538.865264576}, "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-2016-22/segments/1464049276964.14/warc/CC-MAIN-20160524002116-00074-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://readpaper.com/paper/4705295339523948545 | This website requires JavaScript.
# The $X(3872)\rightarrow J/\psi \pi \gamma$ and $X(3872)\rightarrow J/\psi \pi\pi \gamma$ decays
Dec 2022
We study the $\rho$ and $\omega$ meson contribution to the radiative decays$X(3872)\rightarrow J/\psi \pi \gamma$ and $X(3872)\rightarrow J/\psi \pi\pi\gamma$. The $X(3872)\rightarrow J/\psi \pi \gamma$ is dominated by the$\omega$ meson. As for the $X(3872)\rightarrow J/\psi \pi\pi \gamma$, thecontributions of the cascade decays through the $\rho$ and $\omega$ mesons arestrongly suppressed with respect to the diagrams which proceed either throughthe $\psi(2S)$ or the three body decay of $\rho$. The branching ratios of$X(3872)\rightarrow J/\psi \pi \gamma$ and $X(3872)\rightarrow J/\psi \pi\pi\gamma$ are $(8.10^{+3.50}_{-2.88})\times10^{-3}$ and $(2.38\pm1.06)\%$, whichmay be accessible by the BESIII and LHCb Collaborations. Especailly, the$X(3872)\rightarrow J/\psi \pi \gamma$ and $X(3872)\rightarrow J/\psi\pi^+\pi^- \gamma$ decays can be employed to extract the couplings$g_{X\psi\omega}$ and $g_{X\psi\rho}$, which probe the isoscalar and isovectorcomponents of the X(3872) wave function respectively.
Q1论文试图解决什么问题？
Q2这是否是一个新的问题？
Q3这篇文章要验证一个什么科学假设？
0 | {"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": 0.9911640286445618, "perplexity": 1557.0757402284405}, "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/1674764500339.37/warc/CC-MAIN-20230206113934-20230206143934-00275.warc.gz"} |
https://www.physicsforums.com/threads/sets-problem.156406/ | SETS problem
1. Feb 15, 2007
majeedh
this is another problem which i dont know how to start..im not sure which proof i should use to solve this problem
the problem is:
If A,B,C,and D are sets, does it follow
(A Φ B) Φ (C Φ D) = (A Φ C) Φ (B Φ D)
the symbol that is separting the characters is called the oplus symbol, thats the closet symbol i could find
the oplus symbol is a circle with one line going across it and one line going down
Last edited: Feb 15, 2007
2. Feb 15, 2007
HallsofIvy
Staff Emeritus
Darned if I know because I can't tell what symbol you used! Either write it out in letters or use LaTex.
Similar Discussions: SETS problem | {"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": 0.9933027029037476, "perplexity": 3320.5537905903575}, "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-2017-26/segments/1498128323801.5/warc/CC-MAIN-20170628204133-20170628224133-00620.warc.gz"} |
http://blog.jpolak.org/?tag=computation | # Calculation of an Orbital Integral
Posted by Jason Polak on 25. August 2015 · Write a comment · Categories: algebraic-geometry, number-theory · Tags: ,
In the Arthur-Selberg trace formula and other formulas, one encounters so-called ‘orbital integrals’. These integrals might appear forbidding and abstract at first, but actually they are quite concrete objects. In this post we’ll look at an example that should make orbital integrals seem more friendly and approachable. Let $k = \mathbb{F}_q$ be a finite field and let $F = k( (t))$ be the Laurent series field over $k$. We will denote the ring of integers of $F$ by $\mathfrak{o} := k[ [t]]$ and the valuation $v:F^\times\to \mathbb{Z}$ is normalised so that $v(t) = 1$.
Let $G$ be a reductive algebraic group over $\mathfrak{o}$. Orbital integrals are defined with respect to some $\gamma\in G(F)$. Often, $\gamma$ is semisimple, and regular in the sense that the orbit $G\cdot\gamma$ has maximal dimension. One then defines for a compactly supported smooth function $f:G(F)\to \mathbb{C}$ the orbital integral
$$\Ocl_\gamma(f) = \int_{I_\gamma(F)\backslash G(F)} f(g^{-1}\gamma g) \frac{dg}{dg_\gamma}.$$
More »
# Graphing the Mandelbrot Set
Posted by Jason Polak on 13. June 2013 · Write a comment · Categories: analysis, elementary · Tags: , ,
A class of fractals known as Mandelbrot sets, named after Benoit Mandelbrot, have pervaded popular culture and are now controlling us. Well, perhaps not quite, but have you ever wondered how they are drawn? Here is an approximation of one:
From now on, Mandelbrot set will refer to the following set: for any complex number $c$, consider the function $f:\mathbb{C}\to\mathbb{C}$ defined by $f_c(z) = z^2 + c$. We define the Mandelbrot set to be the set of complex numbers $c\in\mathbb{C}$ such that the sequence of numbers $f_c(0), f_c(f_c(0)),f_c(f_c(f_c(0))),\dots$ is bounded.
More » | {"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": 0.965175986289978, "perplexity": 426.02782025700986}, "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-20180123111826-20180123131826-00182.warc.gz"} |
https://proofwiki.org/wiki/Focus_of_Ellipse_from_Major_and_Minor_Axis | # Focus of Ellipse from Major and Minor Axis
## Theorem
Let $K$ be an ellipse whose major axis is $2 a$ and whose minor axis is $2 b$.
Let $c$ be the distance of the foci of $K$ from the center.
Then:
$a^2 = b^2 + c^2$
## Proof
Let the foci of $K$ be $F_1$ and $F_2$.
Let the vertices of $K$ be $V_1$ and $V_2$.
Let the covertices of $K$ be $C_1$ and $C_2$.
Let $P = \tuple {x, y}$ be an arbitrary point on the locus of $K$.
From the equidistance property of $K$ we have that:
$F_1 P + F_2 P = d$
where $d$ is a constant for this particular ellipse.
This is true for all points on $K$.
In particular, it holds true for $V_2$, for example.
Thus:
$\ds d$ $=$ $\ds F_1 V_2 + F_2 V_2$ $\ds$ $=$ $\ds \paren {a + c} + \paren {a - c}$ $\ds$ $=$ $\ds 2 a$
It also holds true for $C_2$:
$F_1 C_2 + F_2 C_2 = d$
Then:
$\ds F_1 C_2^2$ $=$ $\ds O F_1^2 + O C_2^2$ Pythagoras's Theorem $\ds$ $=$ $\ds c^2 + b^2$
and:
$\ds F_1 C_2^2$ $=$ $\ds O F_1^2 + O C_2^2$ Pythagoras's Theorem $\ds$ $=$ $\ds c^2 + b^2$
Thus:
$\ds F_1 C_2 + F_2 C_2$ $=$ $\ds 2 \sqrt {b^2 + c^2}$ $\ds$ $=$ $\ds 2 a$ as $2 a = d$ $\ds \leadsto \ \$ $\ds a$ $=$ $\ds \sqrt {b^2 + c^2}$ $\ds \leadsto \ \$ $\ds b^2 + c^2$ $=$ $\ds a^2$
$\blacksquare$
## Also presented as
This result is also seen presented as:
$c^2 = a^2 - b^2$ | {"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": 0.9318457245826721, "perplexity": 147.68849897958145}, "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-2022-05/segments/1642320304572.73/warc/CC-MAIN-20220124155118-20220124185118-00326.warc.gz"} |
https://www.lebesgue.fr/content/5min-program | • English
• Français
• Conference - Dynamics on representation varieties
Jun 26, 2017 to Jun 30, 2017
Rennes, from June 26th to June 30th
Organization board: Ian Biringer, Ludovic Marquis, Juan Souto
Scientific board: Uri Bader, Jeffrey F. Brock, Jean-Marc Schlenker
Numerous areas of mathematics are touched by what could be called Dynamics on representation varieties. For instance one could mention ergodic theory, Riemannian geometry, low-dimensional topology, Teichmüller theory, and so on... The aim of this workshop is to bring together graduate students, recent graduates and experts in these different areas, giving everybody ample time for discussions and collaborations. Next to a number of research talks, three mini-courses by Tsachik Gelander, Francois Labourie and Julien Marché will take place.
We the organizers of this conference affirm that scientific events must be open to everyone, regardless of race, sex, religion, national origin, sexual orientation, gender identity, disability, age, pregnancy, immigration status, or any other aspect of identity. We believe that such events must be supportive, inclusive, and safe environments for all participants. We believe that all participants are to be treated with dignity and respect. Discrimination and harassment cannot be tolerated. We are committed to ensuring that the Conference Dynamics on representation varieties follows these principles. For more information on the Statement of Inclusiveness, see this dedicated web page.
• School - Analytical aspects of hyperbolic flows
Jul 3, 2017 to Jul 7, 2017
Nantes, from July 3rd to July 7th
Organization board: Sebastien Gouëzel, Laurent Guillopé, Samuel Tapie
Scientific board: Nalini Anantharaman, Viviane Baladi, Colin Guillarmou, Masato Tsujii
Hyperbolic flows are dynamical systems with strong chaotic properties, whose study has been started a long time ago, a crucial example being the geodesic flow on negatively curved manifolds. Whereas the qualitative properties of such flows are well understood, their fine quantitative properties (rate of mixing, spectrum...) require more sophisticated tools. They have been studied both from a dynamical point of view (Dolgopyat's techniques) and more analytically: semi-classical methods, initially introduced to study PDEs, have proven very valuable in this context.
The purpose of this summer school is to make these different techniques accessible to PhD students and young researchers, as well as to give an opportunity for specialists in dynamical systems to learn tools from semi-classical analysis, and conversely. Therefore, the core of this summer school will consist in three introductive mini-courses, completed by a few research talks and question sessions.
Conference brochure
• Conference - Young researcher meeting in dynamics and geometry
Sep 6, 2017 to Sep 8, 2017
Rennes, from September 6th to September 8th
Organization board: Françoise Dal'Bo, Frédéric Paulin, Barbara Schapira, Damien Thomine
Since its creation the Platon network (GDR National Center for Scientific Research n°3341 http: // costia.free.fr / platon/) leads actions towards young researchers in ergodic geometry. The recurrent young researcher meeting is one of the highlights of the year. The goal is to allow about ten PhD students or recent doctors to expose their work and promotes discussions between young and senior researchers. The "Young researcher meeting in dynamics and geometry" follows the spirit of these recurring meetings with an international dimension brought in particular by Swiss and Senegalese networks.
TALKS
Alexander Adam (UPMC) Resonances for Anosov diffeomorphism
Kamel Belarif (Université de Bretagne Occidentale) Genericity of weak mixing in negative curvature
Adrien Boulanger (UPMC) Cascades in affine interval exchanges
Filippo Cerocchi (Max Planck Institute for Mathematics, Bonn) Rigidity and finiteness for compact 3-manifolds with bounded entropy
Maria Cumplido Cabello (Université de Rennes 1) Loxodromic actions of Artin-Tits groups
Nguyen-Bac Dang (Ecole Polytechnique) Degrees of iterates of rational maps
Laurent Dufloux (Oulu University) Hausdorff dimension of limit sets at the boundary of the complex hyperbolic plane
Mikolaj Fraczyk (Université Paris-Sud) Mod p homology growth of locally symmetric spaces
Weikun He (Université Paris-Sud) Sum-product estimates and equidistribution of toral automorphisms
Cyril Lacoste (Université de Rennes 1) Dimension rigidity of lattices in semisimple Lie groups
Erika Pieroni (Università di Roma, Sapienza) Minimal Entropy of 3-manifolds
Fanni M. Selley (Budapest University of Technology) Ergodicity breaking in mean-field coupled map systems
Nasab Yassine (Université de Bretagne Occidentale) Quantitative recurrence of one-dimensional dynamical systems preserving an infinite measure
ABSTRACTS
• Alexander Adam Resonances for Anosov diffeomorphism
The deterministic chaotic behavior of an invertible map T is appropriately described by the existence of expanding and contracting directions of the differential of T. A special class of such maps consist in Anosov diffeomorphisms. Every 2-by-2 hyperbolic matrix M with integer entries induces such a diffeomorphism on the 2-torus. For all pairs of real-analytic functions on the 2-torus, one defines a correlation function for T which captures the asymptotic independence of such a pair under the evolution T^n as n tends to infinity. What is the rate of convergence of the correlation as n tends to infinity, for instance what is its decay rate? The resonances for T are the poles of the Z-transform of the meromorphic continued correlation function. The decay rate is well-understood if T=M. There are no non-trivial resonances of M. In this talk, I consider small real-analytic perturbations T of M where at least one non-trivial resonance of T appears. This affects the decay rate of the correlation.
• Kamel Belarif Genericity of weak mixing in negative curvature
Let M be a manifold with pinched negative sectional curvature. We show that, when M is geometrically finite and the geodesic flow on T^1M is topologically mixing, the set of mixing invariant measures is dense in the set P(T^1M) of invariant probability measures. This implies that the set of weak-mixing measures which are invariant by the geodesic flow is a dense G-delta subset of P(T^1M). We also show how to extend these results to geometrically infinite manifolds with cusps or with constant negative curvature.
• Adrien Boulanger Cascades in affine interval exchanges
Avec un échange d'intervalle affine donné vient naturellement une famille de telles dynamiques indexées par le cercle. En effet, la pré-composition par une rotation de l'application initiale définit un autre échange d'intervalle affine. On étudiera cette famille de dynamiques dans un cas particulier à travers la géométrie de la surface affine associée et son groupe de transformation affine.
An affine interval exchange (AIE) is a piecewise affine map from the circle to itself. Such a map defines a dynamical systems over the circle by iterating it. With an AIE comes naturally a family of AIE indexed by the circle: they are defined by pre-composing the initial AIE by a rotation. The presentation will focus on the study of possible dynamical behaviors of such a family of AIE through a peculiar example.
• Filippo Cerocchi Rigidity and finiteness for compact 3-manifolds with bounded entropy
We present some local topological rigidity results for the set S of non-geometric, compact -- with possibly empty boundary and no spherical boundary components --, orientable Riemannian 3-manifolds having torsionfree fundamental group, with bounded entropy and diameter. By "local", we mean that we consider S endowed with the Gromov-Hausdorff-topology. We shall provide examples to show the necessity of the assumptions and discuss some open problems. Moreover, we shall give a proof of the finiteness of the homeomorphism types of the manifolds in S. These are joint works with A. Sambusetti (Rome, Sapienza).
• Maria Cumplido Cabello Loxodromic actions of Artin-Tits groups
Artin-Tits groups act on a certain delta-hyperbolic complex, called the additional length complex". For an element of the group, acting loxodromically on this complex is a property analogous to the property of being pseudo-Anosov for elements of mapping class groups. A well-known conjecture about mapping class groups claims that "most elements" of the mapping class group of a surface are pseudo-Anosov. In fact, we can prove that a positive proportion is pseudo-Anosov.
By analogy, we conjecture that most'' elements of Artin-Tits groups act loxodromically. More precisely, in the Cayley graph of a subgroup G of an Artin-Tits group, the proportion of loxodromically acting elements in a ball of large radius should tend to one as the radius tends to infinity. We will give a condition guaranteeing that this proportion stays away from zero. This condition is satisfied e.g. for Artin-Tits groups of spherical type, their pure subgroups and some of their commutator subgroups.
• N'Guyen-Bac Dang Degrees of iterates of rational maps
In this talk, I will explain what is a rational map, how to define its k-degrees, and I will study the k-degrees of its iterates. I will explain how the study of the growth of these sequences of numbers helps in understanding the dynamics of these maps.
• Laurent Dufloux Hausdorff dimension of limit sets at the boundary of complex hyperbolic planes
Consider the standard contact structure on the 3-sphere. The associated subriemannian metric has dimension 4. The Gromov comparison problem asks about how the Hausdorff dimension with respect to this subriemannian metric is related tothe Hausdorff dimension with respect to the usual (Riemannian) metric. We will look at this problem in the case of limit sets of discrete groups of complex hyperbolic isometries.
• Mikolaj Fraczyk Mod p homology growth of locally symmetric spaces
I will talk about the growth of the dimension of mod-p homology groups of locally symmetric spaces. Let G be a higher rank Lie group and X its symmetric space and let L be a lattice in G. Results on the rank gradient by Abert, Gelander and Nikolov imply that if L is right angled then for every sequence of subgroups (L_n) of L, the dimensions of the homology groups H_1(X/L_n,Z/pZ) grow sublinearly in the volume of X/L_n. In the special case p=2, I showed that the same statement holds for any sequence of lattices L_n with volume escaping to infinity (even if they are pairwise non-commensurable).
• Weikun He Sum-product estimates and equidistribution of toral automorphisms
Bourgain's sum-product theorem is a metric version of Erdős-Szemerédi sum-product theorem. It asserts that a typical set of real numbers grows fast under addition and multiplication. We will present a generalisation of Bourgain's theorem to matrix algebras and discuss how it is motivated by a ergodic problem, namely, quantitative equidistributions of orbits on the d-dimensional torus under sub-semigroups of SL(d,Z).
• Cyril Lacoste Dimension rigidity of lattices in semisimple Lie groups
We study actions of discrete groups on classifying spaces (or classifying spaces for proper actions). For instance the hyperbolic plane is a classifying space for proper actions of the group PSL(2,Z) (but not of minimal dimension). Such spaces can be used to compute the cohomology of the group, so we want them to have the lowest possible dimension. This leads us to the definitons of the (proper) geometric dimension and the (virtual) cohomological dimension. These two dimensions are not always equal, we will see it is the case for a lattice in the group of isometries G of a symmetric space of non-compact type without Euclidean factors (such a group is a semisimple Lie group but not necessarily connected). This result has an important consequence called "dimension rigidity", that is, the two dimensions are still equal for a group commensurable to a lattice of G.
• Erika Pieroni Minimal Entropy of 3-manifolds
We present the solution of the minimal entropy problem for non-geometric, closed, orientable 3-manifolds (that is, those manifolds which do not admit a com- plete metric locally isometric to one of the eight 3-dimensional model geometries). Together with the results of Besson-Courtois-Gallot for locally symmetric spaces and the work of Soma, Gromov et.al. on the simplicial volume of 3-manifolds and its relation with entropy, this gives a complete picture of the minimal entropy prob- lem for all closed, orientable 3-manifolds. Our work strongly builds on Souto's PhD work (unpublished), filling some gaps in the proof and completing the picture in the case of non-prime manifolds. In detail, we show that the minimal entropy is ad- ditive with respect to the prime decomposition and that for an irreducible manifold X it coincides with the sum of the volume entropies of all the JSJ components of hyperbolic type, each endowed with its complete, hyperbolic metric of nite volume. For the lower bound of MinEnt(X), we adapt Besson-Courtois-Gallot's barycenter method following Souto's ideas; then, we show how this lower bound is realized by producing a sequence of Riemannian metrics gk on X whose volume-entropies tend to
• Fanni M. SelleyErgodicity breaking in mean-field coupled map systems
Coupled map systems are simple models of a finite or infinite network of interacting units. The dynamics of the compound system is given by the composition of the (typically chaotic) individual dynamics and a coupling map representing the characteristics of the interaction. The coupling map usually includes a parameter s in [0,1], representing the strength of interaction. The main interest in such models lies in the emergence of bifurcations when s is varied. We first introduce our results for small finite systems. Then we initiate a new point of view which focuses on the evolution of distributions and allows to incorporate the investigation of a continuum of sites.
• Nasab Yassine Quantitative recurrence of one-dimensional dynamical systems preserving an infinite measure
We are interested in the asymptotic behaviour of the first return time of the orbits of a dynamical system into a small neighbourhood of their starting points. We study this quantity in the context of dynamical systems preserving an infinite measure. More precisely, we consider the case of Z-extensions of subshifts of finite type. We also consider a toy probabilistic model in order to enlighten the strategy of our proofs.
• Conference - Geometric Analysis at Roscoff
Oct 9, 2017 to Oct 13, 2017
Roscoff, from October 9th to October 13th
Organization board: Paul Baird, Gilles Carron, Ali Fardoun, Carl Tipler
Scientific board: Gérard Besson (CNRS, Institut Fourier), Olivier Biquard (ENS Paris), Ahmad El Soufi (Univ. Tours)
Geometric Analysis is the application and development of PDE tools and technics in Riemannian geometry, it is also a fundamental tool in mathematical physics. Recently, important conjectures has been solved: Poincaré's conjecture, Willmore's conjecture, Lawson's conjecture, Yau-Tian-Donaldson's conjecture and a lot of new tools has been introduced and developed : optimal transport, weak formulation of Ricci curvature, Geometric measure theory. This conference will be an opportunity for specialists from theses different areas to meet and exchange ideas, questions and knowledge.
• Conference - Lebesgue PHD meeting 2017
Oct 16, 2017 to Oct 18, 2017
Rennes, from October 16th to October 18th
Organization board: Grégory Boil, Valentin Doli, Caroline Robet, Jérôme Spielmann
Scientific board: Solène Bulteau, Clément Rouffort, Nasab Yassine
Depuis trois ans, le Centre Henri Lebesgue soutient les Rencontres doctorales Lebesgue, initiative des doctorants du Labex. Il s'agit de trois journées de conférences durant lesquelles la parole est donnée à des doctorants de tout horizon géographique et mathématique. L'objectif est ainsi de présenter un panel le plus large possible de la recherche mathématique actuelle telle qu'elle est vue et vécue par les doctorants, mais pas seulement... Lors de ces rencontres, trois chercheurs, appelés 'parrains' de l'évènement, sont invités à exposer et ainsi à partager leur expérience personnelle de la recherche d'aujourd'hui. Cette année, les rencontres sont parrainées par :
Jean-Marc Bardet (SAMM, Université Paris 1);
Jasmin Raissy (Institut mathématique de Toulouse, Université Paul Sabatier);
Gabriel Rivière (Laboratoire Paul Painlevé, Université de Lille 1).
Bien que principalement orientée vers les doctorants, cette conférence se veut également accessible aux étudiants de M2 désireux d’avoir un aperçu des travaux auxquels une thèse en mathématiques peut mener. Il est de plus possible pour des doctorants désireux d'exposer de déposer des propositions sur l'onglet 'Proposer un Exposé'.
• Complex dynamics and quasi-conformal geometry
Oct 23, 2017 to Oct 25, 2017
Our colleague Tan Lei passed away in April 2016. A conference will be held from 23/10/2017 to 25/10/2017 at the University of Angers to honour her memory.
### Scientific Committee
Etienne Ghys (ENS Lyon)
John Milnor (Stony Brook)
Mitsuhiro Shishikura (Kyoto).
### Organizing Committee
Mohammed El Amrani (Angers)
Michel Granger(Angers)
Jean-Jacques Loeb(Angers)
Pascale Roesch(Toulouse).
### Provisional list of speakers
Xavier Buff, Arnaud Cheritat, Nuria Fagella (to be confirmed), Cui Guizhen,Peter Haissinski, John Hamal Hubbard (to be confirmed), Carsten lunde Petersen, Kevin Pilgrim, Mary Rees, Pascale Roesh, Hans Henrik Rugh, Dylan Thurston, Mitsu Shishikura, Giulio Tiozzo. | {"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": 0.8181755542755127, "perplexity": 2574.179785033481}, "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-2017-26/segments/1498128320679.64/warc/CC-MAIN-20170626050425-20170626070425-00694.warc.gz"} |
http://tex.stackexchange.com/questions/13781/multiple-copies-of-short-page-on-one-sheet-of-paper?answertab=oldest | # Multiple copies of short page on one sheet of paper
Suppose you have a short document (including a fancy header) which occupies only a small part of a DIN A4 page, like in this example:
\documentclass[a4paper, 12pt]{article}
\usepackage[top=2cm,hmargin=2.2cm]{geometry}
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}
\fancyfoot[C]{}
\begin{document}
short text short text short text short text short
text short text short text short text short text short
text short text short text short text short text short
text short text short text short text short text short
text short text short text short text short text end of the text.
\end{document}
Suppose I want to print this text 50 times. To save paper I would print it and use a photocopier to copy this text incl. fancy header several times to one sheet of paper.
Now my question is, if there is a simple LaTeX way to do this, i.e. in the above example, have as many copies of all text from fancy header to "end of the text" automatically on a DIN A4 sheet as you can fit on it?
-
I do not know an fully automatically way, but it is possible to do it manually with minimal adjustments.
# First way:
Reduce the height of the page to the minimal height by replacing a4paper with pagewidth=21cm,pageheight=4cm (here 21cm is the A4 width and 4cm depends on the text content and margin requirements).
\usepackage[top=2cm,hmargin=2.2cm,paperwidth=21cm,paperheight=4cm]{geometry}
Then create a second small LaTeX document which includes the PDF of the first using the pdfpages pages and places it multiple times on one page:
\documentclass[a4paper]{article}
\usepackage{pdfpages}
\begin{document}
\includepdf[page={1,1,1,1,1,1,1},nup=1x7]{mainfile}
\end{document}
This places the first document (called mainfile.tex/.pdf here) seven times on page. Adjust both the number of 1, and the number of nup=1x to the real number which fit on one page.
# Second way:
You could create the headers using normal text instead so it can be repeated over the page. A tabular would be adequate here. Putting the main text into one multi-column cell also prevents page breaks inside the text. The the text could be repeated using a loop. Adjust the \vspaces and the repeat numbers to fit your requirements.
\documentclass[a4paper, 12pt]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[pdftex]{graphicx}
\usepackage[top=2cm,hmargin=2.2cm]{geometry}
\pagestyle{empty}
\begin{document}
\newcount\num
\loop\ifnum\num<7
\par\noindent
\begin{tabular}{@{}l@{}r@{}}
\hline\noalign{\vspace{.5cm}}
\multicolumn{2}{@{}p{\linewidth}@{}}{%
\hspace*{\parindent}% if par indent is required
short text short text short text short text short
text short text short text short text short text short
text short text short text short text short text short
text short text short text short text short text short
text short text short text short text short text end of the text
}
\end{tabular}
\par\vspace{1cm}%
\repeat
\end{document}
-
The problem I am finding with this (2nd) solution is, if I put a \ in the text, I get endless stream of errors given below ... see removed non-answer – Tem Pora Nov 24 '12 at 6:56
I found another solution to be simpler, I want to have everything in one file which simplifies everythingh. With the pdfpages solution that is not possible. Simply define the text you want to repeat as a command, with \newcommand{\repeatthis}{text}, and then simply use that command multiple times. As a complete example, I show the complete file for a yahtzee scoresheet, using this technique:
% yahtzee-scoresheet in \LaTeX
\documentclass[a4,12pt,landscape,pdftex]{standalone}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{array}%%%%%Extended, modernized version of tabular
\usepackage[misc]{ifsym}%%%%%\Cube{} command
%%%%% Defining commands for yatzee names:
\newcommand{\one}{\Cube{1}}
\newcommand{\two}{\Cube{2}}
\newcommand{\three}{\Cube{3}}
\newcommand{\four}{\Cube{4}}
\newcommand{\five}{\Cube{5}}
\newcommand{\six}{\Cube{6}}
\newcommand{\onepair}{\Cube{1}\Cube{1}}
\newcommand{\twopair}{\Cube{2}\Cube{2}~\Cube{3}\Cube{3}}
\newcommand{\thrice}{\Cube{3}\Cube{3}\Cube{3}}
\newcommand{\fourequal}{\Cube{4}\Cube{4}\Cube{4}\Cube{4}}
\newcommand{\house}{\onepair{}~\thrice}
\newcommand{\straight}{\Cube{1}\Cube{2}\Cube{3}\Cube{4}\Cube{5}}
\newcommand{\sstraight}{\Cube{2}\Cube{3}\Cube{4}\Cube{5}\Cube{6}}
\newcommand{\yatzee}{\Cube{6}\Cube{6}\Cube{6}\Cube{6}\Cube{6}}
\newcommand{\chance}{\textbf{sjanse}}
\newcommand{\bonus}{\textbf{bonus}}
\newcommand{\total}{\textbf{sum}}
\dimen0=0.6cm{} %%%%% To use within the tabular environment
\newcommand{\onesheet}{%%%%% The complete score sheet:
\begin{tabular}[t]{||l||p{\dimen0}|p{\dimen0}|p{\dimen0}|p{\dimen0}|p{\dimen0}|p{\dimen0}||}\hline \hline
& & & & & & \\ \hline
\one & & & & & & \\ \hline
\two & & & & & & \\ \hline
\three & & & & & & \\ \hline
\four & & & & & & \\ \hline
\five & & & & & & \\ \hline
\six & & & & & & \\ \hline \hline
\total & & & & & & \\ \hline
\bonus & & & & & & \\ \hline \hline
\onepair & & & & & & \\ \hline
\twopair & & & & & & \\ \hline
\thrice & & & & & & \\ \hline
\fourequal & & & & & & \\ \hline
\house & & & & & & \\ \hline
\straight & & & & & & \\ \hline
\sstraight & & & & & & \\ \hline
\yatzee & & & & & & \\ \hline
\chance & & & & & & \\ \hline \hline
\total & & & & & & \\ \hline \hline
\end{tabular}
}
\begin{document}
\onesheet \\[15mm]
\onesheet
\end{document}
-
So how do you replicate the header/footer the way the OP requested? In that sense your approach doesn't solve the problem. – Werner Jan 12 at 20:48 | {"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": 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": 0.9998425245285034, "perplexity": 1342.1355819344358}, "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-2015-35/segments/1440645257063.58/warc/CC-MAIN-20150827031417-00239-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/901849/prove-equilateral-triangle | prove equilateral triangle
Recently I have encountered such a proving problem.
As shown below, given a $\triangle ABC$, $AD$ intersects $BC$ at $D$ so that $AD$ is perpendicular to $BC$, $BE$ intersects $AC$ at $E$ so that $BE$ is the angle bisector of $\angle ABC$, $CF$ intersects $AB$ at $F$ so that $CF$ is the median of $AB$, $AD$, $BE$, $CF$ intersect at a common point.
We need to prove that triangle ABC is an equilateral triangle. I have tried several approaches, but it seems useless in the end.
Any thoughts? Or is it actually not an equilateral triangle?
Start by fixing points $A$ and $B$. We also fix a third point $C^\prime$ and require that $C$ lie on the ray $\vec{BC^\prime}$. Drop a perpendicular from $A$ down to $BC^\prime$ and call the point of intersection $D$. Next construct the angle bisector of $\angle ABC^\prime$; denote the intersection of the angle bisector with the line $AD$ by $G$. Finally, construct the midpoint $F$ of $AB$, and extend a line through $F$ and $G$. The intersection of $FG$ and $BC^\prime$ determines the third vertex of the triangle, $C$.
The result is that for any angle $\angle ABC^\prime$, there is a point $C$ such that the triangle $\triangle ABC$ satisfies all of our hypotheses. And of course, if $\angle ABC^\prime$ is not necessarily $60^\circ$, then $\triangle ABC$ is not necessarily equilateral.
EDIT: The last paragraph is not entirely correct. The first sentence should read: for any $\color{red}{\text{non-obtuse}}$ angle $\angle ABC^\prime$, there is a point $C$ such that the triangle $\triangle ABC$ satisfies all of our hypotheses. | {"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": 0.9503083229064941, "perplexity": 36.383242052319225}, "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-2019-39/segments/1568514575627.91/warc/CC-MAIN-20190922180536-20190922202536-00466.warc.gz"} |
https://community.endnote.com/t/attached-pdf-issues-when-importing-ris-into-x6/292432 | # attached pdf issues when importing .ris into X6
I have been trying to export some references with attached pdfs from Mendeley desktop into Endnote X6 (win7 OS). I’ve tried exporting as endnote .xml, and then importing that file from endnote, but the pdfs are not getting copied into the pdf subfolder in the data folder. So the references have the broken attachment message: "The following file could not be opened. This may occur for a number of reasons including, but not limited to: the file is corrupt or the relative link was deleted.
File D:\References\temp pdf import.Data\PDF\Cothran, Brown, Relyea - 2013 - Proximity to agriculture is correlated with pesticide tolerance evidence for the evolution of amphibian.pdf"
I do have the “copy new file attachments into default folder” option checked in the preferences window. I have also tried exporting the references as a .ris. When I import that file the absolute links to the pdfs are maintained, but no new relative links are created. I’d like the relative links to be created so that I don’t end up with pdfs associated with one library in two different folders.
Any ideas how to make this process work better?
Thanks,
Kevin
If the links are there as absolute (mind you- I have never tried this, so if they are URLs and not attachments this won’t work) then you can ask Endnote to convert Absolute to relative and it will copy the files to the subfolder structure. This option in X7 is under references, file attachments> “convert to relative links”.
That worked (converting to relative links). I’m leaving the post open (not accepting your solution as a solution) in hopes that someone has an idea about how to get the files imported as relative links directly. But thanks for the solution!
Kevin | {"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": 0.811519980430603, "perplexity": 3457.1663898684064}, "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/1656103329963.19/warc/CC-MAIN-20220627073417-20220627103417-00626.warc.gz"} |
http://freakonometrics.hypotheses.org/47962 | # Simple Distributions for Mixtures?
The idea of GLMs is that given some covariates$X$$Y|X$ has a distribution in the exponential family (Gaussian, Poisson, Gamma, etc). But that does not mean that $Y$ has a similar distribution… so there is no reason to test for a Gamma model for $Y$ before running a Gamma regression, for instance. But are there cases where it might work? That the non-conditional distribution is the same (same family at least) than the conditional ones?
For instance, if $(X,Y)$ has a joint Gaussien distribution, then both marginals are Gaussian, but also $Y|X$. So, in that case, if the covariate is normally distributed, it is possible to have a Gaussian distribution also for $Y$. The econometric interpretation is that with a standard Gaussian linear model, if$X$ is normally distributed, not only the conditional distribution $Y|X$ is Gaussian but also the non-conditional distribution of $Y$.
> set.seed(1)
> n=1e3
> X=rnorm(n,10,2)
> Y=1+3*X+rnorm(n)
> plot(X,Y,xlim=c(4,20))
Indeed, here the distribution of $Y$ is also Gaussian
> library(nortest)
Anderson-Darling normality test
data: Y
A = 0.23155, p-value = 0.802
> shapiro.test(Y)
Shapiro-Wilk normality test
data: Y
W = 0.99892, p-value = 0.8293
(not only from a statistical point of view, the thoery of Gaussian random vectors confirms that the non-conditional distribution is Gaussian actually)
Here $X$ is continuous. What if we consider a finite mixture here, i.e.$X$ takes only a finite number of values? Actually, Teicher (1963) proved that it is not possible to have a non-conditional Gaussian distribution for $Y$. But in practice, would we really reject the Gaussian assumption, for $Y$? If the number of classes is to small, yes. But with a large number of classes (a sufficiently large number of mixture components), it is possible,
> pv=function(k=2){
+ n=1e4
+ X=rnorm(n,10,2)
+ Q=quantile(X,(0:k)/k)
+ Q[1]=0
+ Xc=cut(X,Q,labels=1:k)
+ XcN=tapply(X,Xc,mean)
+ Xn=XcN[as.numeric(Xc)]
+ Y=1+3*Xn+rnorm(n)
> plot(2:100,Vectorize(pv)(2:100),type="l")
> abline(h=.05,col="red")
So here, it could be possible to have also a Gaussian distribution, for $Y$. As least to accept that assumption, statistically.
In the context of a Poisson regression, it is well know that it’s not possible to have at the same time $Y|X$ that is Poisson distributed (that’s a Poisson regression) and also $Y$ that is Poisson distributed. That simply comes from the fact that
$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y\vert X]]$
while
$\text{Var}[Y]=\mathbb{E}[\text{Var}[Y\vert X]]+\text{Var}[\mathbb{E}[Y\vert X]]$
and because of the conditional Poisson distribution, then
$\text{Var}[Y\vert X]=\mathbb{E}[Y\vert X]$
Thus,
$\text{Var}[Y]=\mathbb{E}[Y]+\underbrace{\text{Var}[\mathbb{E}[Y\vert X]]}_{>0}$
So $Y$ cannot be Poisson distribution. But again, it could be possible, if heterogeneity is not too large, to accept the null assumption of a Poisson distribution for $Y$.
More generally, it is very difficult to have a distribution family for $Y|X$ that is also the distribution of the non-conditional variable $Y$. In the context of a finite mixture ($X$ takes a finite number of values),Teicher (1963) proved that it was not not possible, neither for the Gaussian distribution nor the Gamma distribution. An to go further, check Monfrini (2002) (thanks Romuald for point out the reference).
Hence, as a keep saying, before running a regression model on$Y|X$ with some given family, it is never a good idea to check if the non-conditional distribution $Y$ has the same distribution. Because there is no reason, usually, to remain in the same family. | {"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": 29, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.8877949714660645, "perplexity": 4756.805699780511}, "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-2016-36/segments/1471982296020.34/warc/CC-MAIN-20160823195816-00106-ip-10-153-172-175.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/75921/what-is-the-tensor-product-for-the-eilenberg-moore-category-of-a-commutative-mona | # What is the tensor product for the Eilenberg-Moore category of a commutative monad?
In Linear logic, monads and the lambda calculus (DRAFT), Proposition 3.0.2 says that the Eilenberg-Moore category for a commutative monad has the structure of a symmetric monoidal closed category. My question is: How do you construct the tensor product for the EM-category? When I followed the reference to Kiegher's paper "Symmetric monoidal closed categories generated by commutative adjoint monads", that paper only seems to give the construction for monads of the form A ⊗ -. I don't see how to generalize the construction to arbitrary commutative monads.
-
One reference that says this explicitly is Gavin J. Seal, "Tensors, monads and actions", arxiv.org/abs/1205.0101, see beginning of section 2.2 and then theorem 2.5.5. On the nLab see here: ncatlab.org/nlab/show/… – Urs Schreiber Feb 12 '14 at 22:42
Let $T \colon C \to C$ be your monad. Being commutative, it comes with maps $\mathrm{dst} \colon T(A) \otimes T(B) \to T(A \otimes B)$. Let $\phi \colon TA \to A$ and $\psi \colon TB \to B$ be algebras. Then $\phi \otimes \psi$ is the coequalizer in $\mathrm{Alg}(T)$ of $T(\phi \otimes \psi)$ and $\mu \circ T(\mathrm{dst})$ (which is a reflexive pair of morphisms from the free algebra on $T(A) \otimes T(B)$ to the free algebra on $A \otimes B$). The unit $I$ in $\mathrm{Alg}(T)$ is the free algebra $\mu \colon T^2(I) \to T(I)$. Moreover, the free functor $C \to \mathrm{Alg}(T)$ preserves monoidal structure.
A good example to keep in mind is where $T$ is the free vector space monad on the category of sets. The coequalizers then is pretty much directly the usual tensor product construction with bilinear maps. | {"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": 0.9501293897628784, "perplexity": 173.00011050521096}, "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-2016-22/segments/1464049278042.30/warc/CC-MAIN-20160524002118-00025-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/confusion-with-time-dialation.609085/ | # Confusion with time dialation.
1. May 26, 2012
### MartinJH
Reading Brian Cox and Jeff Forshaw's book "Why does E=MC2". They mention that satellites speed up with time, but, then reading Wiki it says the crew of the ISS experience the slowing of time. Which one is correct? I'm slightly confused.
Both experience a weaker gravitational pull and high velocities.
Could it be the overall difference in altitude and speed?
Many thanks
http://en.wikipedia.org/wiki/Time_dilation
Last edited: May 26, 2012
2. May 26, 2012
### Naty1
3. May 26, 2012
### yuiop
The velocity based time dilation combined with the gravitational time dilation can be written as:
$$\sqrt{1-\frac{v^2}{c^2}} * \sqrt{1-\frac{2GM}{rc^2}}$$
Since the orbital velocity of a satellite is given by:
$$v = \sqrt{\frac{GM}{r}}$$
then the first equation can be rewritten as:
$$\sqrt{1-\frac{GM}{rc^2}} * \sqrt{1-\frac{2GM}{rc^2}}$$
It can be seen for increasing radius the time dilation reduces due to increased height and due to reduced orbital velocity. Clocks on satellites with large orbits tick faster than clocks on satellites at lower orbits. Clocks on the surface of the Earth are moving much slower than the required orbital velocity at that radius and so tick faster than clocks on satellites with very low orbits. The ISS has a relatively low orbit (its radius is approximately 1.05 times the radius of the Earth), so clocks on the ISS are indeed ticking slower than clocks on the surface of the Earth. The speed up of clocks with increasing radius means that once an orbital radius is larger than 3 times the Earth surface radius (the break even point) the clocks on board a satellite are ticking faster than a clock on the Earth surface. The GPS satellites have an orbital radius of about 4.1 times the radius of the Earth so they are ticking faster.
P.S. The above equations for the time dilation of an orbiting satellite can be fairly accurately approximated in this case by:
$$\sqrt{1-\frac{3GM}{rc^2}} or \left(1-\frac{3GM}{2rc^2}\right)$$
4. May 26, 2012
### MartinJH
Thank you, yuiop. You have hit the nail on the head.
Since I posted the question, a few hours ago, I have been pondering over it and knew their had to be a explanation. From reading your post it looks like I was touching on the reason but couldn't quite grasp it.
Thank you for your post also, Naty1.
Similar Discussions: Confusion with time dialation. | {"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": 0.8799976706504822, "perplexity": 645.3529233371659}, "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-2017-39/segments/1505818694719.89/warc/CC-MAIN-20170926032806-20170926052806-00488.warc.gz"} |
http://math.stackexchange.com/questions/152129/algebra-simplify-question | # Algebra simplify question?
How would I simplify this?
$$5\% \cdot \frac12 \left(3000 + 2x\right)$$
-
My book says it is 165+.11x but I am not sure how – James May 31 '12 at 18:02
First step: write the expression in symbols, then we can give you hints. – rschwieb May 31 '12 at 18:05
You have: $0.05\cdot0.5(3000+2x)=0.025\cdot(3000+2x)=75+0.05x$
-
Assuming this is what you mean:
$$5\%*\frac{1}{2}*(3000+2x)$$
I assume you know that $5\%=\frac{5}{100}=\frac{1}{20}$. So we have
$$\frac{1}{40}*(3000+2x)=\frac{3000}{40}+\frac{2x}{40}=75+\frac{x}{20}$$
-
$$0.11\cdot\frac{1}{2}(3000+2x)$$
I would reduce this by distributing the 1/2 first: $$0.11(1500+x)$$ Then distribute the .11, using the trick $.11(1500)=.11(100)15=11(15)=165$ to get $$165+.11x$$ | {"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": 0.9843727350234985, "perplexity": 1859.6438536437247}, "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-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00328-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/getting-the-wrong-multipole-for-1st-acoustic-peak.923207/ | # I Getting the wrong multipole for 1st acoustic peak
1. Aug 19, 2017
### DoobleD
I'm trying to do a simple calculation, but there must be something wrong.
The wavelength $\lambda_1$ corresponding to first acoustic peak of the CMB is related to the sound horizon at last scattering, $d_{hs}$, by :
$\lambda_1 = 2d_{hs}$ (see for instance slide 14 on Wayne Hu PDF slides).
Now, the multipole $l$ of the first acoustic peak can be related to its wavelength and to the distance to last scattering surface, $D$, by :
$l_1 = \frac{2 \pi}{\lambda_1} D$ (see slide 15)
From that I deduce the following equation :
$l_1 = \frac{\pi}{d_{hs}}D$
I find in the litterature that $D \approx 14000 Mpc$, and $d_{hs} \approx 150 Mpc$. I plug those values into the previous equation, and I find $l_1 \approx 293$, which is quite far from the $l_1 \approx 200$ I should get for the first peak. What's wrong ?
2. Aug 19, 2017
### DoobleD
I get the values for distance to last scattering surface and sound horizon here. I wonder however if 150 Mpc for the sound horizon is not in comoving coordinates, while I should use the physical distance instead (which I don't know) ?
EDIT : I just realized that at the very end of that WMAP values document, they basically give the exact same formula, $l = \frac{\pi}{d_{hs}}D$. And with the values they gives, I get $l = 299$. Why am I not getting 200 ?
Last edited: Aug 19, 2017
3. Aug 19, 2017
### DoobleD
4. Aug 21, 2017
5. Sep 5, 2017
### DoobleD
Thank you for the help. I'm still puzzled by my problem, but great article. | {"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": 0.9197881817817688, "perplexity": 818.2220421091004}, "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/1521257650993.91/warc/CC-MAIN-20180324190917-20180324210917-00123.warc.gz"} |
https://diophantus.org/arxiv/0704.0043 | # diophantus
Hello, this is beta version of diophantus. If you want to report about a mistake, please, write to [email protected]
#### Nonequilibrium entropy limiters in lattice Boltzmann methods
31 Mar 2007 cond-mat.stat-mech, cond-mat.mtrl-sci arxiv.org/abs/0704.0043
Abstract. We construct a system of nonequilibrium entropy limiters for the lattice Boltzmann methods (LBM). These limiters erase spurious oscillations without blurring of shocks, and do not affect smooth solutions. In general, they do the same work for LBM as flux limiters do for finite differences, finite volumes and finite elements methods, but for LBM the main idea behind the construction of nonequilibrium entropy limiter schemes is to transform a field of a scalar quantity - nonequilibrium entropy. There are two families of limiters: (i) based on restriction of nonequilibrium entropy (entropy "trimming") and (ii) based on filtering of nonequilibrium entropy (entropy filtering). The physical properties of LBM provide some additional benefits: the control of entropy production and accurate estimate of introduced artificial dissipation are possible. The constructed limiters are tested on classical numerical examples: 1D athermal shock tubes with an initial density ratio 1:2 and the 2D lid-driven cavity for Reynolds numbers Re between 2000 and 7500 on a coarse 100*100 grid. All limiter constructions are applicable for both entropic and non-entropic quasiequilibria.
# Reviews
There are no reviews yet. | {"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": 0.8167997598648071, "perplexity": 2073.1245532521166}, "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-2021-04/segments/1610703506640.22/warc/CC-MAIN-20210116104719-20210116134719-00449.warc.gz"} |
https://www.physicsforums.com/threads/equations-of-motion-help.79162/ | # Equations of Motion Help
1. Jun 15, 2005
### nissanfreak
Equations of Motion Help!!!
