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https://stats.stackexchange.com/questions/219566/logit-regression-and-poisson-relative-risk-estimators/219616 | # Logit regression and Poisson relative risk estimators
I am running a logistic regression and have determined that Risk Ratios are better to explain my results than odds ratios. I have a dichotomous variable but I have both categorical and continuous predictor variables. My question is, if I use the Relative risk estimation by Poisson regression with robust error variance which coefficients do I report. The ones from the original logistic regression or the ones produced using the Poisson regression, which I only ran to get the RR. If it's the Poisson coefficients have I changed the regression from a logistic to a Poisson regression? or is it just a genearal linear model used to get the RR? And if so what type of diagnostics do I run. I've never used Poisson regressions before but it seems the only way to get the risk ratio's for both my categorical and continuous variables. I am a little confused.
Logistic regression and Poisson regression are modelling different things. In the first one, you are modelling the logit of the probability that your dichotomous variable is 1, where you can estimate probabilities and odds ratios.
With Poisson regression you are modelling expected frequencies in a cell, your output will be expected values. In this case you can compare the expected number of events given one profile versus another one. If your frequencies are events in some interval of space/time, you can model the rate and only in this case you can compare Relative Rates, also named RR. I don't think there's such estimation as a Relative Risk with Poisson Regression.
Logit and Poisson regression are different models that apply to different views of the same scenario - depending on how you define your response variable Y. (With a binomial distribution in the first case and Poisson in the second)
If you use Poisson regression, then provide results for that model, not only Relative Rates but also goodness of fit (use Deviance) and significance of effects (Wald test are ok).
My question is, if I use the Relative risk estimation by Poisson regression with robust error variance which coefficients do I report.
It depends on what you want to convey. If you want to talk about relative risks, then you should report the outputs of the Poisson regression.
If it's the Poisson coefficients have I changed the regression from a logistic to a Poisson regression?
Yes
or is it just a genearal linear model used to get the RR?
No, it is a Poisson regression.
And if so what type of diagnostics do I run.
Since Poisson regression is a GLM, a deviance goodness of fit test will tell you if your model fits the data.
You can also take a look at the deviance residuals. | {"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": 0.9547038078308105, "perplexity": 630.6459543136765}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153739.28/warc/CC-MAIN-20210728154442-20210728184442-00234.warc.gz"} |
http://technet.microsoft.com/de-de/library/dd878513(v=ws.10).aspx | Netsh Commands for Windows Filtering Platform (WFP)
Letzte Aktualisierung: Juni 2009
Betrifft: Windows 7, Windows Server 2008 R2
The Netsh commands for the Windows Filtering Platform (WFP) enable you to perform diagnostics that support Windows Firewall and IPsec. Its primary use is to capture diagnostic data while you reproduce a problem. The tool then exports the collected data into an XML file that you can examine for clues to the cause of the problem.
To run these commands from the command prompt, you must either enter the netsh wfp context or prepend the context to the command. You can enter the netsh wfp context by typing netsh and pressing ENTER, and then typing wfp and pressing ENTER. Alternately, if you are at the command prompt but have not entered the wlan context, you can type:
netsh wfp command
Where command is the command that you want to run, including all of the required parameters for the command.
Hinweis
The netsh wfp context and all of its commands are supported on computers that are running Windows 7 or Windows Server 2008 R2 only.
For computers that are running earlier versions of Windows, you can download the Microsoft IPsec Diagnostic Tool (http://go.microsoft.com/fwlink/?linkid=148386) from the Microsoft Web site. This tool provides some of the same functionality in a form that is compatible with older versions of Windows.
The following section provides Netsh commands for WFP in Windows Server® 2008 R2.
The commands documented in the Netsh WFP reference for Windows Server® 2008 R2 can be run - as documented – on computers running Windows® 7.
Fanden Sie dies hilfreich?
(1500 verbleibende Zeichen)
Vielen Dank für Ihr Feedback.
HINZUFÜGEN
Anzeigen: | {"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": 0.9114968180656433, "perplexity": 4411.258130521958}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500834494.74/warc/CC-MAIN-20140820021354-00079-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://www.ncgeo.nl/index.php/nl/publicaties/groene-serie/item/2326-gs-7-f-a-vening-meinesz-tables-for-regional-and-local-isostatic-reduction-airy-system-for-gravity-values | ## Tables for regional and local isostatic reduction (Airy system) for gravity values
###### F.A. Vening-Meinesz
Nederlandse Commissie voor Geodesie 7, Delft, 1941. 77 pagina's.
ISBN-13: 978 90 6132 013 5. ISBN-10: 90 6132 013 5.
## Summary
The purpose of this publication is to give tables for the gravity effect of the isostatic compensation according to the Airy hypothesis of a floating crust and assuming a regional distribution of the compensating masses. For making the tables more complete and in order to allow a better comparizon, a first column has been added to the tables corresponding to local compensation. The figures of these last mentioned columns are identical to those of the well-known tables of Heiskanen for the Airy system of reduction. The further columns of each table correspond to five different degrees of regional spreading of the compensation, the radii R of the areas of spreading being successively 29.05 km, 58.10 km, 116.20 km, 174.30 km and 232.40 km.
Three sets of tables have been computed, viz for a normal thickness T of the crust of 20 km, of 30 km and of 40 km; these figures represent the thicknesses of the crust assumed for zero elevation. We have based the tables on the supposition of Airy of a rigid crust floating on a denser plastic substratum and we have adopted Heiskanen's figures, 2.67 and 3.27, for the densities of the crust and of the substratum, the crustal density being assumed to be constant over its full height. The differential density of the compensation has, therefore, been assumed to be ± 0.6. We have further adhered to the physical supposition of the hydrostatic equilibrium of the crust and so we assumed equality of pressure in the deeper layers below the crust. Neglecting the increase of gravity with depth, this corresponds to a vertical dimension of the compensating root for local compensation of 2.67/0.6 = 4.450 times the topographic elevation h and of 1.642/0.6= 2.737 times the sea-depth d. Because of the converging of the verticals this assumption does not correspond to the exact equality of mass with opposite sign of the topography and of the compensation, and so we must be aware that the topographic and isostatic reduction according to these tables slightly changes the mass of the Earth. We have to take this e.g. into account in the first factor of the formula for normal gravity, which, strictly speaking, should have to be slightly different for gravity values reduced according to different assumptions of the depth of compensation.
## Contents
• The tables and some general considerations about isostatic compensation 1
• The computation of the tables and of the curves of Plate I 13
• Tables 19
Ga naar boven
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https://math.stackexchange.com/questions/2404526/incomplete-metric-space-or-normed-space-with-only-one-non-convergent-cauchy-sequ | # Incomplete metric space or normed space with only one non-convergent Cauchy sequence
Does there exist an incomplete metric space with exactly one non-convergent Cauchy sequence?
What about normed spaces? If $(x_n)_{n=1}^\infty$ is a non-convergent Cauchy sequence in a normed space, then $(\alpha x_n)_{n=1}^\infty$ is also Cauchy and non-convergent, for every non-zero scalar $\alpha$. Is it possible for a non-Banach space to have only one non-convergent Cauchy sequence, up to multiplication by scalar?
I haven't got enough reputation to comment so:
@Marios Gestas I know that you can always find many convergent Cauchy sequences. I asked if it was possible to find a space with only one non-convergent Cauchy sequence.
@Joey Zou I completely missed the fact that you can take subsequences. It seems so obvious now.
@EugenR This immediately gave an answer to my follow up question: it seems that there also doesn't exist a normed space where all non-convergent Cauchy sequences are subsequences or permutations of each other. I upvoted your answer.
Example: $(x_n)_n$ is nonconvergent, then for any sequence $(a_n)_n$ in $\mathbb{R}$ such that $a_n \to 0$ for $n\to\infty$, new sequence $$\bigl((1+a_n)x_n\bigr)_n$$ is nonconvergent. | {"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": 0.9886491298675537, "perplexity": 307.14502205081254}, "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/1570986684226.55/warc/CC-MAIN-20191018154409-20191018181909-00264.warc.gz"} |
http://slideplayer.com/slide/785465/ | # Equivalent Fractions One Whole 1. Equivalent Fractions Cut them in half.
## Presentation on theme: "Equivalent Fractions One Whole 1. Equivalent Fractions Cut them in half."— Presentation transcript:
Equivalent Fractions One Whole 1
Equivalent Fractions Cut them in half
Equivalent Fractions Shown is one half 1 2 How many pieces we want How many pieces its cut into
Equivalent Fractions Shown is one half 1 2 NUMERATOR DENOMINATOR
Equivalent Fractions I cut my shape again I still show 1 2
Equivalent Fractions But I also show 2 4
Equivalent Fractions One half is EQUIVALENT TO 2 quarters 12 24
Equivalent Fractions 1224 This symbol looks like an equals sign with a third line. It is the mathematical sign for EQUIVALENT TO - which means is worth the same as.
Equivalent Fractions We can use equivalent fraction to make our numbers easier to handle. Smaller numbers are SIMPLE 160200162045 ÷ 10 ÷ 4
Equivalent Fractions This fraction is as simple as we can make it 16020045 We can use different language for making the fraction as small as possible. Watch out for this language in the future. It is often used in Key Stage 3. Simplest form Lowest terms Simplified Cancelled down
Equivalent Fractions 1545 6080 Look for numbers that both the NUMERATOR and the DENOMINATOR can be divided by. We want numbers bigger than 1. We call these COMMON FACTORS
Equivalent Fractions 1545 ÷ 3 6080 ÷ 10 These numbers have 3 as a common factor. This means they can both be shared by 3. A common factor here is 10
Equivalent Fractions 1545515 ÷ 3 608068 ÷ 10
Equivalent Fractions 1545515 ÷ 3 ÷ 5 608068 ÷ 10 ÷ 2
Equivalent Fractions 154551513 ÷ 3 ÷ 5 60806834 ÷ 10 ÷ 2
Equivalent Fractions 154513 608034 If the top number is a 1, we know we can stop. If the top and bottom number are not DIVISIBLE by the same number, we stop.
Equivalent Fractions 154513 608034 They have no FACTORS in common other than 1
Similar presentations | {"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": 0.9845588207244873, "perplexity": 1451.295551020641}, "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-2018-39/segments/1537267160620.68/warc/CC-MAIN-20180924165426-20180924185826-00487.warc.gz"} |
https://sfb1060.iam.uni-bonn.de/project-groups/project-group-a/a06 | A06 – Hysteresis and microstructure in shape memory alloys
Our aim is to develop mathematical tools for the study of microstructure and hysteresis in the martensitic phase transition within the framework of nonlinear elasticity. We focus on a variational formulation, in which the different phases are represented by different eigenstrains, and include the interfacial energy as a singular perturbation to the functional. The key modeling assumption is that hysteresis is largely determined by the height of the energy barrier, which in turn can be estimated by the minimal energy required to generate a nucleus of the critical size. Therefore we shall consider a variational model with boundary conditions which correspond to (part of) a martensitic nucleus in an austenitic matrix, and parameters related to the sample geometry, the interfacial energy, the elastic constants of the material and the eigenstrains.
We shall start from the study of the phase diagram of our model in dependence on the various parameters, in order to identify the different microstructure regimes with two martensite variants near an austenite-martensite interface. We shall then consider in more detail three specific issues. Firstly, the regime with very small volume fraction of one martensite variant, in which we expect this variant to concentrate on a lower-dimensional manifold. Secondly, the regime of needle-like microstructure, and the precise geometry of the needles. Thirdly, the possibility of low-energy martensitic inclusions in an austenitic matrix.
Name Institute Location Phone
Conti, SergioIAMEn60/[email protected]
Zwicknagl, BarbaraTU [email protected]
Contact
Coordinator: Prof. Dr. Stefan Müller | {"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": 0.8386982083320618, "perplexity": 644.8907309549244}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141177566.10/warc/CC-MAIN-20201124195123-20201124225123-00201.warc.gz"} |
https://wiki.octanis.org/octanis1_fta | octanis1_fta
— THIS IS A WORKING DOCUMENT —
Fault tree analysis (FTA) is a top down, deductive failure analysis in which an undesired state of a system is analyzed using Boolean logic to combine a series of lower-level events. | {"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": 0.8792502880096436, "perplexity": 1580.4613311773003}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154099.21/warc/CC-MAIN-20210731172305-20210731202305-00430.warc.gz"} |
http://mathhelpforum.com/math-topics/218232-help-please-print.html | # help please
Printable View
• Apr 26th 2013, 12:16 AM
Trefoil2727
help please
Given that a>0, b>0, log2^1/2 is the arithmetic mean of log4^a and log2^b. The minimum value of 2/a + 1/b?
• Apr 26th 2013, 03:54 AM
Soroban
Re: help please
Hello, Trefoil2727!
First, I had to guess at what you had typed.
Then I've made only partial progress.
Quote:
$\text{Given that }a,b>0\text{ and }\log_2\tfrac{1}{2}\text{ is the arithmetic mean of }\log_4a\text{ and }\log_2b$
. . . $\text{find the minimum value of }\tfrac{2}{a} + \tfrac{1}{b}$
We have: . $\frac{\log_4a + \log_2b}{2} \:=\:\log_2\tfrac{1}{2} \quad\Rightarrow\quad \frac{\frac{1}{2}\log_2a + \log_2b}{2} \:=\:-1$
. . . . . . . . $\tfrac{1}{2}\log_2a + \log_2b \:=\:-2 \quad\Rightarrow\quad \log_2a + 2\log_2b \:=\:-4$
. . . . . . . . $\log_2a + \log_2b^2 \:=\:-4 \quad\Rightarrow\quad \log_2(ab^2) \:=\:-4$
. . . . . . . . $ab^2 \:=\:2^{-4} \quad\Rightarrow\quad ab^2 \:=\:\tfrac{1}{16}$
Now what?
• Apr 26th 2013, 04:06 AM
Trefoil2727
Re: help please
emm, is lg 2^(1/2), lg 4^a, lg 2^b
• Apr 26th 2013, 07:41 AM
Soroban
Re: help please
Hello, Trefoil2727!
Quote:
$\text{Given: }\,a,b>0\text{ and }\log2^{\frac{1}{2}}\text{ is the arithmetic mean of }\log4^a\text{ and }\log2^b.$
$\text{Find the minimum value of: }\,\frac{2}{a} + \frac{1}{b}$
We have: . $\frac{\log4^a + \log2^b}{2} \:=\:\log2^{\frac{1}{2}} \quad\Rightarrow\quad \frac{\log\left(4^a2^b\right)}{2} \:=\:\tfrac{1}{2}\log 2$
. . . . . . . . $\log\left(2^{2a}2^b\right) \;=\;\log 2 \quad\Rightarrow\quad \log\left(2^{2a+b}\right) \;=\;2^1$
. . . . . . . . $2a+b \:=\:1$
Now what? | {"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": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9789789915084839, "perplexity": 2173.0371078484577}, "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-2017-39/segments/1505818688208.1/warc/CC-MAIN-20170922041015-20170922061015-00605.warc.gz"} |
http://mathoverflow.net/questions/136421/beating-kadanes-algorithm | I am seeking some reference on already existing work for the following problem. Given an $n$-dimensional square matrix $A=DP$ where $D$ is a diagonal and $P$ is a permutation matrix (think of Gaussian algorithm) I would like to find a sub-rectangle (very specific sub-matrix) of $A$ that has maximum sum. Kadane's $2d$ algorithm provides $\mathcal{o}(n^3)$ solution. However my matrix $A$ is sparse and even nicer than that so my feeling that this can be done faster. Actually, I have a feeling that the lower bound on this computation should be something like $c_{\epsilon}n^{1+\epsilon}$ but I my coworkers will be happy with anything that beats $n^2\log(n)$ which is what we think we have right now (we are verifying the result).
Edit: We are actively discussing the problem in the office right now and it seems that theoretical complexity of $n^2\log(n)$ can be "easily" improved at least computationally to something like $n^{1+\epsilon}\log(n)$ in particular if we allow approximate solutions which true maximum to something like $1+\epsilon$. We have no clue how to prove this right now. | {"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": 0.9417157173156738, "perplexity": 223.06777205300207}, "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/1424936461944.75/warc/CC-MAIN-20150226074101-00053-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://projecteuclid.org/euclid.aoms/1177730442 | ## The Annals of Mathematical Statistics
### The Probability Function of the Product of Two Normally Distributed Variables
Leo A. Aroian
#### Abstract
Let $x$ and $y$ follow a normal bivariate probability function with means $\bar X, \bar Y$, standard deviations $\sigma_1, \sigma_2$, respectively, $r$ the coefficient of correlation, and $\rho_1 = \bar X/\sigma_1, \rho_2 = \bar Y/\sigma_2$. Professor C. C. Craig [1] has found the probability function of $z = xy/\sigma_1\sigma_2$ in closed form as the difference of two integrals. For purposes of numerical computation he has expanded this result in an infinite series involving powers of $z, \rho_1, \rho_2$, and Bessel functions of a certain type; in addition, he has determined the moments, semin-variants, and the moment generating function of $z$. However, for $\rho_1$ and $\rho_2$ large, as Craig points out, the series expansion converges very slowly. Even for $\rho_1$ and $\rho_2$ as small as 2, the expansion is unwieldy. We shall show that as $\rho_1$ and $\rho_2 \rightarrow \infty$, the probability function of $z$ approaches a normal curve and in case $r = 0$ the Type III function and the Gram-Charlier Type A series are excellent approximations to the $z$ distribution in the proper region. Numerical integration provides a substitute for the infinite series wherever the exact values of the probability function of $z$ are needed. Some extensions of the main theorem are given in section 5 and a practical problem involving the probability function of $z$ is solved.
#### Article information
Source
Ann. Math. Statist. Volume 18, Number 2 (1947), 265-271.
Dates
First available in Project Euclid: 28 April 2007
http://projecteuclid.org/euclid.aoms/1177730442
Digital Object Identifier
doi:10.1214/aoms/1177730442
Mathematical Reviews number (MathSciNet)
MR21284
Zentralblatt MATH identifier
0041.45004
JSTOR | {"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": 0.9593867659568787, "perplexity": 342.38284001348376}, "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/1500549448095.6/warc/CC-MAIN-20170728062501-20170728082501-00076.warc.gz"} |
http://nicklutsko.github.io/blog/2018/10/30/Theories-for-the-Poleward-Shift-of-the-Mid-Latitude-Jets | ## Theories for the Poleward Shifts of the Mid-Latitude Jets
30 Oct 2018
A major open problem in atmospheric dynamics is what causes the eddy-driven mid-latitude jets to shift polewards under climate change and, conversely, to shift equatorwards during strong El Nino events. This is seen in idealized models and in comprehensive climate model simulations of future warming scenarios, though more clearly in the southern hemisphere than in the northern hemisphere (where there’s more land). Understanding these shifts is important scientifically and for society, especially as they affect the location and severity of extreme weather events in mid-latitudes.
A closely related question is what causes the tropical circulation to widen and contract in these scenarios (see here for a recent review). These two phenomena might be related, but I’m not going to get into that here, and only focus on the jet shifts.
There are several different theories for why jets shift, but no consensus. To put these together I find it helpful to use this schematic of the dynamics of the mid-latitude upper troposphere:
This is the classic picture of a baroclinically unstable source region – the jet – where eddies (Rossby waves) are excited and then propagate away. The eddies break at lower and higher latitudes and flux momentum back into the jet. With this picture, jets could shift because of shifts of the source region or of the sink regions, which affect the eddy momentum fluxes that drive the jet.
### Background: the zonal-mean structure of global warming
The jets feel the effect of global warming through the structure of the temperature response in height and latitude. The latitudinal variations in the response are particularly important, since the zonal wind is connected to the meridional temperature gradient by the thermal wind relation: $$\frac{\partial u}{\partial z} \sim \frac{\partial T}{\partial y}$$
With global warming the largest warming is in the tropical upper troposphere and, in northern hemisphere winter, in the high-latitude lower troposphere:
(Figure 6 from Vallis et al. (2015). Sorry for the jet colormap.)
In the upper troposphere the meridional temperature gradient increases because of the enhanced tropical warming, and we expect the mean winds to accelerate in the upper troposphere/lower stratosphere. In the lower troposphere the meridional temperature gradient is weakened because of the high-latitude warming, resulting in a reduction of low level baroclinicity. During El Nino events there is also enhanced warming of the tropical upper troposphere, but over a narrower range of latitudes, and this difference is probably key to the opposing responses of the jet.
### Moving the sinks
The dispersion relation for a Rossby wave is: $$c = u - \frac{\beta}{K^2}$$ with $$u$$ the zonal-mean zonal wind, $$c$$ the phase speed of the wave, $$\beta$$ the meridional gradient of the Coriolis parameter and $$K$$ the total wavenumber. Rossby waves break at their critical latitudes, where $$u \sim c$$ and $$K$$ goes to infinity (these are also called critical layers). So shifts of the sink regions are due to some combination of changes in the waves' phase speeds and changes in the zonal-mean wind.
• Chen and Held (2007) (see also Chen et al. (2008))\) focus on the breaking region on the equatorward side of the jet. This seems like a good starting point because Earth's spherical geometry means that eddies preferentially propagate equatorward, and the largest eddy momentum fluxes are on this side of the jet.
Chen and Held make a relatively simple argument: because the subtropical winds accelerate, the equatorward critical layer (where $$u = c$$) moves polewards, and waves break closer to the jet. This effectively shifts the eddy momentum fluxes into the jet polewards and the jet shifts as well.
• Kidston et al. (2010) make an argument based on changes to eddies’ phase speeds. They focus on the $$K$$ term in the dispersion relation and suggest that, as a result of global warming, eddies get larger ($$K$$ gets smaller). If eddies’ size is set by the first baroclinic Rossby radius: $$\lambda_R = \frac{NH}{f}$$ with $$N$$ the static stability, $$H$$ the height of the tropopause and $$f$$ the coriolis parameter; then since the tropopause rises under global warming (see the Vallis et al. paper for an explanation of why this happens), $$\lambda_R$$ increases and, equivalently, $$K$$ decreases. This slows down the eddies’ phase speeds, so eddies actually break further from the jet.
Kidston et al. then point out that the sink region on the poleward side of the jet overlaps with the source region. Since eddies are able to propagate further in a warmer world, they break further from the jet and this overlap is reduced. This increases the net eddy momentum flux on the poleward side of the jet, which shifts poleward.
This might seem to contradict the Chen mechanism, but Kidston et al. argue that both can be true at the same time. The acceleration of the subtropical winds could cause the critical layer on the equatorward side of the jet to move polewards, while at the same time the deceleration of the eddies could cause the poleward critical layer to move to higher latitudes, so both sinks shift poleward.
• Related to the idea of eddy length-scales mattering, Riviere (2011) focused on how individual eddies break. Eddies which break anti-cyclonically tend to propagate equatorward, producing a poleward eddy momentum flux, while eddies which break cyclonically tend to propagate polewards and flux momentum towards the equator. Since long-wavelength eddies are more likely to break anti-cyclonically, an increase in eddy length-scales could increase anti-cyclonic wave-breaking, which would result in a poleward movement of the jet.
• A different perspective is to focus on wave reflection. This happens when $$\beta / K^2$$ is larger than $$u$$ and $$c$$ changes sign. Several studies have suggested than in a warmer world, particularly if the mid-latitude jet speeds up, more waves are reflected on the poleward side of the jet (see Simpson et al. (2012) and Kidston and Vallis (2012)). This means there are more equatorward propagating waves and a larger net poleward momentum flux across the jet.
### Moving the source region
The other approach is to focus on the baroclinic stirring which is the source for the eddies. The essence is that enhanced warming in the tropical upper troposphere results in a poleward shift of the maximum meridional temperature gradient in the upper troposphere which, combined with an increase in the static stability of the subtropics (see here), pushes the region of maximum baroclinicity polewards. A number of papers have attributed the poleward shift to this movement of the baroclinicity, including Lu et al. (2008), Butler et al. (2010), Wu et al. (2011) and Tandon et al. (2013).
### Teasing out the different mechanisms
All of these mechanisms probably play a role in the jet shifts, with different levels of importance. And their relative importances probably vary across models of different complexity. Figuring out which mechanisms are most responsible for driving the shift isn't easy though. For instance, Robinson (2000) showed how an initial jet shift can cause a shift of the low-level baroclinicity, which feeds back on the jet shift. Barnes and Hartmann (2011) suggest that the poleward jet shift itself actually causes eddy length-scales to increase. There are a lot of things to untangle in this wave-mean flow problem. Nevertheless, some attempts have been made to separate out the different effects:
• David Lorenz used a technique called Rossby wave chromatography to track how individual wavenumbers behave. This let him look at how different eddies influence the jet (though not at changes in the stirring), and Lorenz found that increased reflection is the key factor in the jet shift. The caveat is that these results come from a barotropic model in which the stirring is prescribed.
• Another approach was taken in a series of papers by Gang Chen, Jian Lu and co-authors (e.g., here, here and here), who looked at the responses of the jet to global warming-like perturbations across a large number of transient simulations with different models. These seem to consistently show that shifts of the baroclinic generation zone lag jet-shifts. This implies that the direct cause of jet shifts is shifts of the sink regions, though exactly how this comes about seems to be model-dependent.
More than anything, what we need is a quantitative theory for the baroclinic stirring: what sets the strength and location of the source region. We might expect this to shift poleward because of changing temperature gradients, but we can't predict how changes in jet speeds and eddy properties will affect this, and are forced to rely on model simulations.
Another issue is that these theories are all based on dry dynamics. To fully understand – and predict – the jet shifts we also need to understand the roles of water vapor and clouds (Ceppi and Hartmann (2016) is an example of a study looking at the role of clouds). The stratosphere also influences the jets, and it seems like the recovery of the ozone hole will push the southern hemispheric jet equatorward a bit over the next few decades. Finally, there's the problem of internal variability: how long it takes for jet shifts to be clearly discernible from the noise (if they ever are), and whether the jets' internal variability can be leveraged to predict how the jets will shift, using ideas like the fluctuation-dissipation theorem. | {"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": 0.8317725658416748, "perplexity": 1386.604178336361}, "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-2020-24/segments/1590347400101.39/warc/CC-MAIN-20200528201823-20200528231823-00593.warc.gz"} |
http://tex.stackexchange.com/questions/163745/capital-omega-problem-i-can-not-renewcommand-it | Capital Omega problem, I can not renewcommand it
I want to change all the \Omega in my paper into \varOmega, but always receive an error information. Here is the details.
I have add the line in the preamble \renewcommand{\Omega}{\varOmega} when I compile the TeX file, it said that
Command \Omega' already defined. \begin{document}
Because there are too many \Omega, it is almost impossible to change them one by one. Therefore I want to use \renewcommand{\Omega}{\varOmega} to realize my purpose. What is wrong with my method?
This is my MWE:
\documentclass{article}
\usepackage{amsmath,amssymb}
\renewcommand{\Omega}{\varOmega}
\usepackage{xeCJK}
\begin{document}
$\Omega$
\end{document}
-
That's strange. Can you narrow down the problem in terms of a minimal working example (MWE)? For example, a document with only \Omega in it together with the current preamble? Or even sequentially removing packages from the preamble until the problem goes away? – Werner Mar 5 at 5:16
the following works fine for me: \documentclass{article} \usepackage{amsmath} \renewcommand{\Omega}{\varOmega} \begin{document} $\Omega$ \end{document} – cmhughes Mar 5 at 5:21
The following works fine for me too: \documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \renewcommand{\Omega}{\varOmega} \begin{document} Here is a test for $\Omega$ \end{document} But when I insert the line \renewcommand{\Omega}{\varOmega} into my previous huge tex-file, it does not work fine anymore. It always give the information below: Command \Omega' already defined. It is very strange! – azhi Mar 5 at 5:25
@azhi make a copy of your 'huge' file, add the command, and start stripping it while keeping the error- this will get you close to a MWE that you can post in your question :) – cmhughes Mar 5 at 5:30
I have found the reason! Here is my MWE, you can check if xeCJK package will conflict with that command line I gave before: \documentclass{article} \usepackage{amsmath,amssymb} \renewcommand{\Omega}{\varOmega} \usepackage{xeCJK} \begin{document} $\Omega$ \end{document} – azhi Mar 5 at 6:05
Redefine \Omega at the beginning of the document, that is replace
\renewcommand{\Omega}{\varOmega}
with
\AtBeginDocument{\renewcommand{\Omega}{\varOmega}}
MWE:
\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage{xeCJK}
\AtBeginDocument{\renewcommand{\Omega}{\varOmega}}
\begin{document}
$\Omega$
\end{document}
-
You may want to pass the no-math option to fontspec.
\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage[no-math]{xeCJK}
\setCJKmainfont{SimSun} % this one I have on my machine
\renewcommand{\Omega}{\varOmega}
\begin{document}
$\Omega$
\end{document}
- | {"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": 0.973394513130188, "perplexity": 3503.6737957602018}, "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-2014-52/segments/1419447548525.16/warc/CC-MAIN-20141224185908-00070-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://codecogs.com/library/maths/calculus/differential/first-order.php | I have forgotten
• https://me.yahoo.com
# First Order
First Order Differential Equations with worked examples
View other versions (3)
## Examples With Separable Variables Differential Equations
Definition
Separable Differential Equations are differential equations which respect one of the following forms :
• where F is a two variable function,also continuous.
• , where f and g are two real continuous functions.
### Rational Functions
A rational function is a real function respecting where are polynomials.
Example:
##### Example - Simple Differential Equation
Problem
Solve:
Workings
As the equation is of first order, integrate the function twice, i.e.
and
Solution
### Trigonometric Functions
A trigonometric function is a real function where contains one or more of the trigonometric functions :
Example:
##### Example - Simple Cosine
Problem
Workings
This is the same as
which we integrate in the normal way to yield
Solution
### Physics Examples
Example:
##### Example - Potential example
Problem
If a and b are the radii of concentric spherical conductors at potentials of respectively, then V is the potential at a distance r from the centre. Find the value of V if:
and at r=a and at r=b
Workings
Substituting in the given values for V and r
and
Thus
Solution
## Linear Type Of Differential Equation
Equations of the type
Where P and Q are function of x ( but not of y) are said to be linear of the first order
Example:
##### Example - Rational equation
Problem
Workings
If each side of te equation is multiplied by x the equation becomes:-
i.e
Hence integrating
This equation has been solved by using the obvious integrating factor x. It is possible to find a more general solution by using R as and integrating factor.
Consider the following equation :
By Inspection the left hand side of this equation must reduce to (Ry)
This gives
Thus
This gives the rule that to solve multiply both sides by an integrating factor of:-
Solution
Hence the Method of solving this type of equation is :
• Reduce the equation into the form
• Multiply through by the Integrating Factor:-
• The equation becomes :-
## Equations That Can Be Reduced To The Linear Form
A linear form in 2 variables is given by where
Analogous for n variables .
Example:
##### Example - Simple equations
Problem
Consider the equation:
Workings
Divide through by
Putting
Solution
Hence
Therefore
Or
This example is a particular case of The Bernoulli Equation
## General Solution Of The Bernoulli Equation
Bernoulli equations have an important property :
• they are nonlinear differential equations with known exact solutions.
This section is presenting the Bernoulli Equation.
and are functions of x
This can be reduced to a linear form by putting Therefore
The original equation can be re-written as:
## Homogeneous Equations
Any equation which can be put into the form:
A homogeneous polynomial is a polynomial whose monomials with nonzero coefficients all have the same total degree. For example : is homogenous.
is said to be Homogeneous. To test whether a function of x and y can be written in the form of the right hand side, substitute for . If the result is in the form , i.e. all the x's cancel, then the test is satisfied and the equation is homogeneous.
Example:
##### Example - Testing a function is homogeneous
Problem
Workings
Substitute for y=vx,
or
or
As all the x have cancelled out, the test is satisfied.
Solution
Function is homogeous
## The Method Of Solution For Homogeneous Equations
Substitute in both sides of the equation
Note. If y is a function of x then so is v
Thus the equatican be re-written as:
Re-writing and Separating the variables:
Integrating
But
Example:
##### Example - Homogenous
Problem
Workings
Rearranging
Putting y = vx
i.e.
Integrating
Therefore
Therefore
Solution
Substituting for v Therefore
## The Exceptional Case Of Homogeneous Equations
If the straight lines are parallel there is no finite point of intersection and the method of solving such equations is illustrated by the following example.
Put Z = 3y - 4x and thus
The equation can now be written as:
Integrating
Replacing Z the solution to the differential equation is :
## Exact Equations
A form is said to be exact in a region if there is a function such as .
The expression
is an exact differential.
Thus the equation giving that i.e. is called an exact Equation.
Example:
##### Example - Exact differential
Problem
Solve
Workings
This equation is not exact as it stands but if it is multiplied through by it becomes:
Solution
The solution | {"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": 61, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 791.9097169069232}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401585213.82/warc/CC-MAIN-20200928041630-20200928071630-00365.warc.gz"} |
https://www.physicsforums.com/threads/prove-orthogonality-condition-for-sines-integral.435196/ | # Prove Orthogonality Condition For Sines (Integral)
1. Oct 5, 2010
### maherelharake
1. The problem statement, all variables and given/known data
I need to prove the equation attached. I also have to describe why the integrals vanish.
2. Relevant equations
3. The attempt at a solution
I am not sure how to begin. Our teacher told us this equation is known as the orthogonality condition for sines. I also know that n' is a positive integer. I think if I get help starting this I will be able to proceed.
File size:
10.4 KB
Views:
187
2. Oct 5, 2010
### ╔(σ_σ)╝
Use a trig identity to convert the product to sums.
3. Oct 5, 2010
### maherelharake
I worked through the integral, but I kept getting zero for all cases. I have attached my work. Any help is appreciated.
#### Attached Files:
• ###### Scan 1.jpg
File size:
16.3 KB
Views:
213
Last edited: Oct 5, 2010
4. Oct 5, 2010
### ╔(σ_σ)╝
If n=n' you simply have sine squared blah blah... If you integrate it, you will not get zero. This can be intuitively seen from the fact that the graph of sine squared is strictly greater than or equal zero.
If you keep getting zero then something is wrong.
5. Oct 5, 2010
### maherelharake
Right I understand that completely. I just can't seem to find my mistake. I went over it several times
6. Oct 5, 2010
### ╔(σ_σ)╝
For the case n=n' don't convert the product to a sum. Just integrate sine square whatever by using the half angle identity for cosine.
EDIT
Double angle not half angle.
7. Oct 5, 2010
### maherelharake
But in order to prove that, don't I have to integrate regardless? I feel like it's not the right way to do it if I use the idea of n=n' beforehand.
8. Oct 5, 2010
### ╔(σ_σ)╝
What do you mean ? The question asked you to verify the formula for the two cases. So I see nothing wrong with considering the cases separatly.
What you have done ,thus far, is valid when n'=/=n.
If you look at your solution, when you converted the product to a sum if you let n=n' one of the cosine terms vanish before you integrate. You cannot just plug in n=n' into your final solution because of division by zero :-). Those constants you have in the final step or your solution become infinity when n=n'.
So you have to consider the cases separatly.
9. Oct 5, 2010
### maherelharake
Hmm I see what you are saying. Well I suppose I will ask the professor if this method is acceptable, and let you know tomorrow. Thanks.
10. Oct 5, 2010
### ╔(σ_σ)╝
What exactly makes it seem invalid to you?
11. Oct 5, 2010
### maherelharake
It just kind of seems to me that the teacher phrased the question in a way such that he wants us to completely simplify the left hand side of the equation before we use any of the knowledge on the left. I hope that makes sense in the way I explained it.
12. Oct 5, 2010
### ╔(σ_σ)╝
That is what we are doing.
The only thing is that if n=n' the method you are using won't work directly.
The reason is what I already pointed out. In your result of your integration you have a 1/(blah(n-n')). If you make n=n' you have a problem with dividing by zero.
This is why we have to consider different cases.
13. Oct 5, 2010
### maherelharake
Ohhhh I think I see what you are saying now. Ok I will work on it and post what I have done either tonight or tomorrow. Thanks.
14. Oct 5, 2010
### ╔(σ_σ)╝
No problem :-). Take your time.
15. Oct 7, 2010
### maherelharake
Got it now. Thanks.
16. Feb 3, 2011
### nareto
aargghh why haven't you posted the solution
EDIT: nevermind, I found it here: http://www.mathreference.com/la-xf-four,orth.html and I'm copying it below for reference:
Consider two functions f = sin(mx) and g = sin(nx). Here m and n are distinct positive integers. What is the integral of f×g, as x runs from 0 to 2π?
Use the angle addition formula to write the following trig identity.
sin(mx)×sin(nx) = ½ ( cos(mx-nx) - cos(mx+nx) )
Integrate the right side from 0 to 2π and get 0. Here are two more identities to demonstrate the orthogonality of cos(mx)×cos(nx) and sin(mx)×cos(nx).
cos(mx)×cos(nx) = ½ ( cos(mx-nx) + cos(mx+nx) )
sin(mx)×cos(nx) = ½ ( sin(mx+nx) + sin(mx-nx) )
Last edited: Feb 3, 2011
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http://mathhelpforum.com/math-software/164240-power-set-calculator.html | ## Power Set Calculator
This calculator determines the Power Set for any Set S. It uses binary notation to represent the method to derive each subset term and then sorts the Power Set alphabetically or numerically at the end.
Shortcut commands for sets without numbers are available as well such as a,b,c,d,e. The calculator can handle a set of numbers, but the shortcut for that notation goes to a different lesson. Enjoy.
Power Set | {"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": 0.9807655215263367, "perplexity": 1050.1215952667135}, "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-2014-23/segments/1405997894865.50/warc/CC-MAIN-20140722025814-00230-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://phys.libretexts.org?title=TextBooks_%26_TextMaps/Astronomy_and_Cosmology_TextMaps/Map:_Celestial_Mechanics_(Tatum)/9:_The_Two_Body_Problem_in_Two_Dimensions/9.11:_Mean_Distance_in_an_Elliptic_Orbit | $$\require{cancel}$$
It is sometimes said that “$$a$$” in an elliptic orbit is the “mean distance” of a planet from the Sun. In fact $$a$$ is the semi major axis of the orbit. Whether and it what sense it might also be the “mean distance” is worth a moment of thought.
It was the late Professor C. E. M Joad whose familiar answer to the weighty questions of the day was “It all depends what you mean by...” And the “mean distance” depends on whether you mean the distance averaged over the true anomaly $$v$$ or over the time. The mean distance averaged over the true anomaly is $$\frac{1}{π} \int^π_0 r dv$$, where $$r = l /(1 + e cosv)$$. If you are looking for some nice substitution to help you to integrate this, equation 2.13.6 does very nicely, and you soon find the unexpected result that the mean distance, averaged over the mean anomaly, is $$b$$, the semi minor axis.
On the other hand, the mean distance averaged over the time is $$\frac{1}{2} P \int_0^{\frac{1}{2}P} r dt$$. This one is slightly more tricky, but, following the hint for evaluating $$\frac{1}{π} \int^π_0 r dv$$, you could try expressing $$r$$ and $$v$$ in terms of the eccentric anomaly. It will take you a moment or so, but you should eventually find that the mean distance averaged over the time is $$a(1 + \frac{1}{2}e^2)$$.
It is often pointed out that, because of Kepler’s second law, a planet spends more time far from the Sun that it does near to the Sun, which is why we have longer summers than winters in the northern hemisphere. An easy exercise would be to ask you what fraction of its orbital period does a planet spend on the sunny side of a latus rectum. A slightly more difficult exercise would be to ask: What fraction of its orbital period does a planet spend closer to the Sun than its mean (time-averaged) distance? You’d first have to ask, what is the true anomaly when $$r = a(1 + \frac{1}{2}e^2)$$? Then you need to calculate the fraction of the area of the orbit. Area in polar coordinates is $$\frac{1}{2} ∫ r^2 dv$$. I haven’t tried this, but, if it proves difficult, I’d try and write $$r$$ and $$v$$ in terms of the eccentric anomaly $$E$$ and see if that helped. | {"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": 0.9367769360542297, "perplexity": 249.38179627106757}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794864572.13/warc/CC-MAIN-20180521235548-20180522015548-00007.warc.gz"} |
https://apalache.informal.systems/docs/adr/018adr-inlining.html | authorrevisionlast revised
Jure Kukovec12022-04-21
## Summary
This ADR defines the various kinds of inlining considered in Apalache and discusses the pros and cons of their implementations. Since we have recently reworked the inliner in #1569, we saw it fit to document exactly how inlining is supposed to work and we have chosen the transformations performed in the inlining pass.
## Context
TLA+ allows the user to define their own operators (e.g. A), in addition to the standard ones built into the language itself (e.g. \union). This can be done either globally, where the module directly contains a definition, or locally via LET-IN, where a local operator is defined within the body of another operator. For example:
GlobalA(p, q) ==
LET LocalB(r) == r * r
IN LocalB(p + q)
defines a global operator GlobalA, within which there is a locally defined LocalB.
Suppose we are given an invariant GlobalA(1,2) = 9. How do we evaluate whether or not this invariant holds? To do that, we need to evaluate LET LocalB(r) == r * r IN LocalB(p + q), and to do that, we need to evaluate LocalB(p + q). However, we cannot evaluate LocalB(p + q) in a vacuum, because p and q are not values we can reason about, but instead formal parameters. What we need to do, is determine the value of "LocalB(p + q), if p = 1 and q = 2". In other words, we need to apply the substitution {p -> 1, q ->2} to LocalB(p + q), which gives us LocalB(1 + 2). Repeating this process, we apply the substitution {r -> 1 + 2} to r * r, the body of LocalB, to obtain the following equivalence:
GlobalA(1,2) = 9 <=> (1 + 2) * (1 + 2) = 9
The process of applying these substitutions as syntactic transformations is called inlining.