I have been working on this for awile and still cant figure it out. What equation would I use to solve this problem? Here is the problem:
On a ride called the Detonator at Worlds of Fun in Kansas City, passengers accelerate straight downward from zero to 45 mi/h in 1.0 seconds. What is the average acceleration of the passengers on this ride?
Any and all help would be greatly appreciated!
2. Jun 15, 2005
### OlderDan
This is a direct application of the definition of acceleration. No other equations are needed. You might need to do a unit conversion to match your answer to a given answer. The natural units that come from the information given would be miles per hour per second, but a book answer might be feet per second per second or feet per second squared. Other possibilities exist.
3. Jun 15, 2005
### nissanfreak
The answer in the back of the book is 20.1 m/s^2 I just dont know how they got that? Can you help me figure this out!!! Thanks in advance!!!!
4. Jun 15, 2005
### WhirlwindMonk
It's just unit conversions. The acceleration is 45 miles/hour/second, and when converted, it gives you 20.1 meters/second^2
5. Jun 15, 2005
### whozum
First thing you're going to do is convert to metric units.
edit: I give up, it won't show up right.
Fill in hte proper conversion for each then multiply out. You'll get 20.1m/s^2
Last edited: Jun 15, 2005
6. Jun 15, 2005
### nissanfreak
Yea I was trying it out but it doesnt want to come out right. I believe I need to use one of the constant-acceleration equations of motion. But im not sure which one? And how to plug the info into the equation?
7. Jun 15, 2005
### WhirlwindMonk
I did it and got the correct answer. Well, my calculator did it. You are either using incorrect conversions or doing a conversion wrong.
1 mile = 1609.344 meters
1 hour = 3600 seconds
8. Jun 15, 2005
### nissanfreak
can you show me how you multiplied it? I really appreciate it!!!
9. Jun 15, 2005
### WhirlwindMonk
45 mile/(hr*s) * 1 hr/3600 s = 0.0125 mile/s^2
.0125 mi/s^2 * 1609.344 m/1 mile = 20.1 m/s^2
10. Jun 15, 2005
### nissanfreak
Thank you for your help!!! I do however have one last question. Would you be able find this answer 20.1m/s^2 if you didnt know that it was the answer? Sorry if this sounds dumb but I have only had 4 physics classes. Im taking a 5 week course this summer over physics so I am fairly new to all of this. Thanks again everyone for the help!!!!! :)
11. Jun 15, 2005
### WhirlwindMonk
Yes, I definately could have. All it is is a goofy unit conversion problem. One thing I learned from my AP physics class is that you need to keep with it and ask for help if you need it. Once you get the concepts, physics is really cool and a lot of fun. Until then...let's just say i still have bruises from banging my head against my desk. :tongue:
12. Jun 16, 2005
### OlderDan
Yes you would be able to find that answer, but there is no way you would have known that you were supposed to express it in that form unless the question told you to do so. Miles per hour per second is a prefectly reasonable unit for expressing acceleration. It is the natural unit when speed is given in miles per hour and changing rapidly, such as for objects like the one in your problem or accelerating cars like dragsters. Don't feel bad about not anticipating the form of the answer you were given if the problem did not tell you to express it that way. Just make sure you understand how to do unit conversions when required.
You should notice that WhirlwindMonk accomplished the unit conversion by a series of multiplications by factors equal to 1. All unit conversions should be done that way. The ratio of two equal quantities is always 1. All you need to do is write the fraction in the form that is going to cancel the units you want replaced and leave you with the units you want. Sometimes it takes several steps to accomplish that. I am going to rewrite what you were already told, but in more detail, to emphasize this point.
$$a = \frac{{45\frac{{mi}}{h}}}{{1s}} = \frac{{45mi}}{{hs}}\left( {\frac{{5280ft}}{{mi}}} \right)\left( {\frac{{12in}}{{ft}}} \right)\left( {\frac{{2.54cm}}{{in}}} \right)\left( {\frac{{1m}}{{100cm}}} \right)\left( {\frac{1 h}{{60\min }}} \right)\left( {\frac{{1\min }}{{60{\mathop{\rm s}\nolimits} }}} \right) = 20.1\frac{m}{{s^2 }}$$
Every fraction in parentheses has the value 1 because the numerator and denominator are equal.
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https://proofwiki.org/wiki/Definition:Order_of_Group_Element | # Definition:Order of Group Element
## Definition
Let $G$ be a group whose identity is $e_G$.
Let $x \in G$ be an element of $G$.
### Definition 1
The order of $x$ (in $G$), denoted $\order x$, is the smallest $k \in \Z_{> 0}$ such that $x^k = e_G$.
### Definition 2
The order of $x$ (in $G$), denoted $\order x$, is the order of the group generated by $x$:
$\order x := \order {\gen x}$
### Definition 3
The order of $x$ (in $G$), denoted $\left\vert{x}\right\vert$, is the largest $k \in \Z_{\gt 0}$ such that:
$\forall i, j \in \Z: 0 \le i < j < k \implies x^i \ne x^j$
## Infinite Order
$x$ is of infinite order, or has infinite order if and only if there exists no $k \in \Z_{> 0}$ such that $x^k = e_G$:
$\order x = \infty$
## Finite Order
$x$ is of finite order, or has finite order if and only if there exists $k \in \Z_{> 0}$ such that $x^k = e_G$.
## Examples
### Order of $2$ in $\struct {\R_{\ne 0}, \times}$
Consider the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$.
The order of $2$ in $\struct {\R_{\ne 0}, \times}$ is infinite.
### Order of $i$ in $\struct {\C_{\ne 0}, \times}$
Consider the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$.
The order of $i$ in $\struct {\C_{\ne 0}, \times}$ is $4$.
### Order of $\begin{bmatrix} 1 & 1 \cr 0 & 1 \end{bmatrix}$ in General Linear Group
Consider the general linear group $\GL 2$.
Let $\mathbf A := \begin{bmatrix} 1 & 1 \cr 0 & 1 \end{bmatrix} \in \GL 2$
The order of $\mathbf A$ in $\GL 2$ is infinite.
### Rotation Through the $n$th Part of a Full Angle
Let $G$ denote the group of isometries in the plane under composition of mappings.
Let $r$ be the rotation of the plane about a given point $O$ through an angle $\dfrac {2 \pi} n$, for some $n \in \Z_{> 0}$.
Then $r$ is the generator of a subgroup $\gen r$ of $G$ which is of order $n$.
### Possible Orders of $x$ when $x^2 = x^{12}$
Let $G$ be a group.
Let $x \in G \setminus \set e$ be such that $x^2 = x^{12}$.
Then the possible orders of $x$ are $2$, $5$ and $10$.
## Also known as
Some sources refer to the order of an element of a group as its period.
## Also denoted as
The order of an element $x$ in a group is sometimes seen as $\map o x$.
Some sources render it as $\map {\operatorname {Ord} } x$.
## Also see
• Results about order of group elements can be found here. | {"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": 0.9897925853729248, "perplexity": 165.33487115131513}, "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-2019-39/segments/1568514573368.43/warc/CC-MAIN-20190918213931-20190918235931-00130.warc.gz"} |
http://www.ipam.ucla.edu/abstract/?tid=14342&pcode=ELWS4 | ## Parameter uncertainty, model reduction, and effective theories in physics, biology, and beyond
#### Mark TranstrumBrigham Young University
The success of science is due in large part to the hierarchical nature of physical theories. These effective theories model natural phenomena as if the physics at macroscopic length scales were almost independent of the underlying, shorter-length-scale details. The efficacy of these simplified models can be understood in terms of parameter identifiability. Parameters associated with microscopic degrees of freedom are usually unidentifiable as quantified by the Fisher Information Matrix. I apply an information geometric approach in which a microscopic, mechanistic model is interpreted as a manifold of predictions in data space. Model manifolds are often characterized by a hierarchy of boundaries--faces, edges, corners, hyper-corners, etc. These boundaries correspond to reduced-order models, leading to a model reduction technique known as the Manifold Boundary Approximation Method. In this way, effective models can be systematically derived from microscopic first principles for a variety of complex systems in physics, biology, and other fields.
Back to Workshop IV: Uncertainty Quantification for Stochastic Systems and Applications | {"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": 0.8973988890647888, "perplexity": 648.5058350978362}, "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-2017-47/segments/1510934804666.54/warc/CC-MAIN-20171118055757-20171118075757-00424.warc.gz"} |
https://arxiv.org/abs/1604.00456 | math.MG
(what is this?)
# Title: A note on the affine-invariant plank problem
Abstract: Suppose that $C$ is a bounded convex subset of $\mathbb{R}^n$, and that $P_1, \dots, P_k$ are planks which cover $C$ in respective directions $v_1, \dots, v_k$ and with widths $w_1, \dots, w_k$. In 1951, Bang conjectured that $$\sum_{i=1}^k \frac{w_i}{w_{v_i}(C)} \geq 1,$$ generalizing a previous conjecture of Tarski. Here, $w_{v_i}(C)$ is the width of $C$ in the direction $v_i$. In this note we give a short proof of this conjecture under the assumption that, for every $m$ with $1 \leq m \leq k$, $C \setminus \bigcup_{i = 1}^m P_i$ is a convex set.
Comments: 5 pages, 1 figure Subjects: Metric Geometry (math.MG) MSC classes: 52C17 Cite as: arXiv:1604.00456 [math.MG] (or arXiv:1604.00456v1 [math.MG] for this version)
## Submission history
From: Gregory R. Chambers [view email]
[v1] Sat, 2 Apr 2016 04:07:58 GMT (18kb,D) | {"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": 0.8045207858085632, "perplexity": 1179.690309458952}, "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/1516084891705.93/warc/CC-MAIN-20180123012644-20180123032644-00162.warc.gz"} |
https://manual.frontistr.com/en/theory/theory_04.html | # Heat Conduction Analysis
## Heat Conduction Analysis
In this section, the method of heat conduction analysis for solid bodies with the finite element methods used in this development code is described.
### Basic Equation
The heat conduction equation in a continuous body is as follows:
where,
$\rho = \rho(x)$ mass (density)
$c = c(x, T)$ specific heat
$T=T(x, t)$ temperature
$K=k(x, T)$ thermal conductivity
$Q=Q(x, T, t)$ calorific value
$x$ represents the position, $T$ represents the temperature and $t$ represents the time.
The area being considered is defined as $S$ and its surroundings as $\Gamma$. Assuming that Dirichet or Neumann-type boundary conditions are given throughout $\Gamma$, the boundary conditions become as follows:
The function form of $T_1$ and $q$ is known. $q$ is the outflow heat flux from the boundaries.
With this program, it is possible to consider three types of heat flux.
where $q_s$ is the distributed heat flux, $q_c$ is the heat flux by the convective heat transfer, and $q_r$ is the heat flux by the radiant heat transfer.
Furthermore,
$Tc=Tc(x,t)$ Convective heat transfer coefficient atomospheric temperature
$hc=hc(x,t)$ Convective heat transfer coeffcient
$Tr=Tr(x,t)$ Radiation heat transfer coefficient atmospheric temperature
$hr=\varepsilon\sigma F=hr(x,t)$ Radiation heat transfer coefficient
$\varepsilon$: radiation rate, $\sigma$: Stefan-Boltzmann constant, $F$: shape factor
### Discretization
If Eq.$\eqref{eq:2.4.1}$ is discretized with the Galerkin method,
However,
Eq.$\eqref{eq:2.4.8}$ is a formula of non-linear and non-steady-state. The objective now is to discretize it in time by the backward Euler method and calculate the temperature at time $t = t_0$ (when the temperature at $t = t_0+\Delta t$ is known) with the following equation:
The next step is to improve the temperature vector $\lbrace T \rbrace_{t=t_0+\Delta t}^{(i)}$ which approximately satisfies Eq.$\eqref{eq:2.4.13}$ to determine the solution $\lbrace T \rbrace_{t=t_0+\Delta t}^{(i)+1}$ with a good precision.
Therefore, the temperature vector must be expressed as follows:
The product of the heat transfer matrix and temperature vector, as well as the mass matrix, are expressed approximately by the following equations:
By substituting Eq.$\eqref{eq:2.4.14}$, Eq.$\eqref{eq:2.4.15}$ and Eq.$\eqref{eq:2.4.16}$ into Eq.$\eqref{eq:2.4.13}$, and omitting the terms of second or higher order, the following equation is obtained:
Moreover, the coefficient matrix of the left side is approximately evaluated with the following equation:
where $[K_T]^{(i)}_{t=t_0 + \Delta t}$ is a tangent stiffness matrix.
Finally, it is possible to calculate the temperature at time $t = t_0 + \Delta t$ through iterative calculation using the following equation:
In steady-state analysis, the iterative calculation is performed with the following equation:
In non-steady-state analysis, the discretization in time is done through the implicit method; thus, the analysis is normally not affected by the restriction of the size of the time increment $\Delta t$. However, if the time increment $\Delta t$ is to large, the number of convergences in the iterative calculation increases. Therefore, this program is equipped with an automatic increment function, which constantly monitors the dimension of the residual vector in the iterative calculation process. If the convergence of the iterative calculation is too slow, it decreases the time increment $\Delta t$. Moreover, when the number of iterative calculations is too small, it increases the time increment $\Delta t$. | {"extraction_info": {"found_math": true, "script_math_tex": 41, "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": 6, "equation": 10, "x-ck12": 0, "texerror": 0, "math_score": 0.8779803514480591, "perplexity": 522.9364464842713}, "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-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00290.warc.gz"} |
https://cs.stackexchange.com/questions/116274/floating-point-arithmetic-on-division | # Floating point arithmetic on division
I am trying to figure out how $$(x/y)$$ in floating point arithmetic
$$fl(fl(x) / fl(y))$$ where $$fl(x) = x(1-\delta_1)$$, $$fl(y) = y(1-\delta_2)$$, $$fl = (1-\delta_3)$$
I have:
$$= x/y \cdot ((1-\delta_1)/(1-\delta_2))(1-\delta_3)$$
after arithmetic
$$= x/y \cdot (\delta_3 \delta_1 - \delta_1 - \delta_3 + 1)/(1-\delta_2)$$
Not sure how to fully write the rest I believe i have to shorten them using the info that $$\delta < \epsilon$$ and $$\delta^2 \leq \delta$$
$$\frac{1}{1 - \delta_2} = 1 + \delta_2 + \delta_2^2 + \delta_2^3 + \cdots$$
Then, for example, if $$\epsilon$$ is the machine epsilon and $$\delta_2 > 0$$:
$$1 + \delta_2 \le \frac{1}{1 - \delta_2} \le 1 + \delta_2 + \epsilon$$
That is quite a tight bound, when you think about it. (Exercise: Prove that this is true if $$\delta_2 \le 0$$, too.) | {"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9583632946014404, "perplexity": 293.1297029020646}, "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/1579250606872.19/warc/CC-MAIN-20200122071919-20200122100919-00439.warc.gz"} |
http://ufdc.ufl.edu/AA00006145/00001 | UFDC Home | Search all Groups | World Studies | Federal Depository Libraries of Florida & the Caribbean | Help
General theory of conical flows and its application to supersonic aerodynamics
Material Information
Title:
General theory of conical flows and its application to supersonic aerodynamics
Series Title:
NACA TM
Physical Description:
333 p. : ill. ; 27 cm.
Language:
English
Creator:
Germain, Paul, 1920-
United States -- National Advisory Committee for Aeronautics
Publisher:
NACA
Place of Publication:
Washington, D.C
Publication Date:
Subjects
Subjects / Keywords:
Superposition principle (Physics) ( lcsh )
Harmonic functions ( lcsh )
Aerodynamics ( lcsh )
Genre:
federal government publication ( marcgt )
bibliography ( marcgt )
technical report ( marcgt )
non-fiction ( marcgt )
Notes
Abstract:
The report deals with a method of studying the equation of cylindrical waves particularly indicated for the solution of certain aerodynamic problems. The method reduces problems of a hyperbolic equation to problems of harmonic functions. The study has been applied toward setting up the fundamental principles, to developing their investigation up to calculation of the pressures on the visualized obstacles, and to showing how the initial field of "conical flows" was considerably enlarged by a procedure of integral superposition.
Bibliography:
Includes bibliographic references (p. 261-263).
Statement of Responsibility:
by Paul Germain.
General Note:
"Report date January 1955."
General Note:
"Translation of "La théorie générale des mouvements coniques et ses applications a l̕ aérodynamique supersonique." Office National ď Études et de Recherches Aéronautiques, no. 34, 1949."
Record Information
Source Institution:
University of Florida
Rights Management:
All applicable rights reserved by the source institution and holding location.
Resource Identifier:
aleph - 003778920
oclc - 44857672
sobekcm - AA00006145_00001
System ID:
AA00006145:00001
Full Text
nCA--M -'31
qq^ 74l2'- q1-7-167S
GENERAL THEORY OF CONICAL FLOWS AND ITS APPLICATION
TO SUPERSONIC AERODYNAMICS
By
Paul Germain
Preface
By
M. J. Peres
NOTICE
This report deals with a method of studying the equation of cylin-
drical waves particularly indicated for the solution of certain problems
in aerodynamics. One of the most remarkable aspects of this method is
that it reduces problems of a hyperbolic equation to problems of harmonic
functions. We have applied ourselves here to setting up the fundamental
principles, to developing their investigation up to calculation of the
pressures on the visualized obstacles, and to showing how the initial
field of "conical flows" was considerably enlarged by a procedure of
integral superposition.
Such an undertaking entails certain dangers. In France the exist-
ence of conical flows was not known before 1946. Abroad, this question
has, for a long time, given rise to numerous reports which either were
not published or were published only after a certain delay. Thus it
must be pointed out that some of the results here obtained, original in
France when found, doubtlessly were not original abroad. Nevertheless
it seems possible to me to specify a certain number of points treated
in this report which, even considering the lapse of time, appear as new:
the parts concerning homogeneous flows, the general study of conical
flows with infinitesimal cone angles, the numerical or analogous methods
for the study of flows flattened in one direction, and a certain number
of the results of chapter IV. Moreover, even where the results which we
not always identical.
Another peculiarity should be noted. Since these questions actually
are everywhere the object of numerous investigations, progress has made
very rapid strides. This report edited at the beginning of 1948, risks
appearing, in certain aspects, slightly outmoded in 1949. To extenuate
this inconvenience we have indicated in a brief appendix placed at the
end of this report the progress made in these questions during the last
year. This appendix is followed by a supplementary bibliography which
indicates recent reports concerning our subject, or older ones of which
I should not have been able to successfully terminate this report
without the advice and support of my teacher, Mr. J. Peres, and it is
very important to me to express here my great respect for and gratitude
to him.
I should equally cite all those who directly or less directly have
contributed to my intellectual development and to whom I owe so much:
my teachers of special mathematics and of normal school, Mr. Bouligand
who directed my first reports, Mr. Villat, promoter of the Study of the
Mechanics of Fluids in France whose brilliant instruction has been of
the greatest value to me.
I also feel obliged to thank the directors of the O.N.E.R.A. who
have facilitated my task, and especially Mr. Girerd, director of aero-
dynamic research.
PREFACE
With his research on conical flows and their application, Mr. Paul
Germain has made a major contribution to the very timely study of super-
sonic aerodynamics. The present volume offers a comprehensive expose
which had been still lacking, an expose of elegance and solid construc-
tion containing a number of original developments. The author has fur-
thermore considered very thoroughly the applications and has shown how
one may solve within the scope of linear theory, by combinations of
conical flows, the general problems of the supersonic wing, taking into
account dihedral and sweepback, and also fuselage and control surface
effects. The analysis he develops in this respect leads him to methods
which permit, either by calculation alone or with the support of
electrolytic-tank experimentation, complete and accurate numerical
determinations.
After a few -preliminary developments (particularly on the validity
of the hypothesis of linearization), chapter I is devoted to the gener-
alities concerning conical flows. In such flows the velocity components
depend only on two variables and their determination makes use of har-
monic functions or of functions which verify the wave equation with two
variables according to whether one is inside or outside of the Mach
cone. Mr. Germain specifies the conditions of agreement between func-
tions defined in one domain or in the other and shows that the study of
conical flows amounts in general to boundary problems relative to three
analytical functions connected by differential relationships. He studies,
on the other hand, homogeneous flows which generalize the cone flows and
are no less useful in the applications.
From the viewpoint of the linear theory of supersonic flows one
must maintain two principal types of conical flows, bounded respectively
by an obstacle in the form of a cone with infinitesimal cone angle, and
by an obstacle in the form of a cone flattened in one direction.
The general investigation of the flows of the first type is entirely
Mr. Germain's own and forms the object of chapter II of his book. By a
subtle analysis of the approximations which may be legitimate Mr. Germain
succeeds in simplifying the rather complex boundary problem he had to
deal with; he replaces it by an external Hilbert problem. He shows how
it is possible, after having obtained the solution for an orientation
of the cone in the relative air stream, to pass, in a manner as simple
as it is elegant, to the calculation of the effect of a change in inci-
dence. He gives general formulas for the forces, treats completely
diverse noteworthy special cases and finally applies the method of trigo-
nometric operators which is also his own to the practical numerical
calculation of the flow about an arbitrary cone.
The determination of movements about infinitely flattened cones has
formed the object of numerous reports. The analysis which Mr. Germain
develops for this question (chapter III) contributes simplifications,
specifications, and important supplements. Thus he evolves, in the case
of an obstacle inside the Mach cone, a principle of minimum singularity
which enters into the determination of the solution. Mr. Germain gives
two original methods for treatment of the general case: one utilizes
the electrolytic-tank analogy, surmounting the difficulty arising from
the experimental application of the principle of minimum singularity;
the other, purely numerical, involves the trigonometric operators quoted
above.
In the last chapter, finally, Mr. Germain visualizes the composi-
tion of conical flows with regard to aerodynamic calculation of a super-
sonic aircraft. Concerning this subject he develops a complete theory
which covers most of the known results and incorporates new ones. He
concludes with an outline of the flows past a flat dihedral, with appli-
cation to the fins and control surfaces.
The creation of the National Office for Aeronautical Study and
Research has already made possible the setting up of groups of investi-
gators which do excellent work in several domains that are of interest
to modern aviation and put us on the level of the best research centers
abroad. Mr. Paul Germain inspirits and directs one of those groups in
the most efficient manner. He is one of those, and the present report
will suffice to bear out this statement, on whom we can count for the
development of the study of aerodynamics in France.
Joseph Peres
Member of the Academy of Sciences
NACA TM 1554
Pages
CHAPTER I GENERALITIES ON CONICAL FLOWS . 1
1.1 Equations of Supersonic Linearized Flows ... 1
1.2 Generalities on Conical Flows . .. 10
1.5 Homogeneous Flows . . ... ... .22
CHAPTER II CONICAL FLOWS WITH INFINITESIMAL CONE ANGLES ..... 30
2.1 Solution of the Problem . ... 30
2.2 Applications . . ... 41
Cone of Revolution . . ... .44
Elliptic Cone ....... . .. ..... 47
Study of a Cone With Semicircular Section . .. .58
2.5 Numerical Calculation of Conical Flows With
Infinitesimal Cone Angles . . .. .62
Calculation of the Trigonometric Operators ... 68
CHAPTER III CONICAL FLOWS INFINITELY FLATTENED IN ONE
DIRECTION . . . ... 79
5.1 Cone Obstacle Entirely Inside the Mach Cone ... 80
Study of the Elementary Problems (Symmetrical Cone ..... 80
Flows With Respect to Oxlx)) . .. ..80
Nonsymmetrical Conical Flows . .97
General Problem . . ... ...... 105
Rheo-Electric Method . .... .108
Purely Numerical Method ................. 117
5.2 Case Where the Cone Is Not Inside the Mach Cone () 152
Cone Totally Bisecting the Mach Cone (Fig. 28) ... 154
Cone Partially Inside and Partially Outside of the Mach
Cone (r) (Fig. 50) . . ... .. .142
Cone Entirely Outside of the Cone (C) (Fig. 29) . 152
5.5 Supplementary Remarks on the Infinitely Flattened
Conical Flows . . ... 159
CHAPTER IV THE COMPOSITION OF CONICAL FLOWS AND ITS APPLICATION
TO THE AERODYNAMIC CALCULATION OF SUPERSONIC AIRCRAFT .. 168
4.1 Calculation of the Wings . .. 168
Symmetrical Problems . . ... ... 171
Rectangular Wings . .... ..... 171
Sweptback Wings . . ... 186
Lifting problems . . ... ..... 206
Rectangular Wings . . ... 214
Effect of Ailerons and Flaps . ... .225
Sweptback Wing . . ... .. 229
The Lifting Segments . .... ... 240
4.2 Study of Fuselages . ... ... 243
4.5 Conical Flows Past a Flat Dihedral. Fins and Control
Surfaces . . ... . 251
vi
NACA TM 1554
Page
REFERENCES .. .. .. . ...... 261
APPENDIX ................. ........... 264
Digitized by the Interne[ Archive
in 2011 wilh funding Irom
University of Florida, Geolge A. Smathers Libraries wilh support from LYRASIS and the Sloan Foundation
Illp: www.arciiive.org details generaltheoryolc00unii
TECHNICAL MEMORANDUM 1354
GENERAL THEORY OF CONICAL FLOWS AND ITS APPLICATION
TO SUPERSONIC AERODYNAMICS*
By Paul Germain
CHAPTER I GENERALITIES ON CONICAL FLOWS
1.1 Equations of Supersonic Linearized Flows
1.1.1 General Equation for the Velocity Potential
Let us visualize the permanent irrotational flow of a compressible
perfect fluid for which the pressure p and the density p are mutual
functions. The space in which the flow takes place will be fixed by
three trirectangular axes Oxl, Ox2, Ox3, the coordinates of a fluid
molecule will be xl, x2, x3, the projections on Oxi of the veloc-
ity V and of the acceleration A of a molecule will be denoted by
ui and ai, respectively.
The fundamental equations which permit determination
are the Euler equations
of the flow
-* 1 -
P
ai 1 6P
al P x,
the equation of continuity
*"La theorie ge6nrale des movements coniques et ses applications
a l'aerodynamique supersonique." Office National d'Itudes et de
Recherches Aeronautiques, no. 34, 1949.
1We employ the classic convention of the silent index: --pui
bxi v
is to be read: ~-(uI) + -~(Pu2)
JL d
S (Pu3)
6x
NACA U' 1354
div pV = 0
or x(Pui) = 0
(1.2)
and the equation of compressibility
p = f(p)
If one notes that
6ui
ai = u
1 k 6xk
and introduces the sonic velocity2
c2 dp
dp
the equation (I.1) assumes the form
p1 i
Sxi
_ 1 dp 6p
P dp 6xi
(1.3)
(1.4)
_ 2 ap
P 6xi
We introduce the velocity potential (xl, x2, x3), defined with
the exception of one constant, by
x -
Ui ox,
2The velocity of sound, introduced here by the symbol d_ has a
dp
well-known physical significance; it is the velocity of propagation of
small disturbances. This significance frequently permits an intuitive
interpretation of certain results which we shall encounter later on
(see section 1.1.4).
oui
lk
axk
(1.5)
NACA TM 1354
which is legitimate since we shall assume the flow to be irrotational.
If we make the combination
u.u i N D 620
Uiuk uk- xi xk xi 0xk
one sees, taking into account equations (1.5) and (1.2), that
6' 6c 620 26 24
63 2 c2 62 (1.6)
3xi Oxk 6xi 6xk 6x 2
i
This equation is the general equation for the velocity potential.
One may show, besides, that c is a function of the velocity modulus;
thus one obtains an equation with partial derivatives of the second
order, linear with respect to the second derivatives, but not completely
linear.
The nonlinear character of the equation for the velocity potential
makes the rigorous investigation of compressible flows rather difficult,
at least in the three-dimensional case.
In order to be able to study, at least approximately, the behavior
of wings., fuselages, and other elements of aeronautical structures, at
velocities due to the compressibility, one has been led to introduce
simplifying hypothesis which permit "linearization" of the equation for
the velocity potential.
1.1.2 The Hypotheses of Linearization and Their Consequences
For aerodynamic calculation, one may assume that the body around
which the flow occurs has a position fixed in space nd that the fluid
at infinity upstream is moving with a velocity U, U being a constant
vector, the modulus of which will be taken as velocity unit. We shall
always assume that the axis Oxi has the same direction as U; the
hypotheses of linearization amount to assuming that at every point of
the fluid the velocity is reasonably equivalent to U.
We put in a more precise manner
S= 1+u u2 = v u3 = W
NACA TM 1354
u, v, w are, according to definition, the components of the "pertur-
bation velocity."
(1) u, v, w are quantities which are very small referred to
unity; if one considers these quantities as infinitesimals of the first
order, one makes it at least permissible to neglect3 in the equations
all infinitesimals of the second order such as u2, v2, uv, etc.
(2) All partial derivatives of u, v, w with respect to the
coordinates are equally infinitesimals at least of the first order so
that one is justified in neglecting terms such as u -, etc.
0x1 k fx2
One may deduce from these hypotheses a few immediate consequences:
(a) At every point of the field, the angle of the velocity vector
with the axis Oxi is an infinitesimal of the first order at least.
Hence there results a condition imposed on the body about which the flow
is to be investigated; at every point the tangent plane must make a
small angle with the direction of the nondisturbed flow (this is what
one calls the uniform motion, defined by the velocity U).
If one designates by q the velocity modulus, one has, taking the
hypotheses setup into account
q2 = (1 + u)2 + v2 + w2 = 1 + 2u
whence
q= 1 u
(b) The pressure p and the density p differ from the values p,
and pl which these magnitudes assume at infinity upstream only by an
infinitesimal of the first order; the equation (1.5) is written in effect
au C1
6xl P1 xl
3This signifies that u, v, w may very well not be infinitesimals
of the same order; in this case one takes as the principal infinitesimal
the perturbation velocity component which has the lowest order.
NACA TM 1354
with cl denoting the sonic velocity at infinity upstream; thus
2 -
u 1 (P pl)
5
(1'7)
On the other hand, according to equation (I.4)
P Pl = c2 P) = -lu
If one defines the pressure coefficient Cp by
SP P
p l /2 12
one has
Cp = -2u
p
(1.8)
(c) Finally, an examination of what becomes of the equation for the
velocity potential (equation (1.6)) under these hypotheses shows that it
is reduced to
x2
12
1 2
1 xl
+ _
x22
Let p(x1,x2,x3) be the "disturbance potential," that is, the
potential the gradient of which is identical with the disturbance-
velocity vector; P (x,x2,x3) is the solution of the equation with
partial derivatives of the second order
2
1 Cl2 _2
2 2
c1 6x1
_ i _
2
dX2
(1.9)
+ 32
Ox,2
a completely linear equation.
x \
2
3
NACA fM 1354
The Mach number of the flow is called the dimensionless con-
stant M which, with the velocity unit to be chosen arbitrarily,
cl
is written here M = 1/c1.
We put: (M2 1) = 02, with e being equal +1 or -1 according to
whether M is larger or smaller than unity.
(1) If M < 1, equation (I.9) is written
2 a29 2 29
P2 2 + 2p + 2 =
6x12 6x22 6x 2
an equation which may be easily reduced to the Laplace equation.
This equation applies to flows called "subsonic" because the velocity
of the nondisturbed flow is smaller than the sonic velocity at infinity
upstream. These flows will not be investigated in the course of this
report .
(2) If M >1, equation (1.9) is thus written
2 a2, -2, +629
032 + (I.10)
ax12 6x22 6x32
This equation applies to "supersonic" flows; if one interprets xl
as representing the time t, this equation is identical with the equa-
tion for cylindrical waves, well-known in mathematical physics. Investi-
gation of this equation will form the object of this report.
Remarks.
(1) It should be noted that, in order to write the preceding equa-
tion, it was not necessary to specify the form of the equation for the
state of the fluid. In particular, the formulas written above do not
introduce the value of the exponent 7 of the adiabatic relation p = kpy
which is the form usually assumed by the equation of compressibility.
Investigation of linear subsonic flows has formed the object of
numerous reports. See references 1 and 2.
NACA TM 1354
(2) The preceding analysis shows clearly the very different char-
acter of subsonic flows which lead to an elliptic equation, and of
supersonic flows which are represented by a hyperbolic equation.
(3) When we wrote equation (I.9), we supposed implicitly that
M2 1 was not infinitely small, that is, that the flow was not "tran-
sonic," according to the expression of Von KArman5. Thus it is impossible
to make M tend toward unity in the results we shall obtain, in the
hope to acquire information on the transonic case6.
(4) It may happen, in agreement with the statement made in foot-
note 3, that u is an infinitesimal of an order higher than first. In
this case, one will take up again the analysis made in paragraph (b) of
section 1.1.2, which leads to a formula yielding the Cp, more adequate
than the formula (1.8)
Cp = -2u (v2 + w2) (.11)
1.1.3 Validity of the Hypotheses of Linearization
Any simplifying hypothesis leads necessarily to results different
from those which one would obtain with a rigorous method. Nevertheless,
it was shown in certain numerical investigations on profiles (two-
dimensional flows) where the rigorous method and the method of lineari-
zation were applied simultaneously that the approximation method provided
a very good approximation for the calculation of forces. Besides, it
is well-known that the classic Prandtl equation for the investigation of
5Study of the transonic flows, with simplifying hypotheses analogous
to those that have been made, requires a more compact analysis of the
phenomena. It leads to a nonlinear equation, described for the first
time by Oswatitsch and Wieghart (ref. 3). From it one may very easily
deduce interesting relations of similitude for the transonic flows
(ref. 4). One may find these relations also, in a very simple manner,
by utilizing the hodograph plane.
In a general manner, according to the values of M, one may be led
to neglect certain terms in the final formulas found for the pressure
coefficient Cp. This requires an evaluation, in every particular case,
of the order of magnitude of the terms occurring in the formulas when M
varies. In this report, we shall never enter into such a discussion.
We shall limit ourselves voluntarily to the general formulas. An inter-
esting example of such a discussion may be found in the recent memorandum
of E. Laitone (ref. 5).
NACA TM 1354
wings of finite span in an incompressible fluid furnishes very acceptable
results, and the Prandtl equation results from a linearization of the
rigorous problem.
It happens frequently, we shall have occasion several times to point
it out, that the solution found for u, v, w will not satisfy the
hypotheses of section 1.1.2 in certain regions (for example in the neigh-
borhood of a leading edge); eventually certain ones among these magni-
tudes could even become infinite.
Under rigorous conditions such a solution should not be retained.
Anyhow, if the regions where the hypotheses of linearization are not
satisfied are "sufficiently small," it is permissible to assume that the
expressions found for the forces (obtained by integration of the pres-
sures) will still'remain valid. This constitutes a justification
a posteriori for the linearization method so frequently utilized in
numerous aerodynamic problems7. Therefore, we shall not systematically
discard the solutions found which will not wholly satisfy the hypotheses
we set up.
1.1.4 Limiting Conditions. Existence Theorem
Physically, the definition of sonic velocity leads to the rule
which has been called the "rule of forbidden signals" (see footnote 2
of section 1.1.1) and which can be stated as follows:
A disturbance in a uniform supersonic flow, of the velocity U
produced at a point P, takes effect only inside of a half-cone of
revolution of the axis U and of the apex half-angle a = Arc sin(l/M);
(D cot a) a is called the Mach angle, the half-cone in question Mach
after-cone at P."
Correlatively, one may state that the condition of the fluid at a
point M (pressure, velocity, etc.) depends only on the character of
the disturbances produced in the nondisturbed flow at points situated
inside of the "Mach fore-cone at M;" the Mach fore-cone at a point is
obviously the symmetrical counterpart of the Mach after-cone with respect
to its apex.
If one wants to justify this rule from the mathematical viewpoint,
one must start out from the formulas solving the problem of Cauchy and
take into account the boundary conditions particular to the problem.
Along the obstacle one must write that the velocity is tangent to the
obstacle which gives the value dP/dn. Moreover, at infinity
7
For instance, in the investigation of vibratory motions of infin-
itely small amplitude about slender profiles.
NACA TM 1354
upstream (xl = -c) the first derivatives of ( must be zero, since 9
is, from the aerodynamic viewpoint, only determined to within a constant,
it will be assumed zero.
The characteristic surfaces of the equation (I.10) are the Mach
cones. If one of the Mach cones of the point P cuts off a region (R)
on a surface (Z), the classic study of the problem of Cauchy8 shows
that the value of 9 at P is a continuous linear function of the
values of 9 and of dcp/dn on R.
Let us therefore consider a point M of a supersonic flow such
that its fore-cone does not intersect the obstacle. We take as the
surface E a plane xl = -A, with A being of arbitrary magnitude.
On E, rP and dP/dn, which are continuous functions, will be arbi-
trarily small. Consequently the value of c at M is zero. Thus one
aspect of the rule of "forbidden signal" is justified.
Let ds suppose that the forward-cone of M cuts off a region r(M)
on the obstacle; on r(M), dc/dn is given by the boundary conditions;
thus 91(M) is a linear function of the values of C on r(M).
One sees therefore that, if one makes M tend toward a point MO
of the obstacle, one will obtain a functional equation permitting the
determination of P on the obstacle, at least in the case where the
existence and uniqueness of the solution will be insured9. Consequently,
((M) depends only on the values of CP/dn in the region r(M); this
justifies the fundamental result of the rule of "forbidden signals."10
1.1.5 General Methods for Investigation
of Linearized Supersonic Flows
In a recent articlell dealing with the study of linear supersonic
flows, Von Kirman indicates that two major general procedures exist for
8For the problem of Cauchy, relative to the equation for cylindrical
waves, see for instance references 6 and 7.
9Such a method has been utilized by G. Temple and H. A. Jahn, in
their study of a partial differential equation with two variables (ref. 8).
10A more exact investigation of this question may be found in
appendix 1, at the end of this report.
11See reference 4. A quick expose of the methods in question may
also be found in the text, in reference 2.
NACA TM 1354
the study of these flows, one called "the source method," the other
"the acoustic analogy."
The first is an old method and its theoretical application is
fairly simple. It consists in placing on the outer surface of the
obstacle a continuous distribution of singularities, called sources,
the superposition of which gives at every point of the space the desired
potential; the local strength of the sources may, in general, easily be
determined with the aid of the boundary conditions. The second method
utilizes a fundamental solution of the equation (I.10), the composition
of which permits one to obtain the desired potential; this procedure is
interesting in that it permits utilization of the Fourier integrals and
thus furnishes, at least in certain particular cases, rather simple
expressions for the total energy.
Von Karman also indicates, at the end of his report, a third general
procedure, that of conical flows.
We intend to investigate in this report the conical flows and
the development of this third procedure which utilizes systematically
the composition of the "conical flows" and, more generally, of the flows
which we shall call "homogeneous flows of the order n." We shall see
that this procedure permits one to find very easily, and frequently
with less expenditure, a great number of the results previously obtained
by other methods, and to bring to a successful end the investigation of
certain problems which, to our knowledge, have not yet been solved.
1.2 Generalities on Conical Flows
1.2.1 History and Definition
Conical flows have been introduced by A. Busemann (ref. 9) who has
given the principal characteristics of these flows and has indicated
briefly in what ways they could be utilized in the investigation of
supersonic flows. Busemann gives as examples some results, frequently
without proof. Several authors have supplemented the investigation of
Busemann: Stewart (ref. 10) has studied the case of the lifting wing A
to which we shall come back later on; L. Beschkine (ref. 11) has fur-
nished a certain number of results but generally without demonstration.
We thought it of interest to attempt a summary of the entire problem.
One calls "conical flows" (more precisely, "infinitesimal conical
flows")12 the flows in which there exists a point 0 such that along
12The adjective "infinitesimal" is remindful of the fact that the
flows have been linearizedd;" we shall henceforward omit this qualifica-
tion since no confusion can arise in this report.
NACA TM 1354
every straight line issuing toward one side of 0, the velocity vector
remains of the same value.
Let (i) be a plane not containing 0, normal to the vector U;
let us suppose only that the velocity vector at every point of (i) is
not normal to (r); the projection of these velocity vectors on (i)
determines a field of vectors, the lines of force of which we shall
call (y): the cones (a) of vertex 0 and directrix (7) are "stream
cones" for the flow.
More generally, let (S) be a stream surface of the flow, passing
through O; every surface deduced from (S) by homothety of the center 0
and of k, k being an arbitrary positive number, is a stream surface.
(S) is not necessarily a conical surface of apex 0, but having (S)
given as an obstacle does not permit one to foresee the existence of
such a flow. It is different if a conical obstacle of apex 0 is given;
the designation "conical flow" is thus justified.
Conversely, let us consider a cone of the apex 0, situated entirely
in the region xl >0, and suppose that a linearized supersonic flow
exists around this cone; this flow is necessarily a conical flow such
as has just been defined; in fact, if V(xlX2,x3) denotes this velocity
field, V( xl,Xx2,Xxj) ( being any arbitrary positive number) is
equally a velocity field satisfying all conditions of the problem; con-
sequently, if the uniqueness of the desired flow is admitted, 7 must
be constant along every half-straight line from 0 13.
Let us also point out that according to equations (1.8) or (I.ll),
the surfaces of equal pressure are also cones of the apex 0.
1.2.2 Partial Differential Equations Satisfied
by the Velocity Components
According to definition, the velocity components of a conical flow
depend only on two variables; on the other hand, as functions of xl,
13It should be noted that this argument will no longer be valid
without restriction in the case of a real supersonic flow around a cone
because in this case the principle of "forbidden signals" is no longer
valid in the rigorous form stated. Among other possibilities, a detached
shock wave may form upstream from the cone behind which the motion is
no longer irrotational.
NACA TM 1354
x2, x3, they are naturally the solution of the equation
S2f
6x3 2
O3
Let us first put
x2 = r cos 9
x3 = r sin 0
the equation then assumes the form
2 32f
6xl2
2 2
_ 2f r1 2 1 rf
or2 r2 602 r 3r
(1.12)
The second term of equation (1.12) is actually nothing else but
the Laplacian of f(x1,x2,x3) in the plane x2, x3 (xl being con-
sidered as parameter); naturally f(xl,r,8) is periodic in 8, the
period being equal to 2n.
To make the conical character of the flow evident, let us put
xI = prX
(I.13)
X is a new variable; X < 1 characterizes the exterior of the Mach
cone with the apex 0, X > 1 characterizes the interior of the cone.
Under these conditions, the disturbance-velocity components are func-
tions only of :. and e. Since f is a function of X and 0 only
d2f = 2 dx2
dx2
+ 232 dv dB +
ax 6e
02f
O82
+ 6f d2X + kf d2
5ax e
dX = r (dxl OX dr)
o~r\ 1I
d2x = -L(d2x X d2r -
pr
2dr dx 1 + 20 X dr2)
r r
2 62f
6x 2
_ 2f
bx22
NACA TM 1354
2f
;r2
62f
6e2
6f are the respective coefficients of dxl2,
Tr
dr2,
d2r in the expression of d2f as a function of the vari-
s xl, r, e.
As a consequence, the equation (1.12) becomes under these conditions
r(x2
6-A.O2
(1.14)
+ b2f + f = 0
+02 -
6e2 ax
One may try to simplify this equation further by replacing the
variable X by the variable t, X and t being connected by a rela-
tionship X = X()), and by making a judicious choice for the func-
tion x(S). The first operation gives
(x2 2 i)
6t2
+ y,2 r
602
+ B 0
6 at (x. )x' =
oS X'
with the primes denoting derivatives with respect to E.
this equation, one may make the term in disappear.
realized by putting
(1) If X >1,
For simplifying
This will be
X = ch e
one obtains for f Laplace's equation
62f
6t2
+ 2
S62
(1.16)
(2) If X < 1,
= cos 1 (I.17)
in this case, one obtains the equation for waves with two variables
62f
3,2
62f
682
(I.18)
(1.15)
NACA TM 1354
Geometrical interpretation.- X > 1 corresponds to the interior of
the Mach rearward cone (r) of the point 0; every semi-infinite line,
issuing from 0, inside of this cone, has as image a point 8, E. One
will assume, for instance, -n < 0 < i; k = 0 corresponds to the
cone (r), E = w corresponds to the cone axis (it will always be pos-
sible to assume as positive). The image of the interior of (r)
forms therefore on the region (A) of the plane (0,a) (fig. 1), limited
by the semi-infinite lines AT, A'T' and by the segment AA'. The
correspondence is double valued in the sense that to a semi-infinite
line issuing from 0 there corresponds one point and one only (0e,)
in the bounded region and conversely, to one point of this region there
corresponds one semi-infinite line, and one only, issuing from O,
inside of (r).
Since we shall suppose, in general, that the cone investigated is
entirely in the region xl > 0, only this region will be of interest
(P then being identically zero for xl < 0). The semi-infinite lines
of this region issuing from 0, outside of (r), correspond to 0 < X < 1
(fig. 2), that is, according to equation (I.17), 0 < r < !L; n = 0
2
corresponds to the cone (F), T = n to the plane xI = 0; the semi-
2
infinite lines issuing from 0 correspond biunivocally to the points
of the region (A'), inside of the rectangle AA'B'B in the plane (e,r).
Summing up, the velocity components satisfy the simple equa-
tions (I.16) and (I.18), the first of which is relative to the region (A),
the second to the region (A').
1.2.3 Fundamental Theorem
The equation (1.14) which represents the fundamental equation of
our problem is an equation of mixed type; it is elliptic or hyperbolic
according to whether is larger or smaller than unity. In order to
study this equation in a simpler manner, we have been led to divide the
domain of the variables into two parts and to represent them on two
different planes. How an agreement will be reached between the solutions
obtained for f in the two planes that is the question which will be
completely elucidated by the following theorem which will be fundamental
in the course of our investigation.
Theorem: There exists "agreement" as to X = 1 for all derivatives of
f, defined in either the region (A) or (A'), provided that there is
"agreement" for the function itself.
NACA TM 1354
In fact, let us take two functions fl(e,), f2(6,Ti), the first
satisfying the equation (1.16) in the region (A), the second the equa-
tion (I.18) in the region (A'), both assuming the same values c(e) on
the respective segments ( 0 = 0, -i < 0 < n) (r = 0, -n < 0 < i). Let
Inf
B0 be the abcissa of a point of AA'. If 6nfL(80,0) exists,
8en
Sd-; consequently
den
nf 2 (0,0)
en
exists and
ae oO
aen
a3f(e,, )
- On
Let us now pass to the investigation of the derivatives of the
order n of the form nf the equation (1.14) shows first that
6Xoen-1
f(e,1) =- -f(,1)
OX 602
which proves that all partial derivatives of the order 1 with respect
to X have the same value on (P), whether they are calculated starting
from fl or from f2. The argument develops without difficulty through
recurrence. By deriving equation (1.14) n times with respect to :
and making X = 1, one obtains
(2n + 1) + n2 + -6nf 0
An+1 .a 2" 2n
which finally shows that the values n+f can be uniquely expressed
ae. n
as a function of the derivatives of f(0) with respect to G and that
they, consequently, have the same value, whether calculated starting
from fl or from f2.
Summing up, one may say that it is sufficient for the establishment
of the "agreement" between two solutions defined in (A) and (A'), if
these solutions assume the same value on the segment AA'.
anfl(00
agn
16 NACA TM 1354
1.2.4 Mode of Dependence of the Semi-Infinite
Lines Issuing From 0
If one puts in the plane (e0,)
8 + T = 2X 8 q = 2 (1.19)
one sees that the characteristics of the equation (I.18) are the parallels
to the bisectrices K = cte, = cte. These characteristics are, in
the plane (1,e), the images of the planes
xI = -r cos(2A 0) and xl = or cos(O 2P)
which are the planes tangent to the cone (P). The characteristics
passing through a point s0(O0,0) are the images of two planes tangent
to the cone (r) which one may lay through the semi-infinite A0 cor-
responding to the point 60 of the plane (8,0) (fig. 3). The gener-
atrices of contact are characterized on the cone by the values 81
and 82 of the angle 0. One encounters here a result which seems to
is immediately explained if one notes that, since all points of a semi-
infinite A0 issued from 0 are equivalent, one must consider at the
same time all Mach cones, the apexes of which are situated on A0; the
group of these cones admits as envelope precisely the two planes tangent
to the cone (r) passing through A0. We shall call "Mach dihedron
posterior" to the semi-infinite A0 that one of the dihedra formed by
the two planes which contains the group of the Mach cones to the rear of
the points of A0. The region inside ot this dihedron and outside of
the cone (r) has as image in the plane (0,rj) the triangle 01 5062.
A semi-infinite A1 will be said to be dependent on or independent of
A0 according to whether the image of A1 will be inside or outside of
the triangle 81 6082. This argument also explains why the equa-
tion (1.14) shows elliptic character inside of (r). More precisely,
two semi-infinite lines A1 and 62, inside of (r), are in a state of
neutral dependence (ref. 9). In fact, let M1 be a point of Ak, M2
a point of 2; let us suppose that Mi is outside of the Mach forward
cone of M2; according to the argument of section 1.1.4 the point M2
seems to be independent of Ml; but on the other hand, if one assumes Mi'
NACA TM 1354
to be a point of A1, inside of the Mach forward cone of M2, M1'
and M1 are equivalent which explains that M2 is actually not inde-
pendent of M1 (fig. 4).
1.2.5 The Conditions of Compatibility
Thus one may foresee how the solution of a problem of conical flow
will unfold itself. One will attempt to solve this problem in the
region (A') which will generally be fairly easy since the general solu-
tion of the equation (1.18) is written immediately by adjoining an arbi-
trary function of the variable 6 + n to an arbitrary function of the
variable 6 q. This will have the effect of "transporting" onto the
segment AA' the boundary conditions relative to the region (A').
Applying the fundamental theorem, one will be led to a problem of har-
monic functions in the region (A). But taking as unknown functions the
components u, v, w, of the disturbance velocity, we have introduced
three unknown functions (while there was only one when we dealt with
the function C). One must therefore write certain relationships of
compatibility which express finally that the motion is indeed irrotational.
The motion will be irrotational if u dxI + v dx2 + w dx3 is an
exact differential which will be the case when, and only when
x1 du + x2 dv + x dw = r(py du + cos 8 dv + sin 0 dw)
is an exact differential. This can occur only if this expression
is identically zero, with u, v, w being functions uniquely of 0
and of X (the total differential not containing a term.in dr must
be independent of r):
In a conical flow the potential is written
P = uxl + vx2 + wx3 = r(puX + v cos 0 + w sin e)
with u, v, w being the disturbance-velocity components.
One will note that C, is proportional to r.
Moreover
OX du + cos 0 dv + sin 0 dw = 0
(1.20)
18 NACA TM 1354
This is the relationship which is to be written, and this is the
point in question, on one hand in the plane (C,0), on the other in the
plane (6,).
(a) Relations in the plane (0,q). One may write
u = ul(?) + u2(p)
and analogous formulas for v and w, X and
relations (I.19). One has in particular
dUl 3u + u du2 u
dX 5q 60 du o0
Besides, according to equation (1.20)
- being defined by the
P cos T dul + cos 0 dvl + sin 0 dwl = 0
(1.21)
3 cos T du2 + cos 0 dv2 + sin 0 dw2 = 0
however: 0 = x + P, T = A 1; and consequently the first equa-
tion (1.21) is written
cos r 1 cos K dui + cos K dvI + sin k dwli +
sin psin du sin K dv1 + cos K dw = 0
since the two quantities between brackets are unique functions of
the preceding equality causes
p cos K dul + cos K dvi + sin X dwl = 0
P sin k dul sin X dv1 + cos K dwI = 0
dv1
-p dul = 1
cos 2X
dwl
sin 2X
(1.22)
NACA TM 1354
In the same manner one will show that
dv2 dw2
-p du2 = --
cos 2Ii sin 21
(1.23)
(b) Relations in the plane (e,t).
The calculation is perfectly analogous. The equation (1.16) causes
us to introduce the complex variable 0 = 8 + it and the func-
tions U(.), V(C), W(t), defined with the exception of an imaginary
additive constant, the real parts of which in (A) are, respectively,
identical to u(e,a), v(0,t), w(e,t).
The equation'(I.20) permits one to write
0 ch t dU + cos 0 dV + sin 0 dW = 0
If one puts
e + it =
e iE5 =
one obtains
cos T L
2sin
sin [
cos dU + cos -
2 2
sin dU sin -
2 2
dV + sin dW +
2
dV + cos -
2
thence one concludes as previously
- dU dV dW
cos ( sin c
The formulas (I.22), (1.23), (I.24) express the relationships of
compatibility which we had in mind.
dW = 0
(1.24)
NACA TM 1354
Remark.
We shall utilize frequently the conformal representation for studying
the problems relative to the domain (A). If one puts, in particular
Z = eit = e- ei
one sees that (A) becomes in the plane Z the interior area of the
circle (CO) with the center 0 14 and the radius 1 (fig. 5).
If one puts Z = pe the point Z is the image of a semi-infinite
line, issuing from the origin of the space (xl,x2,x3), characterized
by the angle 6 and the relationship
X1_ 1 + p2
pr 2p
The origin of the plane Z corresponds to the axis of the cone (P),
the circle (Co) to the cone (r) itself. A problem of conical flow
appears in a more intuitive manner in the plane Z than in the plane f.
In the plane Z, the formulas (1.24) are written
SdU = 2Z dV = 2iZ dW (1.25)
Z2 + 1 Z2 1
We shall moreover utilize the plane z defined by
S 2Z
Z2 + 1
The domain (A) corresponds conformably to the plane z notched by
the semi-infinite lines Ax, A'x' (fig. 6), the cone (r) at the edges
of the cuts thus determined, and the axis of the cone (F) at the origin
14
No confusion is possible between the point 0, origin of the sys-
tem of axes xl, x2, x3 and the point 0, here introduced as the
origin of the plane Z.
NACA TM 1354
of the plane z. The relations of compatibility in the plane z then
assume the form
-0 dU = z dV iz dW (1.26)
1 -z2
1.2.6 Boundary Conditions
The Two Main Types of Conical Flows
The boundary conditions are obtained by writing that the velocity
vector is tangent to the cone obstacle. Let, for instance, x2(t), x3(t)
be a parametric representation of the section xI = 0 of the cone;
x3x2' x2x'3, xx3', -PY2' constitute a system of direction parameters
of the normal to the cone obstacle, and the boundary condition reads
x' vx3' = (x3x2' x2x3' (1 + u) (1.27)
It will be possible to simplify this condition according to the
cases. However, the simplification will have to be treated in a dif-
ferent manner according to the conical flows investigated. As set forth
in section 1.1.2, two main types of conical flows may exist.
(1) The flow about cones with infinitesimal cone angles, that is,
cones where every generatrix forms with the vector U an angle which
remains small. Naturally, the cone section may, under these conditions,
be of any arbitrary form; since the flow outside of (P) is undisturbed
(velocity equivalent to U), on the cone (r) u, v, w are zero.
The problem may have to be treated in the plane Z; U(Z), V(Z),
W(Z) will have real parts of zero on (CO). The image (C) of the
obstacle, in the plane Z, is defined by a relation p = f(e); conse-
quently, a parametric representation of the section xi = p will be
obtained by means of the formulas
2p 2p
X2 = 2 cos e X3 sin G
1 + p2 1 + p2
NACA TM 1354
Thus the condition (1.27) becomes
w sin 0 p' cos a + p2(p sin e + p' cos 8) +
v cos 0 + P' sin e + 2(p cos 0 p' sin 8) = 2-(1 + u) (1.28)
with 0 taken as parameter, and p' denoting the derivative of p
with respect to 0. The investigation of conical flows with infinitesimal
cone angles will form the object of chapter II.
(2) The flow about flattened cones, that is, cones, the generatrices
of which deviate only little from a plane containing U. Let us remember
that (section 1.1-.2) the tangent plane is to form a small angle with 6;
consequently, rigorously speaking, the section of such a cone cannot be a
regular closed curve, an ellipse for instance; it must present a lentic-
ular profile (fig. 7). In chapter III we shall study the flows about
such cones.
Remark.
Actually, we have, therewith, not exhausted all types of conical
flows, that is, those for which linearization is legitimate. One may,
for instance, obtain flows about cones, the section of which presents
the form shown in fire 8; the axis of such a cone has infinitely small
inclination toward U.
Before beginning the study of these flows we shall, in order to
terminate these generalities, introduce a generalization of the flows,
the possible utilization of which we shall see in a final chapter.
1.3 Homogeneous Flows
1.3.1 Definition and Properties
The conical flows are flows for which the velocity potential is of
the form
cP = rf(e,X)
as we gad seen in section 1.2.5. One may visualize flows for which
P = rnf(0,x)
NACA TM 1354
We shall call them homogeneous flows of the nth orderl5. The conical
flows defined in section 1.2 are, therefore, homogeneous flows of the
order I. However, we shall maintain the expression "conical flow" to
designate these flows since this term has been used by numerous authors
and gives a good picture.
The derivatives of the velocity potential with respect to the vari-
ables xl, x2, x3 all satisfy the equation (I.10). If one then con-
siders the derivatives of the nth order of the potential of an homogeneous
flow of the nth order, one finds that they depend only on X. and a
and satisfy the equation (1.14); the analysis made in section 1.2.2
remains entirely valid. One may make the changes in variables (I.15)
and (T.17) which lead to the equations (1.16) and (1.18). Thus one has
here a method sufficiently general to obtain solutions of the equa-
tion (I.10) which may prove useful.
The simplest flows are the homogeneous flows of the order 0 which
do not give rise to any particular condition of compatibility. For the
flows of nth order, in contrast, one has to write a certain number of
conditions connecting the derivatives of nth order. We shall examine16
as an example the case of homogeneous flows of 2nd order.
There are six second derivatives which we shall denote Tij (i
and j may assume independently the values 1, 2, 3), Cij designating
4 Outside of (r) we shall put
6xi 6xj
ij ij i
1 2
with P.ij being a function of X only, Pij2 of p only (see for-
mula 1.19). Inside of (r), ij. is the real part .of a function .i.(U).
In order to obtain the desired relations, it is sufficient to note
that
15The definition for homogeneous flows of the nth order has been
given for the first time by L. Beshkine (ref. 11); this author, by the
way, calls them conical flows of the nth order. One may also connect
this question with the article of Hayes (ref. 12).
S6ee appendix 2.
NACA TM 1354
ij dxj = dPi
and to apply the results of section 1.2.5; thus one may write the fol-
lowing six relations between the c 1
-0 dil 1 d 1 d 1 (i = 1,2,3)
11 cos 2X 12 sin 2X i3
which, besides, are reduced to five as one sees immediately. One will
have analogous relations for the functions Cpj2 (it is sufficient to
exchange the role of X and of p).
Finally, one has for the analytic functions oij()
d 1 d 1 d3
ii cos i i2 sin i i3
namely six relations which as before are reduced to five. The written
conditions are not only necessary but also sufficient since the func-
tions Pi necessarily are the components of a gradient. Thus one sees
that there is no difficulty in writing the conditions of compatibility
for a homogeneous flow of nth order.
1.3.2 Relations Between the Homogeneous Flows
of nth and of (n-l)th Order
We shall establish a theorem which can be useful in certain prob-
lems and which specifies the relations existing between homogeneous
flows of nth and of (n-1)th order; we shall examine the case where n = 1.
1.3.2.1.- Let us consider inside of the cone (P) a homogeneous
flow of the order 0 defined by
f = REcZ)]
NACA TM 1354 2
We shall first of all seek the components u, v, w of the dis-
turbance velocity
dC = u dxI + v dx2 +
wdx = R[O'(Z) dz = R[Zs'(Z) d
then
1 + p2
xl = pr 2
2P
thus
dZ = dp + i d0
Z p
x2 = r cos 0
x3 = r sin 8
_ 2 + 1 dx x2 dx2 + x2 dx
p2 1L r2
Sx2 dx3 x3 dx2
r2
whence one deduces
p2 + 1 1
p2 1 1-
v cos P02 + 1
r P2 1
w sin 8 02 + 1
r p2 1
R[Zq'(Zz] + sin e T[ Z'(z]
R-Zo'(Z CS- '(Z
~r -
p2 + 1 sin 0
P
p2-1 sin 0
P
however
Z 1
z
Z
z
p2 + 1
P21
_ 2 -
p
cos 8 + i
cos e + i
NACA 4M 1354
hence the result
1
u
xl
xl
p2 + 1 R Z'(Z)
p2
p2 +1
p2 -1
p2 1
R [- (Z2 + 1)'(Z)
R (2 1)0,(Z)]
R- 2 1)01(z
(1.29)
1.3.2.2.- Let us now consider a point O' (xI = E1, x2 = 0, x3 = 0),
El being a very small quantity. Let M be a point with the coordi-
nates (xl,r,e) with respect to O, inside of (F), and with the para-
meters (p,8) in the plane Z. For the conical flow (homogeneous of
Ist order) with the vertex 0', its coordinates are: (xl- E r, a) and
its parameters in the Z-plane:
( p2 1 1
since
dxI = 1 = Br
p2 1
2p2
dp = x 2 1 dp
do = x p
p + 1
Let us then consider two identical conical fields but with the
apexes 0 and 0', and form their difference. We shall obtain a
velocity field which, due to the linear character of the equation (I.10),
will satisfy this equation. If
U0 = R[(Z)
denotes the component u of the field with the vertex
component u in the "difference field"
0, one has as
u = +RF(Z) R 2 + 1 z
P2 X1 ]
- P2 + 1 1 RFZF'(Z)
p2 1 xl
~~2
(I.30)
NACA TM 1354
El being considered as infinitely small. Moreover, according to the
relations (1.25), the components v and w are written
v = 2 + 1 aE fi (Z2 + 1)F'(Z]
w l p2 +- i PZ2 1)F'(Z)
xi p2 1 2
1.3.2.3.- Let us consider the point 0''(O,E2,0), with E2 being
a small quantity.' Let M be a point with the coordinates (x1,r,O)
with respect to 0, inside of (r), with the parameters (p,8) in the
plane Z. For the flow with apex 0'', the coordinates of M are
(x1, r E cos 8, 0 + E2 sin as can be easily stated by projecting
M in m on the plane x2x3 (fig. 9). But on the other hand
2xd 1 2
dr dp = -C2 cos 0
( (1+p2)2
r de x P dO = 2 sin 0
1 + p2
thus
dZ = e p + ip d] = e2 12 2 x i sin e cos 8 + e1i
with Z + dZ representing the point M in the conical field with the
vertex 0''.
Let us then consider two identical conical flows, but with the
apexes 0 and 0'', and form their difference. We shall obtain a
velocity field which due to the linear character of the equation (I.10)
will satisfy this equation. If
O = R[G(Z)
28 NACA TM 1354
denotes the component v in the field of the vertex 0, one has a com-
ponent v in the "difference field"
v +n[G(Z) R[(Z + dZ) = -R[G'(Z)dZ]
Cx2 P2 +_1 2) + i sin 1(P2
= p-- R G'(Z) cos e ( + p2) + i sin e(p2 l)e
S2 P 2+ 1 R ZG'(Z) Z +1
2x1 2 -1 2
2x p2 1-L Z/j
x1 P2 1 2 (1.32)
besides, according to equation (I.25), the components u and w are
written
E2 02 + 1
1 po' 1
(1.33)
w =2 p2 + 1 R (Z2 1)G'(Z]
x p2 1 2
1.3.2.4.- With these three lemmas established, it is easy to demon-
strate the property we have in mind. Let us call "complex potential" of
a homogeneous flow of zero order the function ((Z) (section 1.3.2.1)
so that
P = R [0(Z)]
so that the function of complex variable, the real part of which gives
insideof (P) the projection of the disturbance velocity in the direc-_
tion i, is the "complex velocity" of a conical field in the direction 1;
so that, finally, the velocity field obtained by the difference of two iden-
tical conical fields, the verties of which are infinitely close and
ranged on a line parallel to 1, is the "field derived from a conical
flow" in the direction 1; then we may state:
NACA TM 1354
Theorem: The field derived from a conical flow in the direction Z is
the velocity field of a homogeneous flow of zero order; the complex
potential of that flow of zero order is proportional to the complex
velocity of the conical field given in the direction i, since the pro-
portionality factor is real.
The proof follows immediately. According to sections 1.1.2
and 1.1.3 one may be satisfied with considering, for definition of a
homogeneous flow, the inside of the cone (r); comparison of the for-
mulas (1.29), (I.30), (I.31), (1.32), (1.33) entails the validity of
the above theorem'if I is parallel or orthogonal to U. Hence the
general case where I is arbitrary may be deduced immediately; if
F(Z), G(Z), H(Z) are the complex velocities in projection on Oxl,
Ox2, Ox3, the expression for the component u of the field derived in
-4
the direction I(EE 2,3) is
u = p2 + 1 R Z F'(Z) + e20'(Z) + 3H'(Z)
u i 02 1 C
Thus, with E1F(Z) + E2G(Z) + E3H(Z) being the complex velocity in
projection on 1, comparison of this formula with the first formula (I.29)
completely demonstrates the theorem.
-4
Corollary: The field derived in the direction 2 of a conical flow,
the complex velocity of which in the direction I is K(Z), is a
velocity field of a homogeneous flow dependent only on K(Z) (not on
the direction I).
The theorem just demonstrated may be extended without difficulty
to the homogeneous flows of nth and (n-l)th order. A statement of this
general theorem would require only specification of a few definitions;
however, since we shall not have to utilize it later on, we shall not
formulate this statement.
NACA TM 1354
CHAPTER II CONICAL FLOWS WITH INFINITESIMAL CONE ANGLES *
2.1 Solution of the Problem
2.1.1 Generalities
We shall now treat the first problem set up in section 1.2.6. We
shall operate in the plane Z. Let us recall that the image of the
cone (r) is the circle (CO) of radius unity centered at the origin,
and that the image of the obstacle is a curve (C), defined by its polar
equation p(e). We shall denote by (D) the annular domain comprised
between (C) and (CO); we shall call (O0) the circle of smallest
radius centered at the origin and containing (A) in its interior, and
we shall call k the radius of the circle (70). In this entire
chapter, k will be considered as the principal infinitesimal.
The problem then consists in finding three functions U(Z), V(Z),
W(Z) defined inside of (D) except for an additive imaginary constant,
so that
(1)
-3 dU 2Z dV 2iZ dW (1.25)
Z2 + 1 Z2 1
(2) the real parts u, v, w, which are uniform become zero on (CO),
(3) on (C), one has the relation
vp cos e + p' sin 8 + p2(p cos e p' sin e) +
r -j 2
p sin e p' cos e + p2( sin 0 + p' cos e) = 2-2( + u)
Put in this manner, the problem is obviously very hard to solve in
its whole generality; however, an analysis of the permissible approxima-
tions will simplify it considerably.
2.1.2 Investigation of the Functions U(Z), V(Z), W(Z)
2.1.2.1.- An analytical function of Z will be the said func-
tion (A) if its real part becomes zero on (Co). Let us designate by
NACA editor's note: Some minor inconsistencies appear in the number
of equations in this chapter and subsequently in chapters III and IV, but
attempt was made to change the numbering as given in the original text.
NACA TM 1354
(70') the circle with the radius 1/k, centered at the origin, and by
(D') the annulus limited by (o0) and (70') (fig. 10).
Lemma I.- A uniform function (A), defined inside the annulus
limited by (70) and (CO) may be continued over the entire domain (D').
This results immediately from Schwartz' principle. Let M and M'
be two symmetrical points with respect to (Co), M being inside of
(CO); 6ne defines the function (A) at the point (M') as having,
respectively, an opposite real and an equal imaginary part compared to
the real and the imaginary part of the function given at the point M.
Lemma II.- A holomorphic function (A) inside of (D') has a
Laurent development of the form17
+ 2IQnn~izn
ip+ Z Kn)
1
Let H(Z) = h + ih' be such a function (A). Let us write its
Laurent development in (D') provisorily in the form
H(Z) = JnZn + -n
0 1
It is an immediate demonstration and yields the formulas defining Jn
and Kn
Kn = (h + ih')o7eind8
21t 0O
17
We remember that Kn denotes the conjugate imaginary of Kn.
NACA TM 1354
(h + ih')70 denoting the value of H on (70); likewise
J -kn 2
I230
(h + ih') ie-inede
(h~ih '70
Consequently, according to the lemma I:
Kn = -Jn
moreover
S2n
H(Z) d
Z 2xJo
S.2n
(h + ih')0 dO -
00 2n 0o
is purely imaginary, and the lemma II is therewith demonstrated.
We shall note that, if H(Z) is limited by M on (70)
one has the inequality
or (70'),
Kn < Mkn (II.]
Lemma III.- A function (A) with a real and uniform part defined
in (D) can be developed inside of (D') in the form
B log Z + ip + (f KnZn (II._
with B being real.
Actually, the derivative of the function (A) is necessarily uni-
form. Thus one knows (see for instance ref. 13) that one may consider
the given function as the sum of a uniform function H(Z) and a loga-
rithmic term; since the critical point of the logarithm is arbitrary
inside of (O), it is particularly indicated to choose this point at
the origin; since the real part of the function is uniform, the coeffi-
cient of log Z is real. Besides, since log Z has a real part zero
h d
Co
Jo i-
2i ,0
co
NACA TM 1354
on (Co), H(Z) is itself a function (A). The given function may
therefore be continued inside of (D') and the development (11.2) is
thus justified.
Remark.
(7o)
form
If one chooses as pole of the logarithmic term a point inside of
but different from the origin, one obtains a development of the
B' log aa -- + i8 +
1 1
- K'nZn
2.1.2.2.- The functions U,
three functions (A) with a real
be developed in the form (II.2).
- e u(z)
2
= A log Z + ia
V(Z) = B log Z + ip
W(Z) = C log Z iy
+
ixz
V, W of the variable Z are all
uniform part and, consequently, can
We shall write henceforward
(n JZ
Znn nZn)
J K n Zn
- n
- L,'n
Lnzn)
(11.3)
A, B, C are real, a, 0, ? are real and also arbitrary; but these
developments are not independent since the relations (I.25) must be
taken into account. For instance, Z dV/dZ must be divisible by Z2 + 1;
otherwise we would have for U logarithmic singularities on the cone (r)
=B-i
1
n(K"n + izZn
Zn|- + K, I
zd
dZ
(+ Ln
1
NACA TM 1354
Hence one deduces the relations
B 1= ( P[K2p + R2]
1
S=3 ( 1)P(p + 1)K2p
0
- Kp+1
obtained by putting in the preceding equality Z = i and
Z = -i.
Z dW/dZ must be divisible by Z2 1 which gives
C = 2p (L2p + L2p)
1
0 => (2p + 1)(L2p+1 + L2p+l)
0
connecting the coefficients of the developments (II.3)
thus one may write the relations
B + 2K2 =-i[C 2L2]
nKn (n 2)Kn2 =
to relationships
among themselves;
K1 K1 = -i + L]
i[(n 2)Ln-2 + nr]
(n >,2)
and on the other hand
B =-(J + J1
K1 = -A + 2J2
nKn = (n 1)Jn-1 +
Likewise,
(n >2)
(n + 1)Jn+1
NACA TM 1354
2.1.2.3. Approximations for the developments (II.3).- Moreover,
the hypotheses of linearization must be taken into account which, as we
shall see, will permit us to simplify the developments (11.3) consider-
ably and will lead us in a very simple manner to the solution of the
problem posed in section 2.1.1.
The qualities (11.6) make V(Z) and W(Z)
order. We shall denote by M an upper limit of
circle (70). M will be equally an upper limit
(70') and hence in the entire domain (D').
seem of the same
their modulus on the
of their modulus on
If one utilizes the inequality (1.1), (11.4) shows that18
B = 0 Mk2)
K1 1 = o(Mk2
If one assumes a, 0, 7 zero in what follows, which does not at
all impair the generality, one may write the second formula (11.3) in
the form
V(Z) R(Kn ( -
and consequently:
2
2
Kn = B log Z -
Zn
In the annulus limited by (70) and
equality is
o(Mk2log k)
Likewise according to equation (11.5)
C = o(Mk2)
(Co), the second term of this
L1 + i =o(Mk3)
W(Z) iT(L1)(. + Z) n = C log Z + RL)
2
- z)
- n
2
180 denotes Landau's symbol, A = 0 Mk2) signifies that A is
Mk2
limited when k tends toward zero.
nZn + iT(K1)(
+ z)
NACA TM 1354
In the annulus comprised between (70) and (CO), the second term
of this equality is also
S(Mk21og k)
Furthermore, according to equation (1.6)
Kn-2 + Ln-2 = 0 (kn)
(n > 2)
Thus
W(Z) iV(Z) = (Mk2log k) + 2iK1Z
in the annulus (70,C).
Finally, according to equation (11.7)
A = -K1 + o(Mkl3)
Jn = n 1 Knl + O n+2
n
Q U(Z) = -R(Kl)log Z 2K2Z + n + 1 Kn+ +1
2 -- n Zn
1
Summing up: If one is satisfied with defining V(Z)
for O(Mk2log k) and U(Z) except for O(Mk3log k),
the corona (o',Co)
O(Mk3log k)
and W(Z) except
one may write in
W(Z) = iV(Z) + 2iKiZ
V(z) = H(Z) K1Z
(11.8)
Thus
NACA TM 1354 37
with
H(Z) = Kn (II.10)
1
and
U(Z) = Z dZ 2K2 (II.11)
The coefficient K1 may be supposed to be real, and the integra-
tion occurring in equation (II.11) must be made in such a manner that
R[U(Z)] will be an infinitely small quantity of the third order at
least on II = 1.
2.1.2.4 Remarks.
(1) The formula (1.8) which is the most important may be estab-
lished immediately from the second formula (1.25). However, the method
followed in the text, even though a little lengthy, seems to us more
S natural; also, it shows more clearly the developments of the func-
tions U, V, W.
(2) Strictly speaking, the hypotheses set forth in the course of
this study must be verified by the solutions found in each particular
case. We shall, however, omit this verification which in the usual
cases is automatically satisfactory.
(3) The results obtained by the preceding analysis and condensed
in the formulas (II.8), (II.9), (II.11) are in all strictness valid only
in the annulus (o0'CO), but not in the domain (D). However, it is
very easy to extend, by analytical continuation, the definition of H
to (D). Let us now first suppose that (C) contains 0 in its
interior; since one may write V(Z) in the form
V(Z) = H(Z) Z KnZ + B log Z
1
one sees that, since V(Z) is defined by hypothesis in (D), and one
can extend KnZn and B log Z inside of (70) up to (C), H(Z)
1
NACA TM 1354
may itself be defined without difficulty inside of (D). The case where
(C) does not contain the origin offers no difficulty; it is then suffi-
cient to utilize the development given at the end of section 2.1.2.1.
As to the order of the terms neglected when one writes the equal-
ity (IT.9) in the domain (D), they are found to be O(Mk2log k) in
(D) in the case where there exists inside of (C) a circle of the
radius Xk (X and 1/k may be considered as 0(1)). Besides, if
that is not the case, one may justify the validity of the results of
the formulas (11.8), (II.9), (11.10), (II.11) by making a conformal
representation of the domain (D) on an annulus; the radius of the
image circle of (Co) may be assumed equal to unity; the image circle
of (C) has a radius infinitely small of first order with respect to
k and the study may be carried out in the new plane of complex variable
thus introduced, without essential complication.
2.1.3 Reduction of the Problem to a Hilbert Problem
If one puts, according to the formula (II.8)
V = v + iv'
with v' denoting the imaginary part of V, one may write on (C) the
relation
w = -v'
Since one may, of course, with the accepted approximations, neglect u
compared to 1 in the second term of the formula (1.28), one sees that
this boundary condition (1.28) affects now only one single analytical
function, the function V(Z); this is a first fundamental consequence
of the preceding study. Formula (II.9) shows that this condition con-
sists in posing a linear relation between the real and the imaginary
part of H(Z) on the obstacle. Now according to equation (II.10) the
function H(Z) is a holomorphic function outside of (C), regular at
infinity; the problem stated which initially referred to an annular
area (D) is thus reduced to a Hilbert problem for the function H
defined in a simply connected region; exactly speaking, one has to solve
an exterior Hilbert problem. This is the second fundamental consequence
of the results of section 2.1.2.
Since we attempt to calculate V(Z) and W(Z) not further than
within O(Mk2log k), and U(Z) within 0(Mk5log k), the relation (1.28)
NACA TM 1354
which is written
R (v iw)2Z2d0 i dz(l p2
may be simplified and reduced to
S- dZ(v iw) 2P d
[ ] 2
On (C),
and therefore
KiZ is, according to equation (II.1), of the order of Mk2,
H = V = v + iv' = v iw
consequently, H satisfies, on (C), the Hilbert condition
R iH(Z) dZ] 2p- d6
(11.12)
2.1.4 Solution of the Hilbert Problem
A function H(Z), holomorphic outside (C), regular and zero at
infinity, satisfying on (C) the relation (II.12) must be-found. Let
a1
z = Z + a+ +
S Z
(11.13)
be the conformal canonical representation of the outside of (C) on
the outside of a circle (7) centered at the origin of the plane z;
the adjective canonical simply signifies that z and Z are equivalent
at infinity.
On (7) we shall put
z = reil
S2p2 d
. -de
NACA TM 1354
r being constant and well determined. Let us put
F(Z) = i log z (II.14)
One has on (C) or on (7)
F'(Z) dZ = i dz = -dP = f(e) dO (II.15)
z
with f being real; consequently
d d F'(Z) d F'(Z)
P-- -,z =~ 4i
and therefore equation (II.12) is written
H(Z) p2
S2 (11.16)
l F'(Z) p dI
H(Z)/F'(Z) is a holomorphic function outside of (C) and regular at
infinity. Following a classical procedure, we thus have reduced the
Hilbert problem to an exterior problem of Dirichlet.
Let G(Z) be the holomorphip function outside of (C), real at
infinity; its real part assumes on (C) the values G(Z) is
0 dQP
determined in a unique manner. According to equation (II.12)
H(Z) = -iG(Z)F'(Z) + iEF'(Z) (11.17)
with c being a real constant.
However, we have seen (section 2.1.2.3) that the coefficient of 1/Z
in the development of H(Z) around the point at infinity (coeffi-
cient KI) was real; now, around the point at infinity
a
iF(Z) =- + +.
Z 2
NACA IT 1354 41
In order to have the development of the second term of the formula (11.17)
admit a real coefficient of 1/Z, E must be zero since G(Z) is real at
infinity. Thus the desired solution is
H(Z) = -iG(Z)F'(Z) (II.18)
With the function H(Z) thus determined, the formulas (11.8),
(11.9), (II.11) permit calculation of the complex velocities U(Z),
V(Z), W(Z) within the scope of the accepted approximations. Thus the
problem posed in section 2.1.1 is solved.
Remarks.
(1) Uniqueness of the solution.- The preceding reasoning shows the
solution of the Hilbert problem satisfying the conditions (II.16) to be
unique. This result will be valid for our problem if one shows that
every function satisfying the condition (II.16) is a solution of the
initially posed problem (condition (II.14)) which is immediate since it
suffices to repeat the calculation.
(2) Calculation of the coefficient K1.- According to what has been
said above, the coefficient K1 is equal to the (real) value assumed by
SG(Z) at infinity. In order to find G(Z), we may solve the Dirichlet
problem in the plane z; according to a classic result of the study of
harmonic functions, K1 is equal to the mean value of 2p2 O on the
circle (7). Hence
K1- = 2f 2p d d9P = i-f p2d = 2S
27 0 d dp PJ() Tr
wherein S represents the area inside the contour (C).
2.2 Applications
2.2.1 General Remark
Let us consider a cone of the apex 0 in the space (Oxl,x2,x3),
the image of which in the plane Z is the curve (C), defined by its
polar equation p(e). According to the definition of p (see the
remark of section 1.2.5) the sections of this cone made by planes par-
allel to Ox2x3 are homothetic to the curve
NACA TM 1354
x2 2p cos e x 2p sin 8 (II.19)
1+ p2 1 + p
In the case of the linear approximations, with grad u, grad v,
grad w being infinitely small (it would even be sufficient that they
should be limited), one sees that one may, within the scope of the
approximations of section 2.1, simplify the formulas (11.19) without
inconvenience and write them
x2 = 2p cos 6 x3 = 2p sin 0
hence the result, essential for the applications.
The curve (C) in the plane Z is homothetic to the sections of
the cone obstacle made by planes normal to the nondisturbed velocity.
Let us likewise consider a cone with variable but small incidences
so that the flow about the cone should always be a flow in accordance
with the hypotheses of this chapter. One sees that if the orientation
of the cone varies with respect to the wind, the curve (C) in the
plane Z undergoes a translation.
2.2.2 Study of a Cone of Variable Incidence
This last remark allows us to foresee that when a thorough investi-
gation of a cone has been made for a certain orientation with respect to
the velocity it will not be necessary to repeat all the work for any
other orientation. This we shall specify after having demonstrated the
following lemma.
2.2.2.1 Lemma.- One may write on (C) that
2P2 de 2 R Z dZ (II.20)
0P d r;, ~d (11.20)
P dT p dz
Actually, let us put
Z = p cos 0 + ip sin e = X + iY
X and Y may be considered as functions of :P.
42
NACA TM 1354
Hence one deduces that
Y
tan e =
X
de
cos2
Y'cpX X'rpY
X2--
x2
and consequently
2 p2 de = 2 Y' X X'rY) = 2 R
P a;P 134 p/
- i dZ 2 R dZ
d;' R dz
which establishes the formula (11.20).
2.2.2.2.- Let us now consider two contours (CO) and (C1) defined
in the plane Z by two functions Z(O)(-') and Z(1)(c) such that
Z(O) = Z(1) + a, a being a complex constant determining the change in
orientation. In the development (II.13) which gives the conformal repre-
sentation, only the coefficient a0 varies when one passes from the
contour (CO) to the contour (Cl). Consequently
,z (1)
dz
dZ(o)
dz
and the Dirichlet condition determining the function GC"'(z) is written
in the plane z
RE()z) =2 R z(l) z] L )z +- Raz
dz_ 0- dz] dz
(we have omitted superscripts for the quantities which retain the same
value, affected by the index 0 or 1). Consequently
G(1)(z) = G(0)(z) + [g(z)
since g(z) is a regular function and real at infinity, holomorphic out-
side of (7), the real part of which on (7) assumes the
NACA TM 1354
values R(az dZ/dz), g(z) is then very easily determined. One has
exactly
g(z) = z dZ + ar2
\dz / z
Thence for the function H(1)(z), (since
F'(Z) = i/z dz/dZ)
H(1)(z) = H(0)(z) + a dz + 2
0 \ dZ( -2 dZ
(11.21)
The formula (11.21) gives immediately the solution of the problem of
change in orientation with respect to the nondisturbed flow.
2.2.3 Cone of Revolution
We shall study first of all the case of the cone of zero incidence.
One may then do without the preceding analysis and obtain the solution
directly; that is what we shall do here. The curve (C) is a circle of
the radius p = cte = r; the relation (1.28) is written
2rO
v cos e + w sin 8 = 2ro
P(1 + r02
On the other hand, for reasons of symmetry
v sin 0 w cos e = O
Hence one deduces immediately the values of v
2r0 cos 0
S=
b0 1 r02
and w on (C)
2r0 sin 0
1 r02)
NACA TM 1354
whence
v(z) = 2ro 2 Z
3(1 -)- r
W(Z) = i 2ro 2
0 1 ro
+ )
(11.22)
Finally the relations (1.25) permit the calculation of U
- dU 2r02 2Z 11 + Z2 _
P(l r) Z2 +1 Z2
4 rO2 1
S1 ro4 z
2
U(Z) 4 r0 log Z
2 1 r4
(11.23)
We shall now study, returning to the method of section 2.1, the
case of a cone of revolution with incidence.
The formula (II.13) is written
z = Z a
a being a constant which may be supposed to be real.
Consequently
F'(Z) = i
On the other hand, an immediate calculation shows that
do r(r + a cos T)
dP p2
whence
NACA TM 1354
and consequently
whence
2p2 dO 2 (2 + r cos rP)
P d'P (\
G(Z) = (r2 + ar2)
P\ Z a
According to equation (11.18)
H(Z) = 2 + ar2 ) 1 2 r2 Z
\ Z aZ a p (z a)2
the calculation is easily accomplished; one finds
V(z) 2r2 Z 1
(Z a)2
U(Z) = 4 og(Z a)
02
3_a a2
Z a (Z a)2
K2 = +a r2
In particular, one finds, if a = 0, by means of the approximate
formulas (11.24) and (11.25), the same result as by the formulas (II.22)
and (II.23) under the condition of neglecting in these formulas the term
in r0 of the denominator.
In order to give to these formulas a directly applicable form it
suffices to again connect the quantities a, r with the geometrical
data; for this purpose, one must use the formula defining p (p. 42).
since
+ 4aZ
(11.25)
NACA 3T 1354
Figure 11 represents the cone section made by the aerodynamic plane
of symmetry; a is the semiangle at the apex, 7 denotes the angle of
the cone axis with the nondisturbed velocity.
One has immediately
2r = pa 2a = py
Finally, we shall utilize for the calculation of Cp the for-
mula (I.11) since the velocity component u is infinitely small com-
pared to the components v and w. This formula is here written
Cp = -2R[U(Z) V(Z) 2 (11.26)
According to equations (II.24) and (11.25) one has
C = 2a21og ~2 2 + 4ay cos 0 + 272cos 20 (II.27)
jp Pa
The case of the cone of revolution of zero incidence is obtained
by making 7 = 0. One finds then again a known result. The for-
mula (II.27) had already been given by Busemann (see ref. 9) without
demonstration.
2.2.4 Elliptic Cone
We assume first of all the simplest hypotheses where the
planes Oxlx2, Oxlx3 are symmetry planes of the flow (U is in the
direction of the cone axis), with the cone flattened out on Oxlx2.
The formula (II.13) may be written in the form
Z = z + a2
z
or
p cos 6 + ip sin = r + a2 cos 9 + ir a)sin 9
r r \
48
Hence one deduces successively
tan B = r2 a2 tan P
r2 + a2
Scos2 r2 a2
cos2m r2 + a2
NACA TM 1354
r2 a 1
r2)2
r
2 2 dO = 2/2 a
0 d' P r2
The Dirichlet problem, which permits calculation of G(Z), is
readily formulated; since G(Z) has a constant real part on the con-
tour (C), G(Z) is constant:
z2
F'(Z) = i 1
z z 2
whence
H(z) = r2
H(Z) 2
(r
_ a2 a2
z
aA 1
r2Z2 4a2
We note besides that
and
K2 = 0.
NACA TM 1354
One calculates V(Z) by the formula (II.9)
(14a
V(Z) = 2(r2 1 Z)
\ r2/ ^" j2_
and U(Z)
by the formula (II.11) which may also be written
U(Z) = H K G 2K2Z
0 1 z '
whence
U(z) = i2 r
p2\(
U(Z) 2
p2
lo z z2 a2
Z + Z2 4a2
2
If one makes a = O, one will find again the expressions already
obtained for U(Z) and V(Z) in the case of a cone of revolution of
zero incidence (formulas (II.24) and (II.25) in which one makes a = 0).
We shall denote by e and by Tj
elliptic cone (see fig. 12). One has
the principal angles of the
E3 = 2(r + a2
T = 2(r a)
whence
r = (E + r1)
4
(11.28)
(11.29)
or
(11.30)
- l
r 2
a2 =
16
(C2 2)
NACA TM 1354
The pressure distribution on the cone circumference is easily cal-
culated. It is sufficient to apply the formula (II.26); besides
IV(Z)2 = E2112
T2cos2p + E2sin2cp
R[U(Z]- = log C+ T
2 2sin2(2
r2cos2C + E2sin2rg
hence the final formula
- 1 +
The case where the velocity is
may be treated equally by utilizing
one must put
HO(z) = (2 a 1--
-< -
One then obtains
H(l)(z) = 2(r2
(r
r z2 a2
not in the direction of the axis
the formula (11.21). In this formula
dZ= 1 a2
a2 dz z2
z2
z2 a2
zr2 z2
z2 z2 a2
2 kr2 E- z + r2 a2]
P(z2 a2) k 2
hence, remarking that
a2
Z = z + -- + a
z
and
(11.31)
NACA TM 1354
H(Z) = 2 2 1
\ r T(Z )2 4a2
(ax2 a2(z a(- (Z -a)2 4a2
2a2 (Z a)2 a2
V(Z) = H(Z) jr2 Z
-2]
On the other hand, we shall calculate U by utilizing the vari-
able z and the formula (II.20). The coefficient K2 is equal to
K2 ( r2 a + r2 a 2a
and U(z) is then given by the formula
a4 og 2a2 + az
_ log z z
r2/ z2 a2
+ a2 2 (22 + a)
z(z2 a2)
One will note that, if
mula (11.30), and that, for
except for the notations.
one puts a = 0, one finds again the for-
a = 0, one finds again the formula (11.25),
Thus one can, without any difficulty other than the lengthy writing
expenditure, calculate the pressure distribution coefficient on the
elliptic cone of any arbitrary orientation with respect to the wind.
U(z) =- 4 (r2
P2
K2
+4 -Z
p
(11.32)
52 NACA TM 1354
2.2.5 Calculation of the Total Forces
We have already seen in section 1.2.6 that the normal to the conical
obstacle directed toward the outside has as direction parameters
i(x3x2' x2x'), x3', -x2'
Let n be the unit vector coincidental with this normal, s be
the area of the section with the abscissa xl, L the length of this
section; one may make correspond to the resultant of the forces acting
on a section (x4, xl + dxl) a dimensionlesss) vector
Cz = C pn ds (11.33)
situated in the plane x2x3, and a dimensionless number
C 1 Cp(nU)ds (11.34)
the vector Cz characterizes the lift, the number C, the drag.
The integrals appearing in the formulas (33) and (34) are taken
along the section. Naturally Cz and Cx are independent of this
section. One may also replace C by a complex number Cz, the real
and iaginary parts of which are equal to the components of the vec-
tor C on Ox2 and Ox3. For calculating equations (11.33) and (1.34
one may utilize the section xl = 8. If we assume I to be the length
of the contour (C) in the plane Z, we may write, taking into account
the habitual approximations
Cz Cp dZ (11.35)
and
Cx [i CpZ dZ (I.36)
51 JC J
NACA TM 1354
with the integrals appearing in equations (11.35) and (11.36) taken in
the plane Z. These integrals present a certain analogy to the Blasius
integrals (ref. 13); Cp is given by the formula (II.26); unfortunately,
it is not possible to give simple formulas for the total forces since
the integrals (11.35) and (11.36) make use of all coefficients of the
conformal representation19.
We shall apply
circular cone; Cp
the formulas (11.35) and (11.36) to the case of the
is given by equation (11.27)
dZ = i fB eide
2
Z dZ = i 24- de
4
2 = xSp
One obtains
C, = -2ay
In the case of the elliptic cone of zero incidence, Cz
ously zero
- -i- a2 eiP
Z = re-i+P + e-
r
Cx = 2( r2
0P
Cx= 2a31og 2- 3 My2
OM
(11.37)
is obvi-
ireicP a2 iPj
dZ = i e e- doP
r
- 21 Cp dCP
Z^O p
being given by formula (II.31). Now
ScrT dP =
2(Tl2cos2% + e2sin 2 )
+ E dt
Do T2 + E2t2
19See appendix No. 7.
whence
with Cp
54 NACA TM 1354
As one can see immediately by putting
t = tan CP
the calculation of this last integral is immediate.
Thus one obtains
c = 2 2l g (E + n) 2 (11.38)
with 2 being the length of the ellipse with the semiaxes C-, -.
2 2
2.2.6 Approximate Formula for the Calculation of Cx
Let us consider the function U(z); according to formula (II.11)
and the remark 2 of section 2.1.4 one may say that the principal term
for U(z) is
U(z) = 4 Si log z
rp2
Consequently, in first approximation
8s
C S log r
with S being the area inside of the contour (C), and r the radius
of the circle (7) on which one makes the conformal canonical repre-
sentation of (C). If one now calculates Cx, taking into account this
approximate formula, one has, according to equation (11.36)
C 1S log rR i dZ
whence
C, =+ 32S2 log_ (II.39)
itp3" r
NACA TM 1354
We shall state: In every first approximation the value of the
drag coefficient C, is given by the formula (II.39).