More precisely, suppose we are given a non-recursive operator A with the following definition:
A(p1,...,pn) == body
The term "inlining" (of A) typically refers to the process of replacing instances of operator application A(e1,...,en) with body[e1/p1,...,en/pn], i.e. the expression obtained by replacing each instance of pi with ei within body. We elect to use the term in a broader sense of "replacing an operator with its definition", and define two "flavors" of inlining:
1. Standard inlining: the instantiation described above
1. Non-nullary inlining: the instantiation described above, except the inlining skips nullary LET-defined operators
2. Pass-by-name inlining: replacing an operator name A used as an argument to a higher-order operator with a local LET-definition: LET A_LOCAL(p1,...,pn) == A(p1,...,pn) IN A_LOCAL
The reason for doing (1) is that, at some point, a rewriting rule would have to generate constraints from A(e1,...,en). To do this, we couldn't just separately encode body and e1,...,en, because the richness of the data structures allowed in TLA+ makes it difficult to combine independently generated constraints, in cases where the operator parameters are complex expressions (e.g. e1 is some record with varied nested constructs). This is mostly due to the fact that there is no 1-to-1 correspondence between TLA+ objects and SMT datatypes, so encoding object equality is more complicated (which would be needed to express that ei instantiates pi). Therefore we must, no later than at the point of the rewriting rule, know body[e1/p1,...,en/pn].
While inlining non-nullary operators is strictly necessary, inlining nullary operators is not, because nullary operators, by definition, do not have formal parameters. Therefore, in a well-constructed expression, all variables appearing in a nullary operator are scoped, i.e. they are either specification-level variables (defined as VARIABLE), or bound in the context within which the operator is defined, if local. An example of the latter would be i in
\E i \in S: LET i2 == i * i in i2 = 0
which is not bound in the nullary operator i2, but it is defined in the scope of the \E operator, under which i2 is defined. Therefore, any analysis of i2 will have i in its scope. The non-nullary variant of (1) is therefore strictly better for performance, because it allows for a sort of caching, which avoids repetition. Consider for example:
A1(p) ==
LET pCached == p
IN F(pCached,pCached)
A2(p) == F(p,p)
If we apply the substitution {p -> e}, for some complex value e, to the bodies of both operators, the results are
LET pCached == e
IN F(pCached,pCached)
and
F(e,e)
In the first case, we can translate pCached to a cell (Apalache's SMT representation of TLA+ values, see this paper for details) and reuse the cell expression twice, whereas in the second, e is rewritten twice, independently. So in the case that we perform (1), we will always perform the non-nullary variant, because it is strictly more efficient in our cell-arena framework fo rewriting rules.
The reason for doing (2) is more pragmatic; in order to rewrite expressions which feature any of the higher-order (HO) built-in operators, e.g. ApaFoldSet(A, v, S), we need to know, at the time of rewriting, how to evaluate an application of A (e.g. A(partial, current) for folding). Performing (2) allows us to make the rewriting rule local, since the definition becomes available where the operator is used, and frees us from having to track scope in the rewriting rules.
### Examples
Suppose we have an operator
A(p,q) == p + 2 * q
Then, the result of performing (1) for A(1, 2) would be 1 + 2 * 2. The constant simplification could take the inlined expression and simplify it to 5, whereas it could not do this across the application boundary of A(1,2).
The result of performing (2) for ApaFoldSet(A, 0, {1,2,3}) would be
ApaFoldSet(LET A_LOCAL(p,q) == p + 2 * q, 0, {1,2,3})
While this resulting expression isn't subject to any further simplification, notice that it does contain all the required information to fully translate to SMT, unlike ApaFoldSet(A, 0, {1,2,3}), which requires external information about A.
## Options
Knowing that we must perform (1) at some point, what remains is to decide whether we perform inlining on-demand as part of rewriting, or whether to isolate it to an independent inlining-pass (or as part of preprocessing), i.e. performing a syntactic transformation on the module, that replaces A(e1,...,en) with body[e1/p1,...,en/pn], or merely generating rewriting rules that encode A(e1,...,en) equivalently as body[e1/p1,...,en/pn], while preserving the specification syntax. Additionally, if we do isolate inlining to a separate pass, we can choose whether or not to perform (2).
1. Perform no inlining in preprocessing and inline only as needed in the rewriting rules.
• Pros: Spec intermediate output remains small, since inlining increases the size of the specificaiton
• Cons:
• Fewer optimizations can be applied, as some are only applicable to the syntactic forms obtained after inlining (e.g. ConstSimplifier can simplify IF TRUE THEN a ELSE b, but not IF p THEN a ELSE b)
• Rewriting rules for different encodings have to deal with operators in their generality.
2. Independently perform only standard (non-nullary) inlining (1), but no pass-by-name inlining (2)
• Pros: Allows for additional optimizations after inlining (simplification, normalization, keramelization, etc.)
• Cons: Rewriting rules still need scope, to resolve higher-order operator arguments in certain built-in operators (e.g. folds)
• Recall that the non-nullary variant of (1) is strictly better than the simple one (while being trivial to implement), because nullary inlining is prone to repetition.
3. Independently perform non-nullary inlining and pass-by-name inlining
• Pros:
• Enables further optimizations (simplification, normalization, keramelization, etc.)
• Using non-nullary inlining has all of the benefits of standard inlining, while additionally being able to avoid repetition (e.g. not inlining A in A + A)
• Pass-by-name inlining allows us to keep rewriting rules local
• Cons: Implementation is more complex
## Solution
We elect to implement option (3), as most of its downsides are developer burdens, not theoretical limitations, and its upsides (in particular the ones of non-nullary inlining) are noticeable performance benefits. Maintaining local rewriting rules is also a major technical simplification, which avoids potential bugs with improperly tracked scope.
TBD | {"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": 0.8140996694564819, "perplexity": 2178.9116251917085}, "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/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00738.warc.gz"} |
https://www.physicsforums.com/threads/antisymmetric-connection-torsion-tensor.574412/ | Antisymmetric connection (Torsion Tensor)
1. psimeson
19
How to show:
Tabc = $\Gamma$abc - $\Gamma$acb
is a Tensor of rank (1,2)
Attempted solution:
1. Using definition of Covariant Derivative:
DbTa= ∂aTa+$\Gamma$abcTc (1)
DcTa= ∂cTa+$\Gamma$acbTb (2)
I subtracted (2) from (1) but I couldn't really get a Tensor out of it. I just got lost in the mess.
Is this is the right way to start it?
Do you have to use covariant derivatives in your problem? Is it a hint in your problem? There are several ways to show your property.
And why do you say "antisymmetric connection?"
3. meldraft
280
I am also in the process of learning tensor calculus, so I may not be right, but wouldn't it work if you raised the indices and made every tensor ab-contravariant?
Which text are you using? There are different ways of showing your property, but the method should be adapted to what you already know.
5. psimeson
19
@arkajad: Covariant derivative is not a hint in the problem. I am just trying to solve that way. I am following various kind of textbooks. So, any way would work for me.
@meldraft: I am sure if that will work. Since the purpose of this exercise is to show how the difference between two Christoffel symbols that are asymmetric gives rise to torsion tensor.
Probably the simples way, for you, is to look at the transformation equations for the connection coefficients, and from that find out how the torsion will transform.
Check Eq. (3.6) in
http://preposterousuniverse.com/grnotes/grnotes-three.pdf
But do not read further than that!!!!
7. psimeson
19
Solved.
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https://www.overleaf.com/learn/latex/Molecular_orbital_diagrams | This article provides a brief introduction to the creation of molecular orbital diagrams in LaTeX using the modiagram package. Readers are strongly encouraged to consult the modiagram package documentation which contains numerous helpful examples to demonstrate its many features—far more than we can address in this short article.
## Introduction
Molecular diagrams are created using the modiagram package which you import to your document by adding the following line to its preamble:
\usepackage{modiagram}[$$\langle$$options$$\rangle$$]
The set of $$\langle$$options$$\rangle$$ are listed, and demonstrated, in the package documentation.
To apply package $$\langle$$options$$\rangle$$ globally you can
• set them when you load the package via \usepackage{modiagram}[$$\langle$$options$$\rangle$$], or
• use the setup command \setmodiagram{$$\langle$$options$$\rangle$$}
MO diagrams are created using the modiagram environment, which supports local use of package $$\langle$$options$$\rangle$$:
\begin{modiagram}[$$\langle$$options$$\rangle$$]
...
\end{modiagram}
The following example demonstrates a minimal modiagram environment without any using any $$\langle$$options$$\rangle$$:
\documentclass{article}
\usepackage{modiagram}
\begin{document}
First example atoms:
\begin{modiagram}
\atom{left}{1s, 2s, 2p}
\end{modiagram}
\end{document}
This example produces the following output:
The basic command to draw MO diagrams is \atom which, as demonstrated in the example above, takes two arguments:
• left: the alignment of the atom.
• 1s, 2s, 2p: the energy sub-levels to be drawn. These can be further customized as you will learn in the next section.
## Atoms
You can pass some extra information about the atomic orbitals to the command presented in the introductory example.
\documentclass{article}
\usepackage{modiagram}
\begin{document}
\begin{modiagram}
\atom{right}{
1s = { 0; pair} ,
2s = { 1; pair} ,
2p = {1.5; up, down }
}
\atom{left}{
1s = { 0; pair} ,
2s = { 1; pair} ,
2p = {1.5; up, down }
}
\end{modiagram}
\end{document}
This example produces the following output:
In this example, two identical atoms are drawn, left- and right-aligned respectively.
Following the style of the modiagram package documentation, the generic syntax to create atoms can be written as:
\atom[$$\langle$$name$$\rangle$$]{$$\langle$$left$$\rangle$$|$$\langle$$right$$\rangle$$}{$$\langle$$AO-spec$$\rangle$$}
where
• $$\langle$$name$$\rangle$$ is an optional name for the atom
• $$\langle$$left$$\rangle$$ and $$\langle$$right$$\rangle$$ determine the placement in the diagram
• $$\langle$$AO-spec$$\rangle$$ is the specification of the Atomic Orbital.
The $$\langle$$AO-spec$$\rangle$$ takes the general form
sub-level = {energy; specifications}
where
• sub-level can be 1s, 2s or 2p
• energy is the energy level, a number that determines the vertical spacing in the diagram. If omitted it is set to 0.
• specifications is a comma-separated list of the spins of the electrons contained in each orbital. The possible values are up, down, pair and empty (only the semicolon is typed) for an empty orbital. If omitted it is set to pair.
Here is a description of the commands used in the previous example:
• 1s = { 0; pair}. The sub-level 1s is in the 0 energy level, the orbital contains two (paired) electrons.
• 2s = { 1; pair}. The sub-level 2s drawn in the 1 energy level, there are two electrons in this orbital.
• 2p = {1.5; up, down}. The sub-level 2p drawn in the energy level 1.5, i.e. in the diagram the vertical spacing is set to 1.5; this sub-energy level has two electrons: one with spin up in the first orbital and another with spin down in the second orbital.
The same commands are repeated for the second atom on the right.
To display the (optional) name of an atom use a modiagram environment with the [names] option:
\documentclass{article}
\usepackage{modiagram}
\begin{document}
\begin{modiagram}[names]
\atom[Atom on the right]{right}{
1s = { 0; pair} ,
2s = { 1; pair} ,
2p = {1.5; up, down }
}
\atom[Atom on the left]{left}{
1s = { 0; pair} ,
2s = { 1; pair} ,
2p = {1.5; up, down }
}
\end{modiagram}
\end{document}
This example produces the following output:
## Molecules
The syntax for molecules is very similar to that of the \atom and, in the style of the documentation, can be written as:
\molecule[$$\langle$$name$$\rangle$$]{$$\langle$$MO-spec$$\rangle$$}
where
• $$\langle$$name$$\rangle$$ is an optional caption of the molecule
• $$\langle$$MO-spec$$\rangle$$ is the specification of the Molecular Orbital
The energy sub-levels 1s, 2s and 2p become 1sMO, 2sMO and 2pMO respectively. Here is a basic example:
\documentclass{article}
\usepackage{modiagram}
\begin{document}
\begin{modiagram}
\atom{left}{1s}
\atom{right}{1s={;up}}
\molecule{
1sMO={0.75;pair,up}
}
\end{modiagram}
\end{document}
This example produces the following output:
In the example above, the molecular orbital specification ($$\langle$$MO-spec$$\rangle$$ ) is
1sMO={0.75;pair,up}
where
• 0.75 is now the ratio (energy gain)/(energy loss).
• pair, up are the spins of the electrons in the bonding and anti-bonding molecular orbitals, respectively.
The next, slightly more elaborate, example should help you understand the syntax:
\documentclass{article}
\usepackage{modiagram}
\begin{document}
\begin{modiagram}
\atom{left}{
1s, 2s, 2p = {;pair,up,up}
}
\atom{right}{
1s, 2s, 2p = {;pair,up,up}
}
\molecule{
1sMO, 2sMO, 2pMO = {;pair,pair,pair,up,up}
}
\end{modiagram}
\end{document}
This example produces the following output:
Three atoms are set on each side of the diagram and the corresponding molecule is in the middle.
## Naming scheme
The following diagram is reproduced from the modiagram package documentation. It contains the names (labels) used for the orbitals, which are nodes in a tikzpicture and thus can be used in standard TikZ drawing commands within a modiagram environment.
Here is an example using the name of the anti-bonding orbital 1sigma* for relative positioning.
\documentclass{article}
\usepackage{modiagram}
\begin{document}
\begin{modiagram}
\atom{left}{1s}
\atom{right}{1s={;up}}
\molecule{
1sMO={;pair,up}
}
\draw[<-,shorten <=8pt,shorten >=15pt,blue]
(1sigma*) --++(2,1) node {anti-bonding MO};
\end{modiagram}
\end{document}
This example produces the following output: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8117316365242004, "perplexity": 3416.1298044512114}, "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-2022-40/segments/1664030337853.66/warc/CC-MAIN-20221006155805-20221006185805-00772.warc.gz"} |
https://www.gbhatnagar.com/2002/11/ | ## Friday, November 15, 2002
### Experience Mathematics # 20 -- The sum of angles in a triangle
Euclidean or plane geometry begins with notions of points and lines, and the notion that a point lies on a line. Think of lines as sets, and a point as an element belonging to a set. Points and lines satisfy certain axioms. In Euclidean Geometry (or plane geometry), the axioms are based on Euclid’s original axioms. From these axioms, we can use the rules of logic to derive theorems (or propositions) that can be regarded as truthful statements that apply to the plane. Here a plane is a model, or a mini-universe where those axioms and theorems hold.
For example, consider the theorem: The sum of angles in a triangle is $180$ degrees. The various terms in this theorem (angle, triangle etc.) are constructs in the plane that we wish to study. The theorem itself is a property that will hold in our mini-universe. The proof should proceed from the axioms, use the definitions of the various constructs, and follow the rules of logic.
Even though the theorem is true, it does not imply that the sum all triangles is $180$ degrees. For example, consider the surface of the earth. Draw a triangle with a right angle at the North Pole. Suppose the two sides of this angle go down to the equator, and the third side of the triangle is the equator. The sum of angles of this triangle—made on the surface of the earth—is $270$ degrees!
In fact, in this non-euclidean geometry, the sum of angles in a triangle is always greater that $180$ degrees.
Can you find a surface where a sum of angles in a triangle is always less than $180$ degrees?
## Friday, November 08, 2002
### Experience Mathematics #19 -- Euclid's axioms
Just like elements and sets, Points and Lines are undefined notions.
We can think of a line as a set of points. These satisfy certain axioms, such as: Given a line $l$ and a point $P$ not on the line, there is only one line that is parallel to $l$ containing the point $P$. Axioms are considered to be self-evident truths.
However, several gaps were found in Euclid’s axioms. For example, consider Euclid’s proof that the base angles of an isosceles triangle are equal. Suppose we have an isosceles triangle $ABC$, where the side $AB$ is equal to the side $AC$. Drop a perpendicular $AD$ from a vertex to the side $BC$. There is nothing in Euclid’s axioms that says that the point $D$ is between the points $B$ and $C$. Nevertheless, Euclid proves that the triangles $ABD$ and $ACD$ are congruent. From this it is easy to see that the base angles of an isosceles triangle are equal.
The great mathematician Hilbert completed Euclid’s work by listing a few more axioms. These included the betweenness axioms. For example, given three points $A, B and C$, one of the axioms said either $B$ is between $A$ and $C$, or $C$ is between $A$ and $B$ or $A$ is between $C$ and $B$.
To return to Euclid’s proof, some steps need to be added to show that $D$ is between $B$ and $C$.
But that is not all. We could consider a geometry where given a line $l$ and a point $P$ not on the line, there are no lines parallel to $l$ containing the point $P$. Such a non-euclidean geometry exists on the surface of the Earth. So one of Euclid’s axioms cannot be considered to be a self-evident truth after all. | {"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": 0.8224278092384338, "perplexity": 137.83524041749718}, "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-2019-47/segments/1573496665575.34/warc/CC-MAIN-20191112151954-20191112175954-00368.warc.gz"} |
https://www.physicsforums.com/threads/hydraulic-press-formula.325347/ | # Hydraulic Press Formula
• Start date
• #1
30
0
Hi, this really isn't a homework problem - it's posed in a text book I am using to revise having not used physics for 15 years - I'm now training as a science teacher.
Can anyone prove the hydraulic formula to raise a mass m by distance d where the force on the piston must be increased by delta F = rho g (A1 + A2) d
where A1 is the area of the piston where the force is being applied, A2 is the area of the piston on the side supporting the mass m. rho is density of hydraulic fluid and g is accn due to gravity.
Have I explained this clearly? Any help to a very rusty physicist greatly appreciated!
Related Other Physics Topics News on Phys.org
• #2
354
0
My guess is that you start equating both sides of the equation for a simple hydraulic system.
You know that the pressure in the system is constant, so the force generated on the platforms is dependant on their areas due to F=P*A
F - Force
P - Pressure
A - Platform area (or mass area, etc)
So F1 = P1A1 and F2=P2A2
Where the 1 and 2 represent platform 1 and 2
but, P1 = P2
a little rearranging and...
F1/A1 = F2/A2
The hydraulic fluid also plays a role itself due to it's weight, rho*g*h1 and rho*g*h2
F due to fluid is rho*g*(h1-h2)
d is h1-h2
so Force due to fluid is rho*g*d
So F1/A1 = F2/A2 + rho*g*d
ummm, then I get stuck... doh!
• #3
30
0
yeah, I got that far then tried equating the forces along a horizontal line level with the pushing piston when the mass had been raised a height h:
P1 = mg/A2 + rho g h2 => F1 = mg + rho g h2 A2 => m = F1/g – rho h2 A2 (1)
Also, conservation of energy:
Work done pushing piston down = Force x distance = F1 A2 h2 / A1
Gain in PE of mass and fluid on other side is m g h2 + rho A2 h2 g h2
= g h2 (m + rho A2 h2)
So F1 A2 h2 / A1 = g h2 (m + rho A2 h2)
cancelling and rearranging => F1 = g A1 (m/A2 + rho h2)
Putting in (1) => F1 = gA1 ( (F1 g – rho h2 A2)/A2 + rho h2) => F1 = g A1 F1 g /A2 which is obviously a load of rubbish!
Thanks for trying redargon - is there anyone else who can help please?
• #4
3
0
Hey Dear All,
I have question regarding Hydraulic Press.
In a closed hydraulic press three piston valves (area A1, A2 und A3) are
pressed by the forces F1, F2 and F3.
a) Determine Δh1 and Δh2 for the given situation (see sketch above).
Given data:
A1=200cm2, F1=2000N, A2=500cm2,F2=4000N, A3=350cm2,F3=2000N, density=1000kg/m3,g=9.81m/s2
b) Give the relation F1 : F2 : F3 such that all pistons at the same height.
#### Attachments
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• #5
sophiecentaur
Gold Member
25,199
4,822
If you don't include consider the weight of the hydraulic fluid, you can say that the work in is the same as the work out - Pressure X volume change for each piston should be the same (neglecting friction). This is equivalent to Force times distance but the pressure is the same both sides so it's an easier calculation.
If you want to include the work done to raise and lower the fluid in each cylinder then the fluid mass displaced on each side must be the same and the height moved up or down will be proportional to
1/Piston Area. So the net work will be equal to the difference between the mgh/2 (just express the m in terms of density and volume) in each case.
More work will be involved in changing the height in the narrower cylinder than in the fatter one. I haven't a pencil and paper handy but that's all you need to do.
• #6
3
0
If you don't include consider the weight of the hydraulic fluid, you can say that the work in is the same as the work out - Pressure X volume change for each piston should be the same (neglecting friction). This is equivalent to Force times distance but the pressure is the same both sides so it's an easier calculation.
If you want to include the work done to raise and lower the fluid in each cylinder then the fluid mass displaced on each side must be the same and the height moved up or down will be proportional to
1/Piston Area. So the net work will be equal to the difference between the mgh/2 (just express the m in terms of density and volume) in each case.
More work will be involved in changing the height in the narrower cylinder than in the fatter one. I haven't a pencil and paper handy but that's all you need to do.
Can you please check the picture!!
• #7
sophiecentaur
Gold Member
25,199
4,822
I was commenting on the OP.
You should start another thread with a new problem, surely.
• #8
3
0
Dear Sir thankx for your kind reply, I have confusion in order to generate the equations for delh1 & h2, and what it mean if you don't include. plz help me
• #9
10
0
-1- no force:
|------|...................|---------| A1-> area of small piston
|_A1__|...................|___A2 __| A2-> area of large piston
|_____|...................|________| no force applied
|------|...................|---------|
|------|...................|---------|
|------|____________|---------|
|-------------------------------|
|___________________________|
-2- force applied:
mass m
|deltaF|...................|___A2 __| a force(delta F is applied to raise the mass m a
|------|...................|________| distance d)
|-↑----|.....↑ ...........|---d----| let y be the distance moved by the small piston
|-↓Y---|.....| h.........|----------| y>d
|_A1__|......↓ ..........|----------|
|__X__|____________|---Z-----|
|-------------------------------|
|___________________________|
pressure is equal in X and Z:
Px = Py
delta F/A1 = roh g h
delta F = roh g h A1 h=d+y
delta F = roh g (A1 d + A1 y) volume of liquid is constant so V=v => A1 y=A2 d
delta F = roh g (A1 d + A2 d)
delta F = roh g d (A1+A2)
it's not that difficult - i'm still in highschool :S
Last edited:
• #10
sophiecentaur
Gold Member
25,199
4,822
it's not that difficult - i'm still in highschool :S
You have the right stuff - good lad!
This isn't Rocket Science, is it? It's just a matter of reading it up and applying it.
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657 | {"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": 0.8846436142921448, "perplexity": 2248.666532067036}, "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-2020-50/segments/1606141216175.53/warc/CC-MAIN-20201130130840-20201130160840-00659.warc.gz"} |
http://mathhelpforum.com/pre-calculus/157245-what-limits-these-functions.html | # Thread: What are the LIMITS of these functions?
1. ## What are the LIMITS of these functions?
What is the limit of sin(x) as x is approaching infinity?
Also, what is the limit of tan(x) when x is approaching pi from the right? (RIGHT HAND LIMIT)
Last, what is the limit of tan(x) when x is approaching to pi?
And last one, what is the limit of f(x) (4/x , x less than 1) and (x^3 - 2x + 5 , x greater than 1.) This function is probably peicewise.
Thank you very much.
Thank you!
2. $\lim_{x \to \infty}\sin{x}$ does not exist, because the function continues to oscillate around $[-1, 1]$.
$\lim_{x \to \pi}\tan{x} = \lim_{x \to \pi}\frac{\sin{x}}{\cos{x}}$
$=\frac{\sin{\pi}}{\cos{\pi}}$
$=\frac{0}{1}$
$=0$.
This is the same coming from the right and the left.
$\lim_{x \to 1}f(x)$ where $f(x) = \begin{cases}\frac{4}{x} \textrm{ if }x<1\\ x^3 - 2x + 5\textrm{ if }x>1\end{cases}$?
4. To evaluate that limit, if it exists, work out what the function approaches from the left (when $x < 1$) and what the function approaches from the right (when $x >1$. If they are the same, then the limit exists and is this value. | {"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": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9488661289215088, "perplexity": 361.7219063417405}, "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-2017-17/segments/1492917125881.93/warc/CC-MAIN-20170423031205-00378-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://www.ams.org/joursearch/servlet/DoSearch?f1=msc&v1=57Q10&jrnl=one&onejrnl=proc | # 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=(57Q10) AND publication=(proc) Sort order: Date Format: Standard display
Results: 1 to 9 of 9 found Go to page: 1
[1] Takahiro Kitayama. Non-commutative Reidemeister torsion and Morse-Novikov theory. Proc. Amer. Math. Soc. 138 (2010) 3345-3360. MR 2653964. Abstract, references, and article information View Article: PDF [2] Joan Porti. Mayberry-Murasugi's formula for links in homology 3-spheres. Proc. Amer. Math. Soc. 132 (2004) 3423-3431. MR 2073320. Abstract, references, and article information View Article: PDF This article is available free of charge [3] Teruaki Kitano. Reidemeister torsion of $T^{2}$-bundles over $S^{1}$ for $SL(2;\mathbf{C})$-representations. Proc. Amer. Math. Soc. 128 (2000) 3075-3079. MR 1664382. Abstract, references, and article information View Article: PDF This article is available free of charge [4] Michael Weiss. Curvature and finite domination. Proc. Amer. Math. Soc. 124 (1996) 615-622. MR 1291795. Abstract, references, and article information View Article: PDF This article is available free of charge [5] John Ewing, Peter Löffler and Erik Kjaer Pedersen. A rational torsion invariant . Proc. Amer. Math. Soc. 102 (1988) 731-736. MR 929012. Abstract, references, and article information View Article: PDF This article is available free of charge [6] James F. Davis and Peter Löffler. A note on simple duality . Proc. Amer. Math. Soc. 94 (1985) 343-347. MR 784190. Abstract, references, and article information View Article: PDF This article is available free of charge [7] Harold M. Hastings. Suspensions of strong shape equivalences are CE equivalences . Proc. Amer. Math. Soc. 87 (1983) 743-745. MR 687654. Abstract, references, and article information View Article: PDF This article is available free of charge [8] Micheal N. Dyer. Simple homotopy types for $(G,\,m)$-complexes . Proc. Amer. Math. Soc. 81 (1981) 111-115. MR 589149. Abstract, references, and article information View Article: PDF This article is available free of charge [9] Steve Ferry. Shape equivalence does not imply CE equivalence . Proc. Amer. Math. Soc. 80 (1980) 154-156. MR 574526. Abstract, references, and article information View Article: PDF This article is available free of charge
Results: 1 to 9 of 9 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": 0.8933757543563843, "perplexity": 2857.4805823463957}, "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/1422122152935.2/warc/CC-MAIN-20150124175552-00157-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://mathhelpforum.com/statistics/194200-hypothesis-testing-print.html | # Hypothesis testing
• December 13th 2011, 02:26 PM
jsmith613
Hypothesis testing
http://www.xtremepapers.com/Edexcel/...%202006-01.pdf
Look at question 7 on the last page
so I know that for (ai)
H0: p = 0.2
H1: p <> 0.2 (does not equal)
Now when I carry out the test, if it was p>0.2 I would say
P(X>=9).... and then work out the answer
equall if it was p<0.2 I would say P(X<=9)
But given that it is p <>0.2 how do I carry out the test. Please can you explain as well!
For (b) I am again faced with the same issue
The distribution is
X-N(20,16)
H1 again is p<>0.2
so for my normal approx to I use
P(x>=18) or p(x<=18)
I will continuty correct once I understand the principle
thanks for the help!
• December 14th 2011, 12:01 AM
CaptainBlack
Re: Hypothesis testing
Quote:
This site is blocked by my company because it is classified as:
"Scam/Questionable/Illegal"
CB
• December 14th 2011, 02:22 AM
jsmith613
Re: Hypothesis testing
• December 14th 2011, 03:41 AM
CaptainBlack
Re: Hypothesis testing
Quote:
Originally Posted by jsmith613
The null hypothesis $H_0$ is that $p=0.2$, and the alternative is that $p \ne 0.2$.
We reject the null hypothesis if the probability of the observed result or more extreme conditioned on the null-hypothesis being true is "small".
That is if:
$P(9{\text{ or more from }}20|H_0)<0.05$
Now since under the null hypothesis the distribution of the number of Deano readers in a sample of size $20$ has a binomial distribution $B(20,0.2)$ we can work out the probability on the left (and it is $\sim 0.01$), which is less than $0.05$ so we reject the null hypothesis.
However exact binomial calculations of the size required by this problem are beyond what is reasonable to expect in an exam, so you will either have to use an approximation or tables of critical values provided for this purpose.
You will note that the alternative hypothesis does not figure in this test.
Part (b) works in essentially the same manner except the sample size is now 100, and large(ish) sample approximations are probably appropriate.
CB
• December 14th 2011, 04:01 AM
jsmith613
Re: Hypothesis testing
well what I was asking was how do I decide if I should use P(X<=9) or P(X>=9)
Is it becasue the E(X) - mean - is 4.
9 is above the mean so I calcuate P(X>=9) NOT P(X<=9)
for normal, mean is 20, value is 18 SO here is say P(X<=18) NOT P(X>=18)
thanks
• December 14th 2011, 05:29 AM
CaptainBlack
Re: Hypothesis testing
Quote:
Originally Posted by jsmith613
well what I was asking was how do I decide if I should use P(X<=9) or P(X>=9)
Is it becasue the E(X) - mean - is 4.
9 is above the mean so I calcuate P(X>=9) NOT P(X<=9)
for normal, mean is 20, value is 18 SO here is say P(X<=18) NOT P(X>=18)
thanks
You always ask what is the probability of observing a result this far or further from what is expected.
CB | {"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": 0.9383168816566467, "perplexity": 1384.1123899488712}, "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-2015-06/segments/1422122123549.67/warc/CC-MAIN-20150124175523-00027-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://math.eretrandre.org/tetrationforum/showthread.php?tid=122&pid=1638&mode=threaded | • 0 Vote(s) - 0 Average
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Zeration James Knight Junior Fellow Posts: 6 Threads: 1 Joined: Mar 2008 03/25/2008, 03:44 PM (This post was last modified: 03/26/2008, 01:08 AM by James Knight.) Here are some Laws that are true by this definition of zeration Law 1 Iterative Property Zeration x o x o ... x = x + n - 1 n Knightation x J x J x J ... x = x - n + 1 or = x - (n-1) Cancelation Property (cancels the +- 1) (x o x o ... x )+ (x J x J x J ... x) = x + m - n .............n.........................m Law 2 Addition / Distributive Law Zeration (x o b) + a = (x + a) o (b+a) = (x) o (b+a) = (a) o (b + a) = b + a + 1 Knightation (x J b) + a = (x+a) o (b+a) = (x+a) o (b) = (x+a) o a = x+a - 1 Law 3 Relativity Law Deltation (x+a)Δ(b+a) = x Δ b also (x-a)Δ(b-a) = x Δ b Multi-Lines ----------- Also something that is interesting is multi-lines using polynomial functions as the operands for Zeration, Deltation and Knightation. One could write: y = -1 +- 1 to show dual horizontal Lines and similarly x = -1 +- 1 to show dual vertical lines Using Zeration, the expression y = x o (y^2 + 3y -1) means the same thing and will produce two horizontal lines The equation can be also written for y in terms of x y = -1 +- sqrt ( 2 J x) but simplifies to y = -1 +- sqrt (2-1) y = -1 +- sqrt (1) y = -1 +- 1 which is what we started with. Also two vertical lines can be made by remembering that in y = a Δ b, y is all values when b = a -1 Therefore if x = a, and b = x^2 + 3x - 1 y = x Δ (x^2 + 3x - 1) gives x = -1 +- 1 Anyway, this can be extended to all polynomial functions which many generate multilines. This may be useful in determining roots of equations but I highly doubt it. The only problem I am having is with deltation. I know it's not commutative because x Δ a = a Δ x Let I be the value for all values (Vertical Line) then x Δ a = I whenever a = x -1 and a Δ x = I whenever x = a-1 therefore a = x + 1 a cannot = x - 1 and x + 1 for then -1 = 1 and that is false However Associativity is another question Because it is difficult to evaluate for hyperreal infinite or infinitesimal operands of deltatation What would I Δ x be? what would +- Infintity Δ x be? However I have somewhat came to a conclusion that it is not associative If associativity were true then (aΔb)Δc = aΔ(bΔc) for all values of a, b and c Rearranging I could get a Δ b = (aΔ(bΔc)) o c by definition of zeration assuming of course that the definition extends to non real values a Δ b = c + 1 therefore a = (c+1) o b a = b + 1 from a Δ b = c + 1 we could also say that since a Δ b gives nonreal values, then a Δ b is never equal to c+1 if a,b and c are real numbers. Only when b + 1 = a is the equation true IE VERTICAL LINE INTERSECTION Therefore, since conditions exists, deltation is not associative which follows the same for left inverses of noncommutative, non associative regular operations( ie exponentiation and rooting, tetration and superooting etc.) Of course there is that assumption I made which may be wrong. So things to look into: Zeration, Deltation and Knightation Values for NonReal operands. It may be interesting... Value of I ---------- 'I' is the value of y when y = x Δ (x-1) I = x Δ (x -1) I have found a resemblance in the following equation b = 0c when b = 0 c is all values however, when b is not 0, c is undefined (see the similarity?) In other words, I = c, when 0c = b and b = 0 Therefore we can define c in terms of x and b as follows; If y = x Δ a when y = c = I a = x -1 b = 0 Therefore a - b = x -1, a = x + b -1 Therefore y = x Δ (x+b -1) Therefore the solution of 0c = b where b is a constant is c = x Δ (x+b -1) we can also use the Relativity Law to simplify by adding 1 - x to both operands (you can also see that it's virtually the same as substituting x =1) Therefore c = 1 Δ b is the simplest solution to 0c = b These values look to me to being hyperreal. They technically are a part of the real numbers but not "officially". Also, if one looks at the limits of 1/x as x -> 0 +- infinity occurs. Something to consider for deltation... Well enjoy! and remember to keep an open mind! James « Next Oldest | Next Newest »
Messages In This Thread Zeration - by GFR - 02/14/2008, 06:38 PM RE: Zeration - by Ivars - 02/14/2008, 08:10 PM RE: Zeration - by GFR - 02/14/2008, 10:41 PM RE: Zeration - by mathamateur - 07/30/2009, 06:31 AM RE: Zeration - by Ivars - 02/21/2008, 07:22 PM RE: Zeration - by quickfur - 02/21/2008, 09:34 PM RE: Zeration - by bo198214 - 02/21/2008, 10:18 PM RE: Zeration - by bo198214 - 02/21/2008, 10:25 PM RE: Zeration - by quickfur - 02/21/2008, 11:04 PM RE: Zeration - by quickfur - 02/21/2008, 11:12 PM RE: Zeration - by KAR - 02/21/2008, 11:04 PM RE: Zeration - by quickfur - 02/21/2008, 11:52 PM RE: Zeration - by GFR - 02/24/2008, 12:39 AM RE: Zeration - by Ivars - 02/24/2008, 02:50 PM RE: Zeration - by marraco - 03/20/2015, 09:59 PM RE: Zeration - by bo198214 - 02/24/2008, 11:02 AM RE: Zeration - by GFR - 03/19/2008, 12:40 PM Zeration - My Research / Investigation - by James Knight - 03/25/2008, 08:28 AM RE: Zeration - My Research / Investigation - by bo198214 - 03/25/2008, 09:09 AM More on Zeration - by James Knight - 03/25/2008, 03:44 PM Exponential Laws - New Zeration Law - by James Knight - 03/25/2008, 07:48 PM Delta Numbers As HyperReals - by James Knight - 03/26/2008, 12:50 AM RE: Delta Numbers As HyperReals - by Ivars - 03/26/2008, 12:15 PM RE: Zeration - by GFR - 03/26/2008, 12:22 AM RE: Zeration - by GFR - 04/05/2008, 08:58 PM RE: Zeration - by Igor M - 01/14/2009, 04:04 PM RE: Zeration - by bo198214 - 01/20/2009, 09:59 PM RE: Zeration - by 73939 - 07/05/2010, 12:00 AM RE: Zeration - by bo198214 - 07/05/2010, 07:37 AM RE: Zeration - by brangelito - 07/20/2010, 05:51 PM RE: Zeration - by bo198214 - 07/21/2010, 02:58 AM RE: Zeration - by JmsNxn - 11/09/2011, 01:40 AM RE: Zeration - by quickfur - 11/09/2011, 04:15 AM RE: Zeration - by JmsNxn - 11/10/2011, 01:20 AM RE: Zeration - by quickfur - 11/10/2011, 02:09 AM RE: Zeration - by marraco - 03/20/2015, 09:44 AM RE: Zeration - by marraco - 03/20/2015, 10:41 PM RE: Zeration - by marraco - 03/21/2015, 12:35 AM RE: Zeration - by marraco - 03/21/2015, 01:44 AM RE: Zeration - by marraco - 03/21/2015, 04:10 AM RE: Zeration - by MphLee - 03/21/2015, 11:53 AM RE: Zeration - by marraco - 03/23/2015, 07:58 AM RE: Zeration - by tommy1729 - 03/21/2015, 11:11 PM RE: Zeration - by marraco - 03/23/2015, 08:05 AM RE: Zeration - by marraco - 03/24/2015, 11:29 AM RE: Zeration - by MphLee - 03/23/2015, 09:00 AM RE: Zeration - by marraco - 03/23/2015, 01:39 PM RE: Zeration - by MphLee - 03/23/2015, 02:31 PM RE: Zeration - by Stanislav - 05/28/2015, 11:12 PM RE: Zeration - by marraco - 05/29/2015, 01:33 AM RE: Zeration - by Stanislav - 05/29/2015, 09:06 PM RE: Zeration - by MphLee - 06/03/2015, 01:40 PM RE: Zeration - by Stanislav - 06/04/2015, 06:44 AM RE: Zeration - by marraco - 06/04/2015, 08:44 PM RE: Zeration - by MphLee - 06/05/2015, 09:10 PM RE: Zeration - by Stanislav - 09/09/2015, 10:04 PM RE: Zeration - by Stanislav - 10/31/2016, 02:57 PM
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=131&t=30075&p=93590 | total entropy of the system
$\Delta S = \frac{q_{rev}}{T}$
Sophie Krylova 2J
Posts: 59
Joined: Fri Sep 29, 2017 7:06 am
total entropy of the system
The second law states that the total entropy of an isolated system always increases. At the same time we talk about how if total entropy is negative, the reaction is not spontaneous or if it's zero, the rxn is at equilibrium. How are these two scenarios possible in the light of the second law, stating that the entropy can only increase?
melissa carey 1f
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Joined: Fri Sep 29, 2017 7:06 am
Re: total entropy of the system
OliviaShearin2E
Posts: 37
Joined: Fri Sep 29, 2017 7:05 am
Re: total entropy of the system
The second law does not mean that entropy always increases, but that entropy can never decrease. If total entropy is negative, the reaction is not spontaneous as you said. This means the reverse reaction is spontaneous so the total entropy increases. When the reaction is at equilibrium, the entropy is zero so it does not decrease.
Troy Tavangar 1I
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Re: total entropy of the system
The second law only says that the entropy can never decrease which is why entropy itself is never negative | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8238416314125061, "perplexity": 866.6591285433428}, "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-2019-51/segments/1575540518337.65/warc/CC-MAIN-20191209065626-20191209093626-00080.warc.gz"} |
http://math.stackexchange.com/questions/283977/how-many-binomial-coefficients-are-equal-to-a-specific-integer-binomnr/284032 | # How many binomial coefficients are equal to a specific integer ($\binom{n}{r} = 2013$ or $\binom{n}{r} = 2014$)?
1. Find the number of ordered pairs $(n,r)$ which satisfy $\binom{n}{r} = 2013$.
2. Find the number of ordered pairs $(n,r)$ which satisfy $\binom{n}{r} = 2014$.
My Attempt for $(1)$:
By simple guessing, we can find two solutions:
$$\binom{n}{r} = \binom{2013}{1}=\binom{2013}{2013-1}=\binom{2013}{2012}$$
So two solutions are $(2013,1),(2013,2012)$.
We also know that $\binom{n}{r} = 3\times 11 \times 61$.
How can I calculate the remaining ordered pairs $(n,r)$ from this point?
-
Factoring $2013$ was a good idea. For (2), you'll want to start by factoring $2014$. – Gerry Myerson Jan 22 '13 at 6:30
$$\binom n {r+1}\ge \binom nr \iff \frac{\binom n {r+1}}{\binom nr}\ge 1\iff\frac{n-r}{r+1}\ge1\iff r\le \frac{n-1}2$$
So, $$\binom n r\le \binom n {r+1}\iff r\le \frac{n-1}2$$ and $$\binom n r\ge\binom n {r+1}\iff r\ge\frac{n-1}2$$
For any integer $u,\binom n1=u\implies n=u$ will always have a solution in integers.
For $r=2,\binom n2=\frac{n(n-1)}{2}=2013\iff n^2-n-2\cdot2013=0$ but the discriminant $1+4\cdot2\cdot2013=16105$ is not a perfect square, hence we don't have any rational solution here.
For $r=3,\binom n3=\frac{n(n-1)(n-2)}{1\cdot2\cdot3},$
one of the term in the numerator $n-s$ (say,) where $0\le s\le 2$ is divisible by $61$
So, $n-s=61m,n=61m+s$ for some integer $m$ then $n-t\text{( where$0\le s\le 2$)}\ge 61m-2\ge 59m$ for $m\ge 1$
So, $\binom n3\ge \frac{(59m)^3}{1\cdot2\cdot3}>2013$ for $m\ge1$
Now, $\binom n{r+1}\ge \binom n3$ for $\frac{n-1}2\ge r\ge 3\implies \binom nr>2013$for $\frac{n-1}2\ge r\ge 3$
also $\binom n{n-3}\le \binom nr$ for $\frac{n-1}2\le r\le n-3\implies \binom nr>2013$ for $\frac{n-1}2\le r\le n-3$
$\implies \binom nr>2013$ for $3\le r\le n-3$
As $\binom nr=\binom n{n-r},$ the only other solution is $r=n-1$ corresponding to $r=1$
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Thanks lab bhattacharjee – juantheron Jan 22 '13 at 17:10
@juantheron, my pleasure. But, I think there should be some smarter way. – lab bhattacharjee Jan 22 '13 at 17:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9135506749153137, "perplexity": 247.98327628781854}, "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-2016-30/segments/1469257823670.44/warc/CC-MAIN-20160723071023-00288-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://zbmath.org/?q=an:0818.46059 | ×
# zbMATH — the first resource for mathematics
Banach algebras, decomposable convolution operators, and a spectral mapping property. (English) Zbl 0818.46059
Function spaces, Proc. Conf., Edwardsville/IL (USA) 1990, Lect. Notes Pure Appl. Math 136, 307-323 (1992).
The paper begins with a discussion of four motivating problems concerning spectral properties of convolution operators given by measures on locally compact Abelian groups. An operator $$T$$ on a complex Banach space $$X$$ is called decomposable if, for every open cover $$\{U, V\}$$ of the complex plane, there exist $$T$$-invariant closed subspaces $$Y$$, $$Z$$ of $$X$$ such that $$\sigma(T| Y)\subset U$$, $$\sigma(T| Z)\subset V$$ and $$Y+ Z= X$$; it is called superdecomposable if there exists an operator $$R$$ commuting with $$T$$ such that $$\sigma(T| \overline{R(X)})\subset U$$, $$\sigma(T|\overline{(I-R)(X)})\subset V$$. Theorems characterizing these properties are proved. The basic idea is to investigate a certain algebra of continuous functions on a compact space, not necessarily Hausdorff. Those elements of a commutative semi-simple Banach algebra are considered, whose Gelfand transform is continuous with respect to the hull-kernel topology on the maximal ideal space. This continuity property characterizes decomposability of the corresponding multiplication operator. There are interesting consequences for representations and applications to regular and non-regular Banach algebras. Finally applications to the measure algebra, and to certain natural subalgebras of it, of a locally compact Abelian group, are given.
For the entire collection see [Zbl 0746.00071].
##### MSC:
46J05 General theory of commutative topological algebras 46H05 General theory of topological algebras 47B40 Spectral operators, decomposable operators, well-bounded operators, etc. 43A10 Measure algebras on groups, semigroups, etc. 46H40 Automatic continuity | {"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": 0.8243394494056702, "perplexity": 417.6588685111663}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363465.47/warc/CC-MAIN-20211208083545-20211208113545-00110.warc.gz"} |
https://www.physicsforums.com/threads/contravariant-components-to-covariant.413881/ | # Homework Help: Contravariant components to covariant
1. Jul 2, 2010
### cavus
Hi, everyone
I was playing with the coordinate transformations and metric tensors to get a feeling of how it all behaves, and got stuck with some basic problem I am hoping you can help me with.
So, I have defined a coordinate system (s,t), with the s axis going along the x axis in the cartesian coordinates, and t axis going along the y=x line:
s = x-y
t = y*sqrt(2)
with inverse transformation:
x = s + t/sqrt(2)
y = t/sqrt(2)
If I am differentiating correctly, the metric tensor in these coordinates looks like:
1 (2+sqrt(2))/2
(2+sqrt(2))/2 1
g11 = g22 = 1,
g21=g12 = (2+sqrt(2))/2
Now, I pick a point (3,1) in cartesian coordinates, and transform it to my new frame, and get the contravariant coordinates as (2, sqrt(2)).
So far so good. What I am trying to do is find out what its covariant coordinates are going to be. I think, that covariant coordinates are supposed to be the lengths of orthogonal projections of the vector on the respective axes. From basic geometry, I get (3, 2*sqrt(2)).
The problem is that when I try to multiply my metric tensor by the contravariant vector, I get a different answer - (3+sqrt(2), 2+2*sqrt(2))
Clearly, there is something I am doing wrong here, but I can't figure out what it is :( Can somebody please help me spot the problem?
Thanks a lot for your help!
2. Jul 3, 2010
### weaselman
Your metric is wrong. The mixed components (g12 and g21) should be 1/sqrt(2).
Because you did not show how you got your metric tensor, I can't say where you went wrong, but if your check your index dropping with the correct metric, you'll see that it fits.
Hope, it helps ... | {"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": 0.8542062640190125, "perplexity": 540.4726853532554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2018-26/segments/1529267864958.91/warc/CC-MAIN-20180623113131-20180623133131-00139.warc.gz"} |
https://en.wikibooks.org/wiki/Linear_Algebra/Length_and_Angle_Measures | # Linear Algebra/Length and Angle Measures
Linear Algebra ← Vectors in Space Length and Angle Measures Reduced Echelon Form →
We've translated the first section's results about solution sets into geometric terms for insight into how those sets look. But we must watch out not to be mislead by our own terms; labeling subsets of ${\displaystyle \mathbb {R} ^{k}}$ of the forms ${\displaystyle \{{\vec {p}}+t{\vec {v}}\,{\big |}\,t\in \mathbb {R} \}}$ and ${\displaystyle \{{\vec {p}}+t{\vec {v}}+s{\vec {w}}\,{\big |}\,t,s\in \mathbb {R} \}}$ as "lines" and "planes" doesn't make them act like the lines and planes of our prior experience. Rather, we must ensure that the names suit the sets. While we can't prove that the sets satisfy our intuition— we can't prove anything about intuition— in this subsection we'll observe that a result familiar from ${\displaystyle \mathbb {R} ^{2}}$ and ${\displaystyle \mathbb {R} ^{3}}$, when generalized to arbitrary ${\displaystyle \mathbb {R} ^{k}}$, supports the idea that a line is straight and a plane is flat. Specifically, we'll see how to do Euclidean geometry in a "plane" by giving a definition of the angle between two ${\displaystyle \mathbb {R} ^{n}}$ vectors in the plane that they generate.
Definition 2.1
The length of a vector ${\displaystyle {\vec {v}}\in \mathbb {R} ^{n}}$ is this.
${\displaystyle |{\vec {v}}\,|={\sqrt {v_{1}^{2}+\cdots +v_{n}^{2}}}}$
Remark 2.2
This is a natural generalization of the Pythagorean Theorem. A classic discussion is in (Pólya 1954).
We can use that definition to derive a formula for the angle between two vectors. For a model of what to do, consider two vectors in ${\displaystyle \mathbb {R} ^{3}}$.
Put them in canonical position and, in the plane that they determine, consider the triangle formed by ${\displaystyle {\vec {u}}}$, ${\displaystyle {\vec {v}}}$, and ${\displaystyle {\vec {u}}-{\vec {v}}}$.
Apply the Law of Cosines, ${\displaystyle |{\vec {u}}-{\vec {v}}\,|^{2}=|{\vec {u}}\,|^{2}+|{\vec {v}}\,|^{2}-2\,|{\vec {u}}\,|\,|{\vec {v}}\,|\cos \theta }$, where ${\displaystyle \theta }$ is the angle between the vectors. Expand both sides
${\displaystyle (u_{1}-v_{1})^{2}+(u_{2}-v_{2})^{2}+(u_{3}-v_{3})^{2}}$
${\displaystyle =(u_{1}^{2}+u_{2}^{2}+u_{3}^{2})+(v_{1}^{2}+v_{2}^{2}+v_{3}^{2})-2\,|{\vec {u}}\,|\,|{\vec {v}}\,|\cos \theta }$
and simplify.
${\displaystyle \theta =\arccos(\,{\frac {u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}}{|{\vec {u}}\,|\,|{\vec {v}}\,|}}\,)}$
In higher dimensions no picture suffices but we can make the same argument analytically. First, the form of the numerator is clear— it comes from the middle terms of the squares ${\displaystyle (u_{1}-v_{1})^{2}}$, ${\displaystyle (u_{2}-v_{2})^{2}}$, etc.
Definition 2.3
The dot product (or inner product, or scalar product) of two ${\displaystyle n}$-component real vectors is the linear combination of their components.
${\displaystyle {\vec {u}}\cdot {\vec {v}}=u_{1}v_{1}+u_{2}v_{2}+\cdots +u_{n}v_{n}}$
Note that the dot product of two vectors is a real number, not a vector, and that the dot product of a vector from ${\displaystyle \mathbb {R} ^{n}}$ with a vector from ${\displaystyle \mathbb {R} ^{m}}$ is defined only when ${\displaystyle n}$ equals ${\displaystyle m}$. Note also this relationship between dot product and length: dotting a vector with itself gives its length squared ${\displaystyle {\vec {u}}\cdot {\vec {u}}=u_{1}u_{1}+\cdots +u_{n}u_{n}=|{\vec {u}}\,|^{2}}$.
Remark 2.4
The wording in that definition allows one or both of the two to be a row vector instead of a column vector. Some books require that the first vector be a row vector and that the second vector be a column vector. We shall not be that strict.
Still reasoning with letters, but guided by the pictures, we use the next theorem to argue that the triangle formed by ${\displaystyle {\vec {u}}}$, ${\displaystyle {\vec {v}}}$, and ${\displaystyle {\vec {u}}-{\vec {v}}}$ in ${\displaystyle \mathbb {R} ^{n}}$ lies in the planar subset of ${\displaystyle \mathbb {R} ^{n}}$ generated by ${\displaystyle {\vec {u}}}$ and ${\displaystyle {\vec {v}}}$.
Theorem 2.5 (Triangle Inequality)
For any ${\displaystyle {\vec {u}},{\vec {v}}\in \mathbb {R} ^{n}}$,
${\displaystyle |{\vec {u}}+{\vec {v}}\,|\leq |{\vec {u}}\,|+|{\vec {v}}\,|}$
with equality if and only if one of the vectors is a nonnegative scalar multiple of the other one.
This inequality is the source of the familiar saying, "The shortest distance between two points is in a straight line."
Proof
(We'll use some algebraic properties of dot product that we have not yet checked, for instance that ${\displaystyle {\vec {u}}\cdot ({\vec {a}}+{\vec {b}})={\vec {u}}\cdot {\vec {a}}+{\vec {u}}\cdot {\vec {b}}}$ and that ${\displaystyle {\vec {u}}\cdot {\vec {v}}={\vec {v}}\cdot {\vec {u}}}$. See Problem 8.) The desired inequality holds if and only if its square holds.
${\displaystyle {\begin{array}{rl}|{\vec {u}}+{\vec {v}}\,|^{2}&\leq (\,|{\vec {u}}\,|+|{\vec {v}}\,|\,)^{2}\\(\,{\vec {u}}+{\vec {v}}\,)\cdot (\,{\vec {u}}+{\vec {v}}\,)&\leq |{\vec {u}}\,|^{2}+2\,|{\vec {u}}\,|\,|{\vec {v}}\,|+|{\vec {v}}\,|^{2}\\{\vec {u}}\cdot {\vec {u}}+{\vec {u}}\cdot {\vec {v}}+{\vec {v}}\cdot {\vec {u}}+{\vec {v}}\cdot {\vec {v}}&\leq {\vec {u}}\cdot {\vec {u}}+2\,|{\vec {u}}\,|\,|{\vec {v}}\,|+{\vec {v}}\cdot {\vec {v}}\\2\,{\vec {u}}\cdot {\vec {v}}&\leq 2\,|{\vec {u}}\,|\,|{\vec {v}}\,|\end{array}}}$
That, in turn, holds if and only if the relationship obtained by multiplying both sides by the nonnegative numbers ${\displaystyle |{\vec {u}}\,|}$ and ${\displaystyle |{\vec {v}}\,|}$
${\displaystyle 2\,(\,|{\vec {v}}\,|\,{\vec {u}}\,)\cdot (\,|{\vec {u}}\,|\,{\vec {v}}\,)\leq 2\,|{\vec {u}}\,|^{2}\,|{\vec {v}}\,|^{2}}$
and rewriting
${\displaystyle 0\leq |{\vec {u}}\,|^{2}\,|{\vec {v}}\,|^{2}-2\,(\,|{\vec {v}}\,|\,{\vec {u}}\,)\cdot (\,|{\vec {u}}\,|\,{\vec {v}}\,)+|{\vec {u}}\,|^{2}\,|{\vec {v}}\,|^{2}}$
is true. But factoring
${\displaystyle 0\leq (\,|{\vec {u}}\,|\,{\vec {v}}-|{\vec {v}}\,|\,{\vec {u}}\,)\cdot (\,|{\vec {u}}\,|\,{\vec {v}}-|{\vec {v}}\,|\,{\vec {u}}\,)}$
shows that this certainly is true since it only says that the square of the length of the vector ${\displaystyle |{\vec {u}}\,|\,{\vec {v}}-|{\vec {v}}\,|\,{\vec {u}}\,}$ is not negative.
As for equality, it holds when, and only when, ${\displaystyle |{\vec {u}}\,|\,{\vec {v}}-|{\vec {v}}\,|\,{\vec {u}}}$ is ${\displaystyle {\vec {0}}}$. The check that ${\displaystyle |{\vec {u}}\,|\,{\vec {v}}=|{\vec {v}}\,|\,{\vec {u}}\,}$ if and only if one vector is a nonnegative real scalar multiple of the other is easy.
This result supports the intuition that even in higher-dimensional spaces, lines are straight and planes are flat. For any two points in a linear surface, the line segment connecting them is contained in that surface (this is easily checked from the definition). But if the surface has a bend then that would allow for a shortcut (shown here grayed, while the segment from ${\displaystyle P}$ to ${\displaystyle Q}$ that is contained in the surface is solid).
Because the Triangle Inequality says that in any ${\displaystyle \mathbb {R} ^{n}}$, the shortest cut between two endpoints is simply the line segment connecting them, linear surfaces have no such bends.
Back to the definition of angle measure. The heart of the Triangle Inequality's proof is the "${\displaystyle {\vec {u}}\cdot {\vec {v}}\leq |{\vec {u}}\,|\,|{\vec {v}}\,|}$" line. At first glance, a reader might wonder if some pairs of vectors satisfy the inequality in this way: while ${\displaystyle {\vec {u}}\cdot {\vec {v}}}$ is a large number, with absolute value bigger than the right-hand side, it is a negative large number. The next result says that no such pair of vectors exists.
Corollary 2.6 (Cauchy-Schwarz Inequality)
For any ${\displaystyle {\vec {u}},{\vec {v}}\in \mathbb {R} ^{n}}$,
${\displaystyle |\,{\vec {u}}\cdot {\vec {v}}\,|\leq |\,{\vec {u}}\,|\,|{\vec {v}}\,|}$
with equality if and only if one vector is a scalar multiple of the other.
Proof
The Triangle Inequality's proof shows that ${\displaystyle {\vec {u}}\cdot {\vec {v}}\leq |{\vec {u}}\,|\,|{\vec {v}}\,|}$ so if ${\displaystyle {\vec {u}}\cdot {\vec {v}}}$ is positive or zero then we are done. If ${\displaystyle {\vec {u}}\cdot {\vec {v}}}$ is negative then this holds.
${\displaystyle |\,{\vec {u}}\cdot {\vec {v}}\,|=-(\,{\vec {u}}\cdot {\vec {v}}\,)=(-{\vec {u}}\,)\cdot {\vec {v}}\leq |-{\vec {u}}\,|\,|{\vec {v}}\,|=|{\vec {u}}\,|\,|{\vec {v}}\,|}$
The equality condition is Problem 9.
The Cauchy-Schwarz inequality assures us that the next definition makes sense because the fraction has absolute value less than or equal to one.
Definition 2.7
The angle between two nonzero vectors ${\displaystyle {\vec {u}},{\vec {v}}\in \mathbb {R} ^{n}}$ is
${\displaystyle \theta =\arccos(\,{\frac {{\vec {u}}\cdot {\vec {v}}}{|{\vec {u}}\,|\,|{\vec {v}}\,|}}\,)}$
(the angle between the zero vector and any other vector is defined to be a right angle).
Thus vectors from ${\displaystyle \mathbb {R} ^{n}}$ are orthogonal (or perpendicular) if and only if their dot product is zero.
Example 2.8
These vectors are orthogonal.
${\displaystyle {\begin{pmatrix}1\\-1\end{pmatrix}}\cdot {\begin{pmatrix}1\\1\end{pmatrix}}=0}$
The arrows are shown away from canonical position but nevertheless the vectors are orthogonal.
Example 2.9
The ${\displaystyle \mathbb {R} ^{3}}$ angle formula given at the start of this subsection is a special case of the definition. Between these two
the angle is
${\displaystyle \arccos({\frac {(1)(0)+(1)(3)+(0)(2)}{{\sqrt {1^{2}+1^{2}+0^{2}}}{\sqrt {0^{2}+3^{2}+2^{2}}}}})=\arccos({\frac {3}{{\sqrt {2}}{\sqrt {13}}}})}$
approximately ${\displaystyle 0.94{\text{radians}}}$. Notice that these vectors are not orthogonal. Although the ${\displaystyle yz}$-plane may appear to be perpendicular to the ${\displaystyle xy}$-plane, in fact the two planes are that way only in the weak sense that there are vectors in each orthogonal to all vectors in the other. Not every vector in each is orthogonal to all vectors in the other.
## Exercises
This exercise is recommended for all readers.
Problem 1
Find the length of each vector.
1. ${\displaystyle {\begin{pmatrix}3\\1\end{pmatrix}}}$
2. ${\displaystyle {\begin{pmatrix}-1\\2\end{pmatrix}}}$
3. ${\displaystyle {\begin{pmatrix}4\\1\\1\end{pmatrix}}}$
4. ${\displaystyle {\begin{pmatrix}0\\0\\0\end{pmatrix}}}$
5. ${\displaystyle {\begin{pmatrix}1\\-1\\1\\0\end{pmatrix}}}$
This exercise is recommended for all readers.
Problem 2
Find the angle between each two, if it is defined.
1. ${\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}},{\begin{pmatrix}1\\4\end{pmatrix}}}$
2. ${\displaystyle {\begin{pmatrix}1\\2\\0\end{pmatrix}},{\begin{pmatrix}0\\4\\1\end{pmatrix}}}$
3. ${\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}},{\begin{pmatrix}1\\4\\-1\end{pmatrix}}}$
This exercise is recommended for all readers.
Problem 3
During maneuvers preceding the Battle of Jutland, the British battle cruiser Lion moved as follows (in nautical miles): ${\displaystyle 1.2}$ miles north, ${\displaystyle 6.1}$ miles ${\displaystyle 38}$ degrees east of south, ${\displaystyle 4.0}$ miles at ${\displaystyle 89}$ degrees east of north, and ${\displaystyle 6.5}$ miles at ${\displaystyle 31}$ degrees east of north. Find the distance between starting and ending positions (O'Hanian 1985).
Problem 4
Find ${\displaystyle k}$ so that these two vectors are perpendicular.
${\displaystyle {\begin{pmatrix}k\\1\end{pmatrix}}\qquad {\begin{pmatrix}4\\3\end{pmatrix}}}$
Problem 5
Describe the set of vectors in ${\displaystyle \mathbb {R} ^{3}}$ orthogonal to this one.
${\displaystyle {\begin{pmatrix}1\\3\\-1\end{pmatrix}}}$
This exercise is recommended for all readers.
Problem 6
1. Find the angle between the diagonal of the unit square in ${\displaystyle \mathbb {R} ^{2}}$ and one of the axes.
2. Find the angle between the diagonal of the unit cube in ${\displaystyle \mathbb {R} ^{3}}$ and one of the axes.
3. Find the angle between the diagonal of the unit cube in ${\displaystyle \mathbb {R} ^{n}}$ and one of the axes.
4. What is the limit, as ${\displaystyle n}$ goes to ${\displaystyle \infty }$, of the angle between the diagonal of the unit cube in ${\displaystyle \mathbb {R} ^{n}}$ and one of the axes?
Problem 7
Is any vector perpendicular to itself?
This exercise is recommended for all readers.
Problem 8
Describe the algebraic properties of dot product.
1. Is it right-distributive over addition: ${\displaystyle ({\vec {u}}+{\vec {v}})\cdot {\vec {w}}={\vec {u}}\cdot {\vec {w}}+{\vec {v}}\cdot {\vec {w}}}$?
2. Is is left-distributive (over addition)?
3. Does it commute?
4. Associate?
5. How does it interact with scalar multiplication?
As always, any assertion must be backed by either a proof or an example.
Problem 9
Verify the equality condition in Corollary 2.6, the Cauchy-Schwarz Inequality.
1. Show that if ${\displaystyle {\vec {u}}}$ is a negative scalar multiple of ${\displaystyle {\vec {v}}}$ then ${\displaystyle {\vec {u}}\cdot {\vec {v}}}$ and ${\displaystyle {\vec {v}}\cdot {\vec {u}}}$ are less than or equal to zero.
2. Show that ${\displaystyle |{\vec {u}}\cdot {\vec {v}}|=|{\vec {u}}\,|\,|{\vec {v}}\,|}$ if and only if one vector is a scalar multiple of the other.
Problem 10
Suppose that ${\displaystyle {\vec {u}}\cdot {\vec {v}}={\vec {u}}\cdot {\vec {w}}}$ and ${\displaystyle {\vec {u}}\neq {\vec {0}}}$. Must ${\displaystyle {\vec {v}}={\vec {w}}}$?
This exercise is recommended for all readers.
Problem 11
Does any vector have length zero except a zero vector? (If "yes", produce an example. If "no", prove it.)
This exercise is recommended for all readers.
Problem 12
Find the midpoint of the line segment connecting ${\displaystyle (x_{1},y_{1})}$ with ${\displaystyle (x_{2},y_{2})}$ in ${\displaystyle \mathbb {R} ^{2}}$. Generalize to ${\displaystyle \mathbb {R} ^{n}}$.
Problem 13
Show that if ${\displaystyle {\vec {v}}\neq {\vec {0}}}$ then ${\displaystyle {\vec {v}}/|{\vec {v}}\,|}$ has length one. What if ${\displaystyle {\vec {v}}={\vec {0}}}$?
Problem 14
Show that if ${\displaystyle r\geq 0}$ then ${\displaystyle r{\vec {v}}}$ is ${\displaystyle r}$ times as long as ${\displaystyle {\vec {v}}}$. What if ${\displaystyle r<0}$?
This exercise is recommended for all readers.
Problem 15
A vector ${\displaystyle {\vec {v}}\in \mathbb {R} ^{n}}$ of length one is a unit vector. Show that the dot product of two unit vectors has absolute value less than or equal to one. Can "less than" happen? Can "equal to"?
Problem 16
Prove that ${\displaystyle |{\vec {u}}+{\vec {v}}\,|^{2}+|{\vec {u}}-{\vec {v}}\,|^{2}=2|{\vec {u}}\,|^{2}+2|{\vec {v}}\,|^{2}.}$
Problem 17
Show that if ${\displaystyle {\vec {x}}\cdot {\vec {y}}=0}$ for every ${\displaystyle {\vec {y}}}$ then ${\displaystyle {\vec {x}}={\vec {0}}}$.
Problem 18
Is ${\displaystyle |{\vec {u}}_{1}+\cdots +{\vec {u}}_{n}|\leq |{\vec {u}}_{1}|+\cdots +|{\vec {u}}_{n}|}$? If it is true then it would generalize the Triangle Inequality.
Problem 19
What is the ratio between the sides in the Cauchy-Schwarz inequality?
Problem 20
Why is the zero vector defined to be perpendicular to every vector?
Problem 21
Describe the angle between two vectors in ${\displaystyle \mathbb {R} ^{1}}$.
Problem 22
Give a simple necessary and sufficient condition to determine whether the angle between two vectors is acute, right, or obtuse.
This exercise is recommended for all readers.
Problem 23
Generalize to ${\displaystyle \mathbb {R} ^{n}}$ the converse of the Pythagorean Theorem, that if ${\displaystyle {\vec {u}}}$ and ${\displaystyle {\vec {v}}}$ are perpendicular then ${\displaystyle |{\vec {u}}+{\vec {v}}\,|^{2}=|{\vec {u}}\,|^{2}+|{\vec {v}}\,|^{2}}$.
Problem 24
Show that ${\displaystyle |{\vec {u}}\,|=|{\vec {v}}\,|}$ if and only if ${\displaystyle {\vec {u}}+{\vec {v}}}$ and ${\displaystyle {\vec {u}}-{\vec {v}}}$ are perpendicular. Give an example in ${\displaystyle \mathbb {R} ^{2}}$.
Problem 25
Show that if a vector is perpendicular to each of two others then it is perpendicular to each vector in the plane they generate. (Remark. They could generate a degenerate plane— a line or a point— but the statement remains true.)
Problem 26
Prove that, where ${\displaystyle {\vec {u}},{\vec {v}}\in \mathbb {R} ^{n}}$ are nonzero vectors, the vector
${\displaystyle {\frac {\vec {u}}{|{\vec {u}}\,|}}+{\frac {\vec {v}}{|{\vec {v}}\,|}}}$
bisects the angle between them. Illustrate in ${\displaystyle \mathbb {R} ^{2}}$.
Problem 27
Verify that the definition of angle is dimensionally correct: (1) if ${\displaystyle k>0}$ then the cosine of the angle between ${\displaystyle k{\vec {u}}}$ and ${\displaystyle {\vec {v}}}$ equals the cosine of the angle between ${\displaystyle {\vec {u}}}$ and ${\displaystyle {\vec {v}}}$, and (2) if ${\displaystyle k<0}$ then the cosine of the angle between ${\displaystyle k{\vec {u}}}$ and ${\displaystyle {\vec {v}}}$ is the negative of the cosine of the angle between ${\displaystyle {\vec {u}}}$ and ${\displaystyle {\vec {v}}}$.
This exercise is recommended for all readers.
Problem 28
Show that the inner product operation is linear: for ${\displaystyle {\vec {u}},{\vec {v}},{\vec {w}}\in \mathbb {R} ^{n}}$ and ${\displaystyle k,m\in \mathbb {R} }$, ${\displaystyle {\vec {u}}\cdot (k{\vec {v}}+m{\vec {w}})=k({\vec {u}}\cdot {\vec {v}})+m({\vec {u}}\cdot {\vec {w}})}$.
This exercise is recommended for all readers.
Problem 29
The geometric mean of two positive reals ${\displaystyle x,y}$ is ${\displaystyle {\sqrt {xy}}}$. It is analogous to the arithmetic mean ${\displaystyle (x+y)/2}$. Use the Cauchy-Schwarz inequality to show that the geometric mean of any ${\displaystyle x,y\in \mathbb {R} }$ is less than or equal to the arithmetic mean.
? Problem 30
A ship is sailing with speed and direction ${\displaystyle {\vec {v}}_{1}}$; the wind blows apparently (judging by the vane on the mast) in the direction of a vector ${\displaystyle {\vec {a}}}$; on changing the direction and speed of the ship from ${\displaystyle {\vec {v}}_{1}}$ to ${\displaystyle {\vec {v}}_{2}}$ the apparent wind is in the direction of a vector ${\displaystyle {\vec {b}}}$.
Find the vector velocity of the wind (Ivanoff & Esty 1933).
Problem 31
Verify the Cauchy-Schwarz inequality by first proving Lagrange's identity:
${\displaystyle \left(\sum _{1\leq j\leq n}a_{j}b_{j}\right)^{2}=\left(\sum _{1\leq j\leq n}a_{j}^{2}\right)\left(\sum _{1\leq j\leq n}b_{j}^{2}\right)-\sum _{1\leq k
and then noting that the final term is positive. (Recall the meaning
${\displaystyle \sum _{1\leq j\leq n}a_{j}b_{j}=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}}$
and
${\displaystyle \sum _{1\leq j\leq n}{a_{j}}^{2}={a_{1}}^{2}+{a_{2}}^{2}+\cdots +{a_{n}}^{2}}$
of the ${\displaystyle \Sigma }$ notation.) This result is an improvement over Cauchy-Schwarz because it gives a formula for the difference between the two sides. Interpret that difference in ${\displaystyle \mathbb {R} ^{2}}$.
## References
• O'Hanian, Hans (1985), Physics, 1, W. W. Norton
• Ivanoff, V. F. (proposer); Esty, T. C. (solver) (Feb. 1933), "Problem 3529", American Mathematical Mothly 39 (2): 118
• Pólya, G. (1954), Mathematics and Plausible Reasoning: Volume II Patterns of Plausible Inference, Princeton University Press
Linear Algebra ← Vectors in Space Length and Angle Measures Reduced Echelon Form → | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 159, "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": 0.940387487411499, "perplexity": 316.615478098008}, "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/1579250593937.27/warc/CC-MAIN-20200118193018-20200118221018-00348.warc.gz"} |
http://math.stackexchange.com/questions/686933/can-subsequences-be-finite | # Can subsequences be finite?
Say I'm given the sequence $\{a_1,a_2,a_3,\dots\}$. Does a subsequence have to be infinite? Or can it be finite too? For example, is $\{a_1,a_2,a_3\}$ a subsequence?
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It depends on the context. – Mariano Suárez-Alvarez Feb 23 '14 at 9:10
Yes the subsequence must be infinite. Any subsequence is itself a sequence, and a sequence is basically a function from the naturals to the reals.
-
Wholeheartedly disagreed. Without more specification, a sequence is simply a function from some well-ordered set to the range of the sequence. $\omega$ (or $\mathbb{N}$, if you prefer) is one such well-ordered set, but transfinite sequences are often spoken of and finite sequences aren't at all uncommon. – Steven Stadnicki Sep 26 '14 at 22:32
Usually, this is the definition of subsequence.
Definition. Let $\{p_n\}_n$ be a sequence (in some set), and let $n_1<n_2<n_3<\ldots$ be a strictly increasing sequence of positive integers. Then the sequence $\{p_{n_j}\}_j$ is called a subsequence of $\{p_n\}_n$.
It follows that $\{p_1,p_2,p_3\}$ is not a subsequence of $\{p_n\}_n$, since $n_1=1$, $n_2=2$, $n_3=3$, but what is $n_4$? This is the common feeling for mathematical analysts.
However, we must be careful, since it all boils down to the very definition of sequence. In analysis, a sequence is a function $\mathbb{N} \to X$, or, more generally, a function $N \to X$, where $N \subset \mathbb{N}$ is such that $N$ contains all sufficiently large integers. In other disciplines, it might be useful to call sequence any function defined on a subset of $\mathbb{N}$, even a finite one.
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A sequence in a set $X$ is a function $f:\mathbb{N}\rightarrow X$. Let $g:\mathbb{N} \rightarrow \mathbb{N}$ be a strictly increasing sequence, then the composition of $f$ and $g$, is called a subsequence of $f$. Thus the domain of a subsequence is always infinite, but the range can certainly be finite.
As an example consider the sequence $f(n):=n$ even $n$ and $f(n):=1$ odd $n$. Let $g(n):=2n+1$. The range of $f\circ g$ is equal to $\{1\}$, which is finite.
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In my experience, the term subsequence has always referred to an infinite subsequence. I believe this is standard. If you want to consider finite subsequences, you should say finite subsequence, or if you want to consider subsequences that may or may not be finite, you can say possibly finite subsequence.
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A sequence (real, sets, etc...) is simply a list. In mathematics, unless specified, a finite list (i.e. sequence) is not denoted to be a sequence. A sequence MUST continue unimpeded.
By definition, a sequence of real numbers is a mapping from N -> R. The nature of this definition demonstrates that a finite sequence cannot exist under our current consideration.
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A subsequence can be finite. Note however, that you can always extend to an infinite sequence by $a_1, a_2, a_3, a_3, a_3, \cdots$
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I really don't think $a_1,a_2,a_3,a_3,\ldots$ is typically considered a subsequence of $\{a_i\}_{i=1}^\infty$. – froggie Feb 23 '14 at 9:10
-1 $a_1, a_2, a_3, a_3, a_3, \cdots$ is not a subsequene – miracle173 Feb 23 '14 at 9:12
I never said that it was a subsequence. I said that if you want to extend a finite sequence (which means it converges!) to an infinite sequence you can do so by simply extending it trivially to the convergent value. Why the downvotes? – Euler....IS_ALIVE Feb 23 '14 at 9:13
This is quite ridiculous. Does the sequence $(1,1,1,1,1 \cdots)$ have a convergent subsequence? Of course it does; namely the sequence $1$. – Euler....IS_ALIVE Feb 23 '14 at 9:15
@Euler....IS_ALIVE, I am sorry but that is ridiculuous in the usual meaning of subsequences. If that were a sensible interpretation of the term subsequence, then there would be no point in Bolzano's theorem, for all sequences $a_1$, $a_2$, $\dots$ would have a convergent subsequence, whether they be bounded or not: namely $a_1$. – Mariano Suárez-Alvarez Feb 23 '14 at 9:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9809883832931519, "perplexity": 237.25443393911897}, "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-2016-18/segments/1461860117914.56/warc/CC-MAIN-20160428161517-00021-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/prep-for-quantum-mechanics.374288/ | # Prep for quantum mechanics
1. Feb 1, 2010
### hover
Hey everyone,
I have a question (duh :P). I am hoping to take a course on quantum mechanics and I was wondering what math is necessary to understand the course material. I assume that the math needed is everything up to calculus and linear algebra. Am I correct?
Thanks
2. Feb 1, 2010
Yes, the underlying math is mostly linear algebra, (vector spaces, operators, eigenvalue problems), calculus (differential equations) and probability theory.
3. Feb 1, 2010
Seconded.
4. Feb 1, 2010
### Klockan3
Partial differential equations are important.
Similar Discussions: Prep for quantum mechanics | {"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": 0.9203563332557678, "perplexity": 2471.0147606407972}, "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/1512948520042.35/warc/CC-MAIN-20171212231544-20171213011544-00040.warc.gz"} |
https://math.stackexchange.com/questions/3667337/how-to-solve-3-lfloor-x-rfloor-lfloor-x2-rfloor-2-x/3667359#3667359 | # How to solve $3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$?
I am trying to solve the following question involving floor/greatest integer functions.
$$3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$$ with the notations $$\lfloor x \rfloor$$ denoting the greatest integer less than or equal to $$x$$ and $$\{x\}$$ to mean the fractional part of $$x$$.
I used the following property for floor functions.
$$n\leq x$$ if and only if $$n \leq \lfloor x \rfloor$$ where $$n\in \mathbb{Z}$$
Let $$p=\lfloor x^{2} \rfloor$$, then
$$p\leq \lfloor x^{2} \rfloor < p+1$$
$$\rightarrow p \leq x^{2} < p+1$$
$$\rightarrow \sqrt{p} \leq x < \sqrt{p+1}$$ , since $$\sqrt{p} \in \mathbb{Z}$$
$$\rightarrow \sqrt{p} \leq \lfloor x \rfloor < \sqrt{p+1}$$ We then have $$\sqrt{p} = \lfloor x \rfloor$$
Since $$\{x\}=x-\lfloor x \rfloor,$$
$$3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2\{x\}= 3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2(x-\lfloor x \rfloor)= 5\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2x=0$$
Substituting $$p$$, $$\sqrt{p}$$ for $$\lfloor x^{2} \rfloor$$ and $$\lfloor x \rfloor$$ respectively, and also letting $$x= \sqrt{p},$$ we get $$p = 3\sqrt{p}$$ solving for $$p$$ gives $$p=0, 9$$, and hence $$x=0, 3$$
The problem is that according to the solution for the problem, $$x$$ also equals to $$\frac{3}{2}$$ for $$\{x\}=\frac{1}{2}$$ since $$2\{x\}\in \mathbb{Z}$$. However, by definition for $$\{x\}$$, $$0 \leq \{x\} < 1$$, then $$0 \leq 2\{x\} < 2$$. How can $$\{x\}=\frac{1}{2}$$ and how do I use this to obtain $$x=\frac{3}{2}$$. I am not sure what I am missing. IF I made any mistakes in my reasoning. Can someone point it out to me please. Thank you in advance.
• It's not true that $\sqrt{p}\in\mathbb{Z}$ Say $x=1.5$ Then $p=\lfloor 2.25\rfloor = 2,$ and $\sqrt{p}=\sqrt{2}$ May 10, 2020 at 0:35
• @saulspatz thank you for pointing that out. May 10, 2020 at 0:45
Let $$x = n + r$$ where $$n = [x]$$ and $$r = \{x\}$$.
Then we have $$3n - [n^2 + 2nr + r^2]=2r$$
$$3n - n^2 - [2nr + r^2] = 2r$$
and.... oh, hey, the LHS is an integer the RHS being $$2\{x\}$$ means $$\{x\} = 0$$ or $$0.5$$.
Two options $$x$$ is an integer and $$x = [x] = n$$ and $$r=\{x\} = 0$$ and we have
$$3n-n^2=0$$ and $$n^2 = 3n$$ and $$n= 0$$ or $$n = 3$$.
So $$x = 0$$ and $$x=3$$ are two solutions.
(Check: $$x=0\implies 3[x] - [x^2] = 3*0 - 0 = 0 = \{0\}$$. Check. And $$x = 3\implies 3[x]-[x^2] = 3[3]- [3^2] = 3*3-9 = 0=\{3\}$$. Check.
And if $$x = n + \frac 12$$ and $$r = \frac 12$$ then
$$3n - n^2 - [2n\frac 12 + \frac 14] = 2\frac 12$$
$$3n - n^2 - [n + \frac 14] = 1$$
$$3n -n^2 - n = 1$$
$$n^2 - 2n + 1 =0$$ so $$(n-1)^2 = 0$$ and $$n = 1$$.
$$x = 1+\frac 12 = 1\frac 12$$.
(Check: If $$x = 1.5$$ then $$3[x] - [x^2] = 3[1.5] - [1.5^2] = 3*1 - [2.25]=3-2=1 = 2*\frac 12 = 2\{1.5\}$$. Check.)
Write $$\{x\}=x-\lfloor x\rfloor$$. Then we have $$5\lfloor x\rfloor - \lfloor x^2\rfloor = 2x$$Since the LHS is an integer, the RHS must be as well. There are two cases: $$x$$ is an integer, or $$x$$ is a half-integer.
• $$x$$ an integer. Drop the brackets: $$5x-x^2=2x;\qquad x=0,3$$
• $$x$$ is a half-integer. Write $$x=y+1/2$$. Then $$x^2 = y^2+y+1/4$$, and again we can drop the brackets: $$5y-(y^2+y)=2y+1; \qquad y=1, x=3/2$$
• may I ask how you arrive at $x$ is a half integer. I mean can't $x$ be anything else in between $0$ and $1$? May 10, 2020 at 0:46
• After we substitute $\{x\}=x-\lfloor x\rfloor$, both sides are integers. Then if $2x$ is an integer, $x$ is either an integer or a half-integer. May 10, 2020 at 0:47
• I think my issue is the following: from $0 \leq \{x\} < 1$, we get $0 \leq 2\{x\} < 2$. So how do I determine where else $\{x\}$ could be. May 10, 2020 at 0:56
• Clearly the LHS is an integer. Then $2\{x\}$ is an integer as well. This means either $2\{x\}=0$, i.e. $x$ is an integer, or $2\{x\}=1$, i.e. $x$ is a half-integer. May 10, 2020 at 1:00
• I think I see it now, Since $3\lfloor x \rfloor - \lfloor x^{2} \rfloor\in \mathbb{Z}$ and $0 \leq \{x\} < 1$ then, $\frac{3\lfloor x \rfloor - \lfloor x^{2} \rfloor }{2} = \{x\}$ implies that $\frac{3\lfloor x \rfloor - \lfloor x^{2} \rfloor }{2} \leq \{x\}<1$ which forces $\{x\}=\frac{1}{2}$ May 10, 2020 at 1:05 | {"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": 75, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9722264409065247, "perplexity": 118.51870714812101}, "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/1659882571147.84/warc/CC-MAIN-20220810040253-20220810070253-00749.warc.gz"} |
https://socratic.org/questions/how-do-you-factor-the-expression-4x-2-7x-15 | Algebra
Topics
# How do you factor the expression 4x^2 - 7x - 15?
Mar 21, 2016
#### Answer:
$x = \frac{7 \pm 12 \sqrt{2}}{8}$
I will let you finish this off
#### Explanation:
Possible factors of 15 are 1 and 15 or 3 and 5
Possible factors of 4 are 1 and 4 or 2 and 2
Let try
The 15 has to be negative that means that one of the constants is positive and the other is negative
$\left(2 x + 3\right) \left(2 x - 5\right) = 4 {x}^{2} - 10 x + 6 x \textcolor{red}{\ldots \ldots \text{Fail }}$
$\left(4 x + 3\right) \left(x - 5\right) = 4 {x}^{2} - 20 x + 3 x \textcolor{red}{\ldots \ldots \ldots . \text{Fail }}$
$\left(4 x - 15\right) \left(x + 1\right) = 4 {x}^{2} - 15 x + 4 x \textcolor{red}{\ldots \ldots \ldots . . \text{Fail }}$
Looks as though we are going to need to use the formula!
$y = a {x}^{2} + b x + c \text{ " -> " } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
a=4; b=-7; c=-15
$\implies x = \frac{7 \pm \sqrt{{\left(- 7\right)}^{2} - 4 \left(4\right) \left(- 15\right)}}{2 \left(4\right)}$
$x = \frac{7 \pm \sqrt{288}}{8}$
$x = \frac{7 \pm \sqrt{2 \times {12}^{2}}}{8}$
$x = \frac{7 \pm 12 \sqrt{2}}{8}$
I will let you finish this off
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If you are not sure what you can use to take the root of; use a factor tree. This is the one I built for $\sqrt{288}$. Look for numbers that are squared.
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http://mathhelpforum.com/calculus/151030-rate-change-directional-derivatives.html | # Math Help - Rate of change / Directional Derivatives
1. ## Rate of change / Directional Derivatives
I can't quite figure out where to begin / find an equation for this problem.... I need a good push in the right direction, thanks ahead of time.
The temperature at a point (x, y, z) is given by the following equation where T is measured in °C and x, y, z in meters. T(x, y, z) = 200e^((-x^2)(-3y^2)(-9z^2))
(a) Find the rate of change of temperature at the point P(2, -1, 2) in the direction towards the point (3, -3, 3).
(b) In which direction does the temperature increase fastest at P?
(c) Find the maximum rate of increase at P.
What equation do I need to use, in terms of the given variables in my problem??
2. Originally Posted by piyourface166
[INDENT]The temperature at a point (x, y, z) is given by the following equation where T is measured in °C and x, y, z in meters. T(x, y, z) = 200e^((-x^2)(-3y^2)(-9z^2))
(a) Find the rate of change of temperature at the point P(2, -1, 2) in the direction towards the point (3, -3, 3).
Let $u=<3,-3,3>-<2,-1,2>$.
For a) you want $\dfrac{{\nabla T(2, - 1,2) \cdot u}}{{\left\| u \right\|}}$.
3. when you have ||u|| does that equal - abs of root(6) - because vector is <1,-2,1>? The ||vector|| always confused me. I know what |vector| asks, but the second '||' throws me off.