2.2.7 Case Where the Cone PresenLs
an Exterior Generatrix
If the contour (C) shows an exterior angular point, the various
functions introduced in the course of the study (first paragraph of
this chapter) present certain singularities. These singularities we
shall specify. Let ZO be the designation angular point of (C), and
B6 the angle of the two semitangents to (C) at the point Z0(O < 6 < 1)
(see fig. 13); if- z0 is the image of the point Z0 in the plane z,
one may write, according to a well-known result, in the neighborhood
of zO
(LdZ\ K= z zo)k
\dz10
with K being a complex constant and k = 1 8; consequently
k
F'(Z)0 = K1z -0 = 2 (Z ZO 1+k
with K1 and K2 being complex constants. F'(Z) thus becomes infinite
at the point Z = Z0.
In contrast, the function G(z) has, according to definition, a
real part which assumes on the circle (7) the values
RzZ -]
This real part thus remains finite on the circle (7) (and it
satisfies there a condition of Holder). According to a known theorem,
Sits imaginary part likewise remains continuous on (7) (and likewise
Satisfies a condition of Holder). Consequently, one sees, if one refers
Sto formula (11.18) that
56 NACA TM 1354
k
H(Z) = K3(Z ZO) l+k
in the neighborhood of ZO; likewise, U, V, W will, in the proximity
of this point, be of the order k with respect to 1
1 + k Z- Z
Thus the analysis made in section 2.1 is no longer applicable to
this case. However, the formulas (11.35) and (II.36) show that if the
pressure coefficient assumes very high values in the neighborhood of
Z = ZO, the total energy remains finite. According to what we have
indicated in section 1.1.3 we consider the solution still valid, with
the understanding that the values of Cp in the surroundings of Z = ZO
are not reliable.
2.2.8 Delta (A) Wing of Small Apex Angle
at an Infinitely Small Incidence
If one puts in the formulas r2 = a2, at the end of section 2.2.4,
one obtains the pressure distribution on a delta wing with small apex
angle. Let us recall that a delta wing is an infinitely small angle.
Its angle, according to definition, is the half-angle w at the vertex
(compare fig. 14). Thus one has
op = 4a
The formulas (11.31) and (11.32) are applicable to a delta wing of
small angle placed at an incidence also rather small.
Let us moreover assume that this opening is infinitely small with
respect to the incidence. Under these conditions, the formulas yielding
U(Z) and V(Z) are written
V(Z) 2 a a Z 42 42
P2 2 4z2 4a2
2 2
U(Z) 4 a 4a + 8a (a a)Z (11.40)
82 2 JZ2 _4a2 02
NACA TM 1354
Actually one is justified in omitting the second-order terms with
respect to a. For calculating Cp it suffices to apply the for-
mula (11.8); the second term of the second formula (II.40) may be
neglected.
With the incidence 7, the delta wing being parallel to
70 = 2ia
Finally, one may put along the A
Z = 2a cos q = R cos (P
2
One then finds
c = 2'uy
C s-in
S sin
Ox2, one has
We remark further that p is related to the angle
213 = mP cos i
I of figure 14
W = n cos
2
One may state: the pressure coefficient on a delta wing of infi-
nitely small opening angle is independent of the Mach number of the flow.
One has
Cp = 2 if 2i
S-_ 2 tu
uI t
if one applies formula (11.35), one finds
Cz = inay
This coefficient Cz has not the same significance as the one
utilized in the theory of the lifting wing. Actually, it is, according
(11.41)
58 NACA TM 1354
to the very manner in which it was obtained, relative to the total area
of the A (pressure side and suction side); if one takes only one of
these areas into account, one must write (neglecting the factor -i)
Cz = 2nwu
This formula has been found by other methods by R. T. Jones
(ref. 14). We shall find it again in chapter III, section 5.1.2.4, when
studying the general problem of the delta wing which is here only touched
on incidentally and for the particular case of a A with infinitely
small opening angle.
2.2.9 Study of a Cone With Semicircular Section
As the last application, we shall treat the case of a cone with
.--
semicircular section, with the velocity U being directed along the
intersection of the symmetry plane and of the face plane of the cone20
(fig. 15).
The contour (C) in the plane Z then is a semicircle, centered
at the origin, of the radius a (fig. 16).
One obtains very easily the conformal canonical representation of
the exterior of this contour, on the outside of a circle (7) of the
radius r, centered at the origin of the plane z, by means of a par-
ticular Karman-Trefftz transformation (ref. 13, p. 128) which is written
-i 2
S6
Z a z re (11.42)
Z + a 5n
-1-i
z re
a and r are connected by the relationship
4a = 3r41
In order to obtain the correspondence between the circle (7) and
the contour (C), one must distinguish two cases. Let us put
z = re
20Such a cone formed the front of supersonic models planned by
German engineers.
NACA TM 1354
(1) < 6 < the corresponding point of
of the circle.
Let us put under these conditions
Z = aeir
and we shall find according to formula (II.42):
(C) is on the arc
tan -
2
(11.43)
(2) 7 < p < n, the
6 0
ment AA'; let us put under
corresponding point of
these conditions
(C) is on the seg-
Z = a cos .
The formula (11.42) shows that
r4
tan -
2 P 5t
tsin( + 12
sin(+ j
(II.44)
The two last formulas define completely the desired conformal
representation. Figures (17) and (18) give the variations of and x
as functions of P.
We shall have to utilize equally the value of dz/dZ. The simplest
method for obtaining this value consists in logarithmic differentiation
of the two terms of formula (II.42). One thus obtains the result
dz z2 + rz r2
dZ 2 2
Z -r
(I1.45)
NACA TM 1354
If one has
- < ( < 6 6
z = rei
Z = aei-
whence
dz r2 1 + 2 sin q ei('-A) 8 1 + 2 sin 'P ei((P-)
dZ 2a2 sin 27 sin *
71r 117
If 'p is comprised between -- and --, one puts
b b
Z = a cos X. Thus one obtains
i(E --A
dz 16 1 + 2 sin r e2 )
dZ 27 sin2X
The function G(Z)
has as its real
part R zZ ]d that is
[- d~dz
2 ar sin 4
8 1 + 2 sin T
0
if <'JP <
6 6
if 6 < < 1
b o
The analytic function
a2 z dZ
Z dz
has a real part which, on (b), assumes these same values. This func-
tion is regular at infinity, holomorphic outside of (7), but with a
pole z = -i, with the corresponding residue being equal to -ia2
pole z = -ir, with the corresponding residue being equal to -ia
(11.46)
z = re ,
(1.47)
(II.48)
NACA HM 1354 61
Let us then consider the function
a2/2 dZ 1 z ir\
\Z dz 2 z + ir)
This function is holomorphic outside of (7). It is regular at
infinity; its value at infinity is equal to a2/2. On (7), these real
and imaginary parts satisfy Holder conditions. This function is there-
fore identical with the desired function G(z).
Hence one deduces according to equation (II.18)
and ac
t:
H(Z) ( r\a2 dz
Z dz 2 z + ir/ z dZ
cording to equation (II.19)
V(Z) = a2l Z 1
\Z 2 2z
Finally, the calculation of U(Z)
of formula (II.29)
G dz = a2log Z a z ir dz
2 jz + ir z
and
ZH a2/ 1 Z dz z ir
ZH2 z dZ z +
2 z dZ z + ir
Sa2 a2 z ir dz
Z 2z z + ir dZ
z ir dz\
z + ir dZj
may be carried out with the aid
= a2 (log Z + .1 log z)
\ z + ir 2 /
S= a2 z ir Z dz\
2) 2 z + ir z dZ/
whence
U(Z) = a2 ir Z dz 1 + + 2 log Z + log
\z + ir z dZ g2 z + ir
The calculation of the coefficients K2 offers no difficulty
does not occur in the calculation of the pressures along the cone.
n
i;
ii
i;
iii
,i
II
tl;
,ii
,,
,;
what-
K2
62 NACA TM 1354
This pressure distribution along the cone calculated with the aid
of equation (II.26) is represented in figure 19.
2.3 Numerical Calculation of Conical Flows With
Infinitesimal Cone Angles
2.3.1 General Remarks
In the preceding paragraph, we have studied a certain number of
particularly simple cases. However, if the cone (C) is arbitrary, it
will be necessary to carry out various operations leading to the solu-
tion by purely numerical procedures.
Let us analyze the various operations necessary for the calculation:
(1) The conformal canonical representation of the exterior of (C)
on the outside of the circle (?) must be made; this calculation per-
mits, in particular, determination of the radius r of (7), corre-
spondence of the points of (C) and of (7), and calculation of the
expression dZ on the contour (7).
(2) The function G(z), holomorphic outside of (7), regular and
real at infinity must be determined, the real part on (7) of which is
known; we shall designate it by g('). In fact, it suffices to know,
on (7), only the imaginary part of G(z), for instance g'('P); g'(P)
is the conjugate function of g(T) and is given by the formula
g'() =_ 1 g( ')cot P- dc'
2n Jo 2
This formula is called "Poisson's integral."
(3) With these two operations accomplished, the values of H(z) on
the circle (7) (formula (11.18)) are known which provides the values
of v and w on the cone; u is obtained by the formula (11.29). The
only new calculation to be made is that of the expression:
[B- f dz I drP
the constant of integration being determined so that u should have a
mean value zero on (M).
MACA 9M 1354
All these operations always amount to the following numerical
problems:
(a) With a function given, to calculate its conjugate function
(Poisson integral)
(b) With a function prescribed, to calculate the derivative of the
conjugated function
(c) With a function prescribed, to calculate its derivative21.
We shall justify this result in the following paragraph by showing
that the operation (1) may be performed by applying the calculations (a),
(b), (c). We shall then indicate a general method, relatively simple
and accurate, which permits solution of these problems. We shall ter-
minate this chapter by giving an application.
2.3.2 Conformal Canonical Representation
of a Contour (C) on a Circle (y)
The numerical problem of determination of the conformal canonical
representation of a contour (C) on a circle (y) has been solved for
the first time by Theodorsen22. We shall briefly summarize the principle
of this method, simplifying, however, the initial expos of that author.
Let us suppose, first of all, that the contour (C) is neighboring
on a circle of the radius a, centered at the origin (fig. O0); in a
more accurate manner, putting on (C)
Z = ae+ie (11.49)
with being a function of e, 0 = 4(e), we shall suppose that *(e)
and d- are functions which assume small values. We shall then call
de
21If the conformal representation of the exterior of (C) on the
outside of (7) is known in explicit form, it will naturally be suffi-
cient to apply operation (a).
22Compare references 15 and 16. One may achieve this conformal
representation also by the elegant method of electrical analogies (ref. 17);
the time expenditure required by the experimental method and by the purely
numerical methods here described as well as the accuracy of these pro-
cedures are of the same order of magnitude.
NACA TM 1354
(C) "quasicircular." Let Q be the angular abscissa of the point of
(7) which corresponds to the point of (C), the polar angle of which
is 0; we put
e = P + c((p)
cP = 0 E(0)
e(e) and E(cp) representing the same function but expressed as a
function of 0 or as a function of qP; we shall put likewise
(CP) = J1(e)
The desired conformal transformation may be written
Z zeh(z)
with h(z) being a holomorphic function outside of (7), regular and
zero at infinity. The equality (II.50) becomes, if one writes it on the
circle (7),
ae())+i T +(' = reiqeh(z)
whence
h(z) = W(~() + iE(cp) + log
r r
(11.51)
Finding the conformal representation of (C) on (T) amounts to cal-
culating the functions (cp) and T(p). First of all, one knows (equa-
tion of (C)) that
(11.52)
(p) = + C(r4) )
On the other hand, according to equation (II.51), e((P)
gate function of (cp), and consequently
is the conju-
-)=(<'P)cot(' 2 (
N(c') ot M P'
k 2
(11.50)
e(') 1 '
2x p n
2v
(II.53)
NACA TM 1354
the integral being taken at its principal value. There is no constant
to add to the second term of equation (11.53), for i(P) has a mean
value zero since h(z) is zero at infinity. For the same reason, if
*0 denotes the mean value of T('o) in an interval of the amplitude 21
r = ae'0
(11.54)
an equality which will permit calculation of r if T(P) is known.
In order to calculate T(P) and T(P), one disposes therefore of the
relations (11.52) and (11.53); one can solve this system by a procedure
of successive approximations.
We shall put first
o(e) = T0() = o
According to equation (11.52)
4(e) = on(0)
and according to equation (11.53)
l(e) = 1
'(S')cot d' e de'
2
Thence a first approximation for 7'
1 = a l(e)
S=' 1 + 1 )
From it one deduces, according to equation (II.52), a first approxima-
tion for Tl(c')
11 1 +
whence a second approximation for the function E
NACA TM 1354
r2 1 02n 1 1cot 1' 2p1
1 = j*2y W (c1)cot %i l ,P
2a 2 2 T
2(e) = c2 [e e1()]
whence
2 = a E(e) e = 2 + 2n
The procedure can be followed indefinitely.
The convergence of the successive approximations forms the subject
of a memorandum by S. E. Warschawski (ref. 18). We refer the reader
who wants to go more deeply into that question to this meritorious
report.
From the practical point of view one may say that the convergence
is very rapid; two approximations suffice very amply in the majority of
cases; the different changes in variables which encumber the preceding
expose are very easily made by graphic method. Thus one sees that the
numerical work essentially consists in calculating twice the inte-
gral (II.53). This calculation is precisely the object of the prob-
lem (a) stated at the end of section 2.3.1.
If the contour (C) is not "quasicircular," one may make, first
of all, a conformal representation which transforms it into the "quasi-
circular" contour (C'); one will then apply the preceding analysis to
the contour (C'). For certain cases it will be quicker to use a direct
method. Let us assume, for instance, that (C) is a contour flattened
on the axis of the X (compare fig. 21) and for simplification that
X'OX is permissible as the axis of symmetry.
Let us suppose that X varies along (C) from -a to +a while
IY| remains bounded by ma (with m being, for instance, of the order
of 1/10); it will then be indicated to operate as follows:
We put along (y).
Z = [f(rP) + ig(
SNACA TM 1354
One has
or also
X('C) = f cos CP g sin r
Y(P) = f sin p + g cos r
f = X cos r + Y sin m
g = Y cos '" X sin cP
f(P) is an even function of r,, g(C') is an odd function
f(0) = +f(n) = a
g(O) = g(n) = 0
The functions X(T) and Y(0) have to be found. Let us take as
starting point
XO(0) = a cos P
an approximation which would be definitive if (C) were an ellipse.
On the contour (C) one reads the corresponding value YO('P), and
by means of the second formula (II.56) one obtains a first approximation
gl(+) = YO(cp)cos rp Xo(p)sin (p
fl(q) will be given by a Poisson integral
2xi
0
gl(r)cot -'- "- dT' + X1
2
with X1 being a constant, such as fl(O) = a.
Owing to the formulas (II.55), one has a first approximation Xl('),
Y1(T) for the functions X(), Y('P). One proceeds in the same manner,
reading off on (C) the functions Y1(mP) corresponding to Xl('), then
(11.55)
(I1.56)
NACA I 1354
calculating
g2 1) = Yl(c)cos (p Xl(p)sin C
and
n')
=f
g2( p')cot P' dTP' + X2
2
When one
approximation
tions; then
has obtained a pair fn(P), gn(') providing a sufficient
Xn(), Yn(p) of X(C'), Y('), one stops the calcula-
r = X
In practice23 it suffices to take
of very slight adaptations) will apply
being flattened on OX, will no longer
axis.
n = 2; the same method (averaging
to the case where (C), although
admit of OX as the symmetry
Finally, for a complete solution of the problem (1) posed at the
beginning of the preceding paragraph, only dZ/dz remains to be calcu-
lated, which will obviously be possible with the aid of the problems (b)
or (c).
2.3.3 Calculation of the Trigonometric Operators24
The method we shall summarize permits calculation of the linear
operators A, transforming a function P(e) into a function Q(e)
23The principle of this method is the one we applied for the study
of profiles in an incompressible fluid. But in the case of the profiles
a few complications (which can, however, easily be eliminated) arise due
to the fact of the "tip."
24We gave the principle of this method for the first time in
March 1945 (ref. 19). Compare also reference 20. In continuation of
this report, M. Watson provided a demonstration of the formulas which
we obtained by a different method (ref. 21). We also point out a "War-
time Report" of Irven Naiman, of September 1945, proposing this same
method of calculation for the Poisson integral (ref. 22).
etc.
NACA TM 1354 69
Q(e) = A[P(e)
and re-entering one or the other of the following categories:
First category: The operator possesses the following properties
A(cos me) = am sin me
A(sin me) = -a cos me (II.57)
A(1) = 0
with am being a nonzero constant, m any arbitrary integral different
from zero.
Second category: A possesses the properties
A(cos me) = bm cos me
SA(sin me) = bm sin me
A(1) = bO
with bm being a nonzero constant, m any arbitrary integral.
We shall call these operators trigonometricc operators." The
operations which form the subject of the problems (a), (b), (c) are,
precisely, particular cases of trigonometricc operators."
With the function P(e) known, one now has to calculate the func-
tion Q(e); the functions P(e) and Q(e) are assumed as periodic, of
the period 2v. P(e) and Q(8) are determined approximately by knowl-
edge of their values for 2n particular values of 8, uniformly dis-
tributed in the interval 0, 2n. One knows that the unknown 2n values
of Q are linear functions of the known 2n values of P. The entire
problem consists in calculating the coefficients of these linear equa-
tions. We shall do this, examining two possible modes of calculation.
NACA IM 1354
2.3.3.1 First mode of calculation.- After having divided the
circle into 2n equal parts, we shall put
f= f
(1) Operators of the first category.- Obvious considerations of
parity show that the Qi are expressed as functions of the Pj by
equations of the form
n-1
i = KpPip i-p)
1
(11.58)
We shall apply the
equations (11.58)
relations (II.57), that is, carry into the
P(8) = cos me
P(O) = sin me
We thus obtain 4n
equation
Q(e) = am sin me
Q(e) = -am cos me
equations which are all reduced to the unique
n-l
SKp sin p m -a
1
(I1.59)
This reduction is the explanation for the success of the method.
We have to determine (n 1) unknown Kp. For this purpose, we shall
write the equation (II.59), for the values of p varying from 1 to
n 1. The system remains to be solved. If one multiplies the first
2!, scn b2 i)th
equation by sin -E, the second by sin L, the (n 1 by
n n
sin(n 1)n, and if one adds term by term, one obtains a linear rela-
tion between the Kp, with the following coefficients of Kp
NACA IM 1354 71
n-1 n-1
Wpn 1 (-nIP Jfn (p + [Jn
sin m sin m -- Co cos m
n n 2 5 n 1 nn
m=l m=l
1 Sn P Cn[ P( + )_n'
= 2 E D n In]J
with
n-1 ( sin x
Cn(x) = cos mx = cos x
2 x
m=O sin
Thus the coefficient of Kp is zero if p / n, and equal to if
2
p= U-
Thence the desired value of Kp
n-l
Kp = a sin m (pI.60)
m=l
Let us apply this result to the calculation of the Poisson integral.
This integral defines an operator Q = A(P) of the first category
for which am = -1.
Consequently, the formula (II.60) is written
n-1
Kp = : sin m = Sn
mn n n (
1
if one puts
n-1 sin nx
n- (n 1)x s 2
Sn(x) = V sin mx = sin -2 l)x
1 sin 2
1 2
NACA TM 1354
Thus
K = 0 if p even
(II1.61)
1 cot pA if p odd
KP n 2n
(2) Operators of the second category.- The considerations of parity
permit one to write the general formula
n-1
Qi = KOPi + (Pi+p + Pi-p) + Kni+n (11.62)
1
Using the same reasoning as before, one is led to determine the coeffi-
cients Kp by the system
n-1
KO + > 2Kp cos m E + ( 1)% = b (11.63)
p=l
with m assuming the values 0, 1, 2, n.
Multiplying the first value by 1/2, the second by cos pu/n, the
third by cos 2?, the nth by cos (n 1)p! and the last by ( )2
n n
and adding them, one obtains a linear relation between the Kp, with the
coefficient of Kp being (p / 0, p / n)
2 + )P+LL n ] + Cn(P- ) 2
that is, n if u = p, and 0 if p.
The coefficient of K0 is
2 2 2 + Cn n
NACA TM 1354
The preceding conclusions remain valid, it is zero for i J 0 and equal
to n if P = 0; the same result is valid for Kn. Finally, one
obtains the general formula of solution
(I..64)
Let us consider, for instance, the operator transforming the func-
tion- P(G) into the function dQ/de, with Q being the conjugate func-
tion of P; it is an operator of the second category for which
bm = -m
Applying formula (11.64), one obtains
KO m
2
n- n
_1
If one notes that
1n
n-l
(x) = m cos mx
0
12 sin n
2 sin2 x
2
one sees that
Kp = 0
1
n cos
n
if p even
if p odd
- 1)P
p / 0
- x sin2 x
2)
74 NACA TM 1354
2.3.3.2 Second mode of calculation.- Examination of an important
particular case will show us that in certain cases it will be advantageous
to consider a second mode of calculation.
The method consists in replacing the function P(e) by a function
of the form
n
D(e) = an cos nO + bn sin nO (11.66)
0
for which the method is applied with the strictest exactness; the con-
stants an and .bn are such that Pi = \$i. One operator of the first
category, one of the most important ones, is the operator of derivation
which makes the function dP/de correspond to the function P(O). If
we apply the first type of calculation, we shall replace ( -) by
\de/i
d(I ; now, it is precisely at the points 8 = in that the deriva-
(dei n
tives and show the greatest deviation. In contrast, we shall
de de
obtain a good approximation of the desired function by replacing
dP (2i + 1)n b d' (2i + 1)t
dO 2n dO 2n
We are thus led to the following mode of calculation: the circle
is divided into 4n equal parts; we shall put
fi = f(il)
\2n/
and we shall express the 2n values Q2i as a function of the 2n value
2j+l"
We shall limit ourselves to the operators of the first category.
The formula expressing the Q2i as a function of P2j+l is written
n
>2i =_ K(P2i+2p-l 2i-2p+l)
p=l
N ACA TM 1354
and we obtain for determination of the Kp the system
Ssin (2p l)m am
l 2n 2
p=l
with m varying from 1 to n.
Multiplying the first equation by sin(2P 1)--, the second by
2n
(20 l)2n th
sin( 1)2, .,the (n 1)th
2n
by sin (2P l)(n 1)n
by sin the
2n
(- 1
last by and adding them, one obtains a linear relation in
2
which the coefficient of Kp is
n-1
Z sin(2p -
m=l
1)mL sin(2p -
2n
n-1
>- cos(p u)L cos(p
1
+.- I )B + + C )P-P
"J 2
S ipn n-p + [L 1) + ( 1)
The coefficient is zero if p 6 I, and equal to E if
2
sin (2p l)mn
sin
2n
( 1)-1
2
p = P. Hence
(11.67)
This procedure may be applied to the calculation of the derivative
of a periodic function. In this case, am = -m. Applying formula (II.67),
one obtains
)n ( 1)P+P
2n 2
NACA IM 1354
Kp= ( 1)P-1 1 (1.68)
[ (2p l)xl
2n 1 cos ( n)
11 2n
2.-.3.4 Remarks on the Employment of the Suggested Methods.- In
order to convey some idea of the accuracy of the proposed methods we
shall give first of all a few examples where the desired results are
theoretically known.
Let us take as the pair of functions P(O), Q(e), the functions
p(e) = 4 cos 20 4 cos 9 + 1 Q(e) -4 sin 0(2 cos 0 1)
(5 4 cos 8)2 (5 4 cos 8)2
which are the real and imaginary parts, respectively, on the circle of
f(z) 1 (z = ei)
(2z 1)2
One will find in figure 22 the graphic representation of the func-
tions P(8), Q(e) and of the derivative Q'(0) of this function, and
also the values of these functions for e = p- (with p ranging
12
between 0 and 12). Furthermore, one will find in figure 23 the values
of Q(6), calculated from P(e) as starting point, by the method just
explained (coefficients Kp, defined by equation (II.61)), and in fig-
ure 24 on one hand the values of Q'(8), calculated from P(e) as
starting point (from coefficients Kp defined by equation (II.65)),
and, on the other, these same values calculated from Q(S) as starting
point (coefficients Kp defined by equation (11.68)). One will see
that the accuracy obtained is excellent although the selected functions
show rather rapid variations. Such calculations by means of customary
calculation methods are a delicate matter; this is particularly obvious
in the case of the Poisson integral which is an integral "of principal
value." Systematic comparisons of the method of trigonometric operators
with those used so far have been made by M. Thwaites (ref. 23); they
have shown that this method gives, in certain calculations, an accuracy
largely superior to any attained before.
The calculation procedure, with the aid of tables like the one
represented (fig. 25) is very easy. One sees that one fills out this
NACA IM 1354 77
table parallel to the main diagonal of the table. With such a table,
about one and a half hours suffice for a Poisson integral if one has a
calculating machine at his disposal.
We have had occasion to point out that the accuracy of the method
obviously increases to the same degree as the functions one operates
with are "regular" and present "rather slight" variations. This leads
in practice to two remarks which are based on the "difference method"
and reasonably improve the result in certain cases. We shall, for
instance, discuss the case of the Poisson integral.
(1) If the function P(G) presents singularities (for instance
discontinuities of the derivative for certain values of 0), it will be
of interest to seek a function P1(O), presenting the same singularities
as the function P(e), for which one knows explicitly the conjugate
function Ql(6). One will make the calculation by means of the func-
tion P(O) P1(0); this function no longer presents a singularity.
(2) If the function P(O) has a very extended range of variations,
one will seek a function Pl(8) for which one knows explicitly the
function Q%(8) so that the difference P(G) P1(0) remains of small
value, and one will operate with this difference.
Finally we note that, if the calculation of the derivative of a
function P(e) as described above necessitates that P(O) be periodic,
method."
2.3.4 Example: Numerical Calculation of a
As an application, we have taken up again the case of the semicir-
cular cone studied in section 2.2.9. The function g(1) is given by
the formula (II.48), and g'(Q) will be calculated by a Poisson inte-
gral. Figure 26 shows the value g'(c) thus calculated compared to the
theoretical value.
25We wanted to test the accuracy of the proposed method by assuming
an extremely unfavorable case, without taking into account the singu-
larities presented by the function g(T). For a numerical operation of
great exactness, this particular case would have required application
of the lemma of Schwartz, with the contour (C) completed symmetrically
with respect to OX.
78 NACA TM 1354
It is then possible to calculate the representation of the pres-
sures, by calculating successively the function H, ZH, and the inte-
gral g' (p).
One will find the pressure distribution thus calculated in fig-
ure 19; one may then compare the result obtained by the calculation
method (for a very unfavorable case) with the result obtained
theoretically.
NACA TM 1354
CHAPTER III CONICAL FLOWS INFINITELY FLATTENED
IN ONE DIRECTION
The purpose of this chapter will be the study of conical flows of
the second type (see chapter I, section 1.2.6). Before starting this
study proper, we shall make a few remarks concerning the boundary con-
ditions. The conical obstacle is flattened in the direction Oxlx2.
Under these conditions, reassuming the formula (1.27)
x2 vx3' = (xx2' x2x3') (1 + u) (1.27)
one may say that it reduces itself, in first approximation, to
w2' = x2x3') (III.)
since x3, x ', v, u are infinitesimals of first order, while x2
and x2' are not infinitesimals. Under these conditions, one may say
that one knows the function w on the contour (C). On the other hand,
one may write, within the scope of the approximations made, this boundary
condition on the surface (d) of the plane Oxlx2, projection of the
cone obstacle on the plane. Let us designate, provisionally, the
value w by w(1)( 1x2x3) if one operates as follows
(1)
w(l) J1x=t),(tx(t = w(l)E1,x2Ct)J] + xj(t) xx,2(t,
With the derivatives of w being, by hypothesis, supposed to be of
first order, and the boundary equation written with neglect of the terms
of second order, the intended simplification is justified.
Various cases may arise, according to whether the cone obstacle is
entirely comprised inside the Mach cone (fig. 27), whether it entirely
bisects the Mach cone (fig. 28), whether the entire obstacle is com-
pletely outside the Mach cone (fig. 29), or whether it is partly inside
and partly outside the Mach cone (fig. 30). In each of these cases
there are two elementary problems, the solution of which is particularly
NACA TM 1354
interesting: the first, where the relation (III.1) is reduced to
w = constant = w0
which we shall call the elementary lifting problem (the corresponding
flow is the flow about a delta wing placed at a certain incidence); the
second, where the relation (III.1) is reduced to
w = wO for x3 = +0
w = -wO for x3 = -0
which we shall call the elementary symmetrical problem. This is the
case of, for instance, the flow about a body consisting essentially of
two delta wings, symmetrical with respect to Oxlx2 and forming an
infinitely small angle with this plane. It is also the case that will
be obtained, the section of which, produced by a plane parallel to Ox2xy,
would be an infinitely flattened rhombus. The fact that one obtains the
same mathematical formulation for two different cases indicates the
relative character of the results which will be obtained. In the case
of the symmetrical problem one may naturally assume that w is zero on
the plane' Oxlx2 at every point situated outside of (d).
Let us finally point out that very frequently the obtained results
do not satisfy the conditions of linearized flows; in particular, the
velocity components and their derivatives will frequently be infinite
along the semi-infinite lines bounding the area (d). However, we admit
once more that the results obtained provide a first approximation of the
problem posed above, in accordance with the remarks made in section 1.1.3
of chapter I.
3.1 Cone Obstacle Entirely Inside the Mach Cone
3.1.1 Study of the Elementary Problems
The case of the lifting cone has already formed the subject of a
memorandum by Stewart (ref. 10); however, the demonstration we are going
to give is more elementary and will permit us to treat simultaneously
the lifting and the symmetrical case.
NACA TM 1354 81
3.1.1.1 Definition of the function F(Z).- We shall make our
study in the plane Z. Let A'A(-a,+a) be the image of the cut of the
surface (d)26, (Co), as usual, the circle of radius 1 (fig. 31).
Naturally, we shall operate with the function W(Z). One of the
conditions to be realized which we shall find again everywhere below is
that dW/dZ must be divisible by (z2 1), unless the compatibility
relations show that U(Z) would admit the points Z = 1 as singular
points which is inadmissible. Thus we introduce the function
F(Z) Z2 dW (III.2)
Z2 1 dZ
iand we shall attempt to determine F(Z) for the symmetrical as well as
for the lifting problem.
F(Z) is a holomorphic function inside of the domain (D), bounded
by the cut and the circle (CO); the only singular points this function
can present on the boundary of (D), are A and A'; on the other hand,
'iF(Z) must be divisible by Z2, unless U, V, W have singularities
at the origin. On the two edges of the cut F(Z) must have a real zero
part. On the circle (Co)
Z 1 1
Z2 1 Z 1 2i sin 0
Z
Z dW = e1 dW dW
dZ dZ dO
Consequently, F(Z) has a real zero part on (Co) as well. The fact
that F(Z) cannot be identically zero, and that its real part is zero
on the boundary of (D), admits A and A' as singular points. We
shall study the nature of these singularities.
3.1.1.2 Singularities of F(Z).- Physically, it is clear that A
Sand A' cannot be essential singular points. Let us therefore suppose
.that, in the neighborhood of Z = a, one has
26One assumes, as a start, that the problem permits the use of the
plane Ox1x3 as the plane of symmetry.
NACA TM 1354
F(Z) ~ K%(Z a)mo
m0 being arbitrary, KM / 0; let us put
Z a = rei
with Pq being equal to +- on the upper edge of the cut, to
the lower edge; for sufficiently small values of r
-it on
orO r" O"
and
KmO e-imo
W)e
must be purely imaginary quantities; thus the same will hold true for
KO cos molt
and for
iKm sin mon;
K 02 = Km02cos2m~ (i0 sin ms)2
is therefore real. On the other hand
2 sin 2mov
2 i2mus = (b cos n)( w ei sin n0g)
must be real which entails
sin 2m0A = 0
Thus there are two possibilities; let us denote by
integral; either
k an arbitrary
KmO is purely imaginary
or else
KmO is real.
mo = k + -,
mo = k,
SNACA TM 1354 83
Let us now consider
F1(Z) = F(Z) K(Z a)m0
In the neighborhood of Z = a
Fl(Z) ~ Kml(Z a)ml
and the same argument shows that 2mI must be an integral. Finally,
one may state the-following theorem:
Theorem: Inside of (Co) the function F(Z) may assume the form
F(Z) = O(Z) + 1 t(Z) (III.3)
a2 Z2
'with Q(Z) and 4(Z) admitting no singularities other than the poles
at A and A'.
The analysis we shall make will be simplified owing to certain
symmetry conditions which F(Z) satisfies. Let us put
W = w + iw'
Obviously, X in w(X,Y) is even (when Y is constant).
Consequently, F(Z) has a real part zero on OY. Applying
Schwartz' principle one may write
F(Z) = -F(-Z) (1I.4)
This equation shows that knowledge of the development of F(Z)
around Z = a immediately entails knowledge of F(Z) around Z = -a.
NACA TM 1354
3.1.1.3 Study of the case where F(Z) is uniform [c(Z) = -O.-
Let us consider the function
iz2P
Ap(Z) = iZ2
(a2 z2)(1 a2z2
with p an integral and >1.
This function satisfies all conditions imposed on F(Z).
Indeed, it satisfies equation (III.4); inside of (Co) it does
not admit singularities other than a and -a which are poles of the
order pl. Its real part is zero on the cut as well as on (CO), as
one can see when writing
Ap(Z)
2 (Z2 + )
a ( +
\ Z2~i
- (1 + a
Finally, the origin should be double zero (at least).
Let us assume F(Z) to be the general solution of the problem
stated; we shall then demonstrate the following theorem:
Theorem: If F(Z) is uniform, one has
n n
F(Z) = XpAp(Z) = i
11 ^a2
(III.6)
- Z2)1 a2Z21
with n being an integral, and the kp being real coefficients.
In case F(Z) is assumed to be a solution of the problem having a
pole of the order n, one can determine a number kn so that
Fl(Z) = F(Z) nAn(Z)
will be a solution admitting the pole Z = a only of an order not
higher than (n 1) at most. But in consequence of equation (III.4),
(III.5)
NACA TM 1354
F1(Z) will allow of Z = -a as pole of, at most, the order (n 1).
Proceeding by recurrence, one finally defines a function
n
Fn(Z) = F(Z) XpAp(Z)
1
which must satisfy all conditions of the problem and be holomorphic
inside of (CO). The boundary conditions on the circle and on the cut
entail Fn(Z) to be a constant which must be zero because Fn(Z) must
become zero at the origin.
3.1.1.4 Case where O(Z) = 0.- We shall study the case where
O(Z) = 0 in a perfectly analogous manner.
Let us put
f(Z) = F(a2 z)(l a2Z2) F(Z)
Z
f(Z) is a uniform function inside of (Co) which admits as poles only
the points (Z = -a, Z = a). Actually, the origin is not a pole since,
according to hypothesis, F(Z) is divisible by Z2. The function f(Z)
possesses the following properties: It is imaginary on the cut, real
on (Co), and real on OY (which entails properties of symmetry if one
changes Z to -Z). Moreover, f(Z) admits the origin as zero of, at
least, the order 1. All these properties appertain equally'to the
functions
\$r =(z 1 iZP-1 2 1)
Bp(Z) = Ap 1) = Z-(Z2 ) p
Z Z) a2 2)(1 a2Z2
p is an integral >1.
Thus one deduces, as before, the theorem:
Theorem: In the case where O(Z)
= 0, one may write
NACA TM 1354
F(Z) = i n Z2pP (III.7)
z21z 2
1 a2 Z2)(1 az2] 2
with n being an integral, the Ap being real.
3.1.1.5 The principle of "minimum singularities".- The for-
mulas (III.6) and (III.7) depend on an arbitrary number of coefficients.
The only datum we know is the wo, the value w assumes on the upper
edge of the cut. Thus we have to introduce a principle which will
guarantee the uniqueness of the solution of the problems we have set
ourselves. This principle which we shall call principle of the "minimum
singularities" may be formulated in the following manner (it is con-
stantly being employed in mathematical physics):
When the conditions of a problem require the introduction of func-
tions presenting singularities, one will, in a case of indeterminite-
ness, be satisfied with introducing the singularities of the lowest
possible order permitting a complete solution of the posed problem.
In the case which is of interest to us, this amounts to assuming
n = 1 in the formulas (III.6) and (III.7). For the problem of interest
to us, this principle has immediate significance; it amounts to stating
that F(Z) and hence dW/dZ must be of an order lower than 2 in
1/Z a, or W(Z) must be of an order lower than 1 with respect to that
same infinity; the considerations set forth in section 2.2.7 show that
these conditions entail the total energy to remain finite.
3.1.1.6 Solution of the elementary symmetrical problem.- Let us
turn again to formula (III.6); one deduces from it, according to equa-
tion (III.2), that in the case where F(Z) is uniform
dW Z2 1
ik
dZ 41 (a2 Z2)(i a2Z2)
and hence
W(Z) = 1 log (a Z)(1 aZ)
2a(l + a2) (a + Z)(1 + aZ)
NACA TM 1354
The determination of the logarithm is just that the real part of W(Z)
is zero on (Co). Besides
2a(l + a2)wo
On the upper edge of the cut
w = w0
and on the lower edge w assumes the opposite value. This shows us that
the case investigated is that of the symmetrical problem. The value W(Z)
for this problem is therefore
W(z) IW- lo(a Z)(1 aZ)
i (a + Z)(l + aZ)
(III.8)
The calculation of the functions U(Z) and V(Z) offers no diffi-
culty whatsoever. It suffices to apply the relationships of compati-
bility (1.25) and to integrate; the only precaution to be taken consists
in choosing the constant of integration in such a manner that the real
parts of U and V on (CO) become zero; one then finds
v(z) -~oW (1 + a2)
u(z) -2
p(1 a2
(a + Z)(1 aZ)
g(Z a)(l + aZ)
Z -a2
S log a! -
) 1 a2Z2]
This last formula is the most interesting one since it permits calcula-
tion of the pressure coefficient (see formula (1.8)). One finds
4C W a [loga2-X2]
p p 1 a2 1 a2x2
(III.9)
(III.10)
(III.11)
88 NACA TM 1354
In order to interpret this formula, one must connect the quanti-
ties a, X, to geometrical quantities, related to the given cone. Fire:
of all
w0 =a
a being the constant
hand
inclination of the cone on
2X = r -= tan u
1 + X2 x
Ox. On the other
whence
= cos w 1 M2sin2
0 sin w
(see fig. 32) and
cos CU \ M2sin2w0
p sin a0
In figure 33 one will find the curves giving the values of
functions of o, for various Mach numbers and various values of
(III.12)
Cp as
CO'
3.1.1.7 Solution of the elementary lifting problem.- If one
starts from the formula (III.7), one obtains
dW =
dZ
(2 1)2
Fa2 Z2)(1 a2z2)] 2
The integration which yields W(Z) introduces elliptic functions (see
section 3.1.1.8); on the other hand, it will (now) be possible to cal-
culate U(Z). We note beforehand that, according to the preceding for-
mula, W(Z) assumes the same value on the two edges of the cut and
that, consequently, this solution corresponds to the lifting problem.
N:ACA TM 1354
The relationships of compatibility show that
dU 2X1
dZ B
z(z2 )
(a2 2) (1 a2z2)] 2
U(z) = 2X1
We s + 1)2 ka
We still have to calculate 1 a
Z2 + 1
Z2)(1 a2Z2 2
s a function of w,
(III.13)
For this
purpose, one may write
-wO = d dZ = iil
0 dZ
We put in this integral Z = iu
T (z2 i)-
Z[ 2 1)2d
[a2 2)(1 a2Z2) 2
Wo = 1 f (+ = Ull(a)
S a2 + u2)( + a2u2) 2
The calculation of I(a) can be made with the aid of the function E
(see ref. 24). We shall put
u+ 2
u t
After a few calculations one obtains
I(a) =4 dt
+ (a t2 2 2 1)2t]2
0 1 .
and hence
NACA TM 1354
Finally, the change in variable
sin = t(a2 + 1)
4a2 + (a2 )2t2
shows that if one puts
1 a2
1 + a2
a2(a2 + 1) O
dT = 1
a2(a2 + 1)
Hence the new formula for U(Z)
u(z) 2 a20w
a (a2 + )E1 + a2
(a2
Z2 + 1
1
_Z2)(1 a272) 2
We still have to
One has (fig. 32)
connect a
and Z to the geometrical quantities.
2a
1 + a2
- 3 tan i'0
2X
1 + X2
One puts
t tan u
tan nLQ
and obtains
w0 tan l 0 1
EL 02tan2 o] t2
E a
1 + a
fi-^
li<*^
(III.14)
= 0 tan J | {"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": 0.9306462407112122, "perplexity": 4204.213215673497}, "config": {"markdown_headings": false, "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-2017-34/segments/1502886103891.56/warc/CC-MAIN-20170817170613-20170817190613-00711.warc.gz"} |
http://mathoverflow.net/questions/114918/borel-group-on-r/114932 | Borel Group on R [closed]
Last week in class we used the fact that if we have a group within R which is also a Borel Set, then it is either R or meagre. Why is it so? Can you direct me to a proof?
-
closed as off topic by Bill Johnson, Andres Caicedo, Qiaochu Yuan, Benoît Kloeckner, Ryan BudneyNov 30 '12 at 19:51
Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question.
This is a version of Steinhaus's theorem: en.wikipedia.org/wiki/Steinhaus_theorem A proof of a natural generalization, Pettis's theorem, is in Kechris's book Classical descriptive set theory (somewhere in chapter 9, I believe): books.google.com/books?id=pE1VMQEACAAJ – Theo Buehler Nov 29 '12 at 21:13
1 Answer
These notes written by Julien Melleray help us to solve the problem. I just state the results which will help in our case.
Lemma 3.3 (Pettis) Let $G$ a Polish group. For $A\subset G$, define $U(A)$ as the biggest open set $V$ such that $A$ is comeagre $V$. For any subsets $A$ and $B$ of $G$, we have $$U(A)\cdot U(B)\subset A\cdot B.$$
As a consequence:
Theorem 3.4 Let $G$ a Polish group, and $A$ a Baire measurable non-meagre subset of $G$. Then $e$, the neutral element, belongs to the interior of $A\cdot A^{-1}$.
Back to the problem. Of course, $\Bbb R$ with the addition is a Polish group. Let $H$ a subgroup of $\Bbb R$ which is non-meagre and Borel measurable. It's Baire measurable. By the last theorem, $e$ belongs to the interior of $H\cdot H^{-1}=H$ as $H$ is a sub-group.
It's well-know that the subgroups of $\Bbb R$ are either of the form $a\Bbb Z$ (hence meagre) or dense. So we have a subgroup $H$ which is dense and has non-empty interior, say $(-r,r)$. Let $x\in \Bbb R$, and $x'\in H$ such that $|x-x'|\lt r$. Then $x-x'\in H$ and $x\in H$.
-
+1 for linking to Julien Melleray's notes, they look very nice at a first glance. I saw that you started with your PhD thesis: Good luck with it! Best wishes (t.b. from math.SE). – Theo Buehler Nov 29 '12 at 21:31
Please keep in mind that the notes have not been carefully re-read and probably contain mistakes (and I'll be grateful, should you find any, if you let me know!)... Also, in the case of $\mathbb R$, probably the easiest way to answer the question is to use the Lebesgue measure, via the following statement (due to Steinhaus, I believe): if $A \subset \mathbb R$ is measurable of positive measure, then $A−A$ contains an interval centered around $0$ (this is a classical consequence of the regularity of the Lebesgue measure) – Julien Melleray Nov 29 '12 at 21:41
Thanks! I hope I will have luck with it, which will be the least we could expect of a probabilist! – Davide Giraudo Nov 29 '12 at 21:43
@Julien I thought about that, and it proves that a subgroup is either $\Bbb R$ or has $0$ measure. Is there a simple way to see it's meagre? – Davide Giraudo Nov 29 '12 at 21:45
A word of warning: Baire measurable is not the same as measurable with respect to the Baire $\sigma$-algebra (the $\sigma$-algebra generated by the compact $G_\delta$'s). It means the $\sigma$-algebra generated by the open sets and the meagre sets. There are meagre sets that aren't Borel: every set of reals can be written as the disjoint union of a meagre set $A$ and a Lebesgue null set $B$. If you partition a Vitali set this way: $V = A \cup B$ then $A$ can't even be Lebesgue measurable but is Baire measurable. – Theo Buehler Nov 29 '12 at 21:46 | {"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": 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": 0.9029480814933777, "perplexity": 238.96939776058443}, "config": {"markdown_headings": false, "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-2015-22/segments/1432207928864.16/warc/CC-MAIN-20150521113208-00085-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/probability-theory-2.213531/ | # Probability theory 2
1. Feb 6, 2008
### Milky
1. The problem statement, all variables and given/known data
The homework question I have is:
Let X be uniform over (0,1). Find E[X|X<1/2].
3. The attempt at a solution
First I found the density function: f(x|x<1/2] = f(x∩x<1/2) / f(x<1/2) = f(x<1/2) / f(x<1/2) = 1.
So, E[X|X<1/2] = ∫ xdx
My question is... are the limits of integration suppose to be from 0 to 1, or 0 to 1/2? Is everything else correct?
2. Feb 6, 2008
### EnumaElish
The limits of integration are {0, 1/2} because f(x|X<1/2) is defined (or is positive) for x<1/2 only.
This information should also alert you to the fact that $$\int_0^{1/2}f(x|X<1/2)dx = 1$$ has to be the case. Does f(x|x<1/2) = 1 satisfy this condition?
3. Feb 6, 2008
### Milky
Okay, new strategy. In my textbook in a similar problem, they did:
f(x|x<1/2) = f(x) / P{x<1/2}
When I do that, I get
f(x) / P{x<1/2} = 1 / (1/2) = 2.
E[x|x<1/2] = $$\int_0^{1/2}2xdx$$
= 1/4
Last edited: Feb 6, 2008
Similar Discussions: Probability theory 2 | {"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": 0.9409748315811157, "perplexity": 3189.2350136631}, "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-2017-47/segments/1510934806438.11/warc/CC-MAIN-20171121223707-20171122003707-00724.warc.gz"} |
https://www.clutchprep.com/chemistry/practice-problems/96424/what-is-the-evidence-that-all-neutral-atoms-and-molecules-exert-attractive-force | # Problem: What is the evidence that all neutral atoms and molecules exert attractive forces on each other?