4. Some authors (older ones particularity) do use $|u|$ for length of a vector.
More modern notation is $||u||$ to distinguish from absolute value as in $||\alpha u||=(|\alpha|)||u||$.
5. thank you so much that has been bothering me. Ok so I got [-518400e^(-432)] / root (6) - for part a, which I'm fairly confident is correct. So for part b - which direction does the temperature increase the fastest at P - we only have two points, which can only give us one vector and two directions, to P from the second point and away from P towards the second point correct? But that doesn't seem correct to me.
6. I got the problem. Thank you for all your help! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8882357478141785, "perplexity": 422.5846538104182}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988922.24/warc/CC-MAIN-20150728002308-00315-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/32059/how-can-i-force-a-compile-error-if-a-ref-does-not-exist | # How can I force a compile error if a ref does not exist?
I was curious if there was a way to force the compile to fail if a ref was not already defined. I do not want the ref to actually output anything into the document, but rather just check the existence condition. For example, I want to be able to do something like:
\begin{definition} \label{def C} \requires{def A} \requires{def B}
A \underline{C} is an A and a B such that... etc etc etc.
\end{definition}
and I do not want the \requires{def A} to actually produce output but just enforce that there was somewhere above a \label{def A}. Ideally, you could use this for theorems and definitions. This way I can just add any dependencies in my theorems/definitions and not have to worry about whether or not I already defined everything above, the compiler will tell me. Does cleveref provide this? If not, how hard would it be to add it? I looked very briefly at the source code but I am not too familiar with the language so it would take me a while.
-
Usually just use \PackageError{<package name>}{<Error message>}{<help text (simply keep empty)>} – Martin Scharrer Oct 19 '11 at 16:34
But the compiler already tells you if there were references that did not exist in the form of a warning: LaTeX Warning: There were undefined references. – Werner Oct 19 '11 at 16:35
If you really want to enforce that the label is before the \requires, then you will have to redefine the \label command, as the current definition writes the label only to the aux-files and so its location relative to the \ref doesn't matter. E.g.
\documentclass{article}
\makeatletter
\let\Orilabel\label
\renewcommand\label[1]{\@namedef{mylabel@#1}{}\Orilabel{#1}}
\newcommand\require[1]{%
\ifcsname mylabel@#1\endcsname
\else
\@latex@error{Reference #1 undefined}
\fi}
\begin{document}
\section{A}
\label{A}
\require{A}\require{B}
\ref{A},\ref{B} \label{B}
\end{document}
-
Yes, this is exactly the sort of thing I am looking for... Do I need to restore that 'at' with a \makeatother after the newcommand? – Tim Schumacher Oct 19 '11 at 18:04
I'd wager you do ;) – mpr Oct 19 '11 at 18:25
@Tim: Yes you should use \makeatother. Nothing will explode if you forget it, but one should be orderly ;-). – Ulrike Fischer Oct 19 '11 at 18:37
@UlrikeFischer: Thanks! The code seems to tell me my reference is undefined regardless of whether or not I put the label above. When I get some more time I will investigate... Thanks again for pointing me in the right direction! – Tim Schumacher Oct 19 '11 at 19:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.910853385925293, "perplexity": 998.5585569750505}, "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/1408500833715.76/warc/CC-MAIN-20140820021353-00371-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://eprint.iacr.org/2017/925 | ## Cryptology ePrint Archive: Report 2017/925
Resettably-Sound Resettable Zero Knowledge in Constant Rounds
Wutichai Chongchitmate and Rafail Ostrovsky and Ivan Visconti
Abstract: In FOCS 2001 Barak et al. conjectured the existence of zero-knowledge arguments that remain secure against resetting provers and resetting verifiers. The conjecture was proven true by Deng et al. in FOCS 2009 under various complexity assumptions and requiring a polynomial number of rounds. Later on in FOCS 2013 Chung et al. improved the assumptions requiring one-way functions only but still with a polynomial number of rounds.
In this work we show a constant-round resettably-sound resettable zero-knowledge argument system, therefore improving the round complexity from polynomial to constant. We obtain this result through the following steps.
1. We show an explicit transform from any $\ell$-round concurrent zero-knowledge argument system into an $O(\ell)$-round resettable zero-knowledge argument system. The transform is based on techniques proposed by Barak et al. in FOCS 2001 and by Deng et al. in FOCS 2009. Then, we make use of a recent breakthrough presented by Chung et al. in CRYPTO 2015 that solved the longstanding open question of constructing a constant-round concurrent zero-knowledge argument system from plausible polynomial-time hardness assumptions. Starting with their construction $\Gamma$ we obtain a constant-round resettable zero-knowledge argument system $\Lambda$.
2. We then show that by carefully embedding $\Lambda$ inside $\Gamma$ (i.e., essentially by playing a modification of the construction of Chung et al. against the construction of Chung et al.) we obtain the first constant-round resettably-sound concurrent zero-knowledge argument system $\Delta$.
3. Finally, we apply a transformation due to Deng et al. to $\Delta$ obtaining a resettably-sound resettable zero-knowledge argument system $\Pi$, the main result of this work.
While our round-preserving transform for resettable zero knowledge requires one-way functions only, both $\Lambda, \Delta$ and $\Pi$ extend the work of Chung et al. and as such they rely on the same assumptions (i.e., families of collision-resistant hash functions, one-way permutations and indistinguishability obfuscation for P/poly, with slightly super-polynomial security).
Category / Keywords: cryptographic protocols / zero knowledge, resettable ZK, resettable soundness, constant-Round, indistinguishability obfuscation
Original Publication (in the same form): IACR-TCC-2017
Date: received 21 Sep 2017, last revised 24 Sep 2017
Contact author: wutichai at cs ucla edu
Available format(s): PDF | BibTeX Citation
Short URL: ia.cr/2017/925
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http://mathhelpforum.com/differential-geometry/142164-differentiable-submanifold-print.html | # Differentiable submanifold
• April 29th 2010, 12:35 PM
AlexanderW
Differentiable submanifold
Hello.
Let $M \subset \mathbb R^n$ be a differentiable submanifold (with the induced metrics from $\mathbb R^n$), $q \in \mathbb R^n\setminus M$,
and $g:M\to \mathbb R, \quad x \mapsto \left \| q-x \right \|$ has a minimum $p \in M$.
Proof, that the vector $q-p$ is orthogonal to $T_p M$.
$T_p M$ is the tangent space at $p \in M$.
- Alexander -
• April 29th 2010, 01:08 PM
Laurent
Quote:
Originally Posted by AlexanderW
Hello.
Let $M \subset \mathbb R^n$ be a differentiable submanifold (with the induced metrics from $\mathbb R^n$), $q \in \mathbb R^n\setminus M$,
and $g:M\to \mathbb R, \quad x \mapsto \left \| q-x \right \|$ has a minimum $p \in M$.
Proof, that the vector $q-p$ is orthogonal to $T_p M$.
$T_p M$ is the tangent space at $p \in M$.
- Alexander -
For any differentiable curve $\gamma$ on $M$ such that $\gamma(0)=p$, the function $s\mapsto \|q-\gamma(s)\|^2$ has a minimum at 0, hence its derivative at this point is 0, which gives $2(q-\gamma(0),-\gamma'(0))=0$, i.e. $(q,\gamma'(0))=0$: $q$ is orthogonal to $\gamma'(0)\in T_pM$. Since $T_pM$ is spanned by the vectors of the form $\gamma'(0)$ (this is even one possible definition), we deduce that $q$ is orthogonal to $T_pM$.
• April 29th 2010, 01:57 PM
AlexanderW
Hello
Quote:
Originally Posted by Laurent
For any differentiable curve $\gamma$ on $M$ such that $\gamma(0)=p$, the function $s\mapsto \|q-\gamma(s)\|^2$ has a minimum at 0, hence its derivative at this point is 0, which gives $2(q-\gamma(0),-\gamma'(0))=0$, i.e. $(q,\gamma'(0))=0$: $q$ is orthogonal to $\gamma'(0)\in T_pM$. Since $T_pM$ is spanned by the vectors of the form $\gamma'(0)$ (this is even one possible definition), we deduce that $q$ is orthogonal to $T_pM$.
I don't understand why do you consider the function $s\mapsto \|q-\gamma(s)\|^2$ and not the function $s\mapsto \|q-\gamma(s)\|$.
And why is the derivation at 0 of $s\mapsto \|q-\gamma(s)\|^2$ the same as $2(q-\gamma(0),-\gamma'(0))$ ? Is $(.,.)$ the dot product?
I don't understand why do you consider the function $s\mapsto \|q-\gamma(s)\|^2$ and not the function $s\mapsto \|q-\gamma(s)\|$.
The minima of $x\mapsto\|q-x\|$ and $x\mapsto\|q-x\|^2$ are the same, and the second one is easier to differentiate, that's why.
And why is the derivation at 0 of $s\mapsto \|q-\gamma(s)\|^2$ the same as $2(q-\gamma(0),-\gamma'(0))$ ? Is $(.,.)$ the dot product?
Yes, it is dot product. The differential of $N:x\mapsto \|x\|^2$ is $dN_x(h)=2(x,h)$. And I used the chain-rule. | {"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": 56, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9949802756309509, "perplexity": 148.86041721857003}, "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/1471982292659.26/warc/CC-MAIN-20160823195812-00285-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://pillowlab.wordpress.com/2013/02/05/lab-meeting-242013-asymptotically-optimal-tuning-curve-in-lp-sense-for-a-poisson-neuron/ | # Lab Meeting 2/4/2013: Asymptotically optimal tuning curve in Lp sense for a Poisson neuron
Optimal tuning curve is the best transformation of the stimulus into neural firing pattern (usually firing rate) under certain constraints and optimality criterion. The following paper I saw at NIPS 2012 was related to what we are doing, so we took a deeper look into it.
Wang, Stocker & Lee (NIPS 2012), Optimal neural tuning curves for arbitrary stimulus distributions: Discrimax, infomax and minimum Lp loss.
The paper assumes a single neuron encoding a 1 dimensional stimulus, governed by a distribution $\pi(s)$. The neuron is assumed to be Poisson (pure rate code). The neuron’s tuning curve $h(s)$ is smooth, monotonically increasing (with $h'(s) > c$), and has a limited minimum and maximum firing rate as its constraint. Authors assume asymptotic regime for MLE decoding where the observation time $T$ is long enough to apply asymptotic normality theory (and convergence of p-th moments) of MLE.
The authors show that there is a 1-to-1 mapping between the tuning curve and the Fisher information $I$ under these constraints. Then for various loss functions, they derive the optimal tuning curve using calculus of variations. In general, to minimize the Lp loss $E\left[ |\hat s - s|^p \right]$ under the constraints, the optimal (squared) tuning curve is:
$\sqrt(h(s)) = \sqrt{h_{min}} + (\sqrt{h_{max}} - \sqrt{h_{min}}) \frac{\int_{-\infty}^s \pi(t)^{1/(p+1)} \mathrm{d}t}{\int_{-\infty}^\infty \pi(t)^{1/(p+1)} \mathrm{d}t}$
Furthermore, in the limit of $p \to 0$, the optimal solution corresponds to the infomax solution (i.e., optimum for mutual information loss). However, all the analysis is only in the asymptotic limit, where the Cramer-Rao bound is attained by the MLE. For the case of mutual information, unlike noise-less case where the optimal tuning curve becomes the stimulus CDF (Laughlin), for Poisson noise, it turns out to be the square of the stimulus CDF. I have plotted the differences below for a normal distribution (left) and a mixture of normals (right):
The results are very nice, and I’d like to see more results with stimulus noise and with population tuning assumptions. | {"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": 8, "mimetex.cgi": 0, "/images/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": 0.971398651599884, "perplexity": 829.7321199331689}, "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-2015-48/segments/1448398447758.91/warc/CC-MAIN-20151124205407-00056-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://mathhelpforum.com/differential-equations/282667-domain-differential-equation.html | ## Domain in differential equation
At the moment, I am struggling to understand how the domain for the general solution of the differential equation can be worked out. I managed to get the correct general solution but I don't understand how the 3 domains are deduced for each of the two restrictions.
I only get the part that t=0 or -1/c as these are the two discontinuity points.
Please could I have some explanation for this?
Thank you | {"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": 0.8222066760063171, "perplexity": 187.90465307151743}, "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/1563195525402.30/warc/CC-MAIN-20190717201828-20190717223828-00279.warc.gz"} |
https://www.physicsforums.com/threads/inelastic-collisions-and-dissipation-of-energy.203764/ | # Inelastic Collisions and Dissipation of Energy
1. Dec 10, 2007
### chiurox
1. The problem statement, all variables and given/known data
Block A of mass, mA = 2.0 kg, is moving on a frictionless surface with a velocity of 5.0 m/s to the right and another block B of mass, mB = 8.0 kg, is moving with a velocity of 3.0 m/s to the left, as shown in the diagram below. The two block eventually collide.
d. If the initial velocity of Block A is 4.0 m/s to the right and -30 J is dissipated in the collision, what was the initial velocity of Block B?
3. The attempt at a solution
mavai + mbvbi = (ma+mb) vf
(0.5)mavai² + (0.5)mbvbi² = Ei
(0.5)(ma+mb)vaf² = Ef
Ef – Ei = -30 J
(4.0) + (4.0)vbi = (5.0) vf
(16) + (4.0)vbi² = Ei
(5.0)vf² = Ef
Ef – Ei = -30 J
I don't know how to go from here. Any help?
2. Dec 10, 2007
### catkin
I don't know where to go from here, either. The question as you have given it gives two velocities for blocks A and B -- and you've made your query harder to understand by missing out the formulae you are using and poor formatting of your attempt.
3. Dec 10, 2007
### chiurox
About the poor formating, yes, I noticed it's pretty bad. I pasted directly from Word because I'm still not familiar with the formating thing we have for this forum.
4. Dec 10, 2007
### dotman
Hello,
Momentum is conserved, as you have shown. This can be used to show how the before and after kinetic energies are related-- see
http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html" [Broken]
I think this may help a bit. If you take a look at the formulas you have written already:
$$E_f - E_i = -30$$
$$m_av_{ai} + m_bv_{bi} = (m_a + m_b)v_f$$
$$\frac{1}{2}m_a{v_{ai}}^2 + \frac{1}{2}m_b{v_{bi}}^2 = E_i$$
$$\frac{1}{2}(m_a + m_b){v_f}^2 = E_f$$
You have four equations and four unknowns ($E_i , E_f, v_{bi}, v_f$) so you should be able to solve this. I would substitute my expressions for $E_i and E_f$ into the first equation, and then use the second equation to substitute in for either $v_f or v_{bi}$.
Finally, check your interpretation of the first equation. You stated that "-30J are dissipated"-- does this mean that 30J are actually gained in the collision? Or were they meaning that 30J were dissipated (lost)? If the former, your equation has a sign error. If the latter (which is most likely), you should be ok.
Solve it with the letters, and put the numbers in last.
Last edited by a moderator: May 3, 2017
5. Dec 10, 2007
### catkin
No amount of conservation of momentum and energy equations is going to solve the riddle of which velocity to use! Has something significant been lost along with parts a, b and c of the question?
6. Dec 10, 2007
### chiurox
Alright, I solved it.
I ended up isolating just the Vf² and the Vf into one equation so I just solved it with the quadratic formula. So I got two final velocities, one positive and one negative. Since the initial velocity of block B has to be negative, Vf had to be negative too.
By dissipated I mean the energy is lost, that's why I put negative.
So the initial velocity for Block B is -1.4 m/s.
I guess it makes sense, could you guys confirm?
Thank you.
7. Dec 10, 2007
### dotman
Hello,
That's not what I got, but I just did the problem real quick on a napkin, and may have made a mistake. What value did you get for Vf? You can at least check if your answer "makes sense".
More importantly, why does the initial velocity of block B have to be negative? And even if it is, this does not imply that Vf is negative, necessarily-- what if Vb is very very slightly negative? (What if its zero?)
Lastly, there is a simpler way to do this problem. Did you check out that link?
EDIT: Actually, re-looking at that link, it looks like its only valid for the case of Block B being motionless. So that's no good for here, necessarily.
Last edited: Dec 10, 2007
8. Dec 11, 2007
### chiurox
Yeah, I checked that link and saw that one of the masses was motionless, so the situation is a bit different from this one.
I redid the problem and got two answers: 0.9 m/s or -13 m/s for the initial velocity of block B and two final velocities for both objects as 1.5 m/s or -9.5 It makes sense I guess, because block A is traveling about twice as faster and hits the bigger block B making Block B travel a bit faster while block A has energy dissipated due to the collision.
Is that what you got?
Last edited: Dec 11, 2007
9. Dec 11, 2007
### dotman
Not quite what I got. From a physical standpoint, you're right, the answers kind of make sense. In the first case, the big block is moving slowly to the right, and the small block comes up and hits it, and the two now move slightly faster than Block B, as expected. In the second case, the massive block flies in fast, to the left, strikes block A, reverses it, and they continue left more slowly. That's all well and good.
You can check, however, to see if your answers are correct or not. They have to satisfy your initial equations-- conservation of momentum, and your energy equation. Let's have a look:
First, vb = 0.9 m/s and vf = 1.5 m/s:
$$m_av_{ai} + m_bv_{bi} = (m_a + m_b)v_f$$
$$\ \Rightarrow (2){(4)} + (8){(0.9)} = (2 + 8){(1.5)}$$
$$\ \Rightarrow 8 + 7.2 = 15$$
$$\ \Rightarrow 15.2 = 15$$
So that's good, the slight difference is due to rounding. Now for the energy equation:
$$\frac{1}{2}m_a{v_{ai}}^2 + \frac{1}{2}m_b{v_{bi}}^2 = E_i$$
$$\frac{1}{2}(m_a + m_b){v_f}^2 = E_f$$
$$E_i - E_f = 30$$
$$\ \Rightarrow \frac{1}{2}m_a{v_{ai}}^2 + \frac{1}{2}m_b{v_{bi}}^2 - [ \frac{1}{2}(m_a + m_b){v_f}^2] = 30$$
$$\ \Rightarrow \frac{1}{2}(2){(4)}^2 + \frac{1}{2}(8){(0.9)}^2 - \frac{1}{2}(2 + 8){(1.5)}^2 = 30$$
$$\ \Rightarrow 16 + 3.24 - 11.25 = 30$$
$$\ \Rightarrow 19.24 - 11.25 = 30$$
$$\ \Rightarrow 7.99 \not= 30$$ So there's a problem here.
You can do the same kind of thing for your other set of answers, to see what fits and what doesn't.
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https://www.physicsforums.com/threads/planet-formation-question.355541/ | # Planet Formation Question
1. Nov 17, 2009
### KevinMWHM
Is it common for rocky planets to form closer to a star and gas planets further?
Why or why not is this?
2. Nov 17, 2009
### Sorry!
Well I don't know how many terrestrial planets have been found around other stars... as far as I know there haven't been any. Reason being is that they are too close to the star and fairly small to be directly observed.
The reason they form closer to the star is simply because they are made out of heavier materials
3. Nov 17, 2009
### qraal
Before the discovery of "Hot Jupiters", "Hot Neptunes" and "Super-Earths" people expected the rocky planets to form inside the "Ice Line" and gaseous/icy planets to form outside of it. The "Ice Line" is the distance around the proto-star at which ice can condense from gaseous into solid form, and it means a lot more material - commonly called "ices" - becomes trapped by the proto-planets and thus they form heavier than the rocky planets.
Since then we've learnt that kinds of planets can form in one place then migrate inwards or outwards from the proto-star. Many of the known exoplanets seem to have migrated from beyond the Ice Line.
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https://www.answers.com/Q/How_is_sin_90_equal_to_1 | 0
# How is sin 90 equal to 1?
Wiki User
2008-08-24 08:08:16
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Buckle up, 'cause we can't draw diagrams here and we have to explain everything. Let's jump. Draw a graph with an x-axis and a y-axis like usual. Don't use graph paper or a ruler unless you have to. Just eyeball the thing. We're going to draw a right triangle on the graph and here's how we'll do it. Start at the origin, (0, 0) and draw a line along the x-axis about "6 or 7 units" long. Now draw a line from the end of the first one straight up (at a right angle to the x-axis) and make it about "2 units" long. Lastly, draw the "slanted" line from the origin up to where the vertical line ended. That last line was the hypotenuse of your nice right triangle. Got a good picture? Super. Let's jump to some review. The trigonometry (trig) function called the "sine" (sin) is the relationship in any right triangle between the length of the opposite side (to an given angle in the triangle) and the length of the hypotenuse of that triangle. It's actually the length of the opposite side of the triangle divided by the length of the hypotenuse of the triangle. This number is a "pure" number without units because the units (inches, feet, miles - whatever) cancel out when the division is made. Now that we've reviewed the sine function, let's take it to our triangle. Look at the angle made by the first line you drew and the last one you drew (which was the hypotenuse). It's the angle with the origin of the graph (0, 0) as the vertex. It's gonna be 25 to 35 degrees or so, ballpark. We don't need to be exact. Now, the sine of that angle is the length of the opposite side divided by the length of the hypotenuse. It's some number between 0 and 1. The hypotenuse is obviously larger, and we'll end up with a fraction or, if you prefer, a decimal number. We don't need to know what it is because we are going to be looking at a "trend" or "shift" as we change our graph. We have some number as the sine, and we're good. Now let's modify our graph and draw a new triangle. Follow closely when we jump. We are going to "keep" the hypotenuse we drew. But we are going to "rotate it up" to make a new triangle. Note that we won't change its length. We're going to "open up" the angle between the x-axis and the hypotenuse. Let's do that by detatching the hypotenuse from the short vertical to the x-axis (which is that little second line we drew). Swing the hypotenuse up (that's counterclockwise from its first position) and put it about "half way" between where it was and where the y-axis is. Got it located? Now "drop a perpendicular" from the end of the hypotenuse to the x-axis, and make the line perpendicular to the x-axis. This forms a new right triangle. And this new triangle has a longer "second side" that is vertical to the x-axis. Let's look at our new triangle. The "new" angle formed by the x-axis and the new location of the hypotenuse is larger than it was. And the sine for that angle has changed. The sine is (again) the length of the opposite side over the length of the hypotenuse, and notice that the "new" opposite side is longer than the old one. (We can call that side, the one perpenducular to the x-axis, the "second side" here.) That means the "new" sine will be a larger fraction or a larger decimal (if you work it that way) than before. We don't know the exact number, but we only need to look at it in comparison to what it was. And it's bigger. So let's rotate the hypotenuse more. Start moving it in a slow but continuous motion in the counterclockwise direction. It's heading for the y-axis as you rotate it. Now focus. The new triangle formed as we rotate the hypotenuse (again, without changing its length) will have a longer and longer "perpendicular" to the x-axis as we move the hypotenuse. Pretend that the second side, the one we keep making longer as we rotate the hypotenuse up, is a rubber band stretching longer and longer as we rotate the hypotenuse. It still has to make a right angle where it is attached to the x-axis, so it must "slide along" that axis toward the origin to keep the angle at 90 degrees. Make sense? The triangle is "getting taller" as we rotate the hypotenuse. And the base is getting shorter and shorter. Through all this, the sine of the angle we are looking at is getting bigger and bigger. See how it works? One more jump. As the hypotenuse is rotated counter clockwise and approaches the y-axis, the length of that "second side" will continue to grow and will actually approach the length of the hypotenuse itself. (The triangle's base gets tinier and tinier through all this.) Our angle is getting bigger and bigger, too, and it is approaching 90 degrees. As the length of the second side approaches the length of the hypotenuse, the sine of the angle, that is, the length of the second side divided by the length of the hypotenuse, actually approaches one. That's because the second side is getting almost as long as the hypotenuse. Closer and closer to vertical we move that hypotenuse. At vertical, that is, when the hypotenuse is rotated to vertical, the triangle "disappears" from view, but imagine what is happeing as we approach this "vanishing point" where the triangle ceases to exist. At 90 degrees, the second side is the exact same length as the hypotenuse. That means the angle formed at the vertex becomes 90 degrees. And the base will be so short as to disappear as well. At the 90 degree point where the hypotenuse has been rotated up to lie along the y-axis, the length of the opposite will equal to the length of the hypotenuse. And the sine of the angle (which is 90 degrees) will be the length of the second side exactly 1 at this point. The sine of an angle varies as the measure of the angle, and as the angle increases in measure from 0 to 90 degrees, the sine of the angle varies from 0 to 1 as we discovered.
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2008-08-24 08:08:16
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0 | {"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": 0.8946481943130493, "perplexity": 306.26886621233587}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153474.19/warc/CC-MAIN-20210727170836-20210727200836-00319.warc.gz"} |
https://www.zbmath.org/?q=an%3A0571.68044 | ×
# zbMATH — the first resource for mathematics
CSP-programs as nets with individual tokens. (English) Zbl 0571.68044
Advances in Petri nets 1984, Lect. Notes Comput. Sci. 188, 169-196 (1985).
[For the entire collection see Zbl 0556.00014.]
We define a subclass of predicate/transition-nets and show how to translate CSP-programs into such nets. We consider a subset of CSP for which M. Hennessy, W. Li and G. Plotkin have given an operational semantics [Proc. 2nd Int. Conf. Distributed Computing, Paris 1981]. We show that the firing sequences of our net translation correspond to this operational semantics. Additionally, we also give a non-interleaving semantics to CSP by considering unfoldings and processes of the net translation. This enables us to analyse CSP-programs applying net theoretic methods.
##### MSC:
68Q85 Models and methods for concurrent and distributed computing (process algebras, bisimulation, transition nets, etc.) 68N25 Theory of operating systems | {"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": 0.8533471822738647, "perplexity": 4208.933591157854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038057476.6/warc/CC-MAIN-20210410181215-20210410211215-00252.warc.gz"} |
https://physics.stackexchange.com/questions/441414/if-a-mu-is-not-determined-uniquely-by-maxwells-equations-what-happens-if-we | # If $A^\mu$ is not determined uniquely by Maxwell's equations, what happens if we solve for it numerically?
Given a solution $$A^{\mu}(x)$$ to Maxwell's equations $$$$\Box A^{\mu}(x)-\partial^{\mu}\partial_{\nu}A^{\nu}=0\tag{1}$$$$ which also satisfies some specified initial conditions at time $$t_0$$ $$$$A^{\mu}(\vec{x},t_0)=f^{\mu}(\vec{x}),\quad \dot{A}^{\mu}(\vec{x},t_0)=g^{\mu}(\vec{x})\tag{2}$$$$ we have that the function $$$$A^{'\mu}(x)=A^{\mu}(x)+\partial^{\mu}\alpha(x)\tag{3}$$$$ also satisfies the equations of motion, and if we arrange that the scalar function $$\alpha$$ also satisfy that $$$$\partial^{\mu}\alpha(\vec{x},t_0)=0,\quad \partial^{\mu}\dot{\alpha}(\vec{x},t_0)=0 \tag{4}$$$$ at the initial time $$t_0$$, then the new solution $$A^{'\mu}$$ also satisfies the initial conditions. For example, the function $$$$\alpha(\vec{x},t)=(t-t_0)^5h(\vec{x})e^{-(t-t_0)^2}$$$$ satisfies the conditions Eq.$$(4)$$ and also vanishes at $$t\rightarrow \pm \infty$$. Therefore, the solution to Eq.$$(1)$$ is not uniquely determined by the initial data Eq.$$(2)$$.
Question: If one simulates Eq.$$(1)$$ numerically on a computer, why is the field configuration at a later time not uniquely determined by the data in Eq.$$(2)$$?
• Try and simulate it yourself. Spoiler alert: you won't be able to, at least not without fixing the gauge first. Numerically solving a PDE requires, for example, inverting a matrix/solving a linear system. This doesn't work when you have gauge invariance, because the matrix is singular. – AccidentalFourierTransform Nov 17 at 0:04
• @AccidentalFourierTransform This isn't quite true. Your numerics may or not converge to a solution, depending on the algorithm. Some techniques involve solving a linear system, and they'll fail, but many techniques will e.g. trivially converge to the solution $\alpha \equiv 0$. The issue is non-uniqueness, not non-existence. – tparker Nov 17 at 2:49
• @tparker I never said anything about non-existence. A linear system with singular matrix has an infinite number of solutions. So we agree the issue is about non-uniqueness, not about non-existence. – AccidentalFourierTransform Nov 17 at 2:53
• – Qmechanic Nov 17 at 3:37
• If one simulates Eq.(1) numerically on a computer, why is the field configuration at a later time not uniquely determined by the data in Eq.(2)? I don't think the assumption is true. Even though the solution is non-unique, your algorithm can converge to a particular solution. Take the ordinary equation $x^ 2=1$. If you apply the bisection method in the interval $[0,2]$, you find the solution $x=1$, although you miss $x=-1$. Other methods might not converge. So I think that without specifying a particular numerical method, answers are going to be very vague. – jinawee Nov 17 at 11:46
Not all initial value problems have unique solution. Your example of $$\alpha$$ function demonstrates that this initial value problem is of such kind.
In this case, the problem is in the system of partial differential equations
$$\partial_\nu\partial^\nu A^{\mu}-\partial^{\mu}\partial_{\nu}A^{\nu}=0$$ itself; it does not put enough constraint on the functions $$\varphi(\mathbf x,t), \mathbf A(\mathbf x,t)$$. It is somewhat similar to a situation in linear algebra that sometimes occurs where a system of $$n$$ linear equations for $$n$$ unknowns has infinity of solutions.
A slightly different way to see this: notice that nowhere in the above system of PDE can we find $$\partial_t^2 A^0$$ or $$\partial_t A^0$$ directly; only a spatial gradient of $$\partial_t A^0$$ is present. The equations for $$A^i$$'s do not relate them directly with time derivatives of $$\varphi$$.
This means that if we have a solution of the initial value problem $$\varphi(x,t),\mathbf A(x,t)$$ and replace the scalar potential by $$\varphi' = \varphi(x,t)+ht^2$$ at time $$t = 0$$ (where $$h$$ is a constant), the equations are still satisfied and at $$t=0$$, initial conditions are satisfied too. This would not be so obviously possible if the system contained directly time derivatives of $$\varphi$$. Consider a slightly different system
$$\partial_\nu\partial^\nu A^{\mu}= 0,$$ (which in EM theory can be derived as a result of the Lorenz gauge choice) - this does constrain $$\partial_t^2 \varphi$$, so the above argument fails. I think this system should have a unique solution, because it is very similar to a set of equations for independent harmonic oscillators. However, for proof better check with mathematicians.
Are you asking for the physical or mathematical explanation? Dan Yand's answer gives the physical explanation.
Regarding the mathematical question: On what basis would you expect the field configuration to be uniquely determined by its initial data? Unlike for (uncoupled) ODE's, there's no theorem to that effect for general linear homogeneous second-order PDEs.
• The issue is not about PDE vs ODE. There are point particle systems with gauge symmetries, and whose time evolution is not uniquely fixed by the equations of motion. And vice versa: there are field systems whose time evolution is uniquely fixed by the equations of motion (say, the heat/Schrödinger equation). The issue is about invertibility of the differential operator, equiv. about existence of a unique Green function. Obstructions may appear whether the system is one-dimensional or not. – AccidentalFourierTransform Nov 17 at 1:22
• A Lagrangian of the form $L(q_1,q_2)=f(q_1-q_2)$, for arbitrary $f$, is invariant under $q_i(t)\to q_i(t)+\eta(t)$. The system has a gauge symmetry. I leave this to you to pick some specific $f$ and compute Euler-Lagrange. You get two redundant equations of motion, so only one independent equation for two degrees of freedom. No unique solution. Etc. (And if we are just going to cite references, let me quote Henneaux, Teitelboim "Quantization of Gauge Systems", which is a book about point particles, not fields). – AccidentalFourierTransform Nov 17 at 2:51
• @AccidentalFourierTransform Oops, you're right. I meant that a function $\mathbb{R} \to \mathbb{R}$ can't have a gauge freedom, but you can get around that by adding more variables at either end of the arrow. Edited to clarify. – tparker Nov 17 at 2:57
• A system with a single degree of freedom, if it has a gauge symmetry, has no effective degrees of freedom at all. So its dynamics are purely topological and/or due to constraints. For example, a relativistic point particle, in the reparametrisation-invariant formalism, has a gauge symmetry, and it is still $\mathbb R\to\mathbb R$. – AccidentalFourierTransform Nov 17 at 3:00
• @AccidentalFourierTransform I would describe a point particle (whether relatvistic or not) with trajectory $(t(\tau), x(\tau))$ as being described by a a function $\mathbb{R} \to \mathbb{R}^2$, not $\mathbb{R} \to \mathbb{R}$. – tparker Nov 17 at 3:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 34, "wp-katex-eq": 0, "align": 0, "equation": 5, "x-ck12": 0, "texerror": 0, "math_score": 0.9266110062599182, "perplexity": 319.01480517555126}, "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-2018-51/segments/1544376823785.24/warc/CC-MAIN-20181212065445-20181212090945-00541.warc.gz"} |
https://asmedigitalcollection.asme.org/FPST/proceedings-abstract/FPMC2018/51968/V001T01A017/271189 | Passive heave compensation (PHC) system is widely applied in offshore equipment because of its superiority in energy conserving and reliability. However, it has poor adaptability to changing sea conditions and the compensation accuracy is low. Hydraulic transformer (HT), working as a pressure-flow control element, can potentially solve the problems mentioned above. In this paper, an HT based PHC (HTPHC) system is proposed for the first time, and a compensation algorithm based on higher-order sliding mode (HOSM) together with a prediction algorithm for the heave motion of the vessel is derived to get good compensation effect using the new PHC system. The prediction algorithm is proved to be effective according to the measured data of sea trials, and reduces the difficulty of designing and parameter tuning process compared with the existing ones. The effectiveness of the proposed control algorithm is evaluated with simulation, moreover, the effectiveness can still be maintained under changing sea conditions which is also verified by simulation.
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": 0.8655844330787659, "perplexity": 1038.7351570886592}, "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/1659882571483.70/warc/CC-MAIN-20220811164257-20220811194257-00200.warc.gz"} |
https://jp.maplesoft.com/support/help/addons/view.aspx?path=RealRange&L=J | ComplexRange - Maple Help
# Online Help
###### All Products Maple MapleSim
RealRange
represent a real interval
ComplexRange
represent a complex interval
Calling Sequence RealRange(a,b) ComplexRange($\mathrm{\alpha }$,$\mathrm{\beta }$)
Parameters
a, b - real numbers, can be $\mathrm{\infty }$, can be Open(a) and/or Open(b) $\mathrm{\alpha }$, $\mathrm{\beta }$ - complex numbers, can be Open($\mathrm{\alpha }$) and/or Open($\mathrm{\beta }$); for ComplexRange, $\mathrm{\alpha }$ and $\mathrm{\beta }$ can also be algebraic expressions
Description
• RealRange(a, b) represents a real interval, that is, a segment of the real line specified by the values of its two extremes a, b, which can be any two real numbers (possibly $\mathrm{\infty }$) satisfying $a\le b$. The interval is closed, i.e., includes its extremes, unless they are represented by Open(a) (and/or Open(b)), in which case the interval is open with respect to a (and/or b).
• Similarly ComplexRange($\mathrm{\alpha }$, $\mathrm{\beta }$) represents a complex interval, that is, a rectangle in the complex plane specified by two points $\mathrm{\alpha }$ = a + b I and $\mathrm{\beta }$ = c + d I, where {a, b, c, d} represent real numbers. This complex interval is closed and so includes the four line segments $\left(a,c\right)$, $\left(b,d\right)$ and the other two parallel segments delimiting the rectangle, unless any of its four real extremes $\left\{a,b,c,d\right\}$ is entered like, for instance, Open(a), in which case the interval is "open" with respect to that point. In the presence of more Open extremes the interval may be open with respect to one or more delimiting segments.
• A rapid picture of what is and what is not included in a given complex range is obtained by converting it to a real range or converting it to a relation - see the examples.
Note: in Maple, by convention, when you say, for instance, $z\le 1$, it is implicitly assumed that $\mathrm{\Im }\left(z\right)=0$.
Examples
RealRange is a Maple program -- it analyzes the input parameters; ComplexRange is not a program.
In this example, the two endpoints are equal and it simplifies to zero.
> $\mathrm{RealRange}\left(0,0\right)$
${0}$ (1)
Here there is no automatic simplification:
> $\mathrm{ComplexRange}\left(0,0\right)$
${\mathrm{ComplexRange}}{}\left({0}{,}{0}\right)$ (2)
The automatic simplification is triggered after the conversion happens:
> $\mathrm{convert}\left(,\mathrm{RealRange}\right)$
${0}$ (3)
A ComplexRange is an object more general than a RealRange in that it contains it as a particular case. When working with a ComplexRange or its RealRange representation, three typical constructions are used. Note the corresponding notation in the following examples. The conversion (when possible) always returns a sequence of two elements.
Case 1: a ComplexRange is itself an object
> $\mathrm{CR}≔\mathrm{ComplexRange}\left(-1-I,\mathrm{Open}\left(0\right)+I\right)$
${\mathrm{CR}}{≔}{\mathrm{ComplexRange}}{}\left({-1}{-}{I}{,}{\mathrm{Open}}{}\left({0}\right){+}{I}\right)$ (4)
> $\mathrm{convert}\left(\mathrm{CR},\mathrm{RealRange}\right)$
$\left[{-1}{,}{0}\right){,}{I}{}\left[{-1}{,}{1}\right]$ (5)
Case 2: a construction indicating that z has values in some ComplexRange, expressed using the :: operator
> $z::\mathrm{CR}$
${z}{::}{\mathrm{ComplexRange}}{}\left({-1}{-}{I}{,}{\mathrm{Open}}{}\left({0}\right){+}{I}\right)$ (6)
> $\mathrm{convert}\left(,\mathrm{RealRange}\right)$
${\mathrm{\Re }}{}\left({z}\right){::}\left[{-1}{,}{0}\right){,}{\mathrm{\Im }}{}\left({z}\right){::}\left[{-1}{,}{1}\right]$ (7)
Case 3: a construction indicating that z has values in some ComplexRange, expressed using the in operator
> $z\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{CR}$
${z}{\in }{\mathrm{ComplexRange}}{}\left({-1}{-}{I}{,}{\mathrm{Open}}{}\left({0}\right){+}{I}\right)$ (8)
> $\mathrm{convert}\left(,\mathrm{RealRange}\right)$
${\mathrm{\Re }}{}\left({z}\right){\in }\left[{-1}{,}{0}\right){,}{\mathrm{\Im }}{}\left({z}\right){\in }\left[{-1}{,}{1}\right]$ (9)
Note that, unlike ComplexRange, RealRange requires numerical arguments, so when the former has not this kind of argument the conversion is not possible.
> $\mathrm{ComplexRange}\left(\mathrm{\alpha },\mathrm{\beta }\right)$
${\mathrm{ComplexRange}}{}\left({\mathrm{\alpha }}{,}{\mathrm{\beta }}\right)$ (10)
Without numerical arguments, this cannot be converted:
> $\mathrm{convert}\left(,\mathrm{RealRange}\right)$
${\mathrm{ComplexRange}}{}\left({\mathrm{\alpha }}{,}{\mathrm{\beta }}\right)$ (11)
Expressing complex and real ranges as relations
> $\mathrm{CR}≔z\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{ComplexRange}\left(-1-I,\mathrm{Open}\left(1\right)+I\right)$
${\mathrm{CR}}{≔}{z}{\in }{\mathrm{ComplexRange}}{}\left({-1}{-}{I}{,}{\mathrm{Open}}{}\left({1}\right){+}{I}\right)$ (12)
> $\mathrm{convert}\left(\mathrm{CR},\mathrm{relation}\right)$
${-1}{\le }{\mathrm{\Re }}{}\left({z}\right){\wedge }{\mathrm{\Re }}{}\left({z}\right){<}{1}{\wedge }{-1}{\le }{\mathrm{\Im }}{}\left({z}\right){\wedge }{\mathrm{\Im }}{}\left({z}\right){\le }{1}$ (13)
> $\mathrm{convert}\left(\mathrm{CR},\mathrm{RealRange}\right)$
${\mathrm{\Re }}{}\left({z}\right){\in }\left[{-1}{,}{1}\right){,}{\mathrm{\Im }}{}\left({z}\right){\in }\left[{-1}{,}{1}\right]$ (14)
> $\mathrm{map}\left(\mathrm{convert},\left[\right],\mathrm{relation}\right)$
$\left[{-1}{\le }{\mathrm{\Re }}{}\left({z}\right){\wedge }{\mathrm{\Re }}{}\left({z}\right){<}{1}{,}{-1}{\le }{\mathrm{\Im }}{}\left({z}\right){\wedge }{\mathrm{\Im }}{}\left({z}\right){\le }{1}\right]$ (15)
ComplexRanges are used for example to express the branch cuts of mathematical functions
> $\mathrm{FunctionAdvisor}\left(\mathrm{branch_cuts},\mathrm{arccot}\right)$
$\left[{\mathrm{arccot}}{}\left({z}\right){,}{z}{\in }{\mathrm{ComplexRange}}{}\left({-}{\mathrm{\infty }}{}{I}{,}{-I}\right){\vee }{z}{\in }{\mathrm{ComplexRange}}{}\left({I}{,}{\mathrm{\infty }}{}{I}\right)\right]$ (16)
When you input $z<1$ or $z\le 1$, it is implicitly assumed that $\mathrm{\Im }\left(z\right)=0$. This is used to simplify the notation in the output of some conversions. For example,
> $\mathrm{z_CR}≔z\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{ComplexRange}\left(0,1\right)$
${\mathrm{z_CR}}{≔}{z}{\in }{\mathrm{ComplexRange}}{}\left({0}{,}{1}\right)$ (17)
> $\mathrm{convert}\left(\mathrm{z_CR},\mathrm{RealRange}\right)$
${z}{\in }\left[{0}{,}{1}\right]$ (18)
> $\mathrm{convert}\left(\mathrm{z_CR},\mathrm{relation}\right)$
${0}{\le }{z}{\wedge }{z}{\le }{1}$ (19)
> $\mathrm{FunctionAdvisor}\left(\mathrm{branch_cuts},\mathrm{arcsin}\right)$
$\left[{\mathrm{arcsin}}{}\left({z}\right){,}{z}{\le }{-1}{\vee }{1}{\le }{z}\right]$ (20)
See Also | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 63, "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": 0.9565074443817139, "perplexity": 1028.376011706292}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446706291.88/warc/CC-MAIN-20221126112341-20221126142341-00409.warc.gz"} |
https://www.mis.mpg.de/calendar/lectures/2018/abstract-24405.html | # Abstract for the talk on 20.06.2018 (11:30 h)
Seminar on Nonlinear Algebra
Jörg Frauendiener (University of Otago)
Computational approach to compact Riemann surfaces
A purely numerical approach to compact Riemann surfaces starting from plane algebraic curves is presented. A set of generators of the fundamental group for the complement of the critical points in the complex plane is constructed from circles around these points and connecting lines obtained from a minimal spanning tree. The monodromies are computed by solving the defining equation of the algebraic curve on collocation points along these contours and by analytically continuing the roots. The collocation points are chosen to correspond to Chebychev collocation points for an ensuing Clenshaw–Curtis integration of the holomorphic differentials which gives the periods of the Riemann surface with spectral accuracy. At the singularities of the algebraic curve, Puiseux expansions computed by contour integration on the circles around the singularities are used to identify the holomorphic differentials. The Abel map is also computed with the Clenshaw–Curtis algorithm and contour integrals. Siegel's algorrithm is applied to approximately identify a symplectic transformation to the fundamental domain in order to allow for an efficient computation of multi-dimensional theta functions.
22.06.2018, 02:30 | {"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": 0.9387640953063965, "perplexity": 420.78382909501227}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400202007.15/warc/CC-MAIN-20200921175057-20200921205057-00210.warc.gz"} |
http://mathoverflow.net/questions/109385/how-does-the-schmidt-decomposition-generalize-to-tensor-products-of-several-fini | # How does the Schmidt decomposition generalize to tensor products of several finite-dimensional systems?
Let $n,d_1,\ldots,d_n > 1$ be integers, and $V_1, \ldots, V_n$ be inner product spaces over $\mathbb C$, having dimensions $d_1, \ldots, d_n$ respectively. We consider the ways in which we may decompose vectors $$\Psi \in \mathbf V := V_1 \otimes \cdots \otimes V_n$$ into a sum of product vectors, $$\Psi = \sum_{t=1}^M s_t \Bigl(\psi^{(1)}_t \otimes \cdots \otimes \psi^{(n)}_t\Bigr)\;,$$ for $s_t \in \mathbb C$, where $\psi^{(j)}_t \in V_j$ are unit vectors for each $1 \leqslant j \leqslant n$ and $1 \leqslant t \leqslant M$. I'm specifically interested in the conditions in which we can bound $M$ from above for all $\Psi$.
1. Let $\alpha \geqslant 0$ be a function of $n$ such that $1/(1 - \alpha)$ increases at most polynomially with $n$. Suppose that for each $1 \leqslant j \leqslant n$ and for all $1 \leqslant t,u \leqslant M$, we require that $$\Bigl\langle \psi^{(j)}_t , \psi^{(j)}_u \Bigr\rangle\Bigl\langle \psi^{(j)}_u , \psi^{(j)}_t \Bigr\rangle \ \in\ [0,\alpha] \cup \{1\} \ .$$ What is the smallest value of $M$ for which all vectors in $\mathbf V$ have such a decomposition?
2. What is the smallest such value of $M$ if we set $\alpha := 0$ for all $n$? Equivalently: if for each $1 \leqslant j \leqslant n$ and for all $1 \leqslant t,u \leqslant M$, we require that $\psi^{(j)}_t$ and $\psi^{(j)}_u$ are either equal or orthogonal?
For the second question above, in the case $n = 2$ we may set $M = \min \{ d_1, d_2 \}$ (and as a corollary restrict each scalar $s_t$ to be a non-negative real); this is just the Schmidt decomposition. And we may always bound $M$ from above by $M \leqslant d_1 d_2 \cdots d_n$ by simply chosing an orthonormal basis $\mathbf B$ for $\mathbf V$ consisting of a tensor product of orthonormal bases for each space $V_j$, and decomposing $\Psi$ with respect to that basis. Are there any good upper bounds on $M$ for $n > 2$, for either for $\alpha = 0$, $\alpha > 0$ some constant, or for some function $\alpha \in 1 - \Omega(1/\mathrm{poly}(n))$?
-
I asked a different but related question here: mathoverflow.net/questions/62994/… – Jess Riedel Sep 26 '13 at 18:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.983704686164856, "perplexity": 77.05018426013851}, "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-2016-30/segments/1469257828009.82/warc/CC-MAIN-20160723071028-00173-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://renalfellow.blogspot.com/2011/04/sodium-interpretation-of-measurements.html | ## Thursday, April 14, 2011
### Sodium – interpretation of measurements Part II
Ok, so here is the brief review of the various clinical methods of detecting sodium concentration (as I understand it):
Flame photometer
The older flame photometer method measured the concentration of sodium in plasma. The preparation of the sample involved dilution of the plasma, and assumed that plasma water was 93% of total plasma.
Imagine that the lab reports a sodium concentration of 143mmol/L. Assume that the plasma water is 93% of total plasma - then the sodium concentration in plasma water would be 143/0.93 = 154mmol/L. Let’s take this as ‘normal plasma water sodium’.
Now imagine that the sodium measurement in another patient came back from the lab at 120mmol/L:
- if plasma water is 93% of total plasma, then the sodium in plasma water is 120/0.93 = 129mmol/L – true hyponatraemia
- if plasma water is only 80% of total plasma, then the sodium in plasma water is 120/0.8 = 150mmol/L. This is closer to what we’d expect in normal circumstances, and so we can call this pseudohyponatraemia.
Indirect Ion Selective Electrode
Ion-selective electrodes (ISEs) only measure free sodium ions in a solution; not the volume in which they are dissolved.
The indirect method involves a dilution of the acquired total plasma sample– and again, the lab assumes that plasma water is 93% of total plasma. Therefore the potential for pseudohyponatraemia unfortunately still exists if your lab performs these types of measurements.
Direct Ion Selective Electrode
The direct method of ISE measurement does not use a dilution step and has no assumptions regarding the percentage plasma water of total plasma. Direct measurements are most commonly used in blood gas analyzers. Remember, that direct ISEs measure activity.
Entropy suggests that as energy is introduced to a system, it will be dispersed in a uniform pattern to allow a state of maximal disorder (the lowest energy state). Introduction of sodium (solute) to a solution (blood) does not follow this classic law, as the sodium can interact with water molecules and other ions. Therefore, not all sodium ions are free to enter a reaction, which explains why the concentration of sodium is different than the measured activity. Biochemists multiply the absolute concentration by an activity coefficient (around 0.93 for sodium) to determine the actual activity of sodium.
The direct ISE will only measure free sodium ions, which are limited to the plasma water fraction. From our assumptions above, we noted that a ‘normal person’ had a plasma water sodium concentration of 154mmol/L. But, because not all these sodium ions are free to react, then the measured activity of sodium in plasma water will be around 143mEq/L. This corresponds almost perfectly with the figure obtained by multiplying the sodium concentration in plasma water by the activity coefficient of sodium (154 x 0.93 = 143mEq/L). Note that this 0.93 comes from the activity coefficient of sodium, not the plasma water fraction – it just so happens that they are similar.
Because the direct ISE does not make any assumptions about the plasma water content, it is immune from the concept of pseudohyponatraemia.
Finally, back to our patient with the lab reported sodium of 120mmol/L, who has 93% plasma water, and from our earlier calculations had a plasma water sodium of 129mmol/L. If we measure the sodium activity using direct ISE, we would find that the sodium activity equals the plasma water sodium multiplied by the activity coefficient (129 x 0.93) = 120mmol/L – again true hyponatraemia.
#### 1 comment:
Matt said...
I would like help with the following.
1. Assume that a plasma sample has a [Na] of 140 mEq/L plasma as measured by flame photometry and 150 mEq/L plasma water when corrected for solids content.
What would be the direct ISE reading on modern blood-gas machines?
2. Assume that a carefully prepared NaCl solution has a [Na] of 150 mEq/L H2O. What would be the [Na] measured on the same machine? | {"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": 0.9116047620773315, "perplexity": 2488.749542202949}, "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-2015-48/segments/1448398449160.83/warc/CC-MAIN-20151124205409-00158-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://yutsumura.com/tag/basis-for-a-vector-space/ | # Tagged: basis for a vector space
## Problem 716
Using Gram-Schmidt orthogonalization, find an orthogonal basis for the span of the vectors $\mathbf{w}_{1},\mathbf{w}_{2}\in\R^{3}$ if
$\mathbf{w}_{1} = \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix} ,\quad \mathbf{w}_{2} = \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix} .$
## Problem 715
Let
$\mathbf{v}_{1} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} ,\; \mathbf{v}_{2} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} .$ Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?
If not, then find an orthonormal basis for $V$.
## Problem 713
Determine bases for $\calN(A)$ and $\calN(A^{T}A)$ when
$A= \begin{bmatrix} 1 & 2 & 1 \\ 1 & 1 & 3 \\ 0 & 0 & 0 \end{bmatrix} .$ Then, determine the ranks and nullities of the matrices $A$ and $A^{\trans}A$.
## Problem 710
Find a basis for $\Span(S)$ where $S= \left\{ \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} , \begin{bmatrix} -1 \\ -2 \\ -1 \end{bmatrix} , \begin{bmatrix} 2 \\ 6 \\ -2 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} \right\}$.
## Problem 705
For a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$
$\Fun ( S , V ) = \{ f : S \rightarrow V \} .$
For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by
$(f+g)(s) = f(s) + g(s) \, \mbox{ and } (cf)(s) = c (f(s)) \, \mbox{ for all } s \in S .$
(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?
(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.
(c) Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.
(d) Let $S = \{ s_1 , s_2 , \cdots , s_m \}$. Construct a linear isomorphism between $\Fun(S, V)$ and the vector space of $n$-tuples of $V$, defined as
$V^m = \{ (v_1 , v_2 , \cdots , v_m ) \mid v_i \in V \mbox{ for all } 1 \leq i \leq m \} .$
(e) Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?
(f) Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.
## Problem 704
Let $A=\begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 &3 & 0 & 5 \\ 1 & 1 & 6 & 3 \end{bmatrix}$.
(a) Find a basis for the nullspace of $A$.
(b) Find a basis for the row space of $A$.
(c) Find a basis for the range of $A$ that consists of column vectors of $A$.
(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.
## Problem 703
Using the definition of the range of a matrix, describe the range of the matrix
$A=\begin{bmatrix} 2 & 4 & 1 & -5 \\ 1 &2 & 1 & -2 \\ 1 & 2 & 0 & -3 \end{bmatrix}.$
## Problem 683
Let $\mathrm{P}_3$ denote the set of polynomials of degree $3$ or less with real coefficients. Consider the ordered basis
$B = \left\{ 1+x , 1+x^2 , x – x^2 + 2x^3 , 1 – x – x^2 \right\}.$ Write the coordinate vector for the polynomial $f(x) = -3 + 2x^3$ in terms of the basis $B$.
## Problem 682
Let $V$ denote the vector space of $2 \times 2$ matrices, and $W$ the vector space of $3 \times 2$ matrices. Define the linear transformation $T : V \rightarrow W$ by
$T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} a+b & 2d \\ 2b – d & -3c \\ 2b – c & -3a \end{bmatrix}.$
Find a basis for the range of $T$.
## Problem 672
For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.
Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, for $f \in \mathrm{P}_n$,
$T (f) (x) = x f(x).$
Prove that $T$ is a linear transformation, and find its range and nullspace.
## Problem 665
Let $\mathbf{P}_2$ be the vector space of polynomials of degree $2$ or less.
(a) Prove that the set $\{ 1 , 1 + x , (1 + x)^2 \}$ is a basis for $\mathbf{P}_2$.
(b) Write the polynomial $f(x) = 2 + 3x – x^2$ as a linear combination of the basis $\{ 1 , 1+x , (1+x)^2 \}$.
## Problem 632
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.
Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.
Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.
## Problem 612
Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$.
(a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.
## Problem 606
Let $V$ be a vector space and $B$ be a basis for $V$.
Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$.
Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form
$\begin{bmatrix} 1 & 0 & 2 & 1 & 0 \\ 0 & 1 & 3 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}.$
(a) What is the dimension of $V$?
(b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
(The Ohio State University, Linear Algebra Midterm)
## Problem 601
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.
Let
$W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.$
(a) Show that $W$ is a subspace of $V$.
(b) Find a basis of $W$.
(c) Find the dimension of $W$.
(The Ohio State University, Linear Algebra Midterm)
## Problem 600
Let $\mathbf{v}_1=\begin{bmatrix} 2/3 \\ 2/3 \\ 1/3 \end{bmatrix}$ be a vector in $\R^3$.
Find an orthonormal basis for $\R^3$ containing the vector $\mathbf{v}_1$.
## Problem 590
Let $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let
$V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}$ be a subset in $C[-1, 1]$.
(a) Prove that $V$ is a subspace of $C[-1, 1]$.
(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.
(c) Prove that
$B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}$ is a basis for $V$.
## Problem 588
Let $P_2$ be the vector space over $\R$ of all polynomials of degree $2$ or less.
Let $S=\{p_1(x), p_2(x), p_3(x)\}$, where
$p_1(x)=x^2+1, \quad p_2(x)=6x^2+x+2, \quad p_3(x)=3x^2+x.$
(a) Use the basis $B=\{x^2, x, 1\}$ of $P_2$ to prove that the set $S$ is a basis for $P_2$.
(b) Find the coordinate vector of $p(x)=x^2+2x+3\in P_2$ with respect to the basis $S$.
## Problem 579
Determine whether each of the following sets is a basis for $\R^3$.
(a) $S=\left\{\, \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 4 \end{bmatrix} \,\right\}$
(b) $S=\left\{\, \begin{bmatrix} 1 \\ 4 \\ 7 \end{bmatrix}, \begin{bmatrix} 2 \\ 5 \\ 8 \end{bmatrix}, \begin{bmatrix} 3 \\ 6 \\ 9 \end{bmatrix} \,\right\}$
(c) $S=\left\{\, \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 7 \end{bmatrix} \,\right\}$
(d) $S=\left\{\, \begin{bmatrix} 1 \\ 2 \\ 5 \end{bmatrix}, \begin{bmatrix} 7 \\ 4 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 8 \\ 6 \end{bmatrix}, \begin{bmatrix} -1 \\ 9 \\ 10 \end{bmatrix} \,\right\}$
## Problem 440
Let $U$ and $V$ be finite dimensional subspaces in a vector space over a scalar field $K$.
Then prove that
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# Discrepancy: a constructive proof via random walks in the hypercube
By
In this post, I'll give two constructive/algorithmic discrepancy upper bounds. The first, by Beck and Fiala, applies to sparse set systems. The second, by Lovett and Meka, improves on the Beck-Fiala result and also matches the guarantees of Spencer's theorem.
### Discrepancy Minimization
Recall that, given a system of subsets of $[n]$, $\cS = S_1,\ldots,S_m \subseteq [n]$, the discrepancy of a coloring $x \in \{\pm 1\}^n$ on $\cS$ is defined to be $\disc(x,\cS) = \max_{S_j \in \cS} \left|\sum_{i \in S_j} x_i\right|.$ In the previous post, we proved Spencer's theorem, which says that for any $\cS$, $\min_{x}\disc(x,\cS) \le O(\sqrt{n\log\frac{m}{n}})$.
The natural associated algorithmic task is discrepancy minimization---given $\cS$, we want to compute $x^* = \argmin_{x \in \{\pm 1\}^n} \disc(x,\cS).$ Spencer's theorem guarantees that some $x$ achieving $\disc(x,\cS) \le O(\sqrt{n\log\frac{m}{n}})$ always exists, but the proof does not provide a natural algorithm for finding the discrepancy minimizer $x^*$. Actually, finding the minimizer $x^*$ is NP-hard.
Theorem 1 (Charikar-Newman-Nikolov) Given a set system $\cS$ with $O(n)$ sets, it is NP-hard to distinguish whether $\disc(\cS) = 0$ or $\disc(\cS) = \Omega(\sqrt{n})$.
Still, it turns out that Spencer's theorem can be made algorithmic---there are efficient algorithms for computing a coloring with discrepancy $O(\sqrt{n})$. The first such algorithm was given by Bansal in 2010, and it was based on semidefinite programming. Later, in 2012, Lovett and Meka gave a simplified and slightly more general version of Bansal's result. The Lovett-Meka algorithm uses some ideas from Bansal's algorithm, but it does not rely on SDPs, using instead only linear algebra and properties of random vectors.
I think it is more natural to see the Lovett-Meka result after seeing the simpler result of Beck and Fiala for the special case when $\cS$ is sparse, and so I will give a brief account of that algorithm first.
### Sparse set systems and Beck-Fiala
Suppose that we have a set system $\cS$ which is sparse, so that every item is in at most $t$ sets. In this case, we can get the following specialized bound:
Theorem 2 (Beck-Fiala) If $\cS = S_1,\ldots,S_m$ is a set system with $S_j \subseteq [n] ~ \forall j \in [m]$, and each $i\in[n]$ is only included it at most $t$ sets of $\cS$, then there is an algorithm that computes a coloring $x \in \{\pm 1\}^n$ with $\disc(x,\cS) \le 2t - 1.$
Beck and Fiala also conjectured that one could obtain a bound of $\disc(\cS) \le O(\sqrt{t})$ for this setting---the Beck-Fiala conjecture is a major open problem in discrepancy theory. Proof: The proof is algorithmic---we'll start with the fractional coloring $x_0 = \vec{0}$, and update $x$ iteratively until we reach an integral point in $\{\pm 1\}^n$, arguing that we cannot do too much damage along the way.
The algorithm is as follows. At step $k$ of the algorithm, say we have the fractional coloring $x_k \in [-1,1]^n$. We keep track of the “live” items, or items for which $|x_i| < 1$. We also keep track of the “dangerous” sets: a set is called dangerous if it contain more than $t$ live items.
Claim 1 At step $k$, if there are $n_k$ live items, then there can be at most $n_k - 1$ dangerous sets.
This is true because each dangerous set has at least $t+1$ live items, but the maximum degree of each item is $t$, and so if we restrict the incidence matrix $A$ to the rows corresponding to dangerous sets, there are at most $n_k\cdot t$ nonzero entries, and therefore there can be at most $\lfloor\frac{t\cdot n_k}{t+1}\rfloor \le n_k-1$ dangerous sets.
So, if we let $A_k$ be the restriction of the incidence matrix to live columns and dangerous rows in the $k$th step, $A_k$ is not full rank, so there must always exist some vector $y_k \in \R^{n_k}$ which is orthogonal to all rows of $A_k$, and furthermore we can find $y_k$ efficiently.
Let $z_k$ be the natural extension of $y_k$ to the space of non-live items (so that $z_k(i) = y_k(i)$ if $i$ is live and $0$ otherwise). We perform the update $x_{k+1} = x_k + \alpha \cdot z_k,$ where $\alpha \in \R_+$ is chosen to be the largest number so that $x_{k+1} \in [-1,1]^n$. In other words, we start with $\alpha = 0$, and grow $\alpha$ until at least one of the entries of $x_{k+1}$ hits $1$ or $-1$. Thus the number of live items decreases by at least one, and the discrepancy of every dangerous set is $0$.
Now we only have to argue that once a set $S_j$ is no longer dangerous, its discrepancy can never grow larger than $2t-1$. If $S_j$ stopped being dangerous at step $k'$, $S_j$ had at most $t$ live items in $x_{k'}$ and $\iprod{x_{k'},a_j} = 0$. In the worst case each live $i \in S_j$ can go from $x_{k'}(i) = 1-\epsilon_i$ to $x(i) = -1$, so a bound of $2t$ on the final discrepancy of $S_j$ is easy. To get $2t-1$, we just notice that because the total discrepancy of $S_j$ was $0$ at step $k'$, the sum over the live $i \in S_j$ must be integral, and for live $|x_{k'}(i)| < 1$, so this gives us a lower bound of $\left|\sum \epsilon_i\right| \ge 1$. $$\tag*{\blacksquare}$$
### Constructive Spencer via guided random walks
As mentioned above, the first algorithmic proof of Spencer's result was given by Bansal in 2010. The proof was a little bit similar to the Beck-Fiala algorithm, in that it starts with the fractional coloring $x_0 = 0$, and makes updates to $x$ iteratively until hitting some integral coloring, bounding the error incurred along the way. The extreme point of departure is the manner in which the iterative updates to $x$ are chosen. Instead of choosing some arbitrary direction orthogonal to the dangerous sets, Bansal's algorithm uses a semidefinite program to take a random step---the semidefinite program makes sure that this random walk will make progress without violating discrepancy constraints too much. This makes the proof non-constructive, since to argue the feasibility of the SDP, Bansal relied on Spencer's result.
In 2012, Lovett and Meka simplified Bansal's approach. They removed the semidefinite programming step, returning to linear algebraic arguments reminiscent of the Beck-Fiala proof. Instead of using the SDP to guide the random walk, they argue that so long as there are not too many integral vertices, there is a high-dimensional subspace of $\R^n$ in which the random walk can proceed without violating the discrepancy constraints by too much, and then they take a random step in this subspace. This gives a truly constructive proof of Spencer's result.
Ignoring variations in the constants chosen, the main theorem of the paper is the following:
Theorem 3 Suppose that for $\lambda \in \R^m$ with $\lambda \ge 0$, $$\sum_j \exp\left(-\frac{\lambda_j^2}{32}\right) \le \frac{n}{16}.\label{cond}$$ Then for any starting point $x \in [-1,1]^n$, there exists a partial coloring $x' \in [-1,1]^n$ with at least $n/2$ entries of $x$ having magnitude $1$ and $|\langle x - x', a_j\rangle | \le \lambda_j \sqrt{|S_j|}$ for all $j \in [m]$, and an algorithm that finds such an $x'$ with probability at least $1/10$.
First let's see that this implies Spencer's result. We'll apply the theorem recursively, like Spencer does, $T = O(\log n)$ times. We'll start with the coloring $x_0 = 0$. At the $t$th iteration of the algorithm, say we have $n_t \le n/2^{t-1}$ items uncolored. For all of $j \in [m]$, we will set $\lambda^{(t)}_j = \sqrt{32\log{\frac{m}{n_t}} + \log 16}$ (it's easy to check that this satisfies the condition of the theorem). Then we use the algorithm to find $x_t = x'$. Letting $a_j$ be the 0/1 indicator vector for $S_j$, by the triangle inequality and the guarantees of the theorem, \begin{align*} \disc(x_T, S_j) ~=~ |\langle x_T, a_j\rangle| ~\le~ \sum_{t=0}^T |\langle x_{t} - x_{t+1}, a_j \rangle| ~&\le ~\sum_{t=0} \lambda_j^{(t)}\sqrt{|S_j^{(t)}|}, \end{align*} And since there are at most $n/2^{t-1}$ active items in $S_j$ at timestep $t$, \begin{align*} &\le \sum_{t=0}\sqrt{n_t\log\frac{m}{n_t}} ~\le~ O(\sqrt{n\log m/n}), \end{align*} where the last inequality follows because the sequence $\frac{n_t\log(m/n_t)}{n\log(m/n)}$ decays at least as fast as $2^{-t}\log 2^t$. So, this recovers Spencer's result.
Remark 1 (Sparse set systems) The algorithms of Bansal and of Lovett and Meka can be generalized to give an upper bound of $O(\sqrt{t}\log n)$ for $t$-sparse set systems. If the $\lambda_j$'s are set to $\lambda_j = c \cdot \sqrt{\frac{t}{|S_j|}}$ for some constant $c$, then by Markov's inequality and by the sparsity of $\cS$ there are at most $2^{-k}n$ sets with $|S_j| \in [2^ktn,2^{k+1}tn]$, and so $\sum_{j}\exp\left(-\frac{\lambda_j^2}{32}\right) \le \sum_{k=0}^{\infty} \frac{n}{2^k}\cdot \exp\left( \frac{-c^2}{2^{k+1}\cdot 32}\right),$ which meets condition (\ref{cond}) of the theorem if $c$ is chosen properly, so the conclusion follows.
Now, we will prove the theorem.
Main idea: Just as Beck and Fiala do, we'll start with some point $x_0$, and update $x$ iteratively, fixing $x(i)$ the moment that $|x(i)| = 1$. We will differ in our updates---we redefine a set to be dangerous when we come close to violating the constraint $|\iprod{x_t, a_j}| \ge \lambda_j\sqrt{|S_j|}$. So unlike Beck-Fiala, by default we start with no dangerous sets, and we add sets to the dangerous list when they become too imbalanced.
Just like Beck-Fiala, we will only make updates orthogonal to the dangerous sets. Our updates will take the form of a random walk in the non-dangerous subspace. The trick will be to argue that by our condition (\ref{cond}) and by properties of Gaussian random walks, with reasonable probability the rank of the dangerous subspace does not become too large as long as there are still many live items to color in.
Proof: The algorithm is as follows: Set the step size $\gamma = 1/100n^2$, and the safety margin $\delta = \gamma \cdot 10\log n$. Initialize the set of non-live items $D_v = \emptyset$ (notationally this is more convenient than keeping track of live items), and initialize the set of dangerous constraints $D_S$. Initialize the starting coloring $x_0 = x$ and the starting subspace $V_0 = \R^n$.
1. For $k = 1,\ldots, K= 8/\gamma^2$:
1. Sample the random vector $g_k$ by sampling $g \sim \cN(0, \Id)$ and projecting $g$ into the subspace $V_k$.
2. Take a random step by setting $x_k = x_{k-1} + \gamma \cdot g_k$.
3. For any $i \in [n]$ such that $|x_k(i)| \ge 1 - \delta$, add $i$ to $D_v$.
4. For any $j \in [m]$ such that $|\langle x_k - x_0, a_j\rangle| \ge \lambda_i\sqrt{|S_j|} - \delta$, add $S_j$ to $D_S$.
5. Set $V_{k+1}$ to be the subspace orthogonal to all $e_i$ for $i \in D_v$ and orthogonal to all $a_i$ for $S_i \in D_S$.
2. If $|x_K(i)| \ge 1-\delta$, set $x'(i) = \sgn(x_K(i))$. Otherwise, set $x'(i) = x_K(i)$.
Each of these steps can be done in polynomial time. Now, for proving correctness, there are several concerns: can we always assume $x_k \in [-1,1]^n$ and $|\iprod{x_k,a_j}|\le \lambda_j\sqrt{|S_j|}$, or does step (b) ever make us jump out of the box? Does the rounding in step 5 change the discrepancy of sets by too much? Will the algorithm ever get stuck in a place where we can't make progress (i.e. $V_k = \emptyset$) before coloring at least $n/2$ items?
The first two concerns are easy to take care of, so here are informal arguments. Since the Gaussian steps have small magnitude $\gamma$, and since we have a reasonable safety margin $\delta$ away from violating any constraint, the probability that we ever violate hard constraints in step (b) is polynomially small. The small safety margin also ensures that with high probability, the rounding we perform in step (c) cannot change the discrepancy of any set by more than $n\delta = O(1/\log n)$ over the course of the entire algorithm.
It remains to argue that with probability at least $1/10$, we won't get stuck before we will color at least $n/2$ items. We'll use a couple of (relatively standard) properties of Gaussian projections:
Claim 2 If $u\in \R^n$ and $g\in \R^n$ is a vector with i.i.d. entries $g_i \sim \cN(0,\sigma^2)$, then $\iprod{g,u}\sim \cN(0,\sigma^2\|u\|_2^2)$.
Claim 3 If $g\in \R^n$ is a vector with i.i.d. entries $g_i \sim \cN(0,\sigma^2)$, and $g'$ is the orthogonal projection of $g$ into a subspace $S \subseteq \R^n$, then $\E[\|g'\|_2^2] = \sigma^2\cdot \dim(S)$.
Claim 4 If $u\in \R^n$ and $g\in \R^n$ is a vector with i.i.d. entries $g_i\sim\cN(0,\sigma^2)$, and $g'$ is the orthogonal projection of $g$ into a subspace $S \subseteq \R^n$, then $\iprod{g',u}\sim \cN(0,\alpha\|u\|_2^2)$ where $\alpha \le \sigma^2$.
The proof of the first claim follows from the additive property of Gaussians. The second and third claims can be proven using the first claim, by considering an orthogonal projection matrix into the subspace $S$.
Now, we are equipped to prove the rest of the theorem. We first relate the progress of the algorithm, as measured by the variance of the Gaussian steps, to the dimension of $V_k$. By the independence of the gaussians $g_k$, have that \begin{align*} \E[\|x_{K} - x_0\|_2^2] ~=~ \E\left[\left\|\gamma \cdot \sum_{k=1}^K g_k\right\|_2^2\right] &= \gamma^2 \cdot \sum_{k=1}^K \E\left[\left\|g_k\right\|_2^2\right]\\ &= \gamma^2 \sum_{k=1}^K \E[\dim(V_k)] \qquad \\ &\ge~ \gamma^2 K \cdot \E[\dim(V_K)], \end{align*} where the second line follows from Claim 2 and the last line is because the dimension of $V_k$ decreases with $k$. Since $\dim(V_K) \ge n - |D_v| - |D_S|$, \begin{align*} &\ge 8 \E[n - |D_v| - |D_S|]. \end{align*} On the other hand, $\E[\|x_{K} - x_0\|_2^2] \le 2n$, because we stop moving in the direction of items once the coordinate magnitude is close to $1$. Thus, \begin{align} 2n &\ge 8 (n - \E[|D_S|] - \E[|D_v|],\nonumber\\ \E[|D_v|] &\ge \frac{3}{4}n - \E[|D_S|],\label{dsbd} \end{align}
Now, all that remains for us to do is argue that the expected number of dangerous sets is not too large---if you like, we are arguing that we don't arrive at $V_k = \emptyset$ before coloring in enough vertices. Recall a set $S_j$ is in $D_S$ only if for some $k$, $|\langle x_0 - x_k, a_j\rangle| \ge \lambda_i\sqrt{|S_j|} - \delta$. Since we only move orthogonal to $a_j$ for $S_j \in D_S$, it suffices to count the number of $S_j$ which are dangerous at the final step when $k = K$.
Let $J \subseteq [m]$ be the set of $j \in [m]$ for which $\lambda_j\sqrt{|S_j|} \ge 2\delta$. By Claim 3, $x_0 - x_K$ is a Gaussian vector supported on $\R^n$ with expected square norm at most $K \cdot \gamma^2 \le 8$, and $a_j$ is a vector of norm $\sqrt{S_j}$. Now, although $x_0-x_K$ is not exactly the orthogonal projection of a Gaussian vector into a subspace of $\R^n$, we can more or less apply Claim 4 to11. If we want to be rigorous, we should break up $x_0 - x_K$ into the independent Gaussian increments $x_k - x_{k+1}$ and apply Claim 4 to each of them, then look at their sum. conclude that $\langle x_0-x_K, a_j\rangle$ is distributed as a Gaussian with variance at most $8|S_j|$. Therefore for sets $j \in J$, $\Pr\left[|\langle x_0 - x_K, a_j\rangle| \ge \lambda_j\sqrt{|S_j|} - \delta\right] \le \Pr\left[|\langle x_0 - x_K, a_j\rangle| \ge \frac{1}{2}\lambda_j\sqrt{|S_j|}\right] \le 2\exp\left(-\frac{\lambda_j^2}{32}\right).$ By condition (\ref{cond}) of the theorem, there are at most $n/16$ sets in $[m]\setminus J$, or sets with $\lambda_j\sqrt{|S_j|} < 2\delta$, since for each such set $\exp(-\lambda_j^2/32) \ge n^{O(1/n^4)} \approx 1$. So the expected number of sets for which $|\langle x_0 - x_K, a_j\rangle| \ge \lambda_j\sqrt{|S_j|} -\delta$ is at most \begin{align*} \E[|D_S|] \le \frac{n}{16} + \sum_{j\in J} \Pr\left[|\langle x_0 - x_K, a_j\rangle| \ge \frac{1}{2}\lambda_j\sqrt{|S_j|} \right] \le \frac{n}{16} + 2\cdot \sum_{j \in J} \exp\left(-\frac{\lambda_i^2}{32}\right) \le \frac{3}{16}n, \end{align*} where for the last inequality we apply condition (\ref{cond}) of the theorem.
Now, plugging back into (\ref{dsbd}), \begin{align*} \E[|D_v|] \ge \left(\frac{3}{4} - \frac{3}{16}\right)n = \frac{9}{16}n. \end{align*} Let $p$ be the probability that $|D_v|$ has fewer than $n/2$ colored items. Since there can be at most $n$ colored items, $\frac{9}{16}n \le \E[|D_v|] < (1-p) \cdot n + p\cdot \frac{n}{2} = \left(1-\frac{p}{2}\right)n.$ From this we have that $p < 7/8$, and so by the union bound the algorithm succeeds with probability at least $1/8 - o(1)$. $$\tag*{\blacksquare}$$
### More sources
The paper of Lovett and Meka, as well as the previously mentioned chapter of Nikhil Bansal are good resources. The original algorithmic result can be found in this paper of Bansal. | {"extraction_info": {"found_math": true, "script_math_tex": 2, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9699000716209412, "perplexity": 259.8898409616565}, "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-2018-30/segments/1531676592861.86/warc/CC-MAIN-20180721223206-20180722003206-00016.warc.gz"} |
https://socratic.org/questions/5620fd7011ef6b50e4c18121 | Physics
Topics
# Question #18121
Oct 16, 2015
I found: $\omega = 28.9 \frac{\text{rad}}{\sec}$
#### Explanation:
Maximum tension is $50 , 000 N$ so using Newton's Second Law, $\Sigma \vec{F} = m \vec{a}$, we get that:
$T = m {a}_{c} = m {v}^{2} / 4$ where ${a}_{c}$ is centripetal acceleration;
$50 , 000 = 15 \cdot {v}^{2} / 4$
and:
$v = 115.5 \frac{m}{s}$
This linear velocity can be converted into rotational, $\omega$, by remembering that:
$v = \omega r$
so:
$\omega = \frac{v}{r} = \frac{115.5}{4} = 28.9 \frac{\text{rad}}{\sec}$
##### Impact of this question
340 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "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": 0.9610437750816345, "perplexity": 4745.201065545612}, "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/1596439735958.84/warc/CC-MAIN-20200805124104-20200805154104-00590.warc.gz"} |
https://workforce.libretexts.org/Bookshelves/Electronics_Technology/Book%3A_Electrical_Fundamentals_Competency_(Industry_Training_Authority_of_BC)/01%3A_Basic_Principles_of_Electricity/01%3A_Fundamentals_of_Electricity/1.01%3A_Basic_principles | # 1.1: Basic principles
• Camosun College
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Explain fundamentals of electricity
Over the centuries scientists have discovered that electricity is predictable and measurable. Being familiar with the fundamentals of electricity will help you to understand how and why electrical circuits work.
Basic principles
Electricity is a form of energy. To understand electricity, it is important that you first understand the structure of matter. Anything that occupies space and has weight is called matter. All liquids, gases, and solids are examples of matter in different forms. Matter is made of smaller units called atoms. Atoms can be grouped together in compounds to form molecules.
Atomic theory
Atoms are the most basic part of matter and differ in atomic structure from each other. The structure of the atom can be described in much the same way as the solar system. Instead of the Sun at the centre, there is a nucleus. This nucleus is made of two basic particles: protons and neutrons.
Neutrons make up the mass (or weight equivalency) of the atom, have no electrical charge, and are considered to be neutral. Protons are particles that have a positive (+) electrical charge and cannot be separated from the nucleus. Surrounding the nucleus in orbits are electrons. These are tiny particles with a negative (–) electrical charge. Figure 1 shows a model of a carbon atom.
1. Carbon atom
2. Hydrogen and copper atoms
3. Electrical attraction
4. Transmission of impulse
Sources of electrical force
You have just learned that if there is a surplus of electrons at one end of a conductor and a deficiency at the other end, a current flows in the conductor. There are devices that create this difference in charge so that a current will flow. These devices are referred to as sources of electromotive force. These sources include:
• chemical
• electromagnetic induction
• friction
• heat
• pressure
• light
Chemical
A battery is a source of electrical force due to the chemical reaction that takes place between plates and an electrolyte. This reaction causes a buildup of positive ions on one plate and negative ions on the other plate. This electrical difference between the plates is also known as potential difference.
Electromagnetic induction
Electric force can be generated by using a magnetic field. This is the method by which most of the electrical energy we use is produced. An example is an alternator or generator.
Friction
Friction can cause free electrons to move from one body to another and be stored there temporarily. When you walk across a carpet, electrons are transferred to the atoms in your body and you return them to other atoms when you touch a metallic object.
Heat
If two unlike metals are placed together and heated, they will produce electrical force. An example is the thermocouple in a furnace.
Pressure
Certain crystals will produce electricity if they are squeezed under extreme pressure. An example is a barbecue starter (also called piezoelectric generator).
Light
Some crystals and semiconductors will produce electrical force when they are exposed to light. An example is the photocell in a calculator.
All six of these sources of EMF achieve the same thing. They separate charge by:
• imparting energy to the electrons
• pushing them against an electrostatic field
• causing a surplus of electrons (negative charge) at one terminal of the source and a deficiency of electrons (positive charge) at the other terminal
In a sense, the process can be likened to compressing a spring. The energy stored in the compressed spring can be used later to do useful work. The same is true of the separated charges: they store energy that can be used to do useful work.
Electrical energy always comes from some other form of energy. The source of EMF is simply the device that makes the conversion from some other form of energy to electrical energy. | {"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": 0.8016343116760254, "perplexity": 489.91443446976103}, "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/1674764499954.21/warc/CC-MAIN-20230202003408-20230202033408-00326.warc.gz"} |
http://physics.stackexchange.com/questions/122105/what-does-the-higgs-boson-have-to-do-with-the-uncertainty-principle-and-quantum | # What does the Higgs boson have to do with the uncertainty principle and quantum oscillations?
I was looking in New Scientist the other day when I saw something to do with the Higgs boson, energy levels, entropy, space/time, quantum oscillations and many other things. It was in a feature to do with symmetry. I have read up about this and I know that the magazine does not accept original works, so I guess that this is mainstream physics.
The above is a diagram of what I'm talking about. This is how it appears in the magazine:
According to NS, a ball (like the blue one above) that is sitting at the top of a piece of space/time warped like this has an unstable energy level and position. Given the slightest nudge, it will "fall" into a lower energy state and quantum oscillations (which occur as is "rolls" up and down the "pit") caused by the uncertainty principle create Higgs bosons in the Higgs field.
I have pondered this repetitively, and it still looks like nonsense (even in the eyes of quantum physics). What I want to know is: has this been regarded as true or is this "complete nonsense"? And if so, how can the uncertainty principle cause such things to happen?
What I do not understand is how a vacuum generates Higgs particles if it has no mass - unless it's trading it with energy.
What is actually happening (the magazine hasn't described it in enough detail for me to understand)?
(It's hard for me to visualise; it's an entirely new concept to visualise for me.)
-
What the picture seems to be describing is the vaccum expectation value. This seems to be one of the (debatable) misuses of the uncertainty principle. – jinawee Jun 27 '14 at 18:22
@jinawee Found this: enigmass.in2p3.fr/IMG/jpg/higgs-2.jpg A bunch of other sources say similar, is this a fundamental misconception - if a misuse of Heisenberg's uncertainty principle? – Graviton Jun 27 '14 at 18:48
As always, quantum fluctuations (embodied by the uncertainty principle) will continue to jiggle the vacuum state back and forth along a radial line, and these oscillations give rise to Higgs particles (the "steepness" of the curve is what gives the Higgs its particular mass, 126 GeV). Notice that there can also be rolling along the circle at the bottom for "free" (it costs no energy). These motions are associated with the massless Goldstone bosons that are "eaten" by the weak vector bosons to give them a mass (the massiveness of the $W^{\pm}$ and $Z$ bosons is why many nuclear decays take a while, and one reason why nuclear forces have such short range). | {"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": 0.8055840730667114, "perplexity": 397.52031488014734}, "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/1432207928414.45/warc/CC-MAIN-20150521113208-00064-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://eng.libretexts.org/Bookshelves/Environmental_Engineering_(Sustainability_and_Conservation)/Book%3A_Introduction_to_Environmental_Science_(Zendher_et_al.)/5%3A_Alternative_Energy/5.1%3A_Introduction | # 5.1: Introduction
learning objectives
By the end of this chapter, students should be able to.
1. Describe arguments for alternative energy
2. Explain the following aspects of solar energy:
a. How passive solar energy works and provide examples of its use.
b. How solar panels (photovoltaic cells) work.
c. The limitations and environmental costs associated with solar energy.
3. Explain the following aspects of biofuels / biomass energy:
a. Describe what is meant by the term “carbon neutral” and explain how biomass energy can and cannot be carbon neutral.
b. Describe current achievements in biofuels and potential of this area for growth.
Energy sources that are more or less continuously made available in a time frame useful to people are called renewable energy. Renewable energy sources are often considered alternative sources because, in general, most industrialized countries do not rely on them as their main energy source. Instead, they tend to rely on the conventional energy sources such as fossil fuels or nuclear power that are non-renewable. Because of the energy crisis in the United States during the 1970s, dwindling supplies of fossil fuels and hazards associated with nuclear power, use of renewable energy sources such as solar energy, hydroelectric, wind, biomass, and geothermal has grown. Renewable energy comes from the sun (considered an "unlimited" supply) or other sources that can theoretically be renewed at least as quickly as they are consumed. If used at a sustainable rate, these sources will be available for consumption for thousands of years or longer. Renewable alternatives derive from wind, water, solar or biomass (Figure $$\PageIndex{1}$$). Note that wind, water and biomass energy sources are indirect sources of solar energy. One limitation currently associated with most forms of renewable energy is that the energy is not concentrated and not easily portable.