⚠️Our tutors found the solution shown to be helpful for the problem you're searching for. We don't have the exact solution yet.
###### Problem Details
What is the evidence that all neutral atoms and molecules exert attractive forces on each other? | {"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": 0.9851852059364319, "perplexity": 474.7514071388468}, "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-2020-29/segments/1593655924908.55/warc/CC-MAIN-20200711064158-20200711094158-00020.warc.gz"} |
https://www.physicsforums.com/threads/why-is-this-integral-positive-or-negative.107898/ | # Why is this integral positive or negative
1. Jan 24, 2006
### Natasha1
I have been asked to explain without evaluating the integrals why
1) the integral of x cos x from 0 to pi/2 is positive and the integral of x cos x from pi/2 to pi is negative. And also would I expect x cos x from 0 to pi to be positive, zero or negative??? And why ???
2) to find the indefinite integral of x cos x dx and hence the exact values of values of x cos x from 0 to pi/2, cos x from pi/2 to pi and x cos x from 0 to pi?
Any help would be must appreciated. I know I haven't done anything so far so no need to have a go at me for it but I just wanted some help with it and even better the solutions obviously as this blimming thing is for in a few days time. Many thanks in advance
Nat
2. Jan 24, 2006
### siddharth
1) What are your thoughts on this? Have you drawn the graph?
2) Have you learnt about integration by parts?
3. Jan 24, 2006
### Jameson
$$\int x\cos(x)dx$$
By parts, u=x dv=cos(x)
du = 1 v = sin(x)
$$\int udv= uv - \int vdu$$
You can take it from there.
4. Jan 25, 2006
### VietDao29
#1, What can you say about the integral of $$\int \limits_{\alpha} ^ {\beta} f(x) \ dx$$ (i.e greater or less than or equal to 0), if:
(i) $f(x) > 0 , \ \forall x \in ( \alpha, \ \beta )$
(ii) $f(x) < 0 , \ \forall x \in ( \alpha, \ \beta )$?
You can draw a graph to see this. Remember that definite integral of some function from a to b will give you the area under the graph of that function from x = a to x = b.
#2, as others have pointed out, this should be done by Integration by parts.
You can either read the article there, or look at your text book. There should be something about Integration by parts.
Can you go from here? If you still have problems, just shout out.
Last edited: Jan 25, 2006 | {"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": 0.9522086381912231, "perplexity": 429.15700231782387}, "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-2019-04/segments/1547583658681.7/warc/CC-MAIN-20190117020806-20190117042806-00333.warc.gz"} |
http://www.apmonitor.com/wiki/index.php/Main/DifferentialVariables?action=diff&source=n&minor=y | Main
## Main.DifferentialVariables History
June 16, 2015, at 06:44 PM by 45.56.3.184 -
Deleted lines 0-1:
## Differential Variables
Changed line 13 from:
(:table border=1 width=50% align=left bgcolor=#EEEEEE cellspacing=0:)
to:
(:table border=1 width=100% align=left bgcolor=#EEEEEE cellspacing=0:)
April 23, 2009, at 04:54 PM by 158.35.225.231 -
Changed lines 29-30 from:
! x is the first order lag of parameter u with time constant tau
to:
! x is the first order lag of parameter u
! with time constant tau
October 09, 2008, at 01:22 PM by 158.35.225.231 -
Changed line 3 from:
Differential states are variables that have the differential operator ($) applied in at least one equation. These differential states are declared in the variables section without the differential operator. For dynamic problems, the differential variables are converted to algebraic variables through orthogonal collocation on finite elements. This form allows dynamic problems to be solved by nonlinear programming (NLP) solvers. to: Differential states are variables that have the differential operator ($) applied in at least one equation. These differential states are declared in the variables section without the differential operator. For dynamic problems, the differential variables are converted to algebraic variables through orthogonal collocation on finite elements. This discretized form allows dynamic problems to be solved by nonlinear programming (NLP) solvers.
October 09, 2008, at 12:58 PM by 158.35.225.231 -
Changed line 3 from:
Differential states are variables that have the differential operator ($) applied in at least one equation. These differential states are declared in the variables section as any other variable. to: Differential states are variables that have the differential operator ($) applied in at least one equation. These differential states are declared in the variables section without the differential operator. For dynamic problems, the differential variables are converted to algebraic variables through orthogonal collocation on finite elements. This form allows dynamic problems to be solved by nonlinear programming (NLP) solvers.
September 25, 2008, at 06:42 PM by 158.35.225.230 -
Changed lines 1-6 from:
### Consistent Initial Conditions
Forward stepping algorithms such as DASSL, DASPK, or CVODE generally require ordinary differential equations (ODEs) or index-1 differential algebraic equations (DAEs) and consistent initial conditions. This is not a restriction with simultaneous methods as used by APMonitor. Also, ODEs or DAEs of any index can be solved.
to:
## Differential Variables
Differential states are variables that have the differential operator () applied in at least one equation. These differential states are declared in the variables section as any other variable. ### DAE Index Restrictions Differential and algebraic equation (DAE) index is the number of times equations must be differentiated to restore them to an ODE form. Higher (>=2) index DAE problems are overcome with techniques used in this software. ODEs or DAEs of any index can be solved. ### Consistent Initial Conditions for Dynamic Problems Sequential ODE and DAE solution approaches require consistent initial conditions. This is not a restriction with APMonitor because of the simultaneous approach. The initial conditions do not have to be consistent to achieve a feabile solution. The time-shifting approach guarantees that the initial conditions are consistent after one cycle. ### Example (:table border=1 width=50% align=left bgcolor=#EEEEEE cellspacing=0:) (:cellnr:) ! Example model with a differential equation Model example Parameters u = 1 ! input tau = 5 ! time constant End Parameters Variables x = 1 ! initial condition End Variables Equations ! x is the first order lag of parameter u with time constant tau tau *x = -x + u
End Equations
End Model
(:tableend:)
September 25, 2008, at 06:32 PM by 158.35.225.230 - | {"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": 0.8409579396247864, "perplexity": 2419.823173287881}, "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-2018-43/segments/1539583511122.49/warc/CC-MAIN-20181017090419-20181017111919-00529.warc.gz"} |
http://www.ams.org/joursearch/servlet/DoSearch?f1=msc&v1=14G05&jrnl=one&onejrnl=bull | # American Mathematical Society
Publications Meetings The Profession Membership Programs Math Samplings Policy and Advocacy In the News About the AMS
You are here: Home > Publications
AMS eContent Search Results
Matches for: msc=(14G05) AND publication=(bull) Sort order: Date Format: Standard display
Results: 1 to 8 of 8 found Go to page: 1
[1] Karl Rubin and Alice Silverberg. Ranks of elliptic curves. Bull. Amer. Math. Soc. 39 (2002) 455-474. MR 1920278. Abstract, references, and article information View Article: PDF [2] János Kollár. Which are the simplest algebraic varieties?. Bull. Amer. Math. Soc. 38 (2001) 409-433. MR 1848255. Abstract, references, and article information View Article: PDF [3] Anand Pillay. Model theory and diophantine geometry. Bull. Amer. Math. Soc. 34 (1997) 405-422. MR 1458425. Abstract, references, and article information View Article: PDF [4] Helmut Völklein. ${\text{PSL}}_2 \left( q \right)$ and extensions of ${\mathbf{Q}}\left( x \right)$. Bull. Amer. Math. Soc. 24 (1991) 145-153. MR 1060151. Abstract, references, and article information View Article: PDF [5] Stephen Kudla. On the ${\mathbf{R}}$-forms of certain algebraic varieties. Bull. Amer. Math. Soc. 81 (1975) 471-473. MR 0379505. Abstract, references, and article information View Article: PDF [6] A. P. Ogg. Diophantine equations and modular forms. Bull. Amer. Math. Soc. 81 (1975) 14-27. MR 0354675. Abstract, references, and article information View Article: PDF [7] Don Goelman. Rational critical points of the reduced norm of an algebra. Bull. Amer. Math. Soc. 80 (1974) 138-141. MR 0337877. Abstract, references, and article information View Article: PDF [8] L. G. Roberts. An algebraic proof of an analytic result of Shuck's. Bull. Amer. Math. Soc. 79 (1973) 754-757. MR 0327764. Abstract, references, and article information View Article: PDF
Results: 1 to 8 of 8 found Go to page: 1 | {"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": 0.9580324292182922, "perplexity": 3010.000951999532}, "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-2015-06/segments/1422115858580.32/warc/CC-MAIN-20150124161058-00231-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://astarmathsandphysics.com/o-level-maths-notes/354-solving-trigonometric-equations.html | ## Solving Trigonometric Equations
The basicandcurves are given on the left below andon the right below:
– blueblack
We have typically to solve equations such as
1. We start by making cosx the subject:
2. We take the inverse cos:
Now is the tricky part. There is more than one solution forWe have found one. The other solutions are given by using the symmetry of the cosine graph. It is symmetric about 180 We are using degrees here. The solutions are
41.41, 360-41.41, 360+41.41, 720-41.41, 720+41.41, 1080-41.41, 1080+41.41 ....degrees
Example: Solve
Now we use the symmetry of the sin curve. The solutions are
17.46, 180-17.46, 360+17.46, 540-17.46, 720+17.46,900-17.46 ....degrees
Example: Solve
Now we we the property of the tan curve that it repeats every 180 degrees. The solutions are
60.26, 180+60.26, 360+60.26, 540+60.26, 720+60.26, 900+60.26.....degrees | {"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": 0.9870372414588928, "perplexity": 4090.8108788658305}, "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-2019-30/segments/1563195527396.78/warc/CC-MAIN-20190721225759-20190722011759-00386.warc.gz"} |
http://math.stackexchange.com/questions/391619/finite-ring-of-sets | # Finite ring of sets
I have some questions about finite rings of sets and I'll be very grateful for any help.
Let E be some fixed non-empty set. Suppose we are given some finite ring of subsets of set E, i.e. some non-empty system $S \subset E$ such that
$\forall A,B \in 2^E ~~ A \vartriangle B \in S$
$\forall A,B \in 2^E ~~ A \cap B \in S$
1) Does it necessarily have a unity? (or in more abstract form: is there necessarily a unity in the finite commutative ring in which multiplication is idempotent?)
2) Suppose we're given some finite system of subsets of E. Is there any algorithm for building the minimal ring of sets which contains this system? I know that from n sets using union, intersection and set difference we mat build at most $~2^{2^n}$ different sets so this ring must be finite.
3) If we know that S is a ring with unity (and so it's a boolean ring) how can we build an isomorphism from our ring of sets to some ring $B^n = (\{0,1\}^n, +, \cdot)$ where
$0 + 0 = 1 + 1 = 0$
$0 + 1 = 1 + 0 = 1$
$0 \cdot 1 = 0 \cdot 0 = 1 \cdot 0 = 0$
$1 \cdot 1 = 1$
-
– user26857 May 14 '13 at 17:24
@YACP I was just going to post the same reference :-) And I believe that my answer there covers both 1) and 3) – Andreas Caranti May 14 '13 at 17:25
Notice that $(A\vartriangle B)\vartriangle(A\cap B)=A\cup B$, so the ring actually contains finite unions of its members. Since the ring is finite, $E:=\cup S\in S$. Therefore, $E$ contains all members of $S$, and so $E\cap A=A$ for all $A$.
This also gives a partial start on item 2. Suppose that $G$ is a collection of sets that you want to generate a ring of sets with. Notice that both operations only produce sets which have elements in the two sets you started with. So, any set you generate will not escape $\cup G$. Then the generated ring is contained inside the powerset $\mathcal{P}(\cup G)$. | {"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": 0.9470853209495544, "perplexity": 151.97612787906198}, "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-2016-07/segments/1454701148475.34/warc/CC-MAIN-20160205193908-00070-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://xxicla.dm.uba.ar/viewAbstract.php?code=1368 | Conference abstracts
Session S08 - Lie Groups and Representations
July 28, 15:40 ~ 16:20
## Representation ring of Levi subgroups versus cohomology ring of flag varieties
### University of North Carolina, Chapel Hill, USA - [email protected]
Recall the classical result that the cup product structure constants for the singular cohomology with integral coefficients of the Grassmannian of $r$-planes coincide with the Littlewood-Richardson tensor product structure constants for $GL(r)$. Specifically, the result asserts that there is an explicit ring homomorphism $\phi: \text{Rep}_{poly}(GL(r)) \to H^*(Gr(r, n))$, where $Gr(r, n)$ denotes the Grassmannian of $r$-planes in $\mathbb{C}^n$ and $\text{Rep}_{poly} (GL(r))$ denotes the polynomial representation ring of $GL(r)$.
This work seeks to achieve one possible generalization of this classical result for $GL(r)$ and the Grassmannian $Gr(r,n)$ to the Levi subgroups of any reductive group $G$ and the corresponding flag varieties. | {"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": 0.8622463941574097, "perplexity": 294.33809513064915}, "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/1542039743184.39/warc/CC-MAIN-20181116194306-20181116220306-00227.warc.gz"} |
https://math.stackexchange.com/questions/2435688/the-construction-of-closed-subscheme | # The construction of closed subscheme
Let $(\varphi,\varphi^{\sharp}):(X,\mathcal{O}_X)\to (Y,\mathcal{O}_Y)$ be a morphism of schemes.
Suppose $Y$ is covered by a family of affine open subschemes $\{U_i\}$ with $U_i=\mathrm{Spec} A_i$.
Consider $\varphi^{\sharp}(Y):\mathcal{O}_Y(Y)\to \mathcal{O}_X(X)$ and $\rho_{YU_i}:\mathcal O_Y(Y)\to A_i$. Let $S$ be a nonempty subset of $\ker(\varphi^{\sharp}(Y))$ and $\mathfrak p_i$ be the ideal of $A_i$ generated by $\rho_{YU_i}(S)$.
Given $f\in A_i,g\in A_j$, let $\mathfrak q_f$ be the ideal of $(A_i)_f$ generated by $\rho_{YD(f)}(S)$, and $\mathfrak p_j$ be the ideal of $A_j$ generated by $\rho_{YU_j}(S)$, and $\mathfrak q_g$ be the ideal of $(A_j)_g$ generated by $\rho_{YD(g)}(S)$. Then we
have the following two commutative diagrames:
$$\begin{array}[c]{ccc} A_i&{\rightarrow}&A_i/\mathfrak p_i\\ \downarrow&&\downarrow\\ (A_i)_f&{\rightarrow}&(A_i)_f/\mathfrak q_f \end{array}$$
$$\begin{array}[c]{ccc} A_j&{\rightarrow}&A_j/\mathfrak p_j\\ \downarrow&&\downarrow\\ (A_j)_g&{\rightarrow}&(A_j)_g/\mathfrak q_g \end{array}$$
We get two closed immersions $\psi_i:\mathrm{Spec}(A_i/\mathfrak p_i)\to \mathrm{Spec}A_i$ and $\psi_j:\mathrm{Spec}(A_j/\mathfrak p_j)\to \mathrm{Spec}A_j$.
Define $\lambda_i:(A_i/\mathfrak p_i)_{f+\mathfrak p_i}\to (A_i)_f/\mathfrak q_f$ as follows：
$\frac{a+\mathfrak p_i}{f^n+\mathfrak p_i}\mapsto \frac{a}{f^n}+\mathfrak q_f.$
If $\frac{a+\mathfrak p_i}{f^n+\mathfrak p_i}=\frac{b+\mathfrak p_i}{f^m+\mathfrak p_i},$
then there exists some $k\in \Bbb N$ such that $f^k(af^m-bf^n)\in\mathfrak p_i$,
we get $\frac{a}{f^n}+\mathfrak q_f=\frac{b}{f^m}+\mathfrak q_f$, so $\lambda_i$ is defined well.
Obviously $\lambda_i$ is surjective.
If $\frac{a}{f^n}+\mathfrak q_f=0$,then $\frac{a}{f^n}\in \mathfrak q_f, \frac{a}{1}\in \mathfrak q_f$,
we have
$\frac{a}{1}=\frac{x_1}{f^{n_1}}\frac{r_1}{1}+\frac{x_2}{f^{n_2}}\frac{r_2}{1}+\cdots+\frac{x_k}{f^{n_k}}\frac{r_k}{1}$ where $x_t\in A_i,r_t\in \rho_{YU_i}(S),1\le t\le k$.
Multiply both sides by $\frac{f^{n_1+n_2+\cdots+n_k}}{1}$, there exists some $h\in \Bbb N$ such that $af^h\in \mathfrak p_i$, so
$\frac{a+\mathfrak p_i}{f^n+\mathfrak p_i}=0$, and $\lambda_i$ is injective, therefore it is an isomorphism. In the same way we get that
$\lambda_j:(A_j/\mathfrak p_j)_{g+\mathfrak p_j}\to (A_j)_g/\mathfrak q_g$ is also an isomorphism.
We know that $U_i\cap U_j$ can be covered by a family of affine open subschemes $\{ W_{\alpha} \}$ of $Y$ such that
$W_{\alpha}=D(f_{\alpha})=D(g_{\alpha})$, where $f_{\alpha}\in A_i,g_{\alpha}\in A_j$.
Denote $Z_i=\mathrm{Spec}(A_i/\mathfrak p_i),Z_j=\mathrm{Spec}(A_j/\mathfrak p_j)$.
$\forall c\in W_{\alpha}$, if $\psi_i^{-1}(c)\neq \emptyset$, then $c\in \psi_i(Z_i)$, because $\lambda_i,\lambda_j$
are isomorpisms, we also have $c \in\psi_j(Z_j)$. We get $\chi_i:\psi_i^{-1}(U_i\cap U_j)\simeq \psi_i(Z_i)\cap \psi_j(Z_j),\chi_j: \psi_j^{-1}(U_i\cap U_j)\simeq \psi_i(Z_i)\cap \psi_j(Z_j)$ as homeomorphisms of topological spaces. Denote $Z_{ij}=\psi_i^{-1}(U_i\cap U_j), Z_{ji}=\psi_j^{-1}(U_i\cap U_j)$, $\xi_{ij}=\chi_j^{-1}\chi_i, \xi_{ji}=\chi_i^{-1}\chi_j$. Because $\lambda_i,\lambda_j$ are isomorpisms, $\xi_{ij}$ and $\xi_{ji}$ are isomorphisms of schemes，and $\xi_{ij}=\xi_{ji}^{-1}$.
Can we apply the Glueing Lemma to get a closed subscheme $Z$ of $Y$ and a closed immersion $(\mu,\mu^{\sharp}):(Z,\mathcal O_Z)\to (Y,\mathcal O_Y)$? Does there exist a morphism of schemes $(\nu,\nu^{\sharp}):(X,\mathcal O_X)\to (Z,\mathcal O_Z)$ such that $(\varphi,\varphi^{\sharp})=(\mu,\mu^{\sharp})(\nu,\nu^{\sharp})$? | {"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": 0.9930118918418884, "perplexity": 53.74603340616606}, "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-2019-26/segments/1560627999800.5/warc/CC-MAIN-20190625051950-20190625073950-00208.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=484519 | # PS and Jpg in Latex Files
by ed2288
Tags: jpeg, jpg, latex, postscript
P: 25 Hi everyone, I was wondering if someone could clarify how to put both .ps and .jpg files in the same latex file. At the minute I'm having to convert all my .jpgs to .ps using photoshop then compiling my latex file by going Latex->PS->PDF. However all my converted jpgs look all pixelated and frankly pretty rubbish. *Surely* there must be a way to include both .jpgs and .ps files, people need to do this all the time. When I try to include jpgs I get error messages saying no Bounding Box. Can anyone help me out? Thanks
Sci Advisor P: 1,724 When you say "include both .jpgs and .ps files" does that mean that you're actually using the \include function? If so, there's your problem: ps is postscript, which LaTeX will include in your actual document (it just so happens that your .ps codes for an image, so things happily chug along). If so, take a look at some of the examples here: http://en.wikibooks.org/wiki/LaTeX/I...ges_as_Figures Don't forget to \usepackage{graphicx}! If not, well, I've never run into that particular problem before--can you post the offending code?
P: 313 There's a program called ebb to Extract Bounding Boxes. It's easy to get for *nix http://linux.die.net/man/1/ebb And for windows is distributed with miktex. Here's a page with tips http://www.math.vanderbilt.edu/~sche...d/tips_pix.htm
PF Gold
P: 144
## PS and Jpg in Latex Files
Just want to post a quick thankyou to contributors in this thread. I've been having problems importing gifs into a pdf via latex and you have fixed my problem(s). I'm just converting jpgs to eps's and following what ed2288 and matlabdude suggest. The resultant image isn't perfect but is fit for purpose. Thanks.
Related Discussions Math & Science Software 9 Academic Guidance 5 Programming & Computer Science 3 Computing & Technology 2 | {"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": 0.9133977890014648, "perplexity": 4237.917325221757}, "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/1397609532374.24/warc/CC-MAIN-20140416005212-00201-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/181931/on-a-property-of-the-grothendieck-group-of-a-smooth-projective-curve | # On a property of the Grothendieck group of a smooth projective curve
Let $K$ be a complete DVR of characteristic $0$, $X$ a smooth projective curve over $K$. Denote by $K^0(X)$ the Grothendieck group of locally free sheaves on $X$ and by $\mbox{det}$ the natural group homomorphism from $K^0(X)$ to $\mbox{Pic}(X)$. Suppose that for some $L \in \mbox{Pic}(X)$, there exists $[V] \in K^0(X)$ such that $\mbox{det}([V])=L$. Let $L'$ be another line bundle on $X$ of the same degree as $L$. The question is, does there exist $[V'] \in K^0(X)$ such that $V'$ is of the same rank and degree as $V$ and $\mbox{det}([V'])=L'$? If not true in general, is there any known condition/example of $K$ (other than algebraic closedness in which case the answer is known to be true via twisting $V$ by a suitable line bundle), for example if $K$ is a $C_1$-field under which we have a positive answer to the question?
• Are we allowed to set $V' = L'$? – S. Carnahan Sep 27 '14 at 13:47
• @Carnahan: Sorry. No, we need $V'$ to have the same rank and degree as $V$. I will edit the question. – user43198 Sep 27 '14 at 13:49
Perhaps I am misunderstanding the question, but it seems to me that $\text{deg}(V)=\text{deg}(\text{det}(V))=\text{deg}(L)$, so we are just asking if for each $n$ there is a (virtual) vector bundle $[V']$ of rank $n$ and degree $\text{deg}(L')$. But $\mathcal{O}_X^{\oplus(n-1)}\oplus L'$ works, right? | {"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": 0.9764226675033569, "perplexity": 54.18290063635328}, "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/1575540506459.47/warc/CC-MAIN-20191208044407-20191208072407-00120.warc.gz"} |
https://www.physicsforums.com/threads/the-geometric-interpretation-of-a-set.901905/ | # The geometric interpretation of a set
Tags:
1. Jan 28, 2017
### Sho Kano
1. The problem statement, all variables and given/known data
Show the geometric interpretation of S is a plane in $R^3$ and find its equation
$S=\left\{ (a+b,2b,-a)\in { R }^{ 3 }|a,b\in R \right\}$
2. Relevant equations
3. The attempt at a solution
I know that S is the set of all points that lie in a plane, but have no idea where to go from there
Can you offer guidance or do you also need help?
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Similar Discussions: The geometric interpretation of a set | {"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": 0.8400712609291077, "perplexity": 439.9267865527213}, "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-20170819070335-20170819090335-00666.warc.gz"} |
http://stackoverflow.com/questions/2561702/how-to-fit-the-paper-size-to-the-content-in-latex | # How to fit the paper-size to the content in LaTeX
I'd like to create a pdf/ps/eps that contains only one single formula. I thought the easiest way would be to use latex. Unfortunately, I found no option to specify, that the paper-size should automatically be set to fit the contents.
I found that dvipng has a "-T tight" option, that actually does the trick, but... I want it in vector-graphics format.
Any suggestions?
Thanks.
-
With the standalone class, you get exactly what you want.
\documentclass{standalone}
\begin{document}
$x + y = z$
\end{document}
-
IMNSHO, this is the best answer; it's pure LaTeX, it doesn't rely on cranky external programs, and it's modern and up-to-date, and can be used with modern TeX processors like pdfLaTeX, XeLaTeX, and LuaLaTeX. I have no idea why some "challenged person" downvoted you, and I apologise on his behalf for the whole thinking TeX community. +1 – Brent.Longborough Jun 23 '13 at 10:13
Try pdfcrop, it crops your pdf to the minimum. You need to have Perl installed.
-
for Mac, there is a little app called LaTeXit which does exactly that. http://chachatelier.fr/programmation/latexit_en.php
-
Fixing the bounding box of generated EPS should help:
epstopdf --gsopt=-dEPSCrop blah.eps
or eps2pdf has a -B option that detects the tightest possible Bounding Box.
- | {"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": 0.9223862886428833, "perplexity": 3988.7319002266872}, "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-2014-23/segments/1404776439014.63/warc/CC-MAIN-20140707234039-00041-ip-10-180-212-248.ec2.internal.warc.gz"} |
http://www.lehigh.edu/~zhh210/research.html | ## Publications
[1] A Globally Convergent Primal-Dual Active-Set Framework for Large- Scale Convex Quadratic Optimization
F. E. Curtis, Z. Han, and D. P. Robinson, Computational Optimization and Applications (COAP), 2014.
[2] Globally Convergent Primal-Dual Active-Set Methods with Inexact Subproblem Solves
F. E. Curtis, Z. Han, Optimization Online, 2014(under review).
## Notes
On Optimization
[1] A Note on Inexact Newton's Method
- A proof of global convergence in solving unconstrained nonlinear optimization problem $$\min_{x\in\mathbb R^n}\ f(x)$$ by applying inexact Newton's method on $$\nabla f(x) = 0$$.
[2] Counterexamples in plain primal-dual active-set method
- For mathematicians, the greatest despair might be proposing a deceptively simple conjecture yet has to disprove it after many months of travail. But coming up empty-handed is a vital and oft-overlooked part of research. Each negative result rules out certain theories and strengthens others, shrinking the conceptual space in which the reality can be hiding.
On Numerical Methods
[1] Bounds on the Inverse of Nonnegative Matrix
- An interactive play to show that $$f_n:=\displaystyle\sum_{i=0}^n (I-H)^n$$ approximates $$H^{-1}\in\mathbb R^{n\times n}$$ in the limit. Certain assumptions on $$H$$ yields the interesting interlacing behavior that $$\cdots < f_{2k+1} < f_{2k+3} <\cdots < H^{-1} < \cdots < f_{2k+2} < f_{2k}<\cdots$$ for $$k\in \mathbb N$$.
[2] Submatrix's Inverse and Inverse's Submatrix: A Note on M-matrix
- Suppose $$H\in\mathbb R^{n\times n}\succ 0$$. Can we decide which one of $$\|(H^{-1})_{\mathcal I\mathcal I}\|$$ and $$\|(H_{\mathcal I\mathcal I})^{-1} \|$$ is larger? Yes, if $$H$$ is an $$M$$-matrix or $$\|\cdot\|$$ is Euclidean norm; No, at least for some positive definite matrices on $$\|\cdot \|_{1}$$ or $$\|\cdot \|_{\infty}$$.
[3] An application of Hölder’s inequality
- Hölder’s inequality is applied to obtain a cheaply computable upper bound of $$|H^{-1}r|$$.
## Slides
[1] A Primal-Dual Active-Set Algorithm for a Class of Large-Scale Convex Quadratic Optimization Problems
To appear for INFORMS Annual Conference. San Francisco, CA. Nov 2014.
List of publications, notes, thoughts, etc. | {"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": 0.9431150555610657, "perplexity": 2835.930823782948}, "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-2015-22/segments/1432207929561.98/warc/CC-MAIN-20150521113209-00094-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/122662-dimension.html | # Math Help - dimension
1. ## dimension
hello
what's the dimension of the subspace $F$ of $\mathfrak{F}(\mathbb{R},\mathbb{R})$ spanned by $f_1(x)=\sin^2(x),f_2(x)=\cos^2(x),f_3(x)=\sin(2x)$ and $f_4(x)=\cos(2x)$
thanks.
2. Originally Posted by Raoh
hello
what's the dimension of the subspace $F$ of $\mathfrak{F}(\mathbb{R},\mathbb{R})$ spanned by $f_1(x)=\sin^2(x),f_2(x)=\cos^2(x),f_3(x)=\sin(2x)$ and $f_4(x)=\cos(2x)$
thanks.
you should explain very clearly what your vector space is. just using a non-standard and weird notation such as $\mathfrak{F}(\mathbb{R},\mathbb{R})$ is not good enough. i guess your vector space is the $\mathbb{R}$-vector space of
all real-valued functions defined on $\mathbb{R}$. then $\dim_{\mathbb{R}}F=2$ and $\{\sin^2 x, \sin (2x) \}$ is a basis for $F.$
Edit: i meant $\dim_{\mathbb{R}} F=3$ and $\{1,\sin^2x, \sin(2x) \}$ is a basis for $F.$ sorry for the mistake.
3. Originally Posted by NonCommAlg
you should explain very clearly what your vector space is. just using a non-standard and weird notation such as $\mathfrak{F}(\mathbb{R},\mathbb{R})$ is not good enough. i guess your vector space is the $\mathbb{R}$-vector space of
all real-valued functions defined on $\mathbb{R}$. then $\dim_{\mathbb{R}}F=2$ and $\{\sin^2 x, \sin (2x) \}$ is a basis for $F.$
your guess was right,anyway would you show me how you got that basis.
thank you.
4. Originally Posted by NonCommAlg
you should explain very clearly what your vector space is. just using a non-standard and weird notation such as $\mathfrak{F}(\mathbb{R},\mathbb{R})$ is not good enough. i guess your vector space is the $\mathbb{R}$-vector space of
all real-valued functions defined on $\mathbb{R}$. then $\dim_{\mathbb{R}}F=2$ and $\{\sin^2 x, \sin (2x) \}$ is a basis for $F.$
I don't think this is right since assume there exist $a,b \in \mathbb{R}$ such that $\cos ^2 (x)= a\sin (2x) + b\sin ^2 (x)=2a \sin (x) \cos (x) + b\sin ^2 (x)$ then evaluation in $0$ gives $1=0$. Thus $A= \{ \sin ^2 (x), \cos ^2 (x) , \sin (2x) \}$ is l.i. and since $\cos ^2 (x) - \sin ^2 (x) = \cos (2x)$ we get that $A$ is a basis for $F$.
5. Originally Posted by Jose27
I don't think this is right since assume there exist $a,b \in \mathbb{R}$ such that $\cos ^2 (x)= a\sin (2x) + b\sin ^2 (x)=2a \sin (x) \cos (x) + b\sin ^2 (x)$ then evaluation in $0$ gives $1=0$. Thus $A= \{ \sin ^2 (x), \cos ^2 (x) , \sin (2x) \}$ is l.i. and since $\cos ^2 (x) - \sin ^2 (x) = \cos (2x)$ we get that $A$ is a basis for $F$.
hi "Jose27" would you please explain the method you used to find that basis.
thanks.
6. Assume $\sin ^2(x) = a \sin (2x)$ for some $a\in \mathbb{R}$ then evaluation in $\frac{\pi }{2}$ will give $1=0$ so these two functions are l.i. Then by my previous post we get that $\cos ^2 (x) \notin span \{ \sin ^2 (x), \sin (2x) \}$ so it must be that these three are l.i. but again by my previous post $\cos (2x) \in span(A)$ so $A$ is a basis for $F$.
7. Originally Posted by Jose27
Assume $\sin ^2(x) = a \sin (2x)$ for some $a\in \mathbb{R}$ then evaluation in $\frac{\pi }{2}$ will give $1=0$ so these two functions are l.i. Then by my previous post we get that $\cos ^2 (x) \notin span \{ \sin ^2 (x), \sin (2x) \}$ so it must be that these three are l.i. but again by my previous post $\cos (2x) \in span(A)$ so $A$ is a basis for $F$.
one final question
what is the first step to do to answer this kind of questions ?
thanks "Jose27"
8. I guess trying to determine if they're l.i. To do this pick one vector then add another, see if it's l.i then add another etc.
9. Originally Posted by Jose27
I guess trying to determine if they're l.i. To do this pick one vector then add another, see if it's l.i then add another etc.
Got that,thanks "Jose27"
10. Originally Posted by Jose27
I don't think this is right since assume there exist $a,b \in \mathbb{R}$ such that $\cos ^2 (x)= a\sin (2x) + b\sin ^2 (x)=2a \sin (x) \cos (x) + b\sin ^2 (x)$ then evaluation in $0$ gives $1=0$. Thus $A= \{ \sin ^2 (x), \cos ^2 (x) , \sin (2x) \}$ is l.i. and since $\cos ^2 (x) - \sin ^2 (x) = \cos (2x)$ we get that $A$ is a basis for $F$.
i actually had $\{1,\sin^2x, \sin(2x) \}$ in my mind but i wrote something else. thanks for noticing that. | {"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": 70, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9952516555786133, "perplexity": 174.70958709725707}, "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-2014-41/segments/1410657104131.95/warc/CC-MAIN-20140914011144-00274-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
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• Stability analysis of a single species logistic model with Allee effect
and feedback control
• Abstract: A single species logistic model with Allee effect and feedback control \begin{aligned}& \frac{dx}{dt} = rx(1-x)\frac{x}{\beta+x}-axu, \\& \frac{du}{dt} = -bu+cx, \end{aligned} where β, r, a, b, and c are all positive constants, is for the first time proposed and studied in this paper. We show that, for the system without Allee effect, the system admits a unique positive equilibrium which is globally attractive. However, for the system with Allee effect, if the Allee effect is limited ( $$\beta<\frac{b^{2}r^{2}}{ac(ac+br)}$$ ), then the system could admit a unique positive equilibrium which is locally asymptotically stable; if the Allee effect is too large ( $$\beta>\frac{br}{ac}$$ ), the system has no positive equilibrium, which means the extinction of the species. The Allee effect reduces the population density of the species, which increases the extinction property of the species. The Allee effect makes the system “unstable” in the sense that the system could collapse under large perturbation. Numeric simulations are carried out to show the feasibility of the main results.
PubDate: 2018-05-18
• Dynamic behaviors of a nonlinear amensalism model
• Abstract: A nonlinear amensalism model of the form \begin{aligned} &\frac{dN_{1}}{dt}= r_{1}N_{1} \biggl(1- \biggl( \frac{N_{1}}{P_{1}} \biggr)^{\alpha _{1}}-u \biggl(\frac{N_{2}}{P_{1}} \biggr)^{\alpha_{2}} \biggr), \\ &\frac{dN_{2}}{dt}= r_{2}N_{2} \biggl(1- \biggl( \frac{N_{2}}{P_{2}} \biggr)^{\alpha_{3}} \biggr), \end{aligned} where $$r_{i}, P_{i}, u, i=1, 2, \alpha_{1}, \alpha_{2}, \alpha_{3}$$ are all positive constants, is proposed and studied in this paper. The dynamic behaviors of the system are determined by the sign of the term $$1-u (\frac{P_{2}}{P_{1}} )^{\alpha_{2}}$$ . If $$1-u (\frac {P_{2}}{P_{1}} )^{\alpha_{2}}>0$$ , then the unique positive equilibrium $$D(N_{1}^{*},N_{2}^{*})$$ is globally attractive, if $$1-u (\frac{P_{2}}{P_{1}} )^{\alpha_{2}}<0$$ , then the boundary equilibrium $$C(0, P_{2})$$ is globally attractive. Our results supplement and complement the main results of Xiong, Wang, and Zhang (Advances in Applied Mathematics 5(2):255–261, 2016).
PubDate: 2018-05-18
• Design disturbance attenuating controller for memristive recurrent neural
networks with mixed time-varying delays
• Abstract: This paper investigates the design of disturbance attenuating controller for memristive recurrent neural networks (MRNNs) with mixed time-varying delays. By applying the combination of differential inclusions, set-valued maps and Lyapunov–Razumikhin, a feedback control law is obtained in the simple form of linear matrix inequality (LMI) to ensure disturbance attenuation of memristor-based neural networks. Finally, a numerical example is given to show the effectiveness of the proposed criteria.
PubDate: 2018-05-18
• Existence and attractivity of solutions for fractional difference
equations
• Abstract: In this paper, we study a kind of difference equations with Riemann–Liouville-like fractional difference. The results on existence and attractivity are obtained by using the Picard iteration method and Schauder’s fixed point theorem. Examples are provided to illustrate the main results.
PubDate: 2018-05-18
• Adaptive neural network synchronization for uncertain strick-feedback
chaotic systems subject to dead-zone input
• Abstract: In this paper, an adaptive neural network (NN) synchronization controller is designed for two identical strict-feedback chaotic systems (SFCSs) subject to dead-zone input. The dead-zone models together with the system uncertainties are approximated by NNs. The dynamic surface control (DSC) approach is applied in the synchronization controller design, and the traditional problem of “explosion of complexity” that usually occurs in the backstepping design can be avoided. The proposed synchronization method guarantees the synchronization errors tend to an arbitrarily small region. Finally, this paper presents two simulation examples to confirm the effectiveness and the robustness of the proposed control method.
PubDate: 2018-05-18
• Positive solutions of fractional differential equations involving the
Riemann–Stieltjes integral boundary condition
• Abstract: In this article, the following boundary value problem of fractional differential equation with Riemann–Stieltjes integral boundary condition $$\textstyle\begin{cases} D_{0+}^{\alpha}u(t)+\lambda f(t, u(t),u(t))=0, \quad0< t< 1, n-1< \alpha \leq n, \\ u^{(k)}(0)=0,\quad 0\leq k\leq n-2, \qquad u(1)= \int_{0}^{1}u(s)\,dA(s) \end{cases}$$ is studied, where $$n-1 < \alpha\le n$$ , $$\lambda>0$$ , $$D_{0+}^{\alpha}$$ is the Riemann–Liouville fractional derivative, A is a function of bounded variation, $$\int_{0}^{1}u(s)\,dA(s)$$ denotes the Riemann–Stieltjes integral of u with respect to A. By the use of fixed point theorem and the properties of mixed monotone operator theory, the existence and uniqueness of positive solutions for the problem are acquired. Some examples are presented to illustrate the main result.
PubDate: 2018-05-16
• Parisian ruin probability for Markov additive risk processes
• Abstract: In this paper, we consider a spectrally negative Markov additive risk process. Using the theory of Jordan chain, a compact formula of Parisian ruin probability is given. The formula depends only on the scale matrix of spectrally negative Markov additive risk processes and the transition rate matrix $$\Lambda^{q}$$ .
PubDate: 2018-05-16
• Monotone iterative method for two-point fractional boundary value problems
• Abstract: In this work, we deal with two-point Riemann–Liouville fractional boundary value problems. Firstly, we establish a new comparison principle. Then, we show the existence of extremal solutions for the two-point Riemann–Liouville fractional boundary value problems, using the method of upper and lower solutions. The performance of the approach is tested through a numerical example.
PubDate: 2018-05-16
• Positive solutions of higher-order Sturm–Liouville boundary value
problems with fully nonlinear terms
• Abstract: In this paper we consider the existence of positive solutions of nth-order Sturm–Liouville boundary value problems with fully nonlinear terms, in which the nonlinear term f involves all of the derivatives $$u',\ldots, u^{(n-1)}$$ of the unknown function u. Such cases are seldom investigated in the literature. We present some inequality conditions guaranteeing the existence of positive solutions. Our inequality conditions allow that $$f(t, x_{0}, x_{1},\ldots, x_{n-1})$$ is superlinear or sublinear growth on $$x_{0}, x_{1},\ldots, x_{n-1}$$ . Our discussion is based on the fixed point index theory in cones.
PubDate: 2018-05-16
• The effect of parameters on positive solutions and asymptotic behavior of
an unstirred chemostat model with B–D functional response
• Abstract: This paper deals with the effect of parameters on properties of positive solutions and asymptotic behavior of an unstirred chemostat model with the Beddington–DeAngelis (denote by B–D) functional response under the Robin boundary condition. Firstly, we establish some a priori estimates and a sufficient condition for the existence of positive solutions (see (Feng et al. in J. Inequal. Appl. 2016(1):294, 2016)). Secondly, we study the effect of the small parameter $$k_{1}$$ and sufficiently large $$k_{2}$$ in B–D functional response, which shows that the model has at least two positive solutions. Thirdly, we investigate the case of sufficiently large $$k_{1}$$ . The results show that if $$k_{1}$$ is sufficiently large, then the positive solution of this model is determined by a limiting equation. Finally, we present an asymptotic behavior of solutions depending on time. The main methods used in this paper include the fixed point index theory, bifurcation theory, perturbation technique, comparison principle, and persistence theorem.
PubDate: 2018-05-16
• Existence of positive solutions for two-point boundary value problems of
nonlinear fractional q -difference equation
• Abstract: This paper is concerned with the two-point boundary value problems of a nonlinear fractional q-difference equation with dependence on the first order q-derivative. We discuss some new properties of the Green function by using q-difference calculus. Furthermore, by means of Schauder’s fixed point theorem and an extension of Krasnoselskii’s fixed point theorem in a cone, the existence of one positive solution and of at least one positive solution for the boundary value problem is established.
PubDate: 2018-05-16
• Dynamical behaviors of a food-chain model with stage structure and time
delays
• Abstract: Incorporating two delays ( $$\tau_{1}$$ represents the maturity of predator, $$\tau_{2}$$ represents the maturity of top predator), we establish a novel delayed three-species food-chain model with stage structure in this paper. By analyzing the characteristic equations, constructing a suitable Lyapunov functional, using Lyapunov–LaSalle’s principle, the comparison theorem and iterative technique, we investigate the existence of nonnegative equilibria and their stability. Some interesting findings show that the delays have great impacts on dynamical behaviors for the system: on one hand, if $$\tau_{1}\in (m_{1},m_{2})$$ and $$\tau_{2}\in(m_{4}, +\infty)$$ , then the boundary equilibrium $$E_{2}(x^{0}, y_{1}^{0}, y_{2}^{0}, 0, 0)$$ is asymptotically stable (AS), i.e., the prey species and the predator species will coexist, the top-predator species will go extinct; on the other hand, if $$\tau_{1}\in(m_{2}, +\infty)$$ , then the axial equilibrium $$E_{1}(k, 0, 0, 0, 0)$$ is AS, i.e., all predators will go extinct. Numerical simulations are great well agreement with the theoretical results.
PubDate: 2018-05-16
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JournalTOCs © 2009- | {"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": 0.8586989045143127, "perplexity": 4096.145917334719}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2018-30/segments/1531676591578.1/warc/CC-MAIN-20180720100114-20180720120114-00384.warc.gz"} |
http://mathhelpforum.com/algebra/218083-simplify-expression-isolating-q.html | # Thread: Simplify expression, isolating Q
1. ## Simplify expression, isolating Q
Thank you i advance :-)
2. ## Re: Simplify expression, isolating Q
Hey mrtn.
Hint: Take the -dS/Q^2 term to the RHS side and then take the reciprocal of both sides which will give Q^2/dS = blah. After this you multiply both sides by dS and take the square root.
Show us your attempts so we can guide you through them.
3. ## Re: Simplify expression, isolating Q
Originally Posted by mrtn
Thank you i advance :-)
Question: What is the subscript on h for?
I would move the fraction with Q^2 to the right hand side of the equation to start.
$\frac{h}{2} - \frac{dh}{2p} = \frac{ dS}{Q^2}$
Combine fractions on the left
$\frac{h(p - d)}{2p} = \frac{dS}{Q^2}$
Divide both terms in the parenthesis by p
$\frac{h(1 - \frac{d}{p})}{2} = \frac{Q^2}{dS}$
Invert both sides
$\frac{2}{h(1 - \frac{d}{p})} = \frac{Q^2}{dS}$
Multiply by dS
$\frac{2Sd}{h(1 - \frac{d}{p})} = Q^2$
Extract the root
$Q = \pm \sqrt{\frac{2Sd}{h(1 - \frac{d}{p})}}$
Question: What is the subscript on h for?
4. ## Re: Simplify expression, isolating Q
Hey chiro.
Thank you for your hints.
This is what I can come up with:
I still cant get there.
Sorry about the h / hl thing - they are the same
5. ## Re: Simplify expression, isolating Q
Originally Posted by mrtn
Hey chiro.
Thank you for your hints.
This is what I can come up with:
I still cant get there.
Sorry about the h / hl thing - they are the same
You cannot invert before you combine because it doesn't work COMBINE FIRST , then invert.
$2 + 3 = 5$ but $\frac{1}{2} + \frac{1}{3} \ne \frac{1}{5}$ | {"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": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9400479197502136, "perplexity": 4015.8645394004634}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2016-40/segments/1474738660916.37/warc/CC-MAIN-20160924173740-00211-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/21377/on-the-convergence-of-sum-mun-ns | # On the convergence of $\sum \mu(n)/n^s$
I arrived at something during my maths ponderings which is really exciting for me.
It is clearly stated in the book on Riemann Hypothesis by Borwein that the convergence of $\sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$ for $\Re(s) > 1/2$ is necessary and sufficient for RH. This is ofcourse valid since, $\sum \mu(n)/n^s = 1/\zeta(s)$ for $\Re(s) > 1$
Having said that, I have reached a point where I got, for $\Re(s) > 1/2$ $$\left| \frac{\eta(s)}{s} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} - \frac{1-2^{1-s}}{s} \right| < \infty$$ $\eta(s)$ is the Dirichlet eta function.
My question is,
What do I interpret out of this formula. Does this result imply RH, or falls short of it?
I believe the second option might be more correct, because this result does not say anything about the zeros of the eta function. But at least it is clear that if $\sum \mu(n)/n^s$ blows up then $\eta(s)$ also must have a zero to bring it down.
Any elaborated answer will be highly appreciated, because it is a current work in progress. :)
-
Let me make sure I understand what you mean by the formula $$\left| \frac{\eta(s)}{s} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} - \frac{1-2^{1-s}}{s} \right| < \infty.$$
Do you actually mean that you have proven that this sum is convergent for $s$ in the critical strip and that, for such $s$, this inequality holds? Because, if you have shown that this sum converges in that range, you have already shown RH.
I can think of other interpretations of your statement, but I will wait for clarification before elaborating on them. Here is a basic point to remember: The statement that $\sum_{n=1}^{\infty} a_n$ converges is a statement about the partial sums $\sum_{n=1}^N a_n$. (Namely, that they form a Cauchy sequence.) If you never say anything about these partial sums, it is unlikely you have proved convergence.
Thanks David. I followed on similar lines as by Beurling. Basically, I showed, that this expression satisfies: $$\left| \frac{\eta(s)}{s}\sum_{k=1}^{\infty} \frac{\mu(n)}{n^s} - \frac{1 - 2^{1-s}}{s} \right| \leq || 1 + f_\mu||_2 . || x^{s-1} ||_2$$ where $f_\mu(1/x) = \sum \mu(n)\nu(1/nx)$, $\nu(n) = \rho(x/2) + 1/2 - \rho(x/2 + 1/2)$ where $\rho(x)$ is the fractional part of $x$, and the norm is on $L^2(0,1/2)$ . Then I proceeded to show that $$\lim_{n\to\infty} \left( \int_{0}^{1/2} \left|1 + \sum_{k=1}^{n} \mu(k) \nu(1/kx)\right|^2 dx \right)^{1/2} < \infty$$ – Roupam Ghosh Feb 10 '11 at 23:11
That part is ready, and it does form a Cauchy sequence and this time the result shows that the expression in the main question is zero for $\Re(s) > 1/2$. :) ( I am saying this with a pessimistic voice since the implications are huge and hence the chances of having a unseen bug in my calculations are high :P ) – Roupam Ghosh Feb 11 '11 at 7:50 | {"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": 0.9122626185417175, "perplexity": 215.543835167079}, "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/1435375097861.54/warc/CC-MAIN-20150627031817-00183-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/maclaurin-series.229471/ | # Maclaurin Series
• #1
270
1
1. Homework Statement
find the Maclaurin Series:
f(x)=(e^x - cos(x))/x
2. Homework Equations
3. The Attempt at a Solution
I'm not really sure what to do I do know the Maclaurin series of cos(x) I was thinking that using this known series then doing e^x - the known series then dividinig the entire thing by x. I'm just not sure if this is allowed. Do you have to do the known series of e^x - the cos(x) series ? Someone please let me know.
## Answers and Replies
Related Calculus and Beyond Homework Help News on Phys.org
• #2
1,703
5
you know i didn't expect that to work but it does. [series(exp(x))-series(cos(x))]/x is the right answer
• #3
Gib Z
Homework Helper
3,344
4
Either your question wanted you to derive the series for Taylors theorem straight, or more likely not, show that each of those functions are analytic, as you can only do that trick for analytic functions. Most functions studied in analysis, however, are analytic =]
• #4
HallsofIvy
Homework Helper
41,738
899
1. Homework Statement
find the Maclaurin Series:
f(x)=(e^x - cos(x))/x
2. Homework Equations
3. The Attempt at a Solution
I'm not really sure what to do I do know the Maclaurin series of cos(x) I was thinking that using this known series then doing e^x - the known series then dividinig the entire thing by x. I'm just not sure if this is allowed. Do you have to do the known series of e^x - the cos(x) series ? Someone please let me know.
Are you saying you don't know the Maclaurin series for ex? Surely not! The crucial point here is that the first (constant) term of the Maclaurin series for both ex and cos(x) is 1- subtracting them cancels the constant term so that you can then divide by x and still have a Maclaurin series left!
• #5
270
1
I do know the maclaurin series for e^x and for cos(x) :) oh yes you are right; but I do not know if it is acceptable to leave my answer in a form with just like the terms of the series or if it is standard to put the answer into a summation. I can't really figure out the pattern:( because the x^4 terms cancel out and then I am left with 0+ 1 + x + x^2/3! + x^5/^6! so what I did and I do not know if this is correct is subtracted the general terms and then divided by x to get: (x^(n-1))/n!)-((-1)^n*(x^2n-1))/(2n)!
• #6
1,703
5
Either your question wanted you to derive the series for Taylors theorem straight, or more likely not, show that each of those functions are analytic, as you can only do that trick for analytic functions. Most functions studied in analysis, however, are analytic =]
1/x isn't analytic
• #7
270
1
well theres an example in the book with (1-cos(x^2))/x and they did it the same way. I don't really know what an "analytic function is" but I am pretty sure you can do it by subtracting the two series I just don't know if I subtract the general terms or should subtract like the first 5 terms from each series? Because when I subtracted the first 5 terms from each I was left with something kind of odd that I couldn't see a pattern in to put into a summation form.
the terms I end up getting are:
1 + x + (x^2 / 3!) + (x^5 / 6!) - (x^7/ 8!)
Last edited:
• #8
HallsofIvy
Homework Helper
41,738
899
1/x isn't analytic
No, it isn't. However, since all of the functions in this problem are I don't see how that is relevant!
• #9
HallsofIvy
Homework Helper
41,738
899
You know that the general term of the Maclaurin series for ex is 1/n! xn and that the general term of the Maclaurin series for cos(x) is (-1)n/(2n)!x2n or: "0 if n is odd (-1)n/2/n! xn if n is even". The general term of the Maclaurin series for ex- cos(x) is "(1/n!)xn if n is odd, [1/n!- (-1)n/2/(2n+1)!]xn if n is even". Notice that when n is 0 (even), that is 1- 1= 0. Dividing each term by x gives "(1/n!)xn-1 if n is odd, [1/n!- (-1)n/2/(2n+1)!]xn-1 if n is even"
• #10
270
1
ah yes I understand what you are saying here would it be acceptable to write it like that though ? I do not understnad how you got (2n +1) however how did it go from (2n)! to (2n+1)! Since the cos part will be 0 for odd terms can you just say the series is the series you wrote for the even terms since when its odd the cos part will drop out anyway and will give you basically the odd series?
Last edited:
• #11
1,703
5
No, it isn't. However, since all of the functions in this problem are I don't see how that is relevant!
how is the function in question analytic if it is (expx - cosx)*1/x? are you saying simply because it is (expx - cosx ) divided by x f(x) is analytic?
• #12
Hootenanny
Staff Emeritus
Gold Member
9,598
6
how is the function in question analytic if it is (expx - cosx)*1/x? are you saying simply because it is (expx - cosx ) divided by x f(x) is analytic?
What I believe Halls is saying is that if each of the individual functions are analytic, then one can find a series representation of the entire function, whether the entire function is analytic or not.
• #13
270
1
well this is definitely the way to do it we went over it today in class I asked for help as I was still a little confused but thanks guys I sorta get it now? lol i know that you need two summations one for the even and one for the odd getting to them is a little weird
• #14
1,703
5
What I believe Halls is saying is that if each of the individual functions are analytic, then one can find a series representation of the entire function, whether the entire function is analytic or not.
no i understand that but my question still stands.
• #15
Dick
Homework Helper
26,258
618
no i understand that but my question still stands.
I'm not sure what the question is, but (exp(x)-cos(x))/x only has a removable singularity at x=0. Once you plug that whole the resulting function is analytic.
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https://math.stackexchange.com/questions/536404/proof-by-induction-for-a-recursive-sequence-and-a-formula | # Proof by Induction for a recursive sequence and a formula
So I have a homework assignment that has brought me great strain over the past 2 days. No video or online example have been able to help me with this issue either and I don't know where to turn.
I’m given
$a_0=0$
$a_n=2a_{n-1}+1$
After writing the first 6 terms of the series: 0, 1, 3, 7, 15, 31, 63 I come up with an alternate formula of
$a_n=2^n-1$
I then have to prove these formulas are the same using Induction in 3 parts:
• Proving the base case
• Stating my Inductive Hypothesis
• Showing the Inductive Step
I have done Inductive proofs before but I don’t know how to show cases or do manipulations on a recursive formula. I don’t know how to represent when n = k then n = k + 1 or showing the approach by using n = k – 1 then n = k.
Any ideas?
For the setup, we need to assume that $a_n = 2^n - 1$ for some $n$, and then show that the formula holds for $n + 1$ instead. That is, we need to show that $$a_{n + 1} = 2^{n + 1} - 1$$
Let's just compute directly:
\begin{align*} a_{n + 1} &= 2a_n + 1 \hspace{1.55in}\text{// recursion relation} \\ &= 2 \cdot (2^n - 1) + 1 \hspace{1in} \text{// induction hypothesis} \\ &= 2^{n + 1} - 2 + 1 \hspace{1.15in} \text{// arithmetic} \\ &= 2^{n + 1} - 1 \end{align*}
which is exactly what we wanted to be true.
• +1, great answer. Just as an aside to this fine answer, I'm not sure what your background is @Chris, but what you've observed here falls under the more general problem of solving linear recurrence relations. Here is a nice resource which describes how you can generally solve recurrence relations of this kind: eecs.yorku.ca/course_archive/2008-09/S/1019/Website_files/…. You can find many more like this by simply searching "solving linear recurrence relations." – Alex Wertheim Oct 23 '13 at 1:37
• You had an oops with your arithmetic when you distributed the 2. But anyway, thank you very much, I can't believe it was so simple or that you could represent things that way. The challenge comes when trying to figure out what side to prove and which side I can manipulate to look like the other side. – Chris Oct 23 '13 at 1:48
Proving the base case should be rather simple.
For the inductive hypothesis, we'll assume that for $k\geq1$, $$a_{k-1}=2^{k-1}-1$$ From this you need to prove that $a_k=2^k-1$. It shouldn't be too tough to get it from here just by following the recurrence relation. | {"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": 0.9968653917312622, "perplexity": 172.2883873624854}, "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-2021-31/segments/1627046154163.9/warc/CC-MAIN-20210801061513-20210801091513-00470.warc.gz"} |
https://www.physicsforums.com/threads/simplifying-complex-numbers.236204/ | # Simplifying complex numbers
• #1
908
0
## Homework Statement
Simplify $(1+i\sqrt{2})^5-(1-i\sqrt{2})^5$
## Homework Equations
$$z=a+bi$$
$$z=r(cos\varphi+isin\varphi)$$
$$tg\varphi=\frac{b}{a}$$
$$r=\sqrt{a^2+b^2}$$
## The Attempt at a Solution
$$(\sqrt{3}(arccos\frac{\sqrt{3}}{3}+iarcsin\frac{\sqrt{6}}{3}))^5-(\sqrt{3}(arccos\frac{\sqrt{3}}{3}+iarcsin\frac{-\sqrt{6}}{3}))^5$$
How will I get integer angle out of here?
$$arccos\frac{\sqrt{3}}{3} \approx 54.74^\circ$$
$$arcsin\frac{\sqrt{-6}}{3} \approx -54.74^\circ$$
Related Precalculus Mathematics Homework Help News on Phys.org
• #3
908
0
Yes, I know de Moivre's theorem, but I don't know how will I get integer at the end...
• #4
63
0
Why do you need an integer at the end? Is this a part of the question that hasn't been specified? ???
• #5
908
0
Ok.
$$\sqrt{3^5}(cos5*54.74^\circ+isin5*54.74^\circ)-\sqrt{3^5}(cos5*54.74^\circ-isin5*54.74^\circ)$$
$$9\sqrt{3}(0.065-0.997i)-9\sqrt{3}(0.065+0.997i)=9\sqrt{3}(0.065-0.997i-0.065-0.997i)=9\sqrt{3}(-2*0.997i)=-i17.946\sqrt{3} \approx -31.08i$$
And in my text book results: $-22i\sqrt{2}$, we both get same result, but the question is how they get integer numbers?
• #6
Defennder
Homework Helper
2,591
5
I can think of one possible way to get that: Expand the complex numbers out by binomial theorem and then simplify the expression. This would be very tedious, no doubt.
• #7
908
0
Defennder, I know that I can solve it like that, but it is far more complicated, and the possibility that you may miss some value is very big...
• #8
Hootenanny
Staff Emeritus
Gold Member
9,621
6
I think it was be best to put the two complex numbers into exponential form for the powers and then convert back to Cartesian to perform the subtraction.
• #9
tiny-tim
Homework Helper
25,832
251
Simplify $(1+i\sqrt{2})^5-(1-i\sqrt{2})^5$
Oh come on, guys!
(a + b)^5 - (a - b)^5 = … ?
• #10
86
0
By binom formula?
• #11
HallsofIvy
Homework Helper
41,833
961
Have you tried it?
• #12
Defennder
Homework Helper
2,591
5
Well actually he's asking for a quicker way to get the answer in terms of radicals apart from the binomial theorem or de Moivre's theorem
• #13
63
0
Oh come on, guys!
(a + b)^5 - (a - b)^5 = … ?
Tiny Tim is right to point this out. But the explanation for a general rule of a difference of this sort is in line, for learning purposes of course ;)
If you have an equation of the above kind e.g. (a+b)^n - (a-b)^n , an expansion shows that there will be n+1 terms for (a+b)^n and (a-b)^n. The difference of the two, however, eliminates all but the even terms: for these terms, the coefficients are doubled.
Let's look at an easier example, (a+b)^3 - (a-b)^3 .
Using the binomial theorem (by writing the coefficients as they would appear in Pascale's triangle and by ordering terms by increasing exponents of b and decreasing of a) we get
[a^3 + 3(a^2)(b)+3(a)(b^2) + b^2] - [a^3 - 3(a^2)(b)+3(a)(b^2) -b^3]
= 6 (a^2)(b)+ 2 b^3
As you can see, this is just twice the even-numbered terms in the expansion of (a+b)^3.
The case for (a+b)^5 is analogous.
This is a succinct way of arriving at the result.
• #14
86
0
Ahh... I understand now. So I should also use the binom formula, right?
• #15
tiny-tim
Homework Helper
25,832
251
Ahh... I understand now. So I should also use the binom formula, right?
Theofilius , you keep answering questions with a question …
(a + b)^5 - (a - b)^5 = … ?
… and don't answer with a question … !
• #16
86
0
$$-22i\sqrt{2}$$. I know that, but I should have do that with De Moivre's formula.
• #17
tiny-tim
Homework Helper
25,832
251
$$-22i\sqrt{2}$$. .
erm … no.
… I know that, but I should have do that with De Moivre's formula
eh? … but this is Physicsissuef's question, not yours!
What makes you think he has to use de Moivre?
• #18
86
0
Since I have same problem in my book. And I solve it correctly, why you say no?
• #19
908
0
Yes, since logically we need to solve this problem as simple as possible, but no problem.
• #20
tiny-tim
Homework Helper
25,832
251
oops! it is -22i√2. Sorry!
If you must do it by de Moivre, just put (1 + i√2) = r(cosθ + i sinθ), but leave putting the numbers in until the end.
Then you want r^5[(cos5θ + i sin5θ) - (cos5θ - i sin5θ)], = 2 i r^5 sin5θ.
You know r = √3, and tanθ = √2, so it's fairly easy to work out from that what sin5θ is.
(But the binomial method is probably a more straightforward way of calculating sin5θ, sin 7θ, etc)
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639 | {"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": 0.8127851486206055, "perplexity": 1933.4518915436329}, "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-2021-04/segments/1610703521987.71/warc/CC-MAIN-20210120182259-20210120212259-00009.warc.gz"} |
http://ezyexamsolution.com/jee-mains-temperature-of-heat/ | # JEE Mains Temperature of Heat
mains mathematics Basic Advanced Physics, IIT JEE Mains Physics, Motion Elasticity, Current Elasticity, Fluids, Thermal Properties of Matter, The Ideal Gas Law of Kinetic Theory of Gases, Laws of Thermodynamics, Calorimetry Heat
### Subtopic of
(i) Heat, (ii) Temperature, (iii) Scales of temperature, (iv) Thermometry, (v) Thermometers, (vi) Thermal expansion, (vii) Variation of density with temperature, (viii) Expansion of liquid, (ix) Effect of temperature on upthrust, (x) Anomalous expansion of water, (xi) Expansion of gases, (xii) Application of thermal expansion, (xiii) Thermal capacity and water equivalent, (xiv) Specific heats of solids, water, (xvi) Latent heats, (xvii) Principle of caloriemetry, (xviii) Heating curve
# Heat
The energy associated with configuration and random motion of the atoms and molecules within a body is called internal energy and the part of this internal energy which is transferred from one body to the other due to temperature difference is called heats.
As heat is a form of energy it can be transformed into others and vice-versa.
e.g. Thermocouple converts heat energy into electrical energy, resistor converts electrical energy into heat energy. Friction converts mechanical energy into heat energy. Heat engine converts heat energy into mechanical energy.
Here it is important that whole of mechanical energy i.e. work can be converted into heats but the whole of heats can never be converted into work.
Generally, the temperature of a body rises when the heat’s supplied to it. However, the following two situations are also found to exist.
(i) When heat is supplied to a body either at its melting point or boiling point, the temperature of the body does not change. In this situation, heat supplied to the body is used up in changing its state.
(ii) When the liquid in a thermos flask is vigorously shaken or gas in a cylinder is suddenly compressed, the temperature of liquid or gas gets raised even without supplying heat. In this situation, work done on the system becomes a source of heats energy.
## 01. Temperature and Heat JEE Mains pattern MCQ Paper (Download here PDF.)
### 02. Temperature and Heat JEE Mains pattern MCQ Paper (Download here PDF.)
Joint Entrance Exam Aircraft Maintenance Engineering
#### Thermometry
An instrument used to measure the temperature of a body is called a thermometer.
The linear variation in some physical property of a substance with the change of temperature is the basic principle of thermometry and these properties are defined as a thermometric property (x) of the substance.
x may be (i) Length of liquid in the capillary
(ii) The pressure of gas at constant volume.
(iii) The volume of gas at constant pressure.
(iv) The resistance of a given platinum wire.
In old thermometry, two arbitrarily fixed points ice and steam point (freezing point and the boiling point at 1 atm) are taken to define the temperature scale. In Celsius scale freezing point of water is assumed to be 0°C while boiling point 100°C and the temperature interval between these is divided into 100 equation.
#### Thermal Expansion
When matter is heated without any change in state, it usually expands. According to atomic theory of matter, asymmetry in potential energy curve is responsible for thermal expansion. As with rising in temperature the amplitude of vibration and hence the energy of atoms increases, hence the average distance between the atoms increases. So the matter as a whole expands.
(1) Thermal expansion is minimum in case of solids but maximum in case of gases because the intermolecular force is maximum in solids but minimum in gases.
(2) Solids can expand in one dimension (linear expansion), two dimensions (superficial expansion) and three dimensions (volume expansion) while liquids and gases usually suffer the change in volume only.
### Latent Heat
(1) When a substance changes from one state to another state (say from solid to liquid or liquid to gas or from liquid to solid or gas to liquid) then energy is either absorbed or liberated. This heat energy is called latent heat.
(2) No change in temperature is involved when the substance changes its state. That is, phase transformation is an isothermal change. Ice at 0°C melts into water at 0°C. Water at 100°C boils to form steam at 100°C.
Freezing mixture : If salt is added to ice, then the temperature of mixture drops down to less than 0°C. This is so because, some ice melts down to cool the salt to 0°C. As a result, salt gets dissolved in the water formed and saturated solution of salt is obtained, but the ice point (freezing point) of the solution formed is always less than that of pure water. So, ice cannot be in the solid state with the salt solution at 0°C. The ice which is in contact with the solution starts melting and it absorbs the required latent heat from the mixture, so the temperature of the mixture falls down.
#### The principle of Calorimetry:
When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat will be transferred from a body at higher temperature to a body at lower temperature till both acquire the same temperature. The body at higher temperature releases heat while the body at lower temperature absorbs it, so that
i.e. the principle of calorimetry represents the law of conservation of heat energy.
Conduction: Transfer of energy due to vibration and collision of medium particles without dislocation from their equilibrium position.
Updated: July 20, 2018 — 2:08 pm
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http://www.physicsforums.com/showthread.php?t=487868 | # Probability problem - birthdays
by Char. Limit
Tags: birthdays, probability
PF Gold
P: 1,951
1. The problem statement, all variables and given/known data
44. What is the probability that at least two people in your class (assume a class of 30 students) have the same birthday?
2. Relevant equations
I'm not sure, to be honest.
3. The attempt at a solution
I was helping out my roommate with his probability homework, and this question came up. I'm not sure how to answer it. How is this done? | {"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": 0.811499297618866, "perplexity": 532.1267789634916}, "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-2014-35/segments/1409535922871.14/warc/CC-MAIN-20140901014522-00121-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://tailieu.vn/doc/kalman-filtering-and-neural-networks-chapter-1-kalman-filters-211170.html | # Kalman Filtering and Neural Networks - Chapter 1: KALMAN FILTERS
Chia sẻ: Dph Dph | Ngày: | Loại File: PDF | Số trang:21
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## Kalman Filtering and Neural Networks - Chapter 1: KALMAN FILTERS
Mô tả tài liệu
The celebrated Kalman filter, rooted in the state-space formulation of linear dynamical systems, provides a recursive solution to the linear optimal filtering problem. It applies to stationary as well as nonstationary environments. The solution is recursive in that each updated estimate of the state is computed from the previous estimate and the new input data, so only the previous estimate requires storage.
Chủ đề:
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Lưu | {"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": 0.9489732384681702, "perplexity": 3274.814974427363}, "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-2018-09/segments/1518891813602.12/warc/CC-MAIN-20180221083833-20180221103833-00166.warc.gz"} |
http://www.emathzone.com/tutorials/geometry/area-of-irregular-figures/ | # Area of Irregular Figures
• ### Introduction to Irregular Figures
A figure which is not uniform and regular in shape is where an irregular figure. There are several practical situations where it is necessary to estimate the area of the irregular figures. For example, the areas of plots of land, areas of indicator diagrams of steam engines and area of ships by naval architects, etc. […]
• ### Trapezoidal Rule
To find the area of the as shown in the figure, the base is divided into number of equal intervals of width . The ordinates are accurately measured. The approximation used in this rule is to assume that each strip is equal to the area of a trapezium. Therefore Area of a Trapezium (Sum of […]
• ### Simpsons Rule
The most important rule, in practice, is the Simpson’s Rule, because of its simplicity and accuracy. When more accuracy is required, this rule should be used. To find the area as shown in figure, the base must be divided into an even number of strips of equal width , producing an odd number of ordinates. […] | {"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": 0.9257878065109253, "perplexity": 442.5505435407912}, "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-2017-09/segments/1487501173866.98/warc/CC-MAIN-20170219104613-00048-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-6th-edition/chapter-p-prerequisites-fundamental-concepts-of-algebra-exercise-set-p-2-page-33/3 | College Algebra (6th Edition)
$64$
An exponential expression has a base (the bottom) and an exponent (the top). The exponent indicates how many factors of the base are multiplied together. When the exponent is even, the solution will be positive, even if you start with a negative base. $$(-2)^{6}$$ $$=(-2)\times(-2)\times(-2)\times(-2)\times(-2)\times(-2)$$ $$=4\times(-2)\times(-2)\times(-2)\times(-2)$$ $$=(-8)\times(-2)\times(-2)\times(-2)$$ $$=16\times(-2)\times(-2)$$ $$=(-32)\times(-2)$$ $$=64$$ | {"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": 0.9076613187789917, "perplexity": 202.33560256779776}, "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-2018-39/segments/1537267155942.15/warc/CC-MAIN-20180919063526-20180919083526-00076.warc.gz"} |
http://mathhelpforum.com/trigonometry/126608-prove-following-trigonometrical-identity.html | # Math Help - Prove the following trigonometrical identity-
1. ## Prove the following trigonometrical identity-
(1- tan x + sec x)/(1+tan x + sec x) = cos x/(1+sin x)
2. ## Hii
quick question, are you allowed to manipulate the right side of the equation? Some teachers don´t appreciate that..
What have you tried so far?
Have you tried writing $tan(x)= \frac{sin(x)}{cos(x)}$
and $sec(x) = \frac{1}{cos(x)}$
Doing so will easily give you a common denominator on the left side of the = sign...
3. well i'm afraid i've never tried that....if u know some other method to get this done then plz try that out...
4. Hello, snigdha!
If you know this identity: . $\sec^2\!x - \tan^2\!x \:=\:1$, here's another approach.
Prove: / $\frac{1- \tan x + \sec x)}{1+\tan x + \sec x} \:=\: \frac{\cos x}{1+\sin x}$
We have: . $\frac{1 + \sec x - \tan x}{1 + \sec x +\tan x}$
Multiply by $\frac{\sec x + \tan x}{\sec x + \tan x}\!:\qquad \frac{\sec x + \tan x}{\sec x + \tan x} \cdot\frac{1 + \sec x - \tan x}{1 + \sec x + \tan x}$ . $\;\;=\;\; \frac{(\sec x + \tan x) + (\sec x + \tan x)(\sec x - \tan x)}{(\sec x + \tan x)(1 + \sec x + \tan x)}$
. . . . . $=\; \frac{(\sec x + \tan x) + \overbrace{(\sec^2\!x - \tan^2\!x)}^{\text{This is 1}}}{(\sec x+\tan x)(1 + \sec x +\tan x)}$ . $=\;\;\frac{\overbrace{\sec x + \tan x + 1}^{\text{These cancel}}}{(\sec x + \tan x)\underbrace{(1 + \sec x +\tan x)}_{\text{These cancel}}}$
. . . . . $=\; \frac{1}{\sec x + \tan x} \;\;=\;\;\frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}$
Multiply by $\frac{\cos x}{\cos x}\!:\qquad \frac{\cos x}{\cos x}\cdot \frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \;=\; \frac{\cos x}{1 + \sin x}$
5. Hello snigdha
Originally Posted by snigdha
(1- tan x + sec x)/(1+tan x + sec x) = cos x/(1+sin x)
... and here's an approach that just uses sines and cosines.
$\frac{1-\tan x +\sec x}{1+\tan x + \sec x}=\frac{1-\tan x +\sec x}{1+\tan x + \sec x}\times\frac{\cos^2x}{\cos^2x}$
$=\frac{\cos x(\cos x-\sin x +1)}{\cos^2x+\cos x(\sin x + 1)}$
$=\frac{\cos x(\cos x-\sin x +1)}{1-\sin^2x+\cos x(\sin x + 1)}$
$=\frac{\cos x(\cos x-\sin x +1)}{(1+\sin x)(1-\sin x)+\cos x(\sin x + 1)}$
$=\frac{\cos x(\cos x-\sin x +1)}{(1+\sin x)(1-\sin x+\cos x)}$
$=\frac{\cos x}{1+\sin x}$
6. Soroban and Grandad did it again wih EASE, wow!
!
7. Originally Posted by pacman
Soroban and Grandad did it again wih EASE, wow!
!
... That's what comes of a lifetime of useless activity! | {"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": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9861385822296143, "perplexity": 3167.722930980416}, "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-2014-35/segments/1409535923940.4/warc/CC-MAIN-20140909030308-00011-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://asmedigitalcollection.asme.org/ICONE/proceedings-abstract/ICONE16/48175/817/331799 | The amendment of nuclear reactors regulation law was established and the clearance system was introduced in 2005 in Japan.
This content is only available via PDF. | {"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": 0.8496190309524536, "perplexity": 1431.2534704107284}, "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/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00760.warc.gz"} |
http://mathhelpforum.com/calculus/214130-integrate-e-x-1-e-2x-1-2-dx-print.html | # Integrate e^x(1-e^2x)^(1/2)dx
• March 2nd 2013, 10:25 PM
verasi
Integrate e^x(1-e^2x)^(1/2)dx
Hi, I was wondering if you guys could check my work for indefinite integral of (e^x)(sqrt{1-e^2x)
Let u=e^x
The integral becomes:
integral of sqrt(1-u^2)
Then I do trigonometric substitution where I let u=sinx, and I get:
1/2sin^-1(x) + (x/2)*(1 - x^2)^1/2 + c.
However, the answer in my textbook is -1/3(1-e^(2x))^(3/2).
I was just wondering what I did wrong for the question.
If my complete process is wrong, what should I have done to solve the question?
• March 2nd 2013, 11:51 PM
JJacquelin
Re: Integrate e^x(1-e^2x)^(1/2)dx
Hi !
there is a mishmash in your answer because the same symbol x is used for two different variables :
u=e^x
u=sinx
leading to e^x=sinx which is false.
• March 3rd 2013, 02:18 AM
ibdutt
Re: Integrate e^x(1-e^2x)^(1/2)dx
You are just right put e^x=u that gives e^x dx = du
the question becomes integral of (1-u^2)^1/2 du and that is a standard for for which the integral is given by
(a^2-x^2)^1/2 = x/2 (a^2-x^2)^1/2 + 1/2 a^2 sin^-1 (x/a) + C
• March 3rd 2013, 07:48 AM
Soroban
Re: Integrate e^x(1-e^2x)^(1/2)dx
Hello, verasi!
Quote:
$\int e^x \sqrt{1-e^{2x}}\,dx$
We have: . $\int \sqrt{1-e^{2x}}\,(e^x\,dx)$
Let $u = e^x \quad\Rightarrow\quad du = e^x\,dx$
Substitute: . $\int \sqrt{1-u^2}\,du$
Let $u = \sin\theta \quad\Rightarrow\quad du = \cos\theta\,d\theta$
Substitute: . $\int \sqrt{1-\sin^2\!\theta}\,\cos\theta\,d\theta \;=\;\int\cos^2\!\theta\,d\theta \;=\;\tfrac{1}{2}\int(1 + \cos2\theta)\,d\theta$
. . . . . . . . $=\;\tfrac{1}{2}\left(\theta + \tfrac{1}{2}\sin2\theta\right) + C \;=\;\tfrac{1}{2}(\theta + \sin\theta\cos\theta) + C$
Back-substitute: . $\tfrac{1}{2}\left(\arcsin u + u\sqrt{1-u^2}\right) + C$
Back-substitute: . $\tfrac{1}{2}\left(\arcsin e^x + e^x\sqrt{1-e^{2x}}\right) + C$
• March 3rd 2013, 03:28 PM
verasi
Re: Integrate e^x(1-e^2x)^(1/2)dx
Thanks for the reply, I guess my initial answer was right then. Apparently the textbook has a different answer. It's -1/3(1-e^(2x))^(3/2).
I'm guessing the textbook used a different method then (integrate by part maybe? I didn't try it seemed way too complicated).
T.T wasted 30min on this question trying to figure out what I did wrong
• March 3rd 2013, 10:14 PM
JJacquelin
1 Attachment(s)
Re: Integrate e^x(1-e^2x)^(1/2)dx
Quote:
Originally Posted by verasi
Thanks for the reply, I guess my initial answer was right then. Apparently the textbook has a different answer. It's -1/3(1-e^(2x))^(3/2).
I'm guessing the textbook used a different method then (integrate by part maybe? I didn't try it seemed way too complicated).
T.T wasted 30min on this question trying to figure out what I did wrong
Hi!
Fortunately, any different method leads to the same result ! The results are different if there is a mistake somewhere (in calculus or if the integrals are not the same) See attachment : | {"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": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9897642731666565, "perplexity": 2315.672615991238}, "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-2016-22/segments/1464049282327.67/warc/CC-MAIN-20160524002122-00217-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://openmdao.org/newdocs/versions/latest/features/core_features/working_with_components/units.html | # Specifying Units for Variables¶
As we saw in Declaring Continuous Variables, we can specify units for inputs, outputs, and residuals. There is a units argument to add_input to specify input units, and there are units and res_units arguments on add_output to specify output and residual units, respectively. A complete listing of all available units is given here.
Note
Residual units, if not specified, default to the same units as the output variable. res_units is very rarely specified.
Specifying units has the following result:
1. Unit conversions occur during data passing. For instance, let’s say we have a TimeComp component that outputs time1 in hours and a SpeedComp component takes time2 as an input but in seconds. If we connect TimeComp.time1 to SpeedComp.time2 with hours/seconds specified during the corresponding add_output/add_input calls, then OpenMDAO automatically converts from hours to seconds.
2. The user always gets/sets the variable in the specified units. Declaring an input, output, or residual to have certain units means that any value ‘set’ into the variable is assumed to be in the given units and any time the user asks to ‘get’ the variable, the value is return in the given units. This is the case not only in <Component> methods such as compute, apply_nonlinear, and apply_linear, but everywhere, including the user’s run script.
3. In add_input and add_output, all arguments are assumed to be given in the specified units. In the case of add_input, if units is specified, then val is assumed to be given in those units. In the case of add_output, if units is specified, then val, lower, upper, ref, and ref0 are all assumed to be given in those units. Also in add_output, if res_units is specified, then res_ref is assumed to be given in res_units.
## Units syntax¶
Units are specified as a string that adheres to the following syntax. The string is a composition of numbers and known units that are combined with multiplication (*), division (/), and exponentiation (**) operators. The known units can be prefixed by kilo (k), Mega (M), and so on. The list of units and valid prefixes can be found in the units library.
For example, each of the following is a valid unit string representing the same quantity:
• N
• 0.224809 * lbf
• kg * m / s ** 2
• kg * m * s ** -2
• kkg * mm / s ** 2
Note
If units are not specified, or are specified as None then the variable is assumed to be unitless. If such a variable is connected to a variable with units, the connection will be allowed, but a warning will be issued.
## Example¶
This example illustrates how we can compute speed from distance and time given in km and h using a component that computes speed using m and s.
We first define the component.
import openmdao.api as om
class SpeedComp(om.ExplicitComponent):
"""Simple speed computation from distance and time with unit conversations."""
def setup(self):
def compute(self, inputs, outputs):
outputs['speed'] = inputs['distance'] / inputs['time']
In the overall problem, the first component, c1, defines distance and time in m and s. OpenMDAO handles the unit conversions when passing these two variables into c2, our ‘SpeedComp’. There is a further unit conversion from c2 to c3 since speed must be converted now to m/s.
import openmdao.api as om
from openmdao.core.tests.test_units import SpeedComp
prob = om.Problem()
prob.model.set_input_defaults('c1.distance', val=1., units='m')
prob.model.set_input_defaults('c1.time', val=1., units='s')
prob.model.connect('c1.speed', 'c2.speed')
prob.setup()
prob.run_model()
print(prob.get_val('c1.distance')) # units: km
[0.001]
print(prob.get_val('c1.time')) # units: h
[0.00027778]
print(prob.get_val('c1.speed')) # units: km/h
[3.6]
print(prob.get_val('c2.f')) # units: m/s
[1.] | {"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": 0.8472608923912048, "perplexity": 3092.2369948174914}, "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-2022-05/segments/1642320305420.54/warc/CC-MAIN-20220128043801-20220128073801-00282.warc.gz"} |
https://aviation.stackexchange.com/questions/26479/what-is-the-status-of-notes-in-icao-annexes | # What is the status of “Notes” in ICAO Annexes?
As per the forward to ICAO Annex 11, the status of a "NOTE" in any annex is - "Notes included in the text, where appropriate, to give factual information or references bearing on the Standards or Recommended Practices in question, but not constituting part of the Standards or Recommended Practices." So, does it mean, a note given in the Annex does not hold any authority as it is not a part of the SARPs ? | {"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": 0.8473207950592041, "perplexity": 2160.630775016607}, "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-2019-43/segments/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00427.warc.gz"} |
https://zbmath.org/?q=ai%3Alukashevich.n-a+cc%3A65 | ×
# zbMATH — the first resource for mathematics
Investigation of an approximate solution of the second Painlevé equation. (Russian) Zbl 0692.34010
The equation (1) $$y''=2x^ 3+xy+\alpha$$, $$y(x_ 0)=y_ 0$$, $$y'(x_ 0)=y_ 1$$ is studied. There is given a method for numerical evaluation of singular points of (1).
Reviewer: M.Bartušek
##### MSC:
65J99 Numerical analysis in abstract spaces 65L05 Numerical methods for initial value problems
##### Keywords:
numerical evaluation of singular points | {"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": 0.8707522749900818, "perplexity": 1512.9818913922063}, "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-2021-04/segments/1610703581888.64/warc/CC-MAIN-20210125123120-20210125153120-00043.warc.gz"} |
https://physicsfromplanetearth.wordpress.com/2016/09/ | # Gravitational Radiation 3: GW151226, Encore from LIGO
LIGO (Laser Interferometer Gravitational-Wave Observatory) consists of two gargantuan Michelson-like interferometers, with arms 4 km long, located 3000 km apart in Livingston LA and Hanford WA. Its first observing session began on September 12, 2015. Incredibly, just two days later, it captured the fleeting signal (designated GW150914) produced by the merger of two inspiraling black holes more than 1 billion light years from Earth. (Ref. 1) That groundbreaking discovery, announced on February 11, 2016, ended a 60 year race to directly detect gravitational waves, and came 100 years after the phenomenon was first predicted by Albert Einstein.
On Christmas Day (in the USA), LIGO scientists detected a second event, GW151226 (Ref. 3), proving that the earlier discovery was not a fluke, and that black hole mergers are frequent events. (Ref. 4,5) Since January, LIGO’s sensitivity has been improved, and it will begin a new observing run in Autumn, 2016. This time, the two US detectors will be joined by VIRGO, a similar instrument with arms 3 km long, located near Pisa, Italy. The addition of VIRGO will greatly assist in localizing – by triangulation – the sources of gravitational radiation. In the near future, when LIGO and VIRGO reach their full design sensitivities, they may be capable of detecting one or more black hole collisions daily! Indeed, the 21st Century seems destined to become the era of gravitational wave astronomy.
In this brief post, we will use the mathematical treatment outlined in Ref. 2 to analyze the published data given in Ref. 3 for GW151226. Our educational goal is to add to the collection of homework exercises related to gravitational waves that are suitable for first year physics students. In this way, we hope to inspire instructors to include this exciting topic in their mechanics syllabi.
## Analyzing GW151226
In Ref. 2, the chirp mass of the binary black hole system was defined as $\large \mathfrak{M}=\frac{(m_{1}m_{2})^{3/5}}{(m_{1}+m_{2})^{1/5}}$, where $\large m_{1}$ and $\large m_{2}$ are the masses of the black holes. This quantity is important because it can be found directly from the time dependence of the gravitational wave’s (GW) frequency (Eqn. 10 of Ref. 2):
$\large \mathfrak{M}=\frac{c^{3}}{G}\left ( \frac{5}{96}\: \pi ^{-8/3}f^{-11/3}\: \frac{df}{dt} \right )$ .
(1)
Following Ref. 2, define $\large A=\frac{c^{5}}{G^{5/3}}\: \frac{5}{96}\: \pi ^{-8/3}=5.45\times 10^{56}$ (SI units), and integrate Eqn. 1 over the time interval $\large \Delta t=t_{2}-t_{1}$ to obtain
$\large \mathfrak{M}^{5/3}\Delta t=A\int_{t_{1}}^{t^{_{2}}}f^{-11/3}df=-\frac{3}{8}Af^{-8/3}\mid _{t_{1}}^{t_{2}}$ .
(2)
1. Referring to Fig. 1 above, at 0.80 s before the two black holes coalesce, the frequency of the GW was 39.2 Hz. Later, 0.40 s before the merger, the frequency was 50.3 Hz. (Ref. 6) Calculate the chirp mass and express your answer in terms of $\large M_{Sun}$. (Ans: $\large 9.7\: M_{Sun}$.)
2. Show that the total mass $\large M=m_{1}+m_{2}$ of the binary system before coalescence was at least $\large 22\: M_{Sun}$. (Hint: Let $\large m_{2}=\alpha m_{1}$. Express $\large M$$\large \mathfrak{M}$, and $\large M/\mathfrak{M}$ in terms of $\large m_{1}$ and $\large \alpha$. Show that the ratio is a minimum when $\large \alpha=1$.)
3. The GW frequency at the moment of coalescence of the two black holes was 420 Hz. (See Fig. 1.) Recall from Ref. 2 that the orbital frequency of the binary system is half the frequency of the GW. Use the total mass given in Ref. 3, $\large M=22\: M_{Sun}$, to find the distance between the bodies just before they merged. (Ans: 120 km)
4. Assume that one black hole was twice as massive as the other, i.e., $\large m_{2}=2m_{1}$, which is about what the LIGO team concluded. If the orbits were circular, what were their radii just before coalescence? What were their speeds? Ignore relativity. (Ans: 40 and 80 km; $\large v_{1}=0.35\, c$)
References:
1. B. P. Abbott et al, “Observation of Gravitational Waves from a Binary Black Hole Merger,”, Phys. Rev. Lett. 116, 061102 (2016)
2. this blog: “Gravitational Radiation 2: The Chirp Heard Round the World”
3. B. P. Abbott et al, “Observation of Gravitational Waves from a 22-Solar-Mass Binary Black Hole Coalescence,”, Phys. Rev. Lett. 116, 241103 (2016)
4. A. Cho, “LIGO Detects Another Black Hole Crash,” Science 352, 1374, 17 June 2016
5. D. Castelvecchi, “LIGO Sees a Second Black Hole Crash,” Nature 534, 448, 23 June 2016
6. We thank Jonah Kanner at the LIGO Open Science Center for providing these numbers, which were calculated using the tutorial on the Center’s webpage: https://losc.ligo.org. The numbers obtained from the tutorial differ slightly from those reported in Ref. 3 because the analysis is not identical to the one used for the paper. (In Ref. 3, $\large \mathfrak{M}=8.9\pm 0.3\: M_{Sun}$.) The LIGO Open Science Center is a service of LIGO Laboratory and the LIGO Scientific Collaboration. LIGO is funded by the U.S. National Science Foundation.
# Proxima Centauri b: An Earth-like Neighbor in our Galactic backyard
In August (2016), astronomers at the European Southern Observatory in Chile announced the discovery of an Earth-sized exoplanet orbiting the red dwarf Proxima Centauri, the nearest star to our Sun (Ref. 1, 2). The planet, called Proxima Centauri b, was detected by the radial velocity method: the orbiting planet causes the star to execute a much smaller orbit of its own (as required by momentum conservation), and light from the star is Doppler-shifted due to its radial motion relative to Earth. The discovery of Proxima Centauri b is exciting because its orbit lies within the “Goldilocks,” or temperate, zone of the star, where the planet’s surface temperature would allow water to exist in liquid form – a likely prerequisite for life. Red dwarfs are the most common stars in our galaxy, so if this particular red dwarf harbors a life-supporting planet, it is likely that the Milky Way, which contains over 100 billion stars, is brimming with life.
In this post, we will use the data reported in Reference 1, plus the strategy presented in Section 8.12 of PPE, to calculate the mass and orbital radius of the exoplanet Proxima Centauri b. Our goal is to help students share the excitement of this new discovery through their understanding of introductory mechanics.
## A. Calculating the exoplanet’s mass and orbital radius
In the following, we will assume $\large m\ll M$, where $\large m$ and $\large M$ are the masses of the planet and star, respectively. We will also assume that their orbits are circular, which is consistent with observations, and begin by treating the simplest case where the orbits are viewed edge-on from Earth. (Later, we’ll relax this restriction.) Radial velocity measurements of Proxima Centauri collected over the past 16 years are compiled in Figure 2. Careful analysis indicates that the period $\large T$ of the star’s orbit is 11.186 ± .002 d, and its orbital speed $\large v$ is 1.38 ± 0.02 m/s.
Kepler’s 3rd law states that
$\large \frac{T^{2}}{r^{3}}=\frac{4\pi ^{2}}{GM}$,
where $\large r$ is the orbital radius of the planet. Observations of the star’s luminosity and color indicate that its mass $\large M=0.120\pm .015\, M_{Sun}$.
1. Find the planetary orbit radius $\large r$ (in AU). Hint: use $\large T=11.186/365.25=.03063\, \mathrm{yr}$ and $\large r^{3}\propto MT^{2}$ to compare this orbit to Earth’s orbit about the Sun. (Ans: $\large 0.048 \: \mathrm{AU} = 7.24\times 10^{9}\: \mathrm{m}$.
2. The orbital radius of the star $\large R$ can be found from its period and radial velocity $\large v$$\large vT=2\pi R$. Using $\ 1\: \mathrm{d}=8.64\times 10^{4}\: \boldsymbol{\mathrm{s}}$, find $\large R$. (Ans: $\large 2.12\times 10^{5}\: \mathrm{m}$)
3. The planet and star co-orbit their center of mass. Find the mass of the exoplanet. (Ans: $\large m=MR/r=2.93\times 10^{-5}\, M=6.99\times 10^{24}\, \mathrm{kg}=1.18\, m_{Earth}$)
If the orbit is not viewed edge-on, but is inclined to the line of sight (LOS), the measured radial speed of the star is less than the actual orbital speed $\large v$ used in the above analysis: $\large v_{meas}=vsin(i)$, where i, the angle of inclination, is unmeasurable. See Figure 3. In this case, $\large 2\pi R=Tv=Tv_{meas}/sin(i))$.
4. Prove that, in the general case, the radial velocity measurements are only sufficient to determine $\large msin(i)$. Is the exoplanet mass you found in Question 3 the maximum, or minimum, mass of the planet?
5. Because red dwarfs are much cooler than the Sun, their temperate zones are much closer to them than the Earth is to the Sun. In addition, red dwarfs are less massive than the Sun. Explain why both of these factors favor the detection of small habitable exoplanets. (See Ref. 2 for a brief discussion.)
6. Proxima Centauri subtends an angle of 1.02 ± .08 mas (milliarcseconds) when viewed from Earth. (See PPE, Section 1.7) Find its radius $\large R_{s}$. (Ans: $\large 1.0\times 10^{8}\: \mathrm{m}=0.14\, R_{\mathrm{Sun}}$)
If Proxima centauri b transits, or passes in front of, its star, as viewed from Earth, it will intercept some of the starlight and allow its radius to be determined. Additionally, if the planet has an atmosphere, it will absorb certain wavelengths of light preferentially, enabling astronomers to detect the presence of atmospheric gases such as water, oxygen, carbon dioxide, and methane. Methane (CH4), in particular, is often associated with the presence of life.
The probability of transit is found from the geometric argument outlined in Figure 4. In the figure, the horizontal dotted line is the LOS from Earth, and the vertical line represents the plane perpendicular to the LOS passing through the star’s center. For an orbit to be transiting, its normal (shown as a red line for each of the two orbits depicted) must lie within $\large \pm \theta _{max}=\pm R_{s}/r$ of the vertical plane, where $\large R_{s}$ is the star’s radius. But the normal to the orbit does not need to lie in the plane of the figure: any transiting orbit rotated by any angle (0 – 2π) about the LOS would remain a transiting orbit. The total solid angle available for transiting orbits is therefore $\large 2\pi \cdot 2\theta _{max}=4\pi R_{s}/r$, whereas the total solid angle associated with all possible orientations is $\large 4\pi$. Therefore, the probability of a transiting orbit is $\large R_{s}/r$. (Check: if $\large r=R_{s}$, then the probability equals 1.) For a fuller discussion, see http://kepler.nasa.gov/Science/about/characteristicsOfTransits/ .