Energy is an important ingredient in all phases of society. We live in a very interdependent world, and access to adequate and reliable energy resources is crucial for economic growth and for maintaining the quality of our lives. However, current levels of energy consumption and production are not sustainable because of the heavy reliance on non-renewable energy sources. The principal energy resources used in the world are shown in Figure $$\PageIndex{2}$$. The fuel mix has changed over the years but now is dominated by oil, although natural gas and solar contributions are increasing. About 80 % of our energy comes from nonrenewable fossil fuels and nuclear. The link between global warming and fossil fuel use, with its production of carbon dioxide through combustion, has made, in the eyes of many scientists, a shift to non-fossil fuels of utmost importance – but it will not be easy. About 40 % of the world’s energy comes from oil, and much of that goes to transportation uses. | {"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": 0.948416531085968, "perplexity": 1226.3245272170252}, "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-2021-10/segments/1614178365454.63/warc/CC-MAIN-20210303042832-20210303072832-00433.warc.gz"} |
https://documen.tv/question/the-distance-between-two-towns-is-120-km-there-are-approimately-8-km-in-five-miles-which-measure-23427584-52/ | ## the distance between two towns is 120 km there are approximately 8 km in five miles which measurement is closest to the number of miles betw
Question
the distance between two towns is 120 km there are approximately 8 km in five miles which measurement is closest to the number of miles between these two towns
in progress 0
6 months 2021-07-15T20:03:50+00:00 1 Answers 0 views 0 | {"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": 0.9737657904624939, "perplexity": 997.6060275309543}, "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/1642320300533.72/warc/CC-MAIN-20220117091246-20220117121246-00462.warc.gz"} |
https://www.physicsforums.com/threads/dot-product.395237/ | # Homework Help: Dot product
1. Apr 14, 2010
### 8614smith
1. The problem statement, all variables and given/known data
what is the dot product of $$G{\bullet}A$$ where A = $$\left(\frac{a_3}{l}-\frac{{a_1}}{h}\right)$$ and G = $$2{\pi}h{\frac{{a_2}}x{{a_3}}}{{a_1}{\bullet}{{a_2}}x{{a_3}}}$$
2. Relevant equations
3. The attempt at a solution
The answer is zero and ive got the worked solution infront of me, i just done see where the $$\frac{{{a_3}}}{l}$$ goes, the dot product of a vector with itself is 1 isnt it? but then where does the denominator go? and what about the denominator of G?
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Apr 14, 2010
### 8614smith
sorry that first bit should say:
G = $$2{\pi}h{\frac{{a_2}x{a_3}}{{a_1}{\bullet}{a_2}x{a_3}}$$
3. Apr 14, 2010
### Staff: Mentor
No, the dot product of a vector is not 1 in general. u $\cdot$ v = |u| |v| cos([itex]\theta[\itex]). | {"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": 0.9896901249885559, "perplexity": 965.4942702466924}, "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-2018-26/segments/1529267867885.75/warc/CC-MAIN-20180625131117-20180625151117-00405.warc.gz"} |
http://www.thefullwiki.org/Gay-Lussac%27s_Law | • Wikis
# Gay-Lussac's Law: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
(Redirected to Gay-Lussac's law article)
The expression Gay-Lussac's law is used for each of the two relationships named after the French chemist Joseph Louis Gay-Lussac and which concern the properties of gases, though it is more usually applied to his law of combining volumes, the first listed here. One law relates to volumes before and after a chemical reaction while the other concerns the pressure and temperature relationship for a sample of gas.
## Law of combining volumes
The law of combining volumes states that, when gases react together to form other gases, and all volumes are measured at the same temperature and pressure:
The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.
This reflects the fact that (by Avogadro's law) equal volumes of gas contain equal numbers of molecules (at the same temperature and pressure), and also that in chemical reactions, the molecules combine in a ratio of whole numbers—this is known as the stoichiometry of the chemical reaction and is expressed via the chemical equation for the reaction.
## Pressure-temperature law
Gay-Lussac's other law, discovered in 1802, states that:
The pressure of a fixed mass and fixed volume of a gas is directly proportional to the gas's temperature.
Simply put, if a gas's temperature increases then so does its pressure, if the mass and volume of the gas are held constant. The law has a particularly simple mathematical form if the temperature is measured on an absolute scale, such as in kelvins. The law can then be expressed mathematically as:
${P}\propto{T}$
or
$\frac{P}{T}=k$
where:
P is the pressure of the gas.
T is the temperature of the gas (measured in kelvins).
k is a constant.
This law holds true because temperature is a measure of the average kinetic energy of a substance; as the kinetic energy of a gas increases, its particles collide with the container walls more rapidly, thereby exerting increased pressure.
For comparing the same substance under two different sets of conditions, the law can be written as:
$\frac{P_1}{T_1}=\frac{P_2}{T_2} \qquad \mathrm{or} \qquad {P_1}{T_2}={P_2}{T_1}.$
Charles's Law was also known as the Law of Charles and Gay-Lussac, because Gay-Lussac published it in 1802 using much of Charles' unpublished data from 1787. However, in recent years the term has fallen out of favor since Gay-Lussac has the second but related law presented here and attributed to him. This related form of Gay-Lussac's Law, Charles's Law, and Boyle's law form the combined gas law. The three gas laws in combination with Avogadro's Law can be generalized by the ideal gas law.
• Castka, Joseph F.; Metcalfe, H. Clark; Davis, Raymond E.; Williams, John E. (2002). Modern Chemistry. Holt, Rinehart and Winston. ISBN 0-03-056537-5.
• Guch, Ian (2003). The Complete Idiot's Guide to Chemistry. Alpha, Penguin Group Inc.. ISBN 1-59257-101-8.
• Mascetta, Joseph A. (1998). How to Prepare for the SAT II Chemistry. Barron's. ISBN 0-7641-0331-8.
The expression Gay-Lussac's law is used for each of the two relationships named after the French chemist Joseph Louis Gay-Lussac and which concern the properties of gases, though it is more usually applied to his law of combining volumes, the first listed here. One law relates to volumes before and after a chemical reaction while the other concerns the pressure and temperature relationship for a sample of gas.
## Law of combining volumes
The law of combining volumes states that, when gases react together to form other gases, and all volumes are measured at the same temperature and pressure:
The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.
This reflects the fact that (Avogadro's law) equal volumes of gas contain equal numbers of molecules ( at the same temperature and pressure), and also that in chemical reactions, the molecules combine in a ratio of whole numbers—this is known as the stoichiometry of the chemical reaction and is expressed via the chemical equation for the reaction. The law of combining gases was published by Joseph Louis Gay-Lussac in 1808.[1]
## Pressure-temperature law
Gay-Lussac's name is also associated — erroneously — with another gas law, the so-called pressure law, which states that:
The pressure of a gas of fixed mass and fixed volume is directly proportional to the gas's absolute temperature.
Simply put, if a gas's temperature increases then so does its pressure, if the mass and volume of the gas are held constant. The law has a particularly simple mathematical form if the temperature is measured on an absolute scale, such as in kelvins. The law can then be expressed mathematically as:
$\left\{P\right\}\propto\left\{T\right\}$
or
$\frac\left\{P\right\}\left\{T\right\}=k$
where:
P is the pressure of the gas.
T is the temperature of the gas (measured in Kelvin).
k is a constant.
This law holds true because temperature is a measure of the average kinetic energy of a substance; as the kinetic energy of a gas increases, its particles collide with the container walls more rapidly, thereby exerting increased pressure.
For comparing the same substance under two different sets of conditions, the law can be written as:
$\frac\left\{P_1\right\}\left\{T_1\right\}=\frac\left\{P_2\right\}\left\{T_2\right\} \qquad \mathrm\left\{or\right\} \qquad \left\{P_1\right\}\left\{T_2\right\}=\left\{P_2\right\}\left\{T_1\right\}.$
Amontons's Law of Pressure-Temperature: The pressure law described above should actually be attributed to Guillaume Amontons, who, in the late 17th century (more accurately between 1700 and 1702[2][3]), discovered that the pressure of a fixed mass of gas kept at a constant volume is proportional to the temperature. Amontons discovered this while building an "air thermometer". Calling it Gay-Lussac's law is simply incorrect as Gay-Lussac investigated the relationship between volume and temperature (i.e. Charles's Law), not pressure and temperature.
Charles's Law was also known as the Law of Charles and Gay-Lussac, because Gay-Lussac published it in 1802 using much of Charles's unpublished data from 1787. However, in recent years the term has fallen out of favor, and Gay-Lussac's name is now generally associated with the law of combining volumes. Amontons's Law, Charles's Law, and Boyle's law form the combined gas law. The three gas laws in combination with Avogadro's Law can be generalized by the ideal gas law.
## References
1. ^ http://www.chemistryexplained.com/Fe-Ge/Gay-Lussac-Joseph-Louis.html
2. ^ Barnett, Martin K. (Aug 1841), [Expression error: Unexpected < operator "A brief history of thermometry"], Journal of Chemical Education 18 (8): 358 . Extract.
3. ^ http://web.fccj.org/~ethall/gaslaw/gaslaw.htm | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.948951780796051, "perplexity": 835.8555474250325}, "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/1539583512395.23/warc/CC-MAIN-20181019103957-20181019125457-00134.warc.gz"} |
http://mathoverflow.net/questions/111568/find-a-minimum-entropy-code-for-a-simple-gibbs-random-field | Find a minimum entropy code for a simple gibbs random field.
Just to make precise what I am talking about, I will include the definition of a minimum entropy code. I will then define the precise markov random field I am asking about.
In the rest of this question, if $b$ is a binary string, I will use $b_i$ to refer to the $i$th bit of this string.
Minimum entropy codes were defined by Barlow, Kaushal, and Mitchison in 1989: http://www.mitpressjournals.org/doi/abs/10.1162/neco.1989.1.3.412?journalCode=neco. The idea is quite simple. Given a discrete random variable $X$, a probability mass function $f$ for $X$, and a positive integer $N$, find a one-to-one mapping $m$ from the set of possible outcomes of $X$ to the set of all binary strings of length $N$ (call it $B^N$ where $B$={0,1}), such that the following is minimized: $$- \sum p_i \log p_i - \sum q_i \log q_i$$
Where $p_i = 1 - q_i$ is the probability of the $i$th bit being 1, i.e.: $$p_i = \sum_x \mathrm{Pr}(X=x) m(x)_i$$
Now, on to the Markov random field (MRF). The specific MRF in this question is the Gibbs random field defined by $N$ binary nodes: $(s_1, s_2, \cdots, s_N) \in B^N$, with the nodes arranged circularly and each one inhibiting its immediate neighbors. In more precise terms, the probability of each state is: $$\mathrm{Pr}(s_1, s_2, \cdots, s_N) = \frac{\exp(-\sum_{i=1}^N (s_i s_{i+1} + s_i s_{i-1})/T)}{Z(T)}$$
Where $T$ is a positive real number that is a free variable, and $Z(T)$ is a normalizing factor that is required so that the sum taken over all states is equal to 1 (the partition function). As mentioned, the nodes are connected circularly. That is, in the subscripts, $i+1$ and $i-1$ are assumed to 'wrap around'. In other words, if $i=N$, then $s_{i+1}=s_1$. And if $i=1$, $s_{i-1}=s_N$.
The question is: find a minimum entropy code for the given MRF. This is a question with very important implications for pattern recognition and neural signal processing.
-
I'm confused by the many minus signs in the formula for the probability of a state. Their overall effect is that the probability is greatest when all the $s_i$ are 1, which seems unlikely to be what you meant by "inhibiting". Did you perhaps intend to have a minus sign either before the summation or on the individual terms but not both? As long as I'm asking naive questions, do you really intend the $s_i$ to be in $B$, i.e., to be 0 or 1 (as opposed to perhaps being $-1$ or 1)? – Andreas Blass Nov 5 '12 at 16:27
Thanks for spotting the typo. The minus signs in the sum were by mistake. The definition for $s_i$ is correct; they binary and not in {-1,1}. I double-checked the rest of the post just to make sure to find any other possible typos. – Alin Nov 5 '12 at 16:44 | {"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": 3, "x-ck12": 0, "texerror": 0, "math_score": 0.9515195488929749, "perplexity": 210.48325413558513}, "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-2015-11/segments/1424936466999.23/warc/CC-MAIN-20150226074106-00080-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://www.eomys.com/produits/manatee/plot-commands/forces/article/plot_fr_fft2-em-pp | # FFT2 contour plot of radial magnetic force (plot_Fr_fft2)
Post-processing documentation
Matlab commandplot_Fr_fft2
Module EM.pp
Description PLOT_FR_FFT2 plots the 2D Fourier transform of airgap magnetic radial force on outer structure (rotor or stator) per unit area including natural frequencies
Parameters
Restrictions
See also plot_Ft_fft2, plot_Fr_fft2_stem, plot_Fr_cfft2
### Result
This MANATEE software post processing gives the 2D Fourier Transform of radial magnetic forces per unit area:
Stator radial pressure FFT2 and natural frequencies
plot_Fr_fft2
The top part of the FFT shows the magnitude of the harmonic force wave while the bottom part shows the phase angle of the harmonic force waves. The phase angle is useful to identify if a pulsating force occurs (e.g. two harmonics with opposite wavenumbers and same phase angle). Negative wavenumbers account for the different propagation direction of force waves.
The largest force harmonics should occur at r=2p (twice the number of pole pairs), f=2fs (twice the fundamental electrical frequency), and at r=0 f=0 (static force). Note that only the lowest force wavenumbers are important in magnetic noise and vibration process.
When spectral leakage occurs (e.g. due to slip in induction machines with a too low number of revolutions, or when importing flux or current data that is not exactlty periodic) one can obtain
Airgap radial pressure FFT2 and nature frequencies (+,x)
plot_Fr_fft2
The reference level of the pressure dB scale can be tuned in
`Input.Plot.PM0ref`
For further plot options (e.g. change axis) see How to change plot options.
This type of graph is very useful to identify which are the main harmonic forces responsible for magnetic noise and vibration of electrical machines. When a large force harmonic is close from a natural frequency, large vibration and noise can be expected. | {"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": 0.848614513874054, "perplexity": 3031.440121204361}, "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-2019-18/segments/1555578526228.27/warc/CC-MAIN-20190418181435-20190418203435-00143.warc.gz"} |
https://infoscience.epfl.ch/record/182812?ln=en | ## The effect of variability of urban systems characteristics in the network capacity
Recent experimental analysis has shown that some types of urban networks exhibit a low scatter reproducible relationship between average network flow and density, known as the macroscopic fundamental diagram (MFD). It has also been shown that heterogeneity in the spatial distribution of density can significantly decrease the network flow for the same value of density. Analytical theories have been developed to explore the connection between network structure and an MFD for urban neighborhoods with cars controlled by traffic signals. However these theories have been applied only in cities with deterministic values of topological and control variables for the whole network and by ignoring the effect of turns. In our study we are aiming to generate an MFD for streets with variable link lengths and signal characteristics and understand the effect of variability for different cities and signal structures. Furthermore, this variability gives the opportunity to mimic the effect of turning movements. Route or network capacity can be significantly smaller than the capacity of a single link, because of the correlations developed through the different values of offsets. The above analysis would not be possible using standard traffic engineering techniques. This will be a key issue in planning the signal regimes such a way that maximizes the network capacity and/or the density range of the maximum capacity.
Published in:
Transportation Research Part B Methodological, 46, 10, 1607-1623
Year:
2012
Publisher:
Elsevier
ISSN:
0191-2615
Keywords:
Laboratories: | {"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": 0.866973876953125, "perplexity": 587.1375576657435}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251801423.98/warc/CC-MAIN-20200129164403-20200129193403-00172.warc.gz"} |
http://www.gradesaver.com/up-from-slavery/q-and-a/why-did-he-take-up-public-speaking-and-how-successful-was-he--why-was-the-atlanta-exposition-speech-so-important--172269 | # Why did he take up public speaking and how successful was he ? Why was the Atlanta exposition speech so important ?
Thank you very much if you can help | {"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": 0.8229223489761353, "perplexity": 1334.0259763091033}, "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/1508187824819.92/warc/CC-MAIN-20171021133807-20171021153807-00222.warc.gz"} |
https://arxiv.org/abs/1110.1263 | cs.FL
# Title:Two-Way Automata Making Choices Only at the Endmarkers
Abstract: The question of the state-size cost for simulation of two-way nondeterministic automata (2NFAs) by two-way deterministic automata (2DFAs) was raised in 1978 and, despite many attempts, it is still open. Subsequently, the problem was attacked by restricting the power of 2DFAs (e.g., using a restricted input head movement) to the degree for which it was already possible to derive some exponential gaps between the weaker model and the standard 2NFAs. Here we use an opposite approach, increasing the power of 2DFAs to the degree for which it is still possible to obtain a subexponential conversion from the stronger model to the standard 2DFAs. In particular, it turns out that subexponential conversion is possible for two-way automata that make nondeterministic choices only when the input head scans one of the input tape endmarkers. However, there is no restriction on the input head movement. This implies that an exponential gap between 2NFAs and 2DFAs can be obtained only for unrestricted 2NFAs using capabilities beyond the proposed new model. As an additional bonus, conversion into a machine for the complement of the original language is polynomial in this model. The same holds for making such machines self-verifying, halting, or unambiguous. Finally, any superpolynomial lower bound for the simulation of such machines by standard 2DFAs would imply L<>NL. In the same way, the alternating version of these machines is related to L =? NL =? P, the classical computational complexity problems.
Comments: 23 pages Subjects: Formal Languages and Automata Theory (cs.FL); Computational Complexity (cs.CC) ACM classes: F.1.1; F.1.3; F.2.3; F.4.3 Cite as: arXiv:1110.1263 [cs.FL] (or arXiv:1110.1263v1 [cs.FL] for this version)
## Submission history
From: Giovanni Pighizzini [view email]
[v1] Thu, 6 Oct 2011 13:52:34 UTC (28 KB) | {"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": 0.8075003027915955, "perplexity": 1989.468651453377}, "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/1573496669755.17/warc/CC-MAIN-20191118104047-20191118132047-00475.warc.gz"} |
http://mathhelpforum.com/algebra/121199-converting-decimals-fractions.html | # Thread: Converting Decimals to Fractions
1. ## Converting Decimals to Fractions
1/34 to decimal=
Change to Fraction
x=.12361236
Changes to Fraction
x=.352171717
Thanks,
Steve
2. Originally Posted by skweres1
Change to Fraction
x=.12361236
For a repeating fraction take the part that repeats and put the same number of 9s as the denominator and cancel if appropriate.
$x = \frac{1236}{9999} = \frac{412}{3333}$
3. Originally Posted by skweres1
1/34 to decimal=
Change to Fraction
x=.12361236
Changes to Fraction
x=.352171717
Thanks,
Steve
x = 0.352171717
This is a repeating decimal:
Take it apart
0.352 + 0.000171717 = 0.352171717
as e^(i*pi) showed in his post:
$\dfrac{17}{99000} = 0.0001717171717171717...$
$\dfrac{352}{1000} + \dfrac{17}{99000}$ = 0.352171717
you can see that 352/1000 can be reduced
$\dfrac{352}{1000} = \dfrac{176}{500} = \dfrac{88}{250} = \dfrac{44}{125}$
$\dfrac{44}{125} + \dfrac{17}{99000}$
need a common denominator
$\dfrac{44}{125} \times \dfrac{99000}{99000} \,+\, \dfrac{17}{99000}\times \dfrac{125}{125}$
$\dfrac{ 44 \cdot 99000 \, + \, 17 \cdot 125 }{ 125 \cdot 99000}$
$\dfrac{ 4356000 \, + \, 2125 }{ 12375000}$
$\dfrac{ 4358125 }{ 12375000}$
and that can be reduced:
divide numerator & denominator by 625 to get the fraction in lowest terms.
Spoiler:
$\dfrac{6973}{19800}$
The continued fraction algorithm is the method of choice for doing this type of conversions.
The above is not that algorithm.
Perhaps someone will explain/demonstrate the continued fraction algorithm for converting decimals to rationals.
4. if the digits are recurring indefinitely
x= 0.12361236...
10000x=1236.12361236...
-
x= 0.12361236...
9999x=1236
$x=\frac{1236}{9999}$
if the fraction doesnt repeat indefinitely
$0.12361236=\frac{12361236}{100000000}$
5. Hello, Steve!
I will assume those are repeating decimals.
$\frac{1}{34}\text{ to decimal}$
Just divide it out until you get a repeating cycle . . .
$\frac{1}{34} \;=\;0.0 \overline{2941176970588235}\, \hdots$
$\text{Change to fraction: }\:x \:=\:0.12361236\hdots$
$\begin{array}{ccccc}\text{Multiply by 10,000:} & 10,000x &=& 1236.12361236\hdots \\
$
Therefore: . $x \;=\;\frac{1236}{9999} \;=\;\frac{412}{3333}$
$\text{Change to fraction: }\:x \:=\:0.362171717\hdots$
$\begin{array}{ccccc}\text{Multiply by 100,000:} & 100,\!000x &=& 36217.171717\hdots \\
\text{Multily by 1,000:} & \;\;\;1,\!000x &=& \quad 362.171717\hdots \\
Therefore: . $x \;=\;\frac{35,\!855}{99,\!000} \;=\;\frac{7,\!171}{19,\!800}$ | {"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": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9647015333175659, "perplexity": 4879.053725488205}, "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-2017-13/segments/1490218190134.25/warc/CC-MAIN-20170322212950-00001-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/adjoint-representation-of-a-lie-algebra-of-a-same-dimension-that-the-basis-representa.292985/ | # Adjoint representation of a Lie algebra of a same dimension that the basis representa
1. Feb 17, 2009
### yorik
Hello,
I hope it's not the wrong forum for my question which is the following:
Is there some list of Lie algebras, whose adjoint representations have the same dimension as their basic representation (like, e.g., this is the case for so(3))? How can one find such Lie algebras? Could you recommend some literature to me?
2. Feb 17, 2009
### CompuChip
Re: Adjoint representation of a Lie algebra of a same dimension that the basis repres
I'm not very well-founded in Lie-algebra's, but the adjoint of an element x is the map
$$\operatorname{ad}_x: y \mapsto [x, y]$$
isn't it?
So the adjoint is linear, i.e.
$$\operatorname{ad}_x + \operatorname{ad}_y = \operatorname{ad}_{x + y}, \operatorname{ad}_x(y) + \operatorname{ad}_x(z) = \operatorname{ad}(y + z)$$
etc. - then isn't the adjoint representation always of the same dimension.
I.e. the basic representation provides the generators $g_i$ of the Lie-algebra, and then $\operatorname{ad}_{g_i}$ generate the adjoint representation?
3. Feb 17, 2009
### yorik
Re: Adjoint representation of a Lie algebra of a same dimension that the basis repres
Well, the basic idea of the Lie groups and lie algebras is the following. A Lie group G is a group, whose elements can be written as exp(i * ak Xk), where k runs from 1 to N and over k is summed (Einstein summation convention), ak are some real-valued parameters and Xk are some linearly independent hermitian operators (so called generators). They form an N-dimensional vector space (however, the dimension of the Hilbert space on which they act is not specified).
Now, the generators satisfy [Xk, Xl] = i fklmXm, where the f's are the so called structure constants. They define the Lie algebra of the Lie group G. It turns out, you can find infinitely many generator spaces (of different), which satisfy the algebra. The smallest irreducible generator space gives us the basic (fundamental) representation of the Lie Algebra.
We can also define the so called adjoint representation. We define
(Tk)lm = - i fklm. It has the same dimension as the Lie group. But, in generally, its dimension is not equal to the dimension of the fundamental representation. However, it's the case for so(3), the Lie Algebra of SO(3): http://math.ucr.edu/home/baez/lie/node5.html.
So my question is: Are there any other Lie groups / algebras for which the dimension of the two representations are equal like it is for SO(3) / so(3)?
Similar Discussions: Adjoint representation of a Lie algebra of a same dimension that the basis representa | {"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": 0.9715163111686707, "perplexity": 481.5786347156568}, "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/1502886105334.20/warc/CC-MAIN-20170819085604-20170819105604-00226.warc.gz"} |
https://www.physicsforums.com/threads/mechanics-of-materials-question-angle-of-twist-in-shaft.716751/ | # Mechanics of materials question, angle of twist in shaft
1. Oct 15, 2013
### davidhansson
Hello, I'm studying machanics of materials.. I'm at the chapter that handles the angle of twist in shafts. There's one exercise that I get the most part of, but there's just a tiny part of it that I don't get!
here's the exercise: http://davidhansson.deviantart.com/art/Wp-20131015-001-407472489
and here's the solution from the book: http://davidhansson.deviantart.com/art/Wp-20131015-005-407472644
what I don't get, is what internal torque really means and how it become (1/2)*25x*x ,,
Where does the "(1/2)" comes from, and where does the extra x comes from..
My intuition is that it should be integrated like: 1/((pi/2)*75*10^9) ∫(15000/0.6)*x integration from 0 to 0.6 ,, which would be the twist in radians for the part AB.
I apologies if it's hard to understand
lot's of thanks/ David Hansson
2. Oct 16, 2013
### SteamKing
Staff Emeritus
It's tricky to explain, but I'll have a go.
The value of 15 kNm/m at point A is not the torque at A but the change in torque per unit length of the shaft from that point. Since the torque is zero at B, the slope of the torque distribution = 15 kNm/m divided by 0.6 m = 25 kNm /m^2. To calculate the value of the distributed torque at at distance x from point B, then
t(x) = 25 kNm/m^2 * x m = (25 x) kNm/m
The total torque applied would be the integral of the function t(x) from x = 0 to x = 0.6 m, so
T = (1/2)*(25 x^2) kNm
It's a weird way to express the torque distribution and I can understand your confusion.
3. Oct 17, 2013
### davidhansson
Hello, lots of thanks for the reply!
What you're saying sounds fully logic and was also my intuition. The weird thing though is that they make one additional integration to the (1/2)*(25 x^2) kNm ,, so we get (1/6)*(25 x^3) kNm (from 0 to 0.6).. and then multiply with the constants ofc,, this will give the final answer: 0.00895 rad that's written in the book. To me, this last integration feels totally unlogical, could it be that they actually have made a mistake in the book?
thanks/ David
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https://mathoverflow.net/questions/378590/homotopy-equivalent-smooth-4-manifolds-which-are-not-stably-diffeomorphic | # Homotopy equivalent smooth 4-manifolds which are not stably diffeomorphic?
Recall that two 4-manifolds $$M$$ and $$N$$ are stably diffeomorphic if there exist $$m,n$$ such that $$M \#_n (S^2 \times S^2) \cong N \#_n (S^2 \times S^2).$$ That is, they become diffeomorphic after taking sufficiently many connected sums with $$S^2 \times S^2$$.
I am interested to find examples $$M$$ and $$N$$ which are homotopy equivalent $$M \simeq N$$, but where $$M$$ and $$N$$ fail to be stably diffeomorphic.
I know of two sources of examples of such manifolds. In Example 5.2.4 of
Topological 4-manifolds with finite fundamental group P. Teichner, PhD Thesis, University of Mainz, Germany, Shaker Verlag 1992, ISBN 3-86111-182-9.
Teichner constructs a pair of $$M$$ and $$N$$ where the fundamental group $$\pi$$ is any finite group with Sylow 2-subgroup a generalized Quaterion group $$Q_{8n}$$ with $$n \geq 2$$.
Another pair of $$M$$ and $$N$$ with fundamental group the infinite dihedral group was constructed in:
On the star-construction for topological 4-manifolds. P. Teichner, Proc. of the Georgia International Topology Conference 1993. Geom. top. AMS/IP Stud. Adv. Math. 2 300-312 A.M.S. (1997)
Are there any other known examples of this phenomenon? I have been unsuccessful in finding any others in the literature, but this is not my area of expertise. Are there any general results about when this can occur?
• Is it possible that exotic 4-spheres could give examples? Or perhaps it is known they are all stably diffeomorphic? – Connor Malin Dec 10 '20 at 17:36
• @ConnorMalin Wall proved many moons ago that homotopy equivalent simply connected closed 4-manifolds are stably diffeomorphic. – Mike Miller Eismeier Dec 10 '20 at 18:25
• If you find an exotic 4-sphere, let me know. – archipelago Dec 10 '20 at 18:25
$$\newcommand{\Z}{\mathbb Z}\newcommand{\RP}{\mathbb{RP}}$$ $$\RP^4$$ and Capell-Shaneson's fake $$\RP^4$$, which I'll denote $$Q$$, are an example with fundamental group $$\Z/2$$. I don't know if this generalizes, but I like this example for TFT reasons: David Reutter proved that semisimple 4d TFTs cannot distinguish oriented, stably diffeomorphic $$4$$-manifolds, but there is a semisimple TFT which distinguishes $$\RP^4$$ from $$Q$$.
Kreck's modified surgery theory determines whether two closed $$4$$-manifolds $$X$$ and $$Y$$ are $$(S^2\times S^2)$$-stably diffeomorphic using bordism. Specifically, $$X$$ and $$Y$$ must have the same stable normal $$1$$-type $$\xi\colon B\to BO$$. (See Kreck for the definition of a stable normal $$1$$-type.) Then, one computes the set $$S(\xi) := \Omega_4^\xi/\mathrm{Aut}(\xi)$$, where $$\mathrm{Aut}(\xi)$$ denotes the fiber homotopy equivalences of $$\xi\colon B\to BO$$. $$X$$ and $$Y$$ determine classes in $$S(\xi)$$; they are stably diffeomorphic iff these classes are equal.
In the case of $$\RP^4$$ and $$Q$$, the stable normal type is $$\xi\colon B\mathit{SO}\times B\Z/2\to BO$$, where the map is classified by the rank-zero virtual vector bundle $$V_{\mathit{SO}}\oplus (\sigma - 1)$$; here $$V_{\mathit{SO}}\to B\mathit{SO}$$ and $$\sigma\to B\Z/2$$ are the tautological bundles. A lift of the classifying map across $$\xi$$ is equivalent to a pin$$^+$$ structure on the tangent bundle, so we look at $$\Omega_4^{\mathit{Pin}^+}\cong\Z/16$$. The $$\mathrm{Aut}(\xi)$$-action on $$\Z/16$$ sends $$x\mapsto \pm x$$.
Kirby-Taylor choose an isomorphism $$\Omega_4^{\mathit{Pin}^+}\to\Z/16$$ and show that under this isomorphism, the two pin$$^+$$ structures on $$\RP^4$$ are sent to $$\pm 1$$, and the two pin$$^+$$ structures on $$Q$$ are sent to $$\pm 9$$. Thus when we send $$x\mapsto -x$$, these two remain distinct.
TFT digression: to construct a 4d unoriented TFT that distinguishes $$\RP^4$$ from $$Q$$, begin with the pin$$^+$$ invertible TFT whose partition function is the $$\eta$$-invariant defining the isomorphism $$\Omega_4^{\mathit{Pin}^+}\to\mu_{16}$$ (here $$\mu_{16}$$ denotes the 16th roots of unity in $$\mathbb C$$). Then perform the finite path integral over pin$$^+$$ structures. Both of these operations are mathematically understood for once-extended TFT, so the result is a once-extended (hence semisimple) unoriented TFT which distinguishes $$\RP^4$$ from $$Q$$. I wrote about this in little more detail in another MO answer.
• This is a good example. I was thinking about oriented manifolds and should have specified that in the OP. So this is one more example of what I was looking for. Is this plus what I listed the complete list of known examples? – Chris Schommer-Pries Dec 10 '20 at 22:04
• @ChrisSchommer-Pries I don't know whether this is a complete list of what's known, unfortunately. I personally suspect this example generalizes to unorientable 4-manifolds with some other fundamental groups, but unfortunately I am not familiar with the state of the literature. Sorry about that. – Arun Debray Dec 10 '20 at 22:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 70, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9144349694252014, "perplexity": 393.74825954873774}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988759.29/warc/CC-MAIN-20210506175146-20210506205146-00040.warc.gz"} |
http://mathhelpforum.com/calculus/43679-solved-area-under-curve-confused.html | # Math Help - [SOLVED] Area under a curve, confused
1. ## [SOLVED] Area under a curve, confused
Find the area under this curve : $y=\frac {1}{x^2-4}$ in the region $x\geq 3$ and $y\geq 0$.
I've already checked that $y\geq 0$ is no restriction, so they simply ask me to find $\int_3^{+\infty} \frac{dx}{x^2-4}$. Looking at a formula, I found the integral's worth $\frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -...$. I gave up here, I'm not able to solve this limit... mostly due to the restriction of the logarithm domain. I'd be very grateful if you could help me!
2. How did you get $\frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -...$?
It should be $\frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{b-2}{b+2}\right) \right] -...$ according to mathcad.
3. It should be according to mathcad.
You're right, sorry about it. I made a mistake looking at my book. Anyway, I'm still stuck .
4. Originally Posted by arbolis
Find the area under this curve : $y=\frac {1}{x^2-4}$ in the region $x\geq 3$ and $y\geq 0$.
I've already checked that $y\geq 0$ is no restriction, so they simply ask me to find $\int_3^{+\infty} \frac{dx}{x^2-4}$. Looking at a formula, I found the integral's worth $\frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -...$. I gave up here, I'm not able to solve this limit... mostly due to the restriction of the logarithm domain. I'd be very grateful if you could help me!
I get the value of the integral to be $\frac{1}{4}\cdot\bigg[\lim_{b\to{\infty}}\ln\bigg[\frac{b-2}{b+2}\bigg]+\ln(5)\bigg]$
The limit converges to zero, so the integral has a value of $\frac{1}{4}\ln(5)$
Does this make sense?
--Chris
5. Originally Posted by arbolis
Find the area under this curve : $y=\frac {1}{x^2-4}$ in the region $x\geq 3$ and $y\geq 0$.
I've already checked that $y\geq 0$ is no restriction, so they simply ask me to find $\int_3^{+\infty} \frac{dx}{x^2-4}$. Looking at a formula, I found the integral's worth $\frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -...$. I gave up here, I'm not able to solve this limit... mostly due to the restriction of the logarithm domain. I'd be very grateful if you could help me!
$\int_3^{\infty}\frac{dx}{x^2-4}=\int_0^{\infty}\frac{-dx}{4-x^2}=\frac{-1}{4}\int\frac{dx}{1-\left(\frac{x}{2}\right)^2}$
Now let $\varphi=\frac{x}{2}\Rightarrow{2\varphi=x}$
So $dx=2d\varphi$
So we get that
$\int\frac{dx}{x^2-4}=\frac{-1}{2}{\rm{arctanh}}\left(\frac{x}{2}\right)\bigg|_ {3}^{\infty}=\frac{\ln(5)}{4}$
6. I get the value of the integral to be
The limit converges to zero, so the integral has a value of
Does this make sense?
Yes this does make sens! I'm too tired, how could have missed it from what Plato said...
Thank you Chris.
7. If we solve the integral:
$\int_{3}^{L}\frac{1}{x^{2}-4}dx$
$=\frac{ln(5(L-2))}{4}-\frac{ln(L+2)}{4}$
Let $t=L-2$
$\lim_{L\to \infty}\left[\frac{ln(5t)}{4}-\frac{ln(t+4)}{4}\right]$
$=\frac{1}{4}\lim_{t\to {\infty}}ln(\frac{5t}{t+4})$
$=\lim_{t\to {\infty}}\frac{ln(5-\frac{20}{t+4})}{4}$
As $t\to {\infty}$, we can see we get
$\frac{ln(5)}{4}$
8. Hello, arbolis!
You did fine ... up to where you quit.
Find the area of this region: . $y\:=\:\frac {1}{x^2-4},\;x\geq 3,\;y\geq 0$
You integral is correct: . $\int^{\infty}_3\frac{dx}{x^2-4}$
The formula is usually written: . $\int\frac{dx}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C$
. . but the important feature is the absolute value.
We have: . $\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^{\infty}_3 \quad\Rightarrow\quad \lim_{b\to\infty}\,\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^b_3$
. $=\;
\lim_{b\to\infty}\:\frac{1}{4}\,\bigg[\ln\left(\frac{b-2}{b+2}\right) - \ln\left(\frac{1}{5}\right)\bigg] \;\;=\;\;\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left( 5\cdot\frac{b-2}{b+2}\right)$
Divide top and bottom by $b$
. . $\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left(5\cdot\fr ac{1 - \frac{2}{b}}{1 + \frac{2}{b }}\right) \;\;=\; \;\frac{1}{4}\ln\left(5\cdot\frac{1-0}{1+0}\right) \;\;=\;\;\boxed{\frac{1}{4}\ln(5)}$
9. Originally Posted by Soroban
Hello, arbolis!
You did fine ... up to where you quit.
You integral is correct: . $\int^{\infty}_3\frac{dx}{x^2-4}$
The formula is usually written: . $\int\frac{dx}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C$
. . but the important feature is the absolute value.
We have: . $\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^{\infty}_3 \quad\Rightarrow\quad \lim_{b\to\infty}\,\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^b_3$
. $=\;$ $
\lim_{b\to\infty}\:\frac{1}{4}\,\bigg[\ln\left(\frac{b-2}{b+2}\right) - \ln\left(\frac{1}{5}\right)\bigg] \;\;=\;\;\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left( 5\cdot\frac{b-2}{b+2}\right)" alt="
\lim_{b\to\infty}\:\frac{1}{4}\,\bigg[\ln\left(\frac{b-2}{b+2}\right) - \ln\left(\frac{1}{5}\right)\bigg] \;\;=\;\;\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left( 5\cdot\frac{b-2}{b+2}\right)" />
Divide top and bottom by $b$
. . $\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left(5\cdot\fr ac{1 - \frac{2}{b}}{1 + \frac{2}{b }}\right) \;\;=\; \;\frac{1}{4}\ln\left(5\cdot\frac{1-0}{1+0}\right) \;\;=\;\;\boxed{\frac{1}{4}\ln(5)}$
Honestly I don't see what the big to do is over this.
$\frac{b-2}{b+2}\sim{1}\quad\text{As }b\to\infty$
$\lim_{b\to\infty}\ln\left(\frac{b-2}{b+2}\right)\sim\ln(1)=0$
Giving us
$\frac{1}{4}\bigg[0-\ln\left(\frac{1}{5}\right)\bigg]=\frac{\ln(5)}{4}$ | {"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": 51, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9891850352287292, "perplexity": 499.42842784906}, "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/1435375095183.13/warc/CC-MAIN-20150627031815-00059-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://gradestack.com/JEE-Main-2015-Complete/Elasticity/Stress/19497-3784-40961-study-wtw | # Stress
When a force is applied on a body there will be a relative displacement of the particles and due to the property of elasticity an internal restoring force is developed which tends to restore the body to its original state. The internal restoring force acting per unit area of cross section of the deformed body is called stress.
If external force F is applied on the area A of a body, then
Stress
Stress developed in a body depends upon how the external forces are applied over it. On this basis, there are two types of stresses: normal and shear or tangential stress
# Normal stress
Here the force is applied normal to the surface. It is again of two types: longitudinal and bulk or volume stress.
Longitudinal stress
• Longitudinal stress occurs only in solids and comes in picture when one of the three dimensions, viz., length, breadth, height is much greater than the other two.
• Deforming force is applied parallel to the length and causes increase in length.
• Area taken for the calculation of stress is area of cross section.
• Longitudinal stress produced due to increase in length of a body under a deforming force is called tensile stress.
• Longitudinal stress produced due to decrease in length of a body under a deforming force is called compressional stress.
Bulk or volume stress
• Bulk stress occurs in solids, liquids, or gases.
• In case of fluids only bulk stress can be found.
• It produces change in volume and density, shape remaining same.
• Deforming force is applied normal to surface at all points.
• Area for calculation of stress is the complete surface area perpendicular to the applied forces.
• It is equal to change in pressure because change in pressure is responsible for change in volume.
# Shear or tangential stress
It comes in picture when successive layers of solid move on each other, i.e., when there is a relative displacement between various layers of solid (Fig. 2).
Fig. 2
• Here deforming force is applied tangential to one of the faces.
• Area for calculation is the area of the face on which force is applied. | {"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": 0.8556626439094543, "perplexity": 795.6670798151132}, "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-2017-22/segments/1495463608004.38/warc/CC-MAIN-20170525063740-20170525083740-00446.warc.gz"} |
http://www.biochempages.com/2017/08/molarity-calculator-using-percentage-purity-density-acids-bases.html | # Molarity calculator using percentage purity and density of acids and bases
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This is a simple online molarity calculator from % purity and density of the concentrated acids and bases. I have developed this calculator for the calculations necessary for the preparation of solutions using a concentrated acid or base solutions. Concentrated acids and bases such as sulfuric acid, hydrochloric acid, nitric acid, and ammonium hydroxides are provided with a certain % purity and a specific density. You can find these details on the label of the container.
Using the details; formula weight, percent purity and density of an acid or base, you can calculate the molarity of the concentrated solution. However, it would be tedious to calculate the initial molarity of the concentrated acids and bases and use that molarity for the dilutions. Therefore, to minimize the time and avoid complex calculations, I developed it. Basically, it uses the molarity formula for the percent purity and density. The relationship of the molarity, formula weight, % purity and density of the concentrated solution is given as $\inline&space;M=\frac{(purity\times&space;density\times10)}{MW}$
So, using the formula above, the given online molarity calculator determines the molarity of the initial concentrated acid or base. Then using the initial concentration (molarity), it calculates the required volume for the preparation of desired molar diluted solutions. Therefore, it contains two formula; 1) molarity calculator for the initially concentrated solutions, 2) dilution calculator for the volume required for the preparation of diluted solutions. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9268222451210022, "perplexity": 2610.046587376131}, "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/1502886106990.33/warc/CC-MAIN-20170820204359-20170820224359-00103.warc.gz"} |
http://physics.stackexchange.com/questions/52625/does-planet-revolution-time-is-always-greater-the-more-its-getting-away-from-th/52629 | # Does Planet revolution time is always greater the more it's getting away from the center?
The entire question is in the title .
It's the case for the solar system but is it always the case ? Can a planet do a revolution faster than another that is closer to the center ?
As far as I searched , it's not but I prefer ask here to be sure.
Of course I'm talking about similar orbits (all with the same eccentricity). (When I'm talking of the center, I mean the body in the center (in general a sun), even if it's not always the center)
-
You're talking about a year period, right? There is Kepler's Third Law:
$${S_1^3 \over S_2^3} = {P_1^2 \over P_2^2} .$$ where
• S - is the distances between the 'center' and a planet
• P - orbital period of the planet
Thus the further the planet is from the center (Sun), the longer its orbital period is. You can check that formula by taking examples of Earth (150million kilometers and 365days) and Mercury (58million kilometers, and you can calculate the length of its year).
-
Kepler's Third Law says that the square of a planet's orbital period is proportional to the cube of its semi-major axis; $$T^2 \propto a^3$$ So if one planet has a larger semi-major axis than another, then its orbital period will necessarily be longer. Assuming that we define "closeness to the center" of a planet by the size of the semi-major axis of its orbit, the answer to your question is no.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8713908195495605, "perplexity": 261.93055081334893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246638571.67/warc/CC-MAIN-20150417045718-00191-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://ham.stackexchange.com/questions/18070/frequency-selective-surfaces-working-principle | # Frequency selective surfaces working principle
I've seen that sometimes frequency selective surfaces are used to load a certain impedance to a certain antenna (for instance for input matching purposes).
These surfaces act like a filter on the equivalent transmission line that describes the radiated wave propagation in the space surrounding the antenna.
In this article an incident electromagnetic plane wave on a frequency selective surface made of a periodic sequence of square metal rings is considered.
The authors say:
When the transverse magnetic field faces the vertical strips, it induces current in the loop. This current leads to a secondary magnetic field around the strips, within which magnetic energy is stored. Thus, the vertical strips have an inductive effect.
Well, there are some things I don't understand:
1. How can a time - varying magnetic field induce current in vertical strips and so in the loops? I have always been using the Faraday law in this way: time - varying magnetic flux across a surface => Induced current along the loop that defines that surface.
In this case, the magnetic field does not enter the surfaces of the loops.
1. Can we say that it's the Electric field of the wave that moves electrons in the vertical strips and causes current to flow?
2. Next, the article says that the magnetic field surrounds two adjacent vertical strips, like in the following picture:
Is it the magnetic field created by induced currents in the loops?
• I deleted my earlier answer. I did not understand that the circumference of each loop is on the order of a wavelength. My response incorrectly assumed the circumference was on the order of $\lambda$/10. Feb 22, 2021 at 21:22
You had three separate questions there, I'll try to clarify some concepts:
How can a time - varying magnetic field induce current in vertical strips
Faraday's law of induction tells us how much there is electromotive force (a kind of sum of voltages that must be dissipated in the loop) along a conductive loop when a magnetic field passing through that loop changes. However, a varying magnetic field can drive a charged particle in a single wire that is orthogonal to the direction of the magnetic field. In this case, there are moving electrons that act as current. For more explanation, see:
https://physics.stackexchange.com/questions/211293/does-a-changing-magnetic-field-impart-a-force-on-a-stationary-charged-particle
Can we say that it's the Electric field of the wave that moves electrons in the vertical strips and causes current to flow?
Yes, but that is not the whole story and is not enough to fully explain how the FSS works.
Next, the article says that the magnetic field surrounds two adjacent vertical strips, like in the following picture: -- Is it the magnetic field created by induced currents in the loops?
This graph seems to be a simplified representation of two parallel wires that carry current in the same direction (either directly in or out of your monitor). A single current-carrying wire has a magnetic field that loops around the wire. If you place two wires next to each other, then exactly between these two wires one wire creates a magnetic field pointing up and the other pointing down, causing them to cancel each other out. The authors of the paper use this to illustrate how they calculate the combined inductance of the two parallel wires from adjacent unit cells.
• Thank you for your answer. I have only another doubt. You've shown me that the changing magnetic field induces currents in the vertical strips. The article then concludes it's an inductive effect and rappresents it as an inductance. But why can't we say that such currents are the effect of the vertical incident electric field? If I'd do this kind of analysis, I'll have no reason to conclude that there is an inductive effect. Feb 21, 2021 at 11:59
• I may answer me by saying that such an electric field is generated by the changing magnetic field and so it's an inductive effect. But this is true for each EM wave incident on whatever object. According to this, whatever object would have an inductive effect, it's not a features of FSS. Feb 21, 2021 at 12:00
• From this analysis it seems that inductive or capacitive is a feature of the wave and not of the object Feb 21, 2021 at 12:07
• You are asking very good questions. I think that the sentence "Thus, the vertical strips have an inductive effect", doesn't refer to what created current in the first place, but the fact that the current in these two parallel strips has a magnetic field which tries to maintain the current. This current maintaining itself through magnetic field is called "inductance". Feb 21, 2021 at 14:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8251267671585083, "perplexity": 306.69459946606094}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00741.warc.gz"} |
http://mathhelpforum.com/calculus/78563-radius-curvature-problem.html | Math Help - Radius of curvature problem
1. Radius of curvature problem
I'm having a little trouble finding the radius of curvature for problems of the following form
$y^n=f(x)$
For examples. I'm asked to find the radius of curvature at the point (0,0) for the curve $y^2=4ax$
So
$2y\, y' = 4a \implies y' = 2ay^{-1}$
and
$y'' = -2ay^{-2}\, y'$
So
$\rho = \frac{[1+(y')^2]^{\frac{3}{2}}}{y''} = \frac{[1+4a^2y^{-2}]^{\frac{3}{2}}}{-2ay^{-2}\, y'}$
it's at this point where I'm sure I've messed up. If anyone could tell me how to solve these types of questions I'd be very grateful
Regards
Stonehambey
2. This is a parabola.
Let y=t, then $x=\frac{t^{2}}{4ax}$
and ${\kappa}(t)=\frac{\frac{1}{|2a|}}{[\frac{t^{2}}{4a^{2}}+1]^{\frac{3}{2}}}$
$t=0 \;\ when \;\ (x,y)=(0,0), \;\ so \;\ {\kappa}(0)=\frac{1}{|2a|}, \;\ {\rho}=2|a|$
3. Originally Posted by galactus
This is a parabola.
Let y=t, then $x=\frac{t^{2}}{4ax}$
and ${\kappa}(t)=\frac{\frac{1}{|2a|}}{[\frac{t^{2}}{4a^{2}}+1]^{\frac{3}{2}}}$
$t=0 \;\ when \;\ (x,y)=(0,0), \;\ so \;\ {\kappa}(0)=\frac{1}{|2a|}, \;\ {\rho}=2|a|$
Hi, thanks for the reply, but I'm afraid I cannot follow your answer (and I would very much like to, since it's correct). Would you mind explaining the steps at all?
Regards,
Stonehambey
4. Ah, so I did a little reading this morning, and I can now work through it parametrically. However this question was in the exercise right after the cartesian method had been explained, the parametric method hadn't even been covered yet. The logical conclusion is that you can solve this using cartesian method. But I'm not sure how since I keep getting situations where I'm dividing by zero! | {"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": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9749398231506348, "perplexity": 322.96954820233424}, "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-2015-32/segments/1438042989043.35/warc/CC-MAIN-20150728002309-00132-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.underwise.com/questions/676192/prove-that-and-ell-satisfying-ellf-ge-0-whenever-f-ge-0-is-bounded | Latest update
# Prove that : and $\ell$ satisfying: $\ell(f)\ge 0$ whenever $f\ge 0$ is bounded .
2017-12-03 03:06:18
While reading some functional analysis note I came across the following theorem.
Riesz-Markov: (for linear forms on Wiener spaces) If $X$ is locally compact topological space and $\ell : C_b(X)\to \Bbb R.$ is a linear and continuous form satisfying $\ell(f)\ge 0$ whenever $f\ge 0$. Then there exists a unique Borel measure $\mu$ on $X$ such that $$\ell(f) = \int_X f d\mu, ~~~~\forall~~f\in C_b(X).$$
Where $C_b(X)$ is the space of bounded functions on $X.$
The document says the following statement: Such operators $\ell$ satisfying: $\ell(f)\ge 0$ whenever $f\ge 0$ is automatically bounded.
How to prove that $\ell$ is bounded on $C_b(X)$.
I though it could be a good idea to share this on MSE.
In fact: we have $$\|f\|_\infty\pm f\ge 0\implies \ell(1)\|f\|_\infty\pm \ell(f) \overset{\text{linearity}}{=} \ell(\|f\|_\infty)\pm \ell(f)\overset{\text{linearity}}{=}\ell(\|f\|_\infty\pm f) \ge 0$$
That is for all $f\in C_b(X)$ we have, $$|\ell(f)| = \pm\ell(f) • In fact: we have$$\|f\|_\infty\pm f\ge 0\implies \ell(1)\|f\|_\infty\pm \ell(f) \overset{\text{linearity}}{=} \ell(\|f\|_\infty)\pm \ell(f)\overset{\text{linearity}}{=}\ell(\|f\|_\infty\pm f) \ge 0$$That is for all f\in C_b(X) we have,$$ |\ell(f)| = \pm\ell(f) \le \ell(1)\|f\|_\infty.
this prove the continuity of $\ell$ and hence $\ell \in (C_b(X))^*$
2017-12-03 03:47:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9922319054603577, "perplexity": 334.5100375691074}, "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-2018-30/segments/1531676589455.35/warc/CC-MAIN-20180716193516-20180716213516-00150.warc.gz"} |
https://mathoverflow.net/questions/294178/combinatorial-computational-problem-about-0-1-vectors-and-sampling-algorithms | # Combinatorial computational problem about 0-1 vectors and sampling algorithms
Let $M \in \{0,1\}^{m\times n}$, where $n\gg 1$ and $m\le n$. A procedure consisting of the following three steps is repeated $t\gg 1$ times:
1. A row $\require{amsmath} \boldsymbol{r}$ of $M$ picked in an adversarial (hidden) way, viz. we do not know the row index of $\boldsymbol{r}$.
2. We receive a set $S$ formed by sampling uniformly at random $\sqrt{n}$ indices from $[n]$ and the sequence of all values $r_i$ for all $i \in S$.
3. We need to find at least one row $\boldsymbol{r}'$ of $M$ such that, for all column indices $j \in S$, we have $\boldsymbol{r}'_j=\boldsymbol{r}_j$.
It is easy to see that the simplest method to accomplish this task requires in the worst case a total number of elementary operations (e.g., verifying if two binary digits are equal) that is order of $t\,m\,\sqrt{n}$, as both $t$ and $n$ approach infinity.
Question: Is there a (perhaps randomized) method to accomplish this task such that, by suitably pre-processing in polynomial (in $n$) time the matrix $M$ in a preliminary phase if necessary, the (expected) number of elementary operations is equal in the worst case to $o(t\,m\,\sqrt{n})$?
For example, consider we pre-process the matrix $M$ by permutating its rows such that, if $N_i$ is the integer corresponding to the binary digit representation of the $i$-th row of $M$, they are sorted in such a way $N_i\le N_j$ whenever $i \le j$. Is there any method able to take advantage of this row (sorting) permutation?
• An answer that matches the letter, but presumably not the spirit, of your question is to do so much preprocessing that all possible questions have been answered already and stored in a convenient table. Or you can go part way by, say, making a table that allows any $n^{1/4}$ columns of $M$ to be looked up immediately. To make the question more useful, I think you need to put some limit on either the time used for preprocessing or the space used to hold the results of preprocessing. – Brendan McKay Mar 2 '18 at 2:25
• Having matrix rows sorted (in lexicographic order) allows to perform exact search of $r$ in $M$ in time about $\log_2(m)n$, which may be better than $m\sqrt{n}$ (depending on relationship between $m$ and $n$). – Max Alekseyev Mar 2 '18 at 9:47
• Brendan, thank you. You are right about defining the pre-processing time complexity bound. I would just say that a polynomial (in n and m) time and space complexity for the preprocessing is OK. About the table, I don't think it can work, because the number of columns that can be sampled from a subset of $n^{1/4}$ columns is exponential in $n$ (it is still exponential $n$ if we consider the expected number of columns sampled by any fixed subset of $n^{1/4}$ columns). I instead believe it would be interesting to analyse the expected time required by verifying u.a.r. each row until finding $r'$. – Penelope Benenati Mar 2 '18 at 17:00
• Max, thank you for your answer. What you say it is clear, but in my problem I took for granted that you have access only to $\sqrt{n}$ elements of the copy of $\boldsymbol{r}$ you receive in step 1. And, of course, as you know even if the $n$-dimensional matrix row vectors are sorted in $M$, the rows of a submatrix obtained by sampling u.a.r. $\sqrt{n}$ from $M$ are not sorted in general. Hence, it is no possible in general to apply dichotomic search methods. I am now rephrasing the problem in order to avoid misunderstanding. Thank you again. – Penelope Benenati Mar 2 '18 at 17:14
This answer shows maybe. Pick a parameter $s$. We will precompute $2n$ Bloom filters with false positive rate $s$ and size $g(s)$ such that $B_{i,0}$, $B_{i,1}$ such that $k \in B_{i,b}$ if $m_{i,k}=b$. This takes time $O(g(s)mn)$ Our algorithm is as follows: we will intersect the $\sqrt{n}$ bloom filters (time $O(g(s)\sqrt{n})$) and then look within in that set for the answer. This takes time $O((1+s)\sqrt{n})$
Total time for $t$ operations is $O(g(s)mn+t(g(s)+1+s)\sqrt{n})$. If you can pick a better $s$ for $t$, then yes.
Why is the false positive rate not larger for the intersection? Because the intersection of the Bloom filters is exactly the Bloom filter of the intersection. I don't think the adversary can do much about picking the worst row, but that takes some more analysis.
• Thank you, this seems an interesting approach. However, I would need some days to calculate the false positive rate in the worst case. – Penelope Benenati Mar 3 '18 at 0:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8894563913345337, "perplexity": 198.8839030987296}, "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-2020-50/segments/1606141188947.19/warc/CC-MAIN-20201126200910-20201126230910-00270.warc.gz"} |
http://eprint.iacr.org/2014/651 | ## Cryptology ePrint Archive: Report 2014/651
A note on CCA2-protected McEliece Cryptosystem with a systematic public key
Pavol Zajac
Abstract: We show that the plaintext of some of the proposed CCA2 conversions of McEliece cryptosystem with a public key in systematic form can be recovered faster than with a general linear decoding. This is due to the fact that an attacker only needs to recover a part of the cleartext to decrypt the relevant plaintext.
Category / Keywords: public-key cryptography / cryptanalysis, public-key cryptography, McEliece cryptosystem | {"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": 0.8529531359672546, "perplexity": 1685.643392057116}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660342.42/warc/CC-MAIN-20160924173740-00302-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/60090-solved-complex-taylor-maclaurin-laurent-series.html | # Math Help - [SOLVED] Complex Taylor, Maclaurin and Laurent series
1. ## [SOLVED] Complex Taylor, Maclaurin and Laurent series
1) Find the Taylor series around the point 2i and specify the radius of convergence:
(a) $f(z)=\frac{1}{z}$
(b) $f(z)=log(z)$
(a): $f^{(n)}(z)=\frac{(-1)^n n!}{z^{n+1}}$ and $c_n= \frac{f^{(n)}(z_0)}{n!}=\frac{(-1)^n}{{z_0}^{n+1}}$ where $z_0 = 2i$
So the Taylor series is: $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^n}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{i}{2}\right)^n(z-2i)^n$
My textbook has a slightly different answer though. And the radius of convergence? Do I need to use the ratio test, or is there some simpler way to deal with it?
2) Find the first three nonvanishing terms for the Maclaurin series of: $f(z)=\frac{(1-z)^{1/2}}{1+z^2}$
3) Determine the Laurent series for $\frac{1}{z+z^2}$, $|z+1/2|> 1/2$
I've solved this for different areas, but I'm having a little trouble with this one.
2. Originally Posted by Spec
1) Find the Taylor series around the point 2i and specify the radius of convergence:
(a) $f(z)=\frac{1}{z}$
(b) $f(z)=log(z)$
(a): $f^{(n)}(z)=\frac{(-1)^n n!}{z^{n+1}}$ and $c_n= \frac{f^{(n)}(z_0)}{n!}=\frac{(-1)^n}{{z_0}^{n+1}}$ where $z_0 = 2i$
So the Taylor series is: $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}(z-2i)^n =\frac{1}{2i}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^n}(z-2i)^n$
Correct as far as there,...
Originally Posted by Spec
$= \frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{i}{2}\right)^n(z-2i)^n$
... but that step is wrong. In fact, $\frac{-1}i = i$, so $\frac{(-1)^n}{(2i)^n} = \left(\frac{i}{2}\right)^n$. The $(-1)^{n+1}$ ought not to be there at all.
Originally Posted by Spec
And the radius of convergence? Do I need to use the ratio test, or is there some simpler way to deal with it?
The ratio test is always good for finding a radius of convergence. In this case, it's a bit quicker if you notice that this is a geometric series.
3. Originally Posted by Spec
1) Find the Taylor series around the point 2i and specify the radius of convergence:
(a) $f(z)=\frac{1}{z}$
(b) $f(z)=log(z)$
(a): $f^{(n)}(z)=\frac{(-1)^n n!}{z^{n+1}}$ and $c_n= \frac{f^{(n)}(z_0)}{n!}=\frac{(-1)^n}{{z_0}^{n+1}}$ where $z_0 = 2i$
So the Taylor series is: $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^n}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{i}{2}\right)^n(z-2i)^n$
My textbook has a slightly different answer though. And the radius of convergence? Do I need to use the ratio test, or is there some simpler way to deal with it?
2) Find the first three nonvanishing terms for the Maclaurin series of: $f(z)=\frac{(1-z)^{1/2}}{1+z^2}$
3) Determine the Laurent series for $\frac{1}{z+z^2}$, $|z+1/2|> 1/2$
I've solved this for different areas, but I'm having a little trouble with this one.
Should not the radius of convergence of a Taylor series simply be the distance to the nearest singularity? In that case, the radius of convergence for (1a) would be 2? I would think the same with (1b) except the series will converge to an analytic sheet of $log(z)$
Also, (3) sounds not right. Should it be $|z+1/2|<1/2$ or no?
ok, I understand now. I' wrong. Sorry. . .
4. I solved (3) myself by using the substitution $w=z+1/2$
Then you have:
$\frac{1}{w^2-1/4}=\frac{1}{w^2}\frac{1}{1-\frac{1/4}{w^2}}=\frac{1}{w^2}\sum_{n=0}^{\infty}\frac{1/4}{w^2}^n = ...$
What about the Taylor series for $log z$ around the point $2i$?
5. Originally Posted by Spec
What about the Taylor series for $\log z$ around the point $2i$?
$\log z = \log(2i + z - 2i) = \log(2i) + \log\bigl(1 + \tfrac{z-2i}{2i}\bigr)$. Now use the usual series for log(1+w). | {"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": 37, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9805086255073547, "perplexity": 326.59049670087813}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246649738.26/warc/CC-MAIN-20150417045729-00097-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.quantumstudy.com/a-rectangular-block-has-dimensions-5-cm-x-5-cm-x-10-cm-calculate-the-resistance-measured-between/ | # A rectangular block has dimensions 5 cm × 5 cm × 10 cm. Calculate the resistance measured between…
Q: A rectangular block has dimensions 5 cm × 5 cm × 10 cm. Calculate the resistance measured between (a) two square ends (b) the opposite rectangular ends. Specific resistance of the material is 3.5 × 10^(-5) Ωm.
Sol. (a) Resistance between two square ends $\large R_1 = \rho \frac{l}{A}$
$\large R_1 = 3.5 \times 10^{-5} \times \frac{10 \times 10^{-2}}{5 \times 5 \times 10^{-4}}$
= 1.4 × 10-3 Ω
(b) Resistance between the opposite rectangular ends $\large R_2 = \rho \frac{l}{A}$
$\large R_2 = 3.5 \times 10^{-5} \times \frac{5 \times 10^{-2}}{5 \times 10 \times 10^{-4}}$
= 1.4 × 10-4 Ω | {"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": 0.9870383739471436, "perplexity": 4403.9487567439455}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058589.72/warc/CC-MAIN-20210928002254-20210928032254-00093.warc.gz"} |
https://tex.stackexchange.com/questions/271626/cases-with-brackets-instead-of-curly-brackets | # Cases with brackets instead of curly brackets
How could I use cases with [] (both sides) instead of curly bracket as the attached picture shows?
Thank you.
• Please provide a bit more context, e.g. a minimal working example. Without anything, I can only guess that Bmatrix could help. cases is intentionally one-sided. Oct 7 '15 at 15:15
• @Chris Sorry for the lack of context. I wrote "cases" because I know it's one-sided. Thanks anyway. Oct 7 '15 at 21:48
\documentclass{article}
a= \left[\begin{aligned} &2 \rightarrow \text{John}\\ &5 \rightarrow \text{Mary}\\ &7 \rightarrow \text{Ann} \end{aligned}\right]
$a= \begin{bmatrix} 2 &\rightarrow &\text{John}\\ 5 &\rightarrow &\text{Mary}\\ 7 &\rightarrow &\text{Ann} \end{bmatrix}$
$a= \left[\begin{array}{@{}l@{\,}l@{\,}c@{}} 2 &\rightarrow &\text{John}\\ 5 &\rightarrow &\text{Mary}\\ 7 &\rightarrow &\text{Ann} \end{array}\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": 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": 0.9863044619560242, "perplexity": 4451.4336732303755}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301063.81/warc/CC-MAIN-20220118213028-20220119003028-00122.warc.gz"} |
https://www.physicsforums.com/threads/vectors-specifically-cross-product-application.427018/ | # Vectors ; specifically cross product application
1. Sep 6, 2010
### kapitanma
1. The problem statement, all variables and given/known data
vector A = 1.5i + 6.7j - 7.4k
vector B= -8.2i + 6.5j + 2.3k
(f) What is the magnitude of the component of vector A perpendicular to the direction of vector B but in the plane of vector A and B.
3. The attempt at a solution
This part of the problem has me kinda stumped. My attempt at the solution was using the application of the cross product : C = ABsin(theta).
I calculated the angle between the two as 82.43 degrees, and realized that this formula gave me the same thing as simply taking the magnitude of the cross product vector which I calculated to be 107.207.
This shows me that my fundamental approach to this problem is incorrect, but I have no idea where to go with it. Any pointers would be appreciated.
2. Sep 6, 2010
### Dick
Magnitudes of vectors isn't all you need, you also need directions. It would be nice to find a vector that is perpendicular to B but in the plane of A and B, right? Then you could just find the magnitude of A along that direction. How about (AxB)xB? Can you see why that works?
3. Sep 6, 2010
### kapitanma
Looking at a graph of AxB, I think I can see why that (AxB)xB would give me a vector perpendicular to B in the plane, and I'd just have to apply the dot product to get the component of A in the direction of B.
4. Sep 6, 2010
### Dick
Sure. (AxB)xB is perpendicular to B, and it's also perpendicular to AxB which is the normal to plane containing A and B. And, yes, from here you can use a dot product.
5. Sep 6, 2010
### kapitanma
I successfully solved this problem, thanks for the nudge in the right direction.
Similar Discussions: Vectors ; specifically cross product application | {"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": 0.9212745428085327, "perplexity": 401.6346458719213}, "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/1502886106754.4/warc/CC-MAIN-20170820131332-20170820151332-00690.warc.gz"} |
http://spie.org/Publications/Proceedings/Paper/10.1117/12.671942 | Share Email Print
Proceedings Paper
A coupled-mode theory for infrared and submillimeter wave detectors
Author(s): George Saklatvala; Stafford Withington; Michael P. Hobson
Format Member Price Non-Member Price
PDF \$14.40 \$18.00
Paper Abstract
We present a new theory for the description of detectors in terms of modes. Although the theory is very general, it is expected to be of particular use in the modelling of far-infrared and submillimeter instruments. Such a theory is needed because at far-infrared frequencies, both optical systems and detectors show partially coherent behaviour. That is to say, even when the instrument is illuminated by an incoherent source, the resultant field at the detector is partially coherent, and the detector itself is sensitive to the coherence properties of the field and not just the intensity. We have previously developed a modal description of optical systems at far-infrared wavelengths; here we describe a modal theory for the detectors themselves. The theories can be combined to provide a complete modal description of far-infrared instruments. The theory presented here applies equally well to pulsed or ergodic radiation, and incorporates polarisation effects. We also show how the statistics of the detector output can be determined from the theory. This is important for instruments such as bolometers, where the internal noise of the detector is very low, and either sky noise or radiation from the optical components dominate. We illustrate our work with a number of simulations, showing signal to noise ratios for different detector types, and we show how how the theory may be applied to the array packing density problem.
Paper Details
Date Published: 27 June 2006
PDF: 12 pages
Proc. SPIE 6275, Millimeter and Submillimeter Detectors and Instrumentation for Astronomy III, 62750W (27 June 2006); doi: 10.1117/12.671942
Show Author Affiliations
George Saklatvala, Cavendish Lab., Univ. of Cambridge (United Kingdom)
Stafford Withington, Cavendish Lab., Univ. of Cambridge (United Kingdom)
Michael P. Hobson, Cavendish Lab., Univ. of Cambridge (United Kingdom)
Published in SPIE Proceedings Vol. 6275:
Millimeter and Submillimeter Detectors and Instrumentation for Astronomy III
Jonas Zmuidzinas; Wayne S. Holland; Stafford Withington; William D. Duncan, Editor(s) | {"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": 0.9138507843017578, "perplexity": 3868.718241438695}, "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/1547583659890.6/warc/CC-MAIN-20190118045835-20190118071835-00334.warc.gz"} |
https://www.physicsforums.com/threads/earths-elliptical-orbit.164387/ | # Earth's elliptical orbit
1. Apr 6, 2007
### aranoff
Why is the earth's orbit elliptical? That is, what force acts on the earth perpendicular to the force of the sun, that causes the tangential accelerations characteristic of an ellipse?
2. Apr 6, 2007
### Garth
Hi aranoff and welcome to these Forums!
The real question is: "Why is the Earth's orbit so nearly circular?"
A circle is the ideal limit, Limit e -> 0, of an ellipse. Keplerian motion, explained by Newton's laws, leads us to expect every freely orbiting body to orbit its parent body on an ellipse.
The fact that the Earth's orbit is so nearly circular, e = 0.017, speaks of the averaging of orbital elements of the millions of random collisions of the planetesimals that made up the Earth. The result was very nearly circular.
Proto-planets left with high ellipticity would also be ejected by close encounters with other planets. The eight planets plus dwarf planets left are the end result of such an accretion process.
Garth
3. Apr 6, 2007
### Phobos
Staff Emeritus
Elliptical orbits are the norm and circular orbits are rare. All the gravitational nudges from the other objects in the solar system can disrupt a perfectly circular orbit. However, the Earth's orbit is fairly circular (http://www.seds.org/billa/tnp/help.html#eccentric [Broken] of only 0.02).
Last edited by a moderator: May 2, 2017
4. Apr 6, 2007
### D H
Staff Emeritus
Since nobody has answered your question directly, I will. There is no tangential acceleration (at least, not in the simple two body problem of Newtonian mechanics). No tangential acceleration is needed.
Central force motion (systems in which force is a function of distance to the origin only) does not mean the body subject to the central force is constrained to move in a circle. When the force is proportional to 1/r^2, the body is constrained to move along a conic section (circle, ellipse, parabola, or hyperbola).
5. Apr 6, 2007
### aranoff
earth's orbit
There is indeed a tangential acceleration. Acceleration is change in velocity. For a circular orbit, the change is perpendicular to the velocity, and parallel to the radius. For an elliptical orbit, the velocity changes in magnitude, and so the acceleration has a component perpendicular to the radius.
Where am I going wrong?
Thanks.
6. Apr 6, 2007
### D H
Staff Emeritus
The acceleration is still directed against the position vector for an elliptical orbit (or a parabolic or hyperbolic orbit, for that matter). What makes you think there has to be a tangential acceleration?
I think you are going wrong by assuming that the velocity vector is normal to the position vector. For a non-circular orbit, the velocity vector almost always has a non-zero radial component. The velocity is normal to the position vector at apofocus and perfocus only.
7. Apr 6, 2007
### D H
Staff Emeritus
It's fairly easy to show there is no tangential acceleration using cylindrical coordinates. The position of an orbital body with miniscule mass in cylindrical coordinates is
$$\vec r = r \hat r$$
Differentiating with respect to time,
$$\dot{\vec r} = \dot r \hat r + r \frac d{dt}\hat r = \dot r \hat r + r \dot{\theta} \hat{\theta}$$
Differentiating once more to get the acceleration,
$$\ddot{\vec r} = (\ddot r -r \dot{\theta}^2) \hat r + (2\dot r \dot{\theta}+ r \ddot{\theta})\hat{\theta}$$
The term $(2\dot r \dot{\theta}+ r \ddot{\theta})\hat{\theta}$ is the tangential acceleration. The next few paragraphs show that this tangential acceleration is zero due to conservation of momentum.
The angular momentum with respect to the origin is
$$\vec h = \vec r \times \dot{\vec r} = r^2\dot {\theta} \hat z$$
In any central force motion problem, the force on the body is, by definition, parallel to the position vector. There is no torque on the body since $\vec \tau = \vec r \times \vec F = 0$. No external torques => constant angular momentum. Thus
$$\frac d{dt}(r^2\dot {\theta}) = r(2\dot r \dot {\theta} + r \ddot {\theta}) = 0$$
and the tangential acceleration is zero.
8. Apr 6, 2007
### tony873004
Earth's eccentricity varies from nearly circular to about 0.06 in cycles of about 400,000 years. Jupiter is the main cause of this. So in addition to the large acceleration vector that is always directed radially towards the sun, perhaps Jupiter supplies the tangental component you're looking for.
9. Apr 6, 2007
### aranoff
"Perhaps Jupiter..."
Has anyone done any calculations on the Jupiter effects?
10. Apr 6, 2007
### aranoff
earth's orbit
Can one explain why the orbit is an ellipse rather than a circle without invoking other planets?
11. Apr 6, 2007
### Staff: Mentor
It is true that Jupiter has an effect, but you don't need another object in order for an orbit to be elliptical.
What you are missing, aranoff, is that the speed is changing, but the direction is not always tangential, so the acceleration is always directly toward the center of mass.
An elliptical orbit is the norm - a circular orbit is a special (read: impossibly specific) case of an elliptical orbit.
Last edited: Apr 6, 2007
12. Apr 7, 2007
### aranoff
Let me phrase my comments in another form.
Consider the two-body problem of the earth and sun, each replaced by point masses. Define let the coordinate system so that the orbit is in the x-y plane. At t=0, let the earth be at its closest distance, and moving in the y direction. Solve the equations of motion, using the initial conditions. The solution is a circular path.
The reality is that the orbit of the earth is an ellipse, with an eccentricity ε, which is a function of time. Since the fact of ε(t) is a fact that does not occur in the initial equations, there is no way to find a solution with ε ≠ 0.
I believe that the eccentricity is a result of the many-body problem, taking into account the large planets. There is no analytical solution to the Newtonian gravitational many-body problem. I do not have the tools to solve this problem. I wonder if any of you know about many-body calculations that result in the correct ε(t)?
13. Apr 7, 2007
### Staff: Mentor
You've set up the problem in such a way as to demand a circular orbit, then asked why the orbit is circular. What if the moon was also moving in the X direction at t=0? What shape would you get then?
14. Apr 7, 2007
### D H
Staff Emeritus
This is incorrect.
I don't know what level of math or physics background you have. Solving the two body problem is covered in part in freshman-level physics classes and in full in upper-level undergraduate classes.
All you can say about a planet's orbit is that it is a conic section. A circular orbit is a very special case. Given the masses of the objects and the separation between them, there is only one velocity that results in circular motion. Any other velocity results in some other conic section.
15. Apr 7, 2007
### aranoff
As a researcher in the N-body problem I welcome your query. You are right to pose the problem. Let me try to make some comments.
Suppose we take Earth at present solar distance and add just Jupiter as perturbing body. This is a stable and well behaved system. Assuming we start with circular Earth, the subsequent behaviour would reveal oscillations. I have not done this recently but would guess eccentricity would reach a maximum around 0.05 or so. Adding Saturn would introduce some irregularity but giving essentially similar result.
The actual situation is complicated by the origin of the Solar System. It is generally accepted that the terrestrial planets formed by collisions of smaller bodies. The last few major collisions would be crucial for determining the Earth's starting eccentricity. And it is even more complicated since we believe the orbit may have become more circular due to the presence of gas and accumulation of minor bodies.
I hope the above is helpful. Please don't hesitate to come back to me.
Best regards, Sverre
16. Apr 7, 2007
### Garth
Which is what I said.
And I concur that the presence of gas, dust and minor bodies adds to the process.
Garth
Last edited: Apr 7, 2007
17. Apr 7, 2007
### D H
Staff Emeritus
Aranoff's difficulties are much more basic than describing how addtional bodies perturb the Earth's orbit. He doesn't understand how the two body problem can result in non-circular motion. All: Please keep the perturbations due to multiple bodies or relativity out of this.
Aranoff: That the two-body problem can result in non-circular motion was known even to Kepler, in an ad-hoc manner. Newton developed the mathematics that describes how the two-body problem results in a conic section.
18. Apr 7, 2007
### aranoff
No, I do not understand how a two body problem can give non-circular orbits, for then the result is more than the input.
It is not possible to get more out of a set of equations than the stuff that goes into the equations and initial conditions.
19. Apr 7, 2007
### Garth
It's all in the initial conditions.
Unless the orbital velocity is initially exactly that of the circular velocity and initially exactly normal to the radial direction for both bodies then the two bodies will be on elliptical or hyperbolic orbits.
It's basic orbital dynamics, it is the exactly circular orbit that is impossible to achieve.
Garth
20. Apr 7, 2007
### D H
Staff Emeritus
Aranoff, could you elaborate on why you think a non-circular orbit is impossible? (It is not impossible; as many have said, it is basic orbital mechanics. In fact, it is the circular orbit that is impossible to achieve.)
It would also help if you could let us know what level of mathematics and physics education you have received.
Similar Discussions: Earth's elliptical orbit | {"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": 0.9153940677642822, "perplexity": 706.501554592557}, "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-2017-22/segments/1495463613738.67/warc/CC-MAIN-20170530011338-20170530031338-00380.warc.gz"} |
https://eccc.weizmann.ac.il/report/2013/153/ | Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > DETAIL:
### Paper:
TR13-153 | 8th November 2013 03:38
#### The Limits of Depth Reduction for Arithmetic Formulas: It's all about the top fan-in
TR13-153
Authors: Mrinal Kumar, Shubhangi Saraf
Publication: 8th November 2013 04:05
In this paper we study the power and limitations of depth reduction and shifted partial derivatives for arithmetic formulas. We do it via studying the class of depth 4 homogeneous arithmetic circuits. We show: (1) the first {\it superpolynomial lower bounds} for the class of homogeneous depth 4 circuits with top fan-in $o(\log n)$. The core of our result is to show {\it improved depth reduction} for these circuits. This class of circuits has received much attention for the problem of polynomial identity testing. We give the first nontrivial lower bounds for these circuits for any top fan-in $\geq 2$. (2) We show that improved depth reduction {\it is not possible} when the top fan-in is $\Omega(\log n)$. In particular this shows that the depth reduction procedure of Koiran and Tavenas [Koi12, Tav13] cannot be improved even for homogeneous formulas, thus strengthening the results of Fournier et al [FLMS13] who showed that depth reduction is tight for circuits, and answering some of the main open questions of [KSS13, FLMS13]. Our results in particular suggest that the method of depth reduction and shifted partial derivatives may not be powerful enough to prove superpolynomial lower bounds for (even homogeneous) arithmetic formulas. | {"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": 0.9077453017234802, "perplexity": 964.5550219941622}, "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/1620243991685.16/warc/CC-MAIN-20210512070028-20210512100028-00559.warc.gz"} |
http://math.stackexchange.com/questions/19713/are-conformal-maps-determined-by-a-finite-number-of-points | # Are conformal maps determined by a finite number of points?
For linear fractional transformations, it suffices to specify where three points are mapped to. I am wondering if some analogue of this holds for general conformal mappings.
-
No. For any finite collection of points $\zeta_1$, $\zeta_2$, ..., $\zeta_N$ in the disc of radius $1$, consider the map $$z \mapsto z+\epsilon \prod(z-\zeta_i).$$ This is a holomorphic map fixing all the $\zeta_i$ and, if $\epsilon$ is small enough, it is conformal.
This depends on the group of conformal automorphisms that your region admits. E.g., the Riemann sphere admits a triply transitive group of conformal automorphisms, so you can prescribe the images of three arbitrary chosen points. If your region is the unit disc $D$ then the group of conformal automorphisms has "three real degrees of freedom". This means that you can prescribe, e.g., $f(0)\in D$ and $\arg(f'(0))$, but not $f(0)$ and $f({1\over2})$ independently. | {"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": 0.9788060784339905, "perplexity": 138.3457923400234}, "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-2015-48/segments/1448398460263.61/warc/CC-MAIN-20151124205420-00052-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/34551/range-of-values-of-fx-using-quadratic-inequalities-need-intuition | # Range of values of f(x) using quadratic inequalities (need intuition)
I'm working on an exercise from a book in the chapter on quadratic inequalities: "Find the set of possible values of the given function $\frac{x - 2}{(x + 2)(x - 3)}$". The answer in the book is "all values". I don't have much intuition on the calculations so would be grateful if someone could explain what happens at the various stages. Corrections, qualifications, etc. to my statements are very much welcomed!
Here's my workings:
Looking for the range of values of $y$, so $y = \frac{x - 2}{(x + 2)(x - 3)} \Rightarrow y = \frac{x - 2}{x^2 - x - 6}$
$\Rightarrow y(x^2 - x - 6) = x - 2$
$\Rightarrow yx^2 -(y + 1)x - 6y + 2 = 0$
As far as I can understand from my book, now that I've got a quadratic in $x$, I can use the discriminant to determine the $y$ range . With $a = y, b = -(y + 1), c = (-6y + 2)$, the discriminant ($b^2 - 4ac$) is:
$(-y-1)^2 - 4(y)(-6y + 2)$
$= 25y^2 - 6y + 1$
When $25y^2 - 6y + 1 \geqslant 0$, the roots of the quadratic in $x$ are real. At this stage, from what I understand, the values satisfying the inequality $25y^2 - 6y + 1 \geqslant 0$ are the range of $y$ values for all valid, real $x$ values plugged into the function $\frac{x - 2}{(x + 2)(x - 3)}$.
Solving the inequality using the quadratic formula with $a = 25, b = -6, c = 1$:
roots: $\frac{6\pm\sqrt{36 - (4)(25)(1)}}{50}$
$= \frac{6\pm\sqrt{-64}}{50}$. Because the discriminant here $< 0$ then there are no real roots. I don't get how this is interpreted as "all values" (i.e. the answer in the book).
-
The calculation shows that for any real number $y$, there is a real number $x$ such that $y=\frac{x-2}{(x+2)(x-3)}$ precisely if $25y^2-6y+1 \ge 0$. But this inequality holds for all real $y$. This is because certainly $25y^2-6y+1$ is positive somewhere, and since it is never $0$ (by your calculation), it must be positive everywhere. You would learn a little more by completing the square. Your expression is equal to $(5y-3/5)^2 -(3/5)^2) +1$, so it is always $\ge 1-(3/5)^2$. – André Nicolas Apr 22 '11 at 16:04
Since there are no real roots, the graph of $25y^2−6y+1$ cannot cross the horizontal axis and has to be always negative or always positive. Since it is positive for $y=0$, it is always positive.
Since this is the expression for your original equation, this means that you can always find a solution $x$ which gives this value of $y$ (although you should check the trivial fact that this never gives a 0-denominator).
-
I get it now. Gone through the problem again and understand what each quadratic and discriminant actually means. Thanks for your succinct and clear explanation. – PeteUK Apr 23 '11 at 9:13
This answer is just to point out another way to find the range, since I believe your specific questions have been answered by user9325.
Consider $f(x)=\frac{x-2}{(x+2)(x-3)}$ on the interval $(-2,3)$, where it is everywhere defined and continuous. Since $x+2$ is always positive on this interval and $x-3$ is always negative, the sign is opposite that of $x-2$.
As $x$ approaches $3$ (from the left), $x-2$ will be positive, so $f(x)$ will be negative. The denominator approaches $0$ while the numerator approaches $1$, so the absolute value of $f$ will go to $\infty$. Thus $f(x)\to-\infty$ as $x\to 3$ from the left.
As $x$ approaches $-2$ (from the right), $x-2$ will be negative, so $f(x)$ will be positive. The denominator approaches $0$ while the numerator approaches $-4$, so the absolute value of $f$ will go to $\infty$. Thus $f(x)\to+\infty$ as $x\to-2$ from the right.
Because $f$ is continuous on $(-2,3)$ and takes on arbitrarily large positive and negative values on that interval, the range of $f$ is all real numbers by the Intermediate Value Theorem.
-
The function $f$ is a typical candidate for partial fractions. The familiar computation gives $f(x)={1\over5}({4\over x+2}+{1\over x-3})$ from which Jonas' findings can be immediately read off. – Christian Blatter Apr 22 '11 at 17:53
@Christian: Thanks, that is a nice way to look at it. However, how "immediate" or "familiar" it is will depend on who's doing the reading or computing :) – Jonas Meyer Apr 22 '11 at 18:02
Not sure I can directly answer your final question, since I've forgotten what I've learn about how to apply discriminants in your case. However...
It is clear from inspection of the factored form of the original "y = ..." equation, that the range of y is indeed all values of y from -infinity to +infinity, by observing the behavior of the function for x = -2 +/- dx, x = 2, and x = 3 +/- dx.
-
This alone is not enough to say that all values are assumed; only that the range must be unbounded. Consider for example $f(x)=(x^2+1)/x$, which has the line $x=0$ as a vertical asymptote, but does not assume any values between $-2$ and $2$. – Hans Lundmark Apr 22 '11 at 17:08
I think it does, given that I included the point for x=2; but I agree that I maybe left too much as an exercise for the reader. The value of the function approaches +infinty for x slightly more positive than -2. The value approaches -infinity for x slightly more negative than +3. The value of the function at x=2 is 0. There are no discontinuities in the range -2 < x < 3, and x goes from +infinity through 0 to -infinity in that range, so all Y are covered by the intermediate value theorem (or is it continuity theorem? I forget...). – Vintage Apr 22 '11 at 17:35
Oops, sorry, I didn't see that "x=2" hidden in there. The argument in your comment of course suffices, but when you're saying that the function has no discontinuities, you are implicitly looking at other points than just close to $-2$, $2$ and $3$. ;-) Anyway, +1 for the clarification, and welcome to this site! – Hans Lundmark Apr 22 '11 at 17:40
You can rewrite
$25y^2−6y+1$
as
$25((y - 3/25)^2 + 16/25^2)$
which is clearly > 0 for any value of y
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Poiseuille's Law Formula
Introduction
Last updated date: 28th Jan 2023
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Poiseuille's Law Formula is actually one of the most important concepts that the students have to be familiar with. It is not only one that they should learn but also understand thoroughly. This is so they can have a better understanding of their subject, the topic, and its many applications. Students can also have a very good idea about how they can solve various problems related to this topic. This can ultimately help them at the time of their exams when they have to get good marks.