7. How probable is it that Proxima Centauri b transits its star? (Ans: 1.5%)
## B. How Warm is Proxima Centauri b?
The following questions are only suitable for students who have studied blackbody radiation in a course of thermal physics.
8. The luminosity of a star (its total radiated power) is proportional to its surface area and the fourth power of its surface temperature: $\large L_{star}\propto R_{star}^{2}T_{star}^{4}$. Proxima Centauri’s surface temperature is about 3050 K, whereas the Sun’s is 5780 K. Show that $\large L_{Proxima}\simeq 0.0015\, L_{Sun}$. This indicates that the temperate zone lies much closer than 1 AU to the star.
9. The fraction of the star’s radiation that is intercepted by a planet is equal to the solid angle subtended by the planet divided by 4π: $\large \pi R_{p}^{2}/4\pi r^{2}$. The absorbed radiation warms the planet, which reaches a steady state temperature $\large T_{p}$ when the blackbody radiation emitted by the planet equals the radiation absorbed from the star:
$\large 4\pi R_{p}^{2}\sigma _{B}T_{p}^{4}=L_{Proxima}\pi R_{p}^{2}/4\pi r^{2}$, where $\large \sigma _{B}$ is called the Stefan-Boltzmann constant. (Its value is not needed to answer this question.) Use this information to compare the surface temperature of Proxima Centauri b to that of the Earth, assuming both planets are perfect black bodies (or have the same albedo). (Ans: $\large T_{p}/T_{Earth}=0.90$)
References
1. G. Anglada-Escudé et al, Nature 536, 437 (25 August, 2016)
2. A. P. Hatzes, Nature 536, 408 (25 August, 2016) | {"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": 60, "mimetex.cgi": 0, "/images/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": 0.8541509509086609, "perplexity": 2746.6967839462477}, "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-2018-09/segments/1518891814311.76/warc/CC-MAIN-20180223015726-20180223035726-00730.warc.gz"} |
https://utsv.net/solid-mechanics/4-stress/cauchy-equation-of-motion | Cauchy Equation of Motion
The equation of motion can be expressed in terms of the applied stress, body forces, mass, and acceleration:
(1)
In index notation: ;
or
(2)
Eq. 2 is the continuum mechanics version of Newton’s Second Law, often called the balance of linear momentum.
Note that the step from eq. 1 to eq. 2 is not trivial, since both and do in fact change with time. The complete derivation, starting with eq. 1 and concluding with eq. 2, is given in Appendix B.2. [Holzapfel] calls this derivation (operation) the “material time derivative of a spatial field.”
Thus,
(3)
Note that the “localization theorem” states that , if and are arbitrary.
(4)
Eq. 4 are the Cauchy Equations of Equilibrium. It may not yet be clear which term contains the applied forces and which term contains the quantities analogous to “.” It turns out that the term will contain applied forces and prescribed displacements, along with all of the internal force and displacement quantities that constitute “” for the element. When an entire system is analyzed, which includes many elements, the global equation of motion should be satisfied, naturally, so long as the geometry is accurately represented by the elements, the stiffness and strength properties of the material are defined for each element, and the boundary conditions are correctly assigned for each element. There could be other issues that arise as well, due to simplifications inherent (but quite necessary) in the finite element analysis method (FEA), but these issues will be left to texts that cover FEA in detail. In fact, among the aforementioned element-related issues, this text will only cover material elastic stiffness in detail. Material behavior at the limit state (failure) is covered in texts on plasticity, for example, and topics relating to element geometries, prescribed degrees of freedom or prescribed forces at “nodes” (i.e. boundary conditions), or other issues related to the “assembly” of finite elements will be left to texts devoted to the topic of FEA implementation.
. then eq. 4 reduces to static equilibrium. An alternative derivation of stress equilibrium, which doesn’t use index notation, can be found in [Ugural].
note: This is the basic differential equation used in FEA, though the connection to FEA will not really be clear until we start developing constitutive equations relating stresses and strains (along with the above equation and strain-displacement relationships previously presented). | {"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": 0.8828898668289185, "perplexity": 635.3041654347481}, "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-2019-04/segments/1547583874494.65/warc/CC-MAIN-20190122202547-20190122224547-00105.warc.gz"} |
https://brilliant.org/problems/estimating-zeta/ | # Estimating Zeta
The Riemann zeta function is defined as <br />
$$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n ^ s} = \frac{1}{1 ^ s} + \frac{1}{2 ^s} + \frac{1}{3^s} + ...$$ <br />
We can estimate Riemann zeta values by using large numbers in place of infinity as the upper limit of the series.
Let Z be an estimation of $$100\sqrt{6\zeta(2)}$$, calculated using 1000000 as infinity.
What is Z, rounded to the nearest integer?
× | {"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": 0.9963352084159851, "perplexity": 1615.678237119505}, "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-2017-30/segments/1500549427749.61/warc/CC-MAIN-20170727062229-20170727082229-00632.warc.gz"} |
https://www.physicsforums.com/threads/networks-on-a-torus.51359/ | # Networks on a Torus
1. Nov 4, 2004
### Zurtex
I'm asked to consider regular networks on a torus. I'm given that V - E + F = 0. I need to show it is impossible to have a regular network on a torus where the faces are pentagons; I don't understand that at all. Surely it is easy to have pentagons as faces… All you would need to is draw a pentagon on it, please tell me where I am not getting this.
2. Nov 4, 2004
### HallsofIvy
Staff Emeritus
I suspect that what they mean is a network that completely covers the torus: every point on the torus in on or inside some pentagon.
Suppose your network consisted of n pentagons. Then there are n faces. How many edges are there? (Each pentagon has 5 edges, but each edge is shared by two pentagons.) How many vertices are there? (Each pentagon has 5 vertices but each vertex is shared by 3 pentagons.)
Now plug those numbers into the Euler equation.
3. Nov 4, 2004
Thanks
4. Nov 7, 2004
### Zurtex
Erm writing this out, I'm confused again. How can all shapes be a pentagon in a regular network anyway? Pentagons don't tessellate. | {"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": 0.8199779987335205, "perplexity": 739.4419311359001}, "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/1487501170569.99/warc/CC-MAIN-20170219104610-00479-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://yuezhuang.me/2018/07/24/llvm/2018-07-24/ | Now, let us look inside CBMC and figure out how cbmc works and how does it generate VC. We first download the slides explaining the first version (v1.0) of cbmc given by Daniel. That will help us understand cbmc’s structure.
## cbmc
### 前端处理 frontend
cbmc 的目标是分析 C/C++ 或者 JAVA 程序。CBMC 的第一步是输入程序，生成 CFG。关于如何生成 CFG，是编译原理的基础课程（我们可以用 CIL 来初步处理程序获得 CFG，另外，frama-c 平台也提供了现成的模块，可以直接获取）。
cbmc aims at the analysis of programs given in a commodity
programming language such as C, C++, or Java As the first step, it transforms the program into a control flow graph (CFG)
When we have the cfg, that means we have the execution paths of the programs. Thus, we can obtain the conjunction ($\phi_{i}$) of the condition in each joint point for each path $\phi_{i}$. The path will be executed if $\phi_{i}$ is valid.
let us look at the example in the above graph. We select one path and obtain the corresponding formula. This formula should be passed to a decision procedure and we obtain a satisfying assignment or the formula is invalid.
### SAT
SAT 的求解技术能力已经有了明显的提升。目前的很多研究是提高 word-level reasoningarray decision procedures 的效率。
## Transition System
A transition system $(S, T, I)$ is defined as follows:
• a set of states $S$;
• a transition relation $T: S \rightarrow S$;
• a set of initial states $S_{0}\subseteq S$.
A program is modelled as a transition system. Each state of the program is an evaluation of all variables in the program. Each state may has many possible successors, i.e., the transition relation is non-deterministic.
Example1 : $S = \{a,b,c\}$, $I = \{a\}$ and $T = \{ \langle a,b\rangle, \langle b,b\rangle, \langle b,c\rangle\}$
## 总结
cbmc 的步骤总共有三步：
1. 前端处理，获得程序的 cfg
2. 展开 unwind cfg，获得路径的公式
3. 利用 SAT/SMT 求解公式 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 22, "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": 0.9105541110038757, "perplexity": 2344.583039534158}, "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-2018-43/segments/1539583510998.39/warc/CC-MAIN-20181017044446-20181017065946-00194.warc.gz"} |
http://www.analyzemath.com/Geometry/sector_circle.html | # Sectors and Circles Problems
Problems on sectors and circles with detailed solutions.
Problem 1: In the figure below, the length of arc AB is equal to twice the length of radius r of the circle. Find the area of sector OAB in terms of the radius r. Solution to Problem 1: The formula for the arc length is given by arc length AB = r * angle AOB arc length AB is given and is equal to 2r, hence 2r = r * angle AOB Solve for angle AOB to obtain angle AOB = 2 radians The area of the sector is given by Area = (1/2) * angle AOB * r 2 = (1/2) * 2 * r 2 = r 2 Problem 2: The angle of a sector in a given circle is 40 degrees and the area of the sector is equal to 20 cm 2. Calculate the arc length of the sector. Solution to Problem 2: The formula for the area of a sector is Area = (1 / 2) * angle of sector * r 2 The area and the angle are given. The angle is given in degrees and has to be converted into radians. 20 = (1 / 2) * (40 * Pi / 180) * r 2 Solve for radius r to obtain r = square root [ 2 * 20 * 180 / (40 Pi) ] = square root [ 180 / Pi ] The arc length is given by arc length of sector = angle of sector * r = [ 40 * Pi / 180 ] * square root [ 180 / Pi ] = [ Pi / 4.5 ] * square root [ 180 / Pi ] = 5.3 cm (rounded to 1 decimal place) Problem 3: In the figure below, AB and DC are arcs of concentric circles with center O. The perimeter of figure ABCD is 22 cm. Calculate a) angle AOB b) The area of figure ABCD. Solution to Problem 3: The perimeter of ABCD is given by perimeter = arc AB + 6 + arc DC + 6 = 22 If we let x be the size of the sector central angle, arc AB and arc DC are given by (formula for arc length of sector) arc AB = 2 * x and arc DC = 8 * x Substitute in the above perimeter formula 2x + 6 + 8x + 6 = 22 Solve for x x = 1 radian The area of figure ABCD is found by subtracting the area of the small sector from the area of the larger sector area of ABCD = (1 / 2) 8 2 * 1 - (1 / 2) 2 2 * 1 = 30 cm 2 More references on geometry tutorials and problems. Geometry Tutorials, Problems and Interactive Applets. Triangle Problems | {"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": 0.8867814540863037, "perplexity": 249.0545543247318}, "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-2017-51/segments/1512948521188.19/warc/CC-MAIN-20171213030444-20171213050444-00157.warc.gz"} |
https://www.physicsforums.com/threads/a-question-on-goldstein-and-dalemberts-principle.834084/ | # A Question on Goldstein and D'Alembert's Principle
1. Sep 23, 2015
### coca-cola
Hey all,
I am reading Goldstein and I am at a point where I can't follow along. He has started with D'Alembert's Principle and he is showing that Lagrange's equation can be derived from it. He states the chain rule for partial differentiation:
$$\frac{d\textbf{r}_i}{dt}=\sum_k \frac{\partial \mathbf{r}_i}{\partial q_k}\dot{q}_k+\frac{\partial \mathbf{r}_i}{\partial t}$$
Then he states, by the equation above, that:
$$\frac{d}{dt}\frac{d\mathbf{r}_i}{dq_j}=\sum_k \frac{\partial^2 \textbf{r}_i}{\partial q_j \partial q_k}\dot{q}_k+\frac{\partial^2 \mathbf{r}_i}{\partial q_j\partial t}$$
He further states from the first equation that:
$$\frac{\partial \mathbf{v}_i}{\partial \dot{q}_j}=\frac{\partial \mathbf{r}_i}{\partial q_j}$$
I have tried to connect the dots but I cannot succeed. Any insight is greatly appreciated. Thanks!
2. Sep 23, 2015
### lautaaf
Think the first relation as a definition for an operator
$$\frac{d}{dt}=\sum_k \frac{\partial}{\partial q_k}\dot{q}_k+\frac{\partial }{\partial t}$$
The second equation follows immdiately from applying $\frac{d}{dt}$ to $\frac{d\mathbf{r}_i}{dq_j}$ and the last one from applying $\frac{\partial}{\partial \dot q_j}$ to the first equation.
3. Sep 23, 2015
### coca-cola
Thanks!
So is the definition simply the chain rule of a function that depends on q_1, q_2,...q_N, and t? If the function had no explicit dependence on t, even though the generalized coordinates did, would you simply drop the partial with respect to t?
4. Sep 24, 2015
### lightarrow
Yes, certainly.
--
lightarrow
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https://www.physicsforums.com/threads/rectilinear-motion-of-particle.288126/ | # Rectilinear Motion of Particle
1. Jan 28, 2009
1. The problem statement, all variables and given/known data
3. The acceleration of a particle is directly proportional to the time t. At t=0, the velocity of a particle = -12 m/s. Knowing that v=0 and x = 15 when t=4s, write the equation of the motion.
Here is my solution, could Somebody check it?
Text is in Polish, sorry for that
2. Jan 28, 2009
### turin
I'm still waiting for the attachment approval. In the meantime, dzien dobry.
3. Jan 29, 2009
Dzien dobry Attachment approval? So you cannot look into it?
4. Jan 29, 2009
### turin
You have explained the problem well enough, and I believe I can solve it, but it is against forum rules for me to provide any help until I see your attempt at a solution. Is there a reason why you need to put it into pdf format? I understand that using the tex feature on this forum is a bit frustrating, but you can just try to write out your solution in plane text here, and I will try to follow it.
5. Jan 30, 2009
### turin
When you integrated v to find x as a function of t, you did something wrong. I advise you to put limits on your integration, and put explicitly what is the differential quantity over which you're integrating.
6. Jul 6, 2009
### RyaniC
do you have any other simplier sulotion for rectilinear motion??
7. Jul 6, 2009
### turin
adashiu's approach is the simplest approach that I'm aware of. In fact, off the top of my head, I can't think of another way to do it.
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http://tex.stackexchange.com/questions/102084/annoying-paragraph-spacing-issue-with-memoir | # Annoying paragraph spacing issue with memoir
I have a spacing issue while using the memoir class. The main problem is that after the first paragraph there is a line skip and an indentation and afterwards, all the paragraphs are indented properly and there is no line skip. Here is a MWE:
\documentclass[11pt,twoside,openright]{memoir}
\usepackage[size=pocket,trim,bleed]{createspace}
%\usepackage[paperwidth=4.25in, paperheight=6.875in,bindingoffset=.75in]{geometry}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{tgtermes}
\usepackage{mathpazo}
\usepackage[protrusion=true,expansion=true]{microtype}
\usepackage{lipsum}
%\usepackage{type1cm}
%\usepackage{lettrine}
%\checkandfixthelayout
% See the Memoir customise'' template for some common customisations
% Don't forget to read the Memoir manual: memman.pdf
%\title{TITLE OF BOOK}
%\author{NAME OF AUTHOR}
%\date{} % Delete this line to display the current date
%% BEGIN TITLE
\makeatletter
\def\maketitle{%
\null
\thispagestyle{empty}%
\vfill
\begin{center}\leavevmode
\normalfont
{\LARGE\raggedleft \@author\par}%
\hrulefill\par
{\huge\raggedright \@title\par}%
\vskip 1cm
% {\Large \@date\par}%
\end{center}%
\vfill
\null
\cleardoublepage
}
\makeatother
\author{AUTHOR}
%\author{NAME OF AUTHOR}
\title{TITLE}
%\date{}
%%% BEGIN DOCUMENT
\begin{document}
\pdftitle{TITLE}
\pdfauthor{AUTHOR}
\pdfsubject{SUBJECT}
\chapterstyle{bianchi}
\let\cleardoublepage\clearpage
\maketitle
\frontmatter
\null\vfill
\begin{flushleft}
Copyright \copyright
%\textit{TITLE}
All rights reserved.
ISBN:
ISBN--13:
\bigskip
\end{flushleft}
\let\cleardoublepage\clearpage
\clearpage
\tableofcontents
\chapter{Introduction}
\lipsum[1]
\lipsum[2]
\lipsum[3]
\mainmatter
\sloppy
%Input all of the chapter here
\backmatter
\end{document}
I also get an underfull \vbox which is something I do not know how to fix. Maybe the two are related?
EDIT:
I changed the document to fit createspace's guidelines but now the \vbox is back. I have attempted to use \vfil option but that is not working. What could be the issue?
\documentclass[11pt,twoside,openright]{memoir}
\usepackage{geometry}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{lipsum}
\geometry{
paperheight=8.00in,
top=.75in,
bottom=.75in,
paperwidth=5.25in,
inner=.38in,
outer=.75in,
bindingoffset=.75in,
}
\begin{document}
\chapterstyle{bianchi}
\tableofcontents*
\let\cleardoublepage\clearpage
\clearpage
\pagenumbering{roman}
\chapter{Introduction Chapter}
\vfil
\lipsum[1-10]
\mainmatter
\sloppy
%Input all of the chapter hermane
\pagenumbering{arabic}
\chapter{Awesome chapter}
\vfil
\lipsum[1-20]
\end{document}
UPDATE
I used \raggedbottom in the preamble as suggested before and that seems to have rid me of all of \vbox errors. I will look through the entire file and update again.
UPDATE 2
\raggedbottom fixed 99.99% of the issues except I get 1 overfull \vbox. The exact error is:
Overfull \vbox (13.59999pt too high) has occurred while \output is active []
-
## 1 Answer
The problem is that your settings and packages define a stretchable dimension for \parskip, i.e. a dimension which can increase or decrease (within some limits) in order of producing always a completely filled page (which is the \vbox).
TeX had troubles finding the point at which to end your first page. At some point, there was not enough space to put another line, but if he breaks the page at that point, there is not enough material to produce a completely filled page (i.e: the last line will not be aligned with the bottom). So it uses the stretchable space he can find in the page. Apparently the only stretchable space he found was \parskip. Even using that, the stretchabilty allowed by \parskip was not enough, so TeX had to stretch it beyond its limit (hence the Underful box message).
The "problem" has an easy solution:
1. Set \parskip=0pt. This disables its stretchability and the distance between paragraphs will be always zero, so when TeX cannot fill the page completely, it will insert space at the bottom and complain about Underfill vbox.
2. Put \raggedbottom in the preamble. This adds an infinitely stretchable space at the end of each page. This solution produces the same results than 2, but no Underfull messages will be produced, because that infinite glue will "absorb" all deficit.
Anyway, I suspect that the problem is that createspace style is trying to set all settings so that all pages have lines in a "grid", i.e. the baselines of all text are the same page after page. Probably first page breaks somehow the layout because of the title.
If I'm right, then you won't get more underfull boxes nor extra parskips in the remaining of the document, until the next chapter heading, which can break the layout again.
If this is the case, the better solution would be to insert stretchable space around the chapter heading. This way in those problematic pages TeX would choose that space to "absorb" the deficit, instead of choosing parskip. To do so:
\chapter{Introduction}
\vfil
For more information about stretchable glue you can read my answer to another question.
# Update
I did some experiments, and I concluded that I guessed correctly the causes and best solution of the problem.
The cause is that createspace tries to typeset all pages on the same grid, by enforcing the alignment of the bottom line of each page. However, the first page is different because of the chapter title, which introduces an spacing which is not multiple of baselineskip, and causes problems of alignment.
The following screenshot shows how, in your original document, the baselines of the first and second page do not match:
Using \raggedbottom technique makes the gap between paragraph to disappear, but worsens the problem of the baseline matching, because now the bottom page is not uniform (and the middle lines still do not match correctly):
Using \vfil after \chapter solves all the issues:
-
Thanks for the fantastic answer! However, I have one question. In the preamble, I should add \raggedbottom and then after the chapter heading, I should add \vfil right? – Jeel Shah Mar 12 '13 at 10:43
@gekkostate Updated answer, use only the \vfil trick and forget about \raggedbottom. – JLDiaz Mar 12 '13 at 10:56
Thanks for the response, it's working now! – Jeel Shah Mar 12 '13 at 11:10
After inserting the original text, the problem is back. With ipsum, it's working perfectly but with the new text it is not. It's evident that there is a problem with my text but what could it be? I have 1 \begin{quote} and that's it. – Jeel Shah Mar 12 '13 at 17:43
@gekkostate quote will introduce glue for example \setlength\partopsep{2\p@ \@plus 1\p@ \@minus 1\p@} in 10pt article class. – David Carlisle Mar 13 '13 at 1:34 | {"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": 0.9129487872123718, "perplexity": 2061.051383177418}, "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-2016-07/segments/1454701158609.98/warc/CC-MAIN-20160205193918-00141-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.qb365.in/materials/stateboard/11th-business-maths-differential-calculus-model-question-paper-4832.html | #### Differential Calculus Model Question Paper
11th Standard
Reg.No. :
•
•
•
•
•
•
Business Maths
Time : 01:00:00 Hrs
Total Marks : 50
5 x 1 = 5
1. Let $f\left( x \right) =\begin{cases} { x }^{ 2 }-4x\quad ifx\ge 2 \\ x+2\quad ifx<2 \end{cases}$, then f(5) is
(a)
-1
(b)
2
(c)
5
(d)
7
2. The graph of y = ex intersect the y-axis at
(a)
(0,0)
(b)
(1,0)
(c)
(0,1)
(d)
(1,1)
3. Which one of the following functions has the property f (x) = $f\left( \frac { 1 }{ x } \right)$
(a)
$f\left( x \right) =\frac { { x }^{ 2 }-1 }{ x }$
(b)
$f\left( x \right) =\frac { 1-{ x }^{ 2 } }{ x }$
(c)
f(x) = x
(d)
$f\left( x \right) =\frac { { x }^{ 2 }+1 }{ x }$
4. The graph of f(x) = ex is identical to that of
(a)
f(x) = ax, a > 1
(b)
f(x) = ax, a < 1
(c)
f(x) = ax, 0 < a < 1
(d)
y = ax +b, a $\ne$ 0
5. $\lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } } =$
(a)
e
(b)
nx(n-1)
(c)
1
(d)
0
6. 5 x 2 = 10
7. Determine whether the following functions are odd or even?
f(x)=x+x2
8. Let f be defined by f(x) =x3-kx2+2x,x ∈ R,Find k,if 'f' is an odd function.
9. For $f(x)=\frac { x-1 }{ 3x+1 }$ ,write the expression of $f\left( \frac { 1 }{ x } \right)$ and $\frac { 1 }{ f(x) }$
10. Evaluate $\underset { h\rightarrow 0 }{ lim } \frac { \sqrt { x+h } -\sqrt { x } }{ h }$
11. Differentiate (sec x -1) (sec x +1)
12. 5 x 3 = 15
13. Differentiate sin2 x with respect to x2.
14. Evaluate the following $\lim _{ x\rightarrow \infty }{ \frac { 2x+5 }{ { x }^{ 2 }+3x+9 } }$
15. Find y2 of the following function x = a cos $\theta$, y = a sin $\theta$
16. Find $\frac{dy}{dx}$ if x = 15(t - sin t); y = 18(1 - cos t).
17. If ey (x + 1) = 1, show that $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ \left( \frac { dy }{ dx } \right) }^{ 2 }$
18. 4 x 5 = 20
19. Draw the graph of the following function f(x)=e2x
20. Draw the graph of the following function f(x)=e-2x
21. If $\begin{matrix} \underset { x\rightarrow 1 }{ lim } & \frac { { x }^{ 4 }-1 }{ x-1 } \end{matrix}=\begin{matrix} \underset { x\rightarrow k }{ lim } & \frac { { x }^{ 3 }-{ k }^{ 3 } }{ { x }^{ 2 }-{ k }^{ 2 } } \end{matrix}$,then find the value of K
22. If $x=a\left( t+\frac { 1 }{ t } \right) ;y=a\left( t-\frac { 1 }{ t } \right)$ show that $\frac { dy }{ dx } =\frac { x }{ y }$ | {"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": 0.8682140111923218, "perplexity": 2323.599239209901}, "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-2019-51/segments/1575540508599.52/warc/CC-MAIN-20191208095535-20191208123535-00442.warc.gz"} |
https://mathematica.stackexchange.com/questions/19638/why-doesnt-mathematica-make-an-obvious-simplification/19643 | # Why doesn't Mathematica make an obvious simplification?
I am doing some calculation with summation and the Kronecker symbol. Here are my steps :
$Assumptions = k1 ∈ Reals && k2 ∈ Reals && k3 ∈ Reals && p1 ∈ Reals && p2 ∈ Reals && p3 ∈ Reals k = {k1, k2, k3}; p = {p1, p2, p3}; d[i_, j_] := KroneckerDelta[i, j] proj[i_, j_, k1_, k2_, k3_] := d[i, j] - (d[i, 1]*k1 + d[i, 2]*k2 + d[i, 3]*k3)* (d[j, 1]*k1 + d[j, 2]*k2 + d[j, 3]*k3)/ (k1^2 + k2^2 + k3^2) test1 = proj[i, j, k1, k2, k3]*proj[i, j, p1, p2, p3]; test2 = Sum[Sum[test1, {i, 1, 3}], {j, 1, 3}] test2 // Expand To explain the steps: 1) I define$\vec{k}$and$\vec{p}$with real components. 2) I define a projector$P_{ij} \left( \vec{k} \right) = \delta_{ij} - \frac{k_i k_j}{k^2}$. 3) I compute a summation on the repeated subscript. After the last step, I have a relatively big expression, the product of k and p components. It looks like $$3-\frac{a}{a+b+c} - \frac{b}{a+b+c} - \frac{c}{a+b+c} +...-...$$ The a, b and c stands for k1, k2 and k3 (or p1, 2, 3). Now the question: why doesn't Mathematica make the simplification because, as anyone can see, the preceding expression can be simplified to$2 +...-...$Is the problem linked to the Expand operation? How can I make the simplification I want?. I thought of using /. to do it, but that doesn't work either. I hope someone will understand my question! • You can try using Simplify and related, but what you think it's a simpler expression will not necessarily match Mathematica's. – b.gates.you.know.what Feb 15 '13 at 10:23 • +1 just for formatting your code:) Simplify@test2 works nicely and returns (2 k2 k3 p2 p3 + 2 k1 p1 (k2 p2 + k3 p3) + k1^2 (2 p1^2 + p2^2 + p3^2) + k2^2 (p1^2 + 2 p2^2 + p3^2) + k3^2 (p1^2 + p2^2 + 2 p3^2))/((k1^2 + k2^2 + k3^2) (p1^2 + p2^2 + p3^2)) – Ajasja Feb 15 '13 at 10:34 ## 4 Answers You have to explicitly tell Mathematica to simplify expressions. You can do this using Simplify or FullSimplify Simplify@test2 (2 k2 k3 p2 p3 + 2 k1 p1 (k2 p2 + k3 p3) + k1^2 (2 p1^2 + p2^2 + p3^2) + k2^2 (p1^2 + 2 p2^2 + p3^2) + k3^2 (p1^2 + p2^2 + 2 p3^2))/((k1^2 + k2^2 + k3^2) (p1^2 + p2^2 + p3^2)) • Thank you for your answers. The problem when I use Simplify is that the form of the result is not "usable". I mean in the sense I want to use it. Because the calculation must lead to$1 - (\vec{k}.\vec{p})^2/(k^2*p^2)$. And with simplify, Mathematica doesn't do the factorization I want. I think I have to find an other way to do it, I may already have an idea. I thought I forget an assumption to enable Mathematica to do automatically this simplification. But it won't do what I didn't ask ! – Lalylulelo Feb 15 '13 at 10:52 I would use symbolic tensors instead of explicit tensors. First, let's reformulate your question using symbolic tensors: proj[k_] /; TensorRank[k] == 1 := With[{d = First @ TensorDimensions[k]}, Inactive[IdentityMatrix][d] - TensorProduct[k, k]/k.k ]$Assumptions = (k|p) ∈ Vectors[3];
test = TensorContract[TensorProduct[proj[k], proj[p]], {{1, 3}, {2, 4}}];
test //TeXForm
$\operatorname{TensorContract}\left[\left(\operatorname{IdentityMatrix}[3]-\frac{k\otimes k}{k.k}\right)\otimes \left(\operatorname{IdentityMatrix}[3]-\frac{p\otimes p}{p.p}\right),\left( \begin{array}{cc} 1 & 3 \\ 2 & 4 \\ \end{array} \right)\right]$
Note that TeXForm doesn't yet support Inactive, which is why it doesn't appear above. Also, I've overriden the default TeXForm formatting of \[TensorProduct] so that it looks like \otimes.
Mathematica doesn't (yet) have a symbolic identity tensor, so I use Inactive[IdentityMatrix][3] as a substitute. This means that TensorReduce can't really be used to simplify the output since it doesn't have support for my identity tensor substitute. I've written a package that helps augment TensorReduce to work with these identity tensors, which can be found here:
PacletInstall[
"TensorSimplify",
"Site" -> "http://raw.githubusercontent.com/carlwoll/TensorSimplify/master"
]
Once installed, you can load the package with:
<<TensorSimplify
Let's see this package in action on the above expression. First we need to use TensorReduce (TensorExpand would also work):
r1 = TensorReduce[test];
r1 //TeXForm
$-\frac{\operatorname{TensorContract}\left[k\otimes k\otimes \operatorname{IdentityMatrix}[3],\left( \begin{array}{cc} 1 & 3 \\ 2 & 4 \\ \end{array} \right)\right]}{k.k}-\frac{\operatorname{TensorContract}\left[p\otimes p\otimes \operatorname{IdentityMatrix}[3],\left( \begin{array}{cc} 1 & 3 \\ 2 & 4 \\ \end{array} \right)\right]}{p.p}+\operatorname{TensorContract}\left[\operatorname{IdentityMatrix}[3]\otimes \operatorname{IdentityMatrix}[3],\left( \begin{array}{cc} 1 & 3 \\ 2 & 4 \\ \end{array} \right)\right]+\frac{\operatorname{TensorContract}\left[k\otimes p,\left( \begin{array}{cc} 1 & 2 \\ \end{array} \right)\right]^2}{k.k p.p}$
Using TensorReduce causes the symbolic identity tensor to be contracted. The package function IdentityReduce can then be used to eliminate these contracted identity tensors:
r2 = IdentityReduce[r1];
r2 //TeXForm
$\frac{\operatorname{TensorContract}\left[k\otimes p,\left( \begin{array}{cc} 1 & 2 \\ \end{array} \right)\right]^2}{k.k p.p}-\frac{\operatorname{TensorContract}\left[k\otimes k,\left( \begin{array}{cc} 1 & 2 \\ \end{array} \right)\right]}{k.k}-\frac{\operatorname{TensorContract}\left[p\otimes p,\left( \begin{array}{cc} 1 & 2 \\ \end{array} \right)\right]}{p.p}+3$
Finally, the above expression is a mix of equivalent Dot and TensorContract objects. Using a common form will enable further simplifications. The less verbose version is obtained using the Dot representation (obtained by using the package function FromTensor):
FromTensor[r2]
1 + (k.p)^2/(k.k p.p)
I believe this is the form you're looking for. The package function TensorSimplify does all of the above steps:
TensorSimplify[test]
1 + (k.p)^2/(k.k p.p)
• Impressive! It was exactly the form I was looking for. I will have to test it. I think I figured out a solution by using replacements but yours is much more elegant. Thanks! – Lalylulelo Nov 12 '17 at 9:04
Not sure precisely what you want, but maybe it's this:
x = test2 // Expand;
x1 = PolynomialReduce[
Numerator[Together[x]], {Denominator[Together[x]]}, Variables[x]];
x2 = PolynomialReduce[x1[[2]], {k1^2 + k2^2 + k3^2}, Variables[x]];
x3 = PolynomialReduce[x2[[2]], {p1^2 + p2^2 + p3^2}, Variables[x]];
y = x1[[1, 1]] + x2[[1, 1]]/(p1^2 + p2^2 + p3^2) +
x3[[1, 1]]/(k1^2 + k2^2 + k3^2) +
x3[[2]]/((p1^2 + p2^2 + p3^2) (k1^2 + k2^2 + k3^2))
Here's the output in reverse order; there's an issue with copy/paste:
$\frac{2 \text{k1} \text{k2} \text{p1} \text{p2}+2 \text{k1} \text{k3} \text{p1} \text{p3}+2 \text{k2}^2 \text{p2}^2+\text{k2}^2 \text{p3}^2+2 \text{k2} \text{k3} \text{p2} \text{p3}+\text{k3}^2 \text{p2}^2+2 \text{k3}^2 \text{p3}^2}{\left(\text{k1}^2+\text{k2}^2+\text{k3}^2\right) \left(\text{p1}^2+\text{p2}^2+\text{p3}^2\right)}+\frac{-\text{k2}^2-\text{k3}^2}{\text{k1}^2+\text{k2}^2+\text{k3}^2}+\frac{-\text{p2}^2-\text{p3}^2}{\text{p1}^2+\text{p2}^2+\text{p3}^2}+2$
If you know approximately what denominators are obtained, you coud use command Collect[expression, factor]`. | {"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": 0.8146632313728333, "perplexity": 4738.26158984011}, "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-2019-47/segments/1573496669431.13/warc/CC-MAIN-20191118030116-20191118054116-00233.warc.gz"} |
https://brilliant.org/problems/placing-satellites/ | # Placing satellites
In a geostationary orbit around the earth satellites are to be placed at equal distances from each other in such a manner that each satellite can see every other during any instant of its motion.Find the maximum no of satellites that can be placed in such a manner.Assume that the space station cant send more than 10 satellites.Assume that the orbits are circular.
× | {"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": 0.8236696124076843, "perplexity": 486.08652891439107}, "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-2017-04/segments/1484560281746.82/warc/CC-MAIN-20170116095121-00015-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://asmedigitalcollection.asme.org/OMAE/proceedings-abstract/OMAE2016/49989/V007T06A017/281050 | In this study, numerical computation was carried out for evaluating the effects of the design parameter variations on the added resistance of Aframax tanker in head seas. The design of experiments (DOE) was used to efficiently conduct the numerical simulations with the hull form variations and save computational resources. A computational fluid dynamics (CFD) code based on the continuity and Reynolds averaged Navier-Stokes (RANS) equation was used for the numerical simulation. The simulation was performed in a short wave condition where the wave length was half of the ship length, which is expected to be most frequent in the vessel operation. Five design parameters of fore-body hull form were selected for the variations: design waterline length (DWL), bulbous bow height (BBH), bulbous bow volume (BBV), bow flare angle (BFA) and bow entrance angle (BEA). Each parameter had two levels in the variations, thus total 32 cases were designed initially. The results of the numerical simulations were analyzed statistically to determine the main effects and correlations in the five design parameters variations. Among them, the most significant parameter that influences on the added resistance in waves was DWL, followed by BBV and BEA. The other parameters had little effects on the added resistance in waves. By the computations, it was revealed that Extending DWL and decreasing BEA promoted the reflection of waves more toward the side than forward. In addition, there existed two-way interactions for the following two-factor combinations: DWL-BFA, DWL-BEA, DWL-BBV, BBH-BBV.
This content is only available via PDF. | {"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": 0.8514020442962646, "perplexity": 1794.9767924695589}, "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/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00374.warc.gz"} |
http://freneticarray.com/painting-gabriels-horn/ | Frenetic Array
The intersection of logic and imagination.
There are many things in the world of mathematics that are really quite wonderful — however, I am not sure there will be anything more wonderful yet unintuitive than Gabriel's Horn.
Gabriel's Horn is thus: suppose you have the function $y = \frac{1}{x}$ where $x \in \mathbb{R}^+, 1 \leq x \leq \infty$, rotated around the $x$ axis; not too difficult to conceptualize, it looks like a horn of sorts. But here's the paradox.
Suppose we want to calculate the volume. Simple enough, using solids of revolution, we can show the volume to be:
$$V = \pi \lim_{t \rightarrow \infty} \int _1 ^t \frac{1}{x^2} dx = \pi \lim _{t \rightarrow \infty} ( 1 - \frac{1}{t} ) = \pi$$
A simple, elegant solution; we can expect the volume to be exactly $\pi$. So, let's see about the surface area.
We know the general definition of the arc length to be $\int _a ^b \sqrt{1 + f'(x)^2}$, so combining this with our solids of revolution, we should get
$$A = 2\pi \lim _{t \rightarrow \infty} \int _1 ^t \frac{1}{x} \sqrt{1 + \left( -\frac{1}{x^2} \right)^2 } dx$$
However, this is not a trivial integral; however, there is a trick we can do. Suppose we take the integral $$2\pi \lim _{t \rightarrow \infty} \int _1 ^t \frac{dx}{x}$$ instead, and we can prove this integral will always be equal to or smaller than the former integral (because of the disappearance of $\sqrt{1 + (-\frac{1}{x^2})}$). So, taking this rather trivial integral, we can see that
$$A \geq 2\pi \lim _{t \rightarrow \infty} \int _1 ^t \frac{dx}{x} \implies A \geq \lim _{t \rightarrow \infty} 2\pi \ln(t)$$
Wait a minute; it's divergent! So we know the volume $V = \pi$, but the surface area $A \geq \infty$. This is no mistake, the math is valid. And that is simply wonderful.
A horn you can fill with paint, but you can't paint the surface. | {"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": 0.9873320460319519, "perplexity": 330.1100098910929}, "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-2019-13/segments/1552912203947.59/warc/CC-MAIN-20190325112917-20190325134917-00067.warc.gz"} |
http://physics.stackexchange.com/questions/78524/boltzmann-distribution-with-interaction-between-particles | # Boltzmann distribution with interaction between particles?
First of all, I would like to apologize in advance if I make stupid mistakes. I am a mathematician and I am trying to apply the Boltzmann distribution to places where I am not sure if it is applicable (albeit I have no choice).
The situation is: I have a system which consists in a discrete line of $M$ positions in which $N$ elements are distributed with a separation of at least $D$ positions between them. The state of each element should be its position in the line. Finally, (here's the fun part) each position in the line has an associated potential (so putting an element in the position $i$ means spending $\epsilon_{i}$ units of Energy). The usual approach to this problem as far as I have seen is just assigning $p_{i} = e^{-\epsilon_{i}/kT}$, where $p_{i}$ is the probability that there is an element in the position $i$.
I don't understand this approach and I am trying to derive that new one, but I am stuck when trying to force the particles to be separated. Any insight or reference would be very much appreciated.
Edit: If it is needed, we can also say that the particles might have a velocity (i.e. they can oscillate), but they should not be able to pass through each other.
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Sure, it's no problem to do this. The thing that has to change is that $i$ should index over all possible configurations of the $N$ elements, and the energy in the Boltzmann distribution has to be the total energy of the system.
So if $M=10$, $N=3$ and $D=2$ then, for example, $$p([1,0,0,1,0,0,1,0,0,0]) = \frac{1}{Z}e^{-\frac{\epsilon_1 + \epsilon_4 + \epsilon_7}{kT}},$$ but $$p([1,0,1,0,0,0,1,0,0,0]) = 0$$ because it's not allowed by the constraint.
To calculate the normalising factor (or "partition function") $Z$, you have to sum over all allowed configurations of the system. It isn't immediately obvious (to me) how to do that analytically in this case, but you're the mathematician so I'm sure you can find an elegant way.
Incidentally, you should be able to see that if there are no interactions between the $M$ positions then this reduces to the formula you originally quoted.
I'm glad to help. Yes, no interactions means that the probability of each of the $M$ slots being occupied is independent of whether any of the other slots are occupied. That means no restriction on the distances $D$ or on the total number of particles $N$. – Nathaniel Sep 26 '13 at 0:42 | {"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": 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": 0.8102367520332336, "perplexity": 116.48337270206305}, "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-2015-32/segments/1438042992543.60/warc/CC-MAIN-20150728002312-00137-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.transtutors.com/questions/the-drag-coefficient-for-a-sphere-at-reynolds-numbers-less-than-100-may-be-approxima-5128768.htm | The drag coefficient for a sphere at Reynolds numbers less than 100 may be approximatedby CD...
The drag coefficient for a sphere at Reynolds numbers less than 100 may be approximatedby CD =bRe-1, where b is a constant. Assuming that the Colburn analogybetween heat transfer and fluid friction applies, derive an expression for the heatloss from a sphere of diameter d and temperature Ts, released from rest and allowedto fall in a fluid of temperature T8. (Obtain an expression for the heat lost betweenthe time the sphere is released and the time it reaches some velocity
v. Assume thatthe Reynolds number is less than 100 during this time and that the sphere remainsat a constant temperature.) | {"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": 0.9227914214134216, "perplexity": 565.1302369464313}, "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-2020-10/segments/1581875143373.18/warc/CC-MAIN-20200217205657-20200217235657-00303.warc.gz"} |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/15%3A_Water/15.03%3A_Physical_Properties_of_Water | # 15.3: Physical Properties of Water
Water loss to the atmosphere is a significant problem in many parts of the world. When water supplies are low, anything that can be done to decrease water loss is important for farmers. An evaporation pan can be used to measure how fast water evaporates in a given location. This information can be used as part of projects to develop ways to cut down on evaporation and increase the amount of usable water in a region.
## Properties of Water
Compared to other molecular compounds of relatively low molar mass, ice melts at a very high temperature. A great deal of energy is required to break apart the hydrogen-bonded network of ice and return it to the liquid state. Likewise, the boiling point of water is very high. Most molecular compounds of similar molar mass are gases at room temperature.
### Surface Tension
Water has a high surface tension (attraction between molecules at the surface of a liquid) because of its hydrogen bonding. Liquids that cannot hydrogen bond do not exhibit nearly as much surface tension. Surface tension can be seen by the curved meniscus that forms when water is in a thin column such as a graduated cylinder or a buret.
### Vapor Pressure
The vapor pressure of a liquid is the pressure of the vapor produced by evaporation of a liquid or solid above the liquid or solid in a closed container. The hydrogen bonding between liquid water molecules explains why water has an unusually low vapor pressure. Relatively few molecules of water are capable of escaping the surface of the liquid and enter the vapor phase. Evaporation is slow and thus the vapor exerts a low pressure in a closed container. Low vapor pressure is an important physical property of water, since lakes, oceans, and other large bodies of water would all tend to evaporate much more quickly otherwise.
Vapor pressure is influenced by temperature. As the temperature increases, more molecules are released from the surface of the liquid. This increases movement above the liquid surface, increasing the pressure in the vapor stage. The image below illustrates the effect of temperature on vapor pressure.
## Summary
• Water has high surface tension because of extensive hydrogen bonding.
• The vapor pressure of water is low due to hydrogen bonding.
• Vapor pressure increases as temperature increases.
## Contributors
• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon. | {"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": 0.8959147930145264, "perplexity": 569.1570954918144}, "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-2019-43/segments/1570987817685.87/warc/CC-MAIN-20191022104415-20191022131915-00491.warc.gz"} |
https://zbmath.org/?q=an:0932.60041 | ×
# zbMATH — the first resource for mathematics
Logarithmic multifractal spectrum of stable occupation measure. (English) Zbl 0932.60041
Summary: For a stable subordinator $$Y_t$$ of index $$\alpha$$, $$0<\alpha < 1$$, the occupation measure $$\mu (A) = |\{t\in [0,1] : Y_t \in A\}|$$ is known to have (with probability 1) the property that $\liminf_{r\downarrow 0} \frac {\ln \mu (x-r,x+r)}{\ln r} = \alpha , \quad \forall x\in Y[0,1].$ In order to obtain an interesting spectrum for the large values of $$\mu (x-r,x+r)$$, we consider the set $B_{\theta} = \Big {\{} x\in Y[0,1]:\limsup_{r\downarrow 0} \frac {\mu (x-r,x+r)}{c_{\alpha} r^{\alpha} (\ln (1/r))^{1-\alpha}} = \theta \Big {\}},$ where $$c_{\alpha}$$ is a suitable constant. It is shown that $$B_{\theta} = \emptyset$$ for $$\theta >1$$, and $$B_{\theta} \not = \emptyset$$ for $$0\leq \theta \leq 1$$; moreover, dim $$B_{\theta} = \text{Dim } B_{\theta} = \alpha (1-\theta ^{1/(1-\alpha)})$$.
##### MSC:
60G17 Sample path properties
##### Keywords:
stable subordinators; multifractals; occupation measures
Full Text:
##### References:
[1] Hawkes, J., A lower lipchitz condition for the stable subordinator, Z. wahr. verw. geb., 17, 23-32, (1971) · Zbl 0193.45002 [2] Hu, X.; Taylor, S.J., The multifractal structure of stable occupation measure, Stochastic process. appl., 66, 283-299, (1997) · Zbl 0888.28004 [3] Orey, S.; Taylor, S.J., How often on a Brownian path does the law of iterated logarithm fail?, Proc. London math. soc., 3, 174-192, (1974) · Zbl 0292.60128 [4] Perkins, E.A.; Taylor, S.J., Uniform measure results for the image of subsets under Brownian motion, Probab. th. rel. fields, 76, 257-289, (1987) · Zbl 0613.60071 [5] Taylor, S.J., Sample path properties of a transient stable processes, J. math. mech., 16, 1229-1246, (1967) · Zbl 0178.19301 [6] Taylor, S.J., The measure theory of random fractals, Math. proc. camb. phi. soc., 100, 383-406, (1986) · Zbl 0622.60021
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | {"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": 0.951069712638855, "perplexity": 2247.012596222839}, "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-2021-21/segments/1620243989614.9/warc/CC-MAIN-20210511122905-20210511152905-00584.warc.gz"} |
http://mathoverflow.net/questions/79177/l2-boundeness-of-a-sequence | $L^2$ boundeness of a sequence
Let $f_n \in C^2(\bar{\Omega})$ be a sequence satisfying
$\Delta f_n - f_n^3 \to 0 \ \ {\rm in} \ \ L^2(\Omega)$
where $\Omega \subset {\mathbb R}^2$ is bounded and open with a smooth boundary. Is it necessarily true that $\|f_n^3\|_{L^2(\Omega)}$ and $\|\Delta f_n\|_{L^2(\Omega)}$ are uniformly bounded? If not, can you give a counterexample?