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FAQs on Poiseuille's Law Formula
1. What does poiseuille's law explain?
Poiseuille's equation for viscosity used to calculate the viscosity of the liquid through the velocity of the fluid flow through a narrow tube, which varies directly with the pressure and the fourth power of the radius of the tube and is inversely proportional to the length of the tube and coefficient of viscosity.
Poiseuille's law formula V = πpr4/8lη
2. What are the Units of Poiseuille's Law?
Derivation of poiseuille's equation can determine the liquid’s viscosity (V) flows through the cylindrical tube. The unit of viscosity is (N/m2)s or Pascal Second (Pa⋅s). The flow of liquid is proportional to the difference between the pressure of two points and inversely proportional to resistance Q=(P2−P1) / R. For the laminar flow of liquid in a tube, Poiseuille’s equation resistance states that R=8ηlπr4.
3. Which factor plays the most important role in Poiseuille's Law Formula?
There is no one, but three most important factors that come into play while using Poiseuille's Law Formula. These are:
• Vessel Diameter,
• Vessel Length, and
• The viscosity of blood.
If we talk about the most important one out of these three, then it would be the vessel diameter. This is because, when it comes to the vessels, their diameter actually contracts or expands owing to a few factors of its own. It is very common for the vessels to contract or relax in the blood vessel wall. Such a change can also bring about a change in the formula as well.
4. What is the importance of Poiseuille's Law?
Poiseuille's Law Formula is used to describe the relationship that is there between the pressure, flow rate, and fluidic resistance. It is also used for the determination of pressure drop pertaining to a constant viscosity fluid. This fluid basically exhibits the laminar flow that goes through a tight pipe.
One more importance of such a topic is the fact that it is one that comes in the exams for Chemistry. This sets a foundation for the students so they can do better in their higher studies.
5. How can Vedantu help me to understand Poiseuille’s Law?
There are many ways in which Vedantu can help one with their exams. It becomes the most beneficial at the time when one has to study and have a better understanding of the various topics that are of significance for them in the subject. They can easily get a hold of the numerous topics such as Poisseuille’s Law formula and get some insightful details relating to the topic. They also get a good step-by-step procedure of the derivations as well as the examples and applications.
6. Write a short note on Poiseuille's?
Poiseuille's Law is an easy way to take out the correct viscosity of the liquid that goes through a narrow tube. The equation uses the velocity of the fluid to calculate. One thing that varies directly during the application of this formula is the pressure with which the liquid flows through the tube. The fourth power of the tube’s radius is also a varying factor. One must keep in mind that there is an inverse proportionality between the velocity of the liquid and the length of the tube. It is also inversely proportional to the coefficient of viscosity.
Poiseuille's law formula
V = πpr4/8lη
7. Name the units in which we measure Poiseuille's Law?
The derivation of the Poiseuille's Law Formula is one that can be determined by the liquid’s viscosity (V) that flows via a narrow cylindrical tube. It is the unit of viscosity that can be explained with the help of the following:
• (N/m2)s, or
• Pascal Second (Pa⋅s).
When it comes to the flow of liquid, it can be stated that there is a direct proportionality between the flow of liquid and the difference present in between the pressure present in two points. There is also an inverse proportionality between the flow of liquid and the presence of resistance Q=(P2−P1) / R.
Comment | {"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": 0.8738999962806702, "perplexity": 464.6170141775331}, "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-2023-06/segments/1674764499953.47/warc/CC-MAIN-20230201211725-20230202001725-00515.warc.gz"} |
http://cms.math.ca/cjm/kw/strongly%20maximal%20TAF%20algebra | Nest Representations of TAF Algebras A nest representation of a strongly maximal TAF algebra $A$ with diagonal $D$ is a representation $\pi$ for which $\lat \pi(A)$ is totally ordered. We prove that $\ker \pi$ is a meet irreducible ideal if the spectrum of $A$ is totally ordered or if (after an appropriate similarity) the von Neumann algebra $\pi(D)''$ contains an atom. Keywords:nest representation, meet irreducible ideal, strongly maximal TAF algebraCategories:47L40, 47L35 | {"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": 0.8914765119552612, "perplexity": 482.6966065335284}, "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/1432207929832.32/warc/CC-MAIN-20150521113209-00158-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2405316/evaluate-lim-n-to-infty-int-0-infty-frac-lnnxne-x-cosx | # Evaluate $\lim_{n\to\infty}{ \int_0^{\infty}} \frac{\ln(n+x)}{n}e^{-x} \cos(x)\, dx$
The following is one of my graduate analysis past year paper questions.
Question: Evaluate $$\lim_{n\to\infty} \int_0^{\infty} \frac{\ln(n+x)}{n}e^{-x} \cos(x)\, dx.$$ All main steps must be clearly shown.
My attempt:
For each $n\in\mathbb{N}$ and each $x\in \mathbb{R},$ define $$f_n(x) = \frac{\ln(n+x)}{n}e^{-x} \cos(x).$$
By L'Hopital rule, for all $x>0,$ $$\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty}\left(\frac{\ln(n+x)}{n}e^{-x} \cos(x)\right) = e^{-x}\cos(x) \lim_{n\to\infty}\left(\frac{\ln(n+x)}{n}\right) = 0.$$
So the sequence $(f_n)_{n\geq 1}$ converges pointwise.
Observe that for each $x>0,$ we have $\ln(n+x) \leq n+x$ for all $n\in\mathbb{N}.$ It follows that $$|f_n(x)| \leq \left| \frac{\ln(n+x)}{n} \right| \leq \left| \frac{n+x}{n} \right|.$$ Since the function $\left| \frac{n+x}{n} \right|$ is continuous, and hence integrable, by the Dominated Convergence Theorem, we have
$$\lim_n \int_0^{\infty} \frac{\ln(n+x)}{n}e^{-x} \cos(x)\, dx = \int_0^\infty \lim_{n\to\infty}\frac{\ln(n+x)}{n}e^{-x} \cos(x)\, dx = \int_0^\infty 0 \, dx = 0.$$
Is my attempt correct?
• $(n+x)/n$ is not integrable on $[0,\infty).$
– zhw.
Aug 25, 2017 at 3:56
By the Cauchy-Schwarz inequality, for any $x\geq 1$ we have $$\log(x)=\int_{1}^{x}\frac{dz}{z}\leq\sqrt{(x-1)\int_{1}^{x}\frac{dz}{z^2}}=\sqrt{x}-\frac{1}{\sqrt{x}}\leq \sqrt{x}$$ hence for any $n\geq 1$ $$\mathcal{J}(n)=\int_{0}^{+\infty}\log(x+n)e^{-x}\cos(x)\,dx\leq \int_{0}^{+\infty}\sqrt{x+n}\,e^{-x}\left|\cos x\right|\,dx$$ and by the Cauchy-Schwarz inequality again $$\left|\mathcal{J}(n)\right|\leq\sqrt{\int_{0}^{+\infty}(x+n)e^{-x}\,dx\int_{0}^{+\infty}e^{-x}\cos^2 x\,dx}=\sqrt{\frac{3}{5}(n+1)}$$ hence the given limit is clearly zero.
• +1 (The upper bound in the second integral of the first line should be $x$ instead of $z$.)
Hint: Use the estimate $\ln (n+x) \le \ln (2n)$ on $[0,n].$ Use the estimate $\ln (n+x)\le \ln (2x)$ on $[n,\infty).$ | {"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": 0.9987950921058655, "perplexity": 149.64418604709326}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00387.warc.gz"} |
http://clay6.com/qa/9401/if-i-2-is-one-root-of-the-equation-ax-bx-c-0-then-the-other-root-is | Browse Questions
# If $-i+2$ is one root of the equation $ax^{2}-bx+c=0$, then the other root is
$\begin{array}{1 1}(1)-i-2&(2)i-2\\(3)2+i&(4)2i+i\end{array}$
Given one root is $-i+2$
Complex root occurs in pairs
The other root is $i+2$
Hence 3 is the correct answer. | {"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": 0.9488592147827148, "perplexity": 644.4806483013333}, "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-2016-50/segments/1480698542828.27/warc/CC-MAIN-20161202170902-00455-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/72835/formal-groups-in-the-supersingular-reduction-case | # Formal groups in the supersingular reduction case
Dear MO,
Let $E/\mathbb{Q}$ be an elliptic curve with potential good supersingular reduction at $p$. Thus, there is a finite extension $K/\mathbb{Q}_p$ such that $E/K$ has good supersingular reduction. Let us choose $K/\mathbb{Q}_p$ of minimal degree such that $E/K$ has good reduction, and let us assume $E/K$ is given by a minimal model. Let $A$ be the ring of integers of $K$, let $\pi$ be a uniformizer for $A$, let $\nu$ be a valuation on $K$ with $\nu(\pi)=1$ and $\nu(p)=e$, and let $\hat{E}/A$ be the formal group associated to $E/K$.
Let $\[p](X) = pX+\cdots=\sum_{k=1}^\infty a_kX^k$ be the formal power series for the multiplication-by-$p$ map on $\hat{E}$. Let $e_0=\nu(a_1)=\nu(p)=e$, and $e_1=\nu(a_p)$. Since $E/K$ is supersingular, we know that $e_2=\nu(a_{p^2})=0$.
If we define points in the plane by $P_0=(1,e)$, $P_1=(p,e_1)$ and $P_2=(p^2,0)$, then $N$, the Newton polygon of $[p](X)$ (for those roots with valuation $>0$) is given either by one single segment $P_0P_2$, or two segments $P_0P_1$ and $P_1P_2$, according to whether $ep/(p+1) \leq e_1$ or $ep/(p+1) > e_1$, respectively.
The number $e$ is a divisor of $12$, and if $p\geq 5$, then $e\leq 6$. If $e>1$, and $N$ has two segments, then $1\leq e_1 < e$. Suppose that we are in this case, i.e., $e_1 < e$.
Question: Are there any further constraints on the values of $e_1$?
More specifically:
Question 1: Are there any divisibility conditions on $e_1$, or further relations between $e$ and $e_1$? Can $e_1$ take any of the values $1\leq e_1< e$ ?
Question 2: Is there a more conceptual way to think about $e_1$?
In all examples I have computed, I get that $e_1=1$, $2$ or $4$ (but I haven't done an exhaustive search either). For instance:
$E=27a4,\quad p=3, \quad [K:\mathbb{Q}]=12, \quad e=12, \quad e_1=2;$
$E=121a2,\quad p=11, \quad K=\mathbb{Q}(\sqrt[6]{11}), \quad e=6, \quad e_1 = 4;$
$E=14450p2,\quad p=17, \quad K=\mathbb{Q}(\sqrt[3]{17}), \quad e=3, \quad e_1=1.$
I am confused about this question. Firstly, you don't seem to make any assumptions on $K$ other than it's a field over which $E$ gets good reduction, so I don't see why $e$ has to divide 12. But more importantly, "the" formal group attached to $E$ depends on a choices of parameter $X$, and there is a huge amount of choice for $X$ as far as I can see, so it's not even clear to me that $e_1$ is well-defined. I think that if $e_1<e$ then it's well-defined, but if it isn't (e.g. if $E$ has good ss reduction over $\mathbf{Q}_p$ then $e_1$ can be pretty much anything. – Kevin Buzzard Aug 13 '11 at 16:38
For more positive comments -- you could look in Katz' Antwerp paper, where he really uses the power series you talk about and explains some things that really are canonical about it (e.g. he shows that $v(a_p)$ is independent of the choice of $X$ if (1) $X$ satisfies some mild conditions ($[\zeta]X=\zeta X$ for all $\zeta\in\mu_{p-1}$) and (2) $v(a_p)<e$). – Kevin Buzzard Aug 13 '11 at 16:40
Thank Kevin. I've modified the question so $K$ is chosen to be of minimal degree such that $E/K$ has good reduction. – Álvaro Lozano-Robledo Aug 13 '11 at 17:17
But there is still this more important issue on how one chooses $X$. The formal group is something isomorphic to $R[[X]]$ with $R$ the integers of $K$. But there's no canonical $X$, so there's no canonical $a_p$ etc. For example one can change $X$ to $X'=X+\pi X^2$ and this will completely change $a_p$. – Kevin Buzzard Aug 13 '11 at 18:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9350271224975586, "perplexity": 101.64247728763996}, "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/1454701150206.8/warc/CC-MAIN-20160205193910-00029-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.pfeiffer-vacuum.com/know-how/introduction-to-vacuum-technology/fundamentals/conductivities/technology.action?chapter=tec1.2.8 | # 1.2.8 Conductance
Generally speaking, vacuum chambers are connected to a vacuum pump via piping. Flow resistance occurs as a result of external friction between gas molecules and the wall surface and internal friction between the gas molecules themselves (viscosity). This flow resistance manifests itself in the form of pressure differences and volume flow rate, or pumping speed, losses. In vacuum technology, it is customary to use the reciprocal, the conductivity of piping $L$ or $C$ (conductance) instead of flow resistance $W$. The conductivity has the dimension of a volume flow rate and is normally expressed in [l s-1] or [m3 h-1].
Gas flowing through piping produces a pressure differential $\Delta p$ at the ends of the piping. The following equation applies:
$C=\frac lW=\frac{q_{pV}}{\Delta p}$
Formula 1-18: Definition of conductance
This principle is formally analogous to Ohm’s law of electrotechnology:
$R=\frac UI\mbox{ or }\frac 1R=\frac IU$
Formula 1-19: Ohm’s law
In a formal comparison of Formula 1-18 with Formula 1-19 $q_{pV}$ represents flow $I$, $C$ the reciprocal of resistance $1/R$ and $\Delta p$ the voltage $U$. If the components are connected in parallel, the individual conductivities are added:
$C_\mbox{ges}=C_1+C_2+\dots+C_n$
Formula 1-20: Parallel connection conductance
and if connected in series, the resistances, i. e. the reciprocals, are added together:
$\frac 1{C_\mbox{ges}}=\frac 1{C_1}+\frac 1{C_2}+\dots+\frac 1{C_n}$
Formula 1-21: Series connection conductivities
The conductance of pipes and pipe bends will differ in the various flow regimes. In viscous flow they are proportional to the mean pressure $\bar p$ and in molecular flow they are independent of pressure. Knudsen flow represents a transition between the two types of flow, and the conductivities vary with the Knudsen number.
Figure 1.8: Conductance of a smooth round pipe as a function of the mean pressure in the pipe
A simple approximation for the Knudsen range can be obtained by adding the laminar and molecular conductivities. We would refer you to special literature for exact calculations of the conductance still in the laminar flow range and already in the molecular flow range as well as conductance calculations taking into account inhomogeneities at the inlet of a pipe.
This publication is restricted to the consideration of the conductivities of orifices and long, round pipes for laminar and molecular flow ranges.
Orifices are frequently flow resistances in vacuum systems. Examples of these are constrictions in the cross-section of valves, ventilation devices or orifices in measuring domes for measuring the pumping speed. In pipe openings in vessel walls the orifice resistance of the inlet opening must also be taken into account in addition to the pipe resistance.
### Blocked flow
Let us consider the venting of a vacuum chamber. When the venting valve is opened, ambient air flows into the vessel at high velocity at a pressure of p. The flow velocity reaches not more than sonic velocity. If the gas has reached sonic velocity, the maximum gas throughput has also been reached at which the vessel can be vented. The throughput flowing through it $q_{pV}$ is not a function of the vessel’s interior pressure $p_i$. The following applies for air:
$q_{pV}=15.7\cdot d^2\cdot p_a$
Formula 1-22: Blocking of an orifice [11]
$d$ Diameter of orifice [cm] $a$ External pressure on the vessel [hPa]
### Gas dynamic flow
If the pressure in the vessel now rises beyond a critical pressure, gas flow is reduced and we can use gas dynamic laws according to Bernoulli and Poiseuille to calculate it. The immersive gas flow $q_{pV}$ and the conductance are dependent on
• Narrowest cross-section of the orifice
• External pressure on the vessel
• Internal pressure in the vessel
• Universal gas constant
• Absolute temperature
• Molar mass
• Adiabatic exponent (= ratio of specific or molar heat capacities at constant pressure $c_p$ or constant volume $c_V$) [12]
### Molecular flow [13]
If an orifice connects two vessels in which molecular flow conditions exist (i.e. if the mean free path is considerably greater than the diameter of the vessel), the following will apply for the displaced gas quantity $q_{pV}$ per unit of time
$q_{pV}=A\cdot \frac{\bar c}4\cdot(p_1-p_2)$
Formula 1-23: Orifice flow
$A$ Cross-section of orifice [cm2] $\bar c$ Mean thermal velocity [m-1]
According to Formula 1-23 the following applies for the orifice conductivity
$C_\mathrm{or,\,mol}=A\cdot \frac{\bar c}4=A\cdot\sqrt{\frac{kT}{2\pi m_0}}$
Formula 1-24: Orifice conductivity
For air with a temperature of 293 K we obtain
$C_\mathrm{or,\,mol}=11.6\cdot A$
Formula 1-25: Orifice conductivity for air
$A$ Cross-section of orifice [cm2] $C$ Conductivity [l s-1]
This formula can be used to determine the maximum possible pumping speed of a vacuum pump with an inlet port A. The maximum pumping speed of a pump under molecular flow conditions is therefore determined by the inlet port.
Let us now consider specific pipe conductivities. In the case of laminar flow, the conductivity of a pipe is proportional to the mean pressure:
$C_\mathrm{pipe,\,lam}=\frac{\pi\cdot d^4}{256\cdot\eta\cdot l}\cdot(p_1+p_2)=\frac{\pi\cdot d^4}{128\cdot\eta\cdot l}\cdot\bar p$
Formula 1-26: Conductance of a pipe in laminar flow
For air at 20°C we obtain
$C_\mathrm{pipe,\,lam}=1.35\cdot\frac{d^4}l\cdot\bar p$
Formula 1-27: Conductance of a pipe in laminar flow for air
$l$ Length of pipe [cm] $d$ Diameter of pipe [cm] $\bar p$ Pressure [Pa] $C$ Conductivity [l s-1]
In the molecular flow regime, conductance is constant and is not a function of pressure. It can be considered to be the product of the orifice conductivity of the pipe opening $C_\mathrm{pipe,\,mol}$ and passage probability $P_\mathrm{pipe,\,mol}$ through a component:
$C_\mathrm{pipe,\,mol}=C_\mathrm{orifice,\,mol}\cdot P_\mathrm{pipe,\,mol}$
Formula 1-28: Molecular pipe flow
The mean probability $P_\mathrm{pipe,\,mol}$ can be calculated with a computer program for different pipe profiles, bends or valves using a Monte Carlo simulation. In this connection, the trajectories of individual gas molecules through the component can be tracked on the basis of wall collisions.
The following applies for long round pipes:
$P_\mathrm{pipe,\,mol}=\frac 43\cdot\frac dl$
Formula 1-29: Passage probability for long round pipes
If we multiply this value by the orifice conductivity (Formula 1-24), we obtain
$C_\mathrm{pipe,\,mol}=\frac{\bar c\cdot\pi\cdot d^3}{12\cdot l}$
Formula 1-30: Molecular pipe conductivity
For air at 20°C we obtain
$C_\mathrm{pipe,\,mol}=12.1\cdot \frac{d^3}l$
Formula 1-31: Molecular pipe conductivity
$l$ Length of pipe [cm] $d$ Diameter of pipe [cm] $C$ Conductivity [l 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": 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": 0.8860498666763306, "perplexity": 1678.7387469983041}, "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/1480698542665.72/warc/CC-MAIN-20161202170902-00410-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://pennstate.pure.elsevier.com/en/publications/inelastic-cross-sections-and-rate-coefficients-for-collisions-bet | # Inelastic cross sections and rate coefficients for collisions between CO and H2
Christina Castro, Kyle Doan, Michael Klemka, Robert C. Forrey, Benhui Yang, Phillip C. Stancil, N. Balakrishnan
Research output: Contribution to journalArticlepeer-review
6 Scopus citations
## Abstract
A five-dimensional coupled states (5D-CS) approximation is used to compute cross sections and rate coefficients for CO+H2 collisions. The 5D-CS calculations are benchmarked against accurate six-dimensional close-coupling (6D-CC) calculations for transitions between low-lying rovibrational states. Good agreement between the two formulations is found for collision energies greater than 10 cm−1. The 5D-CS approximation is then used to compute two separate databases which include highly excited states of CO that are beyond the practical limitations of the 6D-CC method. The first database assumes an internally frozen H2 molecule and allows rovibrational transitions for v ≤ 5 and j ≤ 30, where v and j are the vibrational and rotational quantum numbers of the initial state of the CO molecule. The second database allows H2 rotational transitions for initial CO states with v ≤ 5 and j ≤ 10. The two databases are in good agreement with each other for transitions that are common to both basis sets. Together they provide data for astrophysical models which were previously unavailable.
Original language English (US) 47-58 12 Molecular Astrophysics 6 https://doi.org/10.1016/j.molap.2017.01.003 Published - Mar 1 2017
## All Science Journal Classification (ASJC) codes
• Astronomy and Astrophysics
• Spectroscopy
• Physical and Theoretical Chemistry
• Space and Planetary Science
## Fingerprint
Dive into the research topics of 'Inelastic cross sections and rate coefficients for collisions between CO and H2'. Together they form a unique fingerprint. | {"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": 0.9277195930480957, "perplexity": 4385.470778872645}, "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/1659882570741.21/warc/CC-MAIN-20220808001418-20220808031418-00069.warc.gz"} |
https://neurips.cc/Conferences/2013/ScheduleMultitrack?event=3820 | Poster
Online PCA for Contaminated Data
Jiashi Feng · Huan Xu · Shie Mannor · Shuicheng Yan
Sat Dec 7th 07:00 -- 11:59 PM @ Harrah's Special Events Center, 2nd Floor #None
We consider the online Principal Component Analysis (PCA) for contaminated samples (containing outliers) which are revealed sequentially to the Principal Components (PCs) estimator. Due to their sensitiveness to outliers, previous online PCA algorithms fail in this case and their results can be arbitrarily bad. Here we propose the online robust PCA algorithm, which is able to improve the PCs estimation upon an initial one steadily, even when faced with a constant fraction of outliers. We show that the final result of the proposed online RPCA has an acceptable degradation from the optimum. Actually, under mild conditions, online RPCA achieves the maximal robustness with a $50\%$ breakdown point. Moreover, online RPCA is shown to be efficient for both storage and computation, since it need not re-explore the previous samples as in traditional robust PCA algorithms. This endows online RPCA with scalability for large scale data. | {"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": 0.9227129817008972, "perplexity": 2559.6719993456795}, "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-2020-24/segments/1590347410284.51/warc/CC-MAIN-20200530165307-20200530195307-00062.warc.gz"} |
https://www.storyofmathematics.com/19th.html | # 19TH CENTURY MATHEMATICS
Approximation of a periodic function by the Fourier Series
The 19th Century saw an unprecedented increase in the breadth and complexity of mathematical concepts. Both France and Germany were caught up in the age of revolution which swept Europe in the late 18th Century, but the two countries treated mathematics quite differently.
After the French Revolution, Napoleon emphasized the practical usefulness of mathematics and his reforms and military ambitions gave French mathematics a big boost, as exemplified by “the three L’s”, Lagrange, Laplace and Legendre (see the section on 18th Century Mathematics), Fourier and Galois.
Joseph Fourier’s study, at the beginning of the 19th Century, of infinite sums in which the terms are trigonometric functions were another important advance in mathematical analysis. Periodic functions that can be expressed as the sum of an infinite series of sines and cosines are known today as Fourier Series, and they are still powerful tools in pure and applied mathematics. Fourier (following Leibniz, Euler, Lagrange and others) also contributed towards defining exactly what is meant by a function, although the definition that is found in texts today – defining it in terms of a correspondence between elements of the domain and the range – is usually attributed to the 19th Century German mathematician Peter Dirichlet.
In 1806, Jean-Robert Argand published his paper on how complex numbers (of the form a + bi, where i is √-1) could be represented on geometric diagrams and manipulated using trigonometry and vectors. Even though the Dane Caspar Wessel had produced a very similar paper at the end of the 18th Century, and even though it was Gauss who popularized the practice, they are still known today as Argand Diagrams.
The Frenchman Évariste Galois proved in the late 1820s that there is no general algebraic method for solving polynomial equations of any degree greater than four, going further than the Norwegian Niels Henrik Abel who had, just a few years earlier, shown the impossibility of solving quintic equations, and breaching an impasse which had existed for centuries. Galois‘ work also laid the groundwork for further developments such as the beginnings of the field of abstract algebra, including areas like algebraic geometry, group theory, rings, fields, modules, vector spaces and non-commutative algebra.
Germany, on the other hand, under the influence of the great educationalist Wilhelm von Humboldt, took a rather different approach, supporting pure mathematics for its own sake, detached from the demands of the state and military. It was in this environment that the young German prodigy Carl Friedrich Gauss, sometimes called the “Prince of Mathematics”, received his education at the prestigious University of Göttingen. Some of Gauss’ ideas were a hundred years ahead of their time, and touched on many different parts of the mathematical world, including geometry, number theory, calculus, algebra and probability. He is widely regarded as one of the three greatest mathematicians of all times, along with Archimedes and Newton.
Euclidean, hyperbolic and elliptic geometry
Later in life, Gauss also claimed to have investigated a kind of non-Euclidean geometry using curved space but, unwilling to court controversy, he decided not to pursue or publish any of these avant-garde ideas. This left the field open for János Bolyai and Nikolai Lobachevsky (respectively, a Hungarian and a Russian) who both independently explored the potential of hyperbolic geometry and curved spaces.
The German Bernhard Riemann worked on a different kind of non-Euclidean geometry called elliptic geometry, as well as on a generalized theory of all the different types of geometry. Riemann, however, soon took this even further, breaking away completely from all the limitations of 2 and 3 dimensional geometry, whether flat or curved, and began to think in higher dimensions. His exploration of the zeta function in multi-dimensional complex numbers revealed an unexpected link with the distribution of prime numbers, and his famous Riemann Hypothesis, still unproven after 150 years, remains one of the world’s great unsolved mathematical mysteries and the testing ground for new generations of mathematicians.
British mathematics also saw something of a resurgence in the early and mid-19th century. Although the roots of the computer go back to the geared calculators of Pascal and Leibniz in the 17th Century, it was Charles Babbage in 19th Century England who designed a machine that could automatically perform computations based on a program of instructions stored on cards or tape. His large “difference engine” of 1823 was able to calculate logarithms and trigonometric functions, and was the true forerunner of the modern electronic computer. Although never actually built in his lifetime, a machine was built almost 200 years later to his specifications and worked perfectly. He also designed a much more sophisticated machine he called the “analytic engine“, complete with punched cards, printer and computational abilities commensurate with modern computers.
Another 19th Century Englishman, George Peacock, is usually credited with the invention of symbolic algebra, and the extension of the scope of algebra beyond the ordinary systems of numbers. This recognition of the possible existence of non-arithmetical algebras was an important stepping stone toward future developments in abstract algebra.
In the mid-19th Century, the British mathematician George Boole devised an algebra (now called Boolean algebra or Boolean logic), in which the only operators were AND, OR and NOT, and which could be applied to the solution of logical problems and mathematical functions. He also described a kind of binary system which used just two objects, “on” and “off” (or “true” and “false”, 0 and 1, etc), in which, famously, 1 + 1 = 1. Boolean algebra was the starting point of modern mathematical logic and ultimately led to the development of computer science.
Hamilton’s quaternion
The concept of number and algebra was further extended by the Irish mathematician William Hamilton, whose 1843 theory of quaternions (a 4-dimensional number system, where a quantity representing a 3-dimensional rotation can be described by just an angle and a vector). Quaternions, and its later generalization by Hermann Grassmann, provided the first example of a non-commutative algebra (i.e. one in which a x b does not always equal b x a), and showed that several different consistent algebras may be derived by choosing different sets of axioms.
The Englishman Arthur Cayley extended Hamilton’s quaternions and developed the octonions. But Cayley was one of the most prolific mathematicians in history, and was a pioneer of modern group theory, matrix algebra, the theory of higher singularities, and higher dimensional geometry (anticipating the later ideas of Klein), as well as the theory of invariants.
Throughout the 19th Century, mathematics in general became ever more complex and abstract. But it also saw a re-visiting of some older methods and an emphasis on mathematical rigour. In the first decades of the century, the Bohemian priest Bernhard Bolzano was one of the earliest mathematicians to begin instilling rigour into mathematical analysis, as well as giving the first purely analytic proof of both the fundamental theorem of algebra and the intermediate value theorem, and early consideration of sets (collections of objects defined by a common property, such as “all the numbers greater than 7” or “all right triangles“, etc). When the German mathematician Karl Weierstrass discovered the theoretical existence of a continuous function having no derivative (in other words, a continuous curve possessing no tangent at any of its points), he saw the need for a rigorous “arithmetization” of calculus, from which all the basic concepts of analysis could be derived.
Along with Riemann and, particularly, the Frenchman Augustin-Louis Cauchy, Weierstrass completely reformulated calculus in an even more rigorous fashion, leading to the development of mathematical analysis, a branch of pure mathematics largely concerned with the notion of limits (whether it be the limit of a sequence or the limit of a function) and with the theories of differentiation, integration, infinite series and analytic functions. In 1845, Cauchy also proved Cauchy’s theorem, a fundamental theorem of group theory, which he discovered while examining permutation groups. Carl Jacobi also made important contributions to analysis, determinants and matrices, and especially his theory of periodic functions and elliptic functions and their relation to the elliptic theta function.
Non-orientable surfaces with no identifiable “inner” and “outer” sides
August Ferdinand Möbius is best known for his 1858 discovery of the Möbius strip, a non-orientable two-dimensional surface which has only one side when embedded in three-dimensional Euclidean space (actually a German, Johann Benedict Listing, devised the same object just a couple of months before Möbius, but it has come to hold Möbius’ name). Many other concepts are also named after him, including the Möbius configuration, Möbius transformations, the Möbius transform of number theory, the Möbius function and the Möbius inversion formula. He also introduced homogeneous coordinates and discussed geometric and projective transformations.
Felix Klein also pursued more developments in non-Euclidean geometry, include the Klein bottle, a one-sided closed surface that cannot be embedded in three-dimensional Euclidean space, only in four or more dimensions. It can be best visualized as a cylinder looped back through itself to join with its other end from the “inside”. Klein’s 1872 Erlangen Program, which classified geometries by their underlying symmetry groups (or their groups of transformations), was a hugely influential synthesis of much of the mathematics of the day, and his work was very important in the later development of group theory and function theory.
The Norwegian mathematician Marius Sophus Lie also applied algebra to the study of geometry. He largely created the theory of continuous symmetry, and applied it to the geometric theory of differential equations by means of continuous groups of transformations known as Lie groups.
In an unusual occurrence in 1866, an unknown 16-year old Italian, Niccolò Paganini, discovered the second smallest pair of amicable numbers (1,184 and 1210), which had been completely overlooked by some of the greatest mathematicians in history (including Euler, who had identified over 60 such numbers in the 18th Century, some of them huge).
In the later 19th Century, Georg Cantor established the first foundations of set theory, which enabled the rigorous treatment of the notion of infinity, and which has since become the common language of nearly all mathematics. In the face of fierce resistance from most of his contemporaries and his own battle against mental illness, Cantor explored new mathematical worlds where there were many different infinities, some of which were larger than others.
Venn diagram
Cantor’s work on set theory was extended by another German, Richard Dedekind, who defined concepts such as similar sets and infinite sets. Dedekind also came up with the notion, now called a Dedekind cut which is now a standard definition of the real numbers. He showed that any irrational number divides the rational numbers into two classes or sets, the upper class being strictly greater than all the members of the other lower class. Thus, every location on the number line continuum contains either a rational or an irrational number, with no empty locations, gaps or discontinuities. In 1881, the Englishman John Venn introduced his “Venn diagrams” which become useful and ubiquitous tools in set theory.
Building on Riemann’s deep ideas on the distribution of prime numbers, the year 1896 saw two independent proofs of the asymptotic law of the distribution of prime numbers (known as the Prime Number Theorem), one by Jacques Hadamard and one by Charles de la Vallée Poussin, which showed that the number of primes occurring up to any number x is asymptotic to (or tends towards) xlog x.
Minkowski space-time
Hermann Minkowski, a great friend of David Hilbert and teacher of the young Albert Einstein, developed a branch of number theory called the “geometry of numbers” late in the 19th Century as a geometrical method in multi-dimensional space for solving number theory problems, involving complex concepts such as convex sets, lattice points and vector space. Later, in 1907, it was Minkowski who realized that the Einstein’s 1905 special theory of relativity could be best understood in a four-dimensional space, often referred to as Minkowski space-time.
Gottlob Frege’s 1879 “Begriffsschrift” (roughly translated as “Concept-Script”) broke new ground in the field of logic, including a rigorous treatment of the ideas of functions and variables. In his attempt to show that mathematics grows out of logic, he devised techniques that took him far beyond the logical traditions of Aristotle (and even of George Boole). He was the first to explicitly introduce the notion of variables in logical statements, as well as the notions of quantifiers, universals and existentials. He extended Boole‘s “propositional logic” into a new “predicate logic” and, in so doing, set the stage for the radical advances of Giuseppe Peano, Bertrand Russell and David Hilbert in the early 20th Century.
Henri Poincaré came to prominence in the latter part of the 19th Century with at least a partial solution to the “three body problem”, a deceptively simple problem which had stubbornly resisted resolution since the time of Newton, over two hundred years earlier. Although his solution actually proved to be erroneous, its implications led to the early intimations of what would later become known as chaos theory. In between his important work in theoretical physics, he also greatly extended the theory of mathematical topology, leaving behind a knotty problem known as the Poincaré conjecture which remined unsolved until 2002.
Poincaré was also an engineer and a polymath, and perhaps the last of the great mathematicians to adhere to an older conception of mathematics, which championed a faith in human intuition over rigour and formalism. He is sometimes referred to as the “Last Univeralist” as he was perhaps the last mathematician able to shine in almost all of the various aspects of what had become by now a huge, encyclopedic and incredibly complex subject. The 20th Century would belong to the specialists. | {"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": 0.8038041591644287, "perplexity": 645.7191693926766}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964359073.63/warc/CC-MAIN-20211130201935-20211130231935-00466.warc.gz"} |
http://www.physicsforums.com/showpost.php?p=3180720&postcount=1 | View Single Post
P: 743 1. The problem statement, all variables and given/known data Here is a really simple lin.alg problem that for some reason I'm having trouble doing. Assume that $\left\{ v_i \right\}$ is a set of linearly independent vectors. Take w to be a non-zero vector that can be written as a linear combination of the $v_i$. Show that $\left\{ v_i - w \right\}$ is still linearly independent. 3. The attempt at a solution For some reason I'm quite stuck on this. My first goal was to let $b_i$ be such that we can write $$w = \sum_j b_j v_j$$ and then consider the sum \sum_i a_i (v_i-w) = 0 [/itex] and show that each $a_i$ must necessarily be zero. Substituting the first equation into the other yields [tex] \begin{align*}\sum_i a_i (v_i - \sum_j b_j v_j ) &= \sum_i a_i - \sum_{i,j} a_i b_j v_j \\ &= \sum_i \left( a_i - \sum_j a_j b_i \right) v_i \end{align*} where in the last step I've switched the indices in the double summation. By linear independence of the $v_i$ it follows that $$a_i = \sum_j a_j b_i$$ but that's where I'm stuck. It's possible that I'm doing this the wrong way also. Any help would be appreciated. | {"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": 0.9989332556724548, "perplexity": 109.77871044970888}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394010444750/warc/CC-MAIN-20140305090724-00038-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/157224/normal-vector-field-associated-to-deformations-of-riemannian-submanifolds | Normal vector field associated to deformations of Riemannian submanifolds
Let $(M,g)$ be a Riemannian manifold of dimension $n$ and $X$ be a immersed submanifold in $M$ of dimension $k$ i.e there is a immersion $F_{0}:X \longrightarrow M$.
A deformation of the submanifold $X$ is defined by a smooth family of immersions $F : I \times X \longrightarrow M$ i.e. $F_{t}: X \longrightarrow M$ is an immersion for all $t \in I$ and $F_{0}$ is the immersion of $X$ defined above, so we have a family of immersed submanifolds $F(t,(X))=X_{t}$.
Let $F_{*}( \frac{\partial}{\partial t})$ be the deformation vector field on $M$ associated to this deformation i.e. for any $p\in X$ take the curve $F(\cdot,p):I \longrightarrow M$ and $F_{*}( \frac{\partial}{\partial t})$ at $p$ is the tangent vector to this curve at $t=0$.
Now this is my problem: I have read that if the submanifold $X$ is compact and orientable then we can find a family of diffeomorphisms of $X$ depending on $t$ such that we can assume that the vector field of the deformation is normal to $X_{t}$ for all $t$. It is not clear how this reparametrization is done, if I consider the submanifold $X_{0}$ and the deformation vector field on it, i has 2 components corresponding to the splitting $TM \vert_{X}=T(X) \oplus N(X)$ how I get rid of the tangent component using diffeomorphisms of $X$ as it is stated above?
• Project $\tfrac\partial{\partial t}$ to the normal bundle and integrate. – Anton Petrunin Feb 10 '14 at 19:56
The equation (2.1) referred to is the mean curvature flow equation, $$\frac{\partial x}{\partial t} = \vec H(x)$$
This is not exactly what you are asking, but in algebraic geometry we would say the following: the family of immersions $(F_t)$ is not uniquely defined, we can replace $F_t$ by $F_t\circ\varphi _t$, where $(\varphi _t)$ is a one-parameter family of diffeomeorphisms of $X$ -- this will give the same family of submanifolds. Thus $F_*(\frac{\partial}{\partial t} )$ is well defined only in the quotient $TM_{|X}/TX$, which is canonically isomorphic to the normal bundle.
• add to abx answer: i.e., just take $\phi_t$ be the one-parameter group of diffeomorphisms of M generated by the tangent component of the deformation field. – valeri Feb 10 '14 at 18:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9329189658164978, "perplexity": 89.5295964044104}, "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-2020-29/segments/1593655887046.62/warc/CC-MAIN-20200705055259-20200705085259-00112.warc.gz"} |
https://clay6.com/qa/cbse-xi/cbse%2Cclass11%2C+math%2C | Recent questions and answers in Math
Find the equation of the set of all points whose distance from $(0, 4)$ are $\large\frac{2}{3}$ of their distance from the line $y = 9$.
To see more, click for all the questions in this category. | {"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": 0.8517674207687378, "perplexity": 167.41122550574082}, "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-2020-16/segments/1585370502513.35/warc/CC-MAIN-20200331150854-20200331180854-00076.warc.gz"} |
https://arxiv.org/abs/hep-th/0107057 | hep-th
(what is this?)
# Title:N=2 Gauge theories on systems of fractional D3/D7 branes
Abstract: We study a bound state of fractional D3/D7-branes in the ten-dimensional space R^{1,5}*R^{4}/Z_2 using the boundary state formalism. We construct the boundary actions for this system and show that higher order terms in the twisted fields are needed in order to satisfy the zero-force condition. We then find the classical background associated to the bound state and show that the gauge theory living on a probe fractional D3-brane correctly reproduces the perturbative behavior of a four-dimensional N=2 supersymmetric gauge theory with fundamental matter.
Comments: latex, 22 pages. Typos fixed, appendix expanded, some points clarified and references added Subjects: High Energy Physics - Theory (hep-th) Journal reference: Nucl.Phys. B621 (2002) 157-178 DOI: 10.1016/S0550-3213(01)00568-5 Report number: NORDITA-2001/15 HE, DFTT 20/2001 Cite as: arXiv:hep-th/0107057 (or for this version)
## Submission history
From: Matteo Bertolini [view email]
[v1] Fri, 6 Jul 2001 18:54:38 UTC (23 KB)
[v2] Thu, 19 Jul 2001 14:05:55 UTC (24 KB) | {"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": 0.9236610531806946, "perplexity": 4476.051025976588}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247494424.70/warc/CC-MAIN-20190220024254-20190220050254-00091.warc.gz"} |
https://www.beatthegmat.com/if-d-is-the-standard-deviation-x-y-and-z-what-is-the-standard-deviation-of-x-5-y-5-z-5-t329209.html?sid=90a30787b67132eb837110750acf31bf | ## If d is the standard deviation x, y, and z, what is the standard deviation of x + 5, y + 5, z + 5 ?
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### If d is the standard deviation x, y, and z, what is the standard deviation of x + 5, y + 5, z + 5 ?
by BTGmoderatorDC » Tue Jan 25, 2022 5:48 pm
00:00
A
B
C
D
E
## Global Stats
If d is the standard deviation x, y, and z, what is the standard deviation of x + 5, y + 5, z + 5 ?
A. d
B. 3d
C. 15d
D. d + 5
E. d + 15
OA A
Source: GMAT Prep
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### Re: If d is the standard deviation x, y, and z, what is the standard deviation of x + 5, y + 5, z + 5 ?
by [email protected] » Wed Jan 26, 2022 7:28 am
BTGmoderatorDC wrote:
Tue Jan 25, 2022 5:48 pm
If d is the standard deviation x, y, and z, what is the standard deviation of x + 5, y + 5, z + 5 ?
A. d
B. 3d
C. 15d
D. d + 5
E. d + 15
OA A
Source: GMAT Prep
Adding the same value to each value in a set does not change the standard deviation of the set.
So, the standard deviation of {x + 5, y + 5, z + 5 } will also be d. | {"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": 0.8491678833961487, "perplexity": 1475.7134118526476}, "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/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00734.warc.gz"} |
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