If there is a counterexample, I imagine it would involve $f_n$ becoming unbounded on $\partial \Omega$. If so, is it possible that this statement is true for $f_n \in C^2_c(\bar{\Omega})$ (i.e. if $f_n$ all have compact support in $\Omega$.)
The reason I believe it may be true is that the condition $\Delta f_n - f_n^3 \to 0$ seems incompatible with the sequence $f_n$ having an increasing positive interior maximum or a decreasing negative interior minimium as $\Delta f_n$ and $-f_n^3$ would have the same signs and not cancel out. I have tried to come up with counterexamples in one dimension but have had no luck yet.
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Is "$\Delta$" equal $\partial^2/\partial x^2 + \partial^2/\partial y^2$ or the negative of that? I think both definitions are in use, and the answer to your question might depend on the sign. – Noam D. Elkies Oct 26 '11 at 18:23
Yes, I'm using $\Delta = \partial^2/\partial x^2 + \partial^2/\partial y^2$. I think you are right that the answer would be different with $\Delta$ replaced by $-\Delta$. – Jeff Oct 26 '11 at 19:27
In one dimension $f_n(x) = 2^{1/2} / (x-x_n)$ satisfies $\Delta f_n - f_n^3 = 0$, and if $x_n \rightarrow \partial \Omega$ from the outside then $\|f_n^3\| \rightarrow \infty$. Yes, such $f_n$ are unbounded on $\partial \Omega$. – Noam D. Elkies Oct 26 '11 at 21:26
A counterexample in one dimension: take $\Omega:=(0,1)$ and $f_n(x):=\frac{\sqrt 2}{x+\frac{1}{n}}$. Then $f''_n(x)-f_n^3(x)=0$ while $\| f''_n \| _{2,\Omega}=\|f^3_n\|_{2,\Omega}=O(n^{5/2}) \, .$
Thanks! Nice counterexample. This is kind of what I was expecting, $f_n$ becoming unbounded at the boundary. Any idea what happens if we restrict $f_n$ to have compact support? – Jeff Oct 26 '11 at 21:27
Then if you multiply $\Delta f_n - f_n^3 := h_n\to 0$ in $L^2(\Omega)$ by $f_n$ and integrate on the domain you get $\|\nabla f_n\|^2_2 + \|f_n\|^ 4_ 4 = o(\|f_n\|_2)$, which implies that the LHS goes to zero. – Pietro Majer Oct 26 '11 at 22:01 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9396879076957703, "perplexity": 130.00180800784307}, "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-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00172-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://phys.libretexts.org/TextBooks_and_TextMaps/College_Physics/Book%3A_Spiral_Physics_-_Algebra_Based_(D%E2%80%99Alessandris)/Spiral_Mechanics_(Algebra-Based)/Model_3%3A_The_Particle_Model/04._Conservation/02._Applying_the_Impulse-Momentum_Relation | $$\require{cancel}$$
# 02. Applying the Impulse-Momentum Relation
### Applying the Impulse-Momentum Relation
Let's re-examine the same situation we examined at the beginning of the previous chapter, a rocket launched directly upward with a time-dependent thrust.
A 2.0 kg toy rocket is fitted with an engine that provides a thrust roughly modeled by the function F(t) = (60 N/s) t - (15 N/s2) t2, for 0 < t < 4.0 s, and zero thereafter. The rocket is launched directly upward.
For analysis, we'll apply the impulse-momentum relation between:
Event 1: The instant the rocket leaves the launch-pad
Event 2: The instant the thrust drops to zero.
Remember from last chapter that the rocket does not leave the launch pad until 0.36 s after the engine is ignited.
When the engine shuts off, the rocket is traveling at 42.5 m/s upward.
We could also apply the impulse-momentum relation between:
Event 1: The instant the thrust drops to zero.
Event 2: The instant the rocket reaches its maximum height.
During this interval, the only force acting on the rocket is the force of gravity, and the impulse-momentum relation is:
Thus, the rocket reaches its highest altitude 8.34 s after launch.
It's important to note that when a force is a function of time, it's relatively easy to integrate the function and determine the impulse. However, it should be clear that it would not be easy to determine the work done by a force of this type. Since work is expressed as an integral of a force with respect to a displacement (dr), the force function has to be expressed in terms of position, r. In general, it's not an easy (or sometimes possible) task to "convert" a function of time into a function of position, so work-energy is not a particularly useful way to analyze systems when the forces acting are time dependent. However, if the forces depend upon the position of the object, work-energy is a powerful analysis tool.
### Applying the Work-Energy Relation
A 0.15 kg ball is launched vertically upward by means of a spring-loaded plunger, pulled back 8.0 cm and released. It requires a force of about 10 N to push the plunger back 8.0 cm.
#### The Force Exerted by a Spring
The force that the spring exerts on the ball depends on the amount by which the spring is compressed. The more the spring is compressed, the larger the force it exerts on the ball. A common model is that the force exerted by a spring is directly proportional to the amount of deformation of the spring, in this case compression. Deformation (s) is defined to be the difference between the current length of the spring (L) and the equilibrium length (L0):
If we define the positive coordinate direction to point in the same direction as positive deformation (stretch), then
with the proportionality constant, k, referred to as the spring constant.
For the plunger, since it takes 10 N to compress the spring by 8.0 cm,
For analysis, we'll apply the work-energy relation between:
Event 1: The instant the plunger is released.
Event 2: The instant the ball reaches its maximum height.
For these two events, work-energy looks like this:
To do the integral, we must express Fspring in terms of the variable of integration, r. For the coordinate system chosen, s and r are identical. Also note that the force of the spring and the direction of motion of the ball point in the same direction. Thus, f = 0.
The ball reaches a maximum height of 19 cm above the top of the plunger. | {"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": 0.8989483118057251, "perplexity": 425.32938610563025}, "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-2018-26/segments/1529267863463.3/warc/CC-MAIN-20180620050428-20180620070428-00555.warc.gz"} |
http://link.springer.com/article/10.1007%2Fs10587-008-0081-0 | , Volume 58, Issue 4, pp 1221-1231
# A measure-theoretic characterization of the Henstock-Kurzweil integral revisited
### Purchase on Springer.com
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## Abstract
In this paper we show that the measure generated by the indefinite Henstock-Kurzweil integral is F σδ regular. As a result, we give a shorter proof of the measure-theoretic characterization of the Henstock-Kurzweil integral. | {"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": 0.9663201570510864, "perplexity": 1501.0882159613573}, "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-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00408-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/quantum-mechanics-commuting-operators-very-quick-question.150119/ | # Quantum Mechanics - Commuting Operators (very quick question)
1. Jan 5, 2007
### Brewer
Just a quickie:
If two operators commute, what can be said about their eigenfunctions?
The only thing I can glem from the chapter in my textbook about this is that the eigenfunctions are equal? Is this right, or have I misread it?
2. Jan 5, 2007
### cristo
Staff Emeritus
If two (Hermitian) operators commute, then there exists a complete set of eigenvectors which is common to both operators.
3. Jan 5, 2007
### Sojourner01
Not to interrupt, but if I were asking this question, I'd be doing so exclusively in terms of wave mechanics. It seems to be the 'mdern' way to ignore matrix mechanics when teaching QM - at least it is for my department. | {"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": 0.825728178024292, "perplexity": 1014.774457392819}, "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-2017-04/segments/1484560281574.78/warc/CC-MAIN-20170116095121-00271-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://mathhelpforum.com/differential-geometry/137420-norm-indicator-function-l-2-space.html | # Thread: Norm of indicator function in L^2-space
1. ## Norm of indicator function in L^2-space
I am reading a book by one John Conway, and at one point he seems to imply that $\|\chi_{\Delta}\|_2 = \left(\int |\chi_{\Delta}|^2 d \mu\right)^{1/2} = (\mu({\Delta}))^{1/2}$ (where $\chi_{\Delta}$ is the indicator function).
I can't quite understand this. I feel it should just be $\mu({\Delta})$...
So, does anyone know why?
2. Because ${\chi_\Delta}^2=\chi_{\Delta\bigcap\Delta}=\chi_\D elta, and \int \chi_\Delta d\mu=\mu(\Delta)$
3. Originally Posted by karkusha
Because ${\chi_\Delta}^2=\chi_{\Delta\bigcap\Delta}=\chi_\D elta$
Yes, of course! Thanks!
4. Originally Posted by karkusha
Because ${\chi_\Delta}^2=\chi_{\Delta\bigcap\Delta}=\chi_\D elta, and \int \chi_\Delta d\mu=\mu(\Delta)$
Wait a tick - the square is outside the absolute value, not inside...
5. $|\chi_\Delta|^2=|{\chi_\Delta}^2|=|\chi_\Delta|=\c hi_\Delta$ | {"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": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9270772933959961, "perplexity": 1288.8029071879866}, "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/1368705953421/warc/CC-MAIN-20130516120553-00037-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://learncybers.com/determinant-calculator/ | Home How to Matrix Determinant Calculator – 2 x 2, 3 x 3, 4 x 4
# Matrix Determinant Calculator – 2 x 2, 3 x 3, 4 x 4
Determinant calculator can be automatic or manual. In automatic calculators, you may put values and get answer, but you will not be able to learn. Therefore, in this procedure, we will teach you manual determinant calculator method to understand step by step.
Determinant is considered an algebraic representation of the sum of the products of elements each with an algebraic symbol, typically in a square array and used for the solution of systems of linear equations.
The determinant for linear algebra is a scalar value that can be determined by square matrix elements and which encodes those linear transformation properties defined in a matrix. The determinant of a matrix A is denoted det(A), det A, or |A|.
The volume scaling factor defined in the matrix can be considered geometrically as the linear transformation. This is also the signed n-dimensional volume occupied by the matrix ‘ column or row vectors. The determinant is either positive or negative based on the preservation or reversal of n-space orientation through the linear mapping.
In the case of a 2 × 2 matrix the determinant calculator may be defined as:
In the cast of 3×3 matrix, the determinant calculator may be defined as:
Now, it is time to go through the types of determinant matrixes.
## 2×2 Matrices:
### What is a 2×2 matrix?
A matrix that contains two rows and two columns is known as 2x2matrix. A 2X2 matrix is a tool used to help gain insight and outcomes in a dialogue. On each end of the spectrum designers create a matrix of 2×2 with opposite features (i.e. cheap versus costly).
### Why to use a 2×2 matrix?
A 2×2 matrix is a tool that allows people to think and talk about issues. Use it to help you learn about connections between things or people during your synthesis process. It is expected that a 2×2 will provide input. 2×2 matrices are perfect for expressing a reference you want to communicate visually.
You may be interested in reading more articles on Learn Cybers.
### How to use a 2×2 matrix?
For a 2×2 matrix (2 rows and 2 columns):
The determinant is:
“The determinant of A equals a times d minus b times c”
If the input in the matrix is real, Matrix A can be used in two linear maps: One, which maps the regular base vectors to A rows, and the other, which maps them to A columns- A. In other words, you subtract the top-to-bottom-right diagonal to take the determinants of a 2-2 matrix, from which you extract the product from the bottom-left-to-top-right diagonal.
It is easy to remember when you think of a cross:
• Red is negative (−bc)
Example-1
Find determinant of the following matrix ‘B’:
|B|= 4×8 − 6×3
= 32−18
= 14
### Inverse of a 2×2 matrix:
So, how to calculate the inverse of a 2×2 matrix?
It’s very easy. For a 2×2 matrix:
In other words; switch the positions of a and d, put negatives on b and c, and split all by evaluating (ad-bc). Just for example:
Example-2:
it must be true that: A × A-1 = I
You may be interested in reading more articles on Learn Cybers.
### Multiplication of a 2×2 matices:
That row of the first matrix is taken and every column of the second matrix is multiplied. Together, that contributes. And after that, find the determinant of that resulted matrix.
## 3×3 Matrix:
A matrix that contains three rows and three columns is known as 3×3 matrix.
### How to find/solve a 3×3 matrix?
The traditional method for calculating a matrix of 3×3 is a breakdown of smaller, easy-to-manage, evaluating problems of 2×2.
For Example-1:
For a 3×3 matrix:
### Inverse of a 3×3 matrix:
Here, we have used Elementary Row operation for finding the inverse of a 3×3 matrix.
We begin with matrix A and write it down next to it with an identity matrix I:
(This is called the “Augmented Matrix”)
Here, ’I’ is known as Identity Matrix.
### Identity Matrix:
The “Identity Matrix” is the matrix equivalent of the number “1”:
A 3×3 Identity Matrix
• It is “square” (has same number of rows as columns),
• It has 1s on the diagonal and 0s everywhere else.
• It’s symbol is the capital letter I
But in Elementary Row Operation, we can only do,
• swap rows
• multiply or divide each element in a a row by a constant
• replace a row by adding or subtracting a multiple of another row to it
We do these steps as follows:
No.2- Add row 2 to row 1,
No.3- then divide row 1 by 5,
No.4- Then take 2 times the first row, and subtract it from the second row,
No.5- Multiply the second row by -1/2,
No.6- Now swap the second and third row,
No.7- Last, subtract the third row from the second row,
And we are done!
And matrix A has been made into an Identity Matrix and at the same time an Identity Matrix got made into A-1.
### Multiplication of 3×3 matrix:
The following example explains the multiplication of two 3×3 matixes.
Example-1:
Then find the determinant of the resulted matrix ‘C’, by using the method that is explained earlier.
## 4×4 Matrix:
A matrix having four rows and four columns is known as 4×4 matrix. E.g:this is a 4×4 matrix.
### The inverse of a 4×4 matrix:
The following is the easiest formula that shows the way of finding the inverse of a 4×4 matrix;
Using this formula, you can easily find the determinant of a 4×4 matrix by applying the determinant formula to the resulted matrix.
You may be interested in reading more articles on Learn Cybers.
### Applications of determinants:
Following are the applications of a determinant:
• Linear Independence
• The orientation of a basis
• Volume and Jacobean Determinant
• Vandermonde Determinant
• Circulants
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https://www.physicsforums.com/threads/notation-abuse.73091/ | # Notation abuse?
1. Apr 25, 2005
### quasar987
Is it generally acceptable to write the following:
$$a_0 + \sum_{n=1}^{\infty}a_{-n} + \sum_{n=1}^{\infty}a_{n}$$
as
$$\sum_{n=-\infty}^{\infty}a_{n}$$
?
2. Apr 25, 2005
### T@P
i would accept it. but i speak for myself :rofl:
3. Apr 25, 2005
### HallsofIvy
Staff Emeritus
Seems reasonable to me!
(I remember a classmate saying "by abuse of notation ... " and the professor say "lets not be that abusive!")
4. Apr 25, 2005
### whozum
Series can have negative values for n? I thought the 'n' just denoted the active term?
5. Apr 25, 2005
### quasar987
Haha, good quote.
6. Apr 25, 2005
### robert Ihnot
Last edited: Apr 25, 2005
7. Apr 26, 2005
### whozum
$$\sum_{n=1}^{\infty}a_{-n}$$
I'm talking about that.. if you take a term from that sum, it is term 'a sub -1' for example, that doesnt really make sense.. you cant count -1 things..
8. Apr 26, 2005
### robert Ihnot
whozum: you cant count -1 things..
If you see it that way, then you won't be writing it! But, no one said anything about counting -1 things. The subscript simply identifies the term. One might have used letters, say Greek letters like A sub Alpha.
Again, if you will look through the reference I gave, you will see a sum that runs from -infinity to -1. Are you going to claim that is not possible?
If so, argue with Wolfram Research!
Last edited: Apr 26, 2005
9. Apr 26, 2005
### Lonewolf
Well, there is a one-to-one correspondence between positive and negative integers, so you can if you're willing to be flexible with your definition of counting. There's also a one-to-one correspondence between the positive integers and integers as a whole. So, for that reason, the notation makes sense.
10. Apr 26, 2005
### master_coda
Of course, the notation makes even more sense when you realize that the subscript has absolutely nothing to do with counting...
11. Apr 26, 2005
### whozum
The subscript in the series term a_n denotes the term number, The first term is a_1.. the second term a_2.. the nth term a_n.. It doesnt make sense to say "the negative first term in the sequence is a_{-1}"..
To me atleast.
12. Apr 26, 2005
### dextercioby
But what do you if u wanna write
$$\sum_{n=0}^{+\infty} a_{n}\frac{1}{n^{2}+7}$$
and u want to include the negative values too...?
What's wring with
$$\sum_{-\infty}^{+\infty} a_{n}\frac{1}{n^{2}+7}$$ ?
Daniel.
13. Apr 26, 2005
### Data
I usually use the notation
$$\sum_{n \in \mathbb{Z}} a_n$$
for this. Nothing wrong with writing out the indices explicitly though.
14. Apr 26, 2005
### master_coda
What makes you think that the $a_n$'s are supposed to form a sequence? In fact, even if they did form a sequence, there's no reason to assume that the subscripts have to equal the term number; that's just usually convenient.
15. Apr 26, 2005
### whozum
Well my experience with series and sequences terminates at Calc 2.. so I haven't dealt with them in a year.. but I was introduced with the explicit relationship that
"The first term is a_1.. the second term a_2.. the nth term a_n."
And in plain english having a 'negative xth something' doesnt make sense.. well essentially its a sum of a bunch of terms.. and with the abovep aragraph.. i guess im oging in circles.
16. Apr 26, 2005
### master_coda
I can see where you're saying...but that's not what subscripts mean. You can define a sequence $a_1,a_2,a_3,\dotsc$ where the first term is $a_1$, then second is $a_2$, and so on, but it's not necessary to define a sequence that way. I can define a sequence $a_1,b_1,a_2,b_2,\dotsc$ if that's more convenient. Or I can use any other labelling scheme I want. After all, they're just labels that I'm attaching to terms of the sequence.
So there's nothing wrong with having a sequence $a_{-1},a_{-2},\dotsc$. Of course, the nth term won't be $a_n$, but that might not be important. Or it might be important; for example, if I have the sequence $a_1,b_1,a_2,b_2,\dotsc$ then I might want to relabel the terms as $c_1,c_2,c_3,c_4,\dotsc$. That doesn't change the sequence in any way, it just lets me use a more convenient notation.
And just because you're using subscripts, it doesn't automatically mean you're refering to terms of a sequence anyway. For example, I can define a function as $f_c(x)=x+c$, where $c$ is any real number. Then $f_1$ and $f_2$ and $f_{-0.32}$ and $f_\pi$ are all functions; it's not important that it doesn't make sense for there to be a pi'th term in a sequence, because $f_\pi$ isn't supposed to be the pi'th term of a sequence. It's just a convenient label that I chose.
17. Apr 26, 2005
### Hurkyl
Staff Emeritus
The general notion is of a sequence indexed by an index set I.
In the usual case with which all of you are familiar, the index set is the natural numbers, N.
The indices of your sequence are the elements of the index set. So, in a normal sequence, the indices are 0, 1, 2, ... (or 1, 2, 3, ..., depending on how you define N)
The fact that we like to write sequences as an ordered list is not part of the concept of sequence. It comes from the fact we like to have an ordering on the natural numbers. We consider a0 is the first element of the sequence {a} precisely because we consider 0 to be the first element of the index set.
We can always choose other orderings. For instance, I may choose to order N as:
1 < 0 < 3 < 2 < 5 < 4 < ...
Then, using this ordering, the same sequence {a} would then be written:
a1, a0, a3, a2, ...
When using sequences in calculus, the chosen ordering on the index set is often important. I won't bore you with the details, but the fact we order the integers as:
... < -2 < -1 < 0 < 1 < 2 < ...
is important to the meaning of the statement
$$\sum_{n \ -\infty}^{\infty} a_n = \sum _{n \in \mathbb{Z}} a_n$$
because of where the "..."s occur.
If we chose to order the integers differently, say:
... < -3 < -1 < 1 < 3 < ... < ... < -4 < -2 < 0 < 2 < 4 < ...
Then the infinite sum over the integers would acquire a different meaning.
(In particular, if the former is defined, the latter is the same, but the latter can be defined for more series)
Now, so far I've only used countable sets -- that is mildly misleading. While sequences are most commonly used with countable index sets, that is not always the case. For example, sometimes it is useful to consider a sequence whose indices range over R, or even more complicated sets! Any set whatsoever is permitted to serve as the indices.
I've also been misleading in a different way -- we don't always care about an ordering of the index set. As I mentioned, ordering is not part of the sequence concept, and there are, indeed, applications where we never bother to order the index set.
18. Apr 26, 2005
### master_coda
Actually, I've almost never seen anyone call something a sequence unless it was indexed on the natural numbers. And in fact the ordering is often considered to be relevant to the concept of a sequence; even the rare generalized definitions that I've seen used have required that the index set be well-ordered.
However, I think we're basically making the same argument; most of the time, $a_n$ is just part of an indexed set, and our indicies don't have to be from the natural numbers. Whether or not we chose to call the indexed set a sequence isn't really important.
Last edited: Apr 26, 2005
19. Apr 26, 2005
### whozum
This is probably why I'm having trouble accepting this.
Hurkyl, I feel bad having made you type all that out ot prove such a trivial point.
20. Apr 26, 2005
### master_coda
I don't see what your point of view has to do with the remark I made. Even by the restricted definition I was using, $(a_{-n})_{n\in\mathbb{N}}$ is still a sequence. You were arguing that it was not. | {"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": 0.9072468280792236, "perplexity": 634.453353538718}, "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/1480698544672.33/warc/CC-MAIN-20161202170904-00390-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://lara.epfl.ch/w/sav08/sets_and_relations?rev=1267218189 | # LARA
This is an old revision of the document!
# Sets and Relations
## Sets
Sets are unordered collection of elements.
We denote a finite set containing only elements , and by . The order and number of occurrences does not matter: .
• iff
Empty set: . For every we have .
To denote large or infinite sets we can use set comprehensions: is set of all objects with property . $ y \in \{ x. P(x) \} \ \leftrightarrow\ P(y)$
Notation for set comprehension:
Sometimes the binder can be inferred from context so we write simply . In general there is ambiguity in which variables are bound. (Example: what does the in refer to in the expression: $ \{a \} \cup \{ f(a,b) \mid P(a,b) \}$ does it refer to the outerone as in or is it a newly bound variable? The notation with dot and bar resolves this ambiguity.
Subset: means
$ A \cup B = \{ x. x \in A \lor x \in B \}$ $ A \cap B = \{ x. x \in A \land x \in B \}$ $ A \setminus B = \{ x. x \in A \land x \notin B \}$
Boolean algebra of subsets of some set (we define ):
• are associative, commutative, idempotent
• neutral and zero elements: ,
• absorption: ,
• deMorgan laws: ,
• complement as partition of universal set: ,
• double complement:
Which axioms are sufficient?
### Infinte Unions and Intersections
Note that sets can be nested. Consider, for example, the following set $ S = \{ \{ p, \{q, r\} \}, r \}$ This set has two elements. The first element is another set. We have . Note that it is not the case that
Suppose that we have a set that contains other sets. We define union of the sets contained in as follows: $ \bigcup B = \{ x.\ \exists a. a \in B \land x \in a \}$ As a special case, we have $ \bigcup \{ a_1, a_2, a_3 \} = a_1 \cup a_2 \cup a_3$ Often the elements of the set are computed by a set comprehension of the form . We then write $ \bigcup_{i \in J} f(i)$ and the meaning is $ \bigcup \{ f(i).\ i \in J \}$ Therefore, is equivalent to .
We analogously define intersection of elements in the set: $ \bigcap B = \{ x. \forall a. a \in B \rightarrow x \in a \}$ As a special case, we have $ \bigcap \{ a_1, a_2, a_3 \} = a_1 \cap a_2 \cap a_3$ We similarly define intersection of an infinite family $ \bigcap_{i \in J} f(i)$ and the meaning is $ \bigcap \{ f(i).\ i \in J \}$ Therefore, is equivalent to .
## Relations
Pairs: $ (a,b) = (u,v) \iff (a = u \land b = v)$ Cartesian product: $ A \times B = \{ (x,y) \mid x \in A \land y \in B \}$
Relations is simply a subset of , that is .
Note: $ A \times (B \cap C) = (A \times B) \cap (A \times C)$ $ A \times (B \cup C) = (A \times B) \cup (A \times C)$
#### Diagonal relation
, is given by $ \Delta_A = \{(x,x) \mid x \in A\}$
### Set operations
Relations are sets of pairs, so operations apply.
### Relation Inverse
$ r^{-1} = \{(y,x) \mid (x,y) \in r \}$
### Relation Composition
$ r_1 \circ r_2 = \{ (x,z) \mid \exists y. (x,y) \in r_1 \land (y,z) \in r_2\}$
Note: relations on a set together with relation composition and form a monoid structure: $\begin{array}{l} r_1 \circ (r_2 \circ r_3) = (r_1 \circ r_2) \circ r_3 \\ r \circ \Delta_A = r = \Delta_A \circ r \end{array}$
Moreover, $ \emptyset \circ r = \emptyset = r \circ \emptyset$ $ r_1 \subseteq r_2 \rightarrow r_1 \circ s \subseteq r_2 \circ s$ $ r_1 \subseteq r_2 \rightarrow s \circ r_1 \subseteq s \circ r_2$
### Relation Image
When and we define image of a set under relation as $ S\bullet r = \{ y.\ \exists x. x \in S \land (x,y) \in r \}$
### Transitive Closure
Iterated composition let . $\begin{array}{l} r^0 = \Delta_A \\ r^{n+1} = r \circ r^n \end{array}$ So, is n-fold composition of relation with itself.
Transitive closure: $ r^* = \bigcup_{n \geq 0} r^n$
Equivalent statement: is equal to the least relation (with respect to ) that satisfies $ \Delta_A\ \cup\ (s \circ r)\ \subseteq\ s$ or, equivalently, the least relation (with respect to ) that satisfies $ \Delta_A\ \cup\ (r \circ s)\ \subseteq\ s$ or, equivalently, the least relation (with respect to ) that satisfies $ \Delta_A\ \cup\ r \cup (s \circ s)\ \subseteq\ s$
### Some Laws in Algebra of Relations
$ (r_1 \circ r_2)^{-1} = r_2^{-1} \circ r_1^{-1}$ $ r_1 \circ (r_2 \cup r_3) = (r_1 \circ r_2) \cup (r_1 \circ r_3)$ $ (r^{-1})^{*} = (r^{*})^{-1}$
Binary relation can be represented as a directed graph with nodes and edges
• Graphical representation of , , and
Equivalence relation is relation with these properties:
• reflexive:
• symmetric:
• transitive:
Equivalence classes are defined by $ x/r = \{y \mid (x,y) \in r$ The set is a partition:
• each set non-empty
• sets are disjoint
• their union is
Conversely: each collection of sets that is a partition defines equivalence class by $ r = \{ (x,y) \mid \exists c \in P. x \in c \land y \in c \}$
Congruence: equivalence that agrees with some set of operations.
Partial orders:
• reflexive
• antisymmetric:
• transitive
## Functions
Example: an example function for , is $f = \{ (a,3), (b,2), (c,3) \}$
Definition of function, injectivity, surjectivity.
- the set of all functions from to . For it is a strictly bigger set than .
(think of exponentiation on numbers)
Note that is isomorphic to , they are two ways of representing functions with two arguments.
There is also isomorphism between
• n-tuples and
• functions , where
### Function update
Function update operator takes a function and two values , and creates a new function that behaves like in all points except at , where it has value . Formally, $f[a_0 \mapsto b_0](x) = \left\{\begin{array}{l} b_0, \mbox{ if } x=a_0 \\ f(x), \mbox{ if } x \neq a_0$
### Domain and Range of Relations and Functions
For relation we define domain and range of : $ dom(r) = \{ x.\ \exists y. (x,y) \in r \}$ $ ran(r) = \{ y.\ \exists x. (x,y) \in r \}$ Clearly, and .
### Partial Function
Notation: means .
Partial function is relation such that $ \forall x \in A. \exists^{\le 1} y.\ (x,y)\in f$
Generalization of function update is override of partial functions,
### Range, Image, and Composition
The following properties follow from the definitions: $ (S \bullet r_1) \bullet r_2 = S \bullet (r_1 \circ r_2)$ $ S \bullet r = ran(\Delta_S \circ r)$ | {"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": 0.9916890263557434, "perplexity": 1281.485231858503}, "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-2020-34/segments/1596439739177.25/warc/CC-MAIN-20200814040920-20200814070920-00245.warc.gz"} |
https://www.researchgate.net/publication/228291267_American_Option_Pricing_with_Discrete_and_Continuous_Time_Models_An_Empirical_Comparison | Article
# American Option Pricing with Discrete and Continuous Time Models: An Empirical Comparison
(Impact Factor: 0.84). 06/2011; 18(5). DOI: 10.2139/ssrn.1875847
ABSTRACT
This paper considers discrete time GARCH and continuous time SV models and uses these for American option pricing. We perform a Monte Carlo study to examine their differences in terms of option pricing, and we study the convergence of the discrete time option prices to their implied continuous time values. Finally, a large scale empirical analysis using individual stock options and options on an index is performed comparing the estimated prices from discrete time models to the corresponding continuous time model prices. The results indicate that, while the differences in performance are small overall, for in the money options the continuous time SV models do generally perform better than the discrete time GARCH specifications.
### Full-text
Available from: Lars Stentoft
• Source
• "Among these models, Engle and Ng's (1993) nonlinear asymmetric GARCH model (NGARCH) has proven to be a strong contestant for tting stock returns, pricing options, and predicting volatility. Bollerslev and Mikkelsen (1996), Hsieh and Ritchken (2005), Christoersen et al. (2010), and Stentoft (2011), among others, all show that non ane GARCH models (such as the NGARCH) dominate ane models for both tting returns and option valuation. GARCH models also perform well in forecasting volatility. "
##### Article: This is How You Make a GARCH smile -An Improved Estimation and Calibration Method for a Family of GARCH Models
[Hide abstract]
ABSTRACT: This paper proposes an improved estimation and calibration method to a family of GARCH models. The suggested method xes one parameter such that the unconditional kurtosis of the model matches the sample kurtosis. The method can be used to estimate the model on historical returns, or calibrate it on observed option prices. An empirical analysis using Engle and Ng's (1993) NGARCH(1,1) model shows that the method dominates previous estimation methods on multiple aspects. When estimating on historical returns, the processing time is cut in half, and the out of sample t is improved. When calibrating on observed option prices, the optimization is simplied and the processing time is reduced by 50%, without aecting the quality of the t. Results are robust to various samples and selection of initial values.
Full-text · Article · Oct 2015
• Source
##### Article: American Option Pricing Using Simulation: An Introduction with an Application to the GARCH Option Pricing Model
[Hide abstract]
ABSTRACT: It contains an introduction to how simulation methods can be used to price American options and a discussion of various existing methods. An application using one of these methods, the regression based method, to the GARCH option pricing model is also provided.
Preview · Article · Mar 2012
• ##### Article: Non-Gaussian GARCH option pricing models and their diffusion limits
[Hide abstract]
ABSTRACT: This paper investigates the weak convergence of general non-Gaussian GARCH models together with an application to the pricing of European style options determined using an extended Girsanov principle and a conditional Esscher transform as the pricing kernel candidates. Applying these changes of measure to asymmetric GARCH models sampled at increasing frequencies, we obtain two risk neutral families of processes which converge to different bivariate diffusions, which are no longer standard Hull–White stochastic volatility models. Regardless of the innovations used, the GARCH implied diffusion limit based on the Esscher transform can be obtained by applying the minimal martingale measure under the physical measure. However, we further show that for skewed GARCH driving noise, the risk neutral diffusion limit of the extended Girsanov principle exhibits a non-zero market price of volatility risk which is proportional to the market price of the equity risk, where the constant of proportionality depends on the skewness and kurtosis of the underlying distribution. Our theoretical results are further supported by numerical simulations and a calibration exercise to observed market quotes.
No preview · Article · Jun 2015 · European Journal of Operational Research | {"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": 0.855060338973999, "perplexity": 1654.4146737723352}, "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-2016-07/segments/1454701159985.29/warc/CC-MAIN-20160205193919-00325-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://sarcasticresonance.wordpress.com/2009/11/07/perpetual-winter/ | # Perpetual Winter
Winter is my favorite season. Refreshing rain, cold, comfortable clothes, clouds, and much less sweat, all make it superior to the hot summer. Why, oh why, can’t it be winter forever? It turns out, that if Earth’s astronomical characteristics were a little different, we could have had an eternal winter (at least in some places of the world).
It is more or less common knowledge, that the change of seasons and differences between the hemispheres is caused by Earth’s axial tilt, which is about 23.30 relative to the ecliptic (the plane of revolution around the sun).
Generally, the further away from the equator a place on Earth is, the less energy it receives from the sun; this is because the ratio between its area and its cross section perpendicular to the ecliptic is larger than for lower latitudes. In the above picture, suppose that the sun is directly left of Earth. In this case, the southern hemisphere will occupy more of the area facing the sun, resulting in summer. Furthermore, parts of the northern half of the globe are continually hidden from the sun, creating a long night and a cold, harsh, winter.
The problem is, that once Earth gets to the other side of the sun, the exact same phenomena happen, but with the hemispheres reversed; once the sun is to the right of Earth, its tilt makes the northern hemisphere face more towards the sun, getting more sunlight and causing an awful summer. What we want, therefore, is for the tilt to change along with time. If, after half a year, Earth changes its tilt, so that when it gets to the other side of the sun, the northern hemisphere will still be on the farther and darker side, facing away from the sun, then we will have achieved perpetual winter. Of course, this also means perpetual summer for the southern hemisphere, but I suppose they will get used to it.
This phenomenon – the tilt changing its angle – is called axial precession. It is interesting to note that Earth does exhibit such movement, and its axis completes a single revolution once per about 25,800 years. Normally this is not noticeable, but be aware, that in several hundred years, Polaris will not be the Northern star any more. However, we want this effect to synchronize itself with Earth’s orbit around the sun. If this were so, then Earth’s orientation towards the sun would be “locked” – the same hemisphere would always point towards it.
This claim may seem rather intuitive, but it helps to visualise it. Take a small ball in your hand and draw a mark on the ball. Move only your arm in a circular path around a fixed point in space – this is Earth’s orbit around the sun. Now, rotate your hand as you move your arm, so that the mark on the ball always faces that fixed point in space – this is how we would like Earth’s tilt to act – axial precession.
We will now show mathematically that this effect generates the wanted result. This is a bit technical, but its worth it. Our main goal will be to see how much energy any given place on Earth receives when the axis undergoes precession. This will be done by looking at how much of the energy from the sun actually reaches a given location. We will use the following picture as our reference:
With the x axis pointing at the sun, and the z axis pointing upwards, perpendicular to the ecliptic.
Let us consider what an observer who is orbiting the sun sees. He stands in the center of the world, at the very core of the planet, and circles about the sun just like Earth does; however, he does not rotate about himself at all, but his axes stay in a fixed orientation, the same as given in the picture. From his point of view, he is stationary, and the sun revolves around him. The vector pointing in the direction of the sun, as a function of time, is:
$\mathbf{\hat{r}_{patch} }= \left( \begin{array}{c}\cos{\omega_{y}t} \\ \sin{\omega_{y}t} \\ 0 \end{array} \right)$
where ωy is the angular velocity of Earth’s orbit around the sun (ωy = 2π / 1 year).
Let us take take a certain “tile”, a “patch” on Earth, as shown in the picture, and calculate its position vector. We will start with an Earth that is not moving and is untilted, and slowly build up its astronomical characteristics. Any position on Earth can be described by two angles: latitude, and longitude. For our purposes, as you will see, we may choose any longitude we wish without losing generality, and so choose 00. Hence, for a given latitude φ, the position vector of a patch is:
$\mathbf{\hat{r}_{patch} }= \left( \begin{array}{c}0 \\ \cos{\varphi} \\ \sin{\varphi} \end{array} \right)$
Now, let us consider Earth’s rotation around itself. We have assumed for now that it is untilted, so the axis of rotation and the z axis overlap. Thus, the rotation is simply around the z axis. Given our choice of axial orientation, as seen in the picture, a counter-clockwise rotation looks like:
$\begin{array}{c} x_{new} = x_{old}\cos{\omega_{d}t} - y_{old}\sin{\omega_{d}t} \\ y_{new} = y_{old}\cos{\omega_{d}t} + x_{old}\sin{\omega_{d}t} \\ z_{new} = z_{old} \end{array}$
Where ωd is the angular velocity of Earth’s rotation around itself (ωd = 2π / 1 day). Applying this to our patch vector, we get
$\mathbf{\hat{r}_{patch}} = \left( \begin{array}{c}-\cos{\varphi}\sin{\omega_{d}t} \\ \cos{\varphi} \cos{\omega_{d}t}\\ \sin{\varphi}\end{array} \right)$
This is the position of the patch, if Earth were untilted. Looking at our picture, we see that in order to tilt it, we have to rotate it an angle η around the observer’s y axis. This will give us the position of a path on Earth as it is today. Based on our axial orientation, the rotation is:
$\begin{array}{c} x_{new} = x_{old}\cos{\eta} - z_{old}\sin{\eta} \\ y_{new} = y_{old} \\ z_{new} = z_{old}\cos{\eta} + x_{old}\sin{\eta} \end{array}$
Applying this to our patch vector, we get
$\mathbf{\hat{r}_{patch}} = \left( \begin{array}{c} -\cos{\varphi}\sin{\omega_{d}t} - \sin{\varphi}\sin{\eta} \\ \cos{\varphi} \cos{\omega_{d}t} \\ \sin{\varphi}\cos{\eta} - \cos{\varphi} \sin{\omega_{d}t}\sin{\eta}\end{array} \right)$
Now all we have to do is add precession, and we know the final position of the patch as a function of time. Precession is once again, rotation around the z axis. You may be wondering, “wait, we rotated around the z axis when we considered Earth’s daily rotation around itself. Won’t these two just combine?” The answer is no; when rotating, the order in which you apply your operations matters. In this case, Earth’s axis and the observer’s z axis no longer align, because we already took into consideration Earth’s tilt. We will now apply the rotation, but around the observer’s z axis. The direction of rotation is the same as Earth’s rotation around the sun, so that later on we will be able to match it. We get (breathe deep):
$\mathbf{\hat{r}_{patch}} = \left( \begin{array}{c} ( -\cos{\varphi}\sin{\omega_{d}t} - \sin{\varphi}\sin{\eta} ) \cos{\omega_{p}t} - \cos{\varphi}\cos{\omega_{d}t}\sin{\omega_{p}t} \\ \cos{\varphi} \cos{\omega_{d}t}\cos{\omega_{p}t} + ( -\cos{\varphi}\sin{\omega_{d}t} - \sin{\varphi}\sin{\eta} ) \sin{\omega_{p}t} \\ \sin{\varphi}\cos{\eta} - \cos{\varphi} \sin{\omega_{d}t}\sin{\eta}\end{array} \right)$
Where ωp is the angular velocity of the precession.
We now have the precise position of a certain patch on Earth. As we have previously said, the amount of energy that a certain patch gets depends on the ratio between its surface area and its cross section. In other words, it depends on the angle between its normal and light coming in from the sun. Let j = S ⋅ rsun, where S is the solar constant – the energy density coming from the sun. Its current value is about 1368 [W/m2]. In this case, the total energy per square meter that a patch receives is dΦ = jrpatch. We note, however, that this dot product can generate negative results. This occurs when the cosine of the angle between the patch and the sun is negative. The physical interpretation, is that the patch is hidden, obscured from the sun – if the product is negative, then it is currently nighttime for that patch. In this case, it should receive no energy at all. We shall refine our equation then:
$d\Phi = \left\{ \begin{array}{rcl} \mathbf{j}\cdot\mathbf{\hat{r}_{patch}} & \mbox{if} & \mathbf{j}\cdot\mathbf{\hat{r}_{patch}} > 0 \\ 0 & \mbox{if} & \mathbf{j}\cdot\mathbf{\hat{r}_{patch}} \leq 0 \end{array}\right.$
This gives the energy density for a given time, t. We want, however, to see the average density over a period of time. Then, we would be able to look at how much energy arrived every single day for a year, and compare the results for no precession, and for precession with a yearly period. This is also what lets us ignore the longitude, as mentioned earlier – if we take the average energy over time spans which are integer multiples of one day, there is no difference in longitude. Our final result to calculate, then, is:
$\overline{\Phi} = \frac{1}{t_{2} - t_{1}}\int_{t_{1}}^{t_{2}} d\Phi dt$
I do not know of any way to calculate this integral analytically. However, it is very easy to calculate it with a rather cheap numerical integration method, say the Trapezoidal Rule. For this purpose, a small python script was written, attached below (not intended to be robust or extensible).
So, how does the model fare? The graph given here depicts the energy density per day for a latitude of 31.50 (The Mediterranean’s latitude), with no precession. This reflects today’s status.
The initial season, just like in the picture, is winter, hence the incident energy is very low. Half a year later, it is the peak of summer, and it is much hotter. The energy density almost reaches 500 [W/m2], which is quite impressive, as we might think the maximum is about 700 – half the time it is day, half the time it is night. Adding a precession period equal to exactly one year generates the following:
Just as expected! We get an eternal winter, with very little energy coming from the sun, all year long!
The script, although it is rather crude and simple, actually does a wonderful job of showing all sorts of nice effects. Reversing the latitude to -31.50 gives again a constant energy density, but at a value more than twice as large – meaning an eternal summer in the southern hemisphere. Other than that, one can check a myriad of cool things, and it is very fun to play around with the various graphs and results that can be achieved. For example, one can stop Earth’s rotation around itself, see what happens if it were tilted at 900, and compare the results for different latitudes.
One final note: I have not yet researched what the implications of having such a short precession period are. Axial precession is caused by torque, and makes Earth’s motion rather odd. The effects of such a motion on Earth’s inhabitants might be rather strong. It is also interesting to try and think what would cause such a quick movement, and if it can last for a large amount of time. Perhaps I will make another post of it one day.
The code, with default values simulating today’s Earth. For different values of parameters, just change the values given to the function “generateSolarFluxFunction”. All time units are in seconds, of course.
from math import pi, cos, sin, radians
def numericalIntegration(f, t0, t1, timestep):
"""
Numerically integrates the function f, returning the value
of its definite integral from t0 to t1, using the trapezoidal rule.
The parameter "timestep" determines the size of each trapezoid.
@param f: A function of time.
"""
value1 = f(t0)
currentTime = t0 + timestep
total = 0
while (currentTime <= t1):
value2 = f(currentTime)
total = total + timestep * (value1 + value2) * 0.5
value1 = value2
currentTime += timestep
def generateSolarFluxFunction(solarConstant = 1368.0,
tiltDegrees = 23.3, latitudeDegrees = 0,
dailyPeriod = 86400.0, yearlyPeriod = 365.25 * 86400.0,
precessionPeriod = None):
def solarFlux(t):
if dailyPeriod is not None:
wd = 2.0 * pi / dailyPeriod
else:
wd = 0
if yearlyPeriod is not None:
wy = 2.0 * pi / yearlyPeriod
else:
wy = 0
if precessionPeriod is not None:
wp = 2.0 * pi / precessionPeriod
else:
wp = 0
sinLat = sin(latitude)
cosLat = cos(latitude)
sinTilt = sin(tilt)
cosTilt = cos(tilt)
cosWd = cos(wd * t)
sinWd = sin(wd * t)
cosWp = cos(wp * t)
sinWp = sin(wp * t)
cosWy = cos(wy * t)
sinWy = sin(wy * t)
sunVector = [cosWy, sinWy, 0]
patchVector = [
(-cosLat * sinWd * cosTilt - sinLat * sinTilt) * cosWp
- cosLat * cosWd * sinWp,
cosLat * cosWd * cosWp +
(-cosLat * sinWd * cosTilt - sinLat * sinTilt) * sinWp,
sinLat * cosTilt - cosLat * sinWd * sinTilt,
]
res = sum(sunVector[i] * patchVector[i] for i in xrange(3))
if res < 0:
return 0
return solarConstant * res
return solarFlux
def getDailyFluxResults():
day = 86400.0
time = 0
# Change parameters here in order to get different effects.
mySolarFlux = generateSolarFluxFunction()
results = []
while time < (366.0 * day):
totalEnergyForThisDay = \
numericalIntegration(mySolarFlux, time, time + day, 100.0)
meanEnergyForThisDay = totalEnergyForThisDay / day
results.append(meanEnergyForThisDay)
time += day
return results | {"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": 9, "mimetex.cgi": 0, "/images/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": 0.8252848386764526, "perplexity": 685.2095668540823}, "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/1521257647475.66/warc/CC-MAIN-20180320130952-20180320150952-00362.warc.gz"} |
http://opticaltweezers.org/chapter-10-photonic-force-microscope/ | # Chapter 10 — Photonic Force Microscope
One of the most successful applications of optical tweezers has been the measure- ment of small forces: since the optical restoring force is elastic, at least for most standard optical tweezers configurations, it is possible to measure the force acting on an optically trapped particle by multiplying its measured displacement from its equilibrium position and the optical trap stiffness. As shown in Fig. 10.1, the forces measurable by such a photonic force microscope (PFM) are typically in the range of tens of piconewtons down to few femtonewtons. Since these forces are comparable to most forces acting between biomolecules, the PFM has been extremely success- ful at measuring such forces. In this Chapter, we will review how the PFM can be employed to measure torques as well as forces, how the presence of surfaces or other particles affects the force measurement process and how much is the influence of non-conservative effects. Finally, we will describe how the optical force can be directly measured from the scattering associated with the trapping process.
10.1 Scanning probe techniques
10.2 Photonic torque microscope
10.3 Force measurement near surfaces
10.3.1 Equilibrium distribution method
10.3.2 Drift method
10.4 Relevance of non-conservative effects
10.5 Direct force measurement
Problems
References | {"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": 0.8417906761169434, "perplexity": 2011.6874539711728}, "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-2018-26/segments/1529267864957.2/warc/CC-MAIN-20180623093631-20180623113631-00593.warc.gz"} |
